Metal Structures II Lecture V Crane supporting …footbridge.pl/stud/z/se2/lec205.pdf · or...

Metal Structures II

Lecture V

Crane supporting structures

Beams, columns, bracings

Contents

Structure of lectures → #t / 3

Fatigue resistance → #t / 4

Deformations → #t / 26

Connections → #t / 38

Columns → #t / 42

Example 1 → #t / 78

Example 2 → #t / 102

Bracings → #t / 134

Examination issues → #t / 137

Loads → Lecture #3

Beams: geometry of cross-section, resistance, instability prevention, local

effects, transverse stiffeners bumpers → Lecture #4

Beams: fatigue resistance, deformations; Columns: geometry of cross-section,

resistance, instability, deformations; Bracings → Lecture #5

Parts important for IInd design project

Structure of lectures

→ #3 / 3

DsE = smax - smin

smax = s (dead weight of structure + live load)

smin = s (dead weight of structure)

DtE = tmax - tmin

tmax = t (dead weight of structure + live load)

tmin = t (dead weight of structure)

Fatigue resistance

EN 1993-1-9

gFf DsE / (sR / gMf) ≤ 1,0

gFf DtE / (tR / gMf) ≤ 1,0

Checking of fatugue resistance will be presented on lecture #5, at now only

calculation of load is presented.

Fatigue loads

EN 1991-1-9 (8.2)

→ #3 / 67

DsE / (1,5 fy) ≤ 1,0

DtE / (1,5 fy / √3) ≤ 1,0

gFf DsE / (DsC gMf ) ≤ 1,0

gFf DtE,/ (DtC gMf ) ≤ 1,0

(gMf gFf DsE,/ DsC)3 + (gMf gFf DtE,/ DtC)5 ≤ 1,0

EN 1993-1-9 (8.1), (8.2), (8.3)

DsR = DsCm√ (2 ∙ 106 / NR)

DtR = DtCm√ (2 ∙ 106 / NR)

Resistance depends on number of cycles NR

m = 3 for NR = 100 000 - 5 000 000

m = 5 for NR = 5 000 000 - 100 000 000

DsR , DtR are constant for NR > 100 000 000

EN 1991-1-9 7.1 (2)

Photo: EN 1991-1-9 fig.7.1

→ #2 / 90

gMf

gFf = 1,0

EN 1993-1-9 tab. 3.1

Assessment method Consequence of failure

Low High

Damage tolerant 1,00 1,15

Safe life 1,15 1,35

EN 1993-1-9 3(7)

Damage tolerant method

• selecting details, materials and stress level so that in the event of the formation of

cracks a low rate of crack propagation and a long critical crack length would result

• provision of multiple load path

• provision of crack-arresting details

• provision of readily inspectable details during regular inspections

Safe-life method

• selecting details and stress levels resulting in a fatigue life sufficient to achieve the b-

values to be at least equal to those required for ultimate limit state verifications at the end

of the design service life

Stresses are calculated for cpecific value of live load:

DsE = sE (dead weight of structure + Qe) - sE (dead weight of structure) = sE (Qe)

DtE = tE (dead weight of structure + Qe) - tE (dead weight of structure) = tE (Qe)

Qe → lecture #3 / 70 - 78

EN 1991-3 (2.16)

EN 1991-3 (2.16), (2.19)

→ #3 / 70

DsE = DsE (Qe)

DtE = DtE (Qe)

Qe = Qmax,i jfat li

Qmax,i → #3 / 35, 36

jfat = max ( jfat,1 ; jfat,2)

jfat,1 = (1 + j1) / 2

jfat,2 = (1 + j2) / 2

li → #3 / 68

li damage equivalent factor

Simplified approach

EN 1991-3

Semi-accurate approach

EN 1993-1, EN 13001-1

Accurate approach

EN 1993-1, EN 13001-1

kQ - EN 13001-1

SX EN 1993-1 app. B EN 1993-1 tab. 2.11

li EN 1993-1 tab. 2.12 EN 1993-1 (2.17), (2.18)

EN 1991-3 2.12.1

kQ - load spectrum factor

SX - crane class

→ #3 / 71

Tab. 8.1 for plain members and mechanically fastened joints

Tab. 8.2 for welded built-up sections

Tab. 8.3 for transverse butt welds

Tab. 8.4 for weld attachments and stiffeners

Tab. 8.5 for load carying welded joints

Tab. 8.6 for hollow sections

Tab. 8.7 for lattice girder node joints

Tab. 8.8 for orthotropic decks - closed stringers

Tab. 8.9 for orthotropic decks open stringers

Tab. 8.10 for top flange to web junctions of runway beams

DsC, DtC:

EN 1993-1-9

Number of tables which are needed for calculations (tab. 8.1 –

cutting of elements):

Photo: Author

Tab. 8.1 Cutting of elements

DsC

DsC

DsC

DtC

Tab. 8.2 Web – flange welds

DsC , DtC

DsC , DtC

DsC , DtC

DsC , DtC

Tab. 8.4 Stiffeners – I-beam / C-section welds

DtC

DtC

Tab. 8.5 Cross-bar – stifener weld; addidtional information for web – flange welds

DtC

DtC

DtC

Tab. 8.6 Hollow section

Tab. 8.7 Lattice girder

For lattice surge girder and

truss beam

Tab. 8.10 Web – flange welds for top flange

DsC , DtC

DsC , DtC

DsC , DtC

DsC , DtC

Additionally, for truss run-beams, fatigue stresses from secondary moments for lattice

elements, especially consisting of hollow sections, must be taken into consideration.

