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A. Design of Superstructure
1.0 Design Data
Fig 1.BRIDGE CROSS-SECTION
Pa
1.1. Materials and its Properties:
M25
Fe415
Characteristics Strengthof Concrete fck = 25 MPa
Permissible direct compressive stress, c = 6.2 MPaPermissible flexural compressive stress, cbc = 8.3 MPa
Maximum Permissible shear stress, ax ( 0.07*fck) = 1.75 MPa
Fig 1.BRIDGE CROSS-SECTION
.
Basic Permissible Stresses of Reinforcing Bars as per IRC : 21-1987, Section III:
Permissible Flexural Tensile stress, st = 200 MPa
Permissible direct compressive stress, co = 170 MPa
Self weight of materials as per IRC : 6-2000:
Concrete (cement-Reinforced) = 24 kN/m3
Fig 1.BRIDGE CROSS-SECTION
Macadam (binder premix) = 22 kN/m3
1.2. Geometrical Properties:Effective Span of Bridge = 24.00 m
Total length of span = 24.56 m
Numbers of span = 2
Width of expantion Joint = 40 mm
Total length of Bridge = 49.2 m
Fig 1.BRIDGE CROSS-SECTION
Nos. of longitudinal Girder = 3
Spacing of Girder = 2.4 m
Rib width of main girder = 400 mm
Overall depth of main girder = 2000 mm
Depth of kerb above deck slab = 225 mm
Nos. of cross girder = 6
Spacing of cross girder = 4.8 m
Fig 1.BRIDGE CROSS-SECTION
=
Overall depth of cross girder = 1500 mm
Deck slab thickness = 220 mm
Deck slab thickness at edge = 150 mm
Thickness of wearing coat = 80 mm
Fillet size (horizontal) = 150 mm
Fillet size (vertical) = 150 mm
Bridge Width:
Fig 1.BRIDGE CROSS-SECTION
Carriageway width = 6 m
Footpath width = 0.45 m
Kerb width Outer = 0.15 m
Kerb width Inner = 0 m
Total Width of Deck Slab = 7.2 mTotal depth of Kerb Outer = 0.375 m
Total depth of Kerb Inner = 0 m
Fig 1.BRIDGE CROSS-SECTIONFig 1.BRIDGE CROSS-SECTION
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2.0 Design of Slab
cbc3
280
Fig 1.BRIDGE CROSS-SECTION
57KN 350KN 37.5KN
IRC Class AA Track LoadingIRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
Pa
2.1 Design of Cantilever slab: The cantilever slab is designed by effective width method.
300 mm at junction with rib
150 mm at free end
0.5 kN/m (assumed)
Impact factor = 54 % (for IRC class A loading)
25 % (for IRC class AA loading)
Thickness of slab =
Self weight of Railing =
cbc3
280
Fig 1.BRIDGE CROSS-SECTION
57KN 350KN 37.5KN
IRC Class AA Track LoadingIRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
Dead Load Bending Moment and Shear Force:
S.No. Item Width Depth Unit Wt
1 Railing/Parap
et
0.5 kN 1.525-0.100= 0.9 m 0.45 kN.m
2 Kerb (outer) 0.2 0.225 24 1.08 kN 1.525-0.100= 0.9 m 0.97 kN.m
3 Kerb (inner) 0 0 24 0 kN 0.15+0.175/2= 0.2375 m 0 kN.m
4 Wearin Coat 0.4 0.08 22 0.704 kN 0.150/2= 0.075 m 0.05 kN.m
Assumed
Load / m run
(kN)MomentDistance
cbc3
280
Fig 1.BRIDGE CROSS-SECTION
57KN 350KN 37.5KN
IRC Class AA Track LoadingIRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
. . . . . .
5 Slab 1 0.15 24 3.6 kN 1.525/2= 0.5 m 1.8 kN.m
1 0.075 24 1.8 kN 1.525/3= 0.3333 m 0.6 kN.m
Total kN kN.m
= 7.684 kN
= 3.875 kN.m
7.684
Dead load Shear force at theface of rib
Dead load Bending Moment at the face of rib
3.875
cbc3
280
Fig 1.BRIDGE CROSS-SECTION
57KN 350KN 37.5KN
IRC Class AA Track LoadingIRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
Live Load Bending Moment and Shear Force:
cbc3
280
Fig 1.BRIDGE CROSS-SECTION
57KN 350KN 37.5KN
IRC Class AA Track LoadingIRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
IRC Class AA will not operate on the cantilever slab that shown in fig 2.b & 2.c above and Class A
Loading is to be considered and the load will be as shown in fig 2.a above.
Effective width of dispersion be is computed by equation
be = 1.2X+ bw
Here
X= 0.125 m
bw= 0.41 m
cbc3
280
Fig 1.BRIDGE CROSS-SECTION
57KN 350KN 37.5KN
IRC Class AA Track LoadingIRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
.
Hence
be= 0.56 m
IRC Class A Loading Load = 28.5 kN
Live Load per m width including impact = 76.339 kN
Maximum Moment due to live load = 9.5424 kNm
Average thickness of cantilever slab = 225 mm
Taking pedestrain load (LL) = 5.0 kN/m2
cbc3
280
Fig 1.BRIDGE CROSS-SECTION
57KN 350KN 37.5KN
IRC Class AA Track LoadingIRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
Effective width of slab = 0.45 m
Cantilever length of slab = 1 m
Maximum Bending moment = 1.406 kN.m
Shear force at the face of slab = 2.250 kN
Total Design Shear Force = 86.3 kN
Total Design Bending Moment = 14.82 kN.m
Design of Section:
Modular Ratio, m = = 11.245cbc
3
280
Fig 1.BRIDGE CROSS-SECTION
57KN 350KN 37.5KN
IRC Class AA Track LoadingIRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
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Neutral axis factor, k = = 0.3182
Lever arm factor, j = = 0.8939
cbc3
280
stcbcm
cbcm
+
31
k
stjk 2
1
Rb
M
dj
M
st ..=
dj
M
st ..=
bd
d
tanM-V
=v
coc kk .. 21=5.07.014.1
1 = dk
125.05.02
+= k bd
A s=
=2k
co
4.021
= KKc
v
Pa
Moment of resistance coefficient, R = = 1.1804
Therefore, required effective depth of slab=
d = = 112.06 mm
Effective depth of slab, provided = 254 mm > d reqO.K.
cbc3
280
stcbcm
cbcm
+
31
k
stjk 2
1
Rb
M
dj
M
st ..=
dj
M
st ..=
bd
d
tanM-V
=v
coc kk .. 21=5.07.014.1
1 = dk
125.05.02
+= k bd
A s=
=2k
co
4.021
= KKc
v
Area of steel required, Ast = 326.42 mm2
Provide 10 mm bars @ 200 = 393 mm2
> required, Ok.
Distribution Steel:
Distribution steel is to be provided for 0.3 times live load moment plus 0.2 times dead
load moment.
mm c/c, giving area of steel =
cbc3
280
stcbcm
cbcm
+
31
k
stjk 2
1
Rb
M
dj
M
st ..=
dj
M
st ..=
bd
d
tanM-V
=v
coc kk .. 21=5.07.014.1
1 = dk
125.05.02
+= k bd
A s=
=2k
co
4.021
= KKc
v
.
Moment = 4.06 kN.m
Effective dept 244 mm
Area of steel required, Ast = 93.057 mm
Half reinforcement is to be provided at top and half at bottom.
