Design of Super Structure

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Transcript of Design of Super Structure
A. Design of Superstructure
1.0 Design Data
Fig 1.BRIDGE CROSSSECTION
Page 1
1.1. Materials and its Properties:M25Fe415
Characteristics Strengthof Concrete fck = 25 MPaPermissible direct compressive stress, σc = 6.2 MPa
Permissible flexural compressive stress, σcbc = 8.3 MPa
Maximum Permissible shear stress, τmax ( 0.07*fck) = 1.75 MPa
Fig 1.BRIDGE CROSSSECTION
Maximum Permissible shear stress, τmax ( 0.07 fck) 1.75 MPa
Basic Permissible Stresses of Reinforcing Bars as per IRC : 211987, Section III:Permissible Flexural Tensile stress, σst = 200 MPa
Permissible direct compressive stress, σco = 170 MPa
Self weight of materials as per IRC : 62000:Concrete (cementReinforced) = 24 kN/m3
Fig 1.BRIDGE CROSSSECTION
Macadam (binder premix) = 22 kN/m3
1.2. Geometrical Properties:Effective Span of Bridge = 24.00 mTotal length of span = 24.56 mNumbers of span = 2Width of expantion Joint = 40 mmTotal length of Bridge = 49.2 m
Fig 1.BRIDGE CROSSSECTION
g gNos. of longitudinal Girder = 3Spacing of Girder = 2.4 mRib width of main girder = 400 mmOverall depth of main girder = 2000 mmDepth of kerb above deck slab = 225 mmNos. of cross girder = 6Spacing of cross girder = 4.8 mRib width of cross girder = 300 mm
Fig 1.BRIDGE CROSSSECTION
Rib width of cross girder = 300 mmOverall depth of cross girder = 1500 mmDeck slab thickness = 220 mmDeck slab thickness at edge = 150 mmThickness of wearing coat = 80 mmFillet size (horizontal) = 150 mmFillet size (vertical) = 150 mm
Bridge Width:
Fig 1.BRIDGE CROSSSECTION
Carriageway width = 6 mFootpath width = 0.45 mKerb width Outer = 0.15 mKerb width Inner = 0 mTotal Width of Deck Slab = 7.2 mTotal depth of Kerb Outer = 0.375 mTotal depth of Kerb Inner = 0 m
Fig 1.BRIDGE CROSSSECTIONFig 1.BRIDGE CROSSSECTION
Page 1
2.0 Design of Slabh il l b i d i d b i id h h d
cbcσ3280
Fig 1.BRIDGE CROSSSECTION
57KN 350KN 37.5KN
IRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
Page 2
2.1 Design of Cantilever slab: The cantilever slab is designed by effective width method.
300 mm at junction with rib150 mm at free end0.5 kN/m (assumed)
Impact factor = 54 % (for IRC class A loading)
25 % (for IRC class AA loading)
Thickness of slab =
Self weight of Railing =
cbcσ3280
Fig 1.BRIDGE CROSSSECTION
57KN 350KN 37.5KN
IRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
( g)Dead Load Bending Moment and Shear Force:
S.No. Item Width Depth Unit Wt
1 Railing/Parapet
0.5 kN 1.5250.100= 0.9 m 0.45 kN.m
2 Kerb (outer) 0.2 0.225 24 1.08 kN 1.5250.100= 0.9 m 0.97 kN.m3 Kerb (inner) 0 0 24 0 kN 0.15+0.175/2= 0.2375 m 0 kN.m4 Wearing 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
cbcσ3280
Fig 1.BRIDGE CROSSSECTION
57KN 350KN 37.5KN
IRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
4 Wearing Coat 0.4 0.08 22 0.704 kN 0. 50/ 0.075 m 0.05 kN.m5 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.mTotal kN kN.m
= 7.684 kN= 3.875 kN.m
7.684Dead load Shear force at theface of rib
Dead load Bending Moment at the face of rib
3.875
cbcσ3280
Fig 1.BRIDGE CROSSSECTION
57KN 350KN 37.5KN
IRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
Live Load Bending Moment and Shear Force:
cbcσ3280
Fig 1.BRIDGE CROSSSECTION
57KN 350KN 37.5KN
IRC Class AA Track Loading IRC 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+ bwHere
X= 0.125 mbw= 0.41 m
cbcσ3280
Fig 1.BRIDGE CROSSSECTION
57KN 350KN 37.5KN
IRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
bw= 0.41 mHence
be= 0.56 mIRC Class A Loading Load = 28.5 kNLive Load per m width including impact = 76.339 kNMaximum Moment due to live load = 9.5424 kNmAverage thickness of cantilever slab = 225 mmTaking pedestrain load (LL) = 5.0 kN/m2
Eff ti idth f l b 0 45
cbcσ3280
Fig 1.BRIDGE CROSSSECTION
57KN 350KN 37.5KN
IRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
Effective width of slab = 0.45 mCantilever length of slab = 1 mMaximum Bending moment = 1.406 kN.mShear force at the face of slab = 2.250 kNTotal Design Shear Force = 86.3 kNTotal Design Bending Moment = 14.82 kN.m
Design of Section:Modular Ratio, m = = 11.245
cbcσ3280
Fig 1.BRIDGE CROSSSECTION
57KN 350KN 37.5KN
IRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
Page 2
Neutral axis factor, k = = 0.3182
Lever arm factor, j = = 0.8939
cbcσ3280
stσcbcm σcbcm σ
+
31 k
−
stjk σ×××21
RbM
djM
st ..σ=
djM
st ..σ=
bddtanM
Vβ
τ
×
=v
coc kk ττ .. 21=5.07.014.11 ≥×−= dk
125.05.02 ≥+= ρk bdA s=ρ
≥
=2k
coτ
4.021 ××= KKcτ
vτ
Page 3
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.
cbcσ3280
stσcbcm σcbcm σ
+
31 k
−
stjk σ×××21
RbM
djM
st ..σ=
djM
st ..σ=
bddtanM
Vβ
τ
×
=v
coc kk ττ .. 21=5.07.014.11 ≥×−= dk
125.05.02 ≥+= ρk bdA s=ρ
≥
=2k
coτ
4.021 ××= KKcτ
vτ
p , p q
Area of steel required, Ast = 326.42 mm2
Provide φ 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 deadload moment.
mm c/c, giving area of steel =
cbcσ3280
stσcbcm σcbcm σ
+
31 k
−
stjk σ×××21
RbM
djM
st ..σ=
djM
st ..σ=
bddtanM
Vβ
τ
×
=v
coc kk ττ .. 21=5.07.014.11 ≥×−= dk
125.05.02 ≥+= ρk bdA s=ρ
≥
=2k
coτ
4.021 ××= KKcτ
vτ
load moment.Moment = 4.06 kN.m Effective depth 244 mm
Area of steel required, Ast = 93.057 mm2
Half reinforcement is to be provided at top and half at bottom.Provide ø 10 mm bars 200 mm c/c at both top and bottom, giving area of
t l 392 5 mm2 > required OK
cbcσ3280
stσcbcm σcbcm σ
+
31 k
−
stjk σ×××21
RbM
djM
st ..σ=
djM
st ..σ=
bddtanM
Vβ
τ
×
=v
coc kk ττ .. 21=5.07.014.11 ≥×−= dk
125.05.02 ≥+= ρk bdA s=ρ
≥
=2k
coτ
4.021 ××= KKcτ
vτ
steel = 392.5 mm2 > required, OK.
