Ec9 Ex84 Thin Walled

Ec9_ex84_Torsion_Thin walled.mcd Torsion and bending of 9

y

bl

bl

0

0

bt

bt

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:= z

c

0

0

h

h

h c−

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

:=

0 20 40 600

20

40

60

80

100

0

z

mm

0

y

mm

Comment 1: Local buckling is checked after stress calculations

Comment 2: The following expressions are applicable to open cross sections.

Nodes i 1 rows y( ) 1−..:=

Area of crosssectionelements

dAi ti yi yi 1−−( )2 zi zi 1−−( )2+⋅⎡⎣

⎤⎦

→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

:=

Crosssectionarea A

1

rows y( ) 1−

i

dAi∑=

:= A 537.5 mm2=

First moment of area,gravity center

Sy

1

rows y( ) 1−

i

zi zi 1−+( )dAi

2⋅∑

=

:= zgcSy

A:=

Second moment of area of effectivecross section

zgc 48.088 mm=Iy

1

rows y( ) 1−

i

zi( )2 zi 1−( )2+ zi zi 1−⋅+⎡⎣

⎤⎦

dAi

3⋅∑

=

:=Iy Iy A zgc

2⋅−:=

First moment of area,gravity center

Sz

1

rows y( ) 1−

i

yi yi 1−+( )dAi

2⋅∑

=

:= ygcSz

A:= ygc 14.344 mm=

Bending and torsion of beam with asymmetric thin-walled cross section

1

0

2

3 4

5

y

z

q

bt

bl

hc

c

bl 42 mm⋅:= kN 1000 N⋅≡

Section bt 38 mm⋅:= MPa 106 Pa⋅≡

h 98 mm⋅:= γM1 1.1:=

c 18.5 mm⋅:=

t0 2.5 mm⋅:=

Length L 2800 mm⋅:=

Material 5052 - H18 fo 240 N⋅ mm 2−⋅:=

Load, charac-teristic value

q 0.8 kN⋅ m 1−⋅:= at center of bottom flange

Nodes, coordinates and thickness

L

q

ti t0:=i

0

1

2

3

4

5

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

:=

Torsten Höglund aluMATTER 2007-08-27


Shear center

yscIzω Iz⋅ Iyω Iyz⋅−

Iy Iz⋅ Iyz2

−:= zsc

Iyω− Iy⋅ Izω Iyz⋅+

Iy Iz⋅ Iyz2

−:= ysc 21.774− mm=

zsc 43.649 mm=Warping constant Iw Iωω zsc Iyω⋅+ ysc Izω⋅−:= toi

if ti 0 mm⋅> ti, 100 mm⋅,( ):= Iw 3.642 108× mm6

=

Torsionconstant

It1

rows y( ) 1−

i

dAi

ti( )23

⋅∑=

:= WtIt

min to( ):= It 1.12 103× mm4

=

Wt 447.917 mm3=

Sectorial coordinatewith respect to shear center

i 0 rows y( ) 1−..:=

ωsiω i ωmean− zsc yi ygc−( )⋅+ ysc zi zgc−( )⋅−:= ωmi min ωs( ):= ωma max ωs( ):=

