Man Bridge Night 27-2-2012

8/2/2019 Man Bridge Night 27-2-2012

1/125

H.F.Level = 265.5

Reduced

Distance (H)

Reduced

Level

Segment

depth (D)

Difference in

depth

Sloped

Length

Root of

0 26425 261.75 3.75 3.75 25.28

50 256.2 9.3 5.55 25.61

75 256.6 8.9 -0.4 25.00

100 257.3 8.2 -0.7 25.01

125 257.3 8.2 0 25.00

150 260.71 4.79 -3.41 25.23

175 261.8 3.7 -1.09 25.02

200 262.6 2.9 -0.8 25.01

225 263.7 1.8 -1.1 25.02

250 264.8 0.7 -1.1 25.02

275 265.4 0.1 -0.6 25.01

287.5 -0.1

52.34

Segment width 25

Area 1308.5

Wetted Perimeter 276.22

-0.43056

HFL 97.96

Chainage

m

Distance

'x' m

Bed level

(m)

HFL - BL

(m)

Area

(m2)

Difference

in BL 'y'

(m)

WettedPerimeter

Sqrt(x2+y

2)

(m)60 0 97.96 0 0 0 0

55 5 95.7 2.26 5.65 2.26 5.49

50 5 94.6 3.36 14.05 1.1 5.12

40 10 91.8 6.16 47.6 2.8 10.38

20 20 91.4 6.56 127.2 0.4 20.00

0 20 90.9 7.06 136.2 0.5 20.01

-20 20 91.1 6.86 139.2 -0.2 20.00

-40 20 95.16 2.8 96.6 -4.06 20.41

-50 10 95.8 2.16 24.8 -0.64 10.02-55 5 96.08 1.88 10.1 -0.28 5.01

-62 7 97.96 0 6.58 -1.88 7.25

Total 607.98 123.69

ReducedLevelinmts


2/125

256257258259260261262263264

Reduced Distance in mts


3/125

0.6 0.7 0.6285

m1 0.14714 0.1375 0.144393

m2 0.04282 0.062 0.048286

m1 m2 Pw Moments

0.1443 0.0482 487.1696 73.82134.026

Pw

424.2903 0.499 2.301 487.1697

1 Man d/s bridge Man river 16 m

2 Upper Beda d/s Bridge 15 m

3 Jobat U/s bridge Dohi river 15 m

4 Jobat d/s bridge Hathni river 16 m


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A Preliminary Dimensioning :

Longitudinal girders

16.0 m 2.5 m

3.813 m 3.813 m 3.813 m 3.813 m 2.5 m

Clear roadway 7.5 mProviding three longitudinal T beams at spacing 2.5 m c/cEffective span of T-beam 16 mNumber of cross beams (including end cross beams) 5 NosConcrete grade M 20Density of reinforced concrete 25 kN/m

Density of plain concrete 22 kN/m

Reinforcement - High Yield Strength Deformed bars Fe415Clear cover to reinforcement 40 mmPoisson's ratio for concrete () adopted as 0.15

B Deck Slab :

7500

75 mm Thick wearing coat

215 mm Thick deck slab

Haunch Cantilever slab1400.0

Longitudinal beamA B C

Cross beam(250 mm wide)

2500 2500 1100 475

CROSS SECTION

RCC Road Bridge on River Man (d/s of Man dam)

DESIGN OF DECK SLAB OF T-BEAM

Cross girdersPLAN

Dimensions of Deck Structure

(300 mm wide)

4


5/125

1100Cross beams

250 (250 mm wide)

3813 3813

LONGITUDINAL SECTION

The slabs are taken as supported on four edges by beams

Thickness of slab (H) 215 mmThickness of wearing course (D) 75 mmSpan in transverse direction (c/c of longitudinal beams) 2.5 mAssuming width of longitudinal beams 300 mmEffective span of slabs in transverse direction (2.5 - 0.3) 2.20 mSpan in longitudinal direction (c/c of cross beams) 3.813 mAssuming width of cross beams 250 mmEffective span of slabs in longitudinal direction (3.813 - 0.25) 3.563 mThickness of 300 wide haunches at beam face 150 mmSlab thickness at the beam faces (cantilever & haunches) (215 + 150) 365 mm

(i) Maximum bending moment due to dead loadWeight of deck slab (0.215 x 25) 5.375 kN/m

Weight of wearing course (0.075 x 22) 1.65 kN/mTotal uniform dead weight = 5.375+1.650 = 7.025 kN/mInfluence coefficients for computation of moments in two directions

using Pigeaud's curves for slabs supported on all four sidesAspect ratio K = Short span/Long span (2.2/3.56) K = 0.6175

1/K = 1.6195rom geau s curves m1 = 0.048

m2 = 0.015

Total dead weight of slab and wearing course (P) 55.066 kN(7.025x2.2x3.56)Moment along short span due to dead weight M1 = (m1 + m2).P 2.763 kN.m.

(0.048+0.015x0.016)x55.066

Moment along long span due to dead weight M1 = (m2 + m1).P 1.195 kN.m.(0.016+0.015x0.048)x55.066

(ii) Bending moment due to Live Load

a) (IRC class AA tracked vehicle)Dimension of one panel of deck slab (2.5x3.813 m) 2.50 mEffective span of slabs in transverse direction (B) (=2.5-0.3) 2.20 mLongitudinal dimension of one panel of deck slab 3.813 mEffective span of slabs in longitudinal direction (L) (=3.813- 0.25) 3.563 mOne track of tracked vehicle is placed symmetrically on the slab panel

5


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(300 mm wide)

v =

3813

2200 2500U=1.023 850

3600Cross beams(250 mm wide)

3563

Contact dimensions for IRC class AA loadingBreadth of tracked vehicle across the bridge 0.85 mLength of tracked vehicle along the bridge 3.60 mImpact factor 25%

Dispersion of live load through deck slab:Dimension of load spread along short/long span

=SQRT [(x+2D) + H ]

wherex = Dimension of contact areaD = Thickness of wearing courseH = Thickness of deck slab

HenceWidth of load spread along short span U= 1.023 mSQRT [(0.85+2x0.075)^2+0.215^2]Length of load spread along long span V= 3.756 mSQRT [(3.6+2x0.075)^2+0.215^2]Limiting to effective span V= 3.563 m

K=(2.2/3.563) K = 0.6175U/B=(1.023/2.2) U/B = 0.465

V/L=(3.563/3.563) V/L = 1Using Pigeaud's curves, for aspect ratio K=B/L= 2.2/3.563 = 0.6175

0.6 0.7 0.6175m1 0.0777 0.0775 0.0777 m1 = 0.0777m2 0.02155 0.0311 0.0232 m2 = 0.0232

