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Eurocode 3 Folder : Bolted connections
Bolted angilar connection:
EdS
EdEdEd
Dimensions of connection:
Plate thickness d = 15,00 mmBolt spacing e = 35,00 mmSpacing of bolts e o = 30,00 mmSpacing of bolts b = 70,00 mmSpacing of bolts h = 220,00 mm
Angle of tension force = 45,00
Bolts:Bolt = SEL("steel/bolt"; BS; ) = M 20SC = SEL("steel/bolt"; SC; ) = 8.8f u,b = TAB("steel/bolt"; fubk; SC=SC) = 800,00 N/mmHole diameter d l= TAB("steel/bolt"; d; BS=Bolt) = 20,00 mmShaft diameter d ll= d l+2 = 22,00 mmCross-section areas A s = TAB("steel/bolt"; Asp; BS=Bolt) = 2,45 cm
Plate:steel = SEL("steel/EC"; Name; ) = Fe 510f y = TAB("steel/EC"; fy; Name=steel) = 355,00 N/mmf u = TAB("steel/EC"; fu; Name=steel) = 510,00 N/mm M0 = 1,10 Mb = 1,25
Loads:S Ed = 424,26 kNNEd = COS( ) * S Ed = 300,00 kNVEd = SIN( ) * S Ed = 300,00 kN
Force per bolt:
F t,Ed =NEd
4= 75,00 kN
Fv,Ed =VEd
4= 75,00 kN
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Eurocode 3 Folder : Bolted connections
Limit tension force of bolts:
F t,Rd = *0,9*f u, b A s
* Mb 10= 141,12 kN
F t,Ed
F t ,Rd = 0,53 < 1
Check limit shear force for bolts:
Fv,Rd = 0,6 **f u, b * d l
2
*4000 Mb= 120,64 kN
F v,Ed
F v,Rd= 0,62 < 1
Check combined:
+F t,Ed
*1,4 F t ,Rd
F v,Ed
F v,Rd= 1,00 1
Analysis of bearing strength:as in 6.5.1.3: e o / d ll = 1,36 > 1as in 6.5.1.3: e o / d ll = 1,36 < 1,5as in 6.5.1.2(1): e / d ll = 1,59 > 1,2as in 6.5.1.2(3): b / d ll = 3,18 > 3
= MIN(e
*3 d ll;
b
*3 d ll - 0,25;
f u,bf u
; 1) = 0,53
Tab.6.5.3 F b1,Rd = 2,5 * * dl * d *f u
* Mb 103
= 162,18 kN
6.5.5.(10) F b2,Rd =2
3 * F b1,Rd = 108,12 kN
Fb,Rd = F b2,Rd + (F b1,Rd - F b2,Rd ) *
-e o
d l1,2
0,3= 162,18 kN
F v,Ed
F b,Rd= 0,46 < 1
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Eurocode 3 Folder : Bolted connections
Bolted connection subject to bending
Dimensions of connection:a = 80,00 mma 1 = 45,00 mme = 80,00 mme 1 = 45,00 mme 2 = 50,00 mme 3 = 90,00 mmd1 = 20,00 mmd2 = 8,00 mmb1 = 160,00 mm
h2 = 2 * (e 2 + e 3 ) = 280,00 mm
Flange-Bolts:Bolt f = SEL("steel/bolt"; BS; ) = M 20SC1 = SEL("steel/bolt"; SC; ) = 4.6f u,bo = TAB("steel/bolt"; fubk; SC=SC1) = 400,00 N/mmHole diameter d lo = TAB("steel/bolt"; d; BS=Bolt f ) = 20,00 mmShaft diameter d l1o = d lo + 2 = 22,00 mmNumber of rows n o = 3
Steg-Bolts:Bolt S = SEL("steel/bolt"; BS; ) = M 16SC2 = SEL("steel/bolt"; SC; ) = 4.6f u,bs = TAB("steel/bolt"; fubk; SC=SC2) = 400,00 N/mmHole diameter d ls = TAB("steel/bolt"; d; BS=Bolt S ) = 16,00 mmShaft diameter d l1s = d ls + 2 = 18,00 mmNumber of rows n s = 3Number of columns m s = 2
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Eurocode 3 Folder : Bolted connections
Profil:Profil Typ = SEL("steel/Profils"; Name; ) = IPESelected Profil = SEL("steel/"Typ; Name; ) = IPE 240Column height h = TAB("steel/"Typ; h; Name=Profil) = 240,00 mmWeb thickness s = TAB("steel/"Typ; s; Name=Profil) = 6,20 mmFlange thickness t = TAB("steel/"Typ; t; Name=Profil) = 9,80 mmCross-sectional area A = TAB("steel/"Typ; A; Name=Profil) = 39,10 cmMoment of resistance W y = TAB("steel/"Typ; Wy; Name=Profil) = 324,00 cmMoment of resistance W pl = 1,14 * W y = 369,36 cm
Material and Partial safety factors:steel = SEL("steel/EC"; Name; ) = Fe 360f y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mmf u = TAB("steel/EC"; fu; Name=steel) = 360,00 N/mm M0 = 1,10 M2 = 1,25
Loads:MEd = 157,95 kNmVEd = 60,75 kN
Check beam-web strength without holes:
Mc,Rd =*W pl f y
* M0 103
= 78,91 kNm
MEd
Mc,Rd
= 2,00 < 1
A v = 1,04 **h s
100= 15,48 cm
Vpl,Rd = * A vf y
** 3 M0 10= 190,93 kN
VEd
Vpl,Rd= 0,32 < 1
Check beam-web strength with holes:
A z = 0,5 * A = 19,55 cm
A z,net = - A z
*2 *d l1o +t *-n s 1
2*d l1s s
100= 14,12 cm
*f y
f u
M2 M0
*0,9 A z,net
A z
= 1,14 1
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Eurocode 3 Folder : Bolted connections
Wpl,net = W pl - 2 * d l1o * t *-h t
2000= 319,73 cm
Mc,Rd,C =*W pl,net f y
* M0 103
= 68,31 kNm
MEd
Mc,Rd,C= 2,31 < 1
A v,net = Av - n s * dl1s *s
100= 12,13 cm
*f y
f u
Av
Av,net= 0,83 < 1
Area of holes will be neglected.
Analysis of bolts in flange:
Iw =*2 *d 2 h 2
3
*12 104
= 2926,93 cm 4
If = 2 * b 1 * d1 *( )+h d 12
2
104
= 10816,00 cm 4
Mw = MEd *Iw
+Iw If = 33,64 kNm
Mf = MEd *If
+Iw If = 124,31 kNm
Effective tension force in flange:
NEd =M f
+h d 1 * 10 = 478,12 kN
Limit shear force
Fv,Rd = n o * 2 * 0,6 *
*f u,bo *4d lo 2
*103
M2= 361,91 kN
Limit bearing force:
= MIN(a 1
*3 d l1o;
a
*3 d l1o - 0,25;
f u,bo
f u; 1) = 0,682
Fb,Rd = 2 * no * 2,5 * * dlo * t *f u
* M2 103
= 577,46 kN
MAX(
NEd
F v,Rd;
NEd
F b,Rd ) = 1,32 < 1
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Eurocode 3 Folder : Bolted connections
Analysis of bolts in web:
ns1 =-n s 1
2 + 0,49 = 1
ns2 = n s1 * 2 = 2
m s1 =-m s 1
2 + 0,49 = 1
m s2 = m s1 * 2 = 2
Ip =
*n s *m s2 +( )*-m s 12 e2
*n s2 *m s ( )*-n s 12 e 32
100= 420,00 cm
MEw = Mw +*VEd e 3
103
= 39,11 kNm
REd = +( )*10 *MEw e 3Ip2
( )+VEd*m s n s *10 *MEw *-m s 1
2
e
Ip
2
= 96,27 kN
Fv,Rd = m s * 0,6 * f u,bs *4
*d ls
2
* M2 103
= 77,21 kN
= MIN(e 1
*3 d l1s;
e
*3 d l1s - 0,25;
f u,bo
f u; 1) = 0,833
Fb,Rd = 2,5 * * dls * s *f u
* M2 103
= 59,50 kN
MAX(R Ed
F v,Rd;
R Ed
F b,Rd ) = 1,62 < 1
Analysis of cover plates:
A =*d 1 b 1
100= 32,00 cm
A net = A -*2 *d l1o d 1
100= 23,20 cm
Limit tension force:
Nt,Rd = MIN(A *f y
* M0 10; 0,9 * A net *
f u
* M2 10) = 601,34 kN
N Ed
N t ,Rd= 0,80 < 1
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Eurocode 3 Folder : Bolted connections
Connection subject to moment load:
7
1 2 3
54
86
e
e
e
e
ee
e e
2
2
3
3
1 1
Dimensions of connection:
Bolt spacing e = 50,00 mmEdge distance e1 = 35,00 mmBolt spacing e3 = 70,00 mm
Plate thickness t = 6,00 mmNumber of bolts n = 8
Bolts:Bolt1 = SEL("steel/bolt"; BS; ) = M 30SC1 = SEL("steel/bolt"; SC; ) = 5.6f u,b1 = TAB("steel/bolt"; fubk; SC=SC1) = 500,00 N/mmHole diameter d l1 = TAB("steel/bolt"; d; BS=Bolt1) = 30,00 mm
Bolt2 = SEL("steel/bolt"; BS; ) = M 12SC2 = SEL("steel/bolt"; SC; ) = 5.6f u,b2 = TAB("steel/bolt"; fubk; SC=SC2) = 500,00 N/mm
Hole diameter d s2 = TAB("steel/bolt"; d; BS=Bolt2) = 12,00 mmShaft diameter d l2 = d s2 +1 = 13,00 mm
steel = SEL("steel/EC"; Name; ) = Fe 360f u2 = TAB("steel/EC"; fu; Name=steel) = 360,00 N/mm Mb = 1,25 Mo = 1,25
Loads:
VEd = 58,70 kN
MEd = 910,00 kNcm
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Eurocode 3 Folder : Bolted connections
Analysis of the right bolt:Check limit shear force
Fv,Rd = 0,6 * *4 ( )d l110
2
*f u,b1
*10 Mo= 169,65 kN
VEd
F v,Rd= 0,35 < 1
Limit bearing strength
Fb,Rd = 1,5 * d l1 * t *f u2
*103
Mb= 77,76 kN
VEd
F b,Rd= 0,75 < 1
Check bolts on left:
Ip =*6 ( )+e 2 e 3
2
100= 444,00 cm
Maximum horizontal force in bolt due to moment:
Fh = MEd *e 3
*Ip 10= 14,35 kN
Maximum vertical force:
Fv = *MEd +e
*Ip 10
VEd
n= 17,59 kN
Resulting force
Rr = +F h2
F v2 = 22,70 kN
Fv,Rd = 0,6 * f u,b2 * dl2 *
*4 * Mb 103
= 31,86 kN
R r
F v,Rd= 0,71 < 1
Analysis of bearing strength
= MIN(e 1
*3 d l2; 2 *
e
*3 d l2 - 0,25;
f u,b2
f u2; 1) = 0,90
Fb,Rd = 2,5 * * dl2 * t *f u2
* Mb 103
= 50,54 kN
F h
F b,Rd= 0,28 < 1
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Eurocode 3 Folder : Bolted connections
Bolted shear joint:
Dimensions of connection:Spacing of bolts a = 45,00 mmSpacing of bolts a 0 = 35,00 mmSpacing of bolts a 1 = 30,00 mmSpacing of bolts a 2 = 50,00 mmSpacing of bolts a 3 = 90,00 mmDepth of cope a 4 = 30,00 mmPlate projection a s = 20,00 mmLever-arm x = 45,00 mmNumber of bolts n = 3
Bolts:M16 4.6Bolt = SEL("steel/bolt"; BS; ) = M 16SC = SEL("steel/bolt"; SC; ) = 4.6f u,b = TAB("steel/bolt"; fubk; SC=SC) = 400,00 N/mmHole diameter d l= TAB("steel/bolt"; d; BS=Bolt) = 16,00 mmShaft diameter d 0= d l+2 = 18,00 mm
Angle section P = SEL("steel/WG"; Name; ) = L 80x8 Angle thickness t = TAB("steel/WG"; s; Name=P) = 8,00 mmFactor for group bolts as in 6.5.2..2(3):k = 0,5 for 1 group bolts; =2,5 for 2 rows
steel = SEL("steel/EC"; Name; ) = Fe 360f y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mmf u = TAB("steel/EC"; fu; Name=steel) = 360,00 N/mm
Profil Typ = SEL("steel/Profils"; Name; ) = IPESelected Profil = SEL("steel/"Typ; Name; ) = IPE 270Column height h = TAB("steel/"Typ; h; Name=Profil) = 270,00 mmWeb thickness s = TAB("steel/"Typ; s; Name=Profil) = 6,60 mm
Mb = 1,25 M2 = 1,25 M0 = 1,10
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Eurocode 3 Folder : Bolted connections
Loads:VEd = 60,00 kN
Check bolt spacing as in 6.5.1:1,2 * d 0 / a 1 = 0,72 < 1a 1 / MAX(12 * t; 150) = 0,20 < 11,5 * d 0 / a 0 = 0,77 < 1a 0 / MAX(12 * t; 150) = 0,23 < 12,2 * d 0 / a 2 = 0,79 < 1a 2 / MIN(12 * s; 200) = 0,63 < 1
Check bolts strength:
Bolts in web of main beam F v,Ed =VEd
*2 n= 10,00 kN
Bolts in web of connecting beam MEd
= *x
10V
Ed= 270,00 kNcm
Horizontal force due to M Ed Fh,Ed =MEd
*( )-n 1a 2
10
= 27,00 kN
Distributed shearing force on bolts F v,Ed =VEd
n= 20,00 kN
Maximum bolt strength F Ed = +F h,Ed2
F v,Ed2 = 33,60 kN
Limit strength and analysis of bolts:
Bolts in web of main beam:
Plain in the shear plane F v,Rd = 0,6 * f u,b * *d l
2
*4 * Mb 103
= 38,60 kN
Limit bearing strength
= MIN(a 1
*3 d 0;
a 2
*3 d 0 - 0,25;
f u,b
f u; 1) = 0,556
Fb,Rd = 2,5 * * dl * t *f u
* M2 103
= 51,24 kN
Analysis:
F Ed
F b,Rd= 0,66 < 1
Bolts in web of connecting beam:
Fv,Rd = 2 * 0,6 * f u,b * *d l
2
*4 * Mb 103
= 77,21 kN
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Eurocode 3 Folder : Bolted connections
Limit bearing strength
= MIN(a 1
*3 d 0;
a 2
*3 d 0 - 0,25;
f u,b
f u; 1) = 0,556
Fb,Rd = 2,5 * * dl * s *f u
* M2 103 = 42,27 kN
Analysis:
F Ed
F b,Rd= 0,79 < 1
Check angle section strengthDesign loads:
VEdw =
VEd
2 = 30,00 kN
MEdw =MEd
2= 135,00 kNm
A = (2 * a 2 + 2 * a 1 ) *t
200= 6,40 cm
A net = (2 * a 2 + 2 * a 1 - n * d 0 ) *t
200= 4,24 cm
*f y
f u
M2 M0
*0,9 A net
A
= 1,24 > 1
Subtract bottom bolt holes.
