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

1. TALAT Lecture 2301
Design of Members
Bending Moment
Example 4.4 : Bending moment resistance of welded
hollow section with outstands. Class 4 cross section
10 pages
Advanced Level
prepared by Torsten Höglund, Royal Institute of Technology, Stockholm
Date of Issue: 1999
 EAA - European Aluminium Association
TALAT 2301 – Example 4.4 1

2. Example 4.4. Bending moment resistance of welded hollow
section with outstands. Class 4 cross section
Half top flange a 200 . mm Half hollow width c 96 . mm fo 200 . MPa
Half bottom flange b 96 . mm Width of HAZ haz 30 . mm γ M1 1.1
99 . mm MPa 10 . Pa
6
Section dept h HAZ reduction ρ haz 0.8
Flange thickness d 6 . mm
haz .( c
bz c b)
Web thickness w 8 . mm
2 2
(c b) h
Bottom flange t. t 8 . mm
Type of element haz .h
hz
- Internal element: T = 0
2 2
- Outstand: T > 0 (c b) h
o 0 . mm
Nodes, co-ordinates (y, z), thickness (t) and types (T)
0 a o o 0
Comments:
1 c haz o d 1 Node 10 and 13
indicate shift in neutral axis.
2 c o d 1
Element 1-2, 2-3, 4-5, 5-6, 8-9
3 c haz o d 0 and 14-15 define HAZs.
4 c haz o d 0
5 c o d 0
This solution of the example is
6 c haz o d 1 comprehensive because:
1. Calculations are shown in every
7 a o d 1 detail
i y z t T
8 c o o 0 2. It covers all classes of cross
section
9 bz hz w 0 3. The possibility to increase the
10 0.5 .( c b ) 0.5 . h w 0 effective thickness for elements not
fully stressed is shown. The iteration
11 b h w 0 procedure is then mandatory.
12 b h t 0 However, in this example, the result
is almost exactly the same as if a
13 0.5 . ( c b) 0.5 . h w 0 simplified method was used, that's
because there is no reduction of the
14 bz hz w 0
web thickness due to local buckling.
15 c o w 0
The beam is composed by two extrusions. The weld is
close to the neutral axis why HAZ softening does not 0
influence the moment resistance.
Nodes i 1 .. rows ( y ) 1 rows ( y ) = 16
Area of cross
ti .
2 2 100
section dAi yi yi 1 zi zi 1 200 100 0 100 200
elements
TALAT 2301 – Example 4.4 2

3. Cross rows ( y ) 1
section A dAi
area
A = 5.52 . 10 mm
3 2
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 = 41.752 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 = 1.061 . 10 mm
7 4
I y.gr Iy
Elastic section Iy Iy
modulus W el if max z z gc > min z z gc , ,
max z z gc min z z gc
W el = 1.853 . 10 mm
5 3
Torsion rows ( y ) 1 2
ti
dAi . I tu = 9.536 . 10 mm
constant for 4 4
I tu
open parts 3
i =1
Hollow part, (nodes 2, 5, 11, 12 and 15)
y2 z2 t1
96 0 6
y5 z5 t4
96 0 6
yh y11 y h = 96 mm zh z11 zh = 99 mm th t 11 th= 8 mm
y12 96 z12 99 t 12 8
96 0 8
y15 z15 t 14
rows y h 1
Area within
0.5 . y h . z A tu = 1.901 . 10 mm
4 2
mid-line A tu yh hi zh
i i 1 i 1
i =1
Sum of l/t rows y h 1 2 2
yh yh zh zh
i i 1 i i 1 22
Dn if t i > 0 , , 10 Dn = 80.75
th
i =1 i
4 .A
2
Torsion constant tu
I t = 1.79 . 10 mm
7 4
for closed part It
Dn
Torsion constant
I t = 1.799 . 10 mm
7 4
whole section It It I tu
Torsion
2 . A tu . min t h W v = 2.281 . 10 mm
5 3
resistance Wv min t h = 6 mm
z gc = 41.752 mm
TALAT 2301 – Example 4.4 3

