Dsign of Bridge Superstructure Using Indian Codes

7/22/2019 Dsign of Bridge Superstructure Using Indian Codes

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A. Design of Superstructure

1.0 Design Data

Fig 1.BRIDGE CROSS-SECTION

Pa

1.1. Materials and its Properties:

M25

Fe415

Characteristics Strengthof Concrete fck = 25 MPa

Permissible direct compressive stress, c = 6.2 MPaPermissible flexural compressive stress, cbc = 8.3 MPa

Maximum Permissible shear stress, ax ( 0.07*fck) = 1.75 MPa


.

Basic Permissible Stresses of Reinforcing Bars as per IRC : 21-1987, Section III:

Permissible Flexural Tensile stress, st = 200 MPa

Permissible direct compressive stress, co = 170 MPa

Self weight of materials as per IRC : 6-2000:

Concrete (cement-Reinforced) = 24 kN/m3


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


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


=

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:


Carriageway width = 6 m

Footpath width = 0.45 m

Kerb width Outer = 0.15 m

Kerb width Inner = 0 m

Total Width of Deck Slab = 7.2 mTotal depth of Kerb Outer = 0.375 m

Total depth of Kerb Inner = 0 m

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

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2.0 Design of Slab

cbc3

280


57KN 350KN 37.5KN

IRC Class AA Track LoadingIRC Class A Loading IRC Class AA Wheel Loading

Fig. 2.a, 2.b & 2.c

Pa

2.1 Design of Cantilever slab: The cantilever slab is designed by effective width method.

300 mm at junction with rib

150 mm at free end

0.5 kN/m (assumed)

Impact factor = 54 % (for IRC class A loading)

25 % (for IRC class AA loading)

Thickness of slab =

Self weight of Railing =

cbc3

280


57KN 350KN 37.5KN


Fig. 2.a, 2.b & 2.c

Dead Load Bending Moment and Shear Force:

S.No. Item Width Depth Unit Wt

1 Railing/Parap

et

0.5 kN 1.525-0.100= 0.9 m 0.45 kN.m

2 Kerb (outer) 0.2 0.225 24 1.08 kN 1.525-0.100= 0.9 m 0.97 kN.m

3 Kerb (inner) 0 0 24 0 kN 0.15+0.175/2= 0.2375 m 0 kN.m

4 Wearin Coat 0.4 0.08 22 0.704 kN 0.150/2= 0.075 m 0.05 kN.m

Assumed

Load / m run

(kN)MomentDistance

cbc3

280


57KN 350KN 37.5KN


Fig. 2.a, 2.b & 2.c

. . . . . .

5 Slab 1 0.15 24 3.6 kN 1.525/2= 0.5 m 1.8 kN.m

1 0.075 24 1.8 kN 1.525/3= 0.3333 m 0.6 kN.m

Total kN kN.m

= 7.684 kN

= 3.875 kN.m

7.684

Dead load Shear force at theface of rib

Dead load Bending Moment at the face of rib

3.875

cbc3

280


57KN 350KN 37.5KN


Fig. 2.a, 2.b & 2.c

Live Load Bending Moment and Shear Force:

cbc3

280


57KN 350KN 37.5KN


Fig. 2.a, 2.b & 2.c

IRC Class AA will not operate on the cantilever slab that shown in fig 2.b & 2.c above and Class A

Loading is to be considered and the load will be as shown in fig 2.a above.

Effective width of dispersion be is computed by equation

be = 1.2X+ bw

Here

X= 0.125 m

bw= 0.41 m

cbc3

280


57KN 350KN 37.5KN


Fig. 2.a, 2.b & 2.c

.

Hence

be= 0.56 m

IRC Class A Loading Load = 28.5 kN

Live Load per m width including impact = 76.339 kN

Maximum Moment due to live load = 9.5424 kNm

Average thickness of cantilever slab = 225 mm

Taking pedestrain load (LL) = 5.0 kN/m2

cbc3

280


57KN 350KN 37.5KN


Fig. 2.a, 2.b & 2.c

Effective width of slab = 0.45 m

Cantilever length of slab = 1 m

Maximum Bending moment = 1.406 kN.m

Shear force at the face of slab = 2.250 kN

Total Design Shear Force = 86.3 kN

Total Design Bending Moment = 14.82 kN.m

Design of Section:

Modular Ratio, m = = 11.245cbc

3

280


57KN 350KN 37.5KN


Fig. 2.a, 2.b & 2.c

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Neutral axis factor, k = = 0.3182

Lever arm factor, j = = 0.8939

cbc3

280

stcbcm

cbcm

+

31

k

stjk 2

1

Rb

M

dj

M

st ..=

dj

M

st ..=

bd

d

tanM-V

=v

coc kk .. 21=5.07.014.1

1 = dk

125.05.02

+= k bd

A s=

=2k

co

4.021

= KKc

v

Pa

Moment of resistance coefficient, R = = 1.1804

Therefore, required effective depth of slab=

d = = 112.06 mm

Effective depth of slab, provided = 254 mm > d reqO.K.

cbc3

280

stcbcm

cbcm

+

31

k

stjk 2

1

Rb

M

dj

M

st ..=

dj

M

st ..=

bd

d

tanM-V

=v

coc kk .. 21=5.07.014.1

1 = dk

125.05.02

+= k bd

A s=

=2k

co

4.021

= KKc

v

Area of steel required, Ast = 326.42 mm2

Provide 10 mm bars @ 200 = 393 mm2

> required, Ok.

Distribution Steel:

Distribution steel is to be provided for 0.3 times live load moment plus 0.2 times dead

load moment.

mm c/c, giving area of steel =

cbc3

280

stcbcm

cbcm

+

31

k

stjk 2

1

Rb

M

dj

M

st ..=

dj

M

st ..=

bd

d

tanM-V

=v

coc kk .. 21=5.07.014.1

1 = dk

125.05.02

+= k bd

A s=

=2k

co

4.021

= KKc

v

.

Moment = 4.06 kN.m

Effective dept 244 mm

Area of steel required, Ast = 93.057 mm

Half reinforcement is to be provided at top and half at bottom.

Provide 10 mm bars 200 mm c/c at both top and bottom, giving area of2

cbc3

280

stcbcm

cbcm

+

31

k

stjk 2

1

Rb

M

dj

M

st ..=

dj

M

st ..=

bd

d

tanM-V

=v

coc kk .. 21=5.07.014.1

1 = dk

125.05.02

+= k bd

A s=

=2k

co

4.021

= KKc

v

stee = .5 mm > requ re , .

Check for min. area of Steel:

Min. area of steel @ 0.12 % = 360 mm < Provided. O.K.