A = A * 2 = 12,80 cm A net = Anet * 2 = 8,48 cm
*f y
f u
A
Anet= 0,99 < 1
Bolt holes should not be subtracted
Wpl =
*t -( )*2 +a 2 *2 a 1
2
4*d 0 *t a 2
1000= 44,00 cm
Mpl,Rd = *W plf y
* M0 10= 940,00 kNcm
MEd
Mpl,Rd= 0,29 < 1
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Eurocode 3 Folder : Bolted connections
Double shear joint with bolts:
Dimensions of connection:
Plate thickness d 1 = 6,00 mmNumber of plates n 1 = 2Plate thickness d 2 = 7,10 mmNumber of plates n 2 = 1Plate width b = 210,00 mmBolt spacing e = 40,00 mmEdge distance e1 = 60,00 mmEdge distance e2 = 40,00 mmBolt spacing e3 = 65,00 mm
Bolts:Bolt = SEL("steel/bolt"; BS; ) = M 20SC = SEL("steel/bolt"; SC; ) = 10.9f u,b = TAB("steel/bolt"; fubk; SC=SC) = 1000,00 N/mmHole diameter d l= TAB("steel/bolt"; d; BS=Bolt) = 20,00 mmCross-section areas A s = TAB("steel/bolt"; Asp; BS=Bolt) = 2,45 cmNumber of bolts n= 5
Plate :steel = SEL("steel/EC"; Name; ) = Fe 360f y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mmf u = TAB("steel/EC"; fu; Name=steel) = 360,00 N/mm M0 = 1,10 M2 = 1,25
Loads:NEd = 100,00 kN
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Eurocode 3 Folder : Bolted connections
Joint with tension and shear:Limit tension force of plate links
A = b * d 1 *n 1
100= 25,20 cm
d = d l + 2 = 22,00 mmCross-section area 1:
A net,1 = A - 2 * d * d 1 *n 1
100= 19,92 cm
Cross-section area 2:
A net,2 = A-3 *d *d 1 *n 1
100 + 2 * e / (4*e 3 ) * d1 *
n 1
100= 18,76 cm
A net = MIN(Anet,1 ; A net,2 ) = 18,76 cm
Nt,Rd1 = MIN( * Af y
M0; 0,9 * * Anet
f u
M2)/10 = 486,26 kN
Limit tension force of plate rechts
A = b * d 2 *n 2
100= 14,91 cm
d = d l+ 2 = 22,00 mm
A net,1 = A - 2 * d * d 2 *n 2
100= 11,79 cm
A net,2 = A - 3 * d * d 2 *n 2
100 + 2 * e / (4*e 3 ) * d2 *
n 2
100= 11,10 cm
A net = MIN(Anet,1 ; A net,2 ) = 11,10 cm
Nt,Rd2 = MIN( * Af y
M0; 0,9 * * Anet
f u
M2)/10 = 287,71 kN
Nt,Rd = MIN(N t,Rd2 ; Nt,Rd1 ) = 287,71 kN
Limit tension force of boltsn:
A s =
* ( )d l102
4= 3,14 cm
Thread in the shearing plane:
Fv,Rd = 0,6 * n * (n 1 + n 2 - 1) **f u,b As
*10 M2= 1507,20 kN
Limit bearing strength
= MIN(e 1
*3 d; 2 *
e
*3 d - 0,25;
f u,b
f u; 1) = 0,91
Fb,Rd1 = n 1 * n * 2,5 * * dl* d1 *f u
* M2 103
= 786,24 kN
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Eurocode 3 Folder : Bolted connections
= MIN(e 1
*3 d; 2 *
e
*3 d - 0,25;
f u,b
f u; 1) = 0,91
Fb,Rd2 = n 2 * n * 2,5 * * dl* d2 *f u
* M2 103 = 465,19 kN
Fb,Rd = MIN(F b,Rd1 ; F b,Rd2 ) = 465,19 kN
Maximum allowed force:Nl,Rd = MIN(N t,Rd ; Fv,Rd ; F b,Rd ) = 287,71 kN
Analysis:NEd / N l,Rd = 0,35 < 1
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Eurocode 3 Folder : Bolted connections
High strength friction-grip bolt connection:
Dimensions of connection:
Plate thickness 1 d 1 = 10,00 mmPlate thickness 2 d 2 = 8,00 mmBolt spacing e = 65,00 mmEdge distance e1 = 55,00 mmEdge distance e2 = 50,00 mmBolt spacing e3 = 70,00 mmPlate width b= 240,00 mm
Bolts:Bolt = SEL("steel/bolt"; BS; ) = M 22SC = SEL("steel/bolt"; SC; ) = 10.9f u,b = TAB("steel/bolt"; fubk; SC=SC) = 1000,00 N/mmHole diameter d l= TAB("steel/bolt"; d; BS=Bolt) = 22,00 mmCross-section areas A s= TAB("steel/bolt"; Asp; BS=Bolt) = 3,03 cmNumber of bolts n= 5
normal holes with tolerance as K s = 1,00Number of slip joints as in 6.5.8.1 (1) = 1,00coefficient of friction as in 6.5.8.3 = 0,40as in 6.5.8.1 (3) Ms = 1,25
Plate:steel = SEL("steel/EC"; Name; ) = Fe 360
f y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mmf u = TAB("steel/EC"; fu; Name=steel) = 360,00 N/mm M0 = 1,10 M2 = 1,25
Loads:Tension force N Ed = 100,00 kN
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Eurocode 3 Folder : Bolted connections
Limit tension force of plate:
A = MIN(d 1 ; d 2 ) *b
100= 19,20 cm
as in 7.5.2 d = d l + 2 = 24,00 mm
Cross-section area 1:
A net,1 = A - 2 * MIN(d 1 ; d 2 ) *d
100= 15,36 cm
Cross-section area 2:
A net,2 = A - 3 *d
100 * MIN(d 1 ;d2 ) + *
2
100
e2
*4 e 3*MIN(d1 ;d 2 ) = 15,85 cm
A net = MIN(Anet,1 ; A net,2 ) = 15,36 cm
Friction-grip connection:Plate
Nnet,Rd = Anet *f y
* M0 10= 328,15 kN
Limit tension force of bolts
Limit shank tension F p,cd = 0,7 *f u,b
10 * As = 212,10 kN
Limit slip resistance force F s,Rd = n * K s * * *F p,cd
Ms= 339,36 kN
Limit bearing force:
= MIN(e 1
*3 d; 2 *
e
*3 d - 0,25;
f u,b
f u; 1) = 0,76
Fb,Rd = n * 2,5 * * dl * MIN(d 1 ; d 2 ) *f u
* M2 103
= 481,54 kN
Maximum allowed force:Ng,Rd = MIN(Nnet,Rd ; Fb,Rd ; F s,Rd ) = 328,15 kN
Analysis:N Ed
N g,Rd= 0,30 < 1
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Eurocode 3 Folder : Bolted connections
Joint with tension and shear:
Dimensions of connection:
Plate thickness d 1 = 10,00 mm
Plate thickness d 2 = 8,00 mmBolt spacing e = 65,00 mmEdge distance e1 = 55,00 mmEdge distance e2 = 50,00 mmBolt spacing e3 = 70,00 mmPlate width b = 240,00 mm
Bolts:Bolt = SEL("steel/bolt"; BS; ) = M 22SC = SEL("steel/bolt"; SC; ) = 4.6f u,b = TAB("steel/bolt"; fubk; SC=SC) = 400,00 N/mm
Hole diameter d l = TAB("steel/bolt"; d; BS=Bolt) = 22,00 mmCross-section areas A s = TAB("steel/bolt"; Asp; BS=Bolt) = 3,03 cmNumber of bolts n = 5
Plate:steel = SEL("steel/EC"; Name; ) = Fe 360f y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mmf u = TAB("steel/EC"; fu; Name=steel) = 360,00 N/mm M0 = 1,10 M2 = 1,25
Loads:NEd = 100,00 kN
Joint with tension and shear:
Limit tension force of plate:
A = MIN(d 1 ; d 2 ) *b
100= 19,20 cm
d = d l+ 2 = 24,00 mmCross-section area 1:
A net,1 = A - 2 * MIN(d 1 ; d 2 ) *d
100= 15,36 cm
Cross-section area 2:
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Eurocode 3 Folder : Bolted connections
A net,2 = A - 3 *d
100 * MIN(d 1 ;d2 ) + *
2
100
e2
*4 e 3*MIN(d1 ;d 2 ) = 15,85 cm
A net = MIN(Anet,1 ; A net,2 ) = 15,36 cm
Nt,Rd = MIN( * Af y
M0; 0,9 * * Anet
f u
M2)/10 = 398,13 kN
Limit tension force of boltsn:
A s =
* ( )d102
4= 4,52 cm
Thread in the shearing plane:
Fv,Rd = 0,6 * n **f u,b As
*10 M2= 433,92 kN
Limit bearing strength
= MIN(e 1
*3 d; 2 *
e
*3 d - 0,25;
f u,b
f u; 1) = 0,76
Fb,Rd = n * 2,5 * * d * MIN(d 1 ; d 2 ) *f u
* M2 103
= 525,31 kN
Maximum allowed force:Nl,Rd = MIN(N t,Rd ; Fv,Rd ; F b,Rd ) = 398,13 kN
Analysis:NEd / N l,Rd = 0,25 < 1
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Eurocode 3 Folder : Bolted connections
Check shear failure:
VVEd
VVVVVV
Dimensions of connection:
Spacing of bolts a = 50,00 mmSpacing of bolts a 1 = 43,00 mmSpacing of bolts a 2 = 70,00 mmSpacing of bolts a 3 = 75,00 mmNumber of bolts n = 4
Bolts:Bolt = SEL("steel/bolt"; BS; ) = M 20SC = SEL("steel/bolt"; SC; ) = 5.6f u,b = TAB("steel/bolt"; fubk; SC=SC) = 500,00 N/mmHole diameter d l= TAB("steel/bolt"; d; BS=Bolt) = 20,00 mmShaft diameter d l2 = d l+2 = 22,00 mm
k = 0,5 for 1 group bolts; =2,5 for 2 rows
steel = SEL("steel/EC"; Name; ) = Fe 360f y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mmf u = TAB("steel/EC"; fu; Name=steel) = 360,00 N/mm
Profil Typ = SEL("steel/Profils"; Name; ) = IPESelected Profil = SEL("steel/"Typ; Name; ) = IPE 360Column height h = TAB("steel/"Typ; h; Name=Profil) = 360,00 mmWeb thickness s = TAB("steel/"Typ; s; Name=Profil) = 8,00 mm Mb = 1,25 M0 = 1,10
Loads:VEd = 58,70 kN
Limit shear force of section:
A v = 1,04 **h s
100= 29,95 cm
Vpl,Rdp = * Avf y
* 3 * M0 10= 369,41 kN
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Eurocode 3 Folder : Bolted connections
A v,net = Av -*n *s d l2
100= 22,91
*
f y
f u
Av
Av,net= 0,85 < 1
Holes can be neglected.
