�ݺ�ߣ

Design of welded plate girder
Design example: A simply supported welded plate
girder of an effective span of 24 m subjected to
uniformly distributed load of 35 kN/m throughout the
span excluding the self weight of plate girder. Assume
compression flange laterally supported throughout the
span and yield stress of steel is 250 MPa. Design cross
section of plate girder, stiffeners and connections. Draw
sectional plan and elevation.
Fig. 1. Loading on plate girder
35 kN/m
24 m

Maximum bending moment and shear force
Self weight of plate girder
Factored load = 35 x 1.5 = 52.5 kN/m
Self weight of plate girder, w1 = W/200 = 6.3 kN/m.
Total uniformly distributed load on plate girder,
wT = 58.8 kN/m.
Maximum bending moment
M = 58.8 x 242
/8 = 4233.6 kNm.
Maximum shear force
F = 58.8 x 24/2 = 705.6 kN.

Design of web plate
If stiffener spacing c is in between d and 3d, where d is
depth of web, then serviceability requirement is,
k = d/tw ≤ 200, considering, k = 190
Economical depth,
d = = 1473.29 mm = 1500 mm
tw = 1500/190 = 7.89 mm
Provide 1500 mm deep and 8 mm thick web plate.
Provide intermediate stiffeners at a spacing of 2000 mm
centre to centre.(3d ≥ c ≤ d)
3
1
y
f
Mk







Design of flange plate
Assuming the flanges resist the moment,
,
Af ≥ 12418.56 mm2
.
To keep the flange in semi plastic class, bf ≤ 13.6 tf.
13.6 tf xtf =12418.56 , tf = 30.2 mm
Provide 32 mm thick flange plate.
Width of plate, bf = 12418.56/32 = 388.08 mm
Provide 400 mm wide and 32 mm thick flange plate.
M
1.1
d
x
y
f
x
f
A
 6
10
x
4233.6
1.1
1500
x
250
x
f
A


The cross section of plate girder is as shown in Fig. 2.
400 mm
32 mm
32 mm
1500 mm
8 mm
Fig. 2 Cross section of plate girder

Check for shear buckling of web.
Using simple post critical method as per clause 8.4.2.2
of IS 800: 2007.
If c/d ≥ 1.0 then
7.6
1500
2000
4
5.35
d
c
4
5.35
K 2
2
v 
















   
08
39
8
1500
3
0
1
12
10
2
6
7
1
12
2
2
5
2
2
2
2
.
.
x
x
x π
.
t
d
μ
E
π
K
τ
w
v
cr 
















1.92
39.08
x
3
250
x τ
3
f
cr
yw
w 




If λw > 1.2
Vn = Vcr = Av x τb = 1500 x 8 x 39.15
= 469.8 kN < 705.6 kN.
Since Vn is less than the maximum shear force, hence
intermediate stiffener to be used to improve buckling
strength of slender web and end panel should be check
for shear capacity as per clause 8.5.3 of IS 800: 2007
Mpa
39.15
1.92
x
3
250
λ
3
f
τ 2
2
w
yw
b 



Shear capacity of end panel
3
y
f
x t x
d
p
V  N
3
10
x
1732.05
3
250
x
8
x
1500












p
cr
p
q V
V
1
V
1.25
H
kN
924.13
2
H
R
q
tf


kN
1848.26
1732.05
469.8
1
1732.05
x
1.25 









Mtf =
The end panel to be checked as a beam spanning
between the flanges to resist Rtf and Mtf.
Area resisting shear = tw x d = 8 x 1500
= 12000 mm2
.
kNm
277.239
10
1500
x
1848.26
10
d
Hq


kN
705.6
N
x10
1574.59
1.1
x
3
250
x
12000
γ
3
f
A
V
3
mo
yw
v
d





End panel can carry the shear due to anchoring force.
ymax = c/2 = 1000 mm.
Hence end panel can carry the bending moment due to
anchor force.
.
mm
10
x
5333.3
12
2000
x
8
12
c
t
I 4
6
3
3
w



1.1
x
1000
250
x
10
x
5333.33
x γ
y
f
x
I
M
3
mo
max
y
q 

tf
M
kNm
1212.12 


Design of end stiffeners
Reaction at ends = 705.6 kN
Compressive force due to Mtf ,
Mtf /c = 227.239 x 106
/2000 = 138.62 kN.
Total compression = 705.6 + 138.62
= 844.22 kN.
Strength of the stiffener as per clause 8.7.5.2 of IS 800
: 2007,
Equating strength of stiffener to force to be resisted
2
q
q
3
mm
2972
A
1.1
x
0.8
250
x
A
10
x
844.22 


1.1
x
0.8
250
x
A
0.8
f
A
F q
mo
yq
q
psd 



Provide 180 mm wide and 10 mm thick flats on either
side of web. Then provided is,
2 x 180 x 10 = 3600 mm2
.
Check for outstand: It should not be more than 20
Since it is more than 14 , 14 x 10 = 140 mm.
Core area of each stiffener = 140 x 10 = 1400 mm2
.
Buckling check for stiffener
Effective area = 2 x 140 x 10 = 2800 mm2
.
r = (36.34 x 106
/2800)0.5
= 113.9 mm.
q
t
q
A
q
t
  4
6
3
s mm
10
x
36.34
12
8
-
360
10
I 


KLc = 0.7 x 1500 = 1050 mm.
slenderness ratio = 1050/113.9 = 9.22
from table 9c of IS 800: 2007
fcd = 227 Mpa.
Assuming 20 x web thickness on only one side, effective
area = 2 x 140 x 10 + 20 x 8 x 8 = 6000 mm2
.
Buckling resistance of stiffener = 4080 x 227
= 926.16 kN > 844.22 kN.
Check for load carrying capacity of stiffener
As per clause 8.7.4 of IS 800: 2007, taking stiff bearing,
b1 = 0
n2 = 2.5 x tf = 2.5 x 32 = 80 mm.