Design for Shear:

Dead load shear = 7.68 kN

Live load Shear = 2.250 kN

cbc3

280

stcbcm

cbcm

+

31

k

stjk 2

1

Rb

M

dj

M

st ..=

dj

M

st ..=

bd

d

tanM-V

=v

coc kk .. 21=5.07.014.1

1 = dk

125.05.02

+= k bd

A s=

=2k

co

4.021

= KKc

v

Total = 9.93 kN

tan= 0.150

Shear stress, = 0.005 N/mm2

Percentage area of tension steel, pt = 0.13 %

Allowable shear stress as per code is given by

cbc3

280

stcbcm

cbcm

+

31

k

stjk 2

1

Rb

M

dj

M

st ..=

dj

M

st ..=

bd

d

tanM-V

=v

coc kk .. 21=5.07.014.1

1 = dk

125.05.02

+= k bd

A s=

=2k

co

4.021

= KKc

v

( d being in m)

= 0.986

(where )

= 0.500446 1.00

cbc3

280

stcbcm

cbcm

+

31

k

stjk 2

1

Rb

M

dj

M

st ..=

dj

M

st ..=

bd

d

tanM-V

=v

coc kk .. 21=5.07.014.1

1 = dk

125.05.02

+= k bd

A s=

=2k

co

4.021

= KKc

v

Adopt 1

Value ot = for M25 grade of concrete from code = 0.4 N/mm2

Allowable shear stress

= 0.3944 N/mm2 > , Hence Safe

cbc3

280

stcbcm

cbcm

+

31

k

stjk 2

1

Rb

M

dj

M

st ..=

dj

M

st ..=

bd

d

tanM-V

=v

coc kk .. 21=5.07.014.1

1 = dk

125.05.02

+= k bd

A s=

=2k

co

4.021

= KKc

v

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2.2 Design of Interior Panels: The slab panel is designed by Pigeauds method.

D

Pa

B C

Short span of slab, Bs = 2 m

Long span of slab, Ls = 4.5 m

A

Fig : 3 Bridge Plan

Calculation of Bending moments

a) Due to Dead load:

Self weight of wearing coat = 1.76 kN/m2

Self weight of deck slab = 5.28 kN/m

Total = 7.04 kN/m

Since the slab is supported on all four sides and is continuous, Piegauds curves are used to

calculate bending moments.

a o, = s s = .

As the panel is loaded with UDL,

u/Bs = 1

v/Ls = 1

Where, u & v are the dimensions of the loaded area.

From the Pigeauds curve,

m1 = 0.0457

m2 = 0.0086Total dead load W = 63.36 kN

Moment along short span, M1 = W (m1 +0.15m2) = 2.98 kN-m

Moment along long span, M2 = W (0.15m1 +m2) = 0.98 kN-m

Considering effects of continuity, 0.8

Moment along short span, M1 = 2.38 kN-m

Moment along long span, M2 = 0.78 kN-m

Due to Live load: Class AA Tracked Vehicle

For maximum bending moment one wheel is placed at the center of panel.

Tyre contact length along short span, x = 0.85 m

Tyre contact length along long span, y = 3.6 m

Loaded length, u = 1.034 m

Loaded width, v = 3.766 m

Wheel load, W = 350 kN

Ratio, k = Bs/Ls = 0.44

u/Bs = 0.517

v/Ls = 0.837


m1 = 0.0813

m2 = 0.0147

Moment along short span,

= 29.227 kN-m

Moment along long span,

= 9.413 kN-m

M1= W(m1+0.15m2)

M2= W(0.15m1+m2)

W1=350k

Fig: 4

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Bending moment including impact and continuity,

M1 = 29.227 kN-m

M2 = 9.413 kN-m Fig: 4

Pa

c) Due to Live load: Class AA Wheeled Vehicle

Case-I:When two loads of 37.5 kN each and four loads of 62.5kN are placed such that two loads of

62.5kN lies at center line of pannel.

Tyre contact width (along short span), = 0.30 m

Tyre contact length (along long span), = 0.15 mDisperced width along short span, u = 0.510 m

Disperced width along long span, V = 0.380 m

= .

62.5kN

62.5kN

62.5kN

62.5kN

W1 W4

W2 W5

Y

X X

Bending moment due to load W1: 62.5 kN

Ratio k = Bs/Ls = 0.44

Fig: 5

37.5kN

W3

37.5kN

W6

Y

, .

u/Bs = 0.255

v/Ls = 0.084


m1 = 0.1965

m2 = 0.1383


= 13.578 kN-mM1= W(m1+0.15m2)

omen a ong ong span,

= 10.486 kN-m


M1 = 13.578 kN-m

M2 = 10.486 kN-m

Bending moment due to load W2: 62.5 kN Wheel load is placed unsymmetrical wrt the X-X

Intensity of loading, q = 322.45 kN/m2

M2= W(0.15m1+m2)

Considering loaded area 2.000 x 0.380 m

Loaded area = 0.760 m2

Total applied load = q x area = 245 kN

Ratio, k = Bs/Ls = 0.44u/Bs = 1.000

v/Ls = 0.084

From the Pi eauds curve ,

m1 = 0.0935

m2 = 0.0742


= 25.650 kN-m


= 21.628 kN-m


M1 = 25.650 kN-m

M1= W(m1+0.15m2)

M2= W(0.15m1+m2)

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M2 = 21.628 kN-m

Next, Consider the area between the real and the dummy load i.e., 1.490 m X 0.380 m



Pa


u/Bs = 0.745

v/Ls = 0.084


m1 = 0.1157m2 = 0.0944


-= = . -


= 20.412 kN-m


M1 = 23.718 kN-m

M2 = 20.412 kN-m

Final Moment

M1 = 0.966 kN-m

M2= W(0.15m1+m2)

= .