Joint with shear failure as in Bild 6.5.5:Lv = (n - 1) * a 2 = 210,00 mmL1 = a 1 = 43,00 mm
L1
*5 d l2= 0,39 < 1
L2 = (a - k * d l2 ) *f u
f y= 59,74 mm
L3 = Lv + a 1 + a 3 = 328,00 mmLv,eff = Lv + L1 + L2 = 312,74 mmLv, e f f
L3= 0,95 < 1
L3
+Lv +a 1 -a 3 +n d l2= 0,95 < 1
Effective shear area:
A v,eff = Lv,eff *s
100= 25,02 cm
Vpl,Rd = * A v,eff f y
* M0 * 3 10= 308,60 kN
Maximum design force for shear:VRd = MIN(Vpl,Rdp ; V pl,Rd ) = 308,60 kN
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Eurocode 3 Folder : Columns
Column subject to compression force:
N
H
d
Column heigth H = 7,50 mly = H/2 = 3,75 mlz = H/3 = 2,50 m
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 430f y = TAB("steel/EC"; fy; Name=steel) = 275,00 N/mmE = TAB("steel/EC"; E; Name=steel) = 210000,00 N/mm
= 235f y = 0,92Profil:
Profil Typ = SEL("steel/Profils"; Name; ) = IPESelected Profil = SEL("steel/"Typ; Name; ) = IPE 300Cross-sectional area A = TAB("steel/"Typ; A; Name=Profil) = 53,80 cmColumn height h = TAB("steel/"Typ; h; Name=Profil) = 300,00 mmDepth of web h 1 = TAB("steel/"Typ; h1; Name=Profil) = 248,00 mmWeb thickness s = TAB("steel/"Typ; s; Name=Profil) = 7,10 mmFlange width b = TAB("steel/"Typ; b; Name=Profil) = 150,00 mmFlange thickness t = TAB("steel/"Typ; t; Name=Profil) = 10,70 mm
Radius of gyration i y = TAB("steel/"Typ; iy; Name=Profil) = 12,50 cmRadius of gyration i z= TAB("steel/"Typ; iz; Name=Profil) = 3,35 cm
Section classification As in Table 5.3.1:Web:
h 1
*s *33 = 1,15 < 1
h 1*s *38
= 1,00 < 1
Section class 2.
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Eurocode 3 Folder : Columns
Flange:b
*2 *t *10 = 0,76 < 1
Section class 1.
Section will be classified as class 2.
Analysis:nach 5.5.1.1 (1) A= 1,00 Cross-section class 2nach 5.1.1 (2) M1 = 1,10 Cross-section class 2Check buckling In Y-Y Axis:h / b = 2,00 > 1,2t /10 = 1,07 cm < 4 cmas in Tab 5.5.3 :
y = ly *100
iy= 30,00
1
= 93,9 * = 86,39
trans,y = *y1
A = 0,347
Apply strut curve a As in Table 5.5.1 = 0,21
= 0,5 * (1 + * (trans,y - 0,2) + trans,y ) = 0,576
=1
+ -2 trans,y 2= 0,9655
Nb,y,Rd = * A * A *
f y
*10 M1 = 1298,60 kN
Check buckling In Z-Z Axis:h / b = 2,00 > 1,2t / 10 = 1,07 cm < 4cmas in Tab 5.5.3 :z = lz * 100 / i z = 74,63
trans,z = *z1
A = 0,864
Apply strut curve b As in Table 5.5.1 = 0,34
= 0,5 * (1 + * (trans,z - 0,2) + trans,z ) = 0,986
=1
+ -2 trans,z 2= 0,6844
Nb,z,Rd = * A * A *f y
*10 M1= 920,52 kN
max_N d = MIN(Nb,y,Rd ; N b,z,Rd ) = 920,52 kN
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Eurocode 3 Folder : Columns
Column of section class 4:
t1t y
z
b
2hh
N
N
Load diagram:Column heigth H = 4,00 kNFlange width b = 20,00 cmDepth of web h = 30,00 cmWeb thickness t 1 = 1,20 cm
Flange thickness t 2 = 1,20 cm
Loads:Nd = 700,00 kN
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 360f y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mmE = TAB("steel/EC"; E; Name=steel) = 210000,00 N/mm
= 235f y = 1,001 = 93,90 * = 93,90
Partial safety factors: M = 1,10
Section classification As in Table 5.3.1:
b
*t 2 *14 = 1,19 > 1
Section class 4
h
*t 1 *33 = 0,76 > 1
Section class 1
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Eurocode 3 Folder : Columns
Effective web area:Flangee:
As in Table 5.3.3: = 1,00k = 0,43as in 5.3.5(3)
trans,p =b
*t2 *28,4 * k= 0,89 >0,673
=-trans,p 0,22
trans,p2
= 0,846
beff = * b = 16,92 cm Aeff = 2 * (b eff * t2) + (h + 2 * t 2) * t1 = 79,49 cm
Location of shear centre:
ys,eff =
*b e f f *2 *t2+b e f f t2
2
Ae f f = 4,628 cm
Iz,eff = *beff
3
12*2 +t2 *beff *2 *t2 +( )-+beff t12 ys,eff
2
*( )+h *2 t2 ( )+t1 312 *t1 ys,eff 2 = 2604 cm 4Wz,eff =
Iz,eff
-( )+b eff t 1 -y s,eff t 1
2
= 201,99 cm
Gross area: A = b * 2 * t 2 + (h + 2 * t 2) * t1 = 86,88 cm
Location of center of gravity:
ys =
*b *2 *t2+b t 1
2
A= 5,856 cm
Iz = *b
3
12*2 +t2 *b *2 *t2 +( )-+b t12 ys
2
*( )+h *2 t2 ( )+t1 312 *t1 ys 2 = 4018 cm 4e
Nz= y
s- y
s,eff = 1,228 cm
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Eurocode 3 Folder : Columns
Check buckling:
Ncr = *2
*E
10
Iz
( )*H 1002
= 5204,85 kN
A = A eff
A= 0,915
trans,z = * A * A f y*10 N cr = 0,599 Apply strut curve c As in Table 5.5.3 = 0,49 = 0,5 * (1 + * (trans,z - 0,2) + trans,z ) = 0,777
=1
+ -2
trans,z2
= 0,786
Mz,d = Nd *e Nz
100= 8,60 kNm
= 1,00M = 1,8 - 0,7 * = 1,10z = trans,z * (2 * M - 4) = -1,078 < 0,9
kz1 = 1 -*z Nd
* * A eff f y
10
= 1,514 ~ 1,5
kz2 = 1,500kz = MIN(k z1 ; k z2 ) = 1,500
+Nd
* * A eff f y
* M 10
*100 *k zMz,d
*W z,eff f y
* M 10
= 0,823 < 1
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Eurocode 3 Folder : Columns
Column with compression and moment:
H
N
M
d
d
Load diagram:Column heigth H = 5,00 m
Loads:Nd = 200,00 kNMy,d = 10,00 kN
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 360f y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mmE = TAB("steel/EC"; E; Name=steel) = 210000,00 N/mmG = TAB("steel/EC"; G; Name=steel) = 81000,00 N/mm = (235 / f y) = 1,00Partial safety factors: M = 1,10 g = 1,35platischer Formbeiwert pl y = 1,14platischer Formbeiwert pl z = 1,25
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Eurocode 3 Folder : Columns
Profil:Profil Typ = SEL("steel/Profils"; Name; ) = HEASelected Profil = SEL("steel/"Typ; Name; ) = HEA 160Cross-sectional area A = TAB("steel/"Typ; A; Name=Profil) = 38,80 cmColumn height h = TAB("steel/"Typ; h; Name=Profil) = 152,00 mmDepth of web h 1 = TAB("steel/"Typ; h1; Name=Profil) = 104,00 mmWeb thickness s = TAB("steel/"Typ; s; Name=Profil) = 6,00 mmFlange width b = TAB("steel/"Typ; b; Name=Profil) = 160,00 mmFlange thickness t = TAB("steel/"Typ; t; Name=Profil) = 9,00 mmRadius of gyration i y = TAB("steel/"Typ; iy; Name=Profil) = 6,57 cmRadius of gyration i z= TAB("steel/"Typ; iz; Name=Profil) = 3,98 cmWel y= TAB("steel/"Typ; Wy; Name=Profil) = 220,00 cmWpl y = pl y * W el y = 250,80 cmWel z = TAB("steel/"Typ; Wz; Name=Profil) = 76,90 cmWpl z = pl z * W el z = 96,13 cm
Section classification As in Table5.3.1:Web will be assumed as pressed in
h 1
*s *33 = 0,53 < 1
Section class 1.Flange:
b
**2 t *10 = 0,89 < 1
Section class 1.
Check bending and buckling:as in 5.5.1.1 (1) A = 1,00 Cross-section class 1as in 5.1.1 (2) M1 = 1,10 Cross-section class 1
y = H *100
iy= 76,10
1 = 93,9 * = 93,90
trans,y = *y1
A = 0,810
z = H *100
iz= 125,63
trans,z = z
* Ef y= 1,34
As in Table 5.5.3h / b = 0,95 < 1,2t / 10 = 0,90 cm < 10cm
As in Table:5.5.1Strut curve b for Y-Y axis = 0,34
= 0,5 * (1 + * (trans,y - 0,2) + trans,y ) = 0,932
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Eurocode 3 Folder : Columns
y =1
+ -2 trans,y 2= 0,7179
Apply strut curve c for z-Achse
= 0,49 = 0,5 * (1 + * (trans,z - 0,2) + trans,z ) = 1,677
z =1
+ -2 trans,z 2= 0,3724
= 0,00My = 1,8 - 0,7 * = 1,80
y = trans,y * (2 * My - 4) +-W ply W ely
W ely= -0,184 < 0,9
ky = 1 -
*y Nd* y * A f y = 1,006 < 1,5
= MIN( y; z) = 0,372
+Nd
* * Af y* M 10
*k y *100My,d
*W pl yf y* M 10
= 0,836 < 1
Check torsional-flexural buckling:w = 1,00 Cross-section class 1
As in Table: F.1.1
= 0,00k = 1,00C1 = 1,879
LT = *90H
*iz ( )* C 1 +1 *120 ( )*100 Hizht
24
= 59,208
trans,LT = *LT1
w = 0,631
For rolled sections strut curve a = 0,21 = 0,5 * (1 + * (trans,LT - 0,2) + trans,LT ) = 0,74
LT =1
+ -2 trans,LT 2= 0,89
M,LT = 1,8 - 0,7 * = 1,80
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Eurocode 3 Folder : Columns
quer,z = *z1
A = 1,338
LT = 0,15 * quer,z * M,LT - 0,15 = 0,211 < 0,9h / b = 0,95 < 1,2t / 10 = 0,90 cm < 10cmZ-Z axis Strut curve c = 0,49 = 0,5 * (1 + * (trans,z - 0,2) + trans,z ) = 1,677
z =1
+ -2 trans,z 2= 0,3724
kLT = 1 -*LT Nd
* z * A f y= 0,988 < 1,5
+Nd
* z * Af y
* M 10
*kLT
*100My,d
** LT W pl yf y
* M 10
= 0,855 < 1
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Eurocode 3 Folder : Columns
Column subject to compression force:
N
H
d
Column base is pinned.For buckling, column head is pinned about the Y-Y axis, and fixed about the Z-Z axis .