Fatigue stresses from secondary bending moments in lattice memebers are analysed in EN

1993-1-9 tab. 4.1 and 4.2. Specific values of coefficients for these tables for crane

supporting structures are presented in EN 1993-6 tab. 5.4.

EN 1993-6 8.2

Stiffeners can't be welded to bottom flange. There are two possibilities of execution:

• Height of stiffener is lower than height of web (no contact between stiffener and bottom

flange);

• Between stiffener and bottom flange are inserted additional steel plate by method

"push-in" (thickness of plate a little bigger than wide of gap; full contact between

stiffener and bottom flange).

Photo: Author

Photo: Author

One-span beam is recommended structure - for each

type of loads top flange is always compressed and

bottom flange is always tensed.

Photo: Author

Multi-span beam is not recommended structure - for different types of loads both flanges are

sometimes compressed and sometimes tensed.

Photo: Author

Deformations

There are different values of horizontal and

vertical loads, acts on both run-beam. It

cases different deformations of beams. We

must take into consideration max and min

values of these loads = max and min value

of deformations.

Photo: Author

Photo: EN 1991-3 fig. 2.1

There are very important horizontal and vertical displacements for "normal" structures -

requirements for beams and columns:

Similar conditions exist for run-beams.

Member wmax or w3

Main roof girder (truss or beam) L / 250

Purlin L / 200

Corrugated sheel - roofing L / 150

Floor girder:

primary beam

secondary beam

L / 350

L / 250

Door head or window head L / 500

wmax = netto (total - precamber)

w3 = from variable actions

L -length of beam ot 2x length of cantilever

EN 1993-1-1 N.A. 22

It is recommended that horizontal

displacements do not exceed the

following limits:

• H / 150 for single-storey structures

(without cranes);

• H / 500 for multistorey structures;

H - level of considered girder to the

top of the foundation

Additionally, there must be analysed relative deformations of two run-beams:

Photo: Author

dz ≤ min( L / 600 ; 25 mm)

dpay ≤ L / 500

EN 1993-6 tab. 7.1

Vertical deformation, crane:

Vertical deformation, monorail hoist block:

dy ≤ L / 600

EN 1993-6 tab. 7.1

Horizontal deformation:

Ds = dleft + dright ≤ 10 mm

EN 1993-6 tab. 7.1

Change of spacing between centres of crane rail rail, including thermal changes:

EN 1993-6 tab. 7.1

Horizontal deflections and deviations of crane runways are considered together in crane

design. Acceptable deflections and tolerances demepn on the details and clearances in

guidance means. Provided that the clearance c between the crane wheel flanges and the

crane rail (or between the alternative guidance means and the crane beam) is also sufficent

to accomodate the necessary tolerance, larger deflection limits can be specified for each

project if agreed with the crane supplier and the client.

j4 = 1

provided that the tolerances for rail tracks as specified in EN 1993-6 are observed

othervise

→ EN 13001-2

j4 - dynamic effects induced when the crane is travelling;

EN 1991-3 tab. 2.4

↓

→ #3 / 31

hc ≤ s / 600

EN 1993-6 tab. 7.1

Difference between vertical deformations:

dy ≤ hc / 400

EN 1993-6 tab. 7.1

Horizontal displacement of frame / column at crane

support level:

V [m/s]

1 22

2 26

3 22

20 m /s

EN 1991-3 A.1 (6)

Strong wind - according to EN 1991-1-4:

Medium wind:→ #3 / 53

dy ≤ L / 600

dy ≤ L / 400

EN 1993-6 tab. 7.1

Difference between horizontal

displacements of adjaced frames / columns)

• out-off-service wind load (outdoor crane only):

• indoor crane (max wind load);

• outdoor crane (in-service wind load);

• combination: wind (max or in-service, respectively) + lateral

crane forces;

Connections

Surge connectors attaching the top flange of a runway beam to the supporting structure

should be capable of accommodating the rotation movements and vertical movements;

should take into account the possible need for lateral and vertical adjustment of the runway

beams.


Photo: EN 1993-6 fig 8.2

Photo: Author

Longitudinal bolted connection in compressed part of cross-section

Rail fixings


Joint of rails oblique and offset from the joint of beams

Columns

Photo: konar.eu

Photo: stabud.eu

Photo: zksgrzelak.eu

A B

C DPhoto: Author

→ #3 / 8

Additional part of column above crane. No additional part of column.

No additional part of column. No column.

Photo: Author

Monorail hoist block Overhead underslung crane

Run-beams suspended to structure.

Photo: Author

Photo: EN 1991-3 fig.1.2 Photo: EN 1991-3 fig.1.3

Run-beams supported by columns.

Overhead top-mounted crane

Photo: Author

Photo: EN 1991-3 fig.1.4

Overhead top-mounted crane

I-beam, hot-rolled or welded,

I-beam with reinforced top flange,

I-beam with surge girder,

Photo: Author

Photo: Author

Photo: EN 1991-3 fig.1.4

→ #4 / 10

Photo: crscranesystems.com

Photo: abuscranes.pl

Monorail hoist block and underslung crane - the most often solution is suspension run-

beams to structure of hall or additional portal frame.

Photo: youwaycrane.com

Photo: promag.pl

Top-mounted crane

Hot-rolled I-beam

Welded I-beam (plates)

Hot-rolled I-beam

Welded I-beam (plates)

On cantilever

Different cross-section for boht part of

column

Welded I-beam (C-sections, I-

beams, plates)

Laced members

Battened members

Welded I-beam (C-sections, I-beams,

plates)

Laced members

Battened members

Different cross-section for boht part of

column

Photo: Author

Cantilever, change of cross-section.

Photo: konar.eu

Cantilever, no change of cross-section.