Provide 10 mm bars 200 mm c/c at both top and bottom, giving area of2
cbc3
280
stcbcm
cbcm
+
31
k
stjk 2
1
Rb
M
dj
M
st ..=
dj
M
st ..=
bd
d
tanM-V
=v
coc kk .. 21=5.07.014.1
1 = dk
125.05.02
+= k bd
A s=
=2k
co
4.021
= KKc
v
stee = .5 mm > requ re , .
Check for min. area of Steel:
Min. area of steel @ 0.12 % = 360 mm < Provided. O.K.
Design for Shear:
Dead load shear = 7.68 kN
Live load Shear = 2.250 kN
cbc3
280
stcbcm
cbcm
+
31
k
stjk 2
1
Rb
M
dj
M
st ..=
dj
M
st ..=
bd
d
tanM-V
=v
coc kk .. 21=5.07.014.1
1 = dk
125.05.02
+= k bd
A s=
=2k
co
4.021
= KKc
v
Total = 9.93 kN
tan= 0.150
Shear stress, = 0.005 N/mm2
Percentage area of tension steel, pt = 0.13 %
Allowable shear stress as per code is given by
cbc3
280
stcbcm
cbcm
+
31
k
stjk 2
1
Rb
M
dj
M
st ..=
dj
M
st ..=
bd
d
tanM-V
=v
coc kk .. 21=5.07.014.1
1 = dk
125.05.02
+= k bd
A s=
=2k
co
4.021
= KKc
v
( d being in m)
= 0.986
(where )
= 0.500446 1.00
cbc3
280
stcbcm
cbcm
+
31
k
stjk 2
1
Rb
M
dj
M
st ..=
dj
M
st ..=
bd
d
tanM-V
=v
coc kk .. 21=5.07.014.1
1 = dk
125.05.02
+= k bd
A s=
=2k
co
4.021
= KKc
v
Adopt 1
Value ot = for M25 grade of concrete from code = 0.4 N/mm2
Allowable shear stress
= 0.3944 N/mm2 > , Hence Safe
cbc3
280
stcbcm
cbcm
+
31
k
stjk 2
1
Rb
M
dj
M
st ..=
dj
M
st ..=
bd
d
tanM-V
=v
coc kk .. 21=5.07.014.1
1 = dk
125.05.02
+= k bd
A s=
=2k
co
4.021
= KKc
v
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2.2 Design of Interior Panels: The slab panel is designed by Pigeauds method.
D
Pa
B C
Short span of slab, Bs = 2 m
Long span of slab, Ls = 4.5 m
A
Fig : 3 Bridge Plan
Calculation of Bending moments
a) Due to Dead load:
Self weight of wearing coat = 1.76 kN/m2
Self weight of deck slab = 5.28 kN/m
Total = 7.04 kN/m
Since the slab is supported on all four sides and is continuous, Piegauds curves are used to
calculate bending moments.
a o, = s s = .
As the panel is loaded with UDL,
u/Bs = 1
v/Ls = 1
Where, u & v are the dimensions of the loaded area.
From the Pigeauds curve,
m1 = 0.0457
m2 = 0.0086Total dead load W = 63.36 kN
Moment along short span, M1 = W (m1 +0.15m2) = 2.98 kN-m
Moment along long span, M2 = W (0.15m1 +m2) = 0.98 kN-m
Considering effects of continuity, 0.8
Moment along short span, M1 = 2.38 kN-m
Moment along long span, M2 = 0.78 kN-m
Due to Live load: Class AA Tracked Vehicle
For maximum bending moment one wheel is placed at the center of panel.
Tyre contact length along short span, x = 0.85 m
Tyre contact length along long span, y = 3.6 m
Loaded length, u = 1.034 m
Loaded width, v = 3.766 m
Wheel load, W = 350 kN
Ratio, k = Bs/Ls = 0.44
u/Bs = 0.517
v/Ls = 0.837
From the Pigeauds curve,
m1 = 0.0813
m2 = 0.0147
Moment along short span,
= 29.227 kN-m
Moment along long span,
= 9.413 kN-m
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
W1=350k
Fig: 4
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Bending moment including impact and continuity,
M1 = 29.227 kN-m
M2 = 9.413 kN-m Fig: 4
Pa
c) Due to Live load: Class AA Wheeled Vehicle
Case-I:When two loads of 37.5 kN each and four loads of 62.5kN are placed such that two loads of
62.5kN lies at center line of pannel.
Tyre contact width (along short span), = 0.30 m
Tyre contact length (along long span), = 0.15 mDisperced width along short span, u = 0.510 m
Disperced width along long span, V = 0.380 m
= .
62.5kN
62.5kN
62.5kN
62.5kN
W1 W4
W2 W5
Y
X X
Bending moment due to load W1: 62.5 kN
Ratio k = Bs/Ls = 0.44
Fig: 5
37.5kN
W3
37.5kN
W6
Y
, .
u/Bs = 0.255
v/Ls = 0.084
From the Pigeauds curve,
m1 = 0.1965
m2 = 0.1383
Moment along short span,
= 13.578 kN-mM1= W(m1+0.15m2)
omen a ong ong span,
= 10.486 kN-m
Bending moment including impact and continuity,
M1 = 13.578 kN-m
M2 = 10.486 kN-m
Bending moment due to load W2: 62.5 kN Wheel load is placed unsymmetrical wrt the X-X
Intensity of loading, q = 322.45 kN/m2
M2= W(0.15m1+m2)
Considering loaded area 2.000 x 0.380 m
Loaded area = 0.760 m2
Total applied load = q x area = 245 kN
Ratio, k = Bs/Ls = 0.44u/Bs = 1.000
v/Ls = 0.084
From the Pi eauds curve ,
m1 = 0.0935
m2 = 0.0742
Moment along short span,
= 25.650 kN-m
Moment along long span,
= 21.628 kN-m
Bending moment including impact and continuity,
M1 = 25.650 kN-m
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
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M2 = 21.628 kN-m
Next, Consider the area between the real and the dummy load i.e., 1.490 m X 0.380 m
Loaded area = 0.566 m2
Total applied load = q x area = 183 kN
Pa
Ratio, k = Bs/Ls = 0.44
u/Bs = 0.745
v/Ls = 0.084
From the Pigeauds curve,
m1 = 0.1157m2 = 0.0944
Moment along short span,
-= = . -
Moment along long span,
= 20.412 kN-m
Bending moment including impact and continuity,
M1 = 23.718 kN-m
M2 = 20.412 kN-m
Final Moment
M1 = 0.966 kN-m
M2= W(0.15m1+m2)
= .