Load diagram:Column heigth H = 8,00 m
Loads:Nd = 2000,00 kN
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 360f y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mmE = TAB("steel/EC"; E; Name=steel) = 210000,00 N/mm
= 235f y = 1,00Effective length factor:ky = 1,00 For simply supported endskz = 0,70 fixed ends
Profil:Profil Typ = SEL("steel/Profils"; Name; ) = HEBSelected Profil = SEL("steel/"Typ; Name; ) = HEB 300Cross-sectional area A = TAB("steel/"Typ; A; Name=Profil) = 149,00 cmColumn height h = TAB("steel/"Typ; h; Name=Profil) = 300,00 mmDepth of web h 1 = TAB("steel/"Typ; h1; Name=Profil) = 208,00 mmWeb thickness s = TAB("steel/"Typ; s; Name=Profil) = 11,00 mmFlange width b = TAB("steel/"Typ; b; Name=Profil) = 300,00 mmFlange thickness t = TAB("steel/"Typ; t; Name=Profil) = 19,00 mmMoment of inertia I = TAB("steel/"Typ; Iy; Name=Profil) = 25170,00 cm 4Radius of gyration i y = TAB("steel/"Typ; iy; Name=Profil) = 13,00 cmRadius of gyration i z= TAB("steel/"Typ; iz; Name=Profil) = 7,58 cm
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Eurocode 3 Folder : Columns
Section classification:Web:
h 1
*s *33 = 0,57 > 1
Flange:b
**2 t *10 = 0,79 > 1
Section class 1.
Analysis: A = 1,00 Cross-section class 1 M1 = 1,10 Cross-section class 1
Check buckling In Y-Y Axis:h / b = 1,00 < 1,2t / 10 = 1,90 cm < 10cm
y = k y * H *100
iy= 61,54
1 = 93,9 * = 93,90
trans,y = *y1
A = 0,655
Apply strut curve b As in Table 5.5.1 = 0,34
= 0,5 * (1 + * (trans,y - 0,2) + trans,y ) = 0,792
=1
+ -2 trans,y 2= 0,8083
Nb,Rd = * A * A *f y
*10 M1= 2572,966 kN
N d
N b,Rd= 0,78 < 1
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Eurocode 3 Folder : Columns
Check buckling In Z-Z Axis:h / b = 1,00 < 1,2t / 10 = 1,90 cm < 10cm
z = k z * H *100
iz
= 73,88 m
trans,z = *z1
A = 0,787
Apply strut curve c As in Table 5.5.1 = 0,49
= 0,5 * (1 + * (trans,z - 0,2) + trans,z ) = 0,953
=1
+ -2 trans,z 2= 0,6709
Nb,Rd = * A * A *f y
*10 M1 = 2135,60 kN
N d
N b,Rd= 0,94 < 1
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Eurocode 3 Folder : Columns
Column of channel section with compression and transverse loads:
H
d
dq t1t y
z
b
2
h
N
Load diagram:Column height H = 4,00 mBeam width b = 9,50 cmDepth of web h = 29,40 cmWeb thickness t 1 = 1,00 cmFlange width t 2 = 0,60 cm
Loads:qd = 15,00 kN/mNd = 90,00 kN
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 360f y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mmE = TAB("steel/EC"; E; Name=steel) = 210000,00 N/mm
= 235f y = 1,001 = 93,90 * = 93,90
Partial safety factors: M = 1,10
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Eurocode 3 Folder : Columns
Location of shear centre:
ys,eff =
*b e f f *2 *t2+b e f f t2
2
Ae f f = 1,088 cm
e Nz = y s - y s,eff = 0,337 cm
A = A eff A
= 0,965
Slenderness and reductions:In Z-Z Axis:
Ncr,z = *2
*EIz
( )*H 1002
= 4109,09 kN
trans,z = * A * Af y
Ncr,z= 1,52
Apply strut curve c = 0,49 = 0,5 *(1 + * (trans,z - 0,2) + trans,z ) = 1,979
z =1
+ -2 trans,z 2= 0,308
In Y-Y Axis:
Ncr,y = *2
*EIy
( )*H 1002 = 60663,06 kN
trans,y = * A * A f yN cr,y = 0,40 Apply strut curve c = 0,49
= 0,5 * (1 + * (trans,y - 0,2) + trans,y ) = 0,629
y =1
+ -2 trans,y 2= 0,897
= MIN( z; y) = 0,3080
Nb,Rd = * * Ae ff f y
M= 2667,53 kN
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Eurocode 3 Folder : Columns
Limit bending moment about the Y-Y axis:Cross-section class 4
z' s =
*-b e f f *t2 ++h t 2
2
*
( )+b
t1
2
*t2+h t 2
2 Ae f f
= 0,382 cm
*( )+h t 2
3
12+t 1 *( )+h t 2 *t 1 z' s = 2261,46 cm 4
*b eff *t2 +( )++h t 22 z' s2
*( )+b t12 *t 2 ( )-+h t 2
2z' s
2
= 2457,57 cm 4
Ieff,y = 4719,03 cm4
Weff,y =Ieff,y
++h t 22
z' s
= 306,79 cm
Mc,Rd,y = *W eff ,yf y
M= 65541,50 kNm
Limit bending moment about the Z-Z axis:Cross-section class 4 Tab 5.3.3
=-y s
+b -t1
2y s
= -0,17
k = 0,57 - 0,21 * + 0,07 * = 0,61
+bt1
2
*t 2 *21 * k= 1,02 > 1
As in Table:5.3.5 (3)
trans,p =
+bt 1
2*t2 *28,4 * k
= 0,75 >0,673
=- trans,p 0,22
trans,p2
= 0,942
beff = *+b
t1
2
-1 = 8,05 cm
bt = -1 * *( )+b t 12 -1
= 1,45 cm
c eff = b eff + b t = 9,50 cm
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Eurocode 3 Folder : Columns
A eff = (h + t 2 ) * t1 + 2 * c eff * t2 = 41,40 cm
y' = *2 *c e ff 2 t2
*2 Ae f f = 1,31 cm
Ieff,z = (h + t 2 ) * t1 * y' + 2 * c eff 3
12 * t2 + 2 * c eff * t2 *( )-c eff 2 y'
2
= 272,12 cm 4
Weff,z =Ieff,z
-c eff
2y'
= 79,10 cm
Mc,Rd,z = *W eff ,zf y
M= 16898,64 kNm
Check buckling:
NEd = Nd = 90,00 kN
My,Ed = q d *H
2
8= 30,00 kNm
Mz,Ed = Nd *e Nz
100= 0,30 kNm
My = 1,30y = trans,y * (2 * My - 4) = -0,560 < 0,9
ky = 1 -*y NEd
* y
* Aeff
f y
= 1,006 < 1,5
= 1,000M = 1,8 - 0,7 * = 1,10 < 1,5z = trans,z * (2 * M - 4) = -2,736 < 0,9
kz1 = 1 -*z NEd
* * A eff f y= 1,082
kz2 = 1,500kz = MIN(k z1 ; k z2 ) = 1,082
+NEd
* * A eff f y
M
*ky *100 +My,Ed
*W eff,yf y
M
*k z *100Mz,Ed
*W eff,zf y
M
= 0,081 < 1
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Eurocode 3 Folder : Columns
Laced column made up of channel and angle sections:
l
l y
N d
b
h
H sd
Loads:
Nd = 300,00 kNHd = 45,00 kN
Plan and elevation values:Span length l y = 69,20 cmColumn height l = 10*l y/100 = 6,92 mWidth b = 40,00 cmWinkel 1 = 45,00
U-Profil U = SEL("steel/U"; Name; ) = U 300 A f = TAB("steel/U"; A; Name=U) = 58,80 cme z = TAB("steel/U"; ez; Name=U) = 2,70 cmiz = TAB("steel/U"; iz; Name=U) = 2,90 cm
Winkel W = SEL("steel/WG"; Name; ) = L 50x5 Angle thickness A D = TAB("steel/WG"; A; Name=W) = 4,80 cm Angle thickness i = TAB("steel/WG"; i ; Name=W) = 0,98 cm
steel = SEL("steel/EC"; Name; ) = Fe 360f y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mm
E = TAB("steel/EC"; E; Name=steel) = 210000,00 N/mm
= 235f y = 1,001 = 93,90 * = 93,90Partial safety factors: M = 1,10
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Eurocode 3 Folder : Columns
Maximal tension force in chord:Effective length ratio k = 2,00Imperfection:
wo =*2 l
5
= 2,77 cm
Effective moment of inertia:ho = b - 2 * e z = 34,60 cmIeff = 0,5 * h o * A f = 35196,50 cm 4
Shear strength:
S v = 2 *E
10 * AD * ly *
h o2
*2 *2 h o 223= 71276,36 kN
Ncr =1
*4 +( )*100 l
2
*2
*E
10Ieff
1S v
= 3615,26 kN
max_M E = Hd * l +Nd
-1Nd
Ncr
*w o
100= 320,46 kNm
Nf,Ed = 0,5 * N d + 100 *max_M E
h o= 1076,18 kN
Analysis of channels:
=ly
iz= 23,86
trans =*1
= 0,254
Apply strut curve c = 0,49
= 0,5 * (1 + * (trans - 0,2) + trans ) = 0,545
=1
+ -
2
trans2
= 0,9735
Nb,Rd = * Af *f y
* M 10= 1222,89 kN
N f,Ed
N b,Rd= 0,880 < 1
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Eurocode 3 Folder : Columns
Analysis of lacing angles:
max_V E = Hd + *Nd
-1Nd
Ncr
*wo
*200 l= 47,06 kN
Design force for a diagonal:
NEd =max_V E
*2 cos ( )1= 33,28 kN
= *2 h o
2
i= 49,93
trans =*1
= 0,532
Apply strut curve c = 0,49
= 0,5 * (1 + * (trans - 0,2) + trans ) = 0,723
=1
+ -2 trans 2= 0,8247
Nb,Rd = * AD *f y
* M 10= 84,57 kN
N Ed
N b,Rd= 0,39 < 1
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Eurocode 3 Folder : Columns
Column of section class 4:
ht
t
b
C D
B A2
1h
N
N
Load diagram:Column height H = 6,50 mBeam width b = 50,00 cmColumn height h = 100,00 cmWeb thickness t 1 = 1,00 cmFlange thickness t 2 = 1,00 cm
Loads:Nd = 3500,00 kN
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 360f y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mmE = TAB("steel/EC"; E; Name=steel) = 210000,00 N/mm
= 235f y = 1,00Partial safety factors: M = 1,10
g = 1,35
Section classification: As in Table 5.3.1b' = b - 2 * t 1 = 48,00 cm
b'
*t 2 *42 = 1,14 > 1
Section class 4h' = h - 2 * t 2 = 98,00 cm
h'
*t1
*42 = 2,33 > 1
Section class 4
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Eurocode 3 Folder : Columns
Effective web area:Part A-B; C-D As in Table 5.3.2: = 1,00k = 4,00as in 5.3.5(3)
quer,p =b'
*t 2 *28,4 * k = 0,85 >0,673
=-quer,p 0,22
quer,p2
= 0,872
beff = * b' = 41,86 cmPart A-C; B-D As in Table 5.3.2: = 1,00 cmk = 4,00as in 5.3.5(3)
quer,p =h'
*t 1 *28,4 * k = 1,73 >0,673
=-quer,p 0,22
quer,p2
= 0,505
heff = * h' = 49,49 cm
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Eurocode 3 Folder : Columns
Check buckling: A eff = 2 * (h eff * t1 + b eff * t2 ) + 4 * t 1 * t2 = 186,70 cm A = b * h - (b - 2 * t 1 ) * (h - 2 * t 2 ) = 296,00 cm
A
= A eff
A= 0,631
Ib =*b -h
3*( )-b *2 t 1 ( )-h *2 t 2
3
12= 401898,67 cm 4
Ih =*b
3-h *( )-b *2 t 1
3( )-h *2 t 2
12= 138498,67 cm 4
I = MIN(Ib ; Ih ) = 138498,67 cm 4
Ncr = *2 *E I
*H2
105
= 67941,82 kN
transv = * A * A f y
*10 N cr = 0,25
Apply strut curve b = 0,34 = 0,5 * (1 + * (transv - 0,2) + transv ) = 0,540
=1
+ -2 transv 2= 0,9817
Nb,Rd = * A * A *f y
* M 10 = 3917,19 kNN d
N b,Rd= 0,893 < 1
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Eurocode 3 Folder : Continuous Beam
Purlin / Strut with biaxial bending and torsional-flexural buckling:
qdqd
Plan and elevation values:Span length l= 720,00 cm
Angle of slope = 10,00 Spacing of purlin l'= 4,00 mplatischer Formbeiwert pl y = 1,14
platischer Formbeiwert pl z = 1,25
Loads:From dead load g= 0,20 kN/mFrom snow load s= 0,40 kN/m
Section classification for: IPE 200Profil Typ = SEL("steel/Profils"; Name; ) = IPESelected Profil = SEL("steel/"Typ; Name; ) = IPE 200 Column height h = TAB("steel/"Typ; h; Name=Profil) = 200,00 mmFlange width b f = TAB("steel/"Typ; b; Name=Profil) = 100,00 mm
Flange thickness t f = TAB("steel/"Typ; t; Name=Profil) = 8,50 mmFlange thickness t w = TAB("steel/"Typ; s; Name=Profil) = 5,60 mmMoment of inertia I y = TAB("steel/"Typ; Iy; Name=Profil) = 1940,00 cm 4
Moment of inertia I z = TAB("steel/"Typ; Iz; Name=Profil) = 142,00 cm 4
Moment of inertia I y = TAB("steel/"Typ; IT; Name=Profil) = 6,98 cm 4
I = t f * b f * (h - t f ) / (24*10 6 ) = 12988,09 cm 6
iz = TAB("steel/"Typ; iz; Name=Profil) = 2,24 cmWel y= TAB("steel/"Typ; Wy; Name=Profil) = 194,00 cmWpl y = pl y * W el y = 221,16 cmWel z = TAB("steel/"Typ; Wz; Name=Profil) = 28,50 cm
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 360E s = TAB("steel/EC"; E; Name=steel) = 210000,00 N/mmG = TAB("steel/EC"; G; Name=steel) = 81000,00 N/mmf y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mm
= 235f y = 1,00Partial safety factors: M = 1,10 g = 1,35
p = 1,50
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Eurocode 3 Folder : Continuous Beam
Design calculations:
Factored load pd = g * g + p * s = 0,87 kN/mEffective load qd = p d * 4 = 3,48 kN/mVertical load component q
zd= q
d* COS(
) = 3,43 kN/mHorizontal load component q yd = q d * SIN() = 0,60 kN/m
Forces at 2nd support:
MbyEd = 0,105 * q zd *( )l1002
= 18,67 kNm
MbzEd = 0,105 * q yd *( )l1002
= 3,27 kNm
VbzEd = 0,606 * q zd *l
100= 14,97 kN
VbyEd = 0,606 * q yd *l
100= 2,62 kN
Check biaxial bending strength:About Y-Y axis
A v = 1,04 * h * t w / 10 = 11,65 cm
Vpl,z,Rd = * A vf y
** 3 M 10= 143,69 kN
VbzEd
Vpl,z,Rd= 0,10 < 1
VbzEd
Vpl,z,Rd= 0,10 < 0,5
No reduction in moment strength necessary due to shear.