Check for min. area of Steel:Min. area of steel @ 0.12 % = 360 mm2 < Provided. O.K.
Design for Shear:Dead load shear = 7.68 kNLive load Shear = 2.250 kN
cbcσ3280
stσcbcm σcbcm σ
+
31 k
−
stjk σ×××21
RbM
djM
st ..σ=
djM
st ..σ=
bddtanM
Vβ
τ
×
=v
coc kk ττ .. 21=5.07.014.11 ≥×−= dk
125.05.02 ≥+= ρk bdA s=ρ
≥
=2k
coτ
4.021 ××= KKcτ
vτ
Total = 9.93 kNtanβ = 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
cbcσ3280
stσcbcm σcbcm σ
+
31 k
−
stjk σ×××21
RbM
djM
st ..σ=
djM
st ..σ=
bddtanM
Vβ
τ
×
=v
coc kk ττ .. 21=5.07.014.11 ≥×−= dk
125.05.02 ≥+= ρk bdA s=ρ
≥
=2k
coτ
4.021 ××= KKcτ
vτ
( d being in m) = 0.986
(where )
= 0.500446 1.00
Ad t
cbcσ3280
stσcbcm σcbcm σ
+
31 k
−
stjk σ×××21
RbM
djM
st ..σ=
djM
st ..σ=
bddtanM
Vβ
τ
×
=v
coc kk ττ .. 21=5.07.014.11 ≥×−= dk
125.05.02 ≥+= ρk bdA s=ρ
≥
=2k
coτ
4.021 ××= KKcτ
vτ
Adopt 1
Value ot = for M25 grade of concrete from code = 0.4 N/mm2
Allowable shear stress
= 0.3944 N/mm2 > , Hence Safe
cbcσ3280
stσcbcm σcbcm σ
+
31 k
−
stjk σ×××21
RbM
djM
st ..σ=
djM
st ..σ=
bddtanM
Vβ
τ
×
=v
coc kk ττ .. 21=5.07.014.11 ≥×−= dk
125.05.02 ≥+= ρk bdA s=ρ
≥
=2k
coτ
4.021 ××= KKcτ
vτ
Page 3
2.2 Design of Interior Panels: The slab panel is designed by Pigeaud’s method.
D
Page 4
B C
Short span of slab, Bs = 2 mLong span of slab, Ls = 4.5 m
AFig : 3 Bridge Plan
Calculation of Bending momentsa) Due to Dead load:
Self weight of wearing coat = 1.76 kN/m2
Self weight of deck slab = 5.28 kN/m2
Total = 7.04 kN/m2
Since the slab is supported on all four sides and is continuous, Piegaud’s curves are used to calculate bending moments.Ratio k = Bs/Ls = 0 44Ratio, k = Bs/Ls = 0.44As the panel is loaded with UDL,
u/Bs = 1v/Ls = 1
Where, u & v are the dimensions of the loaded area.From the Pigeaud’s curve,
m1 = 0.0457m2 = 0.0086
Total dead load W = 63.36 kNMoment along short span, M1 = W (m1 +0.15m2) = 2.98 kNmMoment along long span, M2 = W (0.15m1 +m2) = 0.98 kNmConsidering effects of continuity, 0.8Moment along short span, M1 = 2.38 kNmMoment along long span, M2 = 0.78 kNm
b) Due to Live load: Class AA Tracked Vehicleb) Due to Live load: Class AA Tracked VehicleFor maximum bending moment one wheel is placed at the center of panel.Tyre contact length along short span, x = 0.85 mTyre contact length along long span, y = 3.6 m
Loaded length, u = 1.034 mLoaded width, v = 3.766 m
Wheel load, W = 350 kNRatio, k = Bs/Ls = 0.44
/B 0 517 Nu/Bs = 0.517v/Ls = 0.837
From the Pigeaud’s curve,m1 = 0.0813m2 = 0.0147
Moment along short span,= 29.227 kNm
Moment along long span,= 9.413 kNm
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
W1=
350k
N
Fig: 4
Page 4
Bending moment including impact and continuity,M1 = 29.227 kNmM2 = 9.413 kNm Fig: 4
Page 5
c) Due to Live load: Class AA Wheeled VehicleCaseI: 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 mTyre contact length (along long span), = 0.15 mDisperced width along short span, u = 0.510 mDisperced width along long span, V = 0.380 m
K = 0 44K = 0.44
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
W6Y
Ratio, k Bs/Ls 0.44u/Bs = 0.255v/Ls = 0.084
From the Pigeaud’s curve,m1 = 0.1965m2 = 0.1383
Moment along short span,= 13.578 kNm
Moment along long spanM1= W(m1+0.15m2)
Moment along long span,= 10.486 kNm
Bending moment including impact and continuity,M1 = 13.578 kNmM2 = 10.486 kNm
Bending moment due to load W2: 62.5 kN Wheel load is placed unsymmetrical wrt the XX
Intensity of loading, q = 322.45 kN/m2
M2= W(0.15m1+m2)
Considering loaded area 2.000 x 0.380 mLoaded area = 0.760 m2
Total applied load = q x area = 245 kNRatio, k = Bs/Ls = 0.44
u/Bs = 1.000v/Ls = 0.084
From the Pigeaud’s curve,From the Pigeaud s curve,m1 = 0.0935m2 = 0.0742
Moment along short span,= 25.650 kNm
Moment along long span,= 21.628 kNm
Bending moment including impact and continuity,M1 = 25.650 kNm
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
Page 5
M2 = 21.628 kNmNext, Consider the area between the real and the dummy load i.e., 1.490 m X 0.380 m
Loaded area = 0.566 m2Total applied load = q x area = 183 kN
Page 6
Total applied load q x area 183 kNRatio, k = Bs/Ls = 0.44
u/Bs = 0.745v/Ls = 0.084
From the Pigeaud’s curve,m1 = 0.1157m2 = 0.0944
Moment along short span,= 23 718 kNmM1= W(m1+0 15m2) = 23.718 kNm
Moment along long span,= 20.412 kNm
Bending moment including impact and continuity,M1 = 23.718 kNmM2 = 20.412 kNm
Final MomentM1 = 0.966 kNmM2
M2= W(0.15m1+m2)
M1= W(m1+0.15m2)
M2 = 0.608 kNmBending moment due to load W3: 37.5 kN Wheel load is placed unsymmetrical wrt the XX
Intensity of loading, q = 193.47 kN/m2Considering loaded area 1.710 x 0.380 m
Loaded area = 0.650 m2Total applied load = q x area = 126 kN
Ratio, k = Bs/Ls = 0.44u/Bs = 0.855u/Bs 0.855v/Ls = 0.084
From the Pigeaud’s curve,m1 = 0.1036m2 = 0.08325
Moment along short span,= 14.598 kNm
Moment along long span,= 12 423 kNm
M1= W(m1+0.15m2)
M2= W(0 15m1+m2) = 12.423 kNmBending moment including impact and continuity,
M1 = 14.598 kNmM2 = 12.423 kNm
Next, Consider the area between the real and the dummy load i.e., 0.69 m X 0.380 mLoaded area = 0.262 m2
Total applied load = q x area = 51 kNRatio, k = Bs/Ls = 0.44
/B 0 345
M2= W(0.15m1+m2)
u/Bs = 0.345v/Ls = 0.084
From the Pigeaud’s curve,m1 = 0.1765m2 = 0.1312
= 9.957 kNmMoment along long span,
= 8.002 kNm
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)Bending moment including impact and continuity,
M1 = 9.957 kNmM2 = 8.002 kNm
Final MomentM1 = 2.321 kNmM2 = 2.210 kNm
Bending moment due to load W4: 62.5 kN Wheel load is placed unsymmetrical wrt the YY Intensity of loading, q = 322.45 kN/m2
( )
Page 6
Considering loaded area 2.780 x 0.510 mLoaded area = 1.418 m2
Total applied load = q x area = 457 kNRatio, k = Bs/Ls = 0.44
Page 7
Ratio, k Bs/Ls 0.44u/Bs = 0.618v/Ls = 0.255