M2 = 0.608 kN-m

Bending moment due to load W3: 37.5 kN Wheel load is placed unsymmetrical wrt the X-X

Intensity of loading, q = 193.47 kN/m2Considering loaded area 1.710 x 0.380 m




u/Bs = 0.855

v/Ls = 0.084


m1 = 0.1036

m2 = 0.08325


= 14.598 kN-m


= -

M1= W(m1+0.15m2)

= . -


M1 = 14.598 kN-m

M2 = 12.423 kN-m





.

u/Bs = 0.345

v/Ls = 0.084


m1 = 0.1765

m2 = 0.1312= 9.957 kN-m


= 8.002 kN-m

M1= W(m1+0.15m2)

M2= W(0.15m1+m2)


M1 = 9.957 kN-m

M2 = 8.002 kN-m

Final Moment

M1 = 2.321 kN-m

M2 = 2.210 kN-m

Bending moment due to load W4: 62.5 kN Wheel load is placed unsymmetrical wrt the Y-Y


Pa


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Pa

u/Bs = 0.618

v/Ls = 0.255


m1 = 0.1168

m2 = 0.0648Moment along short span,

= 57.832 kN-mM1= W(m1+0.15m2)

,

= 37.628 kN-m


M1 = 57.832 kN-m

M2 = 37.628 kN-m




M2= W(0.15m1+m2)


u/Bs = 0.449

v/Ls = 0.255From the Pigeauds curve,

m1 = 0.1353

m2 = 0.0712


= 48.480 kN-mM1= W(m1+0.15m2)


= 30.386 kN-m


M1 = 48.480 kN-m

M2 = 30.386 kN-m

Final Moment

M1 = 4.676 kN-m

=

M2= W(0.15m1+m2)

= . -

Bending moment due to load W5:

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

For X- X Axis

W = 62.5 kN






u/Bs = 1.000

v/Ls = 0.084

From the Pigeauds curve,m1 = 0.0935

m2 = 0.0742

Moment alon short s an, ,

= 25.650 kN-m


= 21.628 kN-m


M1 = 25.650 kN-m

M2 = 21.628 kN-m



M1= W(m1+0.15m2)

M2= W(0.15m1+m2)

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u/Bs = 0.745

v/Ls = 0.084

Pa


m1 = 0.1157

m2 = 0.0944


= 23.718 kN-mMoment along long span,

= 20.412 kN-mM2= W(0.15m1+m2)

M1= W(m1+0.15m2)

,

M1 = 23.718 kN-m

M2 = 20.412 kN-m

Moment Along X-X

M1 = 0.966 kN-m

M2 = 0.608 kN-m

For Y- Y Axis

W = 62.5 kN






u/Bs = 0.618

v/Ls = 0.255


m1 = 0.1168

m2 = 0.0648


= 57.832 kN-m


= 37.628 kN-m


M1= W(m1+0.15m2)

M2= W(0.15m1+m2)

= . -m

M2 = 37.628 kN-m





u/Bs = 0.449

v/Ls = 0.255


m1 = 0.1353

m2 = 0.0712


= 48.480 kN-mMoment along long span,

= 30.386 kN-m

Bendin moment includin im act and continuit

M1= W(m1+0.15m2)

M2= W(0.15m1+m2)

,

M1 = 48.480 kN-m

M2 = 30.386 kN-m

Final Moment

M1 = 4.676 kN-m

M2 = 3.621 kN-m

Resultent Moment

M1 = 5.642 kN-m

M2 = 4.230 kN-m

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

For X-X Axis

Pa

W = 37.5 kN



Loaded area = 0.650 m2Total applied load = q x area = 126 kN


u s = .

v/Ls = 0.084


m1 = 0.1036

m2 = 0.08325


= 14.598 kN-m


M1= W(m1+0.15m2)

= 12.423 kN-m


M1 = 14.598 kN-mM2 = 12.423 kN-m





M2= W(0.15m1+m2)

u/Bs = 0.345

v/Ls = 0.084


m1 = 0.1765

m2 = 0.1312


= 9.957 kN-mM1= W(m1+0.15m2)

,

= 8.002 kN-m


M1 = 9.957 kN-m

M2 = 8.002 kN-m

Moment Along Y-Y

M1 = 2.321 kN-m

M2 = 2.210 kN-m

M2= W(0.15m1+m2)

For Y- Y Axis

W = 37.5 kN



Loaded area = 1.418 m2Total applied load = q x area = 274 kN


u/Bs = 0.618.

v/Ls = 0.255


m1 = 0.1168

m2 = 0.0648


= 34.699 kN-m


= 22.577 kN-m

M1= W(m1+0.15m2)

M2= W(0.15m1+m2)

Pa


10/30


M1 = 34.699 kN-m

M2 = 22.577 kN-m


Pag




u/Bs = 0.449

v/Ls = 0.255From the Pigeauds curve,

m1 = 0.1353

m = .


= 29.088 kN-m


= 18.231 kN-m


M1 = 29.088 kN-m

M2 = 18.231 kN-m

M1= W(m1+0.15m2)

M2= W(0.15m1+m2)

Final Moment

M1 = 2.806 kN-m

M2 = 2.173 kN-mResultent Moment

M1 = 5.127

M2 = 4.383

Total Moment Due to IRC Class AA Wheeled Vechicle

Moment alon short s an M1 = 32.309 kN-m , .

Moment along long span,M2 = 25.539 kN-m

c) Due to Live load: Class A Loading

IRC Class A Loading: For maximum bending moment one wheel of 57kN should be placed at the

centre of span and other at 1.2 m from it as shown. Neglecting small eccentricity of 80mm.

Tyre contact length along short span, Y = 0.5 m

Tyre contact length along long span, X = 0.25 m

Imaginary load W3 = W2 is placed on the other side of W1 to make loading symmetrical.

Due toloads W2 & W3 Bendin moment at center of anel will be that due to load W1andhalf. .

Pag


11/30

Y

Pag

57kN

W1 W2

Y

X X57kN

Fig: 6


Wheel load, W1 = 57 kN

Loaded length, u = 0.696 m

Loaded width, v = 0.465 m


u/Bs = 0.35

v/Ls = 0.10


m1 = 0.1717

m2 = 0.1245


= 10.851 kN-m


= 8.565 kN-m


M1= W(m1+0.15m2)

M2= W(0.15m1+m2)

= . -m

M2 = 8.565 kN-m


Wheel load is placed unsymmetrical wrt the Y-Y

Wheel load, W2 = 57 kN






u/Bs = 0.637

v/Ls = 0.348


m1 = 0.10843.

m2 = 0.0497


= 40.676 kN-m


= 23.154 kN-m


M1 = 40.676 kN-m

M2= W(0.15m1+m2)

M1= W(m1+0.15m2)

= . -m





u/Bs = 0.430

v/Ls = 0.348


m1 = 0.1313

Pag


12/30

m2 = 0.0552


= 33.081 kN-m


M1= W(m1+0.15m2)

Pag

= 17.751 kN-m


M1 = 33.081 kN-m

M2 = 17.751 kN-m

Final Moment

M1 = 3.798 kN-m

M2 = 2.702 kN-m

M2= W(0.15m1+m2)

Total Bending Moment due to load W1 & W2 will be,

M1 = 14.6 kN-m

M2 = 11.3 kN-m

Design Bending Moment due to LL:

M1 = 32.3 kN-m

M2 = 25.5 kN-m

Calculation of Shear Force

a) Due to Dead load:

Dead load shear force = 7.04 kN

b) Due to Live load: Class AA Tracked Vehicle

Load of Tracked Vehicl = 350 kN

350kN 350kN

Dispertion in the direction of span, = 1.45 m

For maximum shear, load is kept such that whole dispersion is in the span. That is at 0.725 m

from the edge of beam. 0.3

Fig 7.a

Effective width of slab =

Span Ratio (L/B) = 2.25

afor continuous slab

= 2.60x = 0.725 m

bw = 3.76

Therefore effective width of slab = 4.96 m

bwl

x1x +

Load per meter width = 70.54 kN

Shear force at left edge = 44.97 kN

Shear force including impact & continuity = 44.97 kN

c) Due to Live load: Class AA Wheeled Vehicle

37.5kN 37.5kN62.5kN 62.5kN Pag


13/30

37.5kN 37.5kN62.5kN 62.5kN

Pag

Dispersion width in the direction of span = 0.900 m

Loads are placed such that outermost load is at distance of

x = 0.450 m from edge of the beam.

bw = 0.31

Effective width for first wheel = = 1.217 m

.

Wheel Load = 62.50 kN

But the center to center distance of two axel are 1.2 m, thus effective width will overlap.

Average effective width for one wheel = 1.208 mPortion of load in span = 1.000 m

Load per meter width of slab = 51.72 kN

For second wheel Wheel Load = 62.50 kN

x = 0.550 m.

Effective width for second wheel = 1.347 m

But the center to center distance of two axel are 1.2 m, thus effective width will overlap.

Average effective width for one wheel = 1.273 m

Load per meter width of slab = 49.1 kN

For third wheel Wheel Load = 37.50 kN

X = -0.050 m

ect ve w t or t r w ee = .

Load Acting on Span = 16.67 kN

Acting at 0.2 m from Support

Effective width for third wheel = 0.918 m < 1.2 m

Load per meter width of slab = 18 kN

Shear force at left edge = 37.3 kN

Shear force including impact and continuity

= 37.318 kN

b) Due to Live load: IRC Class A loading

Fig 7.c

57kN 57kN

Pag


14/30

Shear force due to load W1: 57 kN

Fig 7.c

Pag

Dispersion width in the direction of short span = 1.1 m

For maximum shear force, the load should be placed at distance of 0.55 m from web

of girder. In this position second load will be as shown.

Effective width for first wheel =

Where, x= 0.55 m

bw= 0.41 m

bw

l

x1x +

.

L/B= 2.3 = 2.6

Therefore, Effective width = 1.447 m

But distance between axels is 1.2 m and hence effective width overlaps.

Average effective width / wheel = m

Load W1 = kN

Load per meter width of slab = kN/m

And shear force = kN

57.00

43.07

32.30

1.32

Shear force including impact & continuity = kN

Shear force due to load W2:

Effective width for second wheel =

Where, x= 0.350 m

bw= 0.41 m

.

bwl

x1x +

L/B= 2.4 = 2.6

Therefore, Effective width = 1.161 m < 1.2 m

Load W2 = 57.0 kN

Effective load = 18.1 kN

15.6 kN

And shear force = 12.89 kN

Shear force including impact & continuity = 12.89 kN

Total shear force = 45.19 kN

Load per meter width of slab =

Total Design Bending Moments,

M1 = 34.69 kN-m

M2 = 26.32 kN-m

Total Design Shear force,

S.F. = 52.23 kN

Effective depth required,

d = 171.4 mm

Use 30 mm Clear cover & 12 mm dia

Effective depth available = 184 mm O.K.

Area of steel required along short span,

Ast = 1055 mm2

Check for minimum area of steel:

=Rb

=

djst

M

Pag


15/30

264 mm2

Provide 12 mm bars 100 mm c/c at both top and bottom, giving area of

steel = 1130.4 mm2 > required.

Effective depth for long span = 173 mm

Min. area of steel @

Pag

Area of steel required along long span,

Ast = 851 mm2

Provide 12 mm bars 100 mm c/c at both top and bottom, giving area of

steel = 1130.4 mm2 > required.

Check for shear:

=

djst

M

Nomin v = 0.284 N/mm2

Provided percentage area of tensile steel = 0.51 %

Permissible shear stress,c= 1.16 x 0.313 = 0.363 N/mm2 > O.K.

=

db

Vu

3.0 Design of Longitudinal Girder

Effective Span of Bridge = 24 m

Slab thickness = 0.22 m

Width of Rib = 0.4 m

Spacing of main Beam = 2.4 m c/c

Over all depth of Beam = 2 m

3.1 Calculation of dead load moment and shear force on longitudinal girder:

Let the over all depth of the longitudinal girder be 2000 mm, the depth of its rib will

= 1.78 m

Weight of Rib per m = 17.09 kN/m

Dead load from each cantilever portion (refer design of cantilever slab)

= 7.68 kN/m

Dead load of slab & Wearing coat = 7.04 kN/m2

Total Dead load per m from deck = 51.98 kN/m

This load is borne by all the three girders

Dead Load per girder due to Deck Slab = 17.33 kN/m

Let the Depth of rib of cross girder to be = 1.28 m

let its width be = 0.3 m

Weight of rib of cross girder = 9.216 kN/m

Length of each cross girder = 4 m

It is assumed that the weight of each cross girder is equally borne by the entire threelongitudinal girders. This weight acts as point load on each girder its value being

= 12.29 kN

Total UDL = 34.413333 kN/m

RA= RB = 449.824 kN

Bending Moment (BM)

12.3 kN

Fig : 8RA RB

34.413 kN/m

12.3 kN12.3 kN12.3 kNkN 12.3 kN

Pag


16/30

BM at Centre of span = 2654.707 kNm

BM at th of span = 2064.758 kNm

BM at 3/8th

Span = 2492.474 kNmth

Pag

BM at 1/8 pan = 1157.748 kNm

Shear Force (SF)

SF at Support = 437.536 kN

SF at 1/8th Span = 334.296 kN

SF at 1/4thSpan = 218.768 kN

SF at 3/8th

Span = 115.528 kN

SF at Center of span = 0 kN

Distance from Support BM SF

At Center of Span 2654.71 0.00

At 3/8th

Span 2492.47 115.53

At 1/4th Span 2064.76 218.77

At Support 0.00 437.54

3.2 Calculation of live load moment and shear force on longitudinal girder:

Impact factor for:

0.150

=

=+ L6

5.4

.

IRC Class AA Wheeled Vehicle = 0.215

Distribution of live loads on longitudinal girder for bending moment:

IRC Class AA Tracked Vehicle:

Reaction on the girder will be maximum when the eccentricity is maximum. Eccentricity will be maximum when the loads

are very near to the kerb. Position of loads for maximum eccentricity is shown in figure.