Mpl,y,Rd =*W ply f y
* M 10= 4724,78 kNcm
About Z-Z axis
A v = 2 * b f * tf / 10 = 17,00 cm
Vpl,y,Rd = * A vf y
** 3 M 10= 209,68 kN
VbyEdVpl,y,Rd
= 0,01 < 1
VbyEd
Vpl,y,Rd= 0,01 < 0,5
No reduction in moment strength necessary due to shear
Mpl,z,Rd = 1,5 **W elz f y
* M 10= 913,30 kNcm
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Eurocode 3 Folder : Continuous Beam
Analysis:as in 5.35: for I- and H- sections:= 2,00= 1,00Or else: 5.4.8.1 (11)
abs ( )+( )*MbyEd 10 2Mpl,y,Rd
( )*MbzEd 102
Mpl,z,Rd
= 0,514 < 1
Check flexural-torsional buckling strength:
iLT = *Iz IW pl y
2
4
= 2,48 cm
Load is applied in top flange.z
a= 10,00 cm
z s = 0,00 cmz g = z a - z s = 10,00 cm
As in Table F.1.2, interpolated.C1 = 1,00C2 = 0,80C3 = 0,70For simply supported ends fixed ends k = 0,70No measures taken toward bending k w = 1,00
hs =-h t f
10= 19,15 cm
5.5.2; Section class 1+2 w = 1,00
1 = * E sf y = 93,91LT =
*k /l iLT
* C 1 - +( )kkw 2 *120 +( )*k /l iLTht f 2
( )*2 *C 2 zgh s2 *2 *C 2 zg
h s
= 172,86
trans,LT = *LT1 w = 1,841
Apply strut curve a: As in Table 5.5.1 = 0,21 = 0,5 * (1 + * (trans,LT -0,2) + trans,LT ) = 2,37
LT =1
+ -2 trans,LT 2= 0,26
No longitudinal forces k z = 1,00
No longitudinal forces k LT = 1,00
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Eurocode 3 Folder : Continuous Beam
*k LT +MbyEd
*LTMpl,y,Rd
100
*k zMbzEd
Mpl,z,Rd
100
= 1,88 < 1
Section is not adequate: Restrain section in quarterly points:l = l / 4 = 180,00 cm
LT =*k /l iLT
* C 1 - +( )kkw 2 *120 +( )*k /l iLTht f 2
( )*2 *C 2 zgh s2 *2 *C 2 zg
h s
= 85,08
trans,LT = *LT1
w = 0,906
Apply strut curve a: = 0,5 * (1 + * (trans,LT -0,2) + trans,LT ) = 0,98
LT =1
+ -2 trans,LT 2= 0,74
*k LT +MbyEd
*LTMpl,y,Rd
100
*k zMbzEd
Mpl,z,Rd
100
= 0,89 < 1
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Eurocode 3 Folder : foot and support
Concentrated point load of a beam in another beam:
Design loads and propertiesP
d = 68,00 kN
MEd1 = 70,00 kNmMEd2 = -22,00 kNm
Top beam (IPE180)Profil TypO = SEL("steel/Profils"; Name; ) = IPESelected ProfilO = SEL("steel/"TypO; Name; ) = IPE 180 Web thickness s o = TAB("steel/"TypO; s; Name=ProfilO) = 5,30 mmFlange thickness t o = TAB("steel/"TypO; t; Name=ProfilO) = 8,00 mmRadius r o = TAB("steel/"TypO; r; Name=ProfilO) = 9,00 mmFlange width b o = TAB("steel/"TypO; b; Name=ProfilO) = 91,00 mmWeb depth h o= TAB("steel/"TypO; h; Name=ProfilO) = 180,00 mmWeb depth h 1o = TAB("steel/"TypO; h1; Name=ProfilO) = 146,00 mmWelo = TAB("steel/"TypO; W y; Name=ProfilO) = 146,00 cmWplo = 1,14 * W elo = 166,44 cm 3
Iyo = TAB("steel/"TypO; Iy; Name=ProfilO) = 1320,00 cm 4
Bottom beam (HEA200)Profil TypO = SEL("steel/Profils"; Name; ) = HEASelected ProfilO = SEL("steel/"TypO; Name; ) = HEA 200 Web thickness of beam s u = TAB("steel/"TypO; s; Name=ProfilO) = 6,50 mm
Flange thickness t u = TAB("steel/"TypO; t; Name=ProfilO) = 10,00 mmRadius r u = TAB("steel/"TypO; r; Name=ProfilO) = 18,00 mmFlange width bu = TAB("steel/"TypO; b; Name=ProfilO) = 200,00 mmWeb depth h u= TAB("steel/"TypO; h; Name=ProfilO) = 190,00 mmWeb depth h 1u = TAB("steel/"TypO; h1; Name=ProfilO) = 134,00 mmWelu = TAB("steel/"TypO; W y; Name=ProfilO) = 389,00 cmWplu = 1,14 * W elu = 443,46 cm 3
Iyu = TAB("steel/"TypO; Iy; Name=ProfilO) = 3690,00 cm 4
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Eurocode 3 Folder : foot and support
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 360E s = TAB("steel/EC"; E; Name=steel) = 210000,00 N/mmG = TAB("steel/EC"; G; Name=steel) = 81000,00 N/mmf y
= TAB("steel/EC"; fy; Name=steel) = 235,00 N/mm = (235 / f y) = 1,00 M = 1,101 = 93,90 * = 93,90f yd = f y/ M = 213,64 kN/cm
Limit plastic bearing:Stiff bearing length of top, bottom beam:
s s1 = s o + 2 * t o + 4 * r o * (1 - 22
) = 31,84 mm
f,Ed1 = 100 *
MEd1
W elu = 17,99 kN/cm
bf1 = tu * 25 = 250,00 mmbf1 = MIN(b f1 ; b u ) = 200,00 mm
s y1 = 2 * t u *b us u * -1 ( )* M *f,Ed1 10f y2
= 59,83 mm
Ry,Rd1 = (s s1 + s y1 ) * s u *f y d
103
= 127,30 kN
P d
R y,Rd1= 0,53 < 1
Stiff bearing length of bottom beam:
s s2 = s u + 2 * t u + 4 * r u * (1 - 22
) = 47,59 mm
f,Ed2 = 100 *MEd2
W elo= -15,07 kN/cm
bf = to * 25 = 200,00 mmbf2 = MIN(b f ; b o ) = 91,00 mm
s y2 = 2 * t o*b
os o
* -1 ( )* M *f,Ed2 10
f y
2
= 46,99 mm
Ry,Rd2 = (s s2 + s y2 ) * s o *f y d
103
= 107,09 kN
P d
R y,Rd2= 0,63 < 1
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Eurocode 3 Folder : foot and support
Check web crushing:Bottom beam:
m1 =s s1
h 1u= 0,238
m2 = 0,200mu = MIN(m 1 ; m 2 ) = 0,200
Ra,Rd1 = *0,5 *s u2
* *E s f y
+tus u *3 *s u
tum u
* M 103
= 219,95 kN
P d
R a,Rd1= 0,31 < 1
Mc,Rd1 = W plu * f y d
103
= 94,74 kNm
MEd1
Mc,Rd1= 0,74 < 1
Top beam:
m1 =s s2
h 1o= 0,326
m2 = 0,200
mo = MIN(m 1 ; m 2 ) = 0,200
Ra,Rd2 = *0,5 *s o2
* *E s f y
+tos o *3 *s o
tom o
* M 103
= 145,85 kN
P d
R a,Rd2= 0,47 < 1
Mc,Rd2 = W plo *f y d
103
= 35,56 kNm
abs ( )MEd2Mc,Rd2
= 0,62 < 1
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Eurocode 3 Folder : foot and support
Check web buckling for whole web:Bottom beam:
Effective web width b eff2 = +h u2
s s12 = 192,65 mm
Effective web area A 2 = b eff2 *
s u
100 = 12,52 cm
Moment of inertia I 2 =
*b eff2
10 ( )s u103
12= 0,441 cm 4
Radius of gyration i 2 = I2 A2 = 0,188 cmEffective slenderness trans2 =
h u
*i2 *1 10= 1,076
Apply strut curve c: = 0,49
= 0,5 * (1 + * (trans2 - 0,2) + trans2 ) = 1,294
2 =1
+ -2 trans2 2= 0,4968
Rb,Rd2 = 2 * A2 *f y d
10= 132,88 kN
P d
R b,Rd2= 0,51 < 1
Top beam:
Effective web width b eff1 = +h o2
s s12 = 182,79 mm
Effective web area A 1 = b eff1 *s o
100= 9,69 cm
Moment of inertia I 1 =
*b eff1
10 ( )s o10
3
12= 0,227 cm 4
Radius of gyration i 1 = I1 A1 = 0,153 cmEffective slenderness trans1 =
h o
*i1 *1 10= 1,253
Apply strut curve c: = 0,49
= 0,5 * (1 + * (trans1 - 0,2) + trans1 ) = 1,543
1 =1
+ -2 trans1 2= 0,4093
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Eurocode 3 Folder : foot and support
Rb,Rd1 = 1 * A1 *f y d
10= 84,73 kN
P d
Rb,Rd1
= 0,80 < 1
Check web yield at load points:Bottom beam:
x,Ed2 = MEd1 * 10 *-
h u
2tu
Iy u= 161,25 kN/cm
z,Ed2 =*10
3P d
*( )+s s1 *2 t u s u= 201,80 kN/cm
+( )x,Ed2f yd2
-( ) z,Ed2f yd2
*x,Ed2
f yd
z,Ed2f yd
= 0,75 < 1
Top beam:
x,Ed1 = -MEd2 * 10 *-
h o
2to
Iy o= 136,67 kN/cm
z,Ed1 =*10
3P d
*( )+s s2 *2 t o s o= 201,76 kN/cm
+( )x,Ed1f yd2
-( ) z,Ed1f yd2
*x,Ed1
f yd
z,Ed1f yd
= 0,70 < 1
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Eurocode 3 Folder : foot and support
Column base subject to compression und bending:
l
Msd
Nsd
a
be1
e1p
e
d
Ed
Ed
Plan and elevation values:Distance to edge of baseplate e = 45,00 mmDistance to edge of baseplate e 1 = 100,00 mm
Spacing of bolts p = 300,00 mmWidth of baseplate b = 500,00 mm
Length of baseplate a = 500,00 mmPlate thickness d = 30,00 mmProjection of plate l = 100,00 mmWeld thickness a w = 10,00 mm
Projection of pad footing a r = 1000,00 mm
Depth of pad footing h = 1400,00 mmas in L.3 (idR) Bearing plane factor j = 2/3 = 0,67
Profil Typ = SEL("steel/Profils"; Name; ) = HEBSelected Profil = SEL("steel/"Typ; Name; ) = HEB 300Moment of inertia I y1 = TAB("steel/"Typ; Iy; Name=Profil) = 25170,00 cm 4
Flange width b f = TAB("steel/"Typ; b; Name=Profil) = 300,00 mm
Flange thickness t f = TAB("steel/"Typ; t; Name=Profil) = 19,00 mm
Column height h t = TAB("steel/"Typ; h; Name=Profil) = 300,00 mm
Mpl,y,Rd = TAB("steel/"Typ; Mplyd; Name=Profil) = 418,00 kNmNpl,Rd = TAB("steel/"Typ; Npld; Name=Profil) = 3250,00 kN
Loads:NEd = 1000,00 kNMEd = 221,75 kNm
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Eurocode 3 Folder : foot and support