From the Pigeaud’s curve,m1 = 0.1168m2 = 0.0648
Moment along short span,= 57.832 kNm
Moment along long spanM1= W(m1+0.15m2)
Moment along long span,= 37.628 kNm
Bending moment including impact and continuity,M1 = 57.832 kNmM2 = 37.628 kNm
Next, Consider the area between the real and the dummy load i.e., 2.02 m X 0.510 mLoaded area = 1.030 m2
Total applied load = q x area = 332 kNR i k B /L 0 44
M2= W(0.15m1+m2)
Ratio, k = Bs/Ls = 0.44u/Bs = 0.449v/Ls = 0.255
From the Pigeaud’s curve,m1 = 0.1353m2 = 0.0712
Moment along short span,= 48.480 kNmM1= W(m1+0.15m2)
Moment along long span,= 30.386 kNm
Bending moment including impact and continuity,M1 = 48.480 kNmM2 = 30.386 kNm
Final MomentM1 = 4.676 kNmM2 = 3 621 kNm
( )
M2= W(0.15m1+m2)
M2 = 3.621 kNmBending moment due to load W5:
Wheel load is placed unsymmetrical wrt the Both XX and YY For X X AxisW = 62.5 kN
Intensity of loading, q = 322.45 kN/m2Considering loaded area 2.000 x 0.380 m
Loaded area = 0.760 m2Total applied load = q x area = 245 kN
Ratio, k = Bs/Ls = 0.44u/Bs = 1.000v/Ls = 0.084
From the Pigeaud’s curve,m1 = 0.0935m2 = 0.0742
Moment along short span,Moment along short span,= 25.650 kNm
Moment along long span,= 21.628 kNm
Bending moment including impact and continuity,M1 = 25.650 kNmM2 = 21.628 kNm
Next, Consider the area between the real and the dummy load i.e., 1.490 m X 0.380 mLoaded area = 0.566 m2
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
Page 7
Total applied load = q x area = 183 kNRatio, k = Bs/Ls = 0.44
u/Bs = 0.745v/Ls = 0.084
Page 8
v/Ls 0.084From the Pigeaud’s curve,
m1 = 0.1157m2 = 0.0944
Moment along short span,= 23.718 kNm
Moment along long span,= 20.412 kNm
Bending moment including impact and continuityM2= W(0.15m1+m2)
M1= W(m1+0.15m2)
Bending moment including impact and continuity,M1 = 23.718 kNmM2 = 20.412 kNm
Moment Along XXM1 = 0.966 kNmM2 = 0.608 kNm
For Y Y AxisW = 62.5 kN
Intensity of loading, q = 322.45 kN/m2Considering loaded area 2.780 x 0.510 m
Loaded area = 1.418 m2Total applied load = q x area = 457 kN
Ratio, k = Bs/Ls = 0.44 u/Bs = 0.618v/Ls = 0.255
From the Pigeaud’s curve,From the Pigeaud s curve,m1 = 0.1168m2 = 0.0648
Moment along short span,= 57.832 kNm
Moment along long span,= 37.628 kNm
Bending moment including impact and continuity,M1 = 57 832 kN m
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
M1 = 57.832 kNmM2 = 37.628 kNm
Next, Consider the area between the real and the dummy load i.e., 2.020 m X 0.51 mLoaded area = 1.030 m2
Total applied load = q x area = 332 kNRatio, k = Bs/Ls = 0.44
u/Bs = 0.449v/Ls = 0.255
From the Pigeaud’s curve,m1 = 0.1353m2 = 0.0712
Moment along short span,= 48.480 kNm
Moment along long span,= 30.386 kNm
Bending moment including impact and continuity,
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)Bending moment including impact and continuity,
M1 = 48.480 kNmM2 = 30.386 kNm
Final MomentM1 = 4.676 kNmM2 = 3.621 kNm
Resultent MomentM1 = 5.642 kNmM2 = 4.230 kNm
Page 8
Bending moment due to load W6:Wheel load is placed unsymmetrical wrt the Both XX and YY
For XX Axis
Page 9
For X X AxisW = 37.5 kN
Intensity of loading, q = 193.47 kN/m2
Considering loaded area 1.710 x 0.380 mLoaded area = 0.650 m2
Total applied load = q x area = 126 kNRatio, k = Bs/Ls = 0.44
u/Bs = 0 855u/Bs = 0.855v/Ls = 0.084
From the Pigeaud’s curve,m1 = 0.1036m2 = 0.08325
Moment along short span,= 14.598 kNm
Moment along long span,12 423M2 W(0 15 1 2)
M1= W(m1+0.15m2)
= 12.423 kNmBending moment including impact and continuity,
M1 = 14.598 kNmM2 = 12.423 kNm
Next, Consider the area between the real and the dummy load i.e., 0.69 m X 0.380 mLoaded area = 0.262 m2
Total applied load = q x area = 51 kNRatio, k = Bs/Ls = 0.44
M2= W(0.15m1+m2)
Ratio, k Bs/Ls 0.44u/Bs = 0.345v/Ls = 0.084
From the Pigeaud’s curve,m1 = 0.1765m2 = 0.1312
Moment along short span,= 9.957 kNm
Moment along long spanM1= W(m1+0.15m2)
Moment along long span,= 8.002 kNm
Bending moment including impact and continuity,M1 = 9.957 kNmM2 = 8.002 kNm
Moment Along YYM1 = 2.321 kNmM2 = 2.210 kNm
M2= W(0.15m1+m2)
For Y Y AxisW = 37.5 kN
Intensity of loading, q = 193.47 kN/m2Considering 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 = 0.618u/Bs 0.618v/Ls = 0.255
From the Pigeaud’s curve,m1 = 0.1168m2 = 0.0648
Moment along short span,= 34.699 kNm
Moment along long span,= 22.577 kNm
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
Page 9
Bending moment including impact and continuity,M1 = 34.699 kNmM2 = 22.577 kNm
Next, Consider the area between the real and the dummy load i.e., 2.020 m X 0.510 m
Page 10
Next, Consider the area between the real and the dummy load i.e., 2.020 m X 0.510 mLoaded area = 1.030 m2
Total applied load = q x area = 199 kNRatio, k = Bs/Ls = 0.44
u/Bs = 0.449v/Ls = 0.255
From the Pigeaud’s curve,m1 = 0.1353m2 = 0 0712m2 = 0.0712
Moment along short span,= 29.088 kNm
Moment along long span,= 18.231 kNm
Bending moment including impact and continuity,M1 = 29.088 kNmM2 = 18.231 kNm
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
Final MomentM1 = 2.806 kNmM2 = 2.173 kNm
Resultent MomentM1 = 5.127M2 = 4.383
Total Moment Due to IRC Class AA Wheeled Vechicle Moment along short span,M1 = 32.309 kNmMoment along short span,M1 32.309 kN mMoment along long span,M2 = 25.539 kNm
c) Due to Live load: Class A LoadingIRC Class A Loading: For maximum bending moment one wheel of 57kN should be placed at thecentre 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 mTyre 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 to loads W2 & W3 Bending moment at center of panel will be that due to load W1 and half.Due to loads W2 & W3 Bending moment at center of panel will be that due to load W1 and half.
