This example provides calculations on bending moment of members based on Eurocode 9.

1. TALAT Lecture 2301
Design of Members
Bending Moment
Example 4.2 : Hollow cross section (polygon)
5 pages
Advanced Level
prepared by Torsten Höglund, Royal Institute of Technology, Stockholm
Date of Issue: 1999
 EAA - European Aluminium Association
TALAT 2301 – Example 4.2 1

2. Example 4.2 Hollow cross section (polygon)
b
Half width b 50 . mm Half width of a
flat parts
Thickness s 1.2 . mm 1 2
O 0 . mm a = 20.711 mm
Co-ordinates b a s
and thickness
a b s
a b s
Comment:
300 . MPa MPa 10 . Pa
b a s 6
Nodes 4 and 9 fo
are added to b O s γ M1 1.1 kN 1000 . newton
indicate shift in y b z a t s
neutral axis Nodes i 1 .. rows ( y ) 1
a b s
a b s 50
b a s 0
b O s
b a s z 0
0
mm
Gross cross section
Area of cross
ti .
section 2 2
dAi yi yi 1 zi zi 1 50
elements
50 0 50
Cross y
rows ( y ) 1
section mm
area A dAi
2
A = 397.6 mm
i =1
First rows ( y ) 1
dAi Sy
moment Sy zi zi 1 . z gc
of area. 2 A
Gravity centre i =1
z gc = 0 mm
z gc.gr z gc
Second rows ( y ) 1
dAi
zi . zi 1 . A . z gc
moment 2 2 2
Iy zi zi 1 Iy Iy
of area 3
i =1
I y = 5.255 . 10 mm
5 4
rows ( y ) 1
Area within
0.5 . yi yi 1 . zi A v = 8.284 . 10 mm
3 2
mid-line Av zi 1
Check:
i =1
( 2 .b ) 2 .( b a ) = 8.284 . 10 mm
2 2 3 2
rows ( y ) 1 2 2
yi yi 1 zi zi 1
Sum of l/t Dn Dn = 276.142
ti
i =1
Torsion
4 .A v
2
I v = 9.941 . 10 mm
constant 5 4
Iv
Dn
2 . A v . min ( t ) W v = 1.988 . 10 mm
4 3
Torsion Wv min ( t ) = 1.2 mm
resistance
TALAT 2301 – Example 4.2 2

3. Effective cross section. First iteration Zero position zd 0 ρ c 0
zi 1 z gc zi z gc zd zi z gc
5.4.3 (1) ψ i
if zi 1 z gc < zi z gc , , i
zi z gc zi 1 z gc max z max z d
0.8
(5.7 or 5.8) gi if ψ i > 1 , 0.7 0.30 . ψ i , max z = 50 mm
1 ψ i
2 2
yi yi 1 zi zi 1 250 . MPa
5.4.3 (1) c) βi if t i > 0 , gi . ,1 ε o ε o 0.913
=
ti fo
max z max z
5.4.4 (5) εi if zi > z gc zi 1 > z gc , if zi 1 z gc < zi z gc , ε o . , ε o. , 99
zi z gc zi 1 z gc
2
5.4.5 (3) c) β ε ε Comment: εi is given a large value (99)
> 22 , 32 . 220 .
i i i
heat-treated, ρ c if , 1.0
i ε i
β i
β i
for elements entirely on the tension side
unwelded
Effective βi t eff
t eff ρ .c
t = i
thickness ε ρ c= =
ψi = gi = βi = εi = i i mm
0.414 0.824 28.452 0.913 31.167 0.8 0.96
1 1 34.518 0.913 37.812 0.692 0.831
0.414 0.824 28.452 0.913 31.167 0.8 0.96
0 0.7 12.081 1.418 8.518 1 1.2
Area of effective 0 0.7 12.081 99 0.122 1 1.2
thickness
t eff .
2 2 0.414 0.824 28.452 99 0.287 1 1.2
elements dAi yi yi 1 zi zi 1
i 1 1 34.518 99 0.349 1 1.2
0.414 0.824 28.452 99 0.287 1 1.2
Cross rows ( y ) 1
0 0.7 12.081 99 0.122 1 1.2
section A eff dAi 0 0.7 12.081 1.418 8.518 1 1.2
area
i =1
2 2
A eff = 362.498 mm A = 397.645 mm
First rows ( y ) 1
moment dAi Sy
of area. Sy zi zi 1 . z gc
2 A eff
i =1
Gravity centre
z gc = 4.046 mm
50
The nodes in the vertical webs are moved to
the neutral axis from first iteration 0
z
mm
i 0 .. rows ( y ) 2 0
0
zi if zi > z gc . zi 1 < z gc , z gc , zi z GC1
mm
i 1 .. rows ( y ) 1
zi if zi > z gc . zi 1 < z gc , z gc , zi
50
50 0 50
Check z4 = 4.046 mm z9 = 4.046 mm OK !
y y GC1
,
mm mm
TALAT 2301 – Example 4.2 3