M2 = 0.608 kN-m
Bending moment due to load W3: 37.5 kN Wheel load is placed unsymmetrical wrt the X-X
Intensity of loading, q = 193.47 kN/m2Considering loaded area 1.710 x 0.380 m
Loaded area = 0.650 m2
Total applied load = q x area = 126 kN
Ratio, k = Bs/Ls = 0.44
u/Bs = 0.855
v/Ls = 0.084
From the Pigeauds curve,
m1 = 0.1036
m2 = 0.08325
Moment along short span,
= 14.598 kN-m
Moment along long span,
= -
M1= W(m1+0.15m2)
= . -
Bending moment including impact and continuity,
M1 = 14.598 kN-m
M2 = 12.423 kN-m
Next, Consider the area between the real and the dummy load i.e., 0.69 m X 0.380 m
Loaded area = 0.262 m2
Total applied load = q x area = 51 kN
Ratio, k = Bs/Ls = 0.44
.
u/Bs = 0.345
v/Ls = 0.084
From the Pigeauds curve,
m1 = 0.1765
m2 = 0.1312= 9.957 kN-m
Moment along long span,
= 8.002 kN-m
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
Bending moment including impact and continuity,
M1 = 9.957 kN-m
M2 = 8.002 kN-m
Final Moment
M1 = 2.321 kN-m
M2 = 2.210 kN-m
Bending moment due to load W4: 62.5 kN Wheel load is placed unsymmetrical wrt the Y-Y
Intensity of loading, q = 322.45 kN/m2
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Considering loaded area 2.780 x 0.510 m
Loaded area = 1.418 m2
Total applied load = q x area = 457 kN
Ratio, k = Bs/Ls = 0.44
Pa
u/Bs = 0.618
v/Ls = 0.255
From the Pigeauds curve,
m1 = 0.1168
m2 = 0.0648Moment along short span,
= 57.832 kN-mM1= W(m1+0.15m2)
,
= 37.628 kN-m
Bending moment including impact and continuity,
M1 = 57.832 kN-m
M2 = 37.628 kN-m
Next, Consider the area between the real and the dummy load i.e., 2.02 m X 0.510 m
Loaded area = 1.030 m2
Total applied load = q x area = 332 kN
M2= W(0.15m1+m2)
Ratio, k = Bs/Ls = 0.44
u/Bs = 0.449
v/Ls = 0.255From the Pigeauds curve,
m1 = 0.1353
m2 = 0.0712
Moment along short span,
= 48.480 kN-mM1= W(m1+0.15m2)
Moment along long span,
= 30.386 kN-m
Bending moment including impact and continuity,
M1 = 48.480 kN-m
M2 = 30.386 kN-m
Final Moment
M1 = 4.676 kN-m
=
M2= W(0.15m1+m2)
= . -
Bending moment due to load W5:
Wheel load is placed unsymmetrical wrt the Both X-X and Y-Y
For X- X Axis
W = 62.5 kN
Intensity of loading, q = 322.45 kN/m2
Considering loaded area 2.000 x 0.380 m
Loaded area = 0.760 m2
Total applied load = q x area = 245 kN
Ratio, k = Bs/Ls = 0.44
u/Bs = 1.000
v/Ls = 0.084
From the Pigeauds curve,m1 = 0.0935
m2 = 0.0742
Moment alon short s an, ,
= 25.650 kN-m
Moment along long span,
= 21.628 kN-m
Bending moment including impact and continuity,
M1 = 25.650 kN-m
M2 = 21.628 kN-m
Next, Consider the area between the real and the dummy load i.e., 1.490 m X 0.380 m
Loaded area = 0.566 m2
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
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Total applied load = q x area = 183 kN
Ratio, k = Bs/Ls = 0.44
u/Bs = 0.745
v/Ls = 0.084
Pa
From the Pigeauds curve,
m1 = 0.1157
m2 = 0.0944
Moment along short span,
= 23.718 kN-mMoment along long span,
= 20.412 kN-mM2= W(0.15m1+m2)
M1= W(m1+0.15m2)
,
M1 = 23.718 kN-m
M2 = 20.412 kN-m
Moment Along X-X
M1 = 0.966 kN-m
M2 = 0.608 kN-m
For Y- Y Axis
W = 62.5 kN
Intensity of loading, q = 322.45 kN/m2
Considering loaded area 2.780 x 0.510 m
Loaded area = 1.418 m2
Total applied load = q x area = 457 kN
Ratio, k = Bs/Ls = 0.44
u/Bs = 0.618
v/Ls = 0.255
From the Pi eauds curve ,
m1 = 0.1168
m2 = 0.0648
Moment along short span,
= 57.832 kN-m
Moment along long span,
= 37.628 kN-m
Bending moment including impact and continuity,
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
= . -m
M2 = 37.628 kN-m
Next, Consider the area between the real and the dummy load i.e., 2.020 m X 0.51 m
Loaded area = 1.030 m2
Total applied load = q x area = 332 kN
Ratio, k = Bs/Ls = 0.44
u/Bs = 0.449
v/Ls = 0.255
From the Pigeauds curve,
m1 = 0.1353
m2 = 0.0712
Moment along short span,
= 48.480 kN-mMoment along long span,
= 30.386 kN-m
Bendin moment includin im act and continuit
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
,
M1 = 48.480 kN-m
M2 = 30.386 kN-m
Final Moment
M1 = 4.676 kN-m
M2 = 3.621 kN-m
Resultent Moment
M1 = 5.642 kN-m
M2 = 4.230 kN-m
Pa
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Bending moment due to load W6:
Wheel load is placed unsymmetrical wrt the Both X-X and Y-Y
For X-X Axis
Pa
W = 37.5 kN
Intensity of loading, q = 193.47 kN/m2
Considering loaded area 1.710 x 0.380 m
Loaded area = 0.650 m2Total applied load = q x area = 126 kN
Ratio, k = Bs/Ls = 0.44
u s = .
v/Ls = 0.084
From the Pigeauds curve,
m1 = 0.1036
m2 = 0.08325
Moment along short span,
= 14.598 kN-m
Moment along long span,
M1= W(m1+0.15m2)
= 12.423 kN-m
Bending moment including impact and continuity,
M1 = 14.598 kN-mM2 = 12.423 kN-m
Next, Consider the area between the real and the dummy load i.e., 0.69 m X 0.380 m
Loaded area = 0.262 m2
Total applied load = q x area = 51 kN
Ratio, k = Bs/Ls = 0.44
M2= W(0.15m1+m2)
u/Bs = 0.345
v/Ls = 0.084
From the Pigeauds curve,
m1 = 0.1765
m2 = 0.1312
Moment along short span,
= 9.957 kN-mM1= W(m1+0.15m2)
,
= 8.002 kN-m
Bending moment including impact and continuity,
M1 = 9.957 kN-m
M2 = 8.002 kN-m
Moment Along Y-Y
M1 = 2.321 kN-m
M2 = 2.210 kN-m
M2= W(0.15m1+m2)
For Y- Y Axis
W = 37.5 kN
Intensity of loading, q = 193.47 kN/m2
Considering loaded area 2.780 x 0.510 m
Loaded area = 1.418 m2Total applied load = q x area = 274 kN
Ratio, k = Bs/Ls = 0.44
u/Bs = 0.618.
v/Ls = 0.255
From the Pigeauds curve,
m1 = 0.1168
m2 = 0.0648
Moment along short span,
= 34.699 kN-m
Moment along long span,
= 22.577 kN-m
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
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Bending moment including impact and continuity,
M1 = 34.699 kN-m
M2 = 22.577 kN-m
Next, Consider the area between the real and the dummy load i.e., 2.020 m X 0.510 m
Pag
Loaded area = 1.030 m2
Total applied load = q x area = 199 kN
Ratio, k = Bs/Ls = 0.44
u/Bs = 0.449
v/Ls = 0.255From the Pigeauds curve,
m1 = 0.1353
m = .
Moment along short span,
= 29.088 kN-m
Moment along long span,
= 18.231 kN-m
Bending moment including impact and continuity,
M1 = 29.088 kN-m
M2 = 18.231 kN-m
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
Final Moment
M1 = 2.806 kN-m
M2 = 2.173 kN-mResultent Moment
M1 = 5.127
M2 = 4.383
Total Moment Due to IRC Class AA Wheeled Vechicle
Moment alon short s an M1 = 32.309 kN-m , .
Moment along long span,M2 = 25.539 kN-m
c) Due to Live load: Class A Loading
IRC Class A Loading: For maximum bending moment one wheel of 57kN should be placed at the
centre of span and other at 1.2 m from it as shown. Neglecting small eccentricity of 80mm.