Load per track of tracked vehicle 350 kNTotal load per track including impact 125% 437.5 kNEffective load on span PT (=437.5x3.563/3.756) 415.002 kNMoment along short span due to track load = (m1 + m2).PT 33.676 kNm

(0.0775+0.15*0.0298)*415.002Moment along long span due to track load = (m2 + m1).PT 14.469 kNm

(0.0298+0.15*0.0775)*415.002

Longitudinal beams

Class AA tracked vehicle positioned for maximum bending moment in deck slab

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b) Bending moment due to Live Load (IRC class AA wheeled vehicle):

(300 mm wide)

62.5 2 62.5kN kN

51000

1 4 2200 250062.5 62.5

kN 600 kN6

37.5 3 37.5Cross beams kN 1200 kN(250 mm wide)

3563

3813

Class AA wheeled vehicle is placed on the deck slab panel with front axle alongcentre line between cross beams and 62.5kN wheel at the centre of panelInner wheel loads (wheels 1,2,4 & 5) 62.5 kNOuter wheel loads (wheels 3 & 6) 37.5 kN

1) B.M. due to wheel 1 :

(300 mm wide)

150

1 2200 2500300

62.5 kN

Cross beam(250 mm wide)

35633813

B= 2.2 L= 3.563Wheel load 62.5 kNTyre contact dimenstions

Longitudinal beams

Class AA wheeled vehicle positioned for maximum bending moment in deck slabFront axle placed at the mid span (longitudinally) with 62.5 kN wheel at centre

Longitudinal beams

Class AA wheeled vehicle positioned for maximum bending moment in deck slab

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Width (transverse direction) 300 mmLength (Longitudinal direction) 150 mmDispersion of load in transverse direction U =SQRT [(0.3+2x0.075) 2+0.215^2) = 0.499 mDispersion of load in longitudinal direction V = SQRT [(0.15+2x0.075)^2+0.215^2) 0.369 m

B = 2.20 mL = 3.56 m

U/B = 0.499/2.20 = 0.2267 mV/L = 0.369/3.56 = 0.1036 m

Using Pigeaud's curves, for aspect ratio K=B/L= 2.2/3.563 K = 0.6175

0.6 0.7 0.6175m1 0.213 0.200 0.2107 m1 = 0.2107m2 0.160 0.150 0.1583 m2 = 0.1583Maximum wheel load allowing for 25% impact Pw = 1.25*62.5 = 78.125 kNMoment along short span due to wheel load = (m1 + m2).Pw = 18.317 kNmMoment along long span due to wheel load = (m2 + m1).Pw = 14.833 kNm


Longitudinal beams

62.5 kNReal load

u=369 2200 B=2200 25001501

Cross beams Dummy load62.5 kN

L=35633813

B= 2.2 L= 3.563Wheel load 62.5 kNMaximum wheel load allowing for 25% impact Pw =1.25* 2.5 = 78.125 kN

Wheel is unsymmetrically placed perpendicular to the longitudinal axis of bridge

Pigeaud's curves are applicable for loads symmetrical about centre

Intensity of loading after dispersion in uxv area

u= 0.499 v= 0.369

Pw/(uxv)=(78.125/0.499x0369) = 424.426 kN/m

Assuming the width of loaded area after dispersion U= 2.20 m

Length of loaded area after dispersion V= 0.369 mU/B=(2.2/2.2) 1

Placing an imaginary load symmetrically on either side of longitudinal axis and calculating the

moments due to entire area loaded with same intensity and deducting the moments due to area

beyond the actual loaded area shall give double the desired moments. Half of these moments

shall be approximately equal to desired moments


v=0.499

Front axle placed at the mid span (longitudinally) with one 62.5 kN wheel (wheel 1)

at centre, the position of the other 62.5 kN wheel (wheel 2) is shown above

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9/125

V/L= (0.369/3.563) 0.1036Using Pigeaud's curves, for aspect ratio K=B/L= 2.2/3.563 K= 0.6175

0.6 0.7 0.6175m1 0.086 0.079 0.0850 m1 = 0.0850m2 0.074 0.082 0.0754 m2 = 0.0754Maximum wheel load allowing for 25% impact Pw (=U*V*Int of loading) 344.549 kN

(2.2*0.369*424.426)Moment along short span due to wheel load = (m1 + m2).Pw = 33.1 7 kNmMoment along long span due to wheel load = (m2 + m1).Pw = 30.370 kNm

Now considering the area between inner edges of real and dummy loadsWidth of loaded area (transverse direction) U = 1.501 mLength of loaded area (longitudinal direction) V = 0.369 m

U/B=(1.501/2.2) 0.682V/L=(0.369/3.563) 0.104

Using Pigeaud's curves, for aspect ratio K=B/L= 2.2/3.563 K= 0.61750.6 0.7 0.6175

m1 0.122 0.112 0.1203 m1 = 0.1203m2 0.110 0.115 0.1109 m2 = 0.1109

Maximum wheel load allowing for 25% impact Pw (=U*V*Int of loading) 235.076 kN(1.501*0.369*235.076)

Moment along short span due to wheel load = (m1 + m2).Pw = 32.178 kNmMoment along long span due to wheel load = (m2 + m1).Pw = 30.304 kNm

Net bending moment along short span =0.5*(31.652-30.951) = 0.495 kNmNet bending moment along long span = 0.5*(32.231-31.221) = 0.033 kNm


Longitudinal beams

369

Dummy load37.5

kN 600 B=2200 2500

600Real load

37.5 3Cross beams kN

L=35633813

B= 2.2 L= 3.563Wheel load 37.5 kNMaximum wheel load allowing for 25% impact Pw = 1.25*37.5= 46.875 kN

Class AA wheeled vehicle positioned for maximum bending moment in deck slabClass AA wheeled vehicle positioned for maximum bending moment in deck slab

Front axle placed at the mid span (longitudinally) with 62.5 kN wheel at centre, the

position of 37.5 kN load is as show above.