Photo: udhavind.com

Welded I-beam, change of cross-section.

Photo: stabud.eu

Battened column, change of cross-section.

Photo:hak.com.pl

Laced column.

Photo: zksgrzelak.eu

Beam

Column

Small horizontal transversal loads, acts on run-beam → small bending moment in columns

→ no necessary high cross-section of column.

Photo: Author

Beam

Column

Big horizontal transversal loads, acts on run-beam → big bending moment in columns →

there is necessary high cross-section of column.

or battened or laced columnPhoto: Author

Change of column cross-section → problem with calculations of stability. There is no

information in Eurocode for this case. There is proposal of calculations according to

"Tablice do projektowania konstrukcji metalowych", W. Bogucki, M. Żyburtowicz, Arkady,

Warszawa 1984

Currently, the case is solved by numerical methods.

There is special way of calculations for laced columns and battened columns according to

Eurocode.

Photo: EN 1993-1-1 fig 6.7

n = 29 n = 41

For these three types of members (laced

compression members, battened

compression members) we should use

special way of calculations.

Of course, we can put full geometry

(chords and lacing system or batten

plates) into computer programm, but

each membes consist of many sub-

members

Photo: Author

Photo: Author

There is possible, that in structure we have more than 100 000 sub-members; it means very long term of calculations.

Photo: s9.flog.pl

h = 7 m

Example – difference between cross-

sectional values for normal cantilever,

battened cantilever and laced

cantilever.

Photo: Author

M [kNm] Q [kN] N [kN]

h0 = 30 cm

Photo: Author


h0 = 100 cm Photo: Author


There are big differences between distance between axes of chords, length of battens

and gap between chords. Accurate computer calculations for full geometry are possible

only for computer programme with possibility of putting „off-sets”.

Additionally, proportion between length of batten and high of batten is as for plate

member, not bar member.

Because of these reasons, we use special type of calculation as follow:

Photo: Author

General information about cross-sectional forces:

Recalculations

Photo: Author

Calculation as for single-bar

member; global values of MEd,

VEd, NEd

Local values of Mch, Ed, Mb, Ed, Vch, Ed,

Vb, Ed, Nch, Ed, NL, Ed

Photo: Author

Variuos modes of instability must be taken into consideration:

SV L / 2 L L / [ 1 + Ad h03 / AV d3 ) ] min {

24 X / [1 + 2 Jch h0 / (n Jb a )] ;

2 p X }

Jeff 0,5 h02 Ach 0,5 h0

2 Ach + 2 meff Jch

L = n E Ad a h02 / d3

X = E Jch / a2

n = 4 n = 2

l = m L / i0

i0 = √ [ J1 / ( 2 Ach ) ]

J1 = 0,5 h02 Ach + 2 Jch

l meff

0

2 - l / 75

1,0

≥ 150

≤ 75

75 - 150

EN 1993-1-1 tab. 6.8

↑ EN 1993-1-1 fig 6.7, 6.9, (6.72), (6.73), (6.74)

n - number of battens planes

h0 - distance between centres of

gravity of chords

Xch - geometical characteristic of

one chord

Jb - moment of inertia for cross-

section of batten

J = 2 zs2 Ach + 2 Jch

Photo: Author

VEd = p MEdII / (n L)

h0 = 2 zs

For chord:

Vch, Ed = VEd / 2

Mch, Ed = a VEd / 4

For batten:

Vb, Ed = VEd a / (2 h0)

Mb, Ed = a VEd / 2

EN 1993-1-1 fig 6.11

Nch, Ed = NEd / 2 + 2 MEdII zs Ach / (2 Jeff)

MEd II = NEd e0 / [1 - (NEd / Ncr) - (NEd / SV)]

e0 = L / 500

Ncr = p2 E Jeff, / (m L)2

EN 1993-1-1 6.4.1

For laces:

VEd = p MEd / (n L)

Nl, Ed = VEd / cos a = p MEd / (n L cos a)

Photo: Author

Influence of own stiffness and Steiner’s theorem on effective stiffness:

Jeff = 2 (h0 / 2)2 Ach + 2 meff Jch = 0,5 h02 Ach + 2 meff Jch

meff = 1

(rigid connection on

each point of chord)

meff = 0

(rigid connection not in each

point of chord, „large” h0)

0 ≤ meff ≤ 1

(rigid connection not in

each point of chord,

„average” h0)

Photo: Author

Photo: EN 1993-1-1 fig 6.7

There is inconsequence in Eurocode: according EN 1993-1-1 p.6.4.1.(1) - general

information - olny compressed members can be calculated acording to presented

procedure. On the other hand, in EN 1993-1-1 p.6.4.1.(6), global bending moment

is analysed. But there is no information, if resistance on global bending moment and

lateral buckling should be taken into consideration. For safety, these aspects should

be taken into account.

Example 1

300 kN

50 kN

Photo: Author

S 235

C 300

Ach = A (C 300) = 58,8 cm2

Jch, y = Jy (C 300) = 7 640 cm4

Jch, z = Jz1 (C 300) = 473 cm4

L = 7,000 m

h0 = 246 mm

a = 1,000 m

NEd = 300,000 kN

VEd = 50,000 kN

Mz, Ed, max = 50 ∙ 7 = 350,000 kNm

n = 2

Algorithm of analysis:

General calculations → #t / 80

Glogal behaviour of column → #t / 82

Local behaviour of one chord → #t / 92

Batten → #t / 99

Welds between battens and chords → #t / 101

General calculations

C 300 → Ist class of cross-section

e0 = L / 500 = 14 mm

J1 = Jz = 0,5 h02 Ach + 2 Jch, z1 = 18 737,704 cm4

i0 = √ [ J1 / ( 2 Ach ) ] = 12,62 cm

Cantilever: my = mz = mLT = 2,0

l = mz L / i0 = 2 ∙ 7 000 / 12,62 = 1 109,35 → #t / 73: meff = 0

Jeff = 0,5 h02 Ach + 2 meff Jch = 0,5 h0

2 Ach + 0 = 17 791,70 cm4

X = E Jch / a2 = E Jeff / a2 = 37 362,57 kN

Jb = 103 ∙ 1 / 12 = 83,33 cm4

SV = min {24 X / [1 + 2 Jch, z1 h0 / (n Jb a )] ; 2 p X } =

= min { 373 310,240 kN ; 234 755,951 kN } = 234 755,951 kN

Ncr = p2 E Jeff / (mz L)2 = 1 881,397 kN

MEdII = (NEd e0 + Mz, Ed, max) / [1 - (NEd / Ncr) - (NEd / SV)] = 422,035 kNm

Nch, Ed = NEd / 2 + MEdII h0 Ach / (2 Jeff) = 1 865,590 kN

Vch, Ed = p MEdII / n L = 94,704 kN

Mch, z1, Ed = Vch, Ed a / 4 = 23,676 kNm

Global behaviour of column:

Resistance:

Resistance for axial force → #t / 86

Resistance for shear force force → #t / 85

Resistance for bending moments → #t / 86

Interactions between shear force and bending moment, axial force and bending

moment → #t / 86

Stability:

Flexural buckling y-y → #t / 87

Flexural buckling z-z → #t / 88

Torsional buckling → #t / 89

Flexural-torsional buckling → #t / 89

Lateral buckling → #t / 90

Interactions between different types of buckling z-z → #t / 91

Analysed cross-section consists of two completely separated parts. There in no information

about:

formulas for interaction between aial force and bending moment;

way of calculations of Jw for torsional and lateral buckling.

Cross-section is rough approximated by I-beam. With of flanges is equal depth of C-section.

Area of flange is equal area of C-section. Flanges centres of gravity are at the same points as

centres of grawity for C-sections. Thickness of web tends to 0. Resistance for shear force

will be calculated for four flanges of C-section.

Photo: Author

J. Żmuda, „Podstawy projektowania konstrukcji metalowych”, TiT Opole 1992

AV = 4 ∙ 0,1 ∙ 0,016 = 64,000 cm2

Jy = 2 Ach (h0/ 2)2 = 17 791,704 cm4

JW = Jy (24,6 + 1,96) 2 / 4 = 3 137 716 cm6

JT = 2 ∙ 30 ∙ 1,963 / 3 = 76,832 cm3

Wz, pl = Wimm, pl = 2 Ach (h0/ 2) = 1 446,480 cm3

NRd = 2 Ach fy / gM0 = 2 763,600 kN

VRd = AV fy / (gM0 √3) = 868,335 kN

Mz, Rd = Wimm, pl fy / gM0 = 339,923 kNm

VEd / VRd = 50,000 / 868,335 = 0,058 < 0,5 → no interaction between VEd and Mz, Ed

a = 2 b = max (5n ; 1,0)

n = NEd / Npl, Rd = 300,000 / 2 763,600 = 0,109 → b = 1,0

a = min [ 0,5 ; (A - 2 b tf) / A]

(A - 2 b tf) / A = (for approximation, thickness of web → 0) = 0,0

a = min [ 0,5 ; 0,0] = 0,0

min ( 0,25 Npl, Rd ; 0,5 hw tw fy / gM0 ) = (for approximation, thickness of web → 0) = 0 kN

NEd > 0 kN → interaction between NEd and Mz, Ed.

MN,z, Rd = min [Mz, Rd ; Mz, Rd (1 - n) / (1 - 0,5 a) ] =

= [339,923 kNm ; 339,923 kNm (1 - 0,109) / (1 - 0,5 ∙ 0,0)] = 302,871 kNm

Mz, Ed / MN, z, Rd = 1,156 WRONG

y- y (material axis) flexural buckling

NEd = 300,000 kN (→ #t / 78)

NRd = 2 763,600 kN (→ #t / 85)

Jy = 2 Jch, y = 15 280 cm4

Lcr = 2 L = 14,000 m

iy = √ (Jy / A) = √ (2 Jch, y / 2 Ach) = 11,40 cm

ly = 1,308

Buckling curve c → a = 0,49

Fy = 1,627

cy = 0,385

NEd / (cy NRd) = 0,262 < 1,000 ok

Photo: Author

z- z (immaterial axis) flexural buckling

NEd = 300,000 kN (→ #t / 78)

NRd = 2 763,600 kN (→ #t / 85)

Jz = Jeff = 17 791,70 cm4

Lcr = 2 L = 14,000 m

iz = √ (Jz / A) = √ (Jeff / 2 Ach) = 12,30 cm

l = 1,212


F = 1,482

c = 0,404

NEd / (c NRd) = 0,250 < 1,000 ok

Photo: Author

Torsional buckling Ncr, T = [p2 EJw / (mT l0T)2 + GJT] / is2

i0 = √ (iy2 + iz

2) = 16,86 cm

is = √ (i02 + zs

2) = 16,86 cm

Ncr, T = 217 397,756 kN

Flexural-torsional buckling Ncr, z-T = {Ncr, i + Ncr, T - √ [(Ncr, i + Ncr, T)2 - 4 Ncr, i Ncr, T x] } / (2 x)

m = min [√ (mz / mLT) ; √ (mLT / mz)] = 1,0

x = 1 - (m zs2 / is

2) = 1

Ncr, i = min (Ncr, y ; Ncr, z) = Ncr, y = 1 615,795 kN

Ncr, z-T = 2 216,683 kN

Critical forces for torsional buckling and flexural-torsional buckling is bigger than

critical force for flexural buckling → flexural buckling is the most dangerous; there is

no need to control other modes.