4. Effective cross section. First iteration
z8 z gc zd zi z gc
5.4.3 (1) ψi 1 j 9 .. 11 ψj ψ j 4
ψ j
i
z11 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 , Outstand gi if Ti > 0 , 1 , gi
1 ψ i Figure 5.2
y2 y0 0.5 . t 13
5.4.3 (1) c) k 1 .. 2 βk βk 5
β k
y2 = 96 mm
t2 y0 = 200 mm
y2 y0
y5 y2 t 13 = 17.333
l 3 .. 5 βl t2
t4
2 2
y11 y8 z11 z8
β j
if t 11> 0 , gj. ,1 βj 4
β j
t 11
250 . MPa max z ε
5.4.4 (5) ε o εi ε o. εi if zi z gc . zi 1 z gc , ε i , 99 ε o 1.118
=
fo z gc
5.4.5 (3) 2 2
a) or c) β ε ε β ε ε
> 4 , 8. 16 . > 15 , 25 . 150 .
i i i i i i
non heat ρ c if Ti > 0 , if , 1.0 , if , 1.0
i ε β β ε β β
treated, welded i i i i i i
Effective
thickness t eff ρ .c
t
Heat Affected Zone
Flange t haz.f .t
ρ haz 2 t haz.f = 4.8 mm
Web .t
t haz.w ρ haz 9 t haz.w = 6.4 mm
n 2 .. 3 t eff if t eff > t haz.f , t haz.f , t eff t eff t eff
n n n 8 n n
t eff if t eff > t haz.w , t haz.w , t eff t eff t eff
9 9 9 15 9
βi t eff
= i
ε ρ c= =
ψi = gi = βi = εi = i i mm
1 1 16.667 1.309 12.731 0.53 3.178
1 1 16.667 1.309 12.731 0.53 3.178
Comment: 1 1 30.667 1.309 23.425 0.794 4.763
εi was given a large value (99) 1 1 30.667 1.309 23.425 0.794 4.763
for elements entirely on 1 1 30.667 1.309 23.425 0.794 4.763
the tension side 1 1 16.667 1.309 12.731 0.53 3.178
1 1 16.667 1.309 12.731 0.53 3.178
1 1 0 1.309 0 1 0
-0.729 0.481 5.955 1.309 4.549 1 6.4
-0.729 0.481 5.955 99 0.06 1 8
-0.729 0.481 5.955 99 0.06 1 8
Area of effective 1 1 0 99 0 1 8
thickness -0.729 0.481 5.955 99 0.06 1 8
t eff .
elements 2 2
dAi yi yi 1 zi zi 1 -0.729 0.481 5.955 99 0.06 1 8
i
-0.729 0.481 5.955 1.309 4.549 1 6.4
TALAT 2301 – Example 4.4)
rows ( y 1 4

5. Cross rows ( y ) 1
section A eff dAi
A = 5.52 . 10 mm
3 2
area
i =1
A eff = 4.6 . 10 mm
3 2
First moment rows ( y ) 1
dAi
of area Sy zi zi 1 . Sy
2 z gc z gc = 49.794 mm
i =1 A eff
Gravity centre
Effective cross section. Second iteration
Node 10 and 13 is moved to the neutral z10 z gc 0.01 . mm z13 z gc 0.01 . mm
axis from the first iteration
z10
y10 y5 . y y11 y13 y10
5
z11
z8 z gc zd zi z gc
5.4.3 (1) ψi 1 j 9 .. 11 ψj ψ j 4
ψ j
i
z11 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 , Outstand gi if Ti > 0 , 1 , gi
1 ψ i Figure 5.2
y2 y0 0.5 . t 13 y5 y2 t 13
5.4.3 (1) c) k 1 .. 2 βk βk 5
β k
l 3 .. 5 βl
t2 t4
2 2
y11 y8 z11 z8
β j
if t 11> 0 , gj. ,1 βj 4
β j
t 11
250 . MPa max z
5.4.4 (5) ε o εi ε o. εi if zi z gc . zi 1 z gc , ε i , 99 ε o 1.118
=
fo z gc
5.4.5 (3) 2 2
a) or c) β ε ε β ε ε
> 4 , 8. 16 . > 15 , 25 . 150 .
i i i i i i
non heat ρ c if Ti > 0 , if , 1.0 , if , 1.0
i ε β β ε β β
treated, welded i i i i i i
Effective
thickness t eff ρ .c
t
TALAT 2301 – Example 4.4 5

6. Heat Affected Zone
t eff if t eff > t haz.f , t haz.f , t eff t eff t eff
n n n 8 n n
t eff if t eff > t haz.w , t haz.w , t eff t eff t eff
9 9 9 15 9
βi t eff
= i
ε ρ c= =
ψ
i = Ti = i = gi = βi = εi = i i mm
1 1 1 1 16.667 1.118 14.907 0.465 2.788
2 1 1 1 16.667 1.118 14.907 0.465 2.788
3 0 1 1 30.667 1.118 27.429 0.712 4.272
4 0 1 1 30.667 1.118 27.429 0.712 4.272
5 0 1 1 30.667 1.118 27.429 0.712 4.272
6 1 1 1 16.667 1.118 14.907 0.465 2.788
7 1 1 1 16.667 1.118 14.907 0.465 2.788
8 0 1 1 0 1.118 0 1 0
9 0 -1.012 0.398 4.921 1.118 4.401 1 6.4
10 0 -1.012 0.398 4.921 1.118 4.401 1 8
11 0 -1.012 0.398 4.921 99 0.05 1 8
β i
Max slender- β e if Ti > 0 , 0 , 12 0 1 1 0 99 0 1 8
ness i ε i 13 0 -1.012 0.398 4.921 99 0.05 1 8
- Internal 14 0 -1.012 0.398 4.921 1.118 4.401 1 8
elements β Imax max β e
15 0 -1.012 0.398 4.921 1.118 4.401 1 6.4
β Imax = 27.429
β i
Max slender- β e if Ti > 0 , ,0
ness i ε i
- Outstands
β Omax max β e
β Omax = 14.907
Cross section class I if β Imax 7 , 1 , if β Imax 11 , 2 , if β Imax 15 , 3 , 4 class I = 4
class
class O if β Omax 2 , 1 , if β Omax 3 , 2 , if β Omax 4 , 3 , 4 class O = 4
(5.15)
class if class I > class O , class I , class O class = 4
Area of effective
t eff .
thickness 2 2
dAi yi yi 1 zi zi 1
elements i
rows ( y ) 1
Area of
A eff dAi
A = 5.52 . 10 mm
effective 3 2
cross section i =1
A eff = 4.424 . 10 mm
3 2
First rows ( y ) 1
moment dAi Sy
of area. Sy zi zi 1 . z gc
2 A eff
Gravity centre i =1
z gc = 51.768 mm
TALAT 2301 – Example 4.4 6

