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Department of Civil and
Structural Engineering
Plastic Design of Portal
frame to Eurocode 3 Worked Example
University of
Sheffield
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University of Sheffield
Department of Civil Structural Engineering
Date16022009 Geometry of the Frame Sheet No
2Reference Calculation
1 Geometry
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Client brief Sheet No
3Reference Calculation
2 Design Brief A client requires a single-storey building having a clear floor area 30 m x 80m with a clear height to the underside of the roof steelwork of 5 m The slopeof the roof member is to be at least 6o
Figure 1‐ Plan view of the frame
Figure 2‐ 3 Dimensional view of the building (Plum 1996)
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Determining load on the frame Sheet No
4Reference Calculation
EN 1991-1-
12002 (E)
Annex A 1
See
Supporting
Notes Sec 64
3 Determining loading on frame
31Combination factors ψThe combination ψ must be found from Eurocode 1 (EN1991-1) or relevant NADNote that because most portal frame designs are governed by gravity (dead + snow)loading so in this worked example only maximum vertical load combination isconsidered Therefore the combination factor ψ is never applied in this examplebut for full analysis the following load combination should be considered
1) Maximum gravity loads without wind causing maximum sagging moment in therafter and maximum hogging moments in the haunches
2) Maximum wind loading with minimum gravity loads causing maximum reversalof moment compared with case 1 The worst wind case might be from eithertransverse wind or longitudinal wind so both must be checked
Basic data
bull Total length b = 72 m
bull Spacing s = 72 m
bull Bay width d = 30 mbull Height (max) h = 7577m
bull Roof slope α = 6o
Figure 3‐ Frame spacing (SX016 Matthias Oppe)
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Determining loading on the frame Sheet No
5Reference Calculation
EN 1991-1-3
Sec 522Eq51
EN 1991-1-3Sec 53Table 51
SeeAppendix ATable A1
EN 1991-1-3Annex C
SeeAppendix ATable A2
32Snow loading
General
Snow loading in the roof should be determined as follow
micro
Wheremicro is the roof shape coefficient is the exposure coefficient usually taken as 1 is the thermal coefficient set to 1 for nominal situations Is the characteristic value of ground snow load for
relevant altitude
Roof shape coefficient
Shape coefficients are needed for an adjustment of the ground snow load to a snowload on the roof taking into account effects caused by non-drifted and drifted snowloading
The roof shape coefficient depends on the roof angle so
0 3 0 micro =08
Snow load on the ground
For the snow load on the ground the characteristic value depends on the climatic regionfor site in the UK the following expression is relevant
Sk=0140z-01+(A501)Where
Z is the( zone number 9 ) depending on the snow load on sea levelhere in Sheffield z=3
A is the altitude above sea level A=175m
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Determining loading on the frame Sheet No
6Reference Calculation
Self-weightestimatedneeded to bechecked athe end
Snow load on the roof
Sk = 08 x 1 x 1 067 = 054 KNm2
Spacing = 72 m
For internal frameUDL by snow = 054 x 72 = 389 m
Figure 4‐ Distributed load due to snow per meter span (SX016 Matthias Oppe)
33Self weight of steel members
Assume the following weight by members
bull Roofing = 02 KNm2
bull Services = 02 KNm2
bull Rafter and column self weight = 025 KNm2
Total self weight _____________
065 KNm2
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Initial sizing if members Sheet No
7Reference Calculation
TP0843EC30816Manual forhe design of
steelworkbuildingstructures toEC3
SeeAppendix Bor themethod
4 Initial Sizing of members
Figure 5‐ Dimensions of portal (The institutionof Structural Engineers TP0843 EC30816)
a) Lh = 306 = 5rL = 157730=00526
b) Loading1) Gravity loading
Snow loading = 054 x 72 = 380KNmSelf weight = 065 x 72 = 468 KNm
2) Factored loadw= (468 x 135 ) + (380 x 15 ) = 120 KNm
c) Finding Mp for the sections
1) Total load on the frame (wL)= 120 x 30 = 3605KN
2) Parameter wl2 = 120 x 302 = 10816 KNm
3) From Graphs (Figure B2) obtain horizontal force ratio (036)
H= 036 x 3605 = 1298 KN
4) From Graphs (Figure B3) obtain rafter Mp ratio (0034)MrafterRd = 0034 x 10816 = 3677 KNm
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Initial sizing if members Sheet No
8Reference Calculation
SectionTables ofUniversalBeams
EN 1993-1-12005 (E)Table 31
5) From Graphs (Figure B4) obtained column Mp ratio (0063)Mcolumn Rd = 0063 x 10816 = 6814 KNm
6) Selecting membersa) Wpl (rafter)required = (3677 x 10
6) 275 = 1337 x 10
3 cm
3
Try UB 457x152x74
b) Wpl(column)required = (6814 x 106)275= 2478 x 10
3 cm
3
Try UB 533 x 210 x 109
bull Properties Rafter Section UB 457x152x74
G=742 Kgm h= 462mm b=1544mmtw=96mm tf =17mm A=9448 x 102 mm2
d=428mm
Iy= 32670 x 104mm
4 Wply=1627 x 10
3 mm
3
iy=186 mm iz = 333 mmIz = 1047 x 104 mm4 Wplz = 2131 x 103 mm3 It = 6618 x 10
4 mm
4 Iw = 5163 x 10
6 mm
6
bull Properties Column Section UB 533x210x109
G=109 Kgm h= 5395mm b=2108mmtw=116mm tf =188mm A=1389 x10
2 mm
2
d=5109mm
Iy= 66820 x 104mm
4 Wply=2828 x 10
3 mm
3
iy=2187 mm iz = 457 mmIz = 2692 x 10
4 mm
4 Wplz = 3994 x 10
3mm
3
It = 1016 x 104 mm4 Iw = 1811 x 106 mm6
Steel grade is S275 Assume Sections Class1 then check
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Ultimate Limit State Analysis Sheet No
9Reference Calculation
EN 1993-1-12005 (E)Sec 532
SeeSupportingNotesSection 9
5 Load Combination (Max vertical Load) (Dead + Snow)
51Frame imperfections equivalent horizontal forces
Oslash Oslash
Oslash = radic 05 1 5
Oslash = 354 x 10-3
The column loads could be calculated by a frame analysis but a simple calculationbased on plan areas is suitable for single storey portals
(i) Permanent loads ( un-factored )
Rafter = (745 x 15 x 98) 103 = 11 KNRoofing = (15 x 02 x 72) = 216 KNServices = (15 x 02 x 72) = 216 KN
_________Total = 542 KN
(ii) Variable loads ( un-factored )
Snow load = 15 x 054 x 72 = 583 KN
Thus the un-factored equivalent horizontal forces are given by
(i) Permanentcolumn = 354 x 10-3
x 542 = 019 KN
(ii) Variablecolumn = 354 x 10-3
x 583 = 021 KN
Note EC3 requires that all loads that could occur at the same time are consideredtogether so the frame imperfection forces and wind loads should be considered asadditive to permanent loads and variable loads with the appropriate load factors
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Ultimate Limit State Analysis Sheet No
10Reference Calculation
For SecondOrder effectsSeeSupportingNotesSection 71 ampSection 72
Figure 6‐Frame imperfections equivalent horizontal forces
52 Partial safety factors and second order effects
Second order effects increases not only the deflections but also the momentsand forces beyond those calculated by the first order Second-order analysis isthe term used to describe analysis method in which the effects of increasingdeflections under increasing load are considered explicitly in the solutionmethod
The effects of the deformed geometry are assessed in EN 1993-1-1 bycalculating alpha crit (αcrit) factor The limitations to the use of the first-orderanalysis are defined in EN 1993-1-1 Section 521 (3) as αcrit
15 for plastic
analysis When a second order analysis is required there are two main methodsto proceed
1) Rigorous 2nd
order analysis (ie using appropriate second order software)2) Approximate 2
nd order analysis (ie hand calculation using first order analysis
with magnification factors) Although the modifications involveapproximations they are sufficiently accurate within the limits given by EN1993-1-1
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Ultimate Limit State Analysis Sheet No
12Reference Calculation
Load Factor Hinge number Member Hinge status102 1 RHC Formed
114 2 LHR Formed
Table 1‐ Position of Hinges and Load factors
Figure 8‐ ndash Member forces (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Member checks - Column Sheet No
23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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Member checks - Column Sheet No
24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Member checks - Column Sheet No
25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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Member checks - Column Sheet No
26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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Date16022009 Member checks ndash Column Sheet No
27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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Date16022009 Member checks - Rafter Sheet No
28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Member checks ndash Rafter Sheet No
30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Date16022009 Geometry of the Frame Sheet No
2Reference Calculation
1 Geometry
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Client brief Sheet No
3Reference Calculation
2 Design Brief A client requires a single-storey building having a clear floor area 30 m x 80m with a clear height to the underside of the roof steelwork of 5 m The slopeof the roof member is to be at least 6o
Figure 1‐ Plan view of the frame
Figure 2‐ 3 Dimensional view of the building (Plum 1996)
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Determining load on the frame Sheet No
4Reference Calculation
EN 1991-1-
12002 (E)
Annex A 1
See
Supporting
Notes Sec 64
3 Determining loading on frame
31Combination factors ψThe combination ψ must be found from Eurocode 1 (EN1991-1) or relevant NADNote that because most portal frame designs are governed by gravity (dead + snow)loading so in this worked example only maximum vertical load combination isconsidered Therefore the combination factor ψ is never applied in this examplebut for full analysis the following load combination should be considered
1) Maximum gravity loads without wind causing maximum sagging moment in therafter and maximum hogging moments in the haunches
2) Maximum wind loading with minimum gravity loads causing maximum reversalof moment compared with case 1 The worst wind case might be from eithertransverse wind or longitudinal wind so both must be checked
Basic data
bull Total length b = 72 m
bull Spacing s = 72 m
bull Bay width d = 30 mbull Height (max) h = 7577m
bull Roof slope α = 6o
Figure 3‐ Frame spacing (SX016 Matthias Oppe)
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Determining loading on the frame Sheet No
5Reference Calculation
EN 1991-1-3
Sec 522Eq51
EN 1991-1-3Sec 53Table 51
SeeAppendix ATable A1
EN 1991-1-3Annex C
SeeAppendix ATable A2
32Snow loading
General
Snow loading in the roof should be determined as follow
micro
Wheremicro is the roof shape coefficient is the exposure coefficient usually taken as 1 is the thermal coefficient set to 1 for nominal situations Is the characteristic value of ground snow load for
relevant altitude
Roof shape coefficient
Shape coefficients are needed for an adjustment of the ground snow load to a snowload on the roof taking into account effects caused by non-drifted and drifted snowloading
The roof shape coefficient depends on the roof angle so
0 3 0 micro =08
Snow load on the ground
For the snow load on the ground the characteristic value depends on the climatic regionfor site in the UK the following expression is relevant
Sk=0140z-01+(A501)Where
Z is the( zone number 9 ) depending on the snow load on sea levelhere in Sheffield z=3
A is the altitude above sea level A=175m
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Determining loading on the frame Sheet No
6Reference Calculation
Self-weightestimatedneeded to bechecked athe end
Snow load on the roof
Sk = 08 x 1 x 1 067 = 054 KNm2
Spacing = 72 m
For internal frameUDL by snow = 054 x 72 = 389 m
Figure 4‐ Distributed load due to snow per meter span (SX016 Matthias Oppe)
33Self weight of steel members
Assume the following weight by members
bull Roofing = 02 KNm2
bull Services = 02 KNm2
bull Rafter and column self weight = 025 KNm2
Total self weight _____________
065 KNm2
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Initial sizing if members Sheet No
7Reference Calculation
TP0843EC30816Manual forhe design of
steelworkbuildingstructures toEC3
SeeAppendix Bor themethod
4 Initial Sizing of members
Figure 5‐ Dimensions of portal (The institutionof Structural Engineers TP0843 EC30816)
a) Lh = 306 = 5rL = 157730=00526
b) Loading1) Gravity loading
Snow loading = 054 x 72 = 380KNmSelf weight = 065 x 72 = 468 KNm
2) Factored loadw= (468 x 135 ) + (380 x 15 ) = 120 KNm
c) Finding Mp for the sections
1) Total load on the frame (wL)= 120 x 30 = 3605KN
2) Parameter wl2 = 120 x 302 = 10816 KNm
3) From Graphs (Figure B2) obtain horizontal force ratio (036)
H= 036 x 3605 = 1298 KN
4) From Graphs (Figure B3) obtain rafter Mp ratio (0034)MrafterRd = 0034 x 10816 = 3677 KNm
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Initial sizing if members Sheet No
8Reference Calculation
SectionTables ofUniversalBeams
EN 1993-1-12005 (E)Table 31
5) From Graphs (Figure B4) obtained column Mp ratio (0063)Mcolumn Rd = 0063 x 10816 = 6814 KNm
6) Selecting membersa) Wpl (rafter)required = (3677 x 10
6) 275 = 1337 x 10
3 cm
3
Try UB 457x152x74
b) Wpl(column)required = (6814 x 106)275= 2478 x 10
3 cm
3
Try UB 533 x 210 x 109
bull Properties Rafter Section UB 457x152x74
G=742 Kgm h= 462mm b=1544mmtw=96mm tf =17mm A=9448 x 102 mm2
d=428mm
Iy= 32670 x 104mm
4 Wply=1627 x 10
3 mm
3
iy=186 mm iz = 333 mmIz = 1047 x 104 mm4 Wplz = 2131 x 103 mm3 It = 6618 x 10
4 mm
4 Iw = 5163 x 10
