This example provides calculations on shear force of members based on Eurocode 9.

1. TALAT Lecture 2301
Design of Members
Shear Force
Example 6.1 – 6.6 : Shear resistance of webs without
and with stiffeners
11 pages
Advanced Level
prepared by Torsten Höglund, Royal Institute of Technology, Stockholm
Date of Issue: 1999
 EAA - European Aluminium Association
TALAT 2301 – Examples 6.1 – 6.6 1

2. Example 6.1 - 6.6 Shear resistance of webs without and
with stiffeners
Dimensions and strength of material Data input in highlighted regions
Web depth hw 2000 . mm MPa 1000000 . Pa
Web thickness tw 15 . mm kN 1000 . newton
Total length of web panel lw 4000 . mm kg
spv 2.4 . (Mass)
355 . MPa ( 0.1 . m )
3
0,2 proof strength of web plate material f ow
Ultimate strength of web plate material f uw 470 . MPa
Elastic modulus E 70000 . MPa
Partial coefficient γ M1 1.1
f uw
Table 5.12 η 0.4 0.2 . η = 0.665
f ow
1) No intermediate stiffener
If "rigid end post" thenendpost = 1 else endpost = 0 endpost 1
a
Length of web panel a lw a = 4000 mm =2
hw
Ref. to Eurocode 9
2 2
a hw hw
(5.97) kτ if > 1.00 , 5.34 4.00 . , 4.00 5.34 .
hw a a k τ = 6.34
0.81 . h w . f ow
(5.96) λ w
tw E λ w 3.055
=
kτ
1.32 0.48
Table 5.12 ρ v if λ w > 0.949 , ,
1.66 λ w λ w ρ v= 0.28
Table 5.12 ρ v if ρ v > η , η , ρ v
ρ v= 0.28
0.48 0.48
Table 5.12 ρ v if endpost 0 , if ρ v > , ,ρ v ,ρ v
λ w λ w ρ v= 0.28
f ow
ρ .t w .h w . V w.Rd = 2.711 . 10 kN
3
(5.95) V w.Rd v
γ M1
TALAT 2301 – Mass1 h w6.1 .–w . spv
Weight of web Examples . t w l 6.6 2 Mass1 = 288 kg V Rd V w.Rd
1

3. 2) Equally spaced flexible transverse stiffeners
If "rigid end post" thenendpost = 1 else endpost = 0 endpost 1
Number of stiffeners n st 3
Single plate stiffener t st 15 . mm
b st
b st 120 . mm =8
t st
Whole web panel. Buckling of stiffener
t st . b st 30 . t w A st = 8.55 . 10 mm
2 3 2
Figure 5.21 A st
t st . b st
2
C.G. e st e st = 12.632 mm
2 . A st
t st . b st
Second 3
A st . e st I st = 7.276 . 10 mm
moment 2 6 4
I st
of area 3
a = 4 . 10 mm
3
Panel length a lw
3
2 4
hw I st
(5.99) k τ st 9. . =
k τ st 2.38
t w .h w
a 3
1
3
2.1 . I st
(5.99) k τ stmin =
k τ stmin 2.153
tw hw
k τ st if k τ st < k τ stmin , k τ stmin , k τ st =
k τ st 2.38
(5.97) 2 2
a hw hw
(5.98) kτ if > 1.00 , 5.34 4.00 . , 4.00 5.34 . k τ = 6.34
hw a a
kτ kτ k τ st kτ = 8.72
0.81 . h w . f ow
(5.96) λ w λ w 2.605
=
tw E
kτ
λ wtot λ w
TALAT 2301 – Examples 6.1 – 6.6 3

4. Sub panel
lw a
a = 1 . 10 mm
3
Length a = 0.5
n st 1 hw
2 2
a hw hw
(5.97 and 5.98) k τ if > 1.00 , 5.34 4.00 . , 4.00 5.34 . k τ = 25.36
hw a a
0.81 . h w . f ow
(5.96) λ w λ wsub λ w λ wsub = 1.527
tw E
kτ
λ wtot 2.605
=
λ λ
The larger of the slenderness parameter wtot for the total panel and wsub for the sub panels is used. If
λwtot > λwsub then the stiffeners are flexible else the stiffeners are rigid.
λ w if λ wtot > λ wsub , λ wtot , λ wsub λ w 2.605
=
1.32 0.48
Table 5.12 ρ v if λ w > 0.949 , , ρ v if ρ v > η , η , ρ ρ v= 0.31
v
1.66 λ w λ w
0.48 0.48
Table 5.12 ρ v if endpost 0 , if ρ v > , ,ρ v ,ρ v ρ v= 0.31
λ w λ w
f ow
ρ .t w .h w . V wRd = 2.997 . 10 kN
3
(5.95) V wRd v
γ M1
Alternative 2 = transverse stiffeners V Rd V wRd
2
Weight Mass2 h w .t w .l w n st . b st . t st . h w . spv Mass2 = 314 kg
TALAT 2301 – Examples 6.1 – 6.6 4