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 430E s = TAB("steel/EC"; E; Name=steel) = 210000,00 N/mmf y = TAB("steel/EC"; fy; Name=steel) = 275,00 N/mmConcrete = SEL("Concrete/EC"; Name; ) = C30/37f ck = TAB("Concrete/EC"; f ck ; Name=Concrete) = 30,00 N/mm
SC = SEL("steel/bolt"; SC; ) = 4.6Bolt = SEL("steel/bolt"; BS; ) = M 24f u = TAB("steel/bolt"; fubk; SC=SC) = 400,00 N/mm
A s = TAB("steel/bolt"; Asp; BS=Bolt) = 3,53 cm
Mb = 1,25 c= 1,50 M0 = 1,10
Design:m = l - e - 0,8 * a w * 2 = 43,69 mmleff1a = MIN(4 * m + 1,25 * e; 2 * m + 0,625 * e + e 1) = 215,51 mm
leff1b = MIN(2 * m + 0,625 * e + 0,5 * p; e 1 + 0,5 * p; b / 2) = 250,00 mm
leff1 = MIN(leff1a ; leff1b ) = 215,51 mm
leff2a = MIN( * 2 * m; * m + 2 * e 1) = 274,51 mm
leff2b = MIN( * m + p; p + 2 * e 1) = 437,26 mm
leff = MIN(leff2a ; leff2b ; leff1 ) = 215,51 mm
Beanspruchbarkeit der Lagerplatte auf der Seite der Zugstangen:
Mpl,1,Rd = *14*leff *( )d10
2
f y* M0 10
= 12,12*10 3 kNmm
n = e = 45,00 mmn
*1,25 m= 0,82 < 1
Bt,Rd = 0,9 * f u * A s
* Mb 10= 101,66 kN
Full flange yield:
FT,Rd1 = 4 *Mpl,1,Rd
m= 1109,64 kN
Bolt failure with flange yielding:
FT,Rd2 =*2 +Mpl,1,Rd *n *2 B t,Rd
+m n= 376,47 kN
Bolt failure without flange yielding:
FT,Rd3 = 2 * B t,Rd = 203,32 kN FT,Rd = MIN(F T,Rd1 ; F T,Rd2 ; F T,Rd3 ) = 203,32 kN
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Eurocode 3 Folder : foot and support
Effective compression area:Limit pressure in the base plate:a 1 = MIN(a + 2 * a r ; 5 * a; a + h) = 1900,00 mm
k j =
a 1
2
*a b= 3,80
f j = j * k j *f ck
c= 50,92 N/mm
A eff = *10+N Ed F T,Rd
f j= 236,32 cm
c = *d *f y 10*3 *f j M0 = 121,36 mmx0 = 100 *
A eff *2 +c b f
= 43,54 mm
x0+t f *2 c
= 0,17 < 1
Limit moment at column base:
r c = +h t2
-cx02
= 249,59 mm
MRd =
*F T,Rd *103
+( )+h t2 -100 e * Ae f f *100 *f j r c10
6= 342,02 kNm
Check permissible strength of column to bending and compression.:
MNy,Rd = 1,1 * M pl,y,Rd * (1-NEd
Npl,Rd) = 318,32 kNm
Analysis:
MAX(MEd
MNy,Rd;
MEd
MRd ) = 0,70 < 1
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Eurocode 3 Folder : Frames
Sway frame with bracing out of plane of frame, plastic theory:
dq
dH
A
B D
C
h
lPlan and elevation values:
Frame width l = 8,00 mFrame depth h = 6,00 mNumber of column n c = 2Number of storeys n s = 1
Beam:Profil Typ1 = SEL("steel/Profils"; Name; ) = IPESelected Profil1= SEL("steel/"Typ1; Name; ) = IPE 240
Column height h 1 = TAB("steel/"Typ1; h; Name=Profil1) = 240,00 mmWeb thickness s 1 = TAB("steel/"Typ1; s; Name=Profil1) = 6,20 mmMoment of inertia I y1 = TAB("steel/"Typ1; Iy; Name=Profil1) = 3890,00 cm 4
Cross-sectional area A 1= TAB("steel/"Typ1; A; Name=Profil1) = 39,10 cmMoment of resistance W y1 = TAB("steel/"Typ1; Wy; Name=Profil1) = 324,00 cmMoment of resistance Wpl y1 = 1,14 * W y1 = 369,36 cm
Column section:Profil Typ2 = SEL("steel/Profils"; Name; ) = HEBSelected Profil2 = SEL("steel/"Typ2; Name; ) = HEB 240Flange width b 2 = TAB("steel/"Typ2; b; Name=Profil2) = 240,00 mmColumn height h
2 = TAB("steel/"Typ2; h; Name=Profil2) = 240,00 mm
Web thickness s 2 = TAB("steel/"Typ2; s; Name=Profil2) = 10,00 mmMoment of inertia I y2 = TAB("steel/"Typ2; Iy; Name=Profil2) = 11260,00 cm 4
Moment of inertia I z2 = TAB("steel/"Typ2; Iz; Name=Profil2) = 3920,00 cm 4
Moment of inertia I t2 = TAB("steel/"Typ2; IT; Name=Profil2) = 103,00 cm 4
Moment of inertia I = 10*TAB("steel/"Typ2; I ; Name=Profil2) = 486900,00 cm 6Cross-sectional area A 2= TAB("steel/"Typ2; A; Name=Profil2) = 106,000 cmiy = TAB("steel/"Typ2; iy; Name=Profil2) = 10,30 cmiz = TAB("steel/"Typ2; iz; Name=Profil2) = 6,08 cmMoment of resistance W ely2 = TAB("steel/"Typ2;Wy; Name=Profil2) = 938,00 cmMoment of resistance W ply2 = 1,25 * W ely2 = 1172,50 cm
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Eurocode 3 Folder : Frames
Loads:Hd = 9,00 kN
qd = 9,00 kN/m
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 360E s = TAB("steel/EC"; E; Name=steel) = 210000,00 N/mmG = TAB("steel/EC"; G; Name=steel) = 81000,00 N/mmf y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mm = (235 / f y) = 1,00 M = 1,101 = 93,90 * = 93,90
Frame classification as in 5.2.5.2.:Frame sways
k = *Iy 1
Iy 2hl
= 0,259
Deflection at top of frame due to horizontal load:H = 1,00 kN
= **( )*100 h
310
*12 *E s Iy2
*2 +k 1
*k H= 0,446 cm
Sumtotal of vertical loads:V = q d * l = 72,00 kN
Bracing is restrained out of plane of frame
*
*h 100
V
H= 0,054 < 0,1
Sway frame, plastic theory, restrained from moving out of plane.
Design with plastic theoryImperfection:
0 =1
200= 0,005
kc = MIN( +0,5 1n c ; 1) = 1,00ks = MIN( +0,2
1
n s; 1) = 1,00
= 0 * kc * ks = 0,0050Hd = * V = 0,36 kNHEd = Hd + Hd = 9,36 kN
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Eurocode 3 Folder : Frames
Design calculations:Reactions at column bases:
Hl=
*q d l2
*4 *h ( )*2 +k 3= 6,82 kN
Vl = q d *l
2= 36,00 kN
MBl = -Hl * h = -40,92 kNmH2 = HEd / 2 = 4,68 kN
V2 = HEd *h
l= 7,02 kN
MB2 = H2 * h = 28,08 kNm
Analysis of column
NEd = Vl + V 2 = 43,02 kNVEd = Hl + H 2 = 11,50 kNMD = MBl - MB2 = -69,00 kNmMC = 0,00 kNm
Buckling in the plane of frame as in section E:
kc =Iy2
*h 100= 18,77 cm
k1 = 1,5 *Iy1
*l 100= 7,29 cm
k2 = 0,02 = 1 (Hinge)
1 =k c
+k c +k 1 k2 = 0,72
Slenderness ratio as in Fig. E.2.1.:
l = 3,2 * h = 19,20 m
y = *100l
iy= 186,41
trans,y =y
*1 = 1,99
h 2
b 2= 1,00 < 1,2
Apply strut curve b: = 0,34 A = 1,000
= 0,5 * (1 + * (trans,y - 0,2) + trans,y ) = 2,784
y =1
+ -2 trans,y 2= 0,2114
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Eurocode 3 Folder : Frames
Buckling outside the plane of frame
z= 100 *h
iz= 98,68
trans,z = z
*1 = 1,05
Apply strut curve c: = 0,49 A = 1,000 = 0,5 * (1 + * (trans,z - 0,2) + trans,z ) = 1,260
z =1
+ -2 trans,z 2= 0,5111
min = MIN( y; z) = 0,2114
Check bending and buckling: = 0,00My = 1,8 - 0,7 * = 1,80
y = trans,y * (2 * My - 4) +-W ply2 W ely2
W ely2= -0,546 < 0,9
ky = 1 -*y NEd
* y * A 2 f y= 1,004 < 1,5
+NEd
* min * A 2 f y
* M 10
*100 *k yabs ( )MD
*W ply2f y* M 10
= 0,37 < 1
Check torsional-flexural buckling:C1 = 1,879
Mcr = *C 1 **
2*E s Iz2
*( )*h 1002
10 +IIz2 *( )*h 100
2*G I t2
*2
*E s Iz2
= 94241,31 kNm
As in Table: F.1.1 = 0,00w = 1,00
k = 1,00
LT = *2 *E s10 W ply2Mcr = 50,78trans,LT = *
LT1
w = 0,541
Apply strut curve a
= 0,21 = 0,5 * (1 + * (trans,LT - 0,2) + trans,LT ) = 0,682
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Eurocode 3 Folder : Frames
LT =1
+ -2 trans,LT 2= 0,911
M,LT = 1,8 - 0,7 * = 1,80
LT = 0,15 * trans,z * M,LT - 0,15 = 0,134 < 0,9
kLT = 1 -*LT NEd
* LT * A 2 f y= 1,000 < 1,5
+NEd
* z * A 2f y
* M 10
*100 *kLTabs ( )MD
** LT W ply2f y
* M 10
= 0,340 < 1
Section analysis at D:Column:
A v = 1,04 * h 2 * s 2 / 10 = 24,96 cm
Vpl,Rd = * Avf y
* 3 * M 10= 307,86 kN
VEd
Vpl,Rd= 0,037 < 0,5
No interaction
Npl,Rd = * A 2f y
* M 10= 2264,55 kN
n =N Ed
N pl,Rd= 0,02
Mpl,y,Rd = *W ply2f y
* M 103
= 250,49 kNm
MNy,Rd = 1,11 * M pl,y,Rd * (1 - n) = 272,48 kNm
MAX(abs ( )MDMpl,y,Rd
;abs ( )MD
MNy,Rd) = 0,28 < 1
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Eurocode 3 Folder : Frames
Beam:
A v = 1,04 * h 1 * s 1 / 10 = 15,48 cm
Vpl,Rd = * Avf y
* 3 * M 10= 190,93 kN
N Ed
Vpl,Rd= 0,225 < 0,5
No interaction
Npl,Rd = A1 * f y / 10 / M = 835,32 kNn = V Ed / N pl,Rd = 0,014
Mpl,y,Rd = *W ply1f y
* M 103
= 78,91 kNm
MNy,Rd = 1,11 * M pl,y,Rd * (1 - n) = 86,36 kNm
MAX(abs ( )MDMpl,y,Rd
;abs ( )MD
MNy,Rd) = 0,87 < 1
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Eurocode 3 Folder : Frames
Frame, plastic with magnifying factor
dP dP
dq
dH
A
B D
C
h
lPlan and elevation values:
Frame width l= 8,00 mFrame depth h= 6,00 mNumber of column n c= 2Number of storeys n s= 1