Page 10
Y
Page 11
57kN
W1 W2
Y
X X57kN
Fig: 6
Bending moment due to load W1:Wheel load, W1 = 57 kN
Loaded length, u = 0.696 mLoaded width, v = 0.465 mRatio, k = Bs/Ls = 0.44
u/Bs = 0.35v/Ls = 0.10
From the Pigeaud’s curve,From the Pigeaud s curve,m1 = 0.1717m2 = 0.1245
Moment along short span,= 10.851 kNm
Moment along long span,= 8.565 kNm
Bending moment including impact and continuity,M1 = 10 851 kN m
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
M1 = 10.851 kNmM2 = 8.565 kNm
Bending moment due to load W2:Wheel load is placed unsymmetrical wrt the YY Wheel load, W2 = 57 kN
Intensity of loading, q = 176.09 kN/m2
Considering loaded area 2.865 x 0.696 mLoaded area = 1.993 m2
Total applied load = q x area = 351 kNRatio, k = Bs/Ls = 0.44
u/Bs = 0.637v/Ls = 0.348
From the Pigeaud’s curve,m1 = 0.10843m1 = 0.10843m2 = 0.0497
Moment along short span,= 40.676 kNm
Moment along long span,= 23.154 kNm
Bending moment including impact and continuity,M1 = 40.676 kNmM2 = 23 154 kN m
M2= W(0.15m1+m2)
M1= W(m1+0.15m2)
M2 = 23.154 kNmNext, Consider the area between the real and the dummy load i.e., 1.935 m X 0.696 m
Loaded area = 1.346 m2Total applied load = q x area = 237 kN
Ratio, k = Bs/Ls = 0.44u/Bs = 0.430v/Ls = 0.348
From the Pigeaud’s curve,m1 = 0.1313
Page 11
m2 = 0.0552Moment along short span,
= 33.081 kNmMoment along long span,
M1= W(m1+0.15m2)
Page 12
= 17.751 kNmBending moment including impact and continuity,
M1 = 33.081 kNmM2 = 17.751 kNm
Final MomentM1 = 3.798 kNmM2 = 2.702 kNm
M2= W(0.15m1+m2)
Total Bending Moment due to load W1 & W2 will be,M1 = 14.6 kNmM2 = 11.3 kNm
Design Bending Moment due to LL:M1 = 32.3 kNmM2 = 25.5 kNm
Calculation of Shear Force
a) Due to Dead load:Dead load shear force = 7.04 kN
b) Due to Live load: Class AA Tracked VehicleLoad of Tracked Vehicle= 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) = 2.25a for continuous slab = 2.60
x = 0.725 mbw = 3.76
Therefore effective width of slab = 4.96 m
bwlx1αx +
−
Load per meter width = 70.54 kNShear force at left edge = 44.97 kNShear force including impact & continuity = 44.97 kN
c) Due to Live load: Class AA Wheeled Vehicle
37.5kN 37.5kN62.5kN 62.5kNPage 12
37.5kN 37.5kN62.5kN 62.5kN
Page 13
Fig 7 bDispersion 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
Fig 7.b
Wheel Load = 62.50 kNBut 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 mLoad per meter width of slab = 51.72 kN
For second wheel Wheel Load = 62.50 kNx = 0.550 mx 0.550 m
Effective width for second wheel = 1.347 mBut 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 mLoad per meter width of slab = 49.1 kN
For third wheel Wheel Load = 37.50 kNX = 0.050 m
Effective width for third wheel = 0 400Effective width for third wheel = 0.400Load Acting on Span = 16.67 kNActing at 0.2 m from SupportEffective width for third wheel = 0.918 m < 1.2 mLoad per meter width of slab = 18 kNShear force at left edge = 37.3 kNShear force including impact and continuity
= 37.318 kN
b) Due to Live load: IRC Class A loading
Fig 7.c
57kN 57kN
Page 13
Shear force due to load W1: 57 kNFig 7.c
Page 14
Dispersion width in the direction of short span = 1.1 mFor maximum shear force, the load should be placed at distance of 0.55 m from webof girder. In this position second load will be as shown.
Effective width for first wheel =
Where, x= 0.55 mbw= 0.41 m
bwlx1αx +
−
bw 0.41 mL/B= 2.3 α = 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 = mLoad W1 = kN
Load per meter width of slab = kN/mAnd shear force = kN
Shear force including impact & continuity = kN
57.00
32 30
43.0732.30
1.32
Shear force including impact & continuity = kN
Shear force due to load W2:
Effective width for second wheel =
Where, x= 0.350 mbw= 0.41 m
32.30
bwlx1αx +
−
L/B= 2.4 α = 2.6Therefore, Effective width = 1.161 m < 1.2 m
Load W2 = 57.0 kNEffective load = 18.1 kN
15.6 kNAnd shear force = 12.89 kNShear force including impact & continuity = 12.89 kN
Total shear force = 45.19 kN
Load per meter width of slab =
Design of Section:Design of Section:Total Design Bending Moments,
M1 = 34.69 kNmM2 = 26.32 kNm
Total Design Shear force,S.F. = 52.23 kN
Effective depth required,
d 171 4Md = 171.4 mm
Use 30 mm Clear cover & 12 mm diaEffective depth available = 184 mm O.K.
Area of steel required along short span,
Ast = 1055 mm2
Check for minimum area of steel:
=RbM
=×× djst
Mσ
Page 14
264 mm2Provide ø 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 @
Page 15
Area of steel required along long span,
Ast = 851 mm2
Provide ø 12 mm bars 100 mm c/c at both top and bottom, giving area of steel = 1130.4 mm2 > required.