ωmax if ωmi ωma> ωmi, ωma,( ):= ωmax 2.064 103× mm2

= WwIw

ωmax:=

Sectorial moment forthe elements

i 1 rows y( ) 1−..:=

Sωi1

i

j

ωs jωs j 1−

+( ) dA j⋅∑=

:= Sω

0.0

129.2

106.0

125.7

139.8

0.0−

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

1000 mm4⋅=

Second moment of area

Iz1

rows y( ) 1−

i

yi( )2 yi 1−( )2+ yi yi 1−⋅+⎡⎣

⎤⎦

dAi

3⋅∑

=

:= Iz Iz A ygc2

⋅−:=

Second moment of area

Iyz

1

rows y( ) 1−

i

2 yi 1−⋅ zi 1−⋅ 2 yi⋅ zi⋅+ yi 1− zi⋅+ yi zi 1−⋅+( )dAi

6⋅∑

=

:= Iyz IyzSy Sz⋅

A−:=

Principal axis

α if Iz Iy−( ) 0 mm4⋅= 0,

12

atan2 Iyz⋅

Iz Iy−

⎛⎜⎝

⎞⎟⎠

⋅,⎡⎢⎣

⎤⎥⎦

:= α180π

⋅ 1.678=

Iξ12

Iy Iz+ Iz Iy−( )2 4 Iyz2

⋅++⎡⎣

⎤⎦⋅:=

Iη12

Iy Iz+ Iz Iy−( )2 4 Iyz2

⋅+−⎡⎣

⎤⎦⋅:=

Sectorial coordinates

ω0 0 mm2⋅:= ω0i

yi 1− zi⋅ yi zi 1−⋅−:= ω i ω i 1− ω0i+:=

ωmean =mean ofsectorialcoordinates

Iω1

rows y( ) 1−

i

ω i 1− ω i+( )dAi

2⋅∑

=

:= ωmeanIωA

:=

Sectorial constant Iyω

1

rows y( ) 1−

i

2 yi 1−⋅ ω i 1−⋅ 2 yi⋅ ω i⋅+ yi 1− ω i⋅+ yi ω i 1−⋅+( )dAi

6⋅∑

=

:= Iyω IyωSz Iω⋅

A−:=

Sectorial constant Izω

1

rows y( ) 1−

i

2 ω i 1−⋅ zi 1−⋅ 2 ω i⋅ zi⋅+ ω i 1− zi⋅+ ω i zi 1−⋅+( )dAi

6⋅∑

=

:= Izω IzωSy Iω⋅

A−:=

Sectorial constant Iωω

1

rows y( ) 1−

i

ω i( )2 ω i 1−( )2+ ω i ω i 1−⋅+⎡⎣

⎤⎦

dAi

3⋅∑

=

:= Iωω IωωIω

2

A−:=



Comment: If the load is acting below the shear center(the point) there is no torsional moment acting on the beam.

Iη 1.447 105× mm4

=

Iξ 8.252 105× mm4

=

α 1.678 deg=Principal axis

ωmax 2.064 103× mm2

=

Ww 1.764 105× mm4

=

Iw 3.642 108× mm6

=Warping constants

Wt 447.917 mm3=

It 1.12 103× mm4

=Torsionconstants

Iyz 1.993− 104× mm4

=

Iz 1.452 105× mm4

=

Iy 8.246 105× mm4

=Second moment of area

zsc 43.649 mm=

ysc 21.774− mm=Shear center

zgc 48.088 mm=

ygc 14.344 mm=Gravity center

40 20 0 20 40 60 8020

0

20

40

60

80

100

120

i 1 rows y( ) 1−..:=A 537.5 mm2=



ν 1.525=

η24

5 ν4

⋅

1cosh ν( )

ν2

2+ 1−

⎛⎜⎝

⎞⎟⎠

⋅:= η 0.513=

Angle of rotation φ35 me⋅ L4

⋅

384 E⋅ Iw⋅η⋅:= φ3 0.331 rad= φ3 18.96 deg=

Ultimate limit state γF 1.5:=

Bi-moment B3γF me⋅ L2

⋅

ψ1 8⋅G It⋅

γF

ψ1⋅ φ3⋅−:= B3 0.0253 kN m2

⋅=

Warping stress σwi

B3

Iw− ωsi

⋅:=

St Venant shear force

χ3

ν3ν tanh ν( )−( ):= Tsv

γF me⋅ L⋅

ψ1 2⋅ν

2

3⋅ χ⋅:= Tsv 0.029 kN m⋅=

St Venant shear stress τsv

Tsv max t( )⋅

It:= τsv 64.7 MPa=

20 0 20 40 60 8020

0

20

40

60

80

100

120

sectorial co-ordinatecross section

Sectorial coordinatewith respect to shear center

i

0

1

2

3

4

5

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

=ωsi

mm2 1000⋅

1.986

0.806

1.027−

1.107

0.958−

2.064−

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

=

L

q

Torsion

Serviceability limit state

Say ψ1 0.6:=

Load qSLS q ψ1⋅:=

qSLS 0.48kNm

=

Span L 2800 mm=

Torsion moment me qSLS ysc− 0.5 y1⋅+( )⋅:= me 0.0205 kN= E 70000 MPa⋅:= G 27000 MPa⋅:=

