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Mg university
**Course:
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B.Tech
**Specialization:
**Civil Engineering**Views:
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Zp 2 (550 x50 x725 700 x 22 x
700
) 5.065 x10 7
2
1x5.065 x10 7 x 250
11511.3 kNm 8826 kNm
1.1x10 6
7.Check for shear capacity
Md
Nominal shear Vn Vp
Av fyw
3 m o
1400 x 22 x 250
3 x1.1x10 3
safe.
4041 1227 kN
8. Check for end bearing
According to the code specification , assuming the slope of dispersion as 1:2.5, the dispersion length
= n2 = 2.5x50 = 125 mm.
Minimum stiffener bearing length provided by support = b1=550/2 =275 mm
Local shear capacity of web = Fw =(b1+n2 )tw(fy/ mo )
Fw =(275+125 ) 22(250/1.10)=(2000x103)N
=2000kN>support reaction (V) = 1276 kN
Hence bearing stiffeners are not required.
Problem 2
The effective span of a through type plate girder two lane highway bridge is 30 m. The R.C.C slab is
250 mm thick including wearing coat. Foot paths of 1.50 m width are provided on both sides of 6.80
m wide carriage way. Cross girders are provided at 3.0 m c/c and the stringers are at 2.45 m c/c. The
spacing between main girders is 9.80 m. Design the maximum section of plate girder to carry IRC
class A loading.
Step 1
Total width of slab = 9.80 m.
Number of cross girders ( 3.0 m cc) = 11 nos
No of Stringers in one panel (2.45 m c/c)= 5 nos.
Step 2 Dead load B.M and S.F.
Total weight of R.C.C slab = 30 x 9.80 x 0.25 x 24 = 1764 kN
(taking unit weight of slab as 24 kN/m3
Weight of stringers (assumed at 1kN/m) = 5x30x1 =150kN
Weight of cross girders(assumed at 3 kN/m) = 11x9.8x3=323.4 kN
Self weight of plate girders by Fuller’s formula = 0.2L+1 =0.2x30+1 =7 kN/m
13

Total weight of P.G = 7x30 =210 kN
Total D.L = 2447.4 kN
D.L per girder = 2448/2=1224 kN
Moment due to D.L = M1 =
S.F due to D.L =
WL 1224 x30
4589 kNm
8
8
W 1224
612 kN
2
2
Step 4.Live Load B.M and S.F
Transverse location of loading
For the design of maximum section of the plate girders, the loads are so placed that the reaction on
one plate girder is the maximum.
Min. clearance from kerb = 0.15+0.25 = 0.40
c/c distance of two wheels of one train = 1.80 m
Minimum distance between adjacent edges of wheels of two trains =
1.20 0.4
g 0.40
(6.80 5.50) 0.92m
2
c/c distance between two wheels =0.92+0.50=1.42 m
Taking moment about B
P
(7.9 6.1 4.68 2.88) 2.2 P
9.8
9
I ..F
0.207
13.5 L
Reaction at A including impact 2.2 x1.207 2.655 P
RA
P represents one wheel load of IRC class A loading
Longitudinal location of loading
14

Af
Mm o 211.56 x10 6 x1.1
2069
fy d
250 x 450
To keep the flange in plastic condition
b
8.4
tf
bf tw
2
bf 8
we get
8.4
2t f
1, outstand, b
Taking
b
8.4
tf
b f 16.8t f 8
A f 16.8t f 8 t f 2069 mm 2
16.8t f 8t f 2069 0
2
t f 10.86
Provide 12 mm thick plate
2069
172
12
Provide 225 x 12 mm flange
bf
Outstand of flange b
225 - 8
112.5 mm
2
b 112.5
9.375 semicompac t
tf
12
Zp 2(225x12x206 225x8x225/2 1.517 x 10 6 mm 3
Step 6. Check for bending strength
Md
b Zp fy 1x1.517 x10 6 x 250
344kNm 211.56 kNm safe.
m o
1.1x10 3
Step 7.check for shear capacity
Nominal shear strengthV n Vp
Avfw
3m o
450 x8 x 250
3x1.1
472kN 313.21kN
saf
Step 8. Check for end bearing
According to the code specification , assuming the slope of dispersion as 1:2.5, the dispersion length
= n2 = 2.5x12 = 30 mm.
Minimum stiffener bearing length provided by support = b1=225/2 =112.5mm
f
Local shear capacity of web = Fw = (b1 n 2 ) t w y
o
Fw =(30+112.5 ) 8(250/1.10)=(259x103)N
Fw=259 kN<313 kN . Provide end bearing stiffener
7. Design of end stiffeners
21

(a) For buckling resistance
225 - 8
108.5 mm
2
Try a pair of 100 x8 mm flats.
Outstand of flange
now 14 t f 14 x8 112 mm
Core area of stiffener along with effective area of web ( 20 t w ), assuming web area is available
only on one side.
Area for buckling rsistance (2 x 100 (112.5 20 x 8 )8 3780 mm 2
1
1
3
x8100 100 8 272.58 3 6.01x10 6 mm 4
12
12
0.7 x3000
52.67
From table 9c in IS 800
39.87
183 1682.67 179 N/mm 2
fcd 183
10
Ix
r
I
A
6.01x10 6
39.87 mm
3780
Buckling resistance 179 x 3780 676620 N 676.62 kn 313 kN
Hence the stiiffener is safe.
(b) Check for bearing capacity of stiffener. (cl. 8.7.5.2)
Fpsd
Fpsd
A q f yq
where A q Area of stiffner in contact with flange 2 x 100 x 8 1600 mm 2
0.8 x1.1
1600 x 250
454.54 x10 3 N 454 kN 313 kN safe
0.8 x1.1
Problem 4
The effective span of a through type plate girder two lane highway bridge is 30 m. The R.C.C slab is
250 mm thick including wearing coat. Foot paths are provided on both sides of the carriage way Cross
girders are provided at 3.0 m c/c and the stringers are at 2.45 m cc. The spacing between main girders
is 9.80 m. Design the maximum sections for the stringers if the bridge is to carry IRC class A loading.
Nov 2016
Problem 5
The effective span of a deck type plate girder two lane highway bridge is 30 m. The reinforced concrete
slab is 250 mm. thick inclusive of wearing coat. Foot paths are provided on either sides of the carriage
way . Design maximum section of plate girder , if the bridge is to carry IRC class A loading.
Nov 2016
Problem 6
The effective span of a through type plate girder two lane highway bridge is 30 m. The R.C.C slab is
250 mm thick including wearing coat. Foot paths are provided on both sides of the carriage way Cross
girders are provided at 3.0 m c/c and the stringers are at 2.45 m cc. The spacing between main girders
22

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