6 mm
6
bull Properties Column Section UB 533x210x109
G=109 Kgm h= 5395mm b=2108mmtw=116mm tf =188mm A=1389 x10
2 mm
2
d=5109mm
Iy= 66820 x 104mm
4 Wply=2828 x 10
3 mm
3
iy=2187 mm iz = 457 mmIz = 2692 x 10
4 mm
4 Wplz = 3994 x 10
3mm
3
It = 1016 x 104 mm4 Iw = 1811 x 106 mm6
Steel grade is S275 Assume Sections Class1 then check
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Ultimate Limit State Analysis Sheet No
9Reference Calculation
EN 1993-1-12005 (E)Sec 532
SeeSupportingNotesSection 9
5 Load Combination (Max vertical Load) (Dead + Snow)
51Frame imperfections equivalent horizontal forces
Oslash Oslash
Oslash = radic 05 1 5
Oslash = 354 x 10-3
The column loads could be calculated by a frame analysis but a simple calculationbased on plan areas is suitable for single storey portals
(i) Permanent loads ( un-factored )
Rafter = (745 x 15 x 98) 103 = 11 KNRoofing = (15 x 02 x 72) = 216 KNServices = (15 x 02 x 72) = 216 KN
_________Total = 542 KN
(ii) Variable loads ( un-factored )
Snow load = 15 x 054 x 72 = 583 KN
Thus the un-factored equivalent horizontal forces are given by
(i) Permanentcolumn = 354 x 10-3
x 542 = 019 KN
(ii) Variablecolumn = 354 x 10-3
x 583 = 021 KN
Note EC3 requires that all loads that could occur at the same time are consideredtogether so the frame imperfection forces and wind loads should be considered asadditive to permanent loads and variable loads with the appropriate load factors
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Ultimate Limit State Analysis Sheet No
10Reference Calculation
For SecondOrder effectsSeeSupportingNotesSection 71 ampSection 72
Figure 6‐Frame imperfections equivalent horizontal forces
52 Partial safety factors and second order effects
Second order effects increases not only the deflections but also the momentsand forces beyond those calculated by the first order Second-order analysis isthe term used to describe analysis method in which the effects of increasingdeflections under increasing load are considered explicitly in the solutionmethod
The effects of the deformed geometry are assessed in EN 1993-1-1 bycalculating alpha crit (αcrit) factor The limitations to the use of the first-orderanalysis are defined in EN 1993-1-1 Section 521 (3) as αcrit
15 for plastic
analysis When a second order analysis is required there are two main methodsto proceed
1) Rigorous 2nd
order analysis (ie using appropriate second order software)2) Approximate 2
nd order analysis (ie hand calculation using first order analysis
with magnification factors) Although the modifications involveapproximations they are sufficiently accurate within the limits given by EN1993-1-1
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Ultimate Limit State Analysis Sheet No
12Reference Calculation
Load Factor Hinge number Member Hinge status102 1 RHC Formed
114 2 LHR Formed
Table 1‐ Position of Hinges and Load factors
Figure 8‐ ndash Member forces (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Date16022009 Geometry of the Frame Sheet No
2Reference Calculation
1 Geometry
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Client brief Sheet No
3Reference Calculation
2 Design Brief A client requires a single-storey building having a clear floor area 30 m x 80m with a clear height to the underside of the roof steelwork of 5 m The slopeof the roof member is to be at least 6o
Figure 1‐ Plan view of the frame
Figure 2‐ 3 Dimensional view of the building (Plum 1996)
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Determining load on the frame Sheet No
4Reference Calculation
EN 1991-1-
12002 (E)
Annex A 1
See
Supporting
Notes Sec 64
3 Determining loading on frame
31Combination factors ψThe combination ψ must be found from Eurocode 1 (EN1991-1) or relevant NADNote that because most portal frame designs are governed by gravity (dead + snow)loading so in this worked example only maximum vertical load combination isconsidered Therefore the combination factor ψ is never applied in this examplebut for full analysis the following load combination should be considered
1) Maximum gravity loads without wind causing maximum sagging moment in therafter and maximum hogging moments in the haunches
2) Maximum wind loading with minimum gravity loads causing maximum reversalof moment compared with case 1 The worst wind case might be from eithertransverse wind or longitudinal wind so both must be checked
Basic data
bull Total length b = 72 m
bull Spacing s = 72 m
bull Bay width d = 30 mbull Height (max) h = 7577m
bull Roof slope α = 6o
Figure 3‐ Frame spacing (SX016 Matthias Oppe)
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Determining loading on the frame Sheet No
5Reference Calculation
EN 1991-1-3
Sec 522Eq51
EN 1991-1-3Sec 53Table 51
SeeAppendix ATable A1
EN 1991-1-3Annex C
SeeAppendix ATable A2
32Snow loading
General
Snow loading in the roof should be determined as follow
micro
Wheremicro is the roof shape coefficient is the exposure coefficient usually taken as 1 is the thermal coefficient set to 1 for nominal situations Is the characteristic value of ground snow load for
relevant altitude
Roof shape coefficient
Shape coefficients are needed for an adjustment of the ground snow load to a snowload on the roof taking into account effects caused by non-drifted and drifted snowloading
The roof shape coefficient depends on the roof angle so
0 3 0 micro =08
Snow load on the ground
For the snow load on the ground the characteristic value depends on the climatic regionfor site in the UK the following expression is relevant
Sk=0140z-01+(A501)Where
Z is the( zone number 9 ) depending on the snow load on sea levelhere in Sheffield z=3
A is the altitude above sea level A=175m
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Determining loading on the frame Sheet No
6Reference Calculation
Self-weightestimatedneeded to bechecked athe end
Snow load on the roof
Sk = 08 x 1 x 1 067 = 054 KNm2
Spacing = 72 m
For internal frameUDL by snow = 054 x 72 = 389 m
Figure 4‐ Distributed load due to snow per meter span (SX016 Matthias Oppe)
33Self weight of steel members
Assume the following weight by members
bull Roofing = 02 KNm2
bull Services = 02 KNm2
bull Rafter and column self weight = 025 KNm2
Total self weight _____________
065 KNm2
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Initial sizing if members Sheet No
7Reference Calculation
TP0843EC30816Manual forhe design of
steelworkbuildingstructures toEC3
SeeAppendix Bor themethod
4 Initial Sizing of members
Figure 5‐ Dimensions of portal (The institutionof Structural Engineers TP0843 EC30816)
a) Lh = 306 = 5rL = 157730=00526
b) Loading1) Gravity loading
Snow loading = 054 x 72 = 380KNmSelf weight = 065 x 72 = 468 KNm
2) Factored loadw= (468 x 135 ) + (380 x 15 ) = 120 KNm
c) Finding Mp for the sections
1) Total load on the frame (wL)= 120 x 30 = 3605KN
2) Parameter wl2 = 120 x 302 = 10816 KNm
3) From Graphs (Figure B2) obtain horizontal force ratio (036)
H= 036 x 3605 = 1298 KN
4) From Graphs (Figure B3) obtain rafter Mp ratio (0034)MrafterRd = 0034 x 10816 = 3677 KNm
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Initial sizing if members Sheet No
8Reference Calculation
SectionTables ofUniversalBeams
EN 1993-1-12005 (E)Table 31
5) From Graphs (Figure B4) obtained column Mp ratio (0063)Mcolumn Rd = 0063 x 10816 = 6814 KNm
6) Selecting membersa) Wpl (rafter)required = (3677 x 10
6) 275 = 1337 x 10
3 cm
3
Try UB 457x152x74
b) Wpl(column)required = (6814 x 106)275= 2478 x 10
3 cm
3
Try UB 533 x 210 x 109
bull Properties Rafter Section UB 457x152x74
G=742 Kgm h= 462mm b=1544mmtw=96mm tf =17mm A=9448 x 102 mm2
d=428mm
Iy= 32670 x 104mm
4 Wply=1627 x 10
3 mm
3
iy=186 mm iz = 333 mmIz = 1047 x 104 mm4 Wplz = 2131 x 103 mm3 It = 6618 x 10
4 mm
4 Iw = 5163 x 10
6 mm
6
bull Properties Column Section UB 533x210x109
G=109 Kgm h= 5395mm b=2108mmtw=116mm tf =188mm A=1389 x10
2 mm
2
d=5109mm
Iy= 66820 x 104mm
4 Wply=2828 x 10
3 mm
3
iy=2187 mm iz = 457 mmIz = 2692 x 10
4 mm
4 Wplz = 3994 x 10
3mm
3
It = 1016 x 104 mm4 Iw = 1811 x 106 mm6
Steel grade is S275 Assume Sections Class1 then check
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Ultimate Limit State Analysis Sheet No
9Reference Calculation
EN 1993-1-12005 (E)Sec 532
SeeSupportingNotesSection 9
5 Load Combination (Max vertical Load) (Dead + Snow)
51Frame imperfections equivalent horizontal forces
Oslash Oslash
Oslash = radic 05 1 5
Oslash = 354 x 10-3
The column loads could be calculated by a frame analysis but a simple calculationbased on plan areas is suitable for single storey portals
(i) Permanent loads ( un-factored )
Rafter = (745 x 15 x 98) 103 = 11 KNRoofing = (15 x 02 x 72) = 216 KNServices = (15 x 02 x 72) = 216 KN
_________Total = 542 KN
(ii) Variable loads ( un-factored )
Snow load = 15 x 054 x 72 = 583 KN
Thus the un-factored equivalent horizontal forces are given by
(i) Permanentcolumn = 354 x 10-3
x 542 = 019 KN
(ii) Variablecolumn = 354 x 10-3
x 583 = 021 KN
Note EC3 requires that all loads that could occur at the same time are consideredtogether so the frame imperfection forces and wind loads should be considered asadditive to permanent loads and variable loads with the appropriate load factors
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Ultimate Limit State Analysis Sheet No
10Reference Calculation
For SecondOrder effectsSeeSupportingNotesSection 71 ampSection 72
Figure 6‐Frame imperfections equivalent horizontal forces
52 Partial safety factors and second order effects
Second order effects increases not only the deflections but also the momentsand forces beyond those calculated by the first order Second-order analysis isthe term used to describe analysis method in which the effects of increasingdeflections under increasing load are considered explicitly in the solutionmethod
The effects of the deformed geometry are assessed in EN 1993-1-1 bycalculating alpha crit (αcrit) factor The limitations to the use of the first-orderanalysis are defined in EN 1993-1-1 Section 521 (3) as αcrit
15 for plastic
analysis When a second order analysis is required there are two main methodsto proceed
1) Rigorous 2nd
order analysis (ie using appropriate second order software)2) Approximate 2
nd order analysis (ie hand calculation using first order analysis
with magnification factors) Although the modifications involveapproximations they are sufficiently accurate within the limits given by EN1993-1-1
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Ultimate Limit State Analysis Sheet No
12Reference Calculation
Load Factor Hinge number Member Hinge status102 1 RHC Formed
114 2 LHR Formed
Table 1‐ Position of Hinges and Load factors
Figure 8‐ ndash Member forces (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
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Client brief Sheet No
3Reference Calculation
2 Design Brief A client requires a single-storey building having a clear floor area 30 m x 80m with a clear height to the underside of the roof steelwork of 5 m The slopeof the roof member is to be at least 6o
Figure 1‐ Plan view of the frame
Figure 2‐ 3 Dimensional view of the building (Plum 1996)
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Determining load on the frame Sheet No
4Reference Calculation
EN 1991-1-
12002 (E)
Annex A 1
See
Supporting
Notes Sec 64
3 Determining loading on frame
31Combination factors ψThe combination ψ must be found from Eurocode 1 (EN1991-1) or relevant NADNote that because most portal frame designs are governed by gravity (dead + snow)loading so in this worked example only maximum vertical load combination isconsidered Therefore the combination factor ψ is never applied in this examplebut for full analysis the following load combination should be considered
1) Maximum gravity loads without wind causing maximum sagging moment in therafter and maximum hogging moments in the haunches
2) Maximum wind loading with minimum gravity loads causing maximum reversalof moment compared with case 1 The worst wind case might be from eithertransverse wind or longitudinal wind so both must be checked
Basic data
bull Total length b = 72 m
bull Spacing s = 72 m
bull Bay width d = 30 mbull Height (max) h = 7577m
bull Roof slope α = 6o
Figure 3‐ Frame spacing (SX016 Matthias Oppe)
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Determining loading on the frame Sheet No
5Reference Calculation
EN 1991-1-3
Sec 522Eq51
EN 1991-1-3Sec 53Table 51
SeeAppendix ATable A1
EN 1991-1-3Annex C
SeeAppendix ATable A2
32Snow loading
General
Snow loading in the roof should be determined as follow
micro
Wheremicro is the roof shape coefficient is the exposure coefficient usually taken as 1 is the thermal coefficient set to 1 for nominal situations Is the characteristic value of ground snow load for
relevant altitude
Roof shape coefficient
Shape coefficients are needed for an adjustment of the ground snow load to a snowload on the roof taking into account effects caused by non-drifted and drifted snowloading
The roof shape coefficient depends on the roof angle so
0 3 0 micro =08
Snow load on the ground
For the snow load on the ground the characteristic value depends on the climatic regionfor site in the UK the following expression is relevant
Sk=0140z-01+(A501)Where
Z is the( zone number 9 ) depending on the snow load on sea levelhere in Sheffield z=3
A is the altitude above sea level A=175m
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Determining loading on the frame Sheet No
6Reference Calculation
Self-weightestimatedneeded to bechecked athe end
Snow load on the roof
Sk = 08 x 1 x 1 067 = 054 KNm2
Spacing = 72 m
For internal frameUDL by snow = 054 x 72 = 389 m
