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8/9/2019 Mnogo zajebana stvar

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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

Pdf-Overview: Templates for Structural Engineering


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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



3/124


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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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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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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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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Connection subject to moment load:

7

1 2 3

54

86

e

e

e

e

ee

e e

2

2

3

3

1 1


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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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



9/124


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



10/124


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


= 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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= 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



12/124


13/124


Double shear joint with bolts:


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



14/124


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


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


= 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



15/124


= 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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High strength friction-grip bolt 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



17/124


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


A net,1 = A - 2 * MIN(d 1 ; d 2 ) *d

100= 15,36 cm


A net,2 = A - 3 *d

100 * MIN(d 1 ;d2 ) + *

2

100

e2

*4 e 3*MIN(d1 ;d 2 ) = 15,85 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



18/124


Joint with tension and shear:


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




19/124


A net,2 = A - 3 *d

100 * MIN(d 1 ;d2 ) + *

2

100

e2

*4 e 3*MIN(d1 ;d 2 ) = 15,85 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


= 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



20/124


Check shear failure:

VVEd

VVVVVV


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



21/124


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



22/124

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.



23/124


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



24/124


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


= 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



25/124


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



26/124


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



27/124


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



28/124


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



29/124


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



30/124


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



31/124


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



= 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



32/124


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



33/124


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


*10 M1 = 2135,60 kN

N d

N b,Rd= 0,94 < 1



34/124


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


= 235f y = 1,001 = 93,90 * = 93,90

Partial safety factors: M = 1,10



35/124


36/124



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



37/124


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



38/124


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



39/124


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



40/124


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



41/124


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



42/124


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



= 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



43/124


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



44/124


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


* M 10 = 3917,19 kNN d

N b,Rd= 0,893 < 1



45/124

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



46/124


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



47/124


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



48/124


*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



49/124

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



50/124


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



51/124


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



52/124


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



53/124


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



54/124


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



55/124


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



56/124


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



57/124

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



58/124


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



59/124


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



60/124


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



61/124


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



62/124


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


* M 103

= 78,91 kNm



;abs ( )MD

MNy,Rd) = 0,87 < 1



63/124


Frame, plastic with magnifying factor

dP dP

dq

dH

A

B D

C

h


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


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 = 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



64/124


Loads:Hd = 9,00 kNqd = 8,00 kN/mP d= 100,00 kN


= 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


= **( )*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



65/124



Hl =*q d l

2

*4 *h ( )*2 +k 3= 6,06 kN

Vl = q d *l

2= 32,00 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



66/124


y =1

+ -2 trans,y 2= 0,8525


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



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


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



67/124


= 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


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


* M 103

= 250,49 kNm



;abs ( )MD

MNy,Rd) = 0,30 < 1



68/124


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


* M 103

= 78,91 kNm



;abs ( )MD

MNy,Rd) = 0,95 < 1



69/124


Sway frame with bracing out of plane of frame, plastic theory:

dq

dH

A

B D

C

h


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



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




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



70/124


Loads:Hd = 9,00 kNqd = 9,00 kN/m


= 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


= **( )*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



71/124



Hl =*q d l

2

*4 *h ( )*2 +k 3= 6,82 kN

Vl = qd *l

2= 36,00 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



72/124



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



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


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



73/124


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


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


* M 103

= 250,49 kNm



;abs ( )MD

MNy,Rd) = 0,28 < 1



74/124


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


* M 103

= 78,91 kNm



;abs ( )MD

MNy,Rd) = 0,89 < 1



75/124

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



76/124


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



77/124



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



78/124


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



79/124


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



80/124


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



81/124


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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