All the girders are assumed to have the same moment of inertia.

W1 W1

Reaction factor for Outer Girder, RA = 0.81 W1

Reaction factor for Inner Girder, RB = 0.667 W1

( ) =

+ 35.04.2

2.42I

I31

3

W2

2

1

=W2

1

.

If W be the axel load, then wheel load W1= W/2

Then reaction factor, RA = 0.406 W

RB = 0.333 W

IRC Class A Loading:

Position of loads for maximum eccentricity is shown in figure.

W1 W1 W1 W1

Pag


17/30

W1 W1 W1 W1

Pag

Reaction on Outer Girder RA,

RA= 1.625 W2( )

=

+ 1.04.2

2.42I

I31

3

W4

2

1

Fig. 10

RB = 1.333 W2

If W be the axel load, then wheel load W2 = W/2

Then reaction factor, RA = 0.813 W

RB = 0.667 W

Bending Moment due to Live load: IRC Class AA Tracked Vehicle

[ ]=+ 013

4W1

The influence line diagram for bending moment is shown in figure.

Effective span of girder, le = 24.0 m

ITC Class AA Tracked Vehicle Load: = 700 kN

Ordinate of Bending Moment at considered section, Mx= xL

x1

Ordinate of Influence line at mid span = 6.0 m 12

Leff= 24 m

5.1 6

.

Leff

700 kN3.6m

Bending Moment = 3885.0 kN-m

Bending Moment including impact and rection factor for outer Girder = 1736 kN-m

Bending Moment including impact and rection factor for Inner Girder = 1425 kN-m

Calculation of bending moment at 3L/8. = 9 m Leff= 24 m


Fig. 11a: ILD for BM at L/ 2RA B

4.5 4.95

5.63

Leff

Fig. 11b: ILD for BM at 3L/8RA R

700 kN3.6m

3*Leff / 8

en ng omen = . m

Bending Moment including impact and rection factor for Outer Girder = 829.3 kNm

Bending Moment including impact and rection factor for Inner Girder = 680.4 kNm

Calculation of bending moment at L/4. = 6 m Leff= 24 m


Leff

RB

700 kN3.6m

Leff / 4

Pag


18/30

3.15

4.500 4.05

700 kN

Leff / 4

Pag

Bending Moment = kNm

Bending Moment including impact and rection factor for Outer Girder = kNm

Bending Moment including impact and rection factor for Inner Girder = kNm

Calculation of bending moment at L/8. = 3 m Leff= 24 m


1500.1

670.3460

550.0275

Fig. 11c: ILD for BM at L/4A

Leff

1.050 2.400

2.625

Bendin Moment = kNm881.2

Fig. 11d: ILD for BM at L/ 8RA RB

700 kN3.6m

Leff / 8

Bending Moment including impact and rection factor for Outer Girder = 393.792 kNm

Bending Moment including impact and rection factor for Inner Girder = 323.11 kNm

Bending Moment due to Live load: IRC Class A Loading

The influence line diagram for bending moment is shown in figure.

Effective span of girder, le = 24 m

Loads Values Unit Loads

W1 27 kN W5 68 kN

.

Values

W2 27 kN W6 68 kN

W3 114 kN W7 68 kN

W4 114 kN W8 68 kN

Distances Values Unit DistancesX Varies X5 4.3 m

X1 Varies X6 3 m

X2 1.1 m X7 3 m

Values

X3 3.2 m X8 3 mX4 1.2 m X9

Calculation of bending moment at L/2 = 12 m, when load W4 is at L/2

Ordinate of Influence line at mid span = 6 m Leff= 24 m

Ordinate of Bending Moment at considered section, Mx=

Varies

xL

x1

x1= 6.5

6.00

X= 12

Position from

Maximum

Load Values kN Moment ComponentLoad Nos. IL Ordinate

Fig. 12a: ILD for BM at L/2RA RB

W1=27kN W2=27kN W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN

Total = kN-m

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

57.80

68

10.30 0.850

114 5.400

114

68

W1 27 87.75

W2 102.60

-5.50

27

615.60

7.30

W5

W3

-4.40

3.250

3.800

1969.35

1840.11

684.00

261.80

W4 6.000

3.850

W7

W6

68

-1.20

0.00

4.30

2.350 159.80

Pag


19/30

Total Bending Moment including impact for Inner Girder = kN-m

Calculation of bending moment at 3L/8 = 9.0 m, when load W3 is at 3L/8

Maximum Ordinate of Influence line = 5.625 m Leff= 24 m

1509.84

Pag

x1= 4.7

5.625

X= 9.0 Fig. 12b: ILD for BM at 3L/8RA RB

W1=27kN W2=27kN W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN W8=68kN

114

3.625

5.625

1.20

5.50

97.88

79.31

r nate

641.25

242.25

2.938

os t on rom

Maximum

-4.30

-3.20

0.00

oa os.

W1

W4 589.955.175

oa a ues

W3

W2 27

114

oment omponent

165.75

27

68

8.50

W5

W6 2.43868

3.563

Total = kN-m



W8

1918.39

1792.49

14.5068 12.750.188

.

1470.76

. .



X1= 1.7

4.50

X= 6RA RB


-3.20 2.100 239.40

W3 114 0.00 4.500 513.00

Moment Component

W1 27 -4.30 1.275 34.43

Load Nos. Load Values kN Position from

Maximum

IL Ordinate

W2 114

.

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


W8 68 14.50 0.000 0.00

W6 68 8.50 2.375 161.50

W7 68 11.50 1.625 110.50

. . .

W5 68 5.50 3.125 212.50

1556.931454.75

1193.64



2.625

X= 3

Fig. 12d: ILD for BM at L/ 8RA RB

W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN W8=68kN

Pag


20/30

Load Nos. Load Values kN Position from

Maximum

IL Ordinate Moment Component

Pag

68 1.20 2.475

114 0.00

8.50 1.563

68 11.50

299.25

W1 27 0.00

0.000 0.00

114

W6

68 5.50 1.938

68

1.188

68 14.50 0.813

168.30

131.75

0.000 0.00

W2

W5

W3

W4

0.00 2.625

55.25

80.75

W8

106.25

W7

Total = kN-m



Absolute Maximum BM

841.55

786.32

645.19

Ordinate of Influence line at mid span = 6 m Leff= 24 m

6

x= 12

C.G of the Load system from outer 27 kN Wheel Load

Calculation of bending moment at the load point which is equidistance from resultant

Fig. 12e: ILD for BM at L/ 2 / Absolute Maximum BMRA RB

W1=27kN W2=27kN W3=144kN W4=144kN W5=68kN W6=68kN W7=68kN

R

.