Figure 4‐ Distributed load due to snow per meter span (SX016 Matthias Oppe)
33Self weight of steel members
Assume the following weight by members
bull Roofing = 02 KNm2
bull Services = 02 KNm2
bull Rafter and column self weight = 025 KNm2
Total self weight _____________
065 KNm2
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Initial sizing if members Sheet No
7Reference Calculation
TP0843EC30816Manual forhe design of
steelworkbuildingstructures toEC3
SeeAppendix Bor themethod
4 Initial Sizing of members
Figure 5‐ Dimensions of portal (The institutionof Structural Engineers TP0843 EC30816)
a) Lh = 306 = 5rL = 157730=00526
b) Loading1) Gravity loading
Snow loading = 054 x 72 = 380KNmSelf weight = 065 x 72 = 468 KNm
2) Factored loadw= (468 x 135 ) + (380 x 15 ) = 120 KNm
c) Finding Mp for the sections
1) Total load on the frame (wL)= 120 x 30 = 3605KN
2) Parameter wl2 = 120 x 302 = 10816 KNm
3) From Graphs (Figure B2) obtain horizontal force ratio (036)
H= 036 x 3605 = 1298 KN
4) From Graphs (Figure B3) obtain rafter Mp ratio (0034)MrafterRd = 0034 x 10816 = 3677 KNm
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Initial sizing if members Sheet No
8Reference Calculation
SectionTables ofUniversalBeams
EN 1993-1-12005 (E)Table 31
5) From Graphs (Figure B4) obtained column Mp ratio (0063)Mcolumn Rd = 0063 x 10816 = 6814 KNm
6) Selecting membersa) Wpl (rafter)required = (3677 x 10
6) 275 = 1337 x 10
3 cm
3
Try UB 457x152x74
b) Wpl(column)required = (6814 x 106)275= 2478 x 10
3 cm
3
Try UB 533 x 210 x 109
bull Properties Rafter Section UB 457x152x74
G=742 Kgm h= 462mm b=1544mmtw=96mm tf =17mm A=9448 x 102 mm2
d=428mm
Iy= 32670 x 104mm
4 Wply=1627 x 10
3 mm
3
iy=186 mm iz = 333 mmIz = 1047 x 104 mm4 Wplz = 2131 x 103 mm3 It = 6618 x 10
4 mm
4 Iw = 5163 x 10
6 mm
6
bull Properties Column Section UB 533x210x109
G=109 Kgm h= 5395mm b=2108mmtw=116mm tf =188mm A=1389 x10
2 mm
2
d=5109mm
Iy= 66820 x 104mm
4 Wply=2828 x 10
3 mm
3
iy=2187 mm iz = 457 mmIz = 2692 x 10
4 mm
4 Wplz = 3994 x 10
3mm
3
It = 1016 x 104 mm4 Iw = 1811 x 106 mm6
Steel grade is S275 Assume Sections Class1 then check
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Ultimate Limit State Analysis Sheet No
9Reference Calculation
EN 1993-1-12005 (E)Sec 532
SeeSupportingNotesSection 9
5 Load Combination (Max vertical Load) (Dead + Snow)
51Frame imperfections equivalent horizontal forces
Oslash Oslash
Oslash = radic 05 1 5
Oslash = 354 x 10-3
The column loads could be calculated by a frame analysis but a simple calculationbased on plan areas is suitable for single storey portals
(i) Permanent loads ( un-factored )
Rafter = (745 x 15 x 98) 103 = 11 KNRoofing = (15 x 02 x 72) = 216 KNServices = (15 x 02 x 72) = 216 KN
_________Total = 542 KN
(ii) Variable loads ( un-factored )
Snow load = 15 x 054 x 72 = 583 KN
Thus the un-factored equivalent horizontal forces are given by
(i) Permanentcolumn = 354 x 10-3
x 542 = 019 KN
(ii) Variablecolumn = 354 x 10-3
x 583 = 021 KN
Note EC3 requires that all loads that could occur at the same time are consideredtogether so the frame imperfection forces and wind loads should be considered asadditive to permanent loads and variable loads with the appropriate load factors
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Ultimate Limit State Analysis Sheet No
10Reference Calculation
For SecondOrder effectsSeeSupportingNotesSection 71 ampSection 72
Figure 6‐Frame imperfections equivalent horizontal forces
52 Partial safety factors and second order effects
Second order effects increases not only the deflections but also the momentsand forces beyond those calculated by the first order Second-order analysis isthe term used to describe analysis method in which the effects of increasingdeflections under increasing load are considered explicitly in the solutionmethod
The effects of the deformed geometry are assessed in EN 1993-1-1 bycalculating alpha crit (αcrit) factor The limitations to the use of the first-orderanalysis are defined in EN 1993-1-1 Section 521 (3) as αcrit
15 for plastic
analysis When a second order analysis is required there are two main methodsto proceed
1) Rigorous 2nd
order analysis (ie using appropriate second order software)2) Approximate 2
nd order analysis (ie hand calculation using first order analysis
with magnification factors) Although the modifications involveapproximations they are sufficiently accurate within the limits given by EN1993-1-1
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Ultimate Limit State Analysis Sheet No
12Reference Calculation
Load Factor Hinge number Member Hinge status102 1 RHC Formed
114 2 LHR Formed
Table 1‐ Position of Hinges and Load factors
Figure 8‐ ndash Member forces (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Member checks - Column Sheet No
23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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Member checks - Column Sheet No
24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Member checks - Column Sheet No
25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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Member checks - Column Sheet No
26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Member checks ndash Rafter Sheet No
30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Determining load on the frame Sheet No
4Reference Calculation
EN 1991-1-
12002 (E)
Annex A 1
See
Supporting
Notes Sec 64
3 Determining loading on frame
31Combination factors ψThe combination ψ must be found from Eurocode 1 (EN1991-1) or relevant NADNote that because most portal frame designs are governed by gravity (dead + snow)loading so in this worked example only maximum vertical load combination isconsidered Therefore the combination factor ψ is never applied in this examplebut for full analysis the following load combination should be considered
1) Maximum gravity loads without wind causing maximum sagging moment in therafter and maximum hogging moments in the haunches
2) Maximum wind loading with minimum gravity loads causing maximum reversalof moment compared with case 1 The worst wind case might be from eithertransverse wind or longitudinal wind so both must be checked
Basic data
bull Total length b = 72 m
bull Spacing s = 72 m
bull Bay width d = 30 mbull Height (max) h = 7577m
bull Roof slope α = 6o
Figure 3‐ Frame spacing (SX016 Matthias Oppe)
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Determining loading on the frame Sheet No
5Reference Calculation
EN 1991-1-3
Sec 522Eq51
EN 1991-1-3Sec 53Table 51
SeeAppendix ATable A1
EN 1991-1-3Annex C
SeeAppendix ATable A2
32Snow loading
General
Snow loading in the roof should be determined as follow
micro
Wheremicro is the roof shape coefficient is the exposure coefficient usually taken as 1 is the thermal coefficient set to 1 for nominal situations Is the characteristic value of ground snow load for
relevant altitude
Roof shape coefficient
Shape coefficients are needed for an adjustment of the ground snow load to a snowload on the roof taking into account effects caused by non-drifted and drifted snowloading
The roof shape coefficient depends on the roof angle so
0 3 0 micro =08
Snow load on the ground
For the snow load on the ground the characteristic value depends on the climatic regionfor site in the UK the following expression is relevant
Sk=0140z-01+(A501)Where
Z is the( zone number 9 ) depending on the snow load on sea levelhere in Sheffield z=3
A is the altitude above sea level A=175m
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Determining loading on the frame Sheet No
6Reference Calculation
Self-weightestimatedneeded to bechecked athe end
Snow load on the roof
Sk = 08 x 1 x 1 067 = 054 KNm2
Spacing = 72 m
For internal frameUDL by snow = 054 x 72 = 389 m
Figure 4‐ Distributed load due to snow per meter span (SX016 Matthias Oppe)
33Self weight of steel members
Assume the following weight by members
bull Roofing = 02 KNm2
bull Services = 02 KNm2
bull Rafter and column self weight = 025 KNm2
Total self weight _____________
065 KNm2
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Initial sizing if members Sheet No
7Reference Calculation
TP0843EC30816Manual forhe design of
steelworkbuildingstructures toEC3
SeeAppendix Bor themethod
4 Initial Sizing of members
Figure 5‐ Dimensions of portal (The institutionof Structural Engineers TP0843 EC30816)
a) Lh = 306 = 5rL = 157730=00526
b) Loading1) Gravity loading
Snow loading = 054 x 72 = 380KNmSelf weight = 065 x 72 = 468 KNm
2) Factored loadw= (468 x 135 ) + (380 x 15 ) = 120 KNm
c) Finding Mp for the sections
1) Total load on the frame (wL)= 120 x 30 = 3605KN
2) Parameter wl2 = 120 x 302 = 10816 KNm
3) From Graphs (Figure B2) obtain horizontal force ratio (036)
H= 036 x 3605 = 1298 KN
4) From Graphs (Figure B3) obtain rafter Mp ratio (0034)MrafterRd = 0034 x 10816 = 3677 KNm
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Initial sizing if members Sheet No
8Reference Calculation
SectionTables ofUniversalBeams
EN 1993-1-12005 (E)Table 31
5) From Graphs (Figure B4) obtained column Mp ratio (0063)Mcolumn Rd = 0063 x 10816 = 6814 KNm
6) Selecting membersa) Wpl (rafter)required = (3677 x 10
6) 275 = 1337 x 10
3 cm
3
Try UB 457x152x74
b) Wpl(column)required = (6814 x 106)275= 2478 x 10
3 cm
3
Try UB 533 x 210 x 109
bull Properties Rafter Section UB 457x152x74
G=742 Kgm h= 462mm b=1544mmtw=96mm tf =17mm A=9448 x 102 mm2
d=428mm
Iy= 32670 x 104mm
4 Wply=1627 x 10
3 mm
3
iy=186 mm iz = 333 mmIz = 1047 x 104 mm4 Wplz = 2131 x 103 mm3 It = 6618 x 10
4 mm
4 Iw = 5163 x 10
6 mm
6
bull Properties Column Section UB 533x210x109
G=109 Kgm h= 5395mm b=2108mmtw=116mm tf =188mm A=1389 x10
2 mm
2
d=5109mm
Iy= 66820 x 104mm
4 Wply=2828 x 10
3 mm
3
iy=2187 mm iz = 457 mmIz = 2692 x 10
4 mm
4 Wplz = 3994 x 10
3mm
3
It = 1016 x 104 mm4 Iw = 1811 x 106 mm6
Steel grade is S275 Assume Sections Class1 then check
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Ultimate Limit State Analysis Sheet No
9Reference Calculation
EN 1993-1-12005 (E)Sec 532
SeeSupportingNotesSection 9
5 Load Combination (Max vertical Load) (Dead + Snow)
51Frame imperfections equivalent horizontal forces
Oslash Oslash
Oslash = radic 05 1 5
Oslash = 354 x 10-3
The column loads could be calculated by a frame analysis but a simple calculationbased on plan areas is suitable for single storey portals
(i) Permanent loads ( un-factored )
Rafter = (745 x 15 x 98) 103 = 11 KNRoofing = (15 x 02 x 72) = 216 KNServices = (15 x 02 x 72) = 216 KN
_________Total = 542 KN
(ii) Variable loads ( un-factored )
Snow load = 15 x 054 x 72 = 583 KN
Thus the un-factored equivalent horizontal forces are given by
(i) Permanentcolumn = 354 x 10-3
x 542 = 019 KN
(ii) Variablecolumn = 354 x 10-3
x 583 = 021 KN
Note EC3 requires that all loads that could occur at the same time are consideredtogether so the frame imperfection forces and wind loads should be considered asadditive to permanent loads and variable loads with the appropriate load factors
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Ultimate Limit State Analysis Sheet No
10Reference Calculation
For SecondOrder effectsSeeSupportingNotesSection 71 ampSection 72
Figure 6‐Frame imperfections equivalent horizontal forces
52 Partial safety factors and second order effects
Second order effects increases not only the deflections but also the momentsand forces beyond those calculated by the first order Second-order analysis isthe term used to describe analysis method in which the effects of increasingdeflections under increasing load are considered explicitly in the solutionmethod
The effects of the deformed geometry are assessed in EN 1993-1-1 bycalculating alpha crit (αcrit) factor The limitations to the use of the first-orderanalysis are defined in EN 1993-1-1 Section 521 (3) as αcrit
15 for plastic
analysis When a second order analysis is required there are two main methodsto proceed
1) Rigorous 2nd
order analysis (ie using appropriate second order software)2) Approximate 2
nd order analysis (ie hand calculation using first order analysis
with magnification factors) Although the modifications involveapproximations they are sufficiently accurate within the limits given by EN1993-1-1
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Ultimate Limit State Analysis Sheet No
12Reference Calculation
Load Factor Hinge number Member Hinge status102 1 RHC Formed
114 2 LHR Formed
Table 1‐ Position of Hinges and Load factors
Figure 8‐ ndash Member forces (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Date
16022009
Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Member checks ndash Rafter Sheet No
30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Determining loading on the frame Sheet No
5Reference Calculation
EN 1991-1-3
Sec 522Eq51
EN 1991-1-3Sec 53Table 51
SeeAppendix ATable A1
EN 1991-1-3Annex C
SeeAppendix ATable A2
32Snow loading
General
Snow loading in the roof should be determined as follow
micro
Wheremicro is the roof shape coefficient is the exposure coefficient usually taken as 1 is the thermal coefficient set to 1 for nominal situations Is the characteristic value of ground snow load for
relevant altitude
Roof shape coefficient
Shape coefficients are needed for an adjustment of the ground snow load to a snowload on the roof taking into account effects caused by non-drifted and drifted snowloading
The roof shape coefficient depends on the roof angle so
0 3 0 micro =08
Snow load on the ground
For the snow load on the ground the characteristic value depends on the climatic regionfor site in the UK the following expression is relevant
Sk=0140z-01+(A501)Where
Z is the( zone number 9 ) depending on the snow load on sea levelhere in Sheffield z=3