Check for shear:
=×× djst
Mσ
Check for shear:
Nomin τv = 0.284 N/mm2
Provided percentage area of tensile steel = 0.51 %Permissible shear stress, Κ×τc = 1.16 x 0.313 = 0.363 N/mm2 >
O.K.
=× db
Vu
3.0 Design of Longitudinal GirderEffective Span of Bridge = 24 m
Slab thickness = 0.22 mWidth of Rib = 0.4 m
Spacing of main Beam = 2.4 m c/cOver 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/mDead load due to deckDead load due to deck
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/mThis 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 mlet its width be = 0.3 m
Weight of rib of cross girder = 9.216 kN/mLength of each cross girder = 4 m
It is assumed that the weight of each cross girder is equally borne by the entire three longitudinal girders. This weight acts as point load on each girder its value being
= 12.29 kN
Total UDL = 34.413333 kN/mRA= RB = 449.824 kN
Bending Moment (BM)
12.3 kN
Fig : 8RA RB
34.413 kN/m
12.3 kN12.3 kN12.3 kN3 kN 12.3 kN
Page 15
BM at Centre of span = 2654.707 kNmBM at ¼ th of span = 2064.758 kNmBM at 3/8th Span = 2492.474 kNmBM t 1/8th Span 1157 748 kN
Page 16
BM at 1/8th Span = 1157.748 kNm
Shear Force (SF)SF at Support = 437.536 kN SF at 1/8th Span = 334.296 kNSF at 1/4th Span = 218.768 kNSF at 3/8th Span = 115.528 kNSF at Center of span = 0 kNSF at Center of span 0 kN
Distance from Support BM SFAt Center of Span 2654.71 0.00At 3/8th Span 2492.47 115.53At 1/4th Span 2064.76 218.77At 1/4th Span 2064.76 218.77At Support 0.00 437.54
3.2 Calculation of live load moment and shear force on longitudinal girder:
Impact factor for:0.150
IRC Class AA Tracked Vehicle = 0 100
=+ L6
5.4
IRC Class AA Tracked Vehicle = 0.100
IRC Class AA Wheeled Vehicle = 0.215
Distribution of live loads on longitudinal girder for bending moment:IRC Class AA Tracked Vehicle:
All th i d d t h th t f i ti
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.
Fig 9
W1 W1
Reaction factor for Outer Girder, RA = 0.81 W1
Reaction factor for Inner Girder, RB = 0.667 W1
( )=
××
×+ 35.04.2
2.42II31
3W2
21
=3W2 1
Fig. 9
If W be the axel load, then wheel load W1 = W/2Then reaction factor, RA = 0.406 W
RB = 0.333 W
IRC Class A Loading:Position of loads for maximum eccentricity is shown in figure.
3
W1 W1 W1 W1Page 16
W1 W1 W1 W1
Page 17
Reaction on Outer Girder RA,
RA= 1.625 W2( )
=
××
×+ 1.04.2
2.42II31
3W4
21
Fig. 10
RB = 1.333 W2
If W be the axel load, then wheel load W2 = W/2Then reaction factor, RA = 0.813 W
RB = 0.667 W
Bending Moment due to Live load: IRC Class AA Tracked Vehicle
[ ] =+ 013
4W 1
g
The influence line diagram for bending moment is shown in figure.Effective span of girder, le = 24.0 mITC Class AA Tracked Vehicle Load: = 700 kN
Ordinate of Bending Moment at considered section, Mx =
Calculation of bending moment at L/2.
xLx1
−
Ordinate of Influence line at mid span = 6.0 m 12
Leff= 24 m
5.1 6
Calculation of bending moment at L/2.
Leff
RB
700 kN3.6m
Bending Moment = 3885.0 kNmBending Moment including impact and rection factor for outer Girder = 1736 kNmBending Moment including impact and rection factor for Inner Girder = 1425 kNm
Calculation of bending moment at 3L/8. = 9 m Leff= 24 mOrdinate of Influence line at mid span = 5.625 m 9
Fig. 11a: ILD for BM at L/ 2RA RB
4.5 4.955.63
Bending Moment = 1855 7 kNm
Leff
Fig. 11b: ILD for BM at 3L/ 8RA RB
700 kN3.6m
3*Leff / 8
Bending Moment = 1855.7 kNmBending Moment including impact and rection factor for Outer Girder = 829.3 kNmBending Moment including impact and rection factor for Inner Girder = 680.4 kNm
Calculation of bending moment at L/4. = 6 m Leff= 24 mOrdinate of Influence line at mid span = 4.500 m 6
Leff
RA RB
700 kN3.6m
Leff / 4
Page 17
3.154.500 4.05
R RB
700 kN
Leff / 4
Page 18
Bending Moment = kNmBending Moment including impact and rection factor for Outer Girder = kNmBending Moment including impact and rection factor for Inner Girder = kNm
Calculation of bending moment at L/8. = 3 m Leff= 24 mOrdinate of Influence line at mid span = 2.625 m 3
1500.1670.3460550.0275
Fig. 11c: ILD for BM at L/ 4RA RB
Leff
1.050 2.4002.625
Bending Moment = kNm881.2
Leff
Fig. 11d: ILD for BM at L/ 8RA RB
700 kN3.6m
Leff / 8
Bending Moment = kNmBending Moment including impact and rection factor for Outer Girder = 393.792 kNmBending Moment including impact and rection factor for Inner Girder = 323.11 kNm
Bending Moment due to Live load: IRC Class A LoadingThe influence line diagram for bending moment is shown in figure.Effective span of girder, le = 24 m
Loads Values Unit LoadsW1 27 kN W5 68 kNW2 27 kN W6 68 kN
881.2
Values
W2 27 kN W6 68 kNW3 114 kN W7 68 kNW4 114 kN W8 68 kN
Distances Values Unit DistancesX Varies X5 4.3 mX1 Varies X6 3 mX2 1.1 m X7 3 m
Values
1.1 3 mX3 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/2Ordinate of Influence line at mid span = 6 m Leff= 24 m
Ordinate of Bending Moment at considered section, Mx =
Varies
xLx1
−
x1= 6.5
6.00X= 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 = kNmTotal Bending Moment including impact for Outer Girder = kNm
57.8068
10.30 0.850
114 5.40011468
W1 27 87.75W2 102.60
5.5027
615.60
7.30W5
W34.40
3.2503.800
1969.351840.11
684.00261.80
W4 6.0003.850
W7W6
68
1.200.004.30
2.350 159.80
Page 18
Total Bending Moment including impact for Inner Girder = kNm
Calculation of bending moment at 3L/8 = 9.0 m, when load W3 is at 3L/8Maximum Ordinate of Influence line = 5.625 m Leff= 24 m
1509.84
Page 19
x1= 4.7
5.625X= 9.0
IL OrdinatePosition fromLoad Nos Load Values kN Moment Component
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
3.6255.625
1.205.50
97.8879.31
IL Ordinate
641.25
242.25
2.938
Position from Maximum
4.30
89 25
3.200.00
Load Nos.