= 6.420 m

The heavier wheel load near C.G. of load System is 114kN which lies at a distance of

6.42-(1.1+3.2+1.2)= 0.92 m from CGX = 0.46 m

6.04

7.14 3.570 406.98

Load Values kN Position from Left

support

IL Ordinate

27 3.020

114W2

W1 81.54

Load Nos. Moment Component

Total = kN-m


392.36

277.44

. .

0.00 0.000 0.00

.114 10.34 5.170

2.580

68

68

68 11.54 5.770

68 15.84

68 18.84

0.000

W3 589.38

W4

W6 175.44

0.00

4.080

W7 21.84

1796.93

1923.14

W8

W5

Total Bending Moment including impact for Inner Girder = kN-m1474.41

Pag


21/30

Pag

Shear Force due to Live load: IRC Class AA Tracked Vehicle.

At Support

Effective span of girder, le = 24 m

Load Class AA Tracked vehicle W1= 350 kN

Formaximum shear at support, load should be as near the support as possible.

The length of the load is 3.6m, the SF will be max. when the C. G. of the load is placed at a distance of 1/2*3.6=

1.8m From the support along its length, thus the load will lies between the support & the Ist Intermediate

X-girder, the width of track being 0.85m, the CG of load will thus lie at a distance of 1.2+0.85/2=1.625 m

from kerb of footpath Load act at a distance of 1.8 m from support A, B and C

L= 4.8 m C/C Distance of L Girder= 2.4 m

X= 1.8 X1= 3.0 a= 0.6 b= 1.025

C= 1.625 d= 2.05 f= 2.050 e= 0.35

g= 1.375 h= 0.675 i= 1.725

Loads on Girders,PA= 0.573 W1

kerb Line

'

L

PB= 1.146 W1PC= 0.281 W1

Reaction at support,RA= 0.358 W1RQ= 0.215 W1RB= 0.716 W1RR= 0.430 W1R =

A

B

CL of outer L Girder

CL of inner L Girder

c

C/C

dIstX-G

irder

ediateX-Girder

Q

R

Q

R

b

RR'

C/C

C/Cf

e

g

h

.RS= 0.105 W1

The loads on the cross girder i.e. RQ, RR& RS are to be distributed by normal Courbon's theory.

Total load, W = W1 = 0.750

C.G. of loads from Q = 2.05 m

kerb Line

CL of outer L GirderC

a

Inter

S SRS'

i

X X1

Fig. 13

Eccentricity, e = 0.35 m

Reaction factor for outer girder,FQ= 0.305 W1in this case xi=2.4 m

and xi =(2.4) +(0) +(2.4) = 2 x (2.4)

Reaction factor for inner girder,FR= 0.250 W1

in this case xi=0 m

RAdue to FQ= 0.24375 W1RBdue to FR= 0.200 W1 `

Total reaction on outer Girder = 0.602 W1Total reaction on inner Girder = 0.916 W1

Max shear at support including impact for outer girder = 231.7 kN

Max shear at su ort includin im act for inner irder = 352.7 kN

It may be seen that the reaction FQand FRact as load at 1/3 span of outer longitudinal girder and inner longitudinal

girder respectively. The reactions at support A and B due to those loads are

.

At Intermediate Section


At Left =

At Right =

Ordinate of Bending SF at considered section, SFx

L

x1

L

x11

Pag


22/30

Shear at 1/8th span

Calculation of bending moment at L/8 = 3 m, when load Placed at Just Right of L/8

Ordinate of Influence line at Left = 0.875 m

Pag

Ordinate of Influence line at Right = 0.125 m Leff= 24 m

x= 3 a'= 0.125 a= 0.875 b= 0.725

Fig. 14a: ILD of SF at L/8 of SpanRA RB

350 kN3.6m

a

a'

b

S.F. = 280 kN

S.F. including impact for outer girder = 125.13 kN

S.F. including impact for inner girder = 102.667 kN

Shear at 1/4th span

Calculation of bending moment at L/4 = 6 m, when load Placed at Just Right of L/4


Ordinate of Influence line at Right = 0.25 m Leff= 24 m

x= 6 a'= 0.25 a= 0.750 b= 0.600

Fig. 14b ILD of SF at L/4 of SpanRA RB

350 kN3.6m

a

a'

b

S.F. = 236.25 kN



Shear at 3/8th span

Calculation of bending moment at 3L/8 = 9 m, when load Placed at Just Right of 3L/8

Ordinate of Influence line at Left = 0.625 mOrdinate of Influence line at Right = 0.375 m Leff= 24 m

x= 9 a'= 0.375 a= 0.625 b= 0.475

Fi . 14cILDofSFat3L/8 ofS anRA RB

350 kN3.6m

a

a'

b

S.F. = 192.5 kN



Shear at 1/2th spanCalculation of bending moment at L/2 = 12 m, when load Placed at Just Right of L/2


Ordinate of Influence line at Ri ht = 0.5 m Leff= 24 m

.

x= 12 a'= 0.5 a= 0.5 b= 0.350

Fig. 14d ILD of SF at L/2 of SpanRA RB

350 kN3.6m

a

a'

b

Pag


23/30

Shear Force due to Live load: IRC Class A Load.

The influence line diagram for shear force is shown in figure.


Loads Values Loads Values

Pag

W1 27 kN W5 68 kN

W2 27 kN W6 68 kN

W3 114 kN W7 68 kN

W4 114 kN W8 68 kN

Distances Distances Values

X Varies X5 4.3 m

Values

X1 Varies X6 3 m

X2 1.1 m X7 3 mX3 3.2 m X8 3 mX4 1.2 m X9 Varies

At Left = At Right =

Ordinate of Bending SF at considered section, SFx

L

x1

L

x11

Pag


24/30

Calculation of Shear Force at Support = 24.0 m, when load W3

Ordinate of Influence line y3= 1 Leff= 24.0 m


Pag

Load PositionLoad Nos. IL Ordinate Moment Component

Fig. 15a: ILD for SF at Support

RA RB

Y3 Y4Y5 Y6 Y7

Y8

a ues

kN

rom e

support

114 0 Y3=

114 1.20 Y4=

68 5.50 Y5=

68 8.50 Y6=

68 11.50 Y7=

68 14.50 Y8=

Total = kN

W4 108.300.950

W7

W6

1.000

380.97

52.420.771

0.646

0.521

0.396

43.92

W3 114.00

W5

35.42

W8 26.92

Total SF including impact for Outer Girder = kN

Total SF including impact for Inner Girder = kN

Calculation of SF at L/8 = 3.000 m, when load W3

Ordinate of Influence line At Right 0.875

At Left 0.125 Leff= 24.0 m x= 3

292.07

355.97

.