A is the altitude above sea level A=175m
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Determining loading on the frame Sheet No
6Reference Calculation
Self-weightestimatedneeded to bechecked athe end
Snow load on the roof
Sk = 08 x 1 x 1 067 = 054 KNm2
Spacing = 72 m
For internal frameUDL by snow = 054 x 72 = 389 m
Figure 4‐ Distributed load due to snow per meter span (SX016 Matthias Oppe)
33Self weight of steel members
Assume the following weight by members
bull Roofing = 02 KNm2
bull Services = 02 KNm2
bull Rafter and column self weight = 025 KNm2
Total self weight _____________
065 KNm2
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Initial sizing if members Sheet No
7Reference Calculation
TP0843EC30816Manual forhe design of
steelworkbuildingstructures toEC3
SeeAppendix Bor themethod
4 Initial Sizing of members
Figure 5‐ Dimensions of portal (The institutionof Structural Engineers TP0843 EC30816)
a) Lh = 306 = 5rL = 157730=00526
b) Loading1) Gravity loading
Snow loading = 054 x 72 = 380KNmSelf weight = 065 x 72 = 468 KNm
2) Factored loadw= (468 x 135 ) + (380 x 15 ) = 120 KNm
c) Finding Mp for the sections
1) Total load on the frame (wL)= 120 x 30 = 3605KN
2) Parameter wl2 = 120 x 302 = 10816 KNm
3) From Graphs (Figure B2) obtain horizontal force ratio (036)
H= 036 x 3605 = 1298 KN
4) From Graphs (Figure B3) obtain rafter Mp ratio (0034)MrafterRd = 0034 x 10816 = 3677 KNm
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Initial sizing if members Sheet No
8Reference Calculation
SectionTables ofUniversalBeams
EN 1993-1-12005 (E)Table 31
5) From Graphs (Figure B4) obtained column Mp ratio (0063)Mcolumn Rd = 0063 x 10816 = 6814 KNm
6) Selecting membersa) Wpl (rafter)required = (3677 x 10
6) 275 = 1337 x 10
3 cm
3
Try UB 457x152x74
b) Wpl(column)required = (6814 x 106)275= 2478 x 10
3 cm
3
Try UB 533 x 210 x 109
bull Properties Rafter Section UB 457x152x74
G=742 Kgm h= 462mm b=1544mmtw=96mm tf =17mm A=9448 x 102 mm2
d=428mm
Iy= 32670 x 104mm
4 Wply=1627 x 10
3 mm
3
iy=186 mm iz = 333 mmIz = 1047 x 104 mm4 Wplz = 2131 x 103 mm3 It = 6618 x 10
4 mm
4 Iw = 5163 x 10
6 mm
6
bull Properties Column Section UB 533x210x109
G=109 Kgm h= 5395mm b=2108mmtw=116mm tf =188mm A=1389 x10
2 mm
2
d=5109mm
Iy= 66820 x 104mm
4 Wply=2828 x 10
3 mm
3
iy=2187 mm iz = 457 mmIz = 2692 x 10
4 mm
4 Wplz = 3994 x 10
3mm
3
It = 1016 x 104 mm4 Iw = 1811 x 106 mm6
Steel grade is S275 Assume Sections Class1 then check
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Ultimate Limit State Analysis Sheet No
9Reference Calculation
EN 1993-1-12005 (E)Sec 532
SeeSupportingNotesSection 9
5 Load Combination (Max vertical Load) (Dead + Snow)
51Frame imperfections equivalent horizontal forces
Oslash Oslash
Oslash = radic 05 1 5
Oslash = 354 x 10-3
The column loads could be calculated by a frame analysis but a simple calculationbased on plan areas is suitable for single storey portals
(i) Permanent loads ( un-factored )
Rafter = (745 x 15 x 98) 103 = 11 KNRoofing = (15 x 02 x 72) = 216 KNServices = (15 x 02 x 72) = 216 KN
_________Total = 542 KN
(ii) Variable loads ( un-factored )
Snow load = 15 x 054 x 72 = 583 KN
Thus the un-factored equivalent horizontal forces are given by
(i) Permanentcolumn = 354 x 10-3
x 542 = 019 KN
(ii) Variablecolumn = 354 x 10-3
x 583 = 021 KN
Note EC3 requires that all loads that could occur at the same time are consideredtogether so the frame imperfection forces and wind loads should be considered asadditive to permanent loads and variable loads with the appropriate load factors
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Ultimate Limit State Analysis Sheet No
10Reference Calculation
For SecondOrder effectsSeeSupportingNotesSection 71 ampSection 72
Figure 6‐Frame imperfections equivalent horizontal forces
52 Partial safety factors and second order effects
Second order effects increases not only the deflections but also the momentsand forces beyond those calculated by the first order Second-order analysis isthe term used to describe analysis method in which the effects of increasingdeflections under increasing load are considered explicitly in the solutionmethod
The effects of the deformed geometry are assessed in EN 1993-1-1 bycalculating alpha crit (αcrit) factor The limitations to the use of the first-orderanalysis are defined in EN 1993-1-1 Section 521 (3) as αcrit
15 for plastic
analysis When a second order analysis is required there are two main methodsto proceed
1) Rigorous 2nd
order analysis (ie using appropriate second order software)2) Approximate 2
nd order analysis (ie hand calculation using first order analysis
with magnification factors) Although the modifications involveapproximations they are sufficiently accurate within the limits given by EN1993-1-1
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Ultimate Limit State Analysis Sheet No
12Reference Calculation
Load Factor Hinge number Member Hinge status102 1 RHC Formed
114 2 LHR Formed
Table 1‐ Position of Hinges and Load factors
Figure 8‐ ndash Member forces (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Member checks - Column Sheet No
23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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Member checks - Column Sheet No
24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Member checks - Column Sheet No
25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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Member checks - Column Sheet No
26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Member checks ndash Rafter Sheet No
30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Determining loading on the frame Sheet No
6Reference Calculation
Self-weightestimatedneeded to bechecked athe end
Snow load on the roof
Sk = 08 x 1 x 1 067 = 054 KNm2
Spacing = 72 m
For internal frameUDL by snow = 054 x 72 = 389 m
Figure 4‐ Distributed load due to snow per meter span (SX016 Matthias Oppe)
33Self weight of steel members
Assume the following weight by members
bull Roofing = 02 KNm2
bull Services = 02 KNm2
bull Rafter and column self weight = 025 KNm2
Total self weight _____________
065 KNm2
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Initial sizing if members Sheet No
7Reference Calculation
TP0843EC30816Manual forhe design of
steelworkbuildingstructures toEC3
SeeAppendix Bor themethod
4 Initial Sizing of members
Figure 5‐ Dimensions of portal (The institutionof Structural Engineers TP0843 EC30816)
a) Lh = 306 = 5rL = 157730=00526
b) Loading1) Gravity loading
Snow loading = 054 x 72 = 380KNmSelf weight = 065 x 72 = 468 KNm
2) Factored loadw= (468 x 135 ) + (380 x 15 ) = 120 KNm
c) Finding Mp for the sections
1) Total load on the frame (wL)= 120 x 30 = 3605KN
2) Parameter wl2 = 120 x 302 = 10816 KNm
3) From Graphs (Figure B2) obtain horizontal force ratio (036)
H= 036 x 3605 = 1298 KN
4) From Graphs (Figure B3) obtain rafter Mp ratio (0034)MrafterRd = 0034 x 10816 = 3677 KNm
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Initial sizing if members Sheet No
8Reference Calculation
SectionTables ofUniversalBeams
EN 1993-1-12005 (E)Table 31
5) From Graphs (Figure B4) obtained column Mp ratio (0063)Mcolumn Rd = 0063 x 10816 = 6814 KNm
6) Selecting membersa) Wpl (rafter)required = (3677 x 10
6) 275 = 1337 x 10
3 cm
3
Try UB 457x152x74
b) Wpl(column)required = (6814 x 106)275= 2478 x 10
3 cm
3
Try UB 533 x 210 x 109
bull Properties Rafter Section UB 457x152x74
G=742 Kgm h= 462mm b=1544mmtw=96mm tf =17mm A=9448 x 102 mm2
d=428mm
Iy= 32670 x 104mm
4 Wply=1627 x 10
3 mm
3
iy=186 mm iz = 333 mmIz = 1047 x 104 mm4 Wplz = 2131 x 103 mm3 It = 6618 x 10
4 mm
4 Iw = 5163 x 10
6 mm
6
bull Properties Column Section UB 533x210x109
G=109 Kgm h= 5395mm b=2108mmtw=116mm tf =188mm A=1389 x10
2 mm
2
d=5109mm
Iy= 66820 x 104mm
4 Wply=2828 x 10
3 mm
3
iy=2187 mm iz = 457 mmIz = 2692 x 10
4 mm
4 Wplz = 3994 x 10
3mm
3
It = 1016 x 104 mm4 Iw = 1811 x 106 mm6
Steel grade is S275 Assume Sections Class1 then check
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Ultimate Limit State Analysis Sheet No
9Reference Calculation
EN 1993-1-12005 (E)Sec 532
SeeSupportingNotesSection 9
5 Load Combination (Max vertical Load) (Dead + Snow)
51Frame imperfections equivalent horizontal forces
Oslash Oslash
Oslash = radic 05 1 5
Oslash = 354 x 10-3
The column loads could be calculated by a frame analysis but a simple calculationbased on plan areas is suitable for single storey portals
(i) Permanent loads ( un-factored )
Rafter = (745 x 15 x 98) 103 = 11 KNRoofing = (15 x 02 x 72) = 216 KNServices = (15 x 02 x 72) = 216 KN
_________Total = 542 KN
(ii) Variable loads ( un-factored )
Snow load = 15 x 054 x 72 = 583 KN
Thus the un-factored equivalent horizontal forces are given by
(i) Permanentcolumn = 354 x 10-3
x 542 = 019 KN
(ii) Variablecolumn = 354 x 10-3
x 583 = 021 KN
Note EC3 requires that all loads that could occur at the same time are consideredtogether so the frame imperfection forces and wind loads should be considered asadditive to permanent loads and variable loads with the appropriate load factors
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Ultimate Limit State Analysis Sheet No
10Reference Calculation
For SecondOrder effectsSeeSupportingNotesSection 71 ampSection 72
Figure 6‐Frame imperfections equivalent horizontal forces
52 Partial safety factors and second order effects
Second order effects increases not only the deflections but also the momentsand forces beyond those calculated by the first order Second-order analysis isthe term used to describe analysis method in which the effects of increasingdeflections under increasing load are considered explicitly in the solutionmethod
The effects of the deformed geometry are assessed in EN 1993-1-1 bycalculating alpha crit (αcrit) factor The limitations to the use of the first-orderanalysis are defined in EN 1993-1-1 Section 521 (3) as αcrit
15 for plastic
analysis When a second order analysis is required there are two main methodsto proceed
1) Rigorous 2nd
order analysis (ie using appropriate second order software)2) Approximate 2
nd order analysis (ie hand calculation using first order analysis
with magnification factors) Although the modifications involveapproximations they are sufficiently accurate within the limits given by EN1993-1-1
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Ultimate Limit State Analysis Sheet No
12Reference Calculation
Load Factor Hinge number Member Hinge status102 1 RHC Formed
114 2 LHR Formed
Table 1‐ Position of Hinges and Load factors
Figure 8‐ ndash Member forces (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Initial sizing if members Sheet No
7Reference Calculation
TP0843EC30816Manual forhe design of
steelworkbuildingstructures toEC3
SeeAppendix Bor themethod
4 Initial Sizing of members
Figure 5‐ Dimensions of portal (The institutionof Structural Engineers TP0843 EC30816)
a) Lh = 306 = 5rL = 157730=00526
b) Loading1) Gravity loading
Snow loading = 054 x 72 = 380KNmSelf weight = 065 x 72 = 468 KNm
2) Factored loadw= (468 x 135 ) + (380 x 15 ) = 120 KNm
c) Finding Mp for the sections
1) Total load on the frame (wL)= 120 x 30 = 3605KN
2) Parameter wl2 = 120 x 302 = 10816 KNm
3) From Graphs (Figure B2) obtain horizontal force ratio (036)
H= 036 x 3605 = 1298 KN
4) From Graphs (Figure B3) obtain rafter Mp ratio (0034)MrafterRd = 0034 x 10816 = 3677 KNm
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Initial sizing if members Sheet No
8Reference Calculation
SectionTables ofUniversalBeams
EN 1993-1-12005 (E)Table 31
5) From Graphs (Figure B4) obtained column Mp ratio (0063)Mcolumn Rd = 0063 x 10816 = 6814 KNm
6) Selecting membersa) Wpl (rafter)required = (3677 x 10
6) 275 = 1337 x 10
3 cm
3
Try UB 457x152x74
b) Wpl(column)required = (6814 x 106)275= 2478 x 10
3 cm
3
Try UB 533 x 210 x 109
bull Properties Rafter Section UB 457x152x74
G=742 Kgm h= 462mm b=1544mmtw=96mm tf =17mm A=9448 x 102 mm2
d=428mm
Iy= 32670 x 104mm
4 Wply=1627 x 10
3 mm
3
iy=186 mm iz = 333 mmIz = 1047 x 104 mm4 Wplz = 2131 x 103 mm3 It = 6618 x 10
4 mm
4 Iw = 5163 x 10
6 mm
6
bull Properties Column Section UB 533x210x109
G=109 Kgm h= 5395mm b=2108mmtw=116mm tf =188mm A=1389 x10
2 mm
2
d=5109mm
Iy= 66820 x 104mm
4 Wply=2828 x 10
3 mm
3
iy=2187 mm iz = 457 mmIz = 2692 x 10
4 mm
4 Wplz = 3994 x 10
3mm
3
It = 1016 x 104 mm4 Iw = 1811 x 106 mm6
Steel grade is S275 Assume Sections Class1 then check
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Ultimate Limit State Analysis Sheet No
9Reference Calculation
EN 1993-1-12005 (E)Sec 532
SeeSupportingNotesSection 9
5 Load Combination (Max vertical Load) (Dead + Snow)
51Frame imperfections equivalent horizontal forces
Oslash Oslash
Oslash = radic 05 1 5
Oslash = 354 x 10-3
The column loads could be calculated by a frame analysis but a simple calculationbased on plan areas is suitable for single storey portals
(i) Permanent loads ( un-factored )
Rafter = (745 x 15 x 98) 103 = 11 KNRoofing = (15 x 02 x 72) = 216 KNServices = (15 x 02 x 72) = 216 KN
_________Total = 542 KN
(ii) Variable loads ( un-factored )
Snow load = 15 x 054 x 72 = 583 KN
Thus the un-factored equivalent horizontal forces are given by
(i) Permanentcolumn = 354 x 10-3
x 542 = 019 KN
(ii) Variablecolumn = 354 x 10-3
x 583 = 021 KN
Note EC3 requires that all loads that could occur at the same time are consideredtogether so the frame imperfection forces and wind loads should be considered asadditive to permanent loads and variable loads with the appropriate load factors
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Ultimate Limit State Analysis Sheet No
10Reference Calculation
For SecondOrder effectsSeeSupportingNotesSection 71 ampSection 72