Beam:Profil Typ1 = SEL("steel/Profils"; Name; ) = IPE
Selected Profil1= SEL("steel/"Typ1; Name; ) = IPE 240Column height h 1 = TAB("steel/"Typ1; h; Name=Profil1) = 240,00 mmFlange thickness s 1 = TAB("steel/"Typ1; s; Name=Profil1) = 6,20 mmMoment of inertia I y1 = TAB("steel/"Typ1; Iy; Name=Profil1) = 3890,00 cm 4
Cross-sectional area A 1= TAB("steel/"Typ1; A; Name=Profil1) = 39,10 cmMoment of resistance W y1 = TAB("steel/"Typ1; Wy; Name=Profil1) = 324,00 cmMoment of resistance Wpl y1 = 1,14 * W y1 = 369,36 cm
Column section:Profil Typ2 = SEL("steel/Profils"; Name; ) = HEBSelected Profil2 = SEL("steel/"Typ2; Name; ) = HEB 240Flange width b 2 = TAB("steel/"Typ2; b; Name=Profil2) = 240,00 mmColumn height h 2 = TAB("steel/"Typ2; h; Name=Profil2) = 240,00 mmFlange thickness s 2 = TAB("steel/"Typ2; s; Name=Profil2) = 10,00 mmMoment of inertia I y2 = TAB("steel/"Typ2; Iy; Name=Profil2) = 11260,00 cm 4
Moment of inertia I z2 = TAB("steel/"Typ2; Iz; Name=Profil2) = 3920,00 cm 4
Moment of inertia I t2 = TAB("steel/"Typ2; IT; Name=Profil2) = 103,00 cm 4
Moment of inertia I = 10*TAB("steel/"Typ2; I ; Name=Profil2) = 486900,00 cm 6Cross-sectional area A 2= TAB("steel/"Typ2; A; Name=Profil2) = 106,000 cmiy = TAB("steel/"Typ2; iy; Name=Profil2) = 10,30 cmiz = TAB("steel/"Typ2; iz; Name=Profil2) = 6,08 cmMoment of resistance W ely2 = TAB("steel/"Typ2;Wy; Name=Profil2) = 938,00 cm
Moment of resistance W ply2 = 1,25 * W ely2 = 1172,50 cm
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Eurocode 3 Folder : Frames
Loads:Hd = 9,00 kNqd = 8,00 kN/mP d= 100,00 kN
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 360E s = TAB("steel/EC"; E; Name=steel) = 210000,00 N/mmG = TAB("steel/EC"; G; Name=steel) = 81000,00 N/mmf y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mm
= 235f y = 1,00 M = 1,101 = 93,90 * = 93,90
Frame classification as in 5.2.5.2.:Frame swaysClassification of stiffness:
k = *Iy 1
Iy 2
h
l= 0,259
Deflection at top of frame due to horizontal load:H = 1,00 kN
= **( )*100 h
310
*12 *E s Iy2
*2 +k 1
*k H= 0,446 cm
Sumtotal of vertical loads:V = P d * 2 + q d * l = 264,00 kN
*
*h 100
V
H= 0,1962 < 0,1
Frame, laterally unrestrained!;restrained from moving out of plane. mit Vergrerungsfaktor!
Design with plastic theory and magnifying factor Imperfection:
0 =1
200= 0,005
kc = MIN( +0,51
n c ; 1) = 1,00
ks = MIN( +0,2 1n s ; 1) = 1,00 = 0 * kc * ks = 0,0050Hd = * V = 1,32 kNHEd = Hd + Hd = 10,32 kN
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Eurocode 3 Folder : Frames
Design calculations:Reactions at column bases:
Hl =*q d l
2
*4 *h ( )*2 +k 3= 6,06 kN
Vl = q d *l
2= 32,00 kN
MBl = -Hl * h = -36,36 kNmH2 = HEd / 2 = 5,16 kN
V2 = HEd *h
l= 7,74 kN
MB2 = H2 * h = 30,96 kNm
Analysis of columnDischinger factor:
D = 1
-1 *
*h 100
V
H
= 1,244
MD = MBl - D * M B2 = -74,87 kNmNEd = Vl + D * V 2 + P d = 141,63 kNVEd = Hl + D * H 2 = 12,48 kN
Buckling in the plane of frame:
kc =Iy2*h 100
= 18,77 cm
k1 =Iy1
*l 100= 4,86 cm
k2 = 0,02 = 1 (Hinge)
1 =k c
+k c +k1 k2 = 0,79
Slenderness ratio as in Fig. E.2.1.:l = 0,92 * h = 5,52 m
y = *100l
iy= 53,59
trans,y =y
*1 = 0,57
h 2b 2
= 1,00 < 1,2
Apply strut curve b: = 0,34 A = 1,000 = 0,5 * (1 + * (trans,y - 0,2) + trans,y ) = 0,725
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Eurocode 3 Folder : Frames
y =1
+ -2 trans,y 2= 0,8525
Buckling outside the plane of frame
z= 100 *h
iz= 98,68
trans,z =z
*1 = 1,05
Apply strut curve c: = 0,49 A = 1,000 = 0,5 * (1 + * (trans,z - 0,2) + trans,z ) = 1,260
z =1
+ -2 trans,z 2= 0,5111
min = MIN( y; z) = 0,5111
Check bending and buckling: = 0,00My = 1,8 - 0,7 * = 1,80
y = trans,y * (2 * My - 4) +-W ply2 W ely2
W ely2= 0,022 < 0,9
ky = 1 -*y NEd
* y * A 2 f y= 1,000 < 1,5
+NEd
* min * A 2f y* M 10
*100 *k yabs ( )MD
*W ply2f y* M 10
= 0,42 < 1
Check torsional-flexural buckling:C1 = 1,879
Mcr = *C 1 **
2*E s Iz2
*( )*h 1002
10 +
I
Iz2
*( )*h 1002
*G I t2
*2 *E s Iz2= 94241,31 kNm
As in Table: F.1.1 = 0,00w = 1,00k = 1,00
LT = *2 *E s10 W ply2Mcr = 50,78trans,LT = *
LT1
w = 0,541
Apply strut curve a = 0,21
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Eurocode 3 Folder : Frames
= 0,5 * (1 + * (trans,LT - 0,2) + trans,LT ) = 0,682
LT =1
+ -2 trans,LT 2= 0,911
M,LT = 1,8 - 0,7 * = 1,80LT = 0,15 * trans,z * M,LT - 0,15 = 0,134 < 0,9
kLT = 1 -*LT NEd
* LT * A 2 f y= 0,999 < 1,5
+NEd
* z * A 2f y
* M 10
*100 *kLTabs ( )MD
** LT W ply2f y
* M 10
= 0,450 < 1
Section analysis at D:Column:
A v = 1,04 * h 2 * s 2 / 10 = 24,96 cm
Vpl,Rd = * Avf y
* 3 * M 10= 307,86 kN
VEd
Vpl,Rd= 0,041 < 0,5
No interaction
Npl,Rd = * A 2f y* M 10
= 2264,55 kN
n =N Ed
N pl,Rd= 0,063
Mpl,y,Rd = *W ply2f y
* M 103
= 250,49 kNm
MNy,Rd = 1,11 * M pl,y,Rd * (1 - n) = 260,53 kNm
MAX(abs ( )MDMpl,y,Rd
;abs ( )MD
MNy,Rd) = 0,30 < 1
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Eurocode 3 Folder : Frames
Beam:
A v = 1,04 * h 1 * s 1 / 10 = 15,48 cm
Vpl,Rd = * Avf y
* 3 * M 10= 190,93 kN
N Ed
Vpl,Rd= 0,742 < 0,5
No interactionNpl,Rd = A1 * f y / 10 / M = 835,32 kNn = V Ed / N pl,Rd = 0,015
Mpl,y,Rd = *W ply1f y
* M 103
= 78,91 kNm
MNy,Rd = 1,11 * M pl,y,Rd * (1 - n) = 86,28 kNm
MAX(abs ( )MDMpl,y,Rd
;abs ( )MD
MNy,Rd) = 0,95 < 1
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Eurocode 3 Folder : Frames
Sway frame with bracing out of plane of frame, plastic theory:
dq
dH
A
B D
C
h
lPlan and elevation values:
Frame width l= 8,00 mFrame depth h= 6,00 mNumber of column n c= 2Number of storeys n s= 1
Beam:Profil Typ1 = SEL("steel/Profils"; Name; ) = IPESelected Profil1= SEL("steel/"Typ1; Name; ) = IPE 240Column height h 1 = TAB("steel/"Typ1; h; Name=Profil1) = 240,00 mm
Web thickness s 1 = TAB("steel/"Typ1; s; Name=Profil1) = 6,20 mmMoment of inertia I y1 = TAB("steel/"Typ1; Iy; Name=Profil1) = 3890,00 cm 4
Cross-sectional area A 1= TAB("steel/"Typ1; A; Name=Profil1) = 39,10 cmMoment of resistance W y1 = TAB("steel/"Typ1; Wy; Name=Profil1) = 324,00 cmMoment of resistance Wpl y1 = 1,14 * W y1 = 369,36 cm
Column section:Profil Typ2 = SEL("steel/Profils"; Name; ) = HEBSelected Profil2 = SEL("steel/"Typ2; Name; ) = HEB 240Flange width b 2 = TAB("steel/"Typ2; b; Name=Profil2) = 240,00 mmColumn height h 2 = TAB("steel/"Typ2; h; Name=Profil2) = 240,00 mm
Web thickness s 2 = TAB("steel/"Typ2; s; Name=Profil2) = 10,00 mmMoment of inertia I y2 = TAB("steel/"Typ2; Iy; Name=Profil2) = 11260,00 cm 4
Moment of inertia I z2 = TAB("steel/"Typ2; Iz; Name=Profil2) = 3920,00 cm 4
Moment of inertia I t2 = TAB("steel/"Typ2; IT; Name=Profil2) = 103,00 cm 4
Moment of inertia I = 10*TAB("steel/"Typ2; I ; Name=Profil2) = 486900,00 cm 6Cross-sectional area A 2= TAB("steel/"Typ2; A; Name=Profil2) = 106,000 cmiy = TAB("steel/"Typ2; iy; Name=Profil2) = 10,30 cmiz = TAB("steel/"Typ2; iz; Name=Profil2) = 6,08 cmMoment of resistance W ely2 = TAB("steel/"Typ2;Wy; Name=Profil2) = 938,00 cmMoment of resistance W ply2 = 1,25 * W ely2 = 1172,50 cm
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Eurocode 3 Folder : Frames
Loads:Hd = 9,00 kNqd = 9,00 kN/m
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 360E s = TAB("steel/EC"; E; Name=steel) = 210000,00 N/mmG = TAB("steel/EC"; G; Name=steel) = 81000,00 N/mmf y = TAB("steel/EC"; fy; Name=steel) = 235,00 N/mm
= 235f y = 1,00 M = 1,101 = 93,90 * = 93,90
Frame classification as in 5.2.5.2.:
Frame sways
k = *Iy 1
Iy 2
h
l= 0,259
Deflection at top of frame due to horizontal load:H = 1,00 kN
= **( )*100 h
310
*12 *E s Iy2
*2 +k 1
*k H= 0,446 cm
Sumtotal of vertical loads:V = q d * l = 72,00 kNBracing is restrained out of plane of frame
**h 100
VH
= 0,0535 < 0,1
Sway frame, plastic theory;restrained from moving out of plane.
