Total width of the bridge = 7.000 m

Clear span of the bridge = 15.000 m

No. of main Girders = 3

Spacing of the T-beams/main girders(c/c) = 2.250 m

Thickness of kerb 0.300 m

Width of kerb(including parapet wall) 0.300 m

Width of parapet wall 0.150 m

Width of the bearing 1.000 m

Effective span of the bridge = 16.000 m

spacing of the cross beam 4.050 m

Provide one at each end of the span, total no.

of cross beams= 6.704

No.of cross beams provided= 7

Spacing of beams 2.667 m c/c

say 2.600 m

Thickness of girder 0.800 m

Thickness of deck slab 21 cm

Thickness of wearing coat 7.6 cm

unit weight of concrete 2400 kg/cum

Unit weight of wearing coat 2200 kg/cum

Depth of cross beams 120 cm

Thickness of cross beam 30 cm

Effective span of cross beams in transverse direction 1.450 m

Effective span of cross beams in longitudinal direction 2.300 m

Ratio of long span to short span L/B= 1.586

Length of catelever portion 0.850 m THE SLAB IS DESIGNED AS TWO WAY

CALCULATION OF DEAD LOAD

Dead load due to wearing coat 167.200 kg/sq.m

Dead load due to deck slab 504.000 kg/sq.m

671.200 kg/sq.m

Slab supported on all four sides and continuous.

From Pigaud's curve, for k= 0.630 for 1/k= 1.586207

m1= 0.049 m2= 0.018

Total dead weight of panel 2238.452 kg =

Moment due along short span 5674.476 kg-cm

Moment due along long span 11572.8 kg-cm

Live load

Placing the track symentrically

Impact Load:

Give type of vehicle (1 for wheeled, 2 for tracked) 1

span= 15.000 m

I= 21.44 %

DESIGN OF CANTELEVER PORTION

Thickness of slab at the end 15 cm

Thicknes of slab at girder 36 cm

sl.no L B D unit wt.t load dist. Moment

m m m kg/cum kg m kg-m

1 parapet 174.00 0.775 134.85

2 kerb 0.300 1.00 0.30 2400 216.00 0.700 151.20

3 weraing coat 0.076 1.00 0.55 2200 91.96 0.275 25.29

4 Slab(rect.) 0.150 1.00 0.85 2400 306.00 0.425 130.05

Item

5 Slab(triangular) 0.105 1.00 0.85 2400 214.20 0.283 60.69

1002.16 502.079

Momenyt due to live load

For Class AA loading, the minimum clearence shall be 1.2 m for carriage width of 5.5 m and above

In the present case the cantelever width excluding the kerb works out 0.550 m

Hence IRC Class A loading shall be considered

The loading will be as shown in the fig 1.

Effective width of dispersion 'e' is computed by Code clause 305.13.2

be=1.2x+bw

where be=effective width

x=dist. of the C.G. of conc. Load from the face of the the support= 0.150 m

bw=breadth of concentration area of the laod= 0.402 m

Therefore effective width= 0.582 m

When the wheel load is at the edge of the slab near abutment, the net effective width of

dispersion= 0.416 m

Live load/m width including impact= Wl x100x(1+I)/be 20552.9 kg

Maximum moment due to live load 3082.93 kg-m

4372.677

Reinforcement

Total moment due to dead load and live load= 3585.012 kg-m

Effective depth required 18.69 cm

Cover to reinforcement 5.00 cm

Dia of main steel 20 mm a_st= 3.14 sq.cm

Provide overall depth of D= 36.00 cm

Effective depth provided =d= 30.00 cm

Area of steel=Ast=[ M/t jd]= 7.07 sq.cm

spacing 44.47 cm c/c

Provide spacing of 30 cm c/c

Area of steel providede= 10.47 sq.cm

Bending moment for distribution steel:

=0.2 Mw + 0.3 Ml = 1025.296 kg-m

Dia of bar 12 mm a_st= 1.13 sq.cm

Effective depth =de= 30.40 cm

Area of steel=Ast=[ M/t jd]= 2.02 sq.cm

spacing 55.97 cm c/c

Provide spacing of 30 cm c/c

Area of steel providede= 3.77 sq.cm

DESIGN OF INTERMEDIATE LONGITUDINAL GIRDER

Bending moment due to dead load:

sl.no Item No. Factor L B D unit wt.t load

m m m kg/cum kg

1 weraing coat 1 1.00 1.00 2.25 0.076 2200 376.20

2 Deck Slab 1 1.00 1.00 2.25 0.21 2400 1134.00

3 T-Rib 1 1.00 1.00 0.80 1.71 2400 3290.88

4 Bottom flange a 2 0.50 1.00 0.15 0.15 2400 54.00

5 b 2 1.00 1.00 0.15 0.30 2400 216.00

6 Top fillet 2 0.50 1.00 0.15 0.15 2400 54.00

7 Cross beams 7 1.00 1.45 0.30 1.20 2400 548.10 *Divided by total length

7 Fillets 32 0.50 1.45 0.15 0.15 2400 78.30 *Divided by total length

5673.18

Maximum bending moment =WL^2/8= 181542 kg-m

Bending moment due to Live load

Maximum live load B.M would occur under Class A two lane loading

Impact factor fraction=A/[B+L]

Where A=constant factor= 4.5 For RCC bridges

B=constant factor= 6 For RCC bridges

L=Span in metres= 16.000 m

Therefore I= 20.5%

P P P P

1800 1700 1800 700

('g+w)

f g

w 550 1250 1000 As per IRC

950 f= 150

2250 g= 1200

w= 500

Transverse disposition of two trains of Class A loading for determination of

reactions on longitudinal beam

Rx=SW[1+SI dx e/(Sdx^2I)]

Live load B.M. by Courbons method

C.G of load from kerb 3050

C.G of of bridge 3200

Eccentricity of loading= 150 mm

n= 3

SW=n x W =where n= 4

P=W/2

wheel Location Total load Ra Rb Rc with impact factor

axle load Load of first SP 1 2 3 Ra Rb Rc

P wheel

1 1.35 1 4 2.25 0.00 2.25 1.467P 1.333P 1.200P 1.77 1.61 1.45

2 1.35 1 4 2.25 0.00 2.25 1.467P 1.333P 1.200P 1.77 1.61 1.45

3 5.7 1 4 2.25 0.00 2.25 1.467P 1.333P 1.200P 1.77 1.61 1.45

4 5.7 1 4 2.25 0.00 2.25 1.467P 1.333P 1.200P 1.77 1.61 1.45

5 3.4 1 4 2.25 0.00 2.25 1.467P 1.333P 1.200P 1.77 1.61 1.45

6 3.4 1 4 2.25 0.00 2.25 1.467P 1.333P 1.200P 1.77 1.61 1.45

7 3.4 1 4 2.25 0.00 2.25 1.467P 1.333P 1.200P 1.77 1.61 1.45

x

Dist. To axis from girder

70

For tracked wheels

For outer girder

1.625 2050

0.55

3600

Rp Rq

0.95 2.25

Ra Rb Rc

intensity of load= 19.44444 t/sq.m

Rx=(SP/n){1+SI/[SX2 I] X e}

Minimum clereance required between the road face of the kerb and outer edge of the track

= 1.2+width of track/2

e= 0.55 = 1.625 m

Ra= 0.9111 0.455556

Rb= 0.6667 0.333333

Rc= 0.4222 0.211111

Maximum bending moment at the centre of the span