Tyre contact length along short span, Y = 0.5 m
Tyre contact length along long span, X = 0.25 m
Imaginary load W3 = W2 is placed on the other side of W1 to make loading symmetrical.
Due toloads W2 & W3 Bendin moment at center of anel will be that due to load W1andhalf. .
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Y
Pag
57kN
W1 W2
Y
X X57kN
Fig: 6
Bending moment due to load W1:
Wheel load, W1 = 57 kN
Loaded length, u = 0.696 m
Loaded width, v = 0.465 m
Ratio, k = Bs/Ls = 0.44
u/Bs = 0.35
v/Ls = 0.10
From the Pi eauds curve ,
m1 = 0.1717
m2 = 0.1245
Moment along short span,
= 10.851 kN-m
Moment along long span,
= 8.565 kN-m
Bending moment including impact and continuity,
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
= . -m
M2 = 8.565 kN-m
Bending moment due to load W2:
Wheel load is placed unsymmetrical wrt the Y-Y
Wheel load, W2 = 57 kN
Intensity of loading, q = 176.09 kN/m2
Considering loaded area 2.865 x 0.696 m
Loaded area = 1.993 m2
Total applied load = q x area = 351 kN
Ratio, k = Bs/Ls = 0.44
u/Bs = 0.637
v/Ls = 0.348
From the Pigeauds curve,
m1 = 0.10843.
m2 = 0.0497
Moment along short span,
= 40.676 kN-m
Moment along long span,
= 23.154 kN-m
Bending moment including impact and continuity,
M1 = 40.676 kN-m
M2= W(0.15m1+m2)
M1= W(m1+0.15m2)
= . -m
Next, Consider the area between the real and the dummy load i.e., 1.935 m X 0.696 m
Loaded area = 1.346 m2
Total applied load = q x area = 237 kN
Ratio, k = Bs/Ls = 0.44
u/Bs = 0.430
v/Ls = 0.348
From the Pigeauds curve,
m1 = 0.1313
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m2 = 0.0552
Moment along short span,
= 33.081 kN-m
Moment along long span,
M1= W(m1+0.15m2)
Pag
= 17.751 kN-m
Bending moment including impact and continuity,
M1 = 33.081 kN-m
M2 = 17.751 kN-m
Final Moment
M1 = 3.798 kN-m
M2 = 2.702 kN-m
M2= W(0.15m1+m2)
Total Bending Moment due to load W1 & W2 will be,
M1 = 14.6 kN-m
M2 = 11.3 kN-m
Design Bending Moment due to LL:
M1 = 32.3 kN-m
M2 = 25.5 kN-m
Calculation of Shear Force
a) Due to Dead load:
Dead load shear force = 7.04 kN
b) Due to Live load: Class AA Tracked Vehicle
Load of Tracked Vehicl = 350 kN
350kN 350kN
Dispertion in the direction of span, = 1.45 m
For maximum shear, load is kept such that whole dispersion is in the span. That is at 0.725 m
from the edge of beam. 0.3
Fig 7.a
Effective width of slab =
Span Ratio (L/B) = 2.25
afor continuous slab
= 2.60x = 0.725 m
bw = 3.76
Therefore effective width of slab = 4.96 m
bwl
x1x +
Load per meter width = 70.54 kN
Shear force at left edge = 44.97 kN
Shear force including impact & continuity = 44.97 kN
c) Due to Live load: Class AA Wheeled Vehicle
37.5kN 37.5kN62.5kN 62.5kN Pag
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37.5kN 37.5kN62.5kN 62.5kN
Pag
Dispersion width in the direction of span = 0.900 m
Loads are placed such that outermost load is at distance of
x = 0.450 m from edge of the beam.
bw = 0.31
Effective width for first wheel = = 1.217 m
.
Wheel Load = 62.50 kN
But the center to center distance of two axel are 1.2 m, thus effective width will overlap.
Average effective width for one wheel = 1.208 mPortion of load in span = 1.000 m
Load per meter width of slab = 51.72 kN
For second wheel Wheel Load = 62.50 kN
x = 0.550 m.
Effective width for second wheel = 1.347 m
But the center to center distance of two axel are 1.2 m, thus effective width will overlap.
Average effective width for one wheel = 1.273 m
Load per meter width of slab = 49.1 kN
For third wheel Wheel Load = 37.50 kN
X = -0.050 m
ect ve w t or t r w ee = .
Load Acting on Span = 16.67 kN
Acting at 0.2 m from Support
Effective width for third wheel = 0.918 m < 1.2 m
Load per meter width of slab = 18 kN
Shear force at left edge = 37.3 kN
Shear force including impact and continuity
= 37.318 kN
b) Due to Live load: IRC Class A loading
Fig 7.c
57kN 57kN
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Shear force due to load W1: 57 kN
Fig 7.c
Pag
Dispersion width in the direction of short span = 1.1 m
For maximum shear force, the load should be placed at distance of 0.55 m from web
of girder. In this position second load will be as shown.
Effective width for first wheel =
Where, x= 0.55 m
bw= 0.41 m
bw
l
x1x +
.
L/B= 2.3 = 2.6
Therefore, Effective width = 1.447 m
But distance between axels is 1.2 m and hence effective width overlaps.
Average effective width / wheel = m
Load W1 = kN
Load per meter width of slab = kN/m
And shear force = kN
57.00
43.07
32.30
1.32
Shear force including impact & continuity = kN
Shear force due to load W2:
Effective width for second wheel =
Where, x= 0.350 m
bw= 0.41 m
.
bwl
x1x +
L/B= 2.4 = 2.6
Therefore, Effective width = 1.161 m < 1.2 m
Load W2 = 57.0 kN
Effective load = 18.1 kN
15.6 kN
And shear force = 12.89 kN
Shear force including impact & continuity = 12.89 kN
Total shear force = 45.19 kN
Load per meter width of slab =
Total Design Bending Moments,
M1 = 34.69 kN-m
M2 = 26.32 kN-m
Total Design Shear force,
S.F. = 52.23 kN
Effective depth required,
d = 171.4 mm
Use 30 mm Clear cover & 12 mm dia
Effective depth available = 184 mm O.K.
Area of steel required along short span,
Ast = 1055 mm2
Check for minimum area of steel:
=Rb
=
djst
M
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264 mm2
Provide 12 mm bars 100 mm c/c at both top and bottom, giving area of
steel = 1130.4 mm2 > required.
Effective depth for long span = 173 mm
Min. area of steel @
Pag
Area of steel required along long span,
Ast = 851 mm2
Provide 12 mm bars 100 mm c/c at both top and bottom, giving area of
steel = 1130.4 mm2 > required.
Check for shear:
=
djst
M
Nomin v = 0.284 N/mm2
Provided percentage area of tensile steel = 0.51 %
Permissible shear stress,c= 1.16 x 0.313 = 0.363 N/mm2 > O.K.