Design with plastic theory and magnifying factor Imperfection:
0 =1
200= 0,005
kc = MIN( +0,5 1n c ; 1) = 1,00ks = MIN( +0,2 1n s ; 1) = 1,00 = 0 * kc * ks = 0,0050Hd = * V = 0,36 kNHEd = Hd + Hd = 9,36 kN
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Eurocode 3 Folder : Frames
Design calculations:Reactions at column bases:
Hl =*q d l
2
*4 *h ( )*2 +k 3= 6,82 kN
Vl = qd *l
2= 36,00 kN
MBl = -Hl * h = -40,92 kNmH2 = HEd / 2 = 4,68 kN
V2 = HEd *h
l= 7,02 kN
MB2 = H2 * h = 28,08 kNm
Analysis of columnNEd = Vl + V 2 = 43,02 kN
VEd = Hl + H 2 = 11,50 kNDischingerfaktor:
D =1
-1 *
*h 100
V
H
= 1,057
MD = MBl - D * M B2 = -70,60 kNmBuckling in the plane of frame:
kc =Iy2
*h 100= 18,77 cm
k1 =
Iy1
*l 100 = 4,86 cm
k2 = 0,02 = 1 (Hinge)
1 =k c
+k c +k 1 k 2 = 0,79
Slenderness ratio as in Fig. E.2.1.:l = 0,92 * h = 5,52 m
y = *100l
iy= 53,59
trans,y =y
*1 = 0,57
h 2b 2
= 1,00 < 1,2
Apply strut curve b: = 0,34 A = 1,000= 0,5*(1+ *(trans,y -0,2)+ trans,y ) = 0,725
y =1
+ -2 trans,y 2= 0,8525
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Eurocode 3 Folder : Frames
Buckling outside the plane of frame
z= 100 *h
iz= 98,68
trans,z =
z*1 = 1,05
Apply strut curve c:= 0,49 A = 1,000 = 0,5 * (1 + * (trans,z - 0,2) + trans,z ) = 1,260
z =1
+ -2 trans,z 2= 0,5111
min = MIN(y; z) = 0,5111
Check bending and buckling: = 0,00My = 1,8 - 0,7 * = 1,80
y = trans,y * (2 * My - 4) +-W ply2 W ely2
W ely2= 0,022 < 0,9
ky = 1 -*y NEd
*y * A 2 f y= 1,000 < 1,5
+NEd
* min * A 2f y
* M 10
*100 *k yabs ( )MD
*W ply2f y
* M 10
= 0,32 < 1
Check torsional-flexural buckling:C1 = 1,879
Mcr = *C 1 **
2*E s Iz2
*( )*h 1002
10 +I Iz2 *( )*h 1002
*G I t2
*2
*E s Iz2
= 94241,31 kNm
As in Table: F.1.1 = 0,00w = 1,00
k = 1,00
LT = * 2 *E s10 W ply2Mcr = 50,78trans,LT = *
LT1
w = 0,541
Apply strut curve a = 0,21
= 0,5 * (1 + * (trans,LT - 0,2) + trans,LT ) = 0,682
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Eurocode 3 Folder : Frames
LT =1
+ -2 trans,LT 2= 0,911
M,LT = 1,8 - 0,7 * = 1,80LT = 0,15 * trans,z * M,LT - 0,15 = 0,134 < 0,9
kLT = 1 -*LT NEd
*LT * A 2 f y= 1,000 < 1,5
+NEd
* z * A 2f y
* M 10
*100 *k LTabs ( )MD
** LT W ply2f y
* M 10
= 0,347 < 1
Section analysis at D:Column:
A v = 1,04 * h 2 * s 2 / 10 = 24,96 cm
Vpl,Rd = * Avf y
* 3 * M 10= 307,86 kN
VEd
Vpl,Rd= 0,037 < 0,5
No interaction
Npl,Rd = * A 2f y
* M 10= 2264,55 kN
n = NEd
N pl,Rd= 0,02
Mpl,y,Rd = *W ply2f y
* M 103
= 250,49 kNm
MNy,Rd = 1,11 * M pl,y,Rd * (1 - n) = 272,48 kNm
MAX(abs ( )MDMpl,y,Rd
;abs ( )MD
MNy,Rd) = 0,28 < 1
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Eurocode 3 Folder : Frames
Beam: A v = 1,04 * h 1 * s 1 / 10 = 15,48 cm
Vpl,Rd = * Avf y
* 3 * M 10= 190,93 kN
N Ed
Vpl,Rd= 0,225 < 0,5
No interaction
Npl,Rd = A1 * f y / 10 / M = 835,32 kNn = V Ed / N pl,Rd = 0,014
Mpl,y,Rd = *W ply1f y
* M 103
= 78,91 kNm
MNy,Rd = 1,11 * M pl,y,Rd * (1 - n) = 86,36 kNm
MAX(abs ( )MDMpl,y,Rd
;abs ( )MD
MNy,Rd) = 0,89 < 1
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Eurocode 3 Folder : Single-span beam
Pos.: Flexural-torsional buckling of a single-span beam:End supports are laterally fixed.
s
b
h
b
t
t
o
o
u
u
ts
qd
Load diagram:Span length l = 24,00 mtop width b o= 40,00 cmbottom width b u= 50,00 cmDepth of web h= 100,00 cmtop Flange thickness t o= 3,50 cmbottom Flange thickness t u= 3,50 cmWeb thickness s= 2,00 cmWeld thickness t s= 1,00 cm
Loads:q = 4,04 kN/m
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 510E s= TAB("steel/EC"; E; Name=steel) = 210000,00 N/mmG= TAB("steel/EC"; G; Name=steel) = 81000,00 N/mmf y= TAB("steel/EC"; fy; Name=steel) = 355,00 N/mm
= 235f y = 0,81Partial safety factors: M= 1,10 g= 1,35
Section classification as in Table 5.3.1: A = (b o * to + b u * tu + h * s) = 515,00 cmPlastic neutral axis:
x =
- A
2*b o to
s= 58,75 cm
xs = h - x = 41,25 cm
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Eurocode 3 Folder : Single-span beam
Flange:
c =
--b o s
2*t s 2
to= 5,02 cm
c
*9 = 0,69 < 1
Section class 1Web:
=x
h= 0,59 cm
h
*s *456
*13 - 1
= 0,90 < 1
h
*s *396
*13 - 1
= 1,04 > 1
Section class 2
Check bending moment strength:
MEd = g **q l
2
8= 392,69 kNm
Critical torsional-buckling moment:
Iz= *to +b o
3
12*tu
b u3
12= 55125,00 cm 4
As in Table: F.1.2(2)k= 1,00
As in Table: F.1.1(4)kw= 1,00
As in Table: F.1.2C1= 1,132C2= 0,459C3= 0,525Location of center of gravity:
e=
*b u *t u +( )+h t o *h *s( )+h t o
2
A= 55,27 cm
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Eurocode 3 Folder : Single-span beam
Location of shear centre:
s s= *( )+h +to t u2b u
3
+b o3
b u3
= 68,45 cm
z s= -(s s - e) = -13,18 cm
It=1
3 * (bu * tu + b o * to + h * s) = 1552,92 cm 4
z a = 0,00 cmz g= z a -z s = 13,18 cmas in F 1.4:
f =b o
3
+b o3
b u3
= 0,339 < 0,5
z j= (2 * f - 1) *
+h+to tu
22
= -16,66 cm
Iw= f * (1 - f ) * Iz *( )+h +to tu22
= 132,32*10 6 cm 6
P 1= *C 1**
2E s Iz
( )*k *l 1002
= 22453,85
P 2=
*( )kk w
2
+IwIz
+*( )*k *l 100
2*G I t
*2
*E s Iz
( )*C 2 -zg *C 3 z j2
= 94,66
P 3= C 2 * zg - C 3 * z j = 14,80
Mcr = *P 1-P 2 P 3
103
= 1793,16 kNm
Wpl = b u * tu * (xs +tu
2 ) + x s *
s
2 + b o * to * (x +
to
2 ) + x *
s
2= 21148,13 cm
Section class 2 w= 1,00
trans,LT = *w *W pl f y*10 3 Mcr = 2,046 As in Table 5.5.1 = 0,49= 0,5 * (1 + * (trans,LT -0,2) + trans,LT ) = 3,05
LT =1
+ -2 trans,LT 2= 0,1883
Mb,Rd = *w *LT *W plf y
* M2
103
= 1168,33 kNm
MEd
Mb,Rd= 0,34 < 1
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Eurocode 3 Folder : Single-span beam
Pos.: Single-span beam with biaxial bending:
N
qd
t,d
qd
Span length l = 5,80 m Angle of slope = 8,53
Loads:qd= 7,00 kN/mN
t,d= 100,00 kN
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 360E s= TAB("steel/EC"; E; Name=steel) = 210000,00 N/mmf y= TAB("steel/EC"; fy; Name=steel) = 235,00 N/mm
= 235f y = 1,00platischer Formbeiwert pl y = 1,14platischer Formbeiwert pl z = 1,25 M= 1,10
Profil Typ = SEL("steel/Profils"; Name; ) = HEASelected Profil = SEL("steel/"Typ; Name; ) = HEA 140
Section classification:Column height h = TAB("steel/"Typ; h; Name=Profil) = 133,00 mmDepth of web h 1 = TAB("steel/"Typ; h1; Name=Profil) = 92,00 mmWeb thickness s = TAB("steel/"Typ; s; Name=Profil) = 5,50 mmFlange width b = TAB("steel/"Typ; b; Name=Profil) = 140,00 mmFlange thickness t = TAB("steel/"Typ; t; Name=Profil) = 8,50 mmMoment of inertia I = TAB("steel/"Typ; Iy; Name=Profil) = 1030,00 cm 4Cross-sectional area A= TAB("steel/"Typ; A; Name=Profil) = 31,40 cmWel y= TAB("steel/"Typ; Wy; Name=Profil) = 155,00 cmWpl y = pl y * W el y = 176,70 cmWel z = TAB("steel/"Typ; Wz; Name=Profil) = 55,60 cmWpl z = pl z * W el z = 69,50 cm
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Eurocode 3 Folder : Single-span beam
Design loads:Plastic limit force:
Npl,Rd = * Af y
* M 10= 670,82 kN
Limit force that plasticizes the web
Npl,w,Rd = (h - 2 * t) * s *f y
* M 103
= 136,30 kN
MAX(Nt,d
*0,5 N pl,w,Rd;
Nt,d*0,25 N pl,Rd
) = 1,47 > 1
Reduce plastic limit force as in 5.4.8.1:
=
- A *2 *bt
100
A= 0,242 < 0,5
=N t, d
N pl,Rd= 0,149
MNy,Rd = *W pl y *f y
* M 103
-1 -1 *0,5
= 36,55 kNm
= 0,62 < 1
MNz,Rd = W plz *f y
* M 103
= 14,85 kNm
Design momente:
qdy = q d * COS( ) = 6,92 kN/mqdz = q d * SIN() = 1,04 kN/m
My,Ed = q dy *l
2
8= 29,10 kNm
Mz,Ed = q dz *l
2
8= 4,37 kNm
Check section strength:for I-Profile gilt = 2,00
1= 5 * = 0,75 < 12= 1,00 = MAX(1 ;2 ) = 1,00
+( )My,EdMNy,Rd
( )Mz,EdMNz,Rd
= 0,93 < 1
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Eurocode 3 Folder : Single-span beam
Pos.: flexural-torsional buckling:End supports are laterally fixed, load is applied in the middle of the bottom flange.
s
b
h
t
tts
qd
Load diagram:
Span length l = 8,00 mBeam width b= 20,00 cmColumn height h= 50,00 cmFlange thickness t= 2,50 cmWeb thickness s= 1,20 cmWeld thickness t s = 0,50 cm
Loads:qd = 50,00 kN/m
Materials and stresses:steel = SEL("steel/EC"; Name; ) = Fe 510
E s= TAB("steel/EC"; E; Name=steel) = 210000,00 N/mmG= TAB("steel/EC"; G; Name=steel) = 81000,00 N/mmf y= TAB("steel/EC"; fy; Name=steel) = 355,00 N/mm
= 235f y = 0,81Partial safety factors: M= 1,10
Section classification:Flange:
c =
-b2
*t s 2
t= 3,72 cm
c
*9 = 0,51 < 1
Section class 1Web:d= -h *2 *t s 2 = 48,59 cm
d
*s *72 = 0,69 < 1
Section class 1
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Eurocode 3 Folder : Single-span beam
Check bending moment strength:
Wpl = *( )*b *t ++h t2 *h2 *s h4 2 = 3375,00 cmM
c,Rd=
*W pl f y
* M 103
= 1089,20 kNm
MEd =*q d l
2
8= 400,00 kNm
MEdMc,Rd
= 0,37 < 1
Check shear strength:
Av= h * s = 60,00 cm
Vpl,Rd = * A vf y
** 3 M 10= 1117,96 kN
VEd = *q dl
2= 200,00 kN
VEdVpl,Rd
= 0,18 < 1
Check torsional buckling strength:
I
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