7. Second rows ( y ) 1
moment of dAi
zi . zi 1 . A eff . z gc
2 2 2
area of effective Iy zi zi 1 I eff Iy
3
cross section i =1
I eff = 8.344 . 10 mm
6 4
Section I eff
W eff = 1.612 . 10 mm
5 3
modulus zd zi z gc max z.eff max z d W eff
i max z.eff
Compare I y.gr
W gr = 1.853 . 10 mm
5 3
gross zd zi z gc.gr max z.gr max z d W gr
i max z.gr
section
Heat Affected Zone
Reduced thickness in HAZ h 2 .. 3 th .t
ρ haz h th 3 th t9 .t
ρ haz h t 15 .t
ρ haz h
ti .
2 2
Area dAi yi yi 1 zi zi 1
rows ( y ) 1
Area of
A ele dAi
A = 5.52 . 10 mm
effective 3 2
cross section i =1
A ele = 5.126 . 10 mm
3 2
First rows ( y ) 1
moment dAi Sy
of area. Sy zi zi 1 . z gc.e
2 A ele
Gravity centre i =1
z gc.e = 44.228 mm
Second rows ( y ) 1
moment of dAi
zi . zi 1 . A ele. z gc.e
2 2 2
area of effective Iy zi zi 1 I ele Iy
3
cross section i =1
I ele = 1.013 . 10 mm
7 4
I ele
W ele = 1.849 . 10 mm
5 3
Section zd zi z gc.e max z.ele max z d W ele
modulus i max z.ele
W el = 1.853 . 10 mm
5 3
TALAT 2301 – Example 4.4 7

8. Plastic section modulus
A = 5.52 . 10 mm
3 2
rows ( y ) 1
Area of
ti .
2 2
A pl yi yi 1 zi zi 1
A pl = 5.126 . 10 mm
effective 3 2
cross section i =1
Cross section 7
ti . 2 . t9.
area of upper 2 2 2 2
A up yi yi 1 zi zi 1 y9 y8 z9 z8
part
i =1
A up = 2.486 . 10 mm
3 2
A z9 = 30 mm
Move node 8 A up
2
to plastic z10 z9 z13 z10 z10 = 47.1 mm
neutral axis t 11 t 13
z10
y10 y5 . y y11 y13 y10 y10 = 96 mm
5
z11
rows ( y ) 1
Plastic section zi zi 1
W ple . t. yi yi 1
2
zi zi 1
2
W ple = 2.267 . 10 mm
5 3
modulus 2 i
i =1 W ple
= 1.224
W el
Shape factor
TALAT 2301 – Example 4.4 8

9. W el
Shape factor
W ele 15 β Imax W ple W ele 4 β Omax W ple
(5.15) α 3I . 1 . 1 α 3O . 1 . 1
W el 15 11 W ele W el 4 3 W ele
α 3I 0.296
= α 3O 1.465
=
α 3 if α 3I < α 3O , α 3I , α 3O α 3 1.465
=
W ple W eff
α if class< 3 , , if class< 4 , α 3 , class = 4 α = 0.87
W el W el
0
50
100
200 150 100 50 0 50 100 150 200
Elastic neutral axis:
- first iteration dashed-dotted line
- second iteration dashed line
Plastic neutral axis dotted line
TALAT 2301 – Example 4.4 9

10. Bending moment resistance
fo W el = 1.853 . 10 mm
5 3
(5.14) M y.Rd α . W el.
γ M1
W ple = 2.267 . 10 mm
5 3
M y.Rd = 29.3 kN . m
Cross section class = 4 α = 0.87
class
A eff = 4.424 . 10 mm
3 2
I eff = 8.344 . 10 mm
6 4
0
50
100
200 150 100 50 0 50 100 150 200
TALAT 2301 – Example 4.4 10