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10/125

Wheel is unsymmetrically placed perpendicular to the longitudinal axis of bridge


Intensity of loading after dispersion in UxV area= Pw/(uxv) 254.574 kN/m

u= 0.499 v= 0.369Width of loaded area (transverse direction) U= 1.699 mLength of loaded area (longitudinal direction) V= 0.369 m

U/B= (1.699/2.2) = 0.7723V/L= (0.369/3.563) = 0.1036


m1 0.108 0.100 0.1067 m1 = 0.1067m2 0.100 0.102 0.1004 m2 = 0.1004Maximum wheel load allowing for 25% impact Pw (=U*V*Int of loading) 159.600 kN

(1.639*0.369*254.574)Moment along short span due to wheel load = (m1 + m2).Pw = 19.424 kNmMoment along long span due to wheel load = (m

2+ m

1).P

w= 18.569 kNm

Now considering the area between inner edges of real and dummy loadsWidth of loaded area (transverse direction) U = 0.701 mLength of loaded area (longitudinal direction) V = 0.369 m

U/B=(0.701/2.2) 0.3186V/L=(0.369/3.563) 0.1036


m1 0.179 0.170 0.1770 m1 = 0.1770m2 0.154 0.140 0.1511 m2 = 0.1511Maximum wheel load allowing for 25% impact Pw (=U*V*Int of loading) 65.850 kNMoment along short span due to wheel load = (m1 + m2).Pw = 13.149 kNm

Moment along long span due to wheel load = (m2 + m1).Pw = 11.701 kNm

Net bending moment along short span 3.138 kNmNet bending moment along long span 3.434 kNm


Longitudinal beams

62.5 kN 62.5 KNDummy load Real load

4 B=2200 2500

Cross beams1200 1200

Placing an imaginary load symmetrically on either side of longitudinal axis and

calculating the moments due to entire area loaded with same intensity and

deducting the moments due to area beyond the actual loaded area shall give double

the desired moments. Half of these moments shall be approximately equal to desired

moments

10


11/125

L=35633813

Front axle placed at the mid span (longitudinally) with 62.5 kN wheel at centre

B= 2.2 L= 3.563Wheel load 62.5 kNMaximum wheel load allowing for 25% impact Pw = 1.25*62.5 = 78.125 kN

Wheel is unsymmetrically placed on one side of transverse axis of bridgePigeaud's curves are applicable for loads symmetrical about centre

Intensity of loading after dispersion in UxV area= Pw/(uxv) 424.290 kN/m


U/B= (0.499/2.2) = 0.2268V/L= (3.563/3.563) = 1.0000


m1 0.093 0.0940 0.0932 m1 = 0.0932m2 0.026 0.0357 0.0277 m2 = 0.0277Maximum wheel load allowing for 25% impact Pw (=U*V*Int of loading) 754.361 kNMoment along short span due to wheel load = (m1 + m2).Pw = 73.421 kNmMoment along long span due to wheel load = (m2 + m1).Pw = 31.433 kNm

Now considering the area between inner edges of real and dummy loadsWidth of loaded area (transverse direction) U= 0.499 mLength of loaded area (longitudinal direction) V= 2.031 m

u/B= (0.499/2.2) 0.2268v/L= (2.031/3.75) 0.5700


m1 0.14714 0.1375 0.1455 m1 = 0.1455m2 0.04282 0.062 0.0462 m2 = 0.0462Maximum wheel load allowing for 25% impact Pw (=U*V*Int of loading) 430.005Moment along short span due to wheel load = (m1 + m2).Pw = 65.524 kNmMoment along long span due to wheel load = (m2 + m1).Pw = 29.238 kNm



Longitudinal beams

62.5 62.5kN kN

Dummy 5Load 1000 Real Load

B=2200 2500


Placing an imaginary load symmetrically on the other side of transverse axis and

calculating the moments due to entire area loaded with same intensity and

deducting the moments due to area beyond the actual loaded area shall give double


moments

11


12/125

Cross beams1200 1200

L=3563

3813

Front axle placed at the mid span (longitudinally) with 62.5 kN wheel at centre

B= 2.2 L= 3.563Wheel load 62.5 kNMaximum wheel load allowing for 25% impact Pw = 1.25*62.5 = 78.125 kN

Wheel is unsymmetrically placed on one side of transverse axisof bridge

Intensity of loading after dispersion in uxv area= Pw/(uxv) 424.290 kN/m


U/B= (0.352/2.2) = 0.1600V/L= (3.563/3.563) = 1.0000

Using Pigeaud's curves, for aspect ratio K=B/L= 2.2/3.563 K 0.61750.6 0.7 0.6175

m1 0.098 0.099 0.0980 m1 = 0.0980m2 0.027 0.037 0.0288 m2 = 0.0288Maximum wheel load allowing for 25% impact Pw (=U*V*Int of loading) 532.135 kNMoment along short span due to wheel load = (m1 + m2).Pw = 54.428 kNmMoment along long span due to wheel load = (m2 + m1).Pw = 23.125 kNm


U/B= (0.352/2.2) 0.1600V/L= (2.031/3.563) 0.5700

Using Pigeaud's curves, for aspect ratio K=B/L= 2.2/3.563 K= 0.6175

0.6 0.7 0.6175m1 0.153 0.144 0.1511 m1 = 0.1511m2 0.042 0.062 0.0455 m2 = 0.0455Maximum wheel load allowing for 25% impact Pw (=U*V*Int of loading) 303.330 kNMoment along short span due to wheel load = (m1 + m2).Pw = 47.893 kNmMoment along long span due to wheel load = (m2 + m1).Pw = 20.675 kNm





Placing an imaginary load symmetrically on either side of transverse axis and

calculating the moments due to entire area loaded with same intensity anddeducting the moments due to area beyond the actual loaded area shall give double


moments

12


13/125

Longitudinal beams

62.5 2 62.5kN kN

51000

1 4 B=2200 2500

62.5 62.5kN 600 kN

637.5 3 37.5

Cross beams kN 1200 kN

L=35633813

Front axle placed at the mid span (longitudinally) with 62.5 kN wheel at centreB= 2.2 L= 3.563

Wheel load 37.5 kNMaximum wheel load allowing for 25% impact Pw = 1.25*37.5 = 46.875 kN

Wheel is unsymmetrically placed to transverse as well as longitudinal axis of bridge

As a reasonable approximation the unsymmetry only aout transverse axishas been consideredPigeaud's curves are applicable for loads symmetrical about centre

Intensity of loading after dispersion in uxv area= Pw/(uxv) 254.574 kN/m


U/B= (0.499/2.2) = 0.2268V/L= (3.563/3.563) = 1.0525


m1 0.09 0.09 0.0932 m1 = 0.0932m2 0.03 0.04 0.0277 m2 = 0.0277

Maximum wheel load allowing for 25% impact Pw (=U*V*Int of loading) 476.372 kNMoment along short span due to wheel load = (m1 + m2).Pw = 46.365 kNmMoment along long span due to wheel load = (m2 + m1).Pw = 19.850 kNm


U/B= (0.499/2.2) = 0.2268V/L= (2.031/3.563) = 0.5700


m1 0.14714 0.1375 0.1455 m1 = 0.1455


Placing an imaginary load symmetrically about transverse axis and calculating the

moments due to entire area loaded with same intensity and deducting the moments

due to area beyond the actual loaded area shall give double the desired moments.