Figure 6‐Frame imperfections equivalent horizontal forces
52 Partial safety factors and second order effects
Second order effects increases not only the deflections but also the momentsand forces beyond those calculated by the first order Second-order analysis isthe term used to describe analysis method in which the effects of increasingdeflections under increasing load are considered explicitly in the solutionmethod
The effects of the deformed geometry are assessed in EN 1993-1-1 bycalculating alpha crit (αcrit) factor The limitations to the use of the first-orderanalysis are defined in EN 1993-1-1 Section 521 (3) as αcrit
15 for plastic
analysis When a second order analysis is required there are two main methodsto proceed
1) Rigorous 2nd
order analysis (ie using appropriate second order software)2) Approximate 2
nd order analysis (ie hand calculation using first order analysis
with magnification factors) Although the modifications involveapproximations they are sufficiently accurate within the limits given by EN1993-1-1
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Ultimate Limit State Analysis Sheet No
12Reference Calculation
Load Factor Hinge number Member Hinge status102 1 RHC Formed
114 2 LHR Formed
Table 1‐ Position of Hinges and Load factors
Figure 8‐ ndash Member forces (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Member checks - Column Sheet No
23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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Member checks - Column Sheet No
24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Initial sizing if members Sheet No
8Reference Calculation
SectionTables ofUniversalBeams
EN 1993-1-12005 (E)Table 31
5) From Graphs (Figure B4) obtained column Mp ratio (0063)Mcolumn Rd = 0063 x 10816 = 6814 KNm
6) Selecting membersa) Wpl (rafter)required = (3677 x 10
6) 275 = 1337 x 10
3 cm
3
Try UB 457x152x74
b) Wpl(column)required = (6814 x 106)275= 2478 x 10
3 cm
3
Try UB 533 x 210 x 109
bull Properties Rafter Section UB 457x152x74
G=742 Kgm h= 462mm b=1544mmtw=96mm tf =17mm A=9448 x 102 mm2
d=428mm
Iy= 32670 x 104mm
4 Wply=1627 x 10
3 mm
3
iy=186 mm iz = 333 mmIz = 1047 x 104 mm4 Wplz = 2131 x 103 mm3 It = 6618 x 10
4 mm
4 Iw = 5163 x 10
6 mm
6
bull Properties Column Section UB 533x210x109
G=109 Kgm h= 5395mm b=2108mmtw=116mm tf =188mm A=1389 x10
2 mm
2
d=5109mm
Iy= 66820 x 104mm
4 Wply=2828 x 10
3 mm
3
iy=2187 mm iz = 457 mmIz = 2692 x 10
4 mm
4 Wplz = 3994 x 10
3mm
3
It = 1016 x 104 mm4 Iw = 1811 x 106 mm6
Steel grade is S275 Assume Sections Class1 then check
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Ultimate Limit State Analysis Sheet No
9Reference Calculation
EN 1993-1-12005 (E)Sec 532
SeeSupportingNotesSection 9
5 Load Combination (Max vertical Load) (Dead + Snow)
51Frame imperfections equivalent horizontal forces
Oslash Oslash
Oslash = radic 05 1 5
Oslash = 354 x 10-3
The column loads could be calculated by a frame analysis but a simple calculationbased on plan areas is suitable for single storey portals
(i) Permanent loads ( un-factored )
Rafter = (745 x 15 x 98) 103 = 11 KNRoofing = (15 x 02 x 72) = 216 KNServices = (15 x 02 x 72) = 216 KN
_________Total = 542 KN
(ii) Variable loads ( un-factored )
Snow load = 15 x 054 x 72 = 583 KN
Thus the un-factored equivalent horizontal forces are given by
(i) Permanentcolumn = 354 x 10-3
x 542 = 019 KN
(ii) Variablecolumn = 354 x 10-3
x 583 = 021 KN
Note EC3 requires that all loads that could occur at the same time are consideredtogether so the frame imperfection forces and wind loads should be considered asadditive to permanent loads and variable loads with the appropriate load factors
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Ultimate Limit State Analysis Sheet No
10Reference Calculation
For SecondOrder effectsSeeSupportingNotesSection 71 ampSection 72
Figure 6‐Frame imperfections equivalent horizontal forces
52 Partial safety factors and second order effects
Second order effects increases not only the deflections but also the momentsand forces beyond those calculated by the first order Second-order analysis isthe term used to describe analysis method in which the effects of increasingdeflections under increasing load are considered explicitly in the solutionmethod
The effects of the deformed geometry are assessed in EN 1993-1-1 bycalculating alpha crit (αcrit) factor The limitations to the use of the first-orderanalysis are defined in EN 1993-1-1 Section 521 (3) as αcrit
15 for plastic
analysis When a second order analysis is required there are two main methodsto proceed
1) Rigorous 2nd
order analysis (ie using appropriate second order software)2) Approximate 2
nd order analysis (ie hand calculation using first order analysis
with magnification factors) Although the modifications involveapproximations they are sufficiently accurate within the limits given by EN1993-1-1
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Ultimate Limit State Analysis Sheet No
12Reference Calculation
Load Factor Hinge number Member Hinge status102 1 RHC Formed
114 2 LHR Formed
Table 1‐ Position of Hinges and Load factors
Figure 8‐ ndash Member forces (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Ultimate Limit State Analysis Sheet No
9Reference Calculation
EN 1993-1-12005 (E)Sec 532
SeeSupportingNotesSection 9
5 Load Combination (Max vertical Load) (Dead + Snow)
51Frame imperfections equivalent horizontal forces
Oslash Oslash
Oslash = radic 05 1 5
Oslash = 354 x 10-3
The column loads could be calculated by a frame analysis but a simple calculationbased on plan areas is suitable for single storey portals
(i) Permanent loads ( un-factored )
Rafter = (745 x 15 x 98) 103 = 11 KNRoofing = (15 x 02 x 72) = 216 KNServices = (15 x 02 x 72) = 216 KN
_________Total = 542 KN
(ii) Variable loads ( un-factored )
Snow load = 15 x 054 x 72 = 583 KN
Thus the un-factored equivalent horizontal forces are given by
(i) Permanentcolumn = 354 x 10-3
x 542 = 019 KN
(ii) Variablecolumn = 354 x 10-3
x 583 = 021 KN
Note EC3 requires that all loads that could occur at the same time are consideredtogether so the frame imperfection forces and wind loads should be considered asadditive to permanent loads and variable loads with the appropriate load factors
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Ultimate Limit State Analysis Sheet No
10Reference Calculation
For SecondOrder effectsSeeSupportingNotesSection 71 ampSection 72
Figure 6‐Frame imperfections equivalent horizontal forces
52 Partial safety factors and second order effects
Second order effects increases not only the deflections but also the momentsand forces beyond those calculated by the first order Second-order analysis isthe term used to describe analysis method in which the effects of increasingdeflections under increasing load are considered explicitly in the solutionmethod
The effects of the deformed geometry are assessed in EN 1993-1-1 bycalculating alpha crit (αcrit) factor The limitations to the use of the first-orderanalysis are defined in EN 1993-1-1 Section 521 (3) as αcrit
15 for plastic
analysis When a second order analysis is required there are two main methodsto proceed
1) Rigorous 2nd
order analysis (ie using appropriate second order software)2) Approximate 2
nd order analysis (ie hand calculation using first order analysis
with magnification factors) Although the modifications involveapproximations they are sufficiently accurate within the limits given by EN1993-1-1
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Ultimate Limit State Analysis Sheet No
12Reference Calculation
Load Factor Hinge number Member Hinge status102 1 RHC Formed
114 2 LHR Formed
Table 1‐ Position of Hinges and Load factors
Figure 8‐ ndash Member forces (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Member checks - Column Sheet No
23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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Member checks - Column Sheet No
24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Member checks - Column Sheet No
25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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Member checks - Column Sheet No
26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Member checks ndash Rafter Sheet No
30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Ultimate Limit State Analysis Sheet No
10Reference Calculation
For SecondOrder effectsSeeSupportingNotesSection 71 ampSection 72
Figure 6‐Frame imperfections equivalent horizontal forces
52 Partial safety factors and second order effects
Second order effects increases not only the deflections but also the momentsand forces beyond those calculated by the first order Second-order analysis isthe term used to describe analysis method in which the effects of increasingdeflections under increasing load are considered explicitly in the solutionmethod
The effects of the deformed geometry are assessed in EN 1993-1-1 bycalculating alpha crit (αcrit) factor The limitations to the use of the first-orderanalysis are defined in EN 1993-1-1 Section 521 (3) as αcrit
15 for plastic
analysis When a second order analysis is required there are two main methodsto proceed
1) Rigorous 2nd
order analysis (ie using appropriate second order software)2) Approximate 2
nd order analysis (ie hand calculation using first order analysis
with magnification factors) Although the modifications involveapproximations they are sufficiently accurate within the limits given by EN1993-1-1
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Ultimate Limit State Analysis Sheet No
12Reference Calculation
Load Factor Hinge number Member Hinge status102 1 RHC Formed
114 2 LHR Formed
Table 1‐ Position of Hinges and Load factors
Figure 8‐ ndash Member forces (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
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Ultimate Limit State Analysis Sheet No
12Reference Calculation
Load Factor Hinge number Member Hinge status102 1 RHC Formed
114 2 LHR Formed
Table 1‐ Position of Hinges and Load factors
Figure 8‐ ndash Member forces (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Member checks - Column Sheet No
23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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Member checks - Column Sheet No
24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Member checks - Column Sheet No
25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
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Ultimate Limit State Analysis Sheet No
12Reference Calculation
Load Factor Hinge number Member Hinge status102 1 RHC Formed
114 2 LHR Formed
Table 1‐ Position of Hinges and Load factors
Figure 8‐ ndash Member forces (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Member checks - Column Sheet No
23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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Member checks - Column Sheet No
24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Member checks - Column Sheet No
25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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Member checks - Column Sheet No
26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Member checks ndash Rafter Sheet No
30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Ultimate Limit State Analysis Sheet No
13Reference Calculation
SeeSupportingNotesSection7321
See Figure 9
521 Sway buckling mode Stability ( αcrsest)
αcrsest = 08 1
is the axial force in rafter see figure 8 (1508KN)
is the Euler load of rafter full span
Where is the in-plan second moment of area of rafterL is the full span length
= 752 KN
is the minimum value for column 1 to n
is the horizontal deflection for top of column as indicated in
Appendix
is the axial force in columns see figure 8 (2075KN 2081KN)
bull As can be seen that
is the lateral deflection at the top of each column
subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can bechosen and using analysis software the deflection at top of each column can beobtained
1) Arbitrary load HEHF=50KN2) = 98mm= 98mm
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
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Ultimate Limit State Analysis Sheet No
14Reference Calculation
So either
OR
Min = min(1475 1475 ) = 1475
Thus
αcrsest = 08 1 1475 95
Figure 9‐ Sway mode check (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Member checks - Column Sheet No
23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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Member checks - Column Sheet No
24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Member checks - Column Sheet No
25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Member checks ndash Rafter Sheet No
30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
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Ultimate Limit State Analysis Sheet No
15Reference Calculation
SeeSupporting NotesSection7322 amp
Section 72
16162710 2753 0 1 0 2386
522 Snap-through buckling stability (αcrrest )
αcrrest = tan2
D cross-section depth of rafter (462mm)
L span of the bay (30m) h mean height of the column (6m) in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 10
4 mm
4 )
nominal yield strength of the rafter (275 Nmm2)
roof slope if roof is symmetrical (6o)
Fr Fo the ratio of the arching effect of the frame whereFr = factored vertical load on the rafter ( 432 KN see section 3)F0 = maximum uniformly distributed load for plastic failure of the rafter
treated as a fixed end beam of span L
= 181
Thus
αcrrest = tan26
αcrrest = 62
bull Henceαcrest =min ( αcrsest αcrrest ) = 62