4. Second iteration
zi 1 z gc zi z gc z gc = 4.046 mm
5.4.3 (1) ψi if zi 1 z gc < zi z gc , ,
zi z gc zi 1 z gc
zd zi z gc
0.8 i
(5.7 or 5.8) gi if ψ i > 1 , 0.7 0.30 . ψ i ,
1 ψ i
max z max z d
2 2 max z = 54.046 mm
yi yi 1 zi zi 1
5.4.3 (1) c) βi if t i > 0 , gi . ,1
ti
max z max z
5.4.4 (5) εi if zi > z gc zi 1 > z gc , if zi 1 z gc < zi z gc , ε o . , ε o. , 99
zi z gc zi 1 z gc
2
5.4.5 (3) c) β ε ε
> 22 , 32 . 220 .
i i i
heat-treated, ρ c if , 1.0
unwelded i ε i
β i
β i
Effective
t eff ρ .c
t βi t eff
thickness = i
ε ρ c= =
β ψi = gi = βi = εi = i i mm
Max slender- i
ness β e β max max β e 0.458 0.837 28.906 0.913 31.665 0.791 0.949
i ε i
- Internal 1 1 34.518 0.913 37.812 0.692 0.831
elements β max = 37.812
0.458 0.837 28.906 0.913 31.665 0.791 0.949
Cross section 0 0.7 14.441 1.349 10.707 1 1.2
class class if β max 22 , 3 , 4 class = 4
0 0.7 9.721 99 0.098 1 1.2
Area of 0.363 0.809 27.918 99 0.282 1 1.2
effective 1 1 34.518 99 0.349 1 1.2
t eff .
thickness 2 2
dAi yi yi 1 zi zi 1 0.363 0.809 27.918 99 0.282 1 1.2
elements i
0 0.7 9.721 99 0.098 1 1.2
Cross rows ( y ) 1 0 0.7 14.441 1.349 10.707 1 1.2
section A eff dAi
area
i =1 2
A eff = 361.596 mm
First rows ( y ) 1
dAi
moment Sy zi zi 1 . Sy
of area. 2 z gc
i =1 A eff
Gravity centre z gc = 4.144 mm
50
0
Gross cross z
section I y.gr Iy mm
I y.gr = 5.255 . 10 mm
5 4 z GC1 0
0
mm
z GC2
mm
50
50 0 50
y y GC1
,
mm mm
TALAT 2301 – Example 4.2 4

5. A . z gc
2
Second rows ( y ) 1 I eff Iy
moment of dAi
zi . zi 1 .
2 2
area of effective I y zi zi 1
I eff = 5.187 . 10 mm
3 5 4
cross section i =1
I eff
W eff = 9.579 . 10 mm
3 3
Section zd zi z gc max z.eff max z d W eff
modulus i max z.eff
Compare I y.gr
W gr = 1.051 . 10 mm
4 3
gross zd zi z gc.gr max z.gr max z d W gr
i max z.gr
section
Moment fo
resistance M Rd W eff . M Rd = 2.613 kN . m
γ M1
Axial force
Move node close zi if zi z GC1 < 1 . mm , zi 1 , zi
to GC1 to
nearest node
2 2
yi yi 1 zi zi 1
5.4.3 (1) c) βi if t i > 0 , ,1
ti
2
5.4.5 (3) c) β ε o ε o
> 22 , 32 . 220 .
i
heat-treated, ρ c if , 1.0 ε o 0.913
=
i ε o β i
β i
unwelded
βi t eff
Effective =
ρ .c
i
t eff t ε o ρ c= =
thickness i = βi = i mm
1 34.518 37.812 0.692 0.831
2 34.518 37.812 0.692 0.831
Area of effective 3 34.518 37.812 0.692 0.831
thickness 4 0 0 1 1.2
t eff .
elements 2 2
dAi yi yi 1 zi zi 1 5 34.518 37.812 0.692 0.831
i
6 34.518 37.812 0.692 0.831
Cross rows ( y ) 1 7 34.518 37.812 0.692 0.831
2
section A eff dAi A eff = 275.335 mm 8 34.518 37.812 0.692 0.831
area 9 0 0 1 1.2
i =1
10 34.518 37.812 0.692 0.831
First rows ( y ) 1
dAi Sy
moment Sy zi zi 1 . z gc
of area. 2 A eff
Gravity centre i =1
z gc = 0 mm 50
0
z
mm
0
0
fo z GC
Axial force N Rd A eff .
resistance γ M1 mm
N Rd = 75.091 kN
50
50 0 50
y y GC
,
mm mm
TALAT 2301 – Example 4.2 5