=
db
Vu
3.0 Design of Longitudinal Girder
Effective Span of Bridge = 24 m
Slab thickness = 0.22 m
Width of Rib = 0.4 m
Spacing of main Beam = 2.4 m c/c
Over all depth of Beam = 2 m
3.1 Calculation of dead load moment and shear force on longitudinal girder:
Let the over all depth of the longitudinal girder be 2000 mm, the depth of its rib will
= 1.78 m
Weight of Rib per m = 17.09 kN/m
Dead load from each cantilever portion (refer design of cantilever slab)
= 7.68 kN/m
Dead load of slab & Wearing coat = 7.04 kN/m2
Total Dead load per m from deck = 51.98 kN/m
This load is borne by all the three girders
Dead Load per girder due to Deck Slab = 17.33 kN/m
Let the Depth of rib of cross girder to be = 1.28 m
let its width be = 0.3 m
Weight of rib of cross girder = 9.216 kN/m
Length of each cross girder = 4 m
It is assumed that the weight of each cross girder is equally borne by the entire threelongitudinal girders. This weight acts as point load on each girder its value being
= 12.29 kN
Total UDL = 34.413333 kN/m
RA= RB = 449.824 kN
Bending Moment (BM)
12.3 kN
Fig : 8RA RB
34.413 kN/m
12.3 kN12.3 kN12.3 kNkN 12.3 kN
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BM at Centre of span = 2654.707 kNm
BM at th of span = 2064.758 kNm
BM at 3/8th
Span = 2492.474 kNmth
Pag
BM at 1/8 pan = 1157.748 kNm
Shear Force (SF)
SF at Support = 437.536 kN
SF at 1/8th Span = 334.296 kN
SF at 1/4thSpan = 218.768 kN
SF at 3/8th
Span = 115.528 kN
SF at Center of span = 0 kN
Distance from Support BM SF
At Center of Span 2654.71 0.00
At 3/8th
Span 2492.47 115.53
At 1/4th Span 2064.76 218.77
At Support 0.00 437.54
3.2 Calculation of live load moment and shear force on longitudinal girder:
Impact factor for:
0.150
=
=+ L6
5.4
.
IRC Class AA Wheeled Vehicle = 0.215
Distribution of live loads on longitudinal girder for bending moment:
IRC Class AA Tracked Vehicle:
Reaction on the girder will be maximum when the eccentricity is maximum. Eccentricity will be maximum when the loads
are very near to the kerb. Position of loads for maximum eccentricity is shown in figure.
All the girders are assumed to have the same moment of inertia.
W1 W1
Reaction factor for Outer Girder, RA = 0.81 W1
Reaction factor for Inner Girder, RB = 0.667 W1
( ) =
+ 35.04.2
2.42I
I31
3
W2
2
1
=W2
1
.
If W be the axel load, then wheel load W1= W/2
Then reaction factor, RA = 0.406 W
RB = 0.333 W
IRC Class A Loading:
Position of loads for maximum eccentricity is shown in figure.
W1 W1 W1 W1
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W1 W1 W1 W1
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Reaction on Outer Girder RA,
RA= 1.625 W2( )
=
+ 1.04.2
2.42I
I31
3
W4
2
1
Fig. 10
RB = 1.333 W2
If W be the axel load, then wheel load W2 = W/2
Then reaction factor, RA = 0.813 W
RB = 0.667 W
Bending Moment due to Live load: IRC Class AA Tracked Vehicle
[ ]=+ 013
4W1
The influence line diagram for bending moment is shown in figure.
Effective span of girder, le = 24.0 m
ITC Class AA Tracked Vehicle Load: = 700 kN
Ordinate of Bending Moment at considered section, Mx= xL
x1
Ordinate of Influence line at mid span = 6.0 m 12
Leff= 24 m
5.1 6
.
Leff
700 kN3.6m
Bending Moment = 3885.0 kN-m
Bending Moment including impact and rection factor for outer Girder = 1736 kN-m
Bending Moment including impact and rection factor for Inner Girder = 1425 kN-m
Calculation of bending moment at 3L/8. = 9 m Leff= 24 m
Ordinate of Influence line at mid span = 5.625 m 9
Fig. 11a: ILD for BM at L/ 2RA B
4.5 4.95
5.63
Leff
Fig. 11b: ILD for BM at 3L/8RA R
700 kN3.6m
3*Leff / 8
en ng omen = . m
Bending Moment including impact and rection factor for Outer Girder = 829.3 kNm
Bending Moment including impact and rection factor for Inner Girder = 680.4 kNm
Calculation of bending moment at L/4. = 6 m Leff= 24 m
Ordinate of Influence line at mid span = 4.500 m 6
Leff
RB
700 kN3.6m
Leff / 4
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3.15
4.500 4.05
700 kN
Leff / 4
Pag
Bending Moment = kNm
Bending Moment including impact and rection factor for Outer Girder = kNm
Bending Moment including impact and rection factor for Inner Girder = kNm
Calculation of bending moment at L/8. = 3 m Leff= 24 m
Ordinate of Influence line at mid span = 2.625 m 3
1500.1
670.3460
550.0275
Fig. 11c: ILD for BM at L/4A
Leff
1.050 2.400
2.625
Bendin Moment = kNm881.2
Fig. 11d: ILD for BM at L/ 8RA RB
700 kN3.6m
Leff / 8
Bending Moment including impact and rection factor for Outer Girder = 393.792 kNm
Bending Moment including impact and rection factor for Inner Girder = 323.11 kNm
Bending Moment due to Live load: IRC Class A Loading
The influence line diagram for bending moment is shown in figure.
Effective span of girder, le = 24 m
Loads Values Unit Loads
W1 27 kN W5 68 kN
.
Values
W2 27 kN W6 68 kN
W3 114 kN W7 68 kN
W4 114 kN W8 68 kN
Distances Values Unit DistancesX Varies X5 4.3 m
X1 Varies X6 3 m
X2 1.1 m X7 3 m
Values
X3 3.2 m X8 3 mX4 1.2 m X9
Calculation of bending moment at L/2 = 12 m, when load W4 is at L/2
Ordinate of Influence line at mid span = 6 m Leff= 24 m
Ordinate of Bending Moment at considered section, Mx=
Varies
xL
x1
x1= 6.5
6.00
X= 12
Position from
Maximum
Load Values kN Moment ComponentLoad Nos. IL Ordinate
Fig. 12a: ILD for BM at L/2RA RB
W1=27kN W2=27kN W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN
Total = kN-m
Total Bending Moment including impact for Outer Girder = kN-m
57.80
68
10.30 0.850
114 5.400
114
68
W1 27 87.75
W2 102.60
-5.50
27
615.60
7.30
W5
W3
-4.40
3.250
3.800
1969.35
1840.11
684.00
261.80
W4 6.000
3.850
W7
W6
68
-1.20
0.00
4.30
2.350 159.80
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Total Bending Moment including impact for Inner Girder = kN-m
Calculation of bending moment at 3L/8 = 9.0 m, when load W3 is at 3L/8
Maximum Ordinate of Influence line = 5.625 m Leff= 24 m
1509.84
Pag
x1= 4.7
5.625
X= 9.0 Fig. 12b: ILD for BM at 3L/8RA RB
W1=27kN W2=27kN W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN W8=68kN
114
3.625
5.625
1.20
5.50
97.88
79.31
r nate
641.25
242.25
2.938
os t on rom
Maximum
-4.30
-3.20
0.00
oa os.
W1
W4 589.955.175
oa a ues
W3
W2 27
114
oment omponent
165.75
27
68
8.50
W5
W6 2.43868
3.563
Total = kN-m
Total Bending Moment including impact for Outer Girder = kN-m
Total Bending Moment including impact for Inner Girder = kN-m
W8
1918.39
1792.49
14.5068 12.750.188
.
1470.76
. .
Calculation of bending moment at L/4 = 6 m, when load W3 is at L/4
Maximum Ordinate of Influence line = 4.500 m Leff= 24 m
X1= 1.7
4.50
X= 6RA RB
W1=27kN W2=27kN W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN W8=68kN
-3.20 2.100 239.40
W3 114 0.00 4.500 513.00
Moment Component
W1 27 -4.30 1.275 34.43
Load Nos. Load Values kN Position from
Maximum
IL Ordinate
W2 114
.