Load Position Moment ComponentLoad Nos. IL Ordinate

Fig. 15b: ILD for SF at L/8RA RB

Y3 Y4Y5 Y6 Y7

Y8

Values

kN

from Left

support

114 0 Y3=

114 1.20 Y4=

68 5.50 Y5=

68 8.50 Y6=

68 11.50 Y7=

68 14.50 Y8=

=

94.05

26.92W7

W5

W4

W8

W3

W6

43.92

0.875

0.825

0.646

0.521

0.396

35.42

99.75

0.271 18.42



=

.

280.36

230.04

. ,


At Left 0.25 Leff= 24.0 m x= 6.00

Fig. 15c: ILD for SF at L/4

Y3

Y3'

= = = = = = = =

Y4 Y5Y6 Y7

Y8Y2Y1

Pag


25/30

Load

Values

Position

from Left/

Load Nos. IL Ordinate Moment Component

Pag

kN Right at X

27 4.3 Y1=

27 3.2 Y2=

114 0 Y3=

114 1.20 Y4=68 5.50 Y5=

68 8.50 Y6=

=

26.92

W5

W6

79.8035.42

W3

W4

W2 -3.15

W1 -0.071 -1.91

85.50

-0.117

0.750

0.7000.521

0.396

.

68 14.50 Y8=

Total = kN



Calculation of SF at 3L/8 = 9.000 m, when load W3


At Left 0.375 Leff= 24.0 m x= 9.0

250.90

234.44

0.146

.

9.92

192.36

W8

.

Fig. 15d: ILD for SF at 3L/8RA RB

Y3

Y3'


Y4 Y5Y6 Y7

Y8Y2Y1

Load

Values

kN

Position

from Left/

Right at X

27 4.3 Y1=

27 3.2 Y2=

114 0 Y3=

114 1.20 Y4=

68 5.50 Y5=

-6.53

71.25W3

-5.29-0.196

W4 65.55


-0.242

0.625

0.575

W1

26.92W5 0.396

W2

.

68 8.50 Y6=

68 11.50 Y7=

68 14.50 Y8=

Total = kN



=

W6 18.42

W7 9.92

W8 1.42

181.65

169.73

139.27

..0.271

0.146

0.021

. ,


At Left 0.5 Leff= 24.0 m x= 12.0

Y3Y3'


Y4 Y5Y6 Y7

Y2Y1

Load

Values

kN

Position

from Left/

Right at X

27 4.3 Y1=

27 3.2 Y2=


W1

W2

-8.66

-9.90

-0.321

-0.367

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

Pag


26/30

114 0 Y3=

114 1.20 Y4=

68 5.50 Y5=

68 8.50 Y6=

W3 57.00

W5 18.42

W6

0.500

0.450

0.271

0.146

W4 51.30

9.92

Pag

.

68 11.50 Y7=Total = kN



Design of Section:

Total Design Bending Moments and Shear Forces for Outer Girder:

Section

119.49

Shear Forces kNBendin Moment kN-m

W7 1.420.021

111.65

91.61

Due to

DL

Due to LL

X = 0 0.00 0.00

X = L/8 1157.75 393.79

X = L/4 2064.76 670.35

X = 3L/8 2492.47 829.26

X = L/2 2654.71 1736.11 111.646

234.439

334.296

Due to DL

111.646

Due to LL

355.966

280.359

169.733

Total

1551.540

0.000

Total

218.768

115.528

437.536

4390.817 0.000

2735.104

3321.735

793.502

614.655

453.207

285.261

o a es gn en ng omen s an ear orces or nner r er:

Section

Due toDL

Due to LL

X = 0 0.00 0.000

X = L/8 1157.75 323.111

X = L/4 2064.76 550.028

X = 3L/8 2492.47 680.419 254.796

2614.79 411.128

3172.89 115.53 139.27

352.720.000

564.334

Shear Forces (kN)

Total Due to DL Due to LL

437.54

334.30

790.252

Bending Moment (kN-m)

1480.859 230.04

192.36218.77

Total

X = L/2 2654.71 1424.50

Design of Outer Girder:

Overall depth of beam, D = 2000 mm

Rib width, bw = 400 mm

Flange width of T-beam will be,

bf = bw + 1/5 x lo 2.5 m > 2.4 m

Therefore width of flange, bf = 2400 mm

-

91.60791.614079.21 0.00

- ,

d = 1830

Area of steel required,

Ast= 13420 mm2

Provide 16 nos. of 32 2000nos. of

mm bars+

mm bars

=

djst

M

.

Provided area of steel = 13843 mm2

Total Provided area of steel = 13843 mm2

Number of bars in bottom row = 4 nos.

Width of beam = 400 mm 400

Side and bottom clear cover to bars = 40 mm

C.G. of the bottom row of bars from bottom = 56 mm

Clear distance between vertical bars = 32 mm

Fig. 16 : Cross Section of Girde

C.G. of the Second row of bars from bottom = 120 mm

C.G. of the third row of bars from bottom = 184 mm

C.G. of the fourth row of bars from bottom = 248 mm

C.G. of the fifth row of bars from bottom = 308.5 mm

C.G. of the bar group from bottom = 163.1 mm

170 mm O.K.

Effective Depth = 1830 mm

Df = 220 mm

Pag


27/30

Check for stresses:

Calculation of depth of neutral axis:

Assuming that the

effective area in of compression and tension sides about neutral axis, we get

Pag

bwxa2+ (bf bw) Df (xa Df/2) = m Ast (d xa)

Solving, we get, xa= 482 mm

Let compressive stress in concrete at top of flange = And compressive stress in concrete at bottom of flange = '

Then, = 0.543 =

a

fa

x

Dx

Position of C.G. of compressive stress in flange from top, x1

= 99.1 mm

Compressive force in flange, C1 = x( + ')x Bf x Df= 407402

Com ressive force in rib C2 = x 'x Xa Df x Bw

=+

+3

D

'

'2 f

,

= 28420

C.G. of compressive force in rib from top, x2

307.2 mm

Total compressive force, C = 435822.3

C.G. of total compressive force from top

112.7 mm=+

=

C2C1

x2C2x1C1

Therefore, lever arm, jd = 1717.3 mm

Critical Neutral axis depth, nd = 582.3 mm < xa

Moment of resistance of the section is given by

Mr = =

699626722.86

=+

cm1

d

st

+

yddb ff2

1

Equating Mr to external B.M we get

Mr =

= 6.28 < 8.3 O.K. Stress Developed in Steel Reinforcement is given by

t = = 197.6 < 200 O.K.