5. 3) Transverse intermediate, rigid stiffeners
Number of transverse stiffeners n st 3 endpost = 1
Web panel lw a
a = 1 . 10 mm
3
length a = 0.5
n st 1 hw
2 2
a hw hw
(5.97) kτ if > 1.00 , 5.34 4.00 . , 4.00 5.34 . k τ = 25.36
hw a a
0.81 . h w . f ow
(5.96) λ w λ w 1.527
=
tw E
kτ
1.32 0.48
Table 5.12 ρ v if λ w > 0.949 , , ρ v= 0.414
1.66 λ w λ w
Table 5.12 ρ v if ρ v > η , η , ρ v ρ v= 0.414
f ow
ρ .t w .h w . V wRd = 4.01 . 10 kN
3
(5.95) V wRd v
γ M1
t st 18 . mm b st 220 . mm
Check rigidity of stiffener
t st . b st 30 . t w A st = 1.071 . 10 mm
2 4 2
Figure 5.21 A st
t st . b st
2
C. G. e st e st = 40.672 mm
2 . A st
t st . b st
Second 3
A st . e st I st = 4.617 . 10 mm
moment 2 7 4
I st
of area 3
h .t
3 3
(5.104) or a . w w , 0.75 . h . t 3 I limit = 4.05 . 10 mm
7 4
(5.105) I limit if < 2 , 1.5 w w
hw 2
a
Ist > Ilimit OK!
Axial force 1
1.4 . t w
2.
in stiffener N st V wRd E . f ow .
γ M1
TALAT
2301.06 N st f ow
(6.03h) σ st σ st = 241 MPa = 323 MPa σst < fow OK!
A st γ M1
Alternative 3 = transverse intermediate, rigid stiffeners V Rd V wRd
3
TALAT 2301 – Examples 6.1 –. 6.6
. 5
Weight Mass3 hw tw lw n st . b st . t st . h w . spv Mass3 = 345 kg

6. 4) Rigid transverse stiffeners and longitudinal stiffeners
Number of longitudinal stiffeners n sl 2
b sl
Longitudinal plate stiffener t sl 15 . mm b sl 120 . mm =8
t sl
Number of rigid transverse stiffeners n st = 3
If "rigid end post" thenendpost = 1 else endpost = 0 endpost 1
Buckling of longitudinal stiffener
t sl. b sl 40 . t w A sl = 1.08 . 10 mm
2 4 2
Area A sl
t sl. b sl
2
GC e sl e sl = 10 mm
2 . A sl
t sl. b sl
Second 3
A sl. e sl I sl = 7.56 . 10 mm
moment 2 6 4
I sl
of area 3
lw
Panel length a
a = 1 . 10 mm
3
n st 1
3
n sl. I sl
2 4
hw
(5.99) k τ st 9 . . =
k τ st 65.916
t w .h w
a 3
1
. 3
2.1 . n sl I sl
(5.99) k τ stmin =
k τ stmin 2.748
tw hw
(5.99) k τ st if k τ st < k τ stmin , k τ stmin , k τ st =
k τ st 65.916
2
hw
(5.97) kτ 5.34 4.00 . k τ st kτ = 87.256
a
0.81 . h w . f ow
(5.96) λ w λ w 0.823
=
tw E
kτ
Whole panel λ wtot λ w
λ wtot = 0.823
TALAT 2301 – Examples 6.1 – 6.6 6

7. Sub panel
hw
Depth h1 h 1 = 667 mm
3
2
h1
(5.98) kτ 4.00 5.34 . k τ = 6.373
a
0.81 . h 1 . f ow λ w 0.823
=
(5.96) λ w
tw E
kτ
Sub-panel λ wsub λ w λ wsub = 1.016
compare λ wtot 0.823
=
λ λ
The larger of the slenderness parameter wtot for the total panel and wsub for the sub panels is used. If
λwtot > λwsub then the stiffeners are flexible else the stiffeners are rigid.
λ w if λ wtot > λ wsub , λ wtot , λ wsub λ w 1.016
=
1.32 0.48
Table 5.12 ρ v if λ w > 0.949 , , ρ v= 0.493
1.66 λ w λ w
Table 5.12 ρ v if ρ v > η , η , ρ v
0.48 0.48
Table 5.12 ρ v if endpost 0 , if ρ v > , ,ρ v ,ρ v ρ v= 0.493
λ w λ w
f ow
ρ .t w .h w . V wRd = 4.78 . 10 kN
3
(5.95) V wRd v
γ M1
Alternative 4 = transverse and longitudinal intermediate stiffeners V Rd V wRd
4
Weight Mass4 h w .t w .l w n st . b st . t st . h w n sl. t sl. b sl. l w . spv Mass4 = 380 kg
TALAT 2301 – Examples 6.1 – 6.6 7