Total = kN-mTotal Bending Moment including impact for Outer Girder = kN-m
Total Bending Moment including impact for Inner Girder = kN-m
W8 68 14.50 0.000 0.00
W6 68 8.50 2.375 161.50
W7 68 11.50 1.625 110.50
. . .
W5 68 5.50 3.125 212.50
1556.931454.75
1193.64
Calculation of bending moment at L/8 = 3 m, when load W4 is at L/8
Maximum Ordinate of Influence line = 2.625 m Leff= 24 m
2.625
X= 3
Fig. 12d: ILD for BM at L/ 8RA RB
W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN W8=68kN
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Load Nos. Load Values kN Position from
Maximum
IL Ordinate Moment Component
Pag
68 1.20 2.475
114 0.00
8.50 1.563
68 11.50
299.25
W1 27 0.00
0.000 0.00
114
W6
68 5.50 1.938
68
1.188
68 14.50 0.813
168.30
131.75
0.000 0.00
W2
W5
W3
W4
0.00 2.625
55.25
80.75
W8
106.25
W7
Total = kN-m
Total Bending Moment including impact for Outer Girder = kN-m
Total Bending Moment including impact for Inner Girder = kN-m
Absolute Maximum BM
841.55
786.32
645.19
Ordinate of Influence line at mid span = 6 m Leff= 24 m
6
x= 12
C.G of the Load system from outer 27 kN Wheel Load
Calculation of bending moment at the load point which is equidistance from resultant
Fig. 12e: ILD for BM at L/ 2 / Absolute Maximum BMRA RB
W1=27kN W2=27kN W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN
R
.
= 6.420 m
The heavier wheel load near C.G. of load System is 114kN which lies at a distance of
6.42-(1.1+3.2+1.2)= 0.92 m from CGX = 0.46 m
6.04
7.14 3.570 406.98
Load Values kN Position from Left
support
IL Ordinate
27 3.020
114W2
W1 81.54
Load Nos. Moment Component
Total = kN-m
Total Bending Moment including impact for Outer Girder = kN-m
392.36
277.44
. .
0.00 0.000 0.00
.114 10.34 5.170
2.580
68
68
68 11.54 5.770
68 15.84
68 18.84
0.000
W3 589.38
W4
W6 175.44
0.00
4.080
W7 21.84
1796.93
1923.14
W8
W5
Total Bending Moment including impact for Inner Girder = kN-m1474.41
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Shear Force due to Live load: IRC Class AA Tracked Vehicle.
At Support
Effective span of girder, le = 24 m
Load Class AA Tracked vehicle W1= 350 kN
Formaximum shear at support, load should be as near the support as possible.
The length of the load is 3.6m, the SF will be max. when the C. G. of the load is placed at a distance of 1/2*3.6=
1.8m From the support along its length, thus the load will lies between the support & the Ist Intermediate
X-girder, the width of track being 0.85m, the CG of load will thus lie at a distance of 1.2+0.85/2=1.625 m
from kerb of footpath Load act at a distance of 1.8 m from support A, B and C
L= 4.8 m C/C Distance of L Girder= 2.4 m
X= 1.8 X1= 3.0 a= 0.6 b= 1.025
C= 1.625 d= 2.05 f= 2.050 e= 0.35
g= 1.375 h= 0.675 i= 1.725
Loads on Girders,PA= 0.573 W1
kerb Line
'
L
PB= 1.146 W1PC= 0.281 W1
Reaction at support,RA= 0.358 W1RQ= 0.215 W1RB= 0.716 W1RR= 0.430 W1R =
A
B
CL of outer L Girder
CL of inner L Girder
c
C/C
dIstX-G
irder
ediateX-Girder
Q
R
Q
R
b
RR'
C/C
C/Cf
e
g
h
.RS= 0.105 W1
The loads on the cross girder i.e. RQ, RR& RS are to be distributed by normal Courbon's theory.
Total load, W = W1 = 0.750
C.G. of loads from Q = 2.05 m
kerb Line
CL of outer L GirderC
a
Inter
S SRS'
i
X X1
Fig. 13
Eccentricity, e = 0.35 m
Reaction factor for outer girder,FQ= 0.305 W1in this case xi=2.4 m
and xi =(2.4) +(0) +(2.4) = 2 x (2.4)
Reaction factor for inner girder,FR= 0.250 W1
in this case xi=0 m
RAdue to FQ= 0.24375 W1RBdue to FR= 0.200 W1 `
Total reaction on outer Girder = 0.602 W1Total reaction on inner Girder = 0.916 W1
Max shear at support including impact for outer girder = 231.7 kN
Max shear at su ort includin im act for inner irder = 352.7 kN
It may be seen that the reaction FQand FRact as load at 1/3 span of outer longitudinal girder and inner longitudinal
girder respectively. The reactions at support A and B due to those loads are
.
At Intermediate Section
Effective span of girder, le = 24.0 m
At Left =
At Right =
Ordinate of Bending SF at considered section, SFx
L
x1
L
x11
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Shear at 1/8th span
Calculation of bending moment at L/8 = 3 m, when load Placed at Just Right of L/8
Ordinate of Influence line at Left = 0.875 m
Pag
Ordinate of Influence line at Right = 0.125 m Leff= 24 m
x= 3 a'= 0.125 a= 0.875 b= 0.725
Fig. 14a: ILD of SF at L/8 of SpanRA RB
350 kN3.6m
a
a'
b
S.F. = 280 kN
S.F. including impact for outer girder = 125.13 kN
S.F. including impact for inner girder = 102.667 kN
Shear at 1/4th span
Calculation of bending moment at L/4 = 6 m, when load Placed at Just Right of L/4
Ordinate of Influence line at Left = 0.75 m
Ordinate of Influence line at Right = 0.25 m Leff= 24 m
x= 6 a'= 0.25 a= 0.750 b= 0.600
Fig. 14b ILD of SF at L/4 of SpanRA RB
350 kN3.6m
a
a'
b
S.F. = 236.25 kN
S.F. including impact for outer girder = 105.574 kN
S.F. including impact for inner girder = 86.625 kN
Shear at 3/8th span
Calculation of bending moment at 3L/8 = 9 m, when load Placed at Just Right of 3L/8
Ordinate of Influence line at Left = 0.625 mOrdinate of Influence line at Right = 0.375 m Leff= 24 m
x= 9 a'= 0.375 a= 0.625 b= 0.475
Fi . 14cILDofSFat3L/8 ofS anRA RB
350 kN3.6m
a
a'
b
S.F. = 192.5 kN
S.F. including impact for outer girder = 86.023 kN
S.F. including impact for inner girder = 70.583 kN
Shear at 1/2th spanCalculation of bending moment at L/2 = 12 m, when load Placed at Just Right of L/2
Ordinate of Influence line at Left = 0.5 m
Ordinate of Influence line at Ri ht = 0.5 m Leff= 24 m
.
x= 12 a'= 0.5 a= 0.5 b= 0.350
Fig. 14d ILD of SF at L/2 of SpanRA RB
350 kN3.6m
a
a'
b
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Shear Force due to Live load: IRC Class A Load.
The influence line diagram for shear force is shown in figure.
