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Page 1: 45793867 Design of Super Structure

A. Design of Superstructure

1.0 Design Data

Fig 1.BRIDGE CROSS-SECTION

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 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

Fig 1.BRIDGE CROSS-SECTIONFig 1.BRIDGE CROSS-SECTION

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Page 2: 45793867 Design of Super Structure

2.0 Design of Slabh il l b i d i d b i id h h d

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

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 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

( g)Dead Load Bending Moment and Shear Force:

S.No. Item Width Depth Unit Wt

1 Railing/Parapet

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.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 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

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 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

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

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Page 3: 45793867 Design of Super Structure

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

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

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

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

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

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

( 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

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

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Page 4: 45793867 Design of Super Structure

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

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Page 5: 45793867 Design of Super Structure

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

Page 6: 45793867 Design of Super Structure

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

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 kN-m

Moment along long span,= 12 423 kN-m

M1= W(m1+0.15m2)

M2= W(0 15m1+m2) = 12.423 kN-mBending 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 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 kN-mMoment 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-mM2 = 8.002 kN-m

Final MomentM1 = 2.321 kN-mM2 = 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

( )

Page 6

Page 7: 45793867 Design of Super Structure

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 kN-m

Moment along long spanM1= W(m1+0.15m2)

Moment along long span,= 37.628 kN-m

Bending moment including impact and continuity,M1 = 57.832 kN-mM2 = 37.628 kN-m

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 kN-mM1= W(m1+0.15m2)

Moment along long span,= 30.386 kN-m

Bending moment including impact and continuity,M1 = 48.480 kN-mM2 = 30.386 kN-m

Final MomentM1 = 4.676 kN-mM2 = 3 621 kN-m

( )

M2= W(0.15m1+m2)

M2 = 3.621 kN-mBending moment due to load W5:

Wheel load is placed unsymmetrical wrt the Both X-X and Y-Y 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 kN-m

Moment along long span,= 21.628 kN-m

Bending moment including impact and continuity,M1 = 25.650 kN-mM2 = 21.628 kN-m

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

Page 8: 45793867 Design of Super Structure

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 kN-m

Moment along long span,= 20.412 kN-m

Bending moment including impact and continuityM2= W(0.15m1+m2)

M1= W(m1+0.15m2)

Bending moment including impact and continuity,M1 = 23.718 kN-mM2 = 20.412 kN-m

Moment Along X-XM1 = 0.966 kN-mM2 = 0.608 kN-m

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 kN-m

Moment along long span,= 37.628 kN-m

Bending moment including impact and continuity,M1 = 57 832 kN m

M1= W(m1+0.15m2)

M2= W(0.15m1+m2)

M1 = 57.832 kN-mM2 = 37.628 kN-m

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 kN-m

Moment along long span,= 30.386 kN-m

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 kN-mM2 = 30.386 kN-m

Final MomentM1 = 4.676 kN-mM2 = 3.621 kN-m

Resultent MomentM1 = 5.642 kN-mM2 = 4.230 kN-m

Page 8

Page 9: 45793867 Design of Super Structure

Bending moment due to load W6:Wheel load is placed unsymmetrical wrt the Both X-X and Y-Y

For X-X 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 kN-m

Moment along long span,12 423M2 W(0 15 1 2)

M1= W(m1+0.15m2)

= 12.423 kN-mBending 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 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 kN-m

Moment along long spanM1= W(m1+0.15m2)

Moment along long span,= 8.002 kN-m

Bending moment including impact and continuity,M1 = 9.957 kN-mM2 = 8.002 kN-m

Moment Along Y-YM1 = 2.321 kN-mM2 = 2.210 kN-m

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 kN-m

Moment along long span,= 22.577 kN-m

M1= W(m1+0.15m2)

M2= W(0.15m1+m2)

Page 9

Page 10: 45793867 Design of Super Structure

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

Page 11: 45793867 Design of Super Structure

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

Page 12: 45793867 Design of Super Structure

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

Page 13: 45793867 Design of Super Structure

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

Page 14: 45793867 Design of Super Structure

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

Page 14

Page 15: 45793867 Design of Super Structure

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

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

Page 16: 45793867 Design of Super Structure

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

Page 17: 45793867 Design of Super Structure

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

Page 18: 45793867 Design of Super Structure

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 = kN-mTotal Bending Moment including impact for Outer Girder = kN-m

57.8068

10.30 0.850

114 5.40011468

W1 27 87.75W2 102.60

-5.5027

615.60

7.30W5

W3-4.40

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

Page 19: 45793867 Design of Super Structure

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/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 = kN-mTotal Bending Moment including impact for Outer Girder = kN-mTotal Bending Moment including impact for Inner Girder = kN-m

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 = kN-mTotal Bending Moment including impact for Outer Girder = kN-mTotal Bending Moment including impact for Inner Girder = kN-m

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

Page 20: 45793867 Design of Super Structure

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 = kN-mTotal Bending Moment including impact for Outer Girder = kN-mTotal Bending Moment including impact for Inner Girder = kN-m

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 = 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: 45793867 Design of Super Structure

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

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 X-G

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

Page 22: 45793867 Design of Super Structure

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

Page 23: 45793867 Design of Super Structure

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

Page 24: 45793867 Design of Super Structure

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

Page 25: 45793867 Design of Super Structure

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

85.50-0.1170.7500.7000.5210.396

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

-5.29-0.196

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

-8.66-9.90

-0.321-0.367

Fig. 15e: ILD for SF at L/ 2RA RB

Page 25

Page 26: 45793867 Design of Super Structure

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 (kN-m)

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 (kN-m)

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 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

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

Page 27: 45793867 Design of Super Structure

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

Page 28: 45793867 Design of Super Structure

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:

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):

2-legged vertical Stirrups providedmm

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

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

2-legged vertical Stirrups providedmm

2-legged vertical Stirrups required

0 mm2

pp φ φ

Page 28

Page 29: 45793867 Design of Super Structure

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

Fig. 19

W1 W1

p y pp pp g

Effective load coming on cross girder = 569 kN=−l

lW 2/8.1

=×× djst

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

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

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 .

1113.37

=−l

lW 2/8.1

=×× djst

Fig. 19

W1 W1

Page 29

Page 30: 45793867 Design of Super Structure

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 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

=× dBV

=××

VusdσAsv st

=××

wb0.4Asv0.87fy

Page 30