4390816575.00

a

a

x

xdm

Check for minimum area of steel

Minimum area of tension steel in beam @ 0.2 % of web area

1600 mm2 < Ast provided = mm

2O.K.

Design for shear:

. N/mm2

13843

=

=dB

Vv

Assuming ####### nos. of 32 will be continued up to support, then provided

percentage area of tension steel = 1.00 %

Permissible shear stress,c= 0.420 N/mm2 < v

Vs = V - tc . bw . D = 457195 N

Assuming 12 mm 2-legged vertical stirrups having area of steel, Asv = 226 mm2

Spacing, S = 181 mm c/c

Shear reinforcement is required. Shear reinforcement shall be provided to carry a shear of,

=

Vs

dAsv st

Pag


28/30

As per minimum shear reinforcement requirements, maximum spacing,

S ax= 283 mm c/c.

Vs

=

st Asv

Pag

.

Hence provide 12 mm, 2-legged vertical stirrups @ 100 mm c/c at support.Design summary:

Tension Reinforcement (Fe 415):

No. Dia.

Area of Steel RequiredSection

Area of steel required and provided at different sections of Outer girder are given in below:

Area

BM Area of Steel Provided

wb0.4

2 25

3L/8 3321.73 KNm 14 32 mm2

L/4 2735.10 KNm 12 32 mm2

L/8 1551.54 KNm 10 32 mm2

Support 0.00 KNm 10 32 mm2

Shear Reinforcement (Fe 415):

10153

8038

11254

9646

8038mm2

mm20

mm2

mm28360

4742

4390.82 13420 mm2KNmL/2 13843 mm2

Section

L/2 111.646 kN 10 @ 1029.4 mm 10 @ 250

3L/8 285.261 kN 10 @ 402.87 mm 10 @ 250

L/4 453.207 kN 10 @ 253.58 mm 10 @ 200

L/8 614.655 kN 12 @ 269.24 mm 12 @ 150

Support 793.502 kN 12 @ 208.56 mm 12 @ 100

2-legged vertical Stirrups provided

mm

mm

mm

mm

SF 2-legged vertical Stirrups required

Area of steel required and provided at different sections of Inner girder are given in below:

mm

ens on e n orcemen e :

No. Dia.

16 32

2 25

3L/8 3172.89 KNm 14 32 mm2

L/4 2614.79 KNm 12 32 mm2

L/8 1480.86 KNm 10 32 mm2

11254

9646

8038

Section BM

7992 mm2

Area of Steel Required

4526 mm2

13843

9698 mm2

Area

Area of Steel Provided

L/2 4079 KNm 12468 mm2 mm2

uppor . m mm

Shear Reinforcement (Fe 415):

Section

L/2 91.607 kN 10 @ 150 mm 10 @ 250

3L/8 254.796 kN 10 @ 150 mm 10 @ 250

L/4 411.128 kN 10 @ 150 mm 10 @ 200

L/8 564.334 kN 12 @ 150 mm 12 @ 150

Support 790.252 kN 12 @ 100 mm 12 @ 100

SF

mm

mm

mm

mm

2-legged vertical Stirrups provided

mm

2-legged vertical Stirrups required

mm

Pag


29/30

Pag

.

Dead Load

Overall Depth of cross girder = 1.5 m 2.4

Width of cross girder = 0.3 m

Self weight of cross girde = 9.216 kN/m 4.8

Dead load from slab = 20.2752 kNFig. 17

This load is assumed as uniformly distributed load per meter run = 8.448 kN/m

Total Dead load per meter run = 17.664 kN/m

Assuming, cross girder as rigid, reaction on main girder = 13.52 kN

Live Load: IRC Class AA Tracked Vehicle

Maximum bending moment occurs when one wheel of a vehicle lies near center of span.

Position for maximum bending moment is shown in figure. Deck Slab is assumed to

be simply supported. The critical supported between two cross girder.

=

l

lW

2/8.1

=

djst

M

Fig. 19

W1 W1

Effective load coming on cross girder = 569 kN=

l

lW

2/8.1

=

djst

M

Fig. 19

W1 W1

Reaction on each longitudinal girder = 189.58 kN

Maximum B.M. occurs under the load, = 260.68 kNm

Bending moment including impact = 286.74 kNm

Dead Load Bending Moment at the section, = 1.8876 kNm

Total bendin moment = 288.6 kNm

=

l

lW

2/8.1

=

djst

M

Fig. 19

W1 W1

.

LL Shear force including impact = 208.5 kN

Total shear force = 222.1 kN

Therefore, Design Moment = 288.6 kNm

Design Shear Force = 222.1 kN

=

l

lW

2/8.1

=

djst

M

Fig. 19

W1 W1

ross g r er s es gne as T-Beam.

Assuming effective depth, d = 1450.00 mm

Area of tension steel required = mm2

Minimum area of tension steel in beam @ 0.2 % of web area =

900 mm2 < 1113.4 mm2

Hence provide 3 nos. of 25 mm bars + 0 nos.of 20

bars having area of steel = 1472 mm2 .

1113.37

=

l

lW

2/8.1

=

djst

M

Fig. 19

W1 W1

Pag


30/30

= dB

V

=

Vus

dAsvst

=

wb0.4

Asv0.87fy

Design for shear:

Nominal shear stress, v= 0.51 N/mm2< max= 1.9 N/mm

2 O.K.

Provided percentage area of tension steel, p = 0.33 %

Permissible shear stress,c= 0.275 N/mm2 < 0.51N/mm O.K.

Shear reinforcement is required. Shear reinforcement shall be provided to carry

a shear force of

= dB

V

=

Vus

dAsvst

=

wb0.4

Asv0.87fy

us = u - c . w . =

Assuming 10 mm 2-legged vertical stirrups having area of steel, Asv = 157 mm2

Spacing, S = 327 mm c/c

As per minimum shear reinforcement requirements, maximum spa 10 mm 2 legged

vertical stirrups, 472 mm

= dB

V

=

Vus

dAsvst

=

wb0.4

Asv0.87fy

Hence provide 10 mm 2-legged vertical stirrups @ 150 mm c/c through out the length of

end cross girder and 10 mm 2-legged vertical stirrups @ 150 mm c/c through out the length

Elastomeric Bearing On Bridge Used

According IRC 83-part II it is reccomended use of elastomeric bearins of size (250X400X50 ) mm embedding 5

of intermediate cross girder.

= dB

V

=

Vus

dAsvst

=

wb0.4

Asv0.87fy

p a es o mm c ness an mm c earance n p an = mm or er ca oa .

= dB

V

=

Vus

dAsvst

=

wb0.4

Asv0.87fy

Dsign of Bridge Superstructure Using Indian Codes

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