From handbook νL2

G It⋅

E Iw⋅⋅:=



Rotation of cross section

Rotation aroundorigin of coordinates Ro

cos φ3( )sin φ3( )−

sin φ3( )cos φ3( )

⎛⎜⎝

⎞⎟⎠

:=yri

zri

⎛⎜⎜⎝

⎞⎟⎟⎠

Royi

zi

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅:=

50 0 50 10050

0

50

100

Cross sectionRotated cross section

0

Displacement of shear center

ys

zs⎛⎜⎝

⎞⎟⎠

Roysc

zsc

⎛⎜⎝

⎞⎟⎠

⋅:=ys

zs⎛⎜⎝

⎞⎟⎠

6.4−

48.4⎛⎜⎝

⎞⎟⎠

mm=ysc

zsc

⎛⎜⎝

⎞⎟⎠

21.8−

43.6⎛⎜⎝

⎞⎟⎠

mm=

Rotation of cross section aroundshear center

yri

zri

⎛⎜⎜⎝

⎞⎟⎟⎠

yriys− ysc+

zrizs− zsc+

⎛⎜⎜⎝

⎞⎟⎟⎠

:=

yr

mm

30.37

24.36

15.36−

16.48

52.42

46.41

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

=zr

mm

0.86−

18.35−

4.71−

87.98

75.63

58.13

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

=

50 0 50 10050

0

50

100

Cross sectionRotated cross section

0

The rotation is not acceptable in the serviceability limit state. Increase the thickness or change shape of cross section.



20 0 20 40 6060

40

20

0

20

40

60

80

100

Cross sectionDeflected cross section

0

yri

zri

⎛⎜⎜⎝

⎞⎟⎟⎠

yrivy−

zrivz−

⎛⎜⎜⎝

⎞⎟⎟⎠

:=Translation of cross section

Not acceptableL200

14.0 mm=The deflection should not be larger than

v 52.1 mm=v vt vz+:=Resulting deflection

vt 14.2 mm=vt φ3 0.5 y1⋅ ysc−( )⋅:=

Vertical deflection due to torsion

vz 37.9 mm=vy

vz

⎛⎜⎝

⎞⎟⎠

0.92−

37.91⎛⎜⎝

⎞⎟⎠

mm=vy

vz

⎛⎜⎝

⎞⎟⎠

RT vξ

vη

⎛⎜⎝

⎞⎟⎠

⋅:=Deflection in vertical and horizontal direction

vη

vξ

vz

vy

vξ

vη

⎛⎜⎝

⎞⎟⎠

0.19

37.92⎛⎜⎝

⎞⎟⎠

mm=vξ

vη

⎛⎜⎝

⎞⎟⎠

5 L4⋅

384 E⋅

qξIξ

qηIη

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

⋅:=

Deflection in principal axis direction

qξ

qη

⎛⎜⎝

⎞⎟⎠

0.014

0.48⎛⎜⎝

⎞⎟⎠

kN m 1−⋅=

qξ

qη

⎛⎜⎝

⎞⎟⎠

Rqy

qz

⎛⎜⎝

⎞⎟⎠

⋅:=Rcos α( )sin α( )−

sin α( )cos α( )

⎛⎜⎝

⎞⎟⎠

:=Principal axis load

qy

qz

⎛⎜⎝

⎞⎟⎠

0

0.48⎛⎜⎝

⎞⎟⎠

kN m 1−⋅=qy 0 kN⋅ m 1−

⋅:=qz ψ1 q⋅:=Loads

Deflection due to biaxial bending

qη

qz

qξ



Sum of stresses σ σξ ση+ σw+:= σmi min σ( ):= σma max σ( ):=

Max stresses σmax if σmi σma> σmi, σma,( ):= σmax 145− MPa=

Principal axis

40 20 0 20 40 6020

0

20

40

60

80

100

Cross sectionAxial stress

0

σsum augment σw σξ, ση, σ,( ):=

σw σξ ση σ

i

0

1

2

3

4

5

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

=σsum

MPa

138−

56−

71

77−

67

143

43

70

68

72−

70−

44−

6−

6−

4

3

6−

6−

101−

7

143

145−

10−

94

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

=

σmax 145− MPa=

Stresses of biaxial bending

Bending moment My γF q⋅L2

8⋅:= Mz 0 kN⋅ m⋅:=

My

Mz

⎛⎜⎝

⎞⎟⎠

1.176

0⎛⎜⎝

⎞⎟⎠

kN m⋅=

Principal axis bending R

cos α( )sin α( )−

sin α( )cos α( )