Effective span of girder, le = 24.0 m
Loads Values Loads Values
Pag
W1 27 kN W5 68 kN
W2 27 kN W6 68 kN
W3 114 kN W7 68 kN
W4 114 kN W8 68 kN
Distances Distances Values
X Varies X5 4.3 m
Values
X1 Varies X6 3 m
X2 1.1 m X7 3 mX3 3.2 m X8 3 mX4 1.2 m X9 Varies
At Left = At Right =
Ordinate of Bending SF at considered section, SFx
L
x1
L
x11
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Calculation of Shear Force at Support = 24.0 m, when load W3
Ordinate of Influence line y3= 1 Leff= 24.0 m
W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN W8=68kN
Pag
Load PositionLoad Nos. IL Ordinate Moment Component
Fig. 15a: ILD for SF at Support
RA RB
Y3 Y4Y5 Y6 Y7
Y8
a ues
kN
rom e
support
114 0 Y3=
114 1.20 Y4=
68 5.50 Y5=
68 8.50 Y6=
68 11.50 Y7=
68 14.50 Y8=
Total = kN
W4 108.300.950
W7
W6
1.000
380.97
52.420.771
0.646
0.521
0.396
43.92
W3 114.00
W5
35.42
W8 26.92
Total SF including impact for Outer Girder = kN
Total SF including impact for Inner Girder = kN
Calculation of SF at L/8 = 3.000 m, when load W3
Ordinate of Influence line At Right 0.875
At Left 0.125 Leff= 24.0 m x= 3
292.07
355.97
.
W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN W8=68kN
Load Position Moment ComponentLoad Nos. IL Ordinate
Fig. 15b: ILD for SF at L/8RA RB
Y3 Y4Y5 Y6 Y7
Y8
Values
kN
from Left
support
114 0 Y3=
114 1.20 Y4=
68 5.50 Y5=
68 8.50 Y6=
68 11.50 Y7=
68 14.50 Y8=
=
94.05
26.92W7
W5
W4
W8
W3
W6
43.92
0.875
0.825
0.646
0.521
0.396
35.42
99.75
0.271 18.42
Total SF including impact for Outer Girder = kN
Total SF including impact for Inner Girder = kN
=
.
280.36
230.04
. ,
Ordinate of Influence line At Right 0.75
At Left 0.25 Leff= 24.0 m x= 6.00
Fig. 15c: ILD for SF at L/4
Y3
Y3'
= = = = = = = =
Y4 Y5Y6 Y7
Y8Y2Y1
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Load
Values
Position
from Left/
Load Nos. IL Ordinate Moment Component
Pag
kN Right at X
27 4.3 Y1=
27 3.2 Y2=
114 0 Y3=
114 1.20 Y4=68 5.50 Y5=
68 8.50 Y6=
=
26.92
W5
W6
79.8035.42
W3
W4
W2 -3.15
W1 -0.071 -1.91
85.50
-0.117
0.750
0.7000.521
0.396
.
68 14.50 Y8=
Total = kN
Total SF including impact for Outer Girder = kN
Total SF including impact for Inner Girder = kN
Calculation of SF at 3L/8 = 9.000 m, when load W3
Ordinate of Influence line At Right 0.625
At Left 0.375 Leff= 24.0 m x= 9.0
250.90
234.44
0.146
.
9.92
192.36
W8
.
Fig. 15d: ILD for SF at 3L/8RA RB
Y3
Y3'
W1=27kN W2=27kN W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN W8=68kN
Y4 Y5Y6 Y7
Y8Y2Y1
Load
Values
kN
Position
from Left/
Right at X
27 4.3 Y1=
27 3.2 Y2=
114 0 Y3=
114 1.20 Y4=
68 5.50 Y5=
-6.53
71.25W3
-5.29-0.196
W4 65.55
Load Nos. IL Ordinate Moment Component
-0.242
0.625
0.575
W1
26.92W5 0.396
W2
.
68 8.50 Y6=
68 11.50 Y7=
68 14.50 Y8=
Total = kN
Total SF including impact for Outer Girder = kN
Total SF including impact for Inner Girder = kN
=
W6 18.42
W7 9.92
W8 1.42
181.65
169.73
139.27
..0.271
0.146
0.021
. ,
Ordinate of Influence line At Right 0.5
At Left 0.5 Leff= 24.0 m x= 12.0
Y3Y3'
W1=27kN W2=27kN W3=144kN W4=144kN W5=68kN W6=68kN
Y4 Y5Y6 Y7
Y2Y1
Load
Values
kN
Position
from Left/
Right at X
27 4.3 Y1=
27 3.2 Y2=
Load Nos. IL Ordinate Moment Component
W1
W2
-8.66
-9.90
-0.321
-0.367
Fig. 15e: ILD for SF at L/2
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114 0 Y3=
114 1.20 Y4=
68 5.50 Y5=
68 8.50 Y6=
W3 57.00
W5 18.42
W6
0.500
0.450
0.271
0.146
W4 51.30
9.92
Pag
.
68 11.50 Y7=Total = kN
Total SF including impact for Outer Girder = kN
Total SF including impact for Inner Girder = kN
Design of Section:
Total Design Bending Moments and Shear Forces for Outer Girder:
Section
119.49
Shear Forces kNBendin Moment kN-m
W7 1.420.021
111.65
91.61
Due to
DL
Due to LL
X = 0 0.00 0.00
X = L/8 1157.75 393.79
X = L/4 2064.76 670.35
X = 3L/8 2492.47 829.26
X = L/2 2654.71 1736.11 111.646
234.439
334.296
Due to DL
111.646
Due to LL
355.966
280.359
169.733
Total
1551.540
0.000
Total
218.768
115.528
437.536
4390.817 0.000
2735.104
3321.735
793.502
614.655
453.207
285.261
o a es gn en ng omen s an ear orces or nner r er:
Section
Due toDL
Due to LL
X = 0 0.00 0.000
X = L/8 1157.75 323.111
X = L/4 2064.76 550.028
X = 3L/8 2492.47 680.419 254.796
2614.79 411.128
3172.89 115.53 139.27
352.720.000
564.334
Shear Forces (kN)
Total Due to DL Due to LL
437.54
334.30
790.252
Bending Moment (kN-m)
1480.859 230.04
192.36218.77
Total
X = L/2 2654.71 1424.50
Design of Outer Girder:
Overall depth of beam, D = 2000 mm
Rib width, bw = 400 mm
Flange width of T-beam will be,
bf = bw + 1/5 x lo 2.5 m > 2.4 m
Therefore width of flange, bf = 2400 mm
-
91.60791.614079.21 0.00
- ,
d = 1830
Area of steel required,
Ast= 13420 mm2
Provide 16 nos. of 32 2000nos. of
mm bars+
mm bars
=
djst
M
.
Provided area of steel = 13843 mm2
Total Provided area of steel = 13843 mm2
Number of bars in bottom row = 4 nos.
Width of beam = 400 mm 400
Side and bottom clear cover to bars = 40 mm
C.G. of the bottom row of bars from bottom = 56 mm
Clear distance between vertical bars = 32 mm
Fig. 16 : Cross Section of Girde
C.G. of the Second row of bars from bottom = 120 mm
C.G. of the third row of bars from bottom = 184 mm
C.G. of the fourth row of bars from bottom = 248 mm
C.G. of the fifth row of bars from bottom = 308.5 mm
C.G. of the bar group from bottom = 163.1 mm
170 mm O.K.