Half of these moments shall be approximately equal to desired moments

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14/125

m2 0.04282 0.062 0.0462 m2 = 0.0462Maximum wheel load allowing for 25% impact Pw (=U*V*Int of loading) 258.003 kNMoment along short span due to wheel load = (m1 + m2).Pw = 39.314 kNmMoment along long span due to wheel load = (m2 + m1).Pw = 17.543 kNm


(c.) Total bending moment due to wheeled loads

Total bending moment along short span due to wheeled loads 32.691 kNm(18.317+0.495+3.138+3.949+3.268+3.525)Total bending moment along long span due to wheeled loads 21.776 kNm(14.833+0.033+3.434+1.098+1.225+1.154)

(d) Total bending moment due to tracked loadMoment along short span due to track load (m1+m*m2)PT 33.676 kNmMoment along long span due to track load (m2 + m*m1)PT 14.469 kNm

(e) Maximum bending moments due to live loadsMaximum bending moment along short span due to live loads 33.676 kNmMaximum bending moment along long span due to live loads 21.776 kNm

(f) Maximum bending moments due to dead loadsMaximum bending moment along short span due to dead loads 2.763 kNmMaximum bending moment along long span due to dead loads 1.195 kNm

(iii) Design bending moments along long and short spansTotal bendng moment along short span = 33.676+2.763 = 36.439 kNmTotal bendng moment along long span = 21.776+1.195 = 22.971 kNm

As the slab panels are continuous over support, multiplying by 0.8to allow for continuityDesign bending moment along short span = 0.8*36.439 = 29.151 kNmDesign bending moment along long span = 0.8*22.971 = 18.377 kNm

(iv) Reinforcement for deck slab: (As per IRC 21- 2000):For M 20 grade concrete and Fe415 steel ck= 20 mm

cb = 7 mm

st = 230 mm

j = 0.9R = 1.1

Effective depth required d= Sqrt(B.M.x1000x1000/Rx1000) 162.79 mm

Effective depth provided assuming diameter of main bars 12 mmand 40 mm clear cover ( =215-40-6) 169 mm> required Hence safe

Area of main reinforcement along short span 865.08 mm(Ast = B.M.x1000x1000/stxjxd) (=29.151*1000*1000)/(230*0.9*162.79)Required spacing of 12 mm HYSD bars (Spacing= astx1.2x1000/Ast) 157 mm

Providing 12mm HYSD bars at 120 mm c/c

Area of main reinforcement along long span 565.45 mm

(Ast = B.M.x1000x1000/stxjxd) (=18.377*1000*1000)/(230*0.9*(169-12))Required spacing of 12 mm HYSD bars (Spacing= astx1.2x1000/Ast) 240 mm

Providing 12mm HYSD bars at 120 mm c/c

(v) Cantilever Slab :

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15/125

225 20057 kN

7500

500 5075 mm th.Thick wearing coat

275

215 Thick deck slab 100

Cantilever slab700

Longitudinal beamsA B Cross beams C (300 wide)

(250 wide)2500 2500 1100 475

(a) Moment due to dead load:Total maximum moment due to dead load per m width of cantilever slab iscomputed in the following table

Load Lever arm MomentkN m kNm

1 Hand rails (approx) 1.74 1.43 2.480

2 Kerb 3.27 1.34 4.368(0.475x0.275x25)3 Wearing course 1.82 0.55 0.998

(0.075x1.1x22)4 Slab rectangular 3.9375 0.79 3.101

(1.575x0.1x25)5 Slab triangular 5.22 0.53 2.739

(1/2x1.575x0.265x25)Total 13.685

(b) Moment due to live load:Due to specified minimum clearance, class AA loading will not operateon the cantilever slab. Class A loading is to be considered and load

will be placed as shown in Sketch aboveEffective width of dispersion be = 1.2x +bw(As per Clause 305.16.2 (2) (i) of IRC:21-2000)

Breadth of concentration area of the load bw= 0.400 m

Position of centre of load from face x= 0.700 mHence be = 1.2*0.7+0.4 = 1.240 m

Live load per m including impact (57x1.5/be) 68.952 kN

Maximum moment due to live load (68.95x0.7) 48.266 kNmTotal moment due to dead load and live load B.M.= 13.685+48.266= 61.951 kNm

CROSS SECTION SHOWING CRITICAL LOAD

POSITION FOR CANTILEVER SLAB

Sl.No. Description

15


16/125

(c.) Reinforcement:Effective depth required d=Sqrt(B.M.x1000x1000/Rx1000) 237.32 mm

sqrt(61.951*1000*1000/(1.1 *1000)Effective depth provided (=215+150-40-8) 317 mm

Hence OK

944.11 mm

(62.082*1000*1000/(230*0.9*317)Required spacing of 16 mm HYSD bars S=(astx1000/Ast) = 255.66 mm c/c

(22/28*16^2*1000/946.10)Providing spacing of 16 mm HYSD bars at 180 mm c/c

Area of main reinforcement required Ast=(B.M.x1000x1000/stxjxd)

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17/125

Preliminary Dimensioning :

Concrete grade M 20

Safe compressive stress in concrete cbc = 7 N/mm

Tensile stress in steel st = 230 N/mm

Permissible shear stress in steel sv = 140 N/mm

Modular ratio for M 20 concrete = 13.33

Density of reinforced concrete = 25 kN/m

Density of plain concrete = 22 kN/m

Reinforcement - High Yield Strength Deformed bars Fe 415

Clear cover to reinforcement = 40 mm

Spacing of longitudinal T beams = 2.5 m

Thickness of wearing coat over the slab = 75 mm

Thickness of deck slab = 215 mm

Thickness of haunches at beam face = 150 mm

Width of the haunch at beam face = 300 mm

Number of cross beams (including end cross beams) = 5 Nos

Longitudinal girders

16.0 m 2.5 m

3.813 m 3.813 m 3.813 m 3.813 m 2.5 m

1 Intermediate longitudinal beam :

(a) Bending moment due to dead load:

Effective span of longitudinal T-beam 16 m

Assuming depth of cross beams 1.1 m

Width of the rib of cross girder 0.25 m

Assuming total depth of longitudinal beams 1.4 m

Width of the rib of longitudinal girder 0.3 m

Effective span of Cross beams= 2.5-0.3 = 2.2 m

Dead load per m run

- Wearing course 4.125 kN

( 2.5 x 0.075 x 22 )