Lateral buckling Mcr = is √ (Ncr, i Ncr, T) = 319,082 kNm

lLT = √ (MN,z, Rd / Mcr) = 0,974

aLT = 0,76

FLT = [1 + aLT (lLT - 0,2) + lLT2] / 2 = 1,267

cLT = min{ 1 / [FLT + √ (FLT2 - lLT

2)] ; 1,0} = 0,481

cLT, mod = 0,592

cLT, mod MN, z, Rd = 179,315 kNm

Cmy = Cmy = 0, 9

CmLT = 0,6

kyy = 1,035

kyz = 0,653

kzy = 0,902

kzz = 1,089

NEd / ( cy NRk / gM1) + kyy (My, Ed + DMy, Ed ) / ( cLT M y, Rk / gM1) +

+ kyz ( Mz, Ed + DMz, Ed ) / (M z, Rk / gM1) ≤ 1,0

0,262 + 1,035 ∙ 1,952 = 2,282 > 1,0 WRONG

NEd / ( cz NRk / gM1) + kzy (My, Ed + DMy, Ed ) / ( cLT M y, Rk / gM1) +

+ kzz ( Mz, Ed + DMz, Ed ) / (M z, Rk / gM1) ≤ 1,0

0,250 + 0,902 ∙ 1,952 = 2,011 > WRONG

Local behaviour of column:

Resistance:


Resistance for shear force → #t / 94



moment → #t / 95

Stability:

Flexural buckling z1-z1 → #t / 96




Analysed cross-section C-section. There is no information about:

formulas for interaction between axial force and bending moment;

Cross-section is rough approximated by I-beam. Resistance for shear force will be calculated

for two flanges of C-section.

Photo: Author

AV = 2 ∙ 0,1 ∙ 0,016 = 32,000 cm2

JW (C 300) = Jy (24,6 + 1,96) 2 / 4 = 73 400 cm6

JT = 2 ∙ 30 ∙ 1,963 / 3 = 40,500 cm3

Wpl, z1, min = 28,812 cm3

NRd = Ach fy / gM0 = 1 381,800 kN

VRd = AV fy / (gM0 √3) = 434,167 kN


Vch, Ed / VRd = 94,704 / 434,167 = 0,218 < 0,5 → no interaction between VEd and Mz, Ed

a = 2 b = max (5n ; 1,0)

n = NEd, ch / Npl, Rd = 1 865,590 / 1 381,800 = 1,350 → b = 6,751

a = min [ 0,5 ; (A - 2 b tf) / A]

(A - 2 b tf) / A = 0,456

a = min [ 0,5 ; 0,456] = 0,456

min ( 0,25 Npl, Rd ; 0,5 hw tw fy / gM0 ) = min (345,450 kN ; 352,500 kN) = 345,450 kN

NEd, ch > 345,450 kN → interaction between NEd and Mz, Ed.


= [6,771 kNm ; 6,771 kNm (1 - 1,350) / (1 - 0,5 ∙ 0,456)] = lower than zero value; nonsense. It is result of NEd, ch > Npl, Rd

Mz, Ed / MN, z, Rd >> 1,0 WRONG

z1- z1 flexural buckling

Nch, Ed = 1 865,590 kN (→ #t / 81)

NRd = 1 381,800 kN (→ #t / 94)

Jch, z = Jz1 (C 300) = 473 cm4

Lcr, z1 = a = 1,000 m

mz1 = 1,0

iz1 = √ (Jz1 / Ach) = 2,84 cm

l = 0,375


F = 0,613

c = 0,911

NEd / (c NRd) = 1,482 > 1,000 WRONG

Photo: Author


i0 = √ (iy2 + iz

2) = 12,05 cm

is = √ (i02 + zs

2) = 15,51 cm

Ncr, T = 141 009,672 kN

Flexural-torsional buckling Ncr, z-T = {Ncr, z1 + Ncr, T - √ [(Ncr, z1 + Ncr, T)2 - 4 Ncr, z1 Ncr, T x] }/(2 x)

m = min [√ (mz / mLT) ; √ (mLT / mz)] = 1,0

x = 1 - (m zs2 / is

2) = 0,954

Ncr, z1 = 9 336,646 kN

Ncr, z-T = 17 756,694 kN




Lateral buckling: bending moment acts about weak axis. Because of this, there is no laeral

buckling and no interaction between flexural and lateral buckling.

Battens:

Resistance:

Resistance for shear force force → #t / 100


Interactions between shear force and bending moment → #t / 100

Stability:

Element is to short to calculate lateral bending

Photo: Author

Nb,Ed = 0

Mb,Ed = Vch, Ed a / 2 = 47,352 kNm

Vb,Ed = Vch, Ed a / h0 = 384,975 kN

Cross section z-z: 2x rectangular 100 mm x 10 mm

Mch, Rd, z = 7,833 kNm

Vch, Rd = 271,354 kN

Calculations of resistances and interaction is made according to the

same way, as for other cross-section under shear force and bending

moment. But, at this example:

Vb,Ed / VRd > 1,0

Mb,Ed / MRd > 1,0

WRONGPhoto: Author

Welds: batten - chord

Nb,Ed = 0

Mb,Ed = Vch, Ed a / 2 = 54,721 kNm

Vb,Ed = Vch, Ed a / h0 = 444,862 kN

Photo: Author

S 235

C 300

Ach = A (C 300) = 58,8 cm2

Jch, y = Jy (C 300) = 7 640 cm4

Jch, z = Jz1 (C 300) = 473 cm4

L = 7,000 m

m = 2

h0 = 1,000 m

a = 1,000 m

NEd = 300,000 kN

VEd = 50,000 kN

Mz, Ed, max = 50 ∙ 7 = 350,000 kNm

n = 2

L 75x75x10

Ad = AV = A (L 75x75x10) =

= 14,1 cm2

d = 1,414 m

Example 2

Photo: Author

Algorithm of analysis:

General calculations → #t / 104

Glogal behaviour of column → #t / 106

Local behaviour of one chord → #t / 116

Laces → #t / 123

Welds between laces and chords → #t / 125

General calculations

C 300 → Ist class of cross-section

L 75x75x10 → Ist class of cross-section

e0 = L / 500 = 14 mm

Jeff = 0,5 h02 Ach = 294 000,000 cm4

L = n E Ad a h02 / d3 = 209 469,200 kN

SV = L / [ 1 + Ad h03 / AV d3 ) ] = 154 736,717 kN

Cantilever: my = mz = mLT = 2,0

Ncr = p2 E Jeff / (mz L)2 = 31 089,254 kN

MEdII = (NEd e0 + Mz, Ed, max) / [1 - (NEd / Ncr) - (NEd / SV)] = 358,353 kNm

Nch, Ed = NEd / 2 + MEdII h0 Ach / (2 Jeff) = 506,524 kN

Vch, Ed = p MEdII / n L = 80,412 kN

Mch, z1, Ed = Vch, Ed a / 4 = 20,103 kNm

Global behaviour of column:

Resistance:





moment → #t / 110

Stability:

Flexural buckling y-y → #t / 111

Flexural buckling z-z → #t / 112




Interactions between different types of buckling → #t / 115

Analysed cross-section consists of two completely separated parts. There in no information

about:

formulas for interaction between aial force and bending moment;

way of calculations of Jw for torsional and lateral buckling.

Cross-section is rough approximated by I-beam. With of flances is equal depth of C-section.

Area of flange is equal area of C-section. Flanges centres of gravity are at the same points as

centres of grawity for C-sections. Thickness of web tends to 0. Resistance for shear force

will be calculated for four flanges of C-section.

Photo: Author

J. Żmuda, „Podstawy projektowania konstrukcji metalowych”, TiT Opole 1992

AV = 4 ∙ 0,1 ∙ 0,016 = 64,000 cm2

Jy = 2 Ach (h0/ 2)2 = 294 000,000 cm4

JW = Jy (100 + 1,96) 2 / 4 = 764 094 358 cm6

JT = 2 ∙ 30 ∙ 1,963 / 3 = 76,832 cm3

Wz, pl = Wimm, pl = 2 Ach (h0/ 2) = 5 880,000 cm3

NRd = 2 Ach fy / gM0 = 2 763,600 kN

VRd = AV fy / (gM0 √3) = 868,335 kN


VEd / VRd = 50,000 / 868,335 = 0,058 < 0,5 → no interaction between VEd and Mz, Ed

a = 2 b = max (5n ; 1,0)

n = NEd / Npl, Rd = 300,000 / 2 763,600 = 0,109 → b = 1,0

a = min [ 0,5 ; (A - 2 b tf) / A]

(A - 2 b tf) / A = (for approximation, thickness of web → 0) = 0,0

a = min [ 0,5 ; 0,0] = 0,0

min ( 0,25 Npl, Rd ; 0,5 hw tw fy / gM0 ) = (for approximation, thickness of web → 0) = 0 kN

NEd > 0 kN → interaction between NEd and Mz, Ed.


= [1381,800 kNm ; 1381,800 kNm(1 - 0,109) / (1 - 0,5 ∙ 0,0)] = 1231,184 kNm

Mz, Ed / MN, z, Rd = 0,284 ok

y- y (material axis) flexural buckling

NEd = 300,000 kN (→ #t / 102)

NRd = 2 763,600 kN (→ #t / 109)

Jy = 2 Jch, y = 15 280 cm4

Lcr = 2 L = 14,000 m

iy = √ (Jy / A) = √ (2 Jch, y / 2 Ach) = 11,40 cm

ly = 1,308


Fy = 1,627

cy = 0,385

NEd / (cy NRd) = 0,262 < 1,000 ok

Photo: Author

z- z (immaterial axis) flexural buckling

NEd = 300,000 kN (→ #t / 102)

NRd = 2 763,600 kN (→ #t / 109)

Jz = Jeff = 294 000,000 cm4

Lcr = 2 L = 14,000 m

iz = √ (Jz / A) = √ (Jeff / 2 Ach) = 50,00 cm

l = 0,298


F = 0,568

c = 0,951

NEd / (c NRd) = 0,106 < 1,000 ok

Photo: Author


i0 = √ (iy2 + iz

2) = 51,28 cm

is = √ (i02 + zs

2) = 51,28 cm

Ncr, T = 33 062,163 kN

Flexural-torsional buckling Ncr, z-T = {Ncr, i + Ncr, T - √ [(Ncr, i + Ncr, T)2 - 4 Ncr, i Ncr, T x] } / (2 x)