Effective Depth = 1830 mm
Df = 220 mm
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Check for stresses:
Calculation of depth of neutral axis:
Assuming that the
effective area in of compression and tension sides about neutral axis, we get
Pag
bwxa2+ (bf bw) Df (xa Df/2) = m Ast (d xa)
Solving, we get, xa= 482 mm
Let compressive stress in concrete at top of flange = And compressive stress in concrete at bottom of flange = '
Then, = 0.543 =
a
fa
x
Dx
Position of C.G. of compressive stress in flange from top, x1
= 99.1 mm
Compressive force in flange, C1 = x( + ')x Bf x Df= 407402
Com ressive force in rib C2 = x 'x Xa Df x Bw
=+
+3
D
'
'2 f
,
= 28420
C.G. of compressive force in rib from top, x2
307.2 mm
Total compressive force, C = 435822.3
C.G. of total compressive force from top
112.7 mm=+
=
C2C1
x2C2x1C1
Therefore, lever arm, jd = 1717.3 mm
Critical Neutral axis depth, nd = 582.3 mm < xa
Moment of resistance of the section is given by
Mr = =
699626722.86
=+
cm1
d
st
+
yddb ff2
1
Equating Mr to external B.M we get
Mr =
= 6.28 < 8.3 O.K. Stress Developed in Steel Reinforcement is given by
t = = 197.6 < 200 O.K.
4390816575.00
a
a
x
xdm
Check for minimum area of steel
Minimum area of tension steel in beam @ 0.2 % of web area
1600 mm2 < Ast provided = mm
2O.K.
Design for shear:
. N/mm2
13843
=
=dB
Vv
Assuming ####### nos. of 32 will be continued up to support, then provided
percentage area of tension steel = 1.00 %
Permissible shear stress,c= 0.420 N/mm2 < v
Vs = V - tc . bw . D = 457195 N
Assuming 12 mm 2-legged vertical stirrups having area of steel, Asv = 226 mm2
Spacing, S = 181 mm c/c
Shear reinforcement is required. Shear reinforcement shall be provided to carry a shear of,
=
Vs
dAsv st
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As per minimum shear reinforcement requirements, maximum spacing,
S ax= 283 mm c/c.
Vs
=
st Asv
Pag
.
Hence provide 12 mm, 2-legged vertical stirrups @ 100 mm c/c at support.Design summary:
Tension Reinforcement (Fe 415):
No. Dia.
Area of Steel RequiredSection
Area of steel required and provided at different sections of Outer girder are given in below:
Area
BM Area of Steel Provided
wb0.4
2 25
3L/8 3321.73 KNm 14 32 mm2
L/4 2735.10 KNm 12 32 mm2
L/8 1551.54 KNm 10 32 mm2
Support 0.00 KNm 10 32 mm2
Shear Reinforcement (Fe 415):
10153
8038
11254
9646
8038mm2
mm20
mm2
mm28360
4742
4390.82 13420 mm2KNmL/2 13843 mm2
Section
L/2 111.646 kN 10 @ 1029.4 mm 10 @ 250
3L/8 285.261 kN 10 @ 402.87 mm 10 @ 250
L/4 453.207 kN 10 @ 253.58 mm 10 @ 200
L/8 614.655 kN 12 @ 269.24 mm 12 @ 150
Support 793.502 kN 12 @ 208.56 mm 12 @ 100
2-legged vertical Stirrups provided
mm
mm
mm
mm
SF 2-legged vertical Stirrups required
Area of steel required and provided at different sections of Inner girder are given in below:
mm
ens on e n orcemen e :
No. Dia.
16 32
2 25
3L/8 3172.89 KNm 14 32 mm2
L/4 2614.79 KNm 12 32 mm2
L/8 1480.86 KNm 10 32 mm2
11254
9646
8038
Section BM
7992 mm2
Area of Steel Required
4526 mm2
13843
9698 mm2
Area
Area of Steel Provided
L/2 4079 KNm 12468 mm2 mm2
uppor . m mm
Shear Reinforcement (Fe 415):
Section
L/2 91.607 kN 10 @ 150 mm 10 @ 250
3L/8 254.796 kN 10 @ 150 mm 10 @ 250
L/4 411.128 kN 10 @ 150 mm 10 @ 200
L/8 564.334 kN 12 @ 150 mm 12 @ 150
Support 790.252 kN 12 @ 100 mm 12 @ 100
SF
mm
mm
mm
mm
2-legged vertical Stirrups provided
mm
2-legged vertical Stirrups required
mm
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Pag
.
Dead Load
Overall Depth of cross girder = 1.5 m 2.4
Width of cross girder = 0.3 m
Self weight of cross girde = 9.216 kN/m 4.8
Dead load from slab = 20.2752 kNFig. 17
This load is assumed as uniformly distributed load per meter run = 8.448 kN/m
Total Dead load per meter run = 17.664 kN/m
Assuming, cross girder as rigid, reaction on main girder = 13.52 kN
Live Load: IRC Class AA Tracked Vehicle
Maximum bending moment occurs when one wheel of a vehicle lies near center of span.
Position for maximum bending moment is shown in figure. Deck Slab is assumed to
be simply supported. The critical supported between two cross girder.
=
l
lW
2/8.1
=
djst
M
Fig. 19
W1 W1
Effective load coming on cross girder = 569 kN=
l
lW
2/8.1
=
djst
M
Fig. 19
W1 W1
Reaction on each longitudinal girder = 189.58 kN
Maximum B.M. occurs under the load, = 260.68 kNm
Bending moment including impact = 286.74 kNm
Dead Load Bending Moment at the section, = 1.8876 kNm
Total bendin moment = 288.6 kNm
=
l
lW
2/8.1
=
djst
M
Fig. 19
W1 W1
.
LL Shear force including impact = 208.5 kN
Total shear force = 222.1 kN
Therefore, Design Moment = 288.6 kNm
Design Shear Force = 222.1 kN
=
l
lW
2/8.1
=
djst
M
Fig. 19
W1 W1
ross g r er s es gne as T-Beam.
Assuming effective depth, d = 1450.00 mm
Area of tension steel required = mm2
Minimum area of tension steel in beam @ 0.2 % of web area =
900 mm2 < 1113.4 mm2
Hence provide 3 nos. of 25 mm bars + 0 nos.of 20
bars having area of steel = 1472 mm2 .
1113.37
=
l
lW
2/8.1
=
djst
M
Fig. 19
W1 W1
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= dB
V
=
Vus
dAsvst
=
wb0.4
Asv0.87fy
Design for shear:
Nominal shear stress, v= 0.51 N/mm2< max= 1.9 N/mm
2 O.K.
Provided percentage area of tension steel, p = 0.33 %
Permissible shear stress,c= 0.275 N/mm2 < 0.51N/mm O.K.
Shear reinforcement is required. Shear reinforcement shall be provided to carry
a shear force of
= dB
V
=
Vus
dAsvst
=
wb0.4
Asv0.87fy
us = u - c . w . =
Assuming 10 mm 2-legged vertical stirrups having area of steel, Asv = 157 mm2
Spacing, S = 327 mm c/c
As per minimum shear reinforcement requirements, maximum spa 10 mm 2 legged
vertical stirrups, 472 mm
= dB
V
=
Vus
dAsvst
=
wb0.4
Asv0.87fy
Hence provide 10 mm 2-legged vertical stirrups @ 150 mm c/c through out the length of
end cross girder and 10 mm 2-legged vertical stirrups @ 150 mm c/c through out the length
Elastomeric Bearing On Bridge Used
According IRC 83-part II it is reccomended use of elastomeric bearins of size (250X400X50 ) mm embedding 5
of intermediate cross girder.
= dB
V
=
Vus
dAsvst
=
wb0.4
Asv0.87fy
p a es o mm c ness an mm c earance n p an = mm or er ca oa .
= dB
V
=
Vus
dAsvst
=
wb0.4
Asv0.87fy