Talat Lecture 2301: Design Of Members Example 8.2: Torsion Constants For Hollow Cross Section
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TALAT Lecture 2301
Design of Members Torsion Example 8.2 : Torsion constants for hollow cross section 3 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm
Date of Issue: 1999 EAA - European Aluminium Association TALAT 2301 – Example 8.2
1
Example 8.2. Torsion constants for hollow cross section a) Polygon Half width
b
50 . mm
Thickness
s
6 . mm
Nodes no., co-ordinates, thickness
i
Half width of flat parts
b
a 1
0
b
a
s
1
a
b
s
2
a
b
s
3
b
a
s
y
4
z
b
a
t
a = 20.711 mm 2
s
5
a
b
s
6
a
b
s
7
b
a
s
8
b
a
s
50 0
Nodes
i
1 .. rows ( y )
1 z
Area of cross section elements Cross section area
dAi
ti .
mm
yi
2
yi 1
zi
zi 1
0 0
2
50
rows ( y )
1
50
0
dAi
A
mm
i =1 First moment of area. Gravity centre
Second moment of area
rows ( y )
dAi zi 1 . 2
zi
1 zi
2
zi 1
2
dAi zi . zi 1 . 3
Iy
mm
A . z gc
2
6
4
1 0.5 . yi
Av
yi 1 . zi
Check: ( 2 .b )
1
Dn i =1
yi
yi 1
2
zi
zi 1
A v = 8.284 . 10 mm 3
zi 1
2
2 .( b
2
2
ti
Dn = 55.228
Iv
3
4 .A v
2
Dn
I v = 4.971 . 10 mm Wv
2 . A v . min ( t )
TALAT 2301 – Example 8.2
min ( t ) = 6 mm
2
2
a ) = 8.284 . 10 mm
6
Torsion resistance
15
I y = 2.627 . 10 mm
i =1
Torsion constant
Iy
i =1
rows ( y )
A
z gc = 1.704 . 10
Iy
Sum of l/t
Sy
z gc
i =1
rows ( y ) Area within mid-line
1
Sy
rows ( y )
50
y
4
W v = 9.941 . 10 mm 4
3
2
TALAT 2301 – Example 8.2
3
b) Symmetric hollow extrusion Half width
b
50 . mm
Indent
c
10 . mm 20 . mm
Half width of a flat parts Thickness
Nodes no., co-ordinates, thickness
6 . mm
s
b
0 b
1
a
s
2
a
3
a
b
s
4
a
b
s
5
a
6
b
7 i
s
c
a
b
b c
b y
8 9
s
s
c a
s
a
s
z
b b
c
t
a a
c
s s
10
a
11
a
b
s
12
a
b
s
13
a b
14
b
s
c
b c
s
c a
s
15
b
a
s
16
b
a
s
50 0 z mm
0 0
50
Nodes
i
1 .. rows ( y )
50
1
0
50
y mm
rows ( y ) Area within mid-line
1 0.5 . yi
Av
yi 1 . zi
A v = 7.2 . 10 mm 3
zi 1
2
i =1 rows ( y ) Sum of l/t
1
Dn i =1
Torsion constant
Iv
Torsion resistance
Wv
4 .A v
yi
yi 1
2
zi
zi 1
2
ti
Dn = 58.856
2
TALAT 2301 – Example 8.2
4
W v = 8.64 . 10 mm
3
6
Dn 2 . A v . min ( t )
I v = 3.523 . 10 mm
min ( t ) = 6 mm
4
4
Design of Members Torsion Example 8.2 : Torsion constants for hollow cross section 3 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm
Date of Issue: 1999 EAA - European Aluminium Association TALAT 2301 – Example 8.2
1
Example 8.2. Torsion constants for hollow cross section a) Polygon Half width
b
50 . mm
Thickness
s
6 . mm
Nodes no., co-ordinates, thickness
i
Half width of flat parts
b
a 1
0
b
a
s
1
a
b
s
2
a
b
s
3
b
a
s
y
4
z
b
a
t
a = 20.711 mm 2
s
5
a
b
s
6
a
b
s
7
b
a
s
8
b
a
s
50 0
Nodes
i
1 .. rows ( y )
1 z
Area of cross section elements Cross section area
dAi
ti .
mm
yi
2
yi 1
zi
zi 1
0 0
2
50
rows ( y )
1
50
0
dAi
A
mm
i =1 First moment of area. Gravity centre
Second moment of area
rows ( y )
dAi zi 1 . 2
zi
1 zi
2
zi 1
2
dAi zi . zi 1 . 3
Iy
mm
A . z gc
2
6
4
1 0.5 . yi
Av
yi 1 . zi
Check: ( 2 .b )
1
Dn i =1
yi
yi 1
2
zi
zi 1
A v = 8.284 . 10 mm 3
zi 1
2
2 .( b
2
2
ti
Dn = 55.228
Iv
3
4 .A v
2
Dn
I v = 4.971 . 10 mm Wv
2 . A v . min ( t )
TALAT 2301 – Example 8.2
min ( t ) = 6 mm
2
2
a ) = 8.284 . 10 mm
6
Torsion resistance
15
I y = 2.627 . 10 mm
i =1
Torsion constant
Iy
i =1
rows ( y )
A
z gc = 1.704 . 10
Iy
Sum of l/t
Sy
z gc
i =1
rows ( y ) Area within mid-line
1
Sy
rows ( y )
50
y
4
W v = 9.941 . 10 mm 4
3
2
TALAT 2301 – Example 8.2
3
b) Symmetric hollow extrusion Half width
b
50 . mm
Indent
c
10 . mm 20 . mm
Half width of a flat parts Thickness
Nodes no., co-ordinates, thickness
6 . mm
s
b
0 b
1
a
s
2
a
3
a
b
s
4
a
b
s
5
a
6
b
7 i
s
c
a
b
b c
b y
8 9
s
s
c a
s
a
s
z
b b
c
t
a a
c
s s
10
a
11
a
b
s
12
a
b
s
13
a b
14
b
s
c
b c
s
c a
s
15
b
a
s
16
b
a
s
50 0 z mm
0 0
50
Nodes
i
1 .. rows ( y )
50
1
0
50
y mm
rows ( y ) Area within mid-line
1 0.5 . yi
Av
yi 1 . zi
A v = 7.2 . 10 mm 3
zi 1
2
i =1 rows ( y ) Sum of l/t
1
Dn i =1
Torsion constant
Iv
Torsion resistance
Wv
4 .A v
yi
yi 1
2
zi
zi 1
2
ti
Dn = 58.856
2
TALAT 2301 – Example 8.2
4
W v = 8.64 . 10 mm
3
6
Dn 2 . A v . min ( t )
I v = 3.523 . 10 mm
min ( t ) = 6 mm
4
4
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