A. Design of Superstructure 1.0 Design Data Fig 1.BRIDGE CROSS-SECTION Page 1 1.1. Materials and its Properties: M25 Fe415 Characteristics Strengthof Concrete fck = 25 MPa Permissible direct compressive stress, σ c = 6.2 MPa Permissible flexural compressive stress, σ cbc = 8.3 MPa Maximum Permissible shear stress, τ max ( 0.07*fck) = 1.75 MPa Fig 1.BRIDGE CROSS-SECTION Maximum Permissible shear stress, τ max ( 0.07 fck) 1.75 MPa Basic Permissible Stresses of Reinforcing Bars as per IRC : 21-1987, Section III: Permissible Flexural Tensile stress, σ st = 200 MPa Permissible direct compressive stress, σ 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 Rib width of cross girder = 300 mm Fig 1.BRIDGE CROSS-SECTION Rib width of cross girder = 300 mm 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 m Total depth of Kerb Outer = 0.375 m Total depth of Kerb Inner = 0 m Fig 1.BRIDGE CROSS-SECTION Fig 1.BRIDGE CROSS-SECTION Page 1
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Maximum Permissible shear stress, τmax ( 0.07*fck) = 1.75 MPa
Fig 1.BRIDGE CROSS-SECTION
Maximum Permissible shear stress, τmax ( 0.07 fck) 1.75 MPa
Basic Permissible Stresses of Reinforcing Bars as per IRC : 21-1987, Section III:Permissible Flexural Tensile stress, σst = 200 MPa
Permissible direct compressive stress, σco = 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 mTotal length of span = 24.56 mNumbers of span = 2Width of expantion Joint = 40 mmTotal length of Bridge = 49.2 m
Fig 1.BRIDGE CROSS-SECTION
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 CROSS-SECTION
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 CROSS-SECTION
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
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 CROSS-SECTION
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 CROSS-SECTION
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 CROSS-SECTION
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 CROSS-SECTION
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.
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 kN-mMoment along long span, M2 = W (0.15m1 +m2) = 0.98 kN-mConsidering effects of continuity, 0.8Moment along short span, M1 = 2.38 kN-mMoment along long span, M2 = 0.78 kN-m
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 kN-m
Moment along long span,= 9.413 kN-m
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 kN-mM2 = 9.413 kN-m Fig: 4
Page 5
c) Due to Live load: Class AA Wheeled VehicleCase-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 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 kN-m
Moment along long spanM1= W(m1+0.15m2)
Moment along long span,= 10.486 kN-m
Bending moment including impact and continuity,M1 = 13.578 kN-mM2 = 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 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 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)
Page 5
M2 = 21.628 kN-mNext, 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 kN-mM1= W(m1+0 15m2) = 23.718 kN-m
Moment along long span,= 20.412 kN-m
Bending moment including impact and continuity,M1 = 23.718 kN-mM2 = 20.412 kN-m
Final MomentM1 = 0.966 kN-mM2
M2= W(0.15m1+m2)
M1= W(m1+0.15m2)
M2 = 0.608 kN-mBending 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 m2Total applied load = q x area = 126 kN
Bending moment including impact and continuity,M1 = 34.699 kN-mM2 = 22.577 kN-m
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 kN-m
Moment along long span,= 18.231 kN-m
Bending moment including impact and continuity,M1 = 29.088 kN-mM2 = 18.231 kN-m
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
Final MomentM1 = 2.806 kN-mM2 = 2.173 kN-m
Resultent MomentM1 = 5.127M2 = 4.383
Total Moment Due to IRC Class AA Wheeled Vechicle Moment along short span,M1 = 32.309 kN-mMoment along short span,M1 32.309 kN mMoment along long span,M2 = 25.539 kN-m
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 kN-m
Moment along long span,= 8.565 kN-m
Bending moment including impact and continuity,M1 = 10 851 kN m
M1= W(m1+0.15m2)
M2= W(0.15m1+m2)
M1 = 10.851 kN-mM2 = 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 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 kN-m
Moment along long span,= 23.154 kN-m
Bending moment including impact and continuity,M1 = 40.676 kN-mM2 = 23 154 kN m
M2= W(0.15m1+m2)
M1= W(m1+0.15m2)
M2 = 23.154 kN-mNext, 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 kN-mMoment along long span,
M1= W(m1+0.15m2)
Page 12
= 17.751 kN-mBending moment including impact and continuity,
M1 = 33.081 kN-mM2 = 17.751 kN-m
Final MomentM1 = 3.798 kN-mM2 = 2.702 kN-m
M2= W(0.15m1+m2)
Total Bending Moment due to load W1 & W2 will be,M1 = 14.6 kN-mM2 = 11.3 kN-m
Design Bending Moment due to LL:M1 = 32.3 kN-mM2 = 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 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 kN-mM2 = 26.32 kN-m
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 kN-mBending Moment including impact and rection factor for outer Girder = 1736 kN-mBending Moment including impact and rection factor for Inner Girder = 1425 kN-m
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
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 = kN-mTotal Bending Moment including impact for Outer Girder = kN-m
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 = kN-m1474.41
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 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 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
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
Overall depth of beam, D = 2000 mmRib width, bw = 400 mmFlange 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
170 mm from bottom of T-beam to the centre of gravity of rod
91.60791.614079.21 0.00
170 mm from bottom of T-beam 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 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,
=××
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, 2-legged 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:
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 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 φ 25 mm bars + 0 nos.of φ 20 bars having area of steel = 1472 mm2 .
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 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
=× dBV
=××
VusdσAsv st
=××
wb0.4Asv0.87fy
Hence provide φ 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 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