This example provides calculations on torsion of members based on Eurocode 9.

1. TALAT Lecture 2301
Design of Members
Torsion
Example 8.1 : Torsion constants for open cross section
6 pages
Advanced Level
prepared by Torsten Höglund, Royal Institute of Technology, Stockholm
Date of Issue: 1999
 EAA - European Aluminium Association
TALAT 2301 – Example 8.1 1

2. Example 8.1. Torsion constants for open cross section
Comment: The following expressions are applicable to
open cross sections. When there are branches, go
back the same way but with thickness = 0.
Example: The co-ordinates for node 7 and 8 are the
same as for 5 and 4 and t = 0 for element 7 and 8.
Nodes, co-ordinates and thickness
0 34 0 6
1 10 0 6
2 8 9 6
3 25 9 6
4 25 3 6
5 59 0 6
6 59 13 15
i y . mm z . mm t . mm
7 59 0 0
300
8 25 3 0
9 25 34 6 0
250
10 0 37 6
11 0 245 6
200
12 62 252 6
13 62 289 10
150
z
Nodes i 1 .. rows ( y ) 1 mm
100
Area of cross
ti .
section 2 2
dAi yi yi 1 zi zi 1
elements 50
Cross rows ( y ) 1 0
section 0
A = 3.199 . 10 mm
3 2
area A dAi
i =1
50
50 0 50 100
First moment rows ( y ) 1 y
dAi Sy
of area, Sy zi zi 1 . z gc mm
gravity centre 2 A
i =1 z gc = 117.14 mm
Second rows ( y ) 1
dAi
zi . zi 1 . A . z gc
moment of 2 2 2
Iy zi zi 1 Iy Iy
area of effective 3
i =1
cross section I y = 3.596 . 10 mm
7 4
rows ( y ) 1
First moment dAi Sz
of area, Sz yi yi 1 . y gc
2 A
gravity centre i =1
y gc = 19.073 mm
TALAT 2301 – Example 8.1 2

3. Second rows ( y ) 1
dAi
yi . yi 1 . A . y gc
moment 2 2 2
Iz yi yi 1 Iz Iz
of area 3
i =1
I z = 2.104 . 10 mm
6 4
rows ( y ) 1
Second dAi S y .S z
moment I yz 2 . yi 1 . zi 1 2 . yi . zi yi 1 . zi yi . zi 1 . I yz I yz
of area 6 A
i =1
I yz = 2.026 . 10 mm
6 4
1 2 . I yz
0 . mm , 0 , . atan
4
Principal α if I z Iy 180
axis 2 Iz Iy α. = 3.414
π
1.
4 . I yz I ξ = 3.608 . 10 mm
2 2 7 4
Iξ Iy Iz Iz Iy
2
1.
4 . I yz I η = 1.983 . 10 mm
2 2 6 4
Iη Iy Iz Iz Iy
2
0 . mm
2
Sectorial ω0
co-ordinates
ω 0 yi .z yi . zi 1
i 1 i
ωi ω i 1
ω 0
i
ωmean = rows ( y ) 1
dAi Iω
mean of Iω ω ω . ω mean
i 1 i
sectorial 2 A
i =1
co-ordinates
ω mean = 799.492 mm
rows ( y ) 1
Sectorial dAi S z.I ω
constant I yω 2 . yi 1 . ω i 1 2 . yi . ω i yi 1 . ω i yi . ω i 1 . I yω I yω
6 A
i =1
I yω = 2.921 . 10 mm
8 5
rows ( y ) 1
Sectorial dA S y .I ω
I zω 2 . ω i 1 . zi 1 2 . ω i . zi ω .z ω i . zi . Ii I zω
constant i 1 i 1
6 zω A
i =1
I zω = 9.181 . 10 mm
8 5
rows ( y ) 1 2
Sectorial dAi Iω
ω i .ω i 1 .
2 2
constant I ωω ω i
ω i 1
I ωω I ωω
3 A
i =1
= 7.377 . 10
10 6
I ωω mm
I zω . I z I yω . I yz I yω . I y I zω . I yz
Shear y sc z sc y sc = 18.728 mm
I y .I z I y .I z
centre 2 2
I yz I yz
z sc = 120.773 mm
Warping
Iw I ωω z sc . I yω y sc . I zω
if t i > 0 . mm , t i , 100 . mm I w = 2.129 . 10
constant 10 6
to mm
i
rows ( y ) 1 2
TALAT 2301 – Example 8.1 3