W1
W4 589.955.175
Load Values kN
W3W2 27
114
Moment Component
165.75
27
68
W7 11 50 1 3138.50
W5W6 2.43868
3.563
68
Total = kNmTotal Bending Moment including impact for Outer Girder = kNmTotal Bending Moment including impact for Inner Girder = kNm
W81918.391792.49
14.5068 12.750.18889.25
1470.76
W7 11.50 1.31368
Calculation of bending moment at L/4 = 6 m, when load W3 is at L/4Maximum Ordinate of Influence line = 4.500 m Leff= 24 m
X1= 1.7
4.50X= 6
Fig 12c: ILD for BM at L/ 4RA RB
W1=27kN W2=27kN W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN W8=68kN
W4 68 1 20 4 200 285 60
3.20 2.100 239.40W3 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
Fig. 12c: ILD for BM at L/ 4
Total = kNmTotal Bending Moment including impact for Outer Girder = kNmTotal Bending Moment including impact for Inner Girder = kNm
W8 68 14.50 0.000 0.00
W6 68 8.50 2.375 161.50W7 68 11.50 1.625 110.50
W4 68 1.20 4.200 285.60W5 68 5.50 3.125 212.50
1556.931454.751193.64
Calculation of bending moment at L/8 = 3 m, when load W4 is at L/8Maximum Ordinate of Influence line = 2.625 m Leff= 24 m
2.625X= 3
Fig. 12d: ILD for BM at L/ 8RA RB
W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN W8=68kN
Page 19
Load Nos. Load Values kN Position from Maximum
IL Ordinate Moment Component
Fig. 12d: ILD for BM at L/ 8
Page 20
68 1.20 2.475
114 0.00
8.50 1.56368 11.50
O di t
299.25
W1 27 0.000.000 0.00
114
W668 5.50 1.93868
1.18868 14.50 0.813
168.30131.75
0.000 0.00W2
W5
W3W4
0.00 2.625
55.2580.75
W8
106.25W7
Total = kNmTotal Bending Moment including impact for Outer Girder = kNmTotal Bending Moment including impact for Inner Girder = kNm
Absolute Maximum BM
841.55
Calc lation of bending moment at the load point hich is eq idistance from res ltant
786.32645.19
Ordinate of Influence line at mid span = 6 m Leff= 24 m
6x= 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=68kNR
C.G of the Load system from outer 27 kN Wheel Load = 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 CG
X = 0.46 m
6.047.14 3.570 406.98
Load Values kN Position from Left support
IL Ordinate
27 3.020114W2
W1 81.54
Load Nos. Moment Component
Total = kNmTotal Bending Moment including impact for Outer Girder = kNm
392.36277.44
7.14 3.570
0.00 0.000 0.00
406.98114 10.34 5.170
2.5806868
68 11.54 5.77068 15.8468 18.84
0.000
W3 589.38W4
W6 175.440.00
4.080
W7 21.84
1796.931923.14
114
W8
W2
W5
g g pTotal Bending Moment including impact for Inner Girder = kNm1474.41
Page 20
Page 21
Shear Force due to Live load: IRC Class AA Tracked Vehicle.At Support
Effective span of girder, le = 24 mLoad Class AA Tracked vehicle W1= 350 kN
For maximum 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 Xgirder, 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 mX= 1.8 X1= 3.0 a= 0.6 b= 1.025C= 1.625 d= 2.05 f= 2.050 e= 0.35g= 1.375 h= 0.675 i= 1.725
Loads on Girders,PA= 0.573 W1 kerb Line
a RQ'
L
PB = 1.146 W1PC = 0.281 W1Reaction at support,RA = 0.358 W1RQ = 0.215 W1RB = 0.716 W1RR = 0.430 W1RC = 0 176 W1
A
B
CL of outer L Girder
CL of inner L Girder
a
c
C/C
d
Ist X
 Gird
er
med
iate XG
irder
Q
R
Q
R
b
RQ'
RR'
C/C
C/Cf
e
gh
RC 0.176 W1RS = 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 = 0.750C.G. of loads from Q = 2.05 m
kerb Line
CL of outer L GirderC
a
Interm
S SRS'
i
X X1
Fig. 13
Eccentricity, e = 0.35 mReaction factor for outer girder,FQ = 0.305 W1 in this case xi=2.4 mand ∑xi
2=(2.4)2+(0)2+(2.4)2= 2 x (2.4)2
Reaction factor for inner girder, FR = 0.250 W1in this case xi=0 m
RA due to FQ = 0.24375 W1RB due to FR = 0.200 W1 `Total reaction on outer Girder = 0.602 W1Total reaction on inner Girder = 0.916 W1Max shear at support including impact for outer girder = 231.7 kNMax shear at support including impact for inner girder = 352.7 kN
It may be seen that the reaction FQ and FR act 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
Max shear at support including impact for inner girder 352.7 kN
At Intermediate SectionEffective span of girder, le = 24.0 m
At Left =
At Right =
Ordinate of Bending SF at considered section, SFx
−
Lx1
−−
Lx11
Page 21
Shear at 1/8th spanCalculation of bending moment at L/8 = 3 m, when load Placed at Just Right of L/8Ordinate of Influence line at Left = 0.875 mO di t f I fl li t Ri ht 0 125 L ff 24
Page 22
Ordinate of Influence line at Right = 0.125 m Leff= 24 mx= 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 kNS.F. including impact for outer girder = 125.13 kNS.F. including impact for inner girder = 102.667 kN
Shear at 1/4th spanCalculation of bending moment at L/4 = 6 m, when load Placed at Just Right of L/4Ordinate of Influence line at Left = 0.75 mOrdinate 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 kNS.F. including impact for outer girder = 105.574 kNS.F. including impact for inner girder = 86.625 kN
Shear at 3/8th spanCalculation of bending moment at 3L/8 = 9 m, when load Placed at Just Right of 3L/8Ordinate 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
Fig. 14cILDof SFat 3L/ 8 of SpanRA RB
350 kN3.6m
a
a'
b
S.F. = 192.5 kNS.F. including impact for outer girder = 86.023 kNS.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/2Ordinate of Influence line at Left = 0.5 mOrdinate of Influence line at Right = 0.5 m Leff= 24 m
Fig. 14c ILD of SF at 3L/ 8 of Span
Ordinate of Influence line at Right 0.5 m Leff 24 mx= 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
Page 22
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
Page 23
Loads Values Loads ValuesW1 27 kN W5 68 kNW2 27 kN W6 68 kNW3 114 kN W7 68 kNW4 114 kN W8 68 kN
Distances Distances ValuesX Varies X5 4.3 m1 6
Values
X1 Varies X6 3 mX2 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
−
Lx1
−−
Lx11
Page 23
Calculation of Shear Force at Support = 24.0 m, when load W3Ordinate of Influence line y3= 1 Leff= 24.0 m
W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN W8=68kN