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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Member checks - Column Sheet No
24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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University of Sheffield
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Revised by EC3 Plastic Portal Frame DesignPrepared by
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
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Ultimate Limit State Analysis Sheet No
16Reference Calculation
SeeSupportingNotes 733
bull Although the snap-through failure mode is critical mode as shown in calculationabove but because this example is for designing single bay portal frames the snap-through mode of failure is irrelevant but included to show complete design steps forsimple portal frame design Snap-through failure mode can be critical mode in threeor more spans as internal bay snap-through may occur because of the spread ofthe columns inversion of the rafter (The institutionof Structural Engineers TP0843
EC30816) see figure 10
Figure 10‐ Snap through failure mode critical for 3 bay or more
532Accounting Second Order effects To account for second order effects the partial safety factors can be modified by thefollowing criteria
1) γG = 150
2) γQ = 168
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Member checks - Column Sheet No
23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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Member checks - Column Sheet No
24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Member checks - Column Sheet No
25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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Member checks - Column Sheet No
26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Member checks ndash Rafter Sheet No
30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Ultimate Limit State Analysis Sheet No
17Reference Calculation
SeeSupportingNotes 733
bull Re-analyze the first order problem with the modified safety factors using same initialsized sections gives the following results
Load Factor Hinge number Member Hinge status
092 1 RHC Formed
102 2 LHR Formed
Table 1‐ Hinges obtained from analysis
bull It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 issuitable although hinge 1 occurs at a load factors le 1 a mechanism is not formeduntil the second hinge is formed Therefore this combination of section sizes issuitable
bull Hence size of member initially estimated is suitable and can withstand second-ordereffects Note that if the load factors in positions 1 and 2 were less than 1 then themembers size needs to be increased to sustain second order effects as the initiallysized members cannot sustain second order effects
Figure 11‐Bending moment diagram for first order analysis (Burgess 20011990)
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Member checks - Column Sheet No
23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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Member checks - Column Sheet No
24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Member checks - Column Sheet No
25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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Member checks - Column Sheet No
26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Member checks ndash Rafter Sheet No
30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
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SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
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Ultimate Limit State Analysis Sheet No
18Reference Calculation
Figure 12‐Member forces for first‐order analysis (Burgess 20011990)
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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Member checks ndash Column Sheet No
20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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16022009
Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
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Member checks ndash Purlins Sheet No
19Reference Calculation
SeeSupporting
notessection 104
Note Herehe safetyactors areused asndicated in
King spanoad table
SeeAppendix C
6 Member checks
61 Purlins
Today the design of the secondary members is dominated by cold formed sectionsThe lsquodesignrsquo of cold formed members consists of looking up the relevant table for thechosen range of sections The choice of a particular manufacturersquos products isdependent on clients or designerrsquos experiences and preferences Table (Appendix C) illustrates a typical purlin load table based on information from manufacturersquoscatalogue (King span) for double span conditions As the overall distance betweencolumns is 30 meters which is assumed to be divided to 18 equal portions wouldgives purlin centers 167 meters (on the slope) The gravity loading (dead (claddingLoad plus snow load) is w= (01x 14) + (054 x 16) = 1004 KNm
2 From the Table
(Appendix C) knowing the purlin length of 72 m purlin spacing of 125m and thegravity load to be supported by purlin 1004KNm
2 the M175065120 section seems
adequate
Figure 13‐ Connection between rafter section and purlins
Figure 14‐ Purlin cross‐section (Kingspan)
Purlin
Rafter
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See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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20Reference Calculation
See Figure11
EN 1993-1-12005 (E)Section 55
Seesupportingnotes section123
62 Column (UB 610 x 229 x 101)
- MEd = 9047 KNm- VEd = 1501 KN- NEd = 2082 KN
621Classification
Web ( Bending + Axial )
ε = 275235 =108
actual (dtw ) = 4404 72ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 4 6 1 9 ε Class1
bull So the column sections are overall class 1
622Cross section resistance
The frame analysis assumed that there is no reduction in the plastic momentresistance from interaction with shear force or axial force This assumption mustbe checked
6221Shear force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 626)
VEd lt 05 VplRd
Av = 104 h tw = 104 x 5395 x 116 = 65085 mm2
VplRd = radic 3 γ VplRd = 65085275radic 3 11 10 9394 KN05 VplRd = 4697 KN
bull VEd lt 05 VplRd so the plastic moment of resistance is not reduced by thecoexistence of axial force
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21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Member checks ndash Column Sheet No
21Reference Calculation
Seesupportingnotes section
124
Seesupportingnotes section134
38 574 7561 235
6222Axial force effects of Plastic moment resistance (EN 1993-1-1 2005 (E) Sec 629)
Checki If
NEd lt 2082 lt
2082 lt 7278
ii If
NEd lt 025 NplRd
NEd lt 025 plastic tensile resisitance of the section
NEd lt
2082 lt
2082 lt 8681
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
623Stability against lateral and torsional buckling (EN 1993-1-1 2005 (E) Sec BB321)
The design of the frame assumes hinge forms at the top of the column
member immediately below the haunch level The plastic hinge position must betorsionally restraint in position by diagonal stays With the hinge position restraintthe hinge stability is ensured by EC3 by limiting distance between hinge and thenext lateral restraint to Lm
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22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Member checks - Column Sheet No
22Reference Calculation
38457
= 153 m
bull Thus there must be a lateral restraint at a distance from the hinge not exceeding(153m)
bull Therefore if 15 meters spacing assumed this would ensure the stability betweenthe intermediate restraints at the top of the column where maximum bendingmoment occurs then the spacing of 18 meters is OK for sheeting rails below 24meters from the top of the column where the moment is lower
Figure 15‐ Column member stability (Plum 1996)
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Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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Member checks - Column Sheet No
24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Member checks - Column Sheet No
23Reference Calculation
Seesupporting
notes section134
It must be checked that the column buckling resistance is sufficient so that the column is
stable between the tensional restraint at S2 and the base This part of the column would be
checked using slenderness calculated
Different countries have different procedure to calculate the slenderness of the column and
check the susceptibility of this part to lateral tensional buckling Thus the designer must refer
to the national Annex In this example the procedure used in for assessing the significance of
the mode of failure is taken from (King Technical Report P164)
Figure 16‐ Column between tensional restraints (King Technical Report P164)
(a) Calculate slenderness λ and λ LT
Assume side rail depth = 200 mm
Figure 17‐ Column Sheeting rails cross‐section (King Technical Report P164)
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24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Member checks ndash Rafter Sheet No
30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Member checks - Column Sheet No
24Reference Calculation
King
Technical
Report P164)
Distance from column shear center to center of the side rail a
a = 60732 + 2002 = 36975 mm
is2 = iy
2 + iz2 + a2
is2 = 21872 + 4572 + 369752 = 186633 mm2
Distance between shear center of flanges
hs = h ndash tf = 5395 ndash 188 = 5207mm
α =
using the simplification for doubly symmetrical I sections
Iw = Iz ( hs 2 )2
α =
α =
= 1122
The slenderness of the column is given by
1
= 6435
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Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Member checks - Column Sheet No
25Reference Calculation
Appendix D
Figure D1
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Where
mt is moment factor obtained from appendix D Because loads combination
considered there is no lateral loads applied to the walls so there are no intermediate
loads
ψ = 0 6031 = 0
y = 82632 ( Lt iz ) = 82632 (4000 457 ) = 0944
mt = 053
c =1 for uniform depth members
053 1
6435= 421
(b) Calculate buckling resistance for axial force
1ФФ Ф 05 1 02 hb = 53952108 = 256
curve b for hot rolled I sections
α= 034
828 6435828 078
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26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
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Member checks - Column Sheet No
26Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
Ф 0 5 1 03407802 078 090
Ф 05 1 02 Ф 0 5 1 021048502 0485 065
Xz = 1090 090 078 = 0741 = (0741 x 1389 x 102 x 275) (11 x 103 )= 257413KN
(c) Calculate buckling resistance for bending
MbRdy =
868 = 421 868 = 0485
1ФФ 1065 065 0485 = 092
MbRdy =
6504
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
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SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
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EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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27Reference Calculation
King
Technical
Report P164)
μLT 00606
1 μLT 10
1 006062082100741138910 275 10
(d) Calculate buckling resistance to combined axial and bending
1
Ψ = 0
βMLT = 18 ndash 07 Ψ = 18 ndash 07 (0) = 18
μLT 015 βMLT 015 but μLT 09μLT 015 078 18015 but μLT 09
0996
= 091
bull The column is OK and stable over the section considered (between restraint S o and S2)
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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28Reference Calculation
EN 1993-1-12005 (E)Section 55
See Figure11 Burgess
20011990)
63Rafter (UB457 x 191 x 89)
631Section Classification
Ensure the section is class 1 to accommodate plastic hinge formation
ε = 275235 =108
Web ( combined axial and bending )
actual (dtw ) = 446
446 le72 ε Class 1
Flanges ( Axial Compressive )
actual (ctf )= 3 6 6 9 ε Class1
bull The rafter section is Class 1
632 Cross-section Resistance
The frame analysis assumed that there is no reduction in the plastic moment
resistance from interaction with shear force or axial force This assumption mustbe checked because it is more onerous than that the cross-sectional resistanceis sufficient
- Max shear force VEd = 1605 KN at haunch tip
- Max axial force NEd = 1669 KN at haunch tip
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Member checks ndash Rafter Sheet No
30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
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Member checks ndash Rafter Sheet No
30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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Date16022009 Appendix A Sheet No
43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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University of Sheffield
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
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Member checks ndash Rafter Sheet No
30Reference Calculation
Seesupportingnotes section132
EN 1993-1-1
2005 (E) Sec
BB322
See section61
38 574 7561 235
38 333 166910574944810 1627 10 275756 1 9448 10 6618 10 235
633 Check rafter buckling in apex region Another highly stressed region is the length of rafter in which the lsquoapexrsquo hingeoccur see fig below Under (dead + snow) loading the outstand flange is intension while compression flange is restrained by purlinrafter connection