8. 5) Shear resistance contribution of the flanges added to girder 4
The panel is at the end of the plate girder. Then the bending moment is neglected
Flange thickness and width tf 50 . mm bf 750 . mm
Yield strength of flange plate f of 355 . MPa
Design shear force V Ed 6000 . kN
V Ed . l w M Ed = 2.4 . 10 kN . m
4
Design bending moment M Ed
f of
(5.101) M fRd b f .t f . h w tf . M Ed
M fRd = 2.481 . 10 kN . m
4
γ M1 = 0.967
M fRd
a = 1 . 10 mm
3
4.4 . b f . t f f of
2.
(5.101) c 0.08 .a
c = 217.5 mm
t w . h w . f ow
2
b f . t f . f of
2 2
M Ed
(5.101) V fRd if M Ed < M fRd , . 1 ,0 V fRd = 178.626 kN
c . γ M1 M fRd
Sum V Rd V wRd V fRd
5
V Rd = 4.955 . 10 kN
3
Example 5 = inclusive shear resistance contribution of the flanges
5
Weight Mass5 h w .t w .l w n st . b st . t st . h w n sl. t sl. b sl. l w . spv Mass5 = 380 kg
TALAT 2301 – Examples 6.1 – 6.6 8

9. 6) Corrugated web
Web depth hw 2000 . mm MPa 1000000 . Pa
Web thickness tw 12 . mm kN 1000 . newton
Total length of web panel lw 4000 . mm
0,2 proof strength of web plate material f ow 355 . MPa
Ultimate strength of web plate material f uw 470 . MPa
Elastic modulus E 70000 . MPa
Trapezoidal web: bo 140 . mm bu 140 . mm
bd 400 . mm hc 100 . mm
b u . 0.5
2 2
sw bd bo hc s w = 116.619 mm
Partial coefficient γ M1 1.1
f uw
Table 5.12 η 0.4 0.2 . η = 0.665
f ow
Local buckling of bo , bu and bu
Max width bm if b o < b u , b u , b o
b m = 140 mm
bm if b m > s w , b m , s w
b m = 140 mm
b m f ow
(5.96) λ w 0.35 . .
tw E λ w 0.291
=
Table 5.12
0.48 0.48
Non-rigid ρ v if λ w < ,η , η = 0.665 ρ v= 0.665
end post η λ w
f ow
0.7 . ρ . t w . h w . V wRd = 3.604 . 10 kN
3
(5.117) V wRd v
γ M1
TALAT 2301 – Examples 6.1 – 6.6 9

10. Trapezoidal web, widthbd
b o .t w b u .t w 2 .s w .t w A = 6.159 . 10 mm
3 2
Area A
Gravity b o .t w .h c 2 . s w . t w . 0.5 . h c
centre e gc e gc = 50 mm
A
Second 2
hc
2 1
b o .t w .h c 2 .s w .t w . A . e gc . I x = 2.683 . 10 mm
moment 2 4 3
Ix
of area 3 bd
3
bd t
(5.122) Iz . w I z = 123.554 mm
3
bu bo 2 . s w 10.9
1
60 . E . 3 4
I z.I x V o.cr = 1.468 . 10 kN
4
(5.121) V o.cr
hw
h w . t w . f ow
(5.120) λ ow λ ow 0.762
=
V o.cr 0.7 . ρ v= 0.465
0.60
χ o
(5.119) 2 χ o if χ o > 0.7 . ρ v , 0.7 . ρ v , χ o χ o 0.435
=
0.8 λ ow
f ow
χ .o w . t w . V o.Rd = 3.366 . 10 kN
3
(5.118) V o.Rd h
γ M1
V Rd = 3.366 . 10 kN
3
Min V Rd if V wRd > V o.Rd , V o.Rd , V wRd
6 6
bu bo 2 .s w
Weight Mass6 t w .l w .h w . . spv Mass6 = 296 kg
bd
TALAT 2301 – Examples 6.1 – 6.6 10

11. Weight /
Summary resistance
Increase in
V Rd resistance Massi Massi 1
i = . =
= Mass1 Mass1 f i
i = 1000 . kN f i =
1) No intermediate stiffeners t w1_5 = 15 mm 1 2.711 1 1 1
2) Transverse flexible stiffeners 2 2.997 1.106 1.09 0.986
3) Transverse rigid stiffeners 3 4.01 1.479 1.198 0.81
4) Transverse rigid + longitudinal stiffeners 4 4.777 1.762 1.318 0.748
5) Transverse rigid + longitudinal stiffeners 5 4.955 1.828 1.318 0.721
+ contribution of flanges 6 3.366 1.242 1.026 0.827
6) Trapezoidal web t w = 12 mm
f ow
"No buckling" V 0.Rd η . h w . t w1_5 . V 0.Rd = 6.436 1000 . kN
γ M1
Comments
The second column gives the increase in shear resistance compared to girder 1 when adding stiffeners.
By adding both transversal and horizontal stiffeners (5) the resistance can be almost doubled.
The last column, "weight per resistance" show that stiffeners give lighter girders, but the comparison does no
pay regard to the cost for welding of the stiffeners.
The contribution of the flanges (girder 5) increase the resistance with = 3.7 %
a
compared to girder 4
Torsten Höglund
TALAT 2301 – Examples 6.1 – 6.6 11