DESIGN OF LONGITUDINAL AND CROSS GIRDERS

Cross girders

PLAN

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- Deck slab 13.438 kN

(0.215 x 2.5 x 25 )

- T-rib 10.50 kN

( 0.3 x 1.4 x 25 )

- Fillets/haunches 1.125 kN

( 2 x 1/2 x 0.3 x 0.15 x 25 )

- Cross beams (Total weight distributed over entire span) 4.727 kN

(5 x 0.25 x 1.1 x 2.2 x 25)/16

Total dead load per metre run 'w' = 4.125+13.438+10.5+1.125+4.727 = 33.914 kN/mMaximum bending moment due to dead load = wL^2/8 1085.25 kNm

(33.914 x 16^2)/8

(b) Bending moment due to live load :

Calculation of Moment of Intertia and Centre of Gravity of Longitudinal girder:2.54

0.2150.3

0.15

1.40

0.30

Longitudinal Girder

Effective flange width as per as per Clause 305.15.2 of IRC:21-2000 for continuous T beams

b = bw + 1/5*lo

where b = effective width for compression flange

bw = thickness of web for beams = 0.3 m

lo = distance between points of zero moments (0.7 x eff. Span)

b = 0.3 +0.2x0.7x16 = 2.54 m

Moment of Inertia of main girder:I = bd

3/12 + A.h

2

Taking moments of each section at the top of flange

I1 = 2.54 x 0.2153/12 + (2.54 x 0.215)(0.215/2)

2= 0.00841 m

4

I2 = 0.3 x 1.43/12 + (0.3 x 1.4)(0.215 +1.4/2)

2= 0.42023 m

4

I3 = 2 x ((0.3 x 0.153/36 + (1/2 x 0.3 x 0.15)(0.215 + 1/3 x 0.15)

2) = 0.00320 m

4

I = I1 + I2 + I3 = 0.43185 m4

Centre of Gravity of main girder:

Taking moments of areas of each section about top of flange

A1 = 2.54x0.215 = 0.5461 m2

A2 = 0.3x1.4 = 0.4200 m2

A3 = 2x1/2x0.3x0.15 = 0.0450 m2

Y1 = 0.215/2 = 0.1075 m

Y2 = 0.215 + 1.4/2 = 0.9150 m

Y3 = 0.215 + 1/3 x 0.15 = 0.2650 m

1

2

3

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Y = (A1Y1 +A2Y2 +A3Y3)/(A1 + A2 + A3) = 0.4499 m

Calculation of Moment of Intertia and Centre of Gravity of Cross girder:

0.60

0.215

1.10

0.25

Longitudinal Girder

bw = thickness of web for beams = 0.25 m

Width of flange of cross girder = bw+1/5*0.7*Io = 0.25+0.2*0.7*2.5 = 0.60 m

Moment of Inertia of cross girder:

I = bd3/12 + A.h

2

Taking moments of each section at the top of flange

I1 = 0.60x 0.2153/12 + (0.60 x 0.215)(0.215/2)

2= 0.00199 m

4

I2 = 0.25 x 1.13/12 + (0.25 x 1.1)(0.215 +1.1/2)

2= 0.18867 m

4

IT = I1 + I2 = 0.19065 m4

Centre of Gravity of cross girder:

Taking moments of areas of each section about top of flange

A1 = 0.60x0.215 = 0.1290 m2

A2 = 0.25x1.1 = 0.2750 m2

Y1 = 0.215/2 = 0.1075 m

Y2 = 0.215 + 1.1/2 = 0.7650 m

Y = (A1Y1 +A2Y2)/(A1 + A2 ) = 0.5551 m

Adopting Hendry - Jaegar method:

Assuming that cross beams can be replaced in the analysis by a uniform continuous

transverse medium of equivalent stiffness

The distribution of loading in an interconnected bridge deck system

depends on the following 3- dimensionless parameters A, F & c as defined below:

A =12/4

.[L/h]3.nEIT/EI

where

Parameter A is function of ratio of span to spacing of longitudinals

and ratio of transverse to longitudinal flexural rigidities.

L = Span of longitudinal T-beam

h = Spacing of longitudinal T-beam

n = No. of Cross beams

EIT and EI = Flexural rigidity of cross girder and longitudinal girder

IT = Moment of Inertia of transverse or cross girders

1

2

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I = Moment of Inertia of longitudinal girders

A = 12/4* (16/2.5)^3*5*E*0.19065/(E*0.43185)

= 71.17

F = 2/2n . [h/L] . CJ/EIT

where

Parameter F is a measure of the relative torsional rigidities of longitudinals.

CJ = Torsional rigidity of longitudinal girder

c = EI1/EI2

where

EI1, EI2 = Flexural rigidities of outer and inner longitudinal girders, if different

( c ) Design Data:

Longitudinal girder:

Effective span L = 16 m

Slab thickness t = 215 mmWidth of rib of longitudinal girder bw = 300 mm

Clear depth of rib = 1400 mm

Spacing of beams = 2.5 m

Overall depth of longitudinal beams D = 1615 mm

Effective flange width (As per Cl. 305.15.2 of IRC:21-2000) = 2.54 m

for continuous beam (b= 0.3 + 0.2 x 0.7 x 16 = 2.54 m)

Area of flange=2.54 x 0.215 = 0.5461 m2

Area of web= 0.3 x 1.4 = 0.42 m2

Distance of CG below top = 0.4499 m

Moment of Inertia of longitudinal T-beam = 0.4319 m4

Cross beams:

Clear depth of cross beams = 1100 mm

Overall depth of cross beam = 1315 mm

Width of cross beams = 250 mm

Width of flange of cross beam (Assumed) (b=0.25+0.2*0.7*2.5) = 600 mm

Area of flange (=0.6*0.215) = 0.129 m2

Area of web (=1.1 x 0.25) = 0.275 m2

Distance of CG below top (as calculated above)= 0.555 m

Moment of Inertia of cross beam (as calculated above) = 0.1907 m4

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(d) Distribution coefficients for longitudinal girders:

For A = 71.17

and F = Infinity

Using standard graphs of Hendry Jaegar :

Girder A Girder B Girder C

0.38 0.32 0.30

(m11) (m21) (m31)0.32 0.35 0.32

(m21) (m22) (m21)

0.30 0.32 0.38

(m31) (m21) (m11)

Ma = -0.85 W

Mb = 0.19 W

Mc = 0 W

Reactions are as below;

Reaction on longitudinal beam A RA = 1.884 W

Reaction on longitudinal beam B RB = 1.352 W

Reaction on longitudinal beam C RC = 0.764 W

(e) Determination of reaction factors:

Load on

girderLoad Girder A Girder B Girder C

0.716 0.603 0.565

(1.884*0.38) (1.884*0.32) (1.884*0.30)

0.433 0.473 0.433

(1.352*0.32) (1.352*0.35) (1.352*0.32)

0.229 0.244 0.290

(0.764*0.30) (0.764*0.32) (0.764*0.38)Net reaction 1.3778 1.3206 1.2882

Reaction factor for girder A = 1.378 W

Reaction factor for girder B = 1.321 W

Reaction factor for girder C = 1.288 W

Moments at supporting girders A, B and C are computed based on Moment

distribution method and are as below:

B

Unit load

on girder

Distribution Coefficient for

Treating the deck slab as continuous in the transverse direction, the support moments at the

locations of the three longitudinal girders due to loading shown inthe sketch below are

computed using method of moment distributuion .