Page 24
Load Values
Position from Left
Load Nos. IL Ordinate Moment Component
Fig. 15a: ILD for SF at SupportRA RB
Y3 Y4 Y5 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=
Total = kN
W4 108.300.950
W7W6
1.000
380.97
52.420.7710.6460.5210.396
43.92
W3 114.00
W5
35.42W8 26.92
Total = kNTotal SF including impact for Outer Girder = kNTotal SF including impact for Inner Girder = kN
Calculation of SF at L/8 = 3.000 m, when load W3Ordinate of Influence line At Right 0.875
At Left 0.125 Leff= 24.0 m x= 3
292.07355.97380.97
W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN W8=68kN
Load V l
Position f L ft
Moment ComponentLoad Nos. IL Ordinate
Fig. 15b: ILD for SF at L/ 8RA RB
Y3 Y4 Y5 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=
Total = kN
94.05
26.92W7
W5W4
W8
W3
W643.92
300 05
0.8750.8250.6460.5210.396
35.42
99.75
0.271 18.42Total = kN
Total SF including impact for Outer Girder = kNTotal SF including impact for Inner Girder = kN
Calculation of SF at L/4 = 6 000 m when load W3
300.05280.36230.04
Calculation of SF at L/4 = 6.000 m, when load W3Ordinate 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/ 4RA RB
Y3
Y3'
W1=27kN W2=27kN W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN W8=68kN
Y4 Y5 Y6 Y7Y8Y2Y1
Page 24
Load Values
Position from Left/
Load Nos. IL Ordinate Moment Component
Page 25
kN Right at X27 4.3 Y1=27 3.2 Y2=114 0 Y3=114 1.20 Y4=68 5.50 Y5=68 8.50 Y6=68 11 50 Y7=
26.9218 42
W5W6W7
79.8035.42
W3W4
W2 3.15W1 0.071 1.91
0 271
85.500.1170.7500.7000.5210.396
68 11.50 Y7=68 14.50 Y8=
Total = kNTotal SF including impact for Outer Girder = kNTotal SF including impact for Inner Girder = kN
Calculation of SF at 3L/8 = 9.000 m, when load W3Ordinate of Influence line At Right 0.625
At Left 0.375 Leff= 24.0 m x= 9.0
250.90234.44
0.14618.42
9.92
192.36
W7W8
0.271
L d P i iL d N IL O di M C
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 Y5 Y6 Y7Y8Y2Y1
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=
6.5371.25W3
5.290.196
W4 65.55
Load Nos. IL Ordinate Moment Component
0.2420.6250.575
W1
26.92W5 0.396
W2
68 5.50 Y568 8.50 Y6=68 11.50 Y7=68 14.50 Y8=
Total = kNTotal SF including impact for Outer Girder = kNTotal SF including impact for Inner Girder = kN
Calculation of SF at L/2 = 12 000 m when load W3
W6 18.42W7 9.92W8 1.42
181.65169.73139.27
26.92W5 0.3960.2710.1460.021
Calculation of SF at L/2 = 12.000 m, when load W3Ordinate 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 Y7Y2Y1
Load Values
kN
Position from Left/ Right at X
27 4.3 Y1=27 3.2 Y2=
Load Nos. IL Ordinate Moment Component
W1W2
8.669.90
0.3210.367
Fig. 15e: ILD for SF at L/ 2RA RB
Page 25
114 0 Y3=114 1.20 Y4=68 5.50 Y5=68 8.50 Y6=
W3 57.00
W5 18.42W6
0.5000.4500.2710.146
W4 51.30
9.92
Page 26
68 8.50 Y668 11.50 Y7=
Total = kNTotal SF including impact for Outer Girder = kNTotal 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 (kN)Bending Moment (kNm)
W7 1.420.021W6 0.146 9.92
111.6591.61
SectionDue to
DLDue to LL
X = 0 0.00 0.00X = L/8 1157.75 393.79X = L/4 2064.76 670.35X = 3L/8 2492.47 829.26X = L/2 2654.71 1736.11
Total Design Bending Moments and Shear Forces for Inner Girder:111.646
234.439334.296
Due to DL
111.646
Due to LL
355.966280.359
169.733
Shear Forces (kN)Bending Moment (kN m)Total
1551.5400.000
Total
218.768115.528
437.536
4390.817 0.000
2735.1043321.735
793.502614.655453.207285.261
Total Design Bending Moments and Shear Forces for Inner Girder:Section
Due to DL
Due to LL
X = 0 0.00 0.000X = L/8 1157.75 323.111X = L/4 2064.76 550.028X = 3L/8 2492.47 680.419 254.796
2614.79 411.1283172.89 115.53 139.27
352.720.000564.334
Shear Forces (kN)Total Due to DL Due to LL
437.54334.30
790.252
Bending Moment (kNm)
1480.859 230.04192.36218.77
Total
X = L/2 2654.71 1424.50Design of Outer Girder:
Overall depth of beam, D = 2000 mmRib width, bw = 400 mmFlange width of Tbeam will be,
bf = bw + 1/5 x lo 2.5 m > 2.4 m Therefore width of flange, bf = 2400 mm
170 mm from bottom of Tbeam to the centre of gravity of rod
91.60791.614079.21 0.00
170 mm from bottom of Tbeam to the centre of gravity of rod, d = 1830
Area of steel required,
Ast= 13420 mm2
Provide 16 nos. of φ 32 20002 nos. of φ 25
mm bars+mm bars
=×× djst
Mσ
2 nos. of φ 25 Provided area of steel = 13843 mm2 Total Provided area of steel = 13843 mm2Number of bars in bottom row = 4 nos.Width of beam = 400 mm 400Side and bottom clear cover to bars = 40 mmC.G. of the bottom row of bars from bottom = 56 mmClear distance between vertical bars = 32 mmC G f th S d f b f b tt 120
mm bars
Fig. 16 : Cross Section of Girde
C.G. of the Second row of bars from bottom = 120 mmC.G. of the third row of bars from bottom = 184 mmC.G. of the fourth row of bars from bottom = 248 mmC.G. of the fifth row of bars from bottom = 308.5 mmC.G. of the bar group from bottom = 163.1 mm
≤ 170 mm O.K.Effective Depth = 1830 mmDf = 220 mm
Page 26
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
Page 27
effective area in of compression and tension sides about neutral axis, we get½ × bw × xa
2 + (bf – bw) × Df × (xa – Df/2) = m × Ast × (d – xa)Solving, we get, xa = 482 mmLet compressive stress in concrete at top of flange = σAnd compressive stress in concrete at bottom of flange = σ'
Then, σ’ = 0.543 σ=×−
σa
fa
xDx
Position of C.G. of compressive stress in flange from top, x1
= 99.1 mm
Compressive force in flange, C1 = ½ x(σ +σ ')x Bf x Df= 407402 σ
Compressive force in rib, C2 = ½ x σ 'x (Xa – Df)x Bw
=×+
+3
D''2 f
σσσσ
Compressive force in rib, C2 ½ x σ x (Xa Df)x Bw= 28420 σ
C.G. of compressive force in rib from top, x2307.2 mm
Total compressive force, C = 435822.3 σC.G. of total compressive force from top
112.7 mm=+
×+×=
C2C1x2C2x1C1
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 = = σ699626722.86
=+
cm1
d
σσ st
−×
+××
−
yddb ff 2
1σσ
Equating Mr to external B.M we getMr =σ = 6.28 < 8.3 O.K.