Therefore the buckling resistance of the rafter member between purlins in theapex region needs to be checked Because the lsquoapexrsquo hinge is the last to form inorder to produce a mechanism (which is true for low pitched portal frame underdead + snow loading) then adequate rotation capacity is not a designrequirement ie hinge is required only to develop Mp not to rotateIt is set by EC3 EN1993-1-1 2005 that if the value of Lm (as defined in BB311)is not exceeded by restraint lateral torsional buckling can be ignored Soassuming that the purlins act as restraint because of their direct attachment to thecompression flanges in the apex hinge region then the purlin spacing should notexceed
Lm = 1221 mm = 1221m
bull As purlin spacing is 167m (on slope) thus because Lm has been smaller than167m then the purlin spacing would have to be reduced in the apex region to 12m
Figure 18‐ Member stability apex region (Plum 1996)
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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Date16022009 Appendix A Sheet No
43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
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Member checks ndash Rafter Sheet No
32Reference Calculation
EN1993-1-
12005
Sec 6322
EN1993-1-
12005
Table 63
1 210000 1047 101670 10 5163 10 1047 10 1670 81000 6618 10 210000 1047 10
6481
(b) Calculate buckling resistance to bending moment
bull
Take C1 = 1 (conservative)
bull λ LT =
λ LT = = 0831
bull ФLT = 05 [ 1 + α ( λ ndash 02 ) + λ2 ]ФLT = 05 [ 1 + 021( 0831 ndash 02 ) + 08312 ] = 0912
bull XLT =
Фradic Ф
XLT = radic = 080
bull MbRdy =
MbRdy =
= 3254KNm
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
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Member checks ndash Rafter Sheet No
33Reference Calculation
(c) Calculate buckling resistance to combined axial and bending
Check that
+ kLT
1
Take kLT = 1 ( conservative )
+ 1 x 1
1 1
The value is slightly greater than one but due to the conservative assumption of
KLT=1 the rafter can be assumed to be stable between intermediate restraint(purlinsheeting rails) and purlin spacing could be increased to 167m between apexand hunch region shown in figure 19 Otherwise if the value was significantly greaterthan 1 the purlin spacing (167m) should be reduced
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EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
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Date16022009 Member checks - Rafter Sheet No
34Reference Calculation
EN 1993-1-12005 (E)Section 55
64Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic raftersections as it is the normal practice to use a UB cutting of the same serial size asthat of the rafter section for the haunch which is welded to the underside of thebasic rafter (UB 457x191x 89) Therefore it is assumed that the haunch has anoverall depth at connection is 090 m
641Classification ε = 275235 =108
WebThe web can be divided into two and classified according to stress
and geometry of each
actual (dtw ) = 446
web 1 ( bending ) -------- 446le72 ε Class 1web 2 ( Compressive) --- 446le38 ε Class 2
Flanges ( Axial Compressive
actual (ctf )= 366 9ε Class1
bull Thus the haunch section is a class 2
Figure 20‐ Haunch region cross‐section classification
(King Technical Report P164)
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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Date16022009 Appendix A Sheet No
43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
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Member checks - Haunch Sheet No
35Reference Calculation
SeeSupportingnotes
section 131 amp 133
642Haunch Stability First check the stability of the haunched portion of the rafter ( from eavesconnection to the haunch rafter intersection) as this represents one of the mosthighly stressed lengths and with its outstand flange (inner) in compression this partof the rafter is the region most likely to fail due to instability As it has alreadydecided to stay the inside corner of the columnhaunch intersection (column hingeposition) assume that the haunchrafter intersection is also effectively torsionallyrestrained be diagonal braces giving an effective length of 3m as indicated in Figbelow
Figure 21‐ Member stability haunch‐rafter region (Plum 1996)
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Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Date
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Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Date
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Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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Date16022009 Appendix A Sheet No
43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
8132019 Plastic Design Portal Frame to Ec3
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Revised by EC3 Plastic Portal Frame DesignPrepared by
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Appendix C Sheet No
49Reference Calculation
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Member checks ndash Haunch Sheet No
36Reference Calculation
EN 1993-1-12005 (E)clauseBB312 (3)B
EN 1993-1-12005 (E)sectionBB313
Seesupportingnotes sectionAppendix B
It would appear that clause BB312 (3)B is the most appropriate creation to checkthe stability of the haunched portion as there is three flanged haunch so thedistance between rotational restraint should be limited to
WhereLk is length limit specified where lateral torsional buckling effects can be
ignored where the length L of the segment of a member between restraintsection at a plastic hinge location and adjacent torsional restraint
Lk
Lk
= 3738 mm
Lk = 3738m
c is the taper factor (shape factor) which accounts for the haunching ofthe restraint length (BB333)
1
1 = 115
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
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Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
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Date
16022009
Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Date
16022009
Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date16022009 Appendix A Sheet No
43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Department of Civil Structural Engineering
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Date
16022009
Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
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Date
16022009
Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Date
16022009
Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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Date
16022009
Appendix C Sheet No
49Reference Calculation
8132019 Plastic Design Portal Frame to Ec3
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Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 3853
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Member checks - Haunch Sheet No
37Reference Calculation
Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐12005)
Is the modification factor for non-linear moment gradient (BB332)
The plastic moduli are determined for five cross-sections indicated on the figurebelow the actual cross-section considered are taken as being normal to the axis ofthe basic rafter (unhaunched member) The plastic moduli together with the relevant
information regarding the evaluation of the ratios NiMi are given in the followingtable The worst stress condition at the hunchrafter intersection (location 5) istaken
8132019 Plastic Design Portal Frame to Ec3
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8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4253
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4453
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix A Sheet No
43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
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Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
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Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix C Sheet No
49Reference Calculation
8132019 Plastic Design Portal Frame to Ec3
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Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
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8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
8132019 Plastic Design Portal Frame to Ec3
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Date
16022009
Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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University of Sheffield
Department of Civil Structural Engineering
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Date
16022009
Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
8132019 Plastic Design Portal Frame to Ec3
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Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date16022009 Appendix A Sheet No
43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
49Reference Calculation
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5353
eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4053
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Member checks - Haunch Sheet No
39Reference Calculation
Seesupporting
notes sectionAppendix A
The modification factor is determined by the form
in which R1 to R5 are the values of R according to equation below at the endsquarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)
In addition only positive values of ( ) should be included where- RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 )
-
Where (a) is the distance between the centroid of the member and the centroid ofrestraining members (such as purlins restraining rafter) Here for simplicity a
conservative value of (a) is found by conservative method of ignoring the middleflange as shown if figure (19)
Figure 24‐ Simplification for distance between
centriod of rafter and purlin sections
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4153
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4253
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4453
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix A Sheet No
43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4653
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4753
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
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University of Sheffield
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Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Date
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Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
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University of Sheffield
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Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix C Sheet No
49Reference Calculation
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Date
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Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
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Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
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Date
16022009
Member checks - Haunch Sheet No
40Reference Calculation
Seesupportingnotes section123
12 3 4 3 2 12086 3084 4089 3079 0862089086
117
radic 34 3
bull Thus this portion of the rafter is stable over the assumed restrained length of 3 m(haunch length) as Ls is around 3m
bull If the value was found to be less than the haunch length then a torsion restraint
should be provided in the haunch region as shown below
643 Cross-section resistance
6431Shear force effects of Plastic moment resistance
The shear in the rafter has been checked above showing that VEd lt 05 VplRd
In the haunch the shear area Av increases more than the applied shear VEdso the shear force has no effect on the plastic moment capacity of the haunch
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Date
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Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
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Date
16022009
Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
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Department of Civil Structural Engineering
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Date16022009 Appendix A Sheet No
43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
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Date
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Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
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Date
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Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
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Date
16022009
Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
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Date
16022009
Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
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Department of Civil Structural Engineering
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Date
16022009
Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
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Date
16022009
Appendix C Sheet No
49Reference Calculation
8132019 Plastic Design Portal Frame to Ec3
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Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
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eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4253
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Member checks - Haunch Sheet No
41Reference Calculation
Seesupportingnotes section124
6432Axial force effects of Plastic moment resistance