m = min [√ (mz / mLT) ; √ (mLT / mz)] = 1,0

x = 1 - (m zs2 / is

2) = 1

Ncr, i = min (Ncr, y ; Ncr, z) = Ncr, y = 1 615,795 kN

Ncr, z-T = 3 231,400 kN




Lateral buckling Mcr = is √ (Ncr, i Ncr, T) = 3 747,952 kNm

lLT = √ (MN,z, Rd / Mcr) = 0,607

aLT = 0,76

FLT = [1 + aLT (lLT - 0,2) + lLT2] / 2 = 0,839

cLT = min{ 1 / [FLT + √ (FLT2 - lLT

2)] ; 1,0} = 0,705

cLT, mod = 0,868

cLT, mod MN, z, Rd = 1 199,198 kNm

Cmy = Cmy = 0, 9

CmLT = 0,6

kyy = 1,141

kyz = 0,605

kzy = 0,302

kzz = 0,900

NEd / ( cy NRk / gM1) + kyy (My, Ed + DMy, Ed ) / ( cLT M y, Rk / gM1) +

+ kyz ( Mz, Ed + DMz, Ed ) / (M z, Rk / gM1) ≤ 1,0

0,262 + 1,141 ∙ 0,292 = 0,595 < 1,0 ok

NEd / ( cz NRk / gM1) + kzy (My, Ed + DMy, Ed ) / ( cLT M y, Rk / gM1) +

+ kzz ( Mz, Ed + DMz, Ed ) / (M z, Rk / gM1) ≤ 1,0

0,106 + 0,302 ∙ 0,292 = 0,194 < 1,0 ok

Local behaviour of column:

Resistance:





moment → #t / 119

Stability:

Flexural buckling z1-z1 → #t / 120




Analysed cross-section C-section. There is no information about:

formulas for interaction between axial force and bending moment;

Cross-section is rough approximated by I-beam. Resistance for shear force will be calculated

for two flanges of C-section.

Photo: Author

AV = 2 ∙ 0,1 ∙ 0,016 = 32,000 cm2

JW (C 300) = Jy (24,6 + 1,96) 2 / 4 = 73 400 cm6

JT = 2 ∙ 30 ∙ 1,963 / 3 = 40,500 cm3

Wpl, z1, min = 28,812 cm3

NRd = Ach fy / gM0 = 1 381,800 kN

VRd = AV fy / (gM0 √3) = 434,167 kN


Vch, Ed / VRd = 80,412 / 434,167 = 0,185 < 0,5 → no interaction between VEd and Mz, Ed

a = 2 b = max (5n ; 1,0)

n = NEd, ch / Npl, Rd = 506,524 / 1 381,800 = 0,367 → b = 1,833

a = min [ 0,5 ; (A - 2 b tf) / A]

(A - 2 b tf) / A = 0,456

a = min [ 0,5 ; 0,456] = 0,456

min ( 0,25 Npl, Rd ; 0,5 hw tw fy / gM0 ) = min (345,450 kN ; 352,500 kN) = 345,450 kN

NEd, ch > 345,450 kN → interaction between NEd and Mz, Ed.


= [6,771 kNm ; 6,771 kNm (1 – 0,367) / (1 - 0,5 ∙ 0,456)] = 5,552 kNm

Mz, Ed / MN, z, Rd = 3,621 > 1,0 WRONG

z1- z1 flexural buckling

Nch, Ed = 506,524 kN (→ #t / 105)

NRd = 1 381,800 kN (→ #t / 118)

Jch, z = Jz1 (C 300) = 473 cm4

Lcr, z1 = a = 1,000 m

mz1 = 1,0

iz1 = √ (Jz1 / Ach) = 2,84 cm

l = 0,375


F = 0,613

c = 0,911

NEd / (c NRd) = 0,402 < 1,000 ok

Photo: Author


i0 = √ (iy2 + iz

2) = 12,05 cm

is = √ (i02 + zs

2) = 15,51 cm

Ncr, T = 141 009,672 kN

Flexural-torsional buckling Ncr, z-T = {Ncr, z1 + Ncr, T - √ [(Ncr, z1 + Ncr, T)2 - 4 Ncr, z1 Ncr, T x] }/(2 x)

m = min [√ (mz / mLT) ; √ (mLT / mz)] = 1,0

x = 1 - (m zs2 / is

2) = 0,954

Ncr, z1 = 9 336,646 kN

Ncr, z-T = 17 756,694 kN




Lateral buckling: bending moment acts about weak axis. Because of this, there is no laeral

buckling and no interaction between flexural and lateral buckling.

Laces resistance

VEd = p MEdII / (n L) = 117,643 kN

Cross section: L 75x75x10

NRd = Ad fy = 331,350 kN

Compression:

Horizontal bar

Nl, Ed = VEd

Lcr = 1,000 m

m = 1,0

Cross bar

Nl, Ed = VEd / cos 45o

Lcr = 1,414 m

m = 1,0

L-section:

Flexural buckling u-u

Flexural buckling v-v

Torsional buckling

Fexural-torsional buckling

Photo: Author

Welds: laces-chord

Nl, Ed

Ml,Ed = 0 kNm

Vl,Ed = 0 kN

l1 d1 = l2 d2Photo: Author

l1

l2

d1

d2

NEd / NRd C → #t / 86 C → #t / 110

VEd / VRd C → #t / 85 C → #t / 109

MEd / MRd D → #t / 86 C → #t / 110

VEd ↔ MEd C → #t / 85 C → #t / 109

NEd ↔ MEd D → #t / 86 C → #t / 110

Flexural buckling y-y C → #t / 87 C → #t / 111

Flexural buckling z-z C → #t / 88 C → #t / 112

Torsional buckling C → #t / 89 C → #t / 113

Flexural-torsional C → #t / 89 C → #t / 113

Lateral buckling D → #t / 90 C → #t / 114

Flexural ↔ lateral D → #t / 91 C → #t / 115

Conclusions – global behaviour

Photo: EN 1993-1-1 fig 6.7

NEd / NRd D → #t / 95 C → #t / 119

VEd / VRd C → #t / 94 C → #t / 118

MEd / MRd D → #t / 95 D → #t / 119

VEd ↔ MEd C → #t / 94 C → #t / 118

NEd ↔ MEd D → #t / 95 D → #t / 119

Flexural buckling z1-z1 D → #t / 96 C → #t / 120

Torsional buckling C → #t / 97 C → #t / 121

Flexural-torsional C → #t / 97 C → #t / 121

Lateral buckling C → #t / 97 C → #t / 122

Flexural ↔ lateral C → #t / 98 C → #t / 122

Conclusions – local behaviourPhoto: EN 1993-1-1 fig 6.7

In analised case, laced column is better than battened column. But C-section is too

weak to bead local bending moment in both cases. This column should be

recalculated to more massive cords (for example I-section).