Stress Developed in Steel Reinforcement is given by
t = = 197.6 < 200 O.K.
4390816575.00
−××
a
a
xxd
m σ
Check for minimum area of steelMinimum area of tension steel in beam @ 0.2 % of web area
1600 mm2 < Ast provided = mm2 O.K.
Design for shear:
1.08 N/mm2
13843
=×
=dB
Vvτ
Assuming ####### nos. of φ 32 will be continued up to support, then provided percentage area of tension steel = 1.00 %Permissible shear stress, τc = 0.420 N/mm2 < τv
Vs = V  tc . bw . D = 457195 NAssuming φ 12 mm 2legged 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,
=××
VsdσAsv st
Page 27
As per minimum shear reinforcement requirements, maximum spacing,
Smax = 283 mm c/c.
Vs
=×st Asvσ
Page 28
Smax 283 mm c/c.
Hence provide φ 12 mm, 2legged vertical stirrups @ 100 mm c/c at support.Design summary:
Tension Reinforcement (Fe 415):
No. Dia.16 32
Area of Steel RequiredSection
Area of steel required and provided at different sections of Outer girder are given in below:
AreaBM Area of Steel Provided
× wb0.4
16 322 25
3L/8 3321.73 KNm 14 32 mm2L/4 2735.10 KNm 12 32 mm2L/8 1551.54 KNm 10 32 mm2
Support 0.00 KNm 10 32 mm2
Shear Reinforcement (Fe 415):
10153
8038
1125496468038mm2
mm20
mm2mm28360
4742
4390.82 13420 mm2KNmL/2 13843 mm2
( )Section
L/2 111.646 kN 10 φ @ 1029.4 mm 10 φ @ 2503L/8 285.261 kN 10 φ @ 402.87 mm 10 φ @ 250L/4 453.207 kN 10 φ @ 253.58 mm 10 φ @ 200L/8 614.655 kN 12 φ @ 269.24 mm 12 φ @ 150
Support 793.502 kN 12 φ @ 208.56 mm 12 φ @ 100
Tension Reinforcement (Fe 415):
2legged vertical Stirrups providedmm
mm
mm
mm
SF 2legged vertical Stirrups required
Area of steel required and provided at different sections of Inner girder are given in below:
mm
Tension Reinforcement (Fe 415):
No. Dia.16 32
2 253L/8 3172.89 KNm 14 32 mm2L/4 2614.79 KNm 12 32 mm2L/8 1480.86 KNm 10 32 mm2
Support 0 000 KNm 10 32 mm2
112549646803880380 mm2
Section BM
7992 mm2
Area of Steel Required
4526 mm2
13843
9698 mm2
AreaArea of Steel Provided
L/2 4079 KNm 12468 mm2 mm2
Support 0.000 KNm 10 32 mm2Shear Reinforcement (Fe 415):
SectionL/2 91.607 kN 10 φ @ 150 mm 10 φ @ 250
3L/8 254.796 kN 10 φ @ 150 mm 10 φ @ 250L/4 411.128 kN 10 φ @ 150 mm 10 φ @ 200L/8 564.334 kN 12 φ @ 150 mm 12 φ @ 150
Support 790.252 kN 12 φ @ 100 mm 12 φ @ 100
SF
mmmm
8038
mmmm
2legged vertical Stirrups providedmm
2legged vertical Stirrups required
0 mm2
pp φ φ
Page 28
Page 29
4 0 Design of Cross Girder:4.0 Design of Cross Girder:
Dead LoadOverall Depth of cross girder = 1.5 m 2.4Width of cross girder = 0.3 mSelf weight of cross girder= 9.216 kN/m 4.8
Dead load from slab = 20.2752 kNThi l d i d if l di ib d l d 8 448 kN/
Fig. 17This load is assumed as uniformly distributed load per meter run = 8.448 kN/mTotal Dead load per meter run = 17.664 kN/mAssuming, cross girder as rigid, reaction on main girder = 13.52 kN
Live Load: IRC Class AA Tracked VehicleMaximum 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
Mσ
Fig. 19
W1 W1
p y pp pp g
Effective load coming on cross girder = 569 kN=−l
lW 2/8.1
=×× djst
Mσ
Fig. 19
W1 W1
Reaction on each longitudinal girder = 189.58 kNMaximum B.M. occurs under the load, = 260.68 kNmBending moment including impact = 286.74 kNmDead Load Bending Moment at the section, = 1.8876 kNm
Total bending moment = 288.6 kNm
=−l
lW 2/8.1
=×× djst
Mσ
Fig. 19
W1 W1
Total bending moment 288.6 kNm
LL Shear force including impact = 208.5 kNTotal shear force = 222.1 kN
Therefore, Design Moment = 288.6 kNmDesign Shear Force = 222.1 kN
C i d i d i d T B
=−l
lW 2/8.1
=×× djst
Mσ
Fig. 19
W1 W1
Cross girder is designed as TBeam.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 φ 25 mm bars + 0 nos.of φ 20 bars having area of steel = 1472 mm2 .
1113.37
=−l
lW 2/8.1
=×× djst
Mσ
Fig. 19
W1 W1
Page 29
D i f h=
× dBV
=××
VusdσAsv st
=××
wb0.4Asv0.87fy
Page 30
Design for shear:Nominal shear stress, τv = 0.51 N/mm2 < τmax = 1.9 N/mm2 O.K.
Provided percentage area of tension steel, p = 0.33 %Permissible shear stress, τc = 0.275 N/mm2 < 0.51 N/mm2 O.K.
Shear reinforcement is required. Shear reinforcement shall be provided to carry a shear force of
Vus = Vu tc bw D = 139433 N
=× dBV
=××
VusdσAsv st
=××
wb0.4Asv0.87fy
Vus = Vu  tc . bw . D = 139433 NAssuming φ 10 mm 2legged 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
=× dBV
=××
VusdσAsv st
=××
wb0.4Asv0.87fy
Hence provide φ 10 mm 2legged vertical stirrups @ 150 mm c/c through out the length of end cross girder and 10 mm 2legged vertical stirrups @ 150 mm c/c through out the length
Elastomeric Bearing On Bridge Used
According IRC 83part II it is reccomended use of elastomeric bearins of size (250X400X50 ) mm embedding 5plates of 3mm thickness and 6 mm clearance in plan G= 1 kN/mm2 For Vertical Load 793 50 kN
of intermediate cross girder.
=× dBV
=××
VusdσAsv st
=××
wb0.4Asv0.87fy
plates of 3mm thickness and 6 mm clearance in plan G= 1 kN/mm2 For Vertical Load 793.50 kN
=× dBV
=××
VusdσAsv st
=××
wb0.4Asv0.87fy
Page 30