4. i i i
rows ( y ) 1 2
ti It
dAi . I t = 5.856 . 10 mm
4 4
Torsion It Wt
constant 3 min t o
i =1
W t = 9.76 . 10 mm
3 3
Sectorial
co-ordinate i 0 .. rows ( y ) 1
with respect ω s ω i ω mean z sc . yi y gc y sc . zi z gc ω mi min ω s ω ma max ω s
to shear centre i
Iw
ω max 7.804 . 10 mm
3 2
ω max if ω mi > ω ma , ω mi , ω ma = Ww
ω max
i 1 .. rows ( y ) 1
A = 3.199 . 10 mm
3 2
y gc = 19.073 mm 300
z gc = 117.14 mm
y sc = 18.728 mm 250
z sc = 120.773 mm
I y = 3.596 . 10 mm
7 4
200
I z = 2.104 . 10 mm
6 4
I yz = 2.026 . 10 mm
6 4
150
I t = 5.856 . 10 mm
4 4
Torsion
constants
W t = 9.76 . 10 mm
3 3 100
Warping
I w = 2.129 . 10
10 6
constants mm
50
W w = 2.729 . 10 mm
6 4
ω max 7.804 . 10 mm
3 2
=
0
α = 3.414 deg
I ξ = 3.608 . 10 mm
7 4
50
50 0 50 100
I η = 1.983 . 10 mm
6 4
Comment:
If the load is acting below the shear centre
(the point) there is no torsional moment
acting on the beam.
TALAT 2301 – Example 8.1 4

5. Sectorial
co-ordinate 300
with respect
to shear centre
250
i 0 .. rows ( y ) 1
ω s 200
i
=
mm . 1000
2
i=
0 -7.804 150
1 -4.906
2 -4.743
3 -0.46
100
4 0.065
5 3.938
6 2.927
50
7 3.938
8 0.065
9 1.42
0
10 -0.618
11 3.278
12 -4.293
13 -1.307 50
100
100 50 0 50 100 150
sectorial co-ordinate
cross section
kN 1000 . N
5 . kN . m 4.8 . m MPa 10 . Pa
1 6
Biaxial bending q L
Moment 2 My 14.4
L
My q. Mz 0 . kN . m = kN . m
8 Mz 0
cos ( α ) sin( α ) Mξ My Mξ 14.374
Principal axis R R. = kN . m
bending sin( α ) cos ( α ) Mη Mz Mη 0.857
Torsion mt 0 . kN m t = 0 kN
m t .L
2
B 3 = 0 kN . m
2
Bi-moment B3
8
St Venants m t .L
torsion Tw T w = 0 kN . m
2
Warping B3
σ w .ω
stress si
i Iw
TALAT 2301 – Example 8.1 5

6. yi yi y gc y gc
Rotation of
co-ordinate i 0 .. rows ( y ) 1 R. R.
zi zi z gc z gc
system
M ξ . yi y gc M η . zi z gc
Bending stresses σ ξ σ η
i Iξ i Iη
Sum of stresses σ σ ξ σ η σ w σ mi min ( σ ) σ ma max ( σ )
Max stresses σ max if σ mi > σ ma , σ mi , σ ma σ max= 74.1 MPa
Principal axis
300
250
σ w σ ξ σ η σi ti
i i i
= = = = =
200 i= MPa MPa MPa MPa mm
1 0 -8.78 51.3 42.52 6
2 0 -7.78 55.133 47.36 6
150 3 0 5.35 54.283 59.63 6
4 0 5.07 49.105 54.17 6
5 0 18.66 49.524 68.18 6
6 0 18.97 55.134 74.1 15
100
7 0 18.66 49.524 0 0
8 0 5.07 49.105 0 0
9 0 4.33 35.727 40.06 6
50
10 0 -5.68 35.076 29.39 6
11 0 -10.62 -54.687 -65.31 6
0 12 0 13.87 -59.304 -45.43 6
0
13 0 13 -75.271 -62.27 10
50
100
50 0 50 100
Cross section
Axial stress
TALAT 2301 – Example 8.1 6