These reactions are treated as loads on interconnected girder system and multiplying these by

respective distribution coefficients and adding the results under each girder, the final

reaction at each girder is obtained as shown in table below:

A

A

C

C

1.352

1.884

0.764

B

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(f) Design of Intermediate Girder (B):

Maximum bending moment would occur under Class A two lane loading

Impact factor fraction for spans 3m to 45m (Clause 211.2 of IRC :6-2000)

= 4.5/(6+L)= 4.5/(6 +16) = 0.2045

Loading is arranged in transverse direction as shown in Sketch

allowing the minimum clearance near the left kerb.

All four wheels are of equal magnitude

Arrangement of Wheel Loads on the Longitudinal beam

for Maximum Bending Moment in intermediate beam is shown below:

27 kN 27 kN 114 kN 114kN 68 kN 68 kN

1.1 3.2 1.2 4.3 3.0

L R

a b c d e f

16.0

Class A Loading on Man bridge (16 m. span)

As per I.S.Code the above loads are halved during calculation.

13.5 kN 13.5 kN 57 kN 57kN 34 kN 34 kN

1.1 3.2 1.2 4.3 3.0

L R

a b c d e f

16.0

Reaction factor for intermediate girder 'B' as per calculation as above = 1.3206

Impact factor = 1+ 0.2045 = 1.2045

21.48 kN 21.48 kN 90.68 kN 90.68 kN W 54.12 kN 54.12 kN

1.1 3.2 1.2 4.3 3.0

L R

a b c d h e f

x

16.0

VL VR

Computation of Absolute B.M.:

332.56 kN

The absolute max. B.M. generally occurs under the heavier wheel load specially

that which is very near to the C.G. of the load system

For computing the distance of resultant (x) from point 'f', take moments of all

loads at point 'f' and divide by the resultant load

W = VL + VR = 21.48 x 2 + 90.68 x 2 + 54.12x2 =

The loads shown in the figure below are corresponding to Class A train loads multiplied by

reaction factor (1.3206) and impact factor (1.2045)

Due to 16m span in the longitudinal girder the first six loads of Class 'A' train are

assumed to be accomodated on the span. The loads are arranged as shown such

that maximum B.M. shall occur under 4th load from left (assumed)

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21.48 21.48 90.68 90.68 54.12 54.12 54.12

1.1 3.2 1.2 4.3 3.0 3.0 0.2

L R

a b c d e f g

16.0

VL VR

Taking moments of all loads at right support R, we have,

Maximum Shear at L:

(VL-21.48)x16 = ( 54.12x0.2+54.12x 3.2+54.1x6.2+ 90.68x10.5+ 90.68x11.7+21.48x14.9)

VL = 199.77 kN

Impact shear force = 199.77 x 0.2045 = 40.86 kN

Dead Load Shear = wL/2 = 32.723x16/2 = 271.31 kN

Total Shear Force= 199.77+40.86+271.31 = 511.95 kN

Nominal shear Stress v = S/bd

= 511.95x1000/300 x 1505

= 1.134 N/mm2

% of steel provided c = 9 Nos. x 804.57 x 100/(300 x 1505)

= 1.604 %

From Table 12 B of IRC : 21 - 2000, c = = 0.458 N/mm2

v > c Hence provide Shear Reinforcement

Therefore, Shear resistance of concrete = cbd

= 0.458 x 300 x 1505

= 206787 N

Therefore Shear to be resisted by Stirrups

Vs =

= 305162.1 N

Spacing of 4-legged - 16 mm diameter bars

Sv = sv x Asv x d / Vs

= ( 140 *4*201.14 x*1505 ) / 305162

= 555.5 mm C/C

Provide 16 mm - 4 legged diameter bars @ 300 c/c

511.95*1000 - 206787

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(g) Design of End girder (A) :

Arrangement of Wheel Loads on the girder fo Maximum bending moment

in end girder (A) or (C.) is shown below:

Maximum bending moment occurs below 4th load (due to live load)

when C.G. of load and heaviest load are equally spaced about centre

27 kN 27 kN 114 kN 114kN 68 kN 68 kN

1.1 3.2 1.2 4.3 3.0

L R

a b c d e f

16.0

Class A Loading on Jobat bridge end girder (16 m. span)

As per I.S.Code the above loads are halved during calculation.

13.5 kN 13.5 kN 57 kN 57kN 34 kN 34 kN

1.1 3.2 1.2 4.3 3.0

L R

a b c d e f

16.0

Reaction factor for Intermediate girder as per calculation as above = 1.3778

Impact factor fraction for spams 3m to 45 m as per Clause 211.2 of IRC:6-2000 = 0.2045

Impact factor = 1.2045

22.4 22.4 94.59 94.59 W 56.42 56.42

1.1 3.2 1.2 4.3 3.0

L R

a b c d h e f

x

16.0

VL VR

Computation of Absolute B.M.:

346.82 kN

x = (22.4x12.8+22.4x11.7+94.59x8.5+94.59x7.3+56.42x3) /(22.4+22.4+94.59+94.59+56.42+56.42)

x = 6.38 m

For computing the distance of resultant (x) from point 'f' take moments of all

loads at point 'f' and divide by the resultant load

The loads shown in the figure below are corresponding to Class A train loads

multiplied by reaction factor (1.3778) and impact factor (1.2045)

W = VL + VR = 22.4 x 2 + 94.59 x 2 + 56.42x2 =

The absolute max. B.M. generally occurs under the heavier wheel load specially

Due to 16m span in the longitudinal girder the first six loads of Class 'A' train are

assumed to be accomodated on the span. The loads are arranged as shown such

that maximum B.M. shall occur under 4th load from left (assumed)

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To find Critical N.A.:

cbc/(st/m) = xc/(d-xc)

Substituting cbc = 7, st = 230, m=13.33 and d = 1505, we get

xc = 434.36 mm

To find Actual N.A.:

Assume x > t

be.t(x-t/2) = m.Ast(d-x)

2540 * 215 (x-215/2) = 13.33*7241.14(1505-x)

Solving for x, we get x = 317.41 mm

Since Actual N.A. is less than Critical N.A. the beam is under reinforced

To find cbc :

Let cbc be the stress in concrete at the top of the flange. Then

'cbc = cbc ((x-t)/x))

Substituting the values of x, t we have

'cbc = 0.3226 cbc

To find Lever arm Z:

Y = ((cbc+2'cbc)/(cbc+'cbc)) x t/3

Substituting the values of 'cbc and t we get Y = 89.15 mm

Z = d-Y = 1505-89.15 Z = 1415.85 mm

Taking moments about tensile steel,

M.R. = be *t (cbc+'cbc)/2 * Z

Substituting the values we get M.R. = 511.31 cbc kNm

Equating M.R. to Design Moment

511.31 cbc = 1944.05

cbc = 3.88 kN/mm2

This is less than 7 kN/mm2

and hence safe.

To find st :

cbc/(st/m) = x/(d-x)

Substituting the values of cbc, m, x and d we getst = 189.63 kN/mm

2which is less than 230 kN/mm

2Hence safe.

Design for Shear:

Width of web bw = 300 mm

Overall depth = 1615 mm

Distance of CG of reinforcement above bottom 110 mm

Effective depth for shear d = = 1505 mm

keeping the first load 22.4 kN just right of L, we have the load diagram as follows:

One more load of 56.42 kN gets added due to shifting of position of first load towards L.

94.59 94.59 56.42 56.42 56.42

1.1 3.2 1.2 4.3 3.0 3.0 0.2

L R

a b c d e f g

VL VR

Maximum S.F. value will be VL. As the C.G. of the load approaches nearer to L

22.4 22.4

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Taking moments of all loads at right support R, we have,

Maximum Shear at left support L:

(VL-22.4)x16 = ( 56.42x0.2+56.42x 3.2+56.42x6.2+ 94.59x10.5+ 94.59x11.7+22.4x14.9)

VL = 208.36 kN

Impact shear force = 208.36 x 0.2045 = 42.61 kN

Dead Load Shear = wL/2 = 33.914x16/2 = 271.31 kN

Total Shear Force= 208.36+42.61+271.31 = 522.28 kN

Nominal shear Stress v = S/bd

= 522.28x1000/300 x 1505

= 1.157 N/mm2

% of steel provided c = 9nos. x 804.57 x 100/(300 x 1505)

= 1.604

From Table 12 B of IRC : 21 - 2000, c = = 0.458 N/mm2

v > c Hence provide Shear Reinforcement

Therefore, Shear resistance of concrete = cbd = 0.458 x 300 x 1505

= 206787 N

Therefore Shear to be resisted by Stirrups

Vs =

= 315490 N

Spacing of 4-legged - 16 mm diameter bars

Sv = sv x Asv x d / Vs

= ( 140 x 4x201.14 x 1505 ) / 315490

= 537 mm C/C

Provide 16 mm - 4 legged diameter bars @ 300 mm C/C

(h) End girder C :

Since load may be placed in either side transversely, design of girder A

has been adopted for girder C

D Design of Cross Beams:

Cross beams are provided for stiffening of longitudinal beams.

Since longitudinal beams are with straight ribs

Minimum desired depth of cross beams are 0.75xD =0.75*1400 = 1050 mm

Clear Depth provided 1100 mm

Hence OK

Slab thickness 215 mm

Total depth= 1100 + 215 = 1315 mm

Width of cross beams = 250 mm

Size of cross beams = 0.25 m x 1.1 m

Spacing of cross beams 4 m

522.28*1000 - 206787

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31/125

Bending moment due to live load

Class AA tracked vehicle produces severer effect than others loadings

Figure below shows the disposition of one track on a cross beam (700/2 = 350kN)

1.675 0.25 1.675

2.5 0.850 Cross beam

3.600

3.037

4.0

Load on cross beam = 2 x [ ( 350 x (1.675/3.6) x (3.037/4.00)] + 0.25/3.6 x 350

= 271.59 kN

Coefficient of maximum positive BM due to concentrated load

= (1/3 ) x 0.203 +( 2/3) x 0.125 0.151

Coefficient of maximum negative BM due to concentrated load

= (1/3 ) x 0.188 +( 2/3) x 0.125 0.146

Positive BM including impact = 0.151 x 271.59 x 2.2 x 1.2045 108.672 kNm

Negative BM including impact = 0.146 x 271.59 x 2.2 x 1.2045 105.074 kNm

Design of Section :

Design positive BM = B.M. due to D.L. + B.M. due to impact. = 4.064+108.672 = 112.737 kNm

Effective depth = 1315 - 73 ( cover ) 1242 mm

Area of steel required = B.M. x 10^6/stxjxd 438.5 mm2

(=112.737/(230*0.9*1242))

Add 0.3 % of area of the beam to give additional stiffness to the beam

Additional area of steel = 0.3/100 x 250 x 1242 931.5 mm2

Total area of the steel required = 438.5 + 931.5 1370 mm2

Provide 20 mm diameter bars, No. of bars required N = Ast/ast

1370/(22/28*20^2) = 4.4 Nos.

Provide 20 mm diameter 6 Nos. in two rowsDesign negative BM = B.M. due to D.L. + B.M. due to impact. = 112.754 kNm

(= 7.680 + 105.074 )

Area of steel required = B.M. x 10^6/sstxjxd 438.6 mm2

(112.754 x 10^6)/(230x 0.9 x 1242)

Provide 20 mm diameter bars, No. of bars required N = Ast/ast

438.6/(22/28*20^2) = 1.4 Nos.

Provide 20 mm diameter 3 Nos.