⎛⎜⎝

⎞⎟⎠

:=Mξ

Mη

⎛⎜⎝

⎞⎟⎠

RMy

Mz

⎛⎜⎝

⎞⎟⎠

⋅:=Mξ

Mη

⎛⎜⎝

⎞⎟⎠

1.175

0.034−⎛⎜⎝

⎞⎟⎠

kN m⋅=

Rotation of coordinatesystem

i 0 rows y( ) 1−..:=yi

zi

⎛⎜⎜⎝

⎞⎟⎟⎠

Ryi

zi

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅:=ygc

zgc

⎛⎜⎝

⎞⎟⎠

Rygc

zgc

⎛⎜⎝

⎞⎟⎠

⋅:=

Bending stresses σξ i

Mξ− zi zgc−( )⋅

Iξ:= σηi

Mη yi ygc−( )⋅

Iη:=

Sum of warping and bending stresses



foγM1

218 MPa=

[1] 6.2.1 (5)σvM.max γM1⋅

fo0.667= < C 1.1= for C 1.2:=

St Venant shear stress at ends of beam

St Venant torsion moment χ

3

ν3ν tanh ν( )−( ):= Tsv

γF me⋅ L⋅

ψ1 2⋅ν

2

3⋅ χ⋅:= Tsv 0.029 kN m⋅=

St Venant shear stress τsv

Tsv max t( )⋅

It:=

[1] 6.2.7.1 (1) τsv 64.7 MPa= < fo

3 γM1⋅126 MPa=

Shear stresses due to warping torsion and combined stresses

Vlasov torsion moment at center of beam

TωγF me⋅ L⋅

ψ1 2⋅:= Tω 0.072 kN m⋅=

B

TTsv

Tw

Vlasov shear stress τ i

Tω Sωi⋅

Iωω ti⋅:= τ

0.00

3.73

3.07

3.63

4.04

0.00−

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

MPa=

τmax max τ

→⎯( ):= τmax 4.04 MPa=

[1] 6.2.7.2 (3)

von Mises stress σvM σ2

3 τ2

⋅+:= σvM.max max σvM→⎯⎯⎯⎛

⎝⎞⎠:=

σvM.max 145.6 MPa=

Compare above σmax 145.5− MPa= <



is less than ρ fo⋅ 229 MPa= so there will be no bucklingσ0 101− MPa=

ρ 0.953=ρ if λp 0.517>0.9λp

10.22λp

−⎛⎜⎝

⎞⎟⎠

⋅, 1,⎡⎢⎣

⎤⎥⎦

:=

λp 0.595=λp 1.052bp

t0⋅

fokσ E⋅

⋅:=

kσ 0.586=kσ 0.57 0.21 ψ⋅− 0.07 ψ2

⋅+:=EN 1993-1-5,Table 4.2

ψ 0.073−=ψσ1

σ0:=

bp 18.5 mm=bp z0 z1−:=

The edge fold of the bottom flange is in compression.

The edge fold of the top flange is in tension

No local bucklingρ 1=ρ if λp 0.517>0.9λp

10.22λp

−⎛⎜⎝

⎞⎟⎠

⋅, 1,⎡⎢⎣

⎤⎥⎦

:=

λp 0.345=λp 1.052bp

t0⋅

fokσ E⋅

⋅:=

kσ 7.356=kσ16

1 ψ+( )2 0.11 1 ψ−( )2⋅+ 1 ψ+( )+

:=

E 7 104× MPa=ψ 0.065=ψ

σ4

σ3:=

fo 240 MPa=bp 38 mm=bp y4 y3−:=Compression flange

Local buckling according to EN 1999-1-4


Ec9 Ex84 Thin Walled

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