The tables provided below gives the axial and moment resistances of thehaunch section at points 1to 5 shown in figures 18 A series of checks iscarried out to determine whether the cross-sectional moment resistance MNRd is reduced by coexistence of axial force
PositionDistance
(mm)NEd
(KN) A
(mm2)
NplRd (KN)
Aweb (mm
2)
(Aweb f y)ymo (KN)
1 0 1711 16092 4023 8216 2054
2 075 1701 15064 3766 7190 1798
3 15 1691 14038 3510 6163 1541
4 225 1681 13010 3253 5136 1284
5 3 1671 11983 2996 4109 1027
bull NplRd = A f y ymo and f y=275Nmm2
Table 3 ndash Axial force at positions indicated in figure 18 for haunch
Position Distance
(mm) MEd
(KNm)
Is NEd gt Does Axial forceaffect plastic bending
resistance05 Aweb f y
ymo 025NplRd
1 0 950 No No No
2 075 850 No No No
3 15 751 No No No
4 225 652 No No No
5 3 553 No No No
Table 4‐ Checking the significance of axial force on plastic moment of resistance
bull Therefore the effect of shear and axial on the plastic moment resistance of thecolumn sections can be neglected according to EC3 EN1993-1-1 2005
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix A Sheet No
43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
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Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix C Sheet No
49Reference Calculation
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
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Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5353
eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4353
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Comparison between outcomes of different codes Sheet No
42Reference Calculation
7 Comparison between Different Codes
As the dimensions of portal frame designed in this worked example were deliberately chosen to beexactly the same as worked-example in (King Technical Report P147) a comparison was donebetween the carried out worked example to (BS EN 1993-1-12005) (BS9590-12000) and(ENV1993-1-11992)
The following is a summary of different outcomes and source of design
Designno
DesignCode
ColumnSection
size
Rafter SectionSize
HaunchLength
(m)
PurlinSpacing
(m)
DesignSource
1BS EN 1993-1-
12005533x210 UB
109457x152UB
743 167
Worked -example
2 BS9590-12000610x229 UB
113533x210 UB
82 3 185
(KingTechnical
ReportP164)
3ENV1993-1-
11992610x229 UB
113457x191UB
743 185
(KingTechnical
ReportP164)
Table 5‐ Comparison between different code outcome
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix A Sheet No
43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4653
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4753
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5053
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
49Reference Calculation
8132019 Plastic Design Portal Frame to Ec3
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University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5253
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5353
eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4453
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix A Sheet No
43Reference Calculation
EN 1991-1-
32003 (E)
Section 532
EN 1991-1-
32003 (E)
Table C1
8Appendix
A1 Roof shape coefficient
The values given in table A1 apply when the snow is not prevented from sliding off theroof Where the snow fences or other obstruction exists or where the lower edge of
the roof is terminated with a parapet then the snow load shape coefficient should notbe reduced below 08
Table A1- Snow load shape coeff icients (BS EN 1991-1-32003)
A2 Snow load relationships
The snow load on ground the characteristic value depends on the climatic region thefollowing table gives different expressions for different regions
Table A2- Altitude-Snow load relationships (BS EN 1991-1-32003)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4553
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4653
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4753
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4853
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4953
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5053
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
49Reference Calculation
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5153
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5253
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5353
eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4553
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix A Sheet No
44Reference Calculation
EN 1991-1-
32003 (E)
Figure C4
WhereSk is the characteristic snow load on the ground (KNm
2)
A is the site altitude above the sea level (m)Z is the zone number given on the map ( see fig A1 )
The following maps gives the zone number Z for UK and republic of Ireland if other Zvalues for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 )
Figure A1 UK Republic of Ireland snow loads at sea level (BS EN 1991-1-32003)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4653
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4753
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4853
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4953
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5053
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
49Reference Calculation
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5153
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5253
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5353
eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4653
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
45Reference Calculation
TP0843EC30816Manual forhe design ofsteelwork
buildingstructures toEC3
B1 Initial sizing using Wellerrsquos charts The method described relies for its simplicity on a series of three charts developed by
Alan Weller The chart has been constructed with the following assumptions and whichleads to reasonably economic solution (Note This is not a rigorous design method it isa set of rules to arrive at initial size)
1) The rafter depth is approximately span 552) The hunch length is approximately span 103) The rafter slope lies between 0o and 20o4) The ratio of span to eaves height is between 2 and 55) The hinges in the mechanism are formed at the level of the underside of the
haunch in the column and close to the apex
Each chart requires a knowledge of the geometry of the frame and the design loading asinput data in order to determine approximate sizes for the column and rafter members
Using of charts
Figure B1 Dimensions of portal (The insti tut ionof Structural Engineers TP0843EC30816)
a) Calculate the spanheight to eaves ratio = Lhb) Calculate the risespan ratio = rLc) Calculate the total design load FL on the frame and then calculate FL
2 where F is
the load per unit length on plan of span L (eg F =qs where q is the total factoredload per m
2 and s is the bay spacing)
d) From figure B2 obtain the horizontal force ratio HFR at the base from rL and Lhe) Calculate the horizontal force at the base of span H=HFR W L
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4753
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4853
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4953
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5053
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
49Reference Calculation
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5153
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5253
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5353
eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4753
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
46Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
f) From figure B3 obtain the rafter Mp ratio MPR from rL and Lhg) Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2h) From figure B4 obtain the column Mp ratio MPL from rL and rhi) Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L
2
j) Determine the plastic moduli for the rafter WplyR and the column WplyC fromWplyR =Mp(rafter) f y WplyC = Mp(column) f y
Where f y is the yield strength
Using the plastic moduli the rafter and column sections may be chosen from the range ofplastic sections as so defined in the section books
01 02 03015 025 035 045040
005
01
015
02
20 25 30 35 40 45 50
R i s e s p a n
HwL
Span to eaves height
Figure B2- Horizontal force at the base (The institutionof StructuralEngineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4853
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4953
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5053
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
49Reference Calculation
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5153
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5253
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5353
eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4853
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix B Sheet No
47Reference Calculation
TP0843EC30816Manual forhe design ofsteelworkbuilding
structures toEC3
2050 4045 35
3025
0
005
01
015
02
002 0025 003 0035 004 0045
Span to eaves height
R i s e s p a n
M wLsup2r
0
005
01
015
02
005 0055 006 007 0080065 0075
50 2045 2540 3035
0045
Span to eaves height
R i s e s p a n
M wLsup2pl
Figure B3- Mp ratio required for the rafter (The
Figure B4- Mp ratio required for the column (The institutionofStructural Engineers TP0843 EC30816)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4953
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5053
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
49Reference Calculation
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5153
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5253
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5353
eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 4953
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
48Reference Calculation
King SpanWebsiteLinkhttpwwwki
ngspanstructur
alcommultibe
amrpload_tableshtm
C1 King Span Multi-beam Purlin (Load tables)
Loading Load FactorDead load 14Dead load restraining uplift or overturning 10Dead load acting with wind and imposed loads combined 12Imposed load 16Imposed load acting with wind load 12Wind load 14Wind load acting with wind and imposed load 12Forces due to temperature effects 12
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5053
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
49Reference Calculation
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5153
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5253
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5353
eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5053
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix C Sheet No
49Reference Calculation
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5153
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5253
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5353
eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5153
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date
16022009
Appendix D Sheet No
50Reference Calculation
KingTechnicalReport P164)
D1 Equivalent uniform moment factor mt for all other cases
This formula derived by (Sinhgh 1969) is applicable in all cases especially when thebending moment diagram is not a straight line between the tensional restraints defining theends of the element
Figure D1- Moment factors (King Technical Report P164)
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5253
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5353
eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5253
University of Sheffield
Department of Civil Structural Engineering
Revised by EC3 Plastic Portal Frame DesignPrepared by
Cia06mh
Date16022009 Appendix D Sheet No
51Reference Calculation
KingTechnicalReport P164)
MEd1 to MEd5 are the values of the applied moments at the ends the quarter pointsand mid- length of the length between effective torsional restraints as shown inFigure D2 Only positive values of MEd should be included MEd is positive when itproduces compression in the unrestrained flange
Figure D2- Intermediate moment (King Technical Report P164)
D2 Equivalent section factor c
For uniform depth members c = 10
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5353
eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
8132019 Plastic Design Portal Frame to Ec3
httpslidepdfcomreaderfullplastic-design-portal-frame-to-ec3 5353
eferences BS EN 1991‐1‐32003 Eurocode 1 mdash Actions on structures Part 1‐3 General actions mdash Snow loads [Book] ‐ 389 Chiswick
High Road London W4 4AL Standards Institution British
BS EN 1993‐1‐12005 Eurocode 3 Design of steel structures BS EN 1993‐1‐12005 [Book] ‐ 389 Chiswick High Road
London W4 4AL Standards Institution British
Burgess Ian PLT Portal frame design Software ‐ 20011990 ‐ Vol Ver 13
King C M Design of Steel Portal for Europe [Book] ‐ [sl] The steel construction Institute Technical Report P164
Kingspan [Online] wwwkingspanstructuralcom ‐ February 10 2009 ‐httpwwwkingspanstructuralcompdfdouble_span_tiled_roofspdf
Plum L J Morris amp D R Structural Steelwrok Design to BS5950 2nd Edition [Book] ‐ [sl] Harlow Longman 1996
Sinhgh KP Ultimate behaviour of laterally supported beams [Book] ‐ University of Manchester [sn] 1969
SX016 Matthias Oppe Determination of loads on a building envelope [Online] wwwaccess‐steelcom ‐ Access Steel ‐
October 20 2008 ‐ httpwwwaccess‐steelcomDiscoveryResourcePreviewaspxID=J6osLkASHmChe7uBKVEzGw==
The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book] ‐
TP0843 EC30816
University of Sheffield
Department of Civil Structural Engineering
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