Local capacity of web
The stiffener is to be designed for a force
844.22 – 145.45 = 698.766 kN.
fcd = 227 Mpa.
Area of stiffener alone, 2 x 180 x 10 = 3600 mm2
.
Bearing capacity of stiffener alone,
Hence the stiffeners are safe. Provide 180 mm wide and
10 mm thick flat as a stiffener.
    kN
145.45
1.1
250
x
8
x
80
0
γ
f
t
n
b
F
mo
yw
w
2
1
w 




kN
698.766
kN
742.9
1.1
3600
x
227 


Design of intermediate stiffeners.
As the shear force reducing towards mid span, the first
stiffener from end is critical. Since first intermediate
stiffener is at c = 2 m from end.
Shear force on the stiffener, V = R – 2w
= 705.6 – 2 x 58.8
= 588 kN.
The ratio c/d = 2000/1500 = 1.33 < √2
Hence required minimum moment of inertia
4
2
3
3
2
3
w
3
s mm
648000
2000
8
x
1500
x
1.5
c
t
1.5d
I 



Provide intermediate stiffener of size 120 mm wide and
10 mm thick which satisfy the condition of outstand.
12.71 x 106
mm4
. > Is required. Hence o. k.
Checking for buckling
Shear buckling resistance of web alone, Vcr = 469.8 kN
Shear strength of stiffener alone required = (V-Vcr)/γmo.
(588 – 469.8)/1.1 = 107.45 kN.
Buckling resistance of intermediate stiffener as per
clause 8.7.15 of IS 800: 2007.
 
12
8
x
10
12
120
8
120
10
I
3
3
s 




Considering 20 tw = 160 mm width of web on both side
along with stiffener.
Area = 2 x 120 x 10 + 2 x 160 x 8 = 4960 mm2
.
r = ( 12.73 x 106
/4960) 0.5
= 50.66 mm
KL/r = (0.7 x 1500)/50.66 = 20.72
From table 9c of IS 800: 2007, fcd = 223.78 Mpa
Buckling resistance = 223.78 x 4960/1000
= 1109.97 kN > 107.45 kN.
Hence safe.
4
6
3
s
x mm
10
x
12.73
12
8
x
160
x
2
I
I 



Design of weld between flange and web plates
Maximum shear force = 705.6 kN
Shear stress at the joint of flange and web
Shear force per mm length = 401.2 N/mm
If s is the size of shop weld, then throat thickness 0.7 s.
Providing weld on both side of web, strength of weld
per mm length is,
Equating shear force to strength of weld per mm length
2
6
3
N/mm
1.003
00
4
x
10
x
17273.14
788
x
32
x
400
x
10
x
705.6
Ib
y
FA
τ 


s
265.1
1.25
1
x
3
410
x
s
0.7
x
2 

s = 1.513 mm.
Provide 5 mm intermittent weld on both side of web
plate with weld length of 40 mm and at an spacing of
120 mm centre to centre.
Design of weld between bearing stiffeners and web
plates
Shear carried by web:
fcd = 24.3 Mpa,
mm
453.75
8
1500
2.42
t
d
2.42
λ 


mm
782
2
D
n1 


Area of web resisting shear, btw = 742 x 8
= 5936 mm2
.
Load transmitted by web = 5936 x 24.3
= 144.24 kN.
Shear to be transferred
through weld = reaction - shear transfer by web
= 705.6 - 144.24
= 561.36 kN
Length of weld = 1500 - 2 x 8 = 1484 mm
Shear stress per mm length = 561.36/1484
= 0.378 kN/mm

Additional shear =
Total shear per mm length = 0.469 kN/mm
= 469 N/mm
If s is the size of fillet weld to be provided on both side
then strength of weld per mm length.
Equating shear to strength of weld per mm length
265.12 s = 469, s = 1.769 mm
kN/mm
0.091
140
x
5
8
5b
tw 2
s
2


N/mm
s
265.12
1.25
1
x
3
410
x
s
0.7
x
2 

Provide 5 mm intermittent weld on both side
of web plate with weld length of 40 mm and at
an spacing of f 120 mm centre to centre.
Design similar welded connection between
intermediate stiffener and web plate.

Bearing Stiffener
Intermediate Stiffener Flange plate
Web Plate
20 t p p p
Sectional plan

Intermediate stiffener
Bearing stiffener Web plate Flange plate
P
P
P
P
Sectional Elevation

�ݺ�ߣ

Design of plate girder by LSM flexural cross section.ppt

Recommended

More Related Content

Similar to Design of plate girder by LSM flexural cross section.ppt (20)

Recently uploaded (20)

Design of plate girder by LSM flexural cross section.ppt