There are two spearated bases of column in case of laced or battened column.

Photo: Author

There are only axial force and shear

force in each base of column’s

chord.

There are different combinations of

loads. It is possible, that there are

various cases of reactions for

various combinations. Resistance of

bases should be calculated for two

different direction of reaction

(contact steel-concrete or tension in

anchor bolts and local bending base

plate).

Photo: quatronsteel.com

Photo: inzynierbudownictwa.pl

There should be massive anchor system in case of tensile force in column’s chord. First of all –base of column’s chord must be able to resist of big value of force from anchor bolts.

Photo: Author

Anchoring in the concrete foundation is solved in various ways. The easiest - for small

forces - it is the friction between anchor bolts and concrete and resistance of the extended

tip of the bolts.

Photo: Post-installed concrete anchors in nuclear power plants: Performance and qualification, Ph.

Mahrenholtz, R. Eligehausen Nuclear Engineering and Design 287 / 2015

https://www.google.pl/url?sa=i&rct=j&q=&esrc=s&source=imgres&cd=&cad=rja&uact=8&ved=0ahUKEwiBkci2k4jSAhXBjCwKHXSgCVwQjRwIBw&url=http%3A%2F%2Fwww.sciencedirect.com%2Fscience%2Farticle%2Fpii%2FS0029549315001089&psig=AFQjCNHAfCCkwc1q_HLtnuCcflDFRg9_gg&ust=1486903753160867

There are used J-anchor bolts for big value of force. These anchors can be weld to the

reinforced steel of base

Photo: peikko.caPhoto: civil-engg-world.blogspot.com

https://www.google.pl/url?sa=i&rct=j&q=&esrc=s&source=imgres&cd=&ved=0ahUKEwj1maiahYjSAhXDIJoKHVzICP4QjRwIBw&url=http%3A%2F%2Fwww.peikko.ca%2Fproduct-category-en_ca%2Fcat%3DCast-in-Place%2520Concrete%2520Connections&psig=AFQjCNHMahC03LEg7GBcRoPLoBUgfXehkw&ust=1486902395285675

https://www.google.pl/url?sa=i&rct=j&q=&esrc=s&source=imgres&cd=&ved=0ahUKEwj4046nhYjSAhWEECwKHaFfDasQjRwIBw&url=https%3A%2F%2Fcivil-engg-world.blogspot.com%2F2011%2F06%2Fsteel-mezzanines.html&psig=AFQjCNGcXvPwNkfSvNNg-PJi1ZbxvJ2CXA&ust=1486902422602204

There are used massive anchor assemby (retaining plates) for extremely value of force.

Photo: homemadetools.netPhoto: strongtie.com

Resistance of bases are calculate constant value of stresses under base plate.

Loading Lever arms Resistance Mj, Rd

Left-C, Right-C, exampe:

MEd > 0 ; NEd < 0 z = zC, l + zC, r

e = MEd / NEd

NEd ≤ 0 0 < e < zC, l NEd ≤ 0 -zC, r < e ≤ 0

min [ -z FC, l, Rd / (1 + zC, r / e)

-z FC, r, Rd / (-1 + zC, l / e)]

min [ -z FC, l, Rd / (1 + zC, r /

e)

-z FC, r, Rd / (-1 + zC, l / e)]

Left-T, Right-T, exampe:

MEd > 0 ; NEd > 0 z = zT, l + zT, r

e = MEd / NEd

NEd > 0 0 < e < zT, l NEd > 0 -zT, r < e ≤ 0

min [ z FT, l, Rd / (1 + zT, r / e)

z FT, r, Rd / (-1 + zT, l / e)]

min [ z FT, l, Rd / (1 + zT, r / e)

z FT, l, Rd / (-1 + zT, l / e)]

EN 1993-1-8 tab. 6.7

Bracings

Recommendation for wall / column bracings: no contact between bracings and run-beam.

Photo: konar.eu

Contact between bracings and run-beam → multi-span run-beam.

There is one-span run-beam recommended for fatigue calculations (#t / 24 - 25).

Additionally, on bracings act directly vertical loads from crane.

Photo: Author

Recommended shape of bracings

Distance between columns ≤ 6,0 m Distance between columns > 6,0 m

Photo: Author

Fatigue calculations for crane runbeams

Displacements and deformations of crane supporting structures

Laced column and battened column – similiarities and differeces

Algorithm of calculation for laced column

Algorithm of calculation for battened column

Examination issues

Surge connectors - podporowe elementy złączneButt weld - spoina czołowaFilled weld - spoina pachwinowaNotch - karb Laced members - słup wielogałęziowy skratowany Battened members - słup wielogałęziowy z przewiązkami Closely spaced build-up members - pręt wielogałęziowyShear stiffeness - sztywność postaciowaEfficency factor - wskaźnik efektywności

Metal Structures II Lecture V Crane supporting …footbridge.pl/stud/z/se2/lec205.pdf · or...

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