Longitudinal beam

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E Reactions on Piers from Deck Structure:

Pier

Intermediate girder

End girder16 m

4 m c/c Cross girder

Pier

2.5 2.5

Total reaction on piers (from one span)

Left end:

With live load and dead load (I+2E) (428.00 + 2x434.75) 1297.51 kN

Dead load) (271.31 + 2x271.31) 813.94 kN

Right end:

With live load and dead load (447.12+2*454.69) 1356.51 kN

Dead load) (271.31+2*271.31) 813.94 kN

Maximum vertical load on intermediate piers with live load 2654.0 kN

(1297.51+1356.51)

Min vertical load on intermediate piers without live load 1627.9 kN(813.94+813.94)

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(i) Data:

HFL for 1 in 100 years discharge 265.50 m

HFL with afflux EL 265.50 m

River bed level NSL EL 255.80 m

Bottom level of the pier foundation EL 253.80 m

Bottom level of longitudinal girders EL 267.15 m

Top level of bridge (deck slab level) EL 268.84 m

(ii) Dimensioning of Pier Cap:

Concrete grade M20

Width of pier cap 1.5 m

Distance between centreline of longitudinal beams 2.5 m

Width of longitudinal beams 0.3 m

Distance between outer faces of end longitudinal beams 5.3 m

( 2x 2.5+0.3)

Width of Elastomeric bearing 0.25 m

Length of Elastomeric bearing 0.50 m

Thickness of Elastomeric bearing( Add 100 mm for pedestals) 0.15 m

(0.05 m for thickness of bearing and 100 mm for pedestal thickness)

(As per standard size index No. 6 Clause 916.2 of IRC: 83 Part-II 1987)

Keeping offset beyond bearing for jacking 0.40 m

(Min. as per Clause 710.8.1 of IRC: 78 -2000)

Minimum required length of rectangular portion 6.30 m

(2.500 x 2 + 2 x0.250 + 2 x0.400 )

Providing width of rectangular portion at bottom of flange = 2.0 m

Radius of rounded ends of pier cap 0.75 m

Thickness of pier cap top 0.6 m

(As per Clause 710.8.2 of IRC: 78 -2000)

Total length of pier cap (nose to nose) 7.8 m

( 6.300 + 2 x0.750 )

(iii) Dimensions of piers:

2500 2500

EL 267.00

(Pier cap top) 1200

EL 265.80

(Pier cap bottom)

9920

EL 255.80

Bed Level 500

500

EL 253.80 1000

Pier bottom


DESIGN OF Pier (4) (deepest foundation)

30008000

2000

7800

750

3500

1500

H.F.L. 265.65

Side view

Elevation

Plan

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Keeping minimum projection of pier cap beyond pier

Width of RCC pier at top 1.50 m

Length of pier (rectangular portion) 2.0 m

Radius of rounded ends of pier 0.75 m

Total length of pier (nose to nose) 2.0+2*0.75 = 3.5 m

Assuming the depth of pier cap 1.2 m

ELevation of pier cap top 267.00 m

ELevation of pier cap bottom 265.80 m

ELevation of pier bottom 253.80 m

Height of pier (265.80-253.80) 12.00 mWidth of pier at base 1.5 m

Radius of rounded ends of pier at base 0.75 m

Total length of pier at base (nose to nose) (2.0+2x0.75) 3.5 m

(iv) Calculation of Horizontal Forces, Vertical Forces and Moments on Piers:Longitudinal forces due to tractive effort and braking effect:

As per foot note under Clause 214.1-c of IRC : 6-2000, the Braking effect is

invariably greater than the tractive effort.

Total load on the span with class - 'A' loading = 418 kN

(=27+27+114+114+68+68)Longitudinal force on one span (=2x418) 836 kN

Longitudinal force on one span due to breaking effect 0.20*836 = 167.2 kN

(20% of maximum wheeL load on span as per Clause 214.2 of IRC:6-2000)

Point of application above roadway 1.2 m

Reaction on bearings due to braking effect = 836.0+167.2 = 1003.2 kN

Additional reaction on piers due to braking effect 13.6 kN

(10% of the load of the suceeding trains or part thereof i.e 10% of 2*68)

Moment at base of pier about long axis due to braking effect 2006.40 kN.m

(Long. Force due to breaking effect x Ht. of Pier = 167.2x12.00)

Moment due to resistance in bearings on account of movement due to temperature:

Reaction due to live load on one span (=1297.51-813.94) 483.57 kN

Dead load from each span 813.94 kN

Total resistance by left side bearings 0.25x(483.57+813.94) 324.38 kN

Total resistance by right side bearings 0.225x813.94 183.14 kN

Unbalanced force at bearing (324.38-183.14) 141.24 kN

Moment at base (Unbalanced forcexHt. of pier=141.24x12.0) 1694.89 kNm

Total Moment at base about long axis (2006.40+1694.89) My = 3701.29 kNm

(Moment due to breaking effect+Moment due to unbalanced force)

(a) Live load on both spans:

Maximum load from both spans of bridge due to Dead Loads 1627.88 kN

(813.94 x 2 )

Maximum load from both spans of bridge due to Live Loads 1026.14 kN

(2654.0-1627.9)

Total maximum load from two adjacent spans of bridge 2654.02 kN

(1627.88+1026.14)

Force due to dead load, live load on both span and braking effect 2834.82 kN

(1627.88+1026.14+167.2+13.6)

Eccentricity of live loads in transverse direction 0.70 m

Total live load acting on both spans 836.0 kN

ssum ng ass tra n o ve c es an presum ng t e rst

six loads are accommodated on the 16 m. span,

ssum ng or severest e ect, t e ve oa to e on e t span

and frictional coefficients of bearings to be 0.25 and 0.225 onthe left and right sides respectively

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Weight of pier ((5.25+5.25)/2)x12.0x25) 1575.0 kN

Thickness of pier bottom 1.0 m

Weight of pier bottom ( 8x3x1.0x25) 600.0 kN

Total vertical load in dry condition P = 5360.82 kN

(2834.82+351.0+1575+600)

Moments due to horizontal forces

About short axis

Due to eccentric transverse position of LL on both spans 585.20 kN.m

Due to wind on live load on both span 1443.84 kN.mDue to wind on super structure 731.55 kN.m

Due to wind load on pier in dry condition 162.72 kN.m

Total moment about short axis Mx= 2923.31 kN.m

About long axis

Due to braking effect on both span My= 3701.29 kN.m

P/A= (5360.82/5.25) = 1021.11 kN/m2

Mx/Zx= (2924.23/32) = 91.35 kN/m

My/Zy= (3701.29/12) = 308.44 kN/m2

Maximum compressive stress at base = P/A+Mx/Zx+My/Zy

(1021.11+91.40+308.44) = 1420.90 kN/m2

< 7000 kN/m2

Hence Safe

Minimum compressive stress at base = P/A - Mx/Zx -My/Zy

(1021.11-91.40-308.44) = 621.31 kN/m2

Positive hence no tensile force acting on piles

2) Submerged condition with no live load on span:

- Vertical forces

Load from bridge deck (DL only) 1627.88 kNWeight of pier cap (8x1.2x1.5x25) 351.0 kN

Area of pier at top (3.5x1.5) A = 5.25 m2

Area of pier at base (3.5x1.5) 5.25 m2

Height of pier 12.0 m

Unit weight of (RCC-Water) 15.0 kN/m3

Submerged weight of pier=(5.25x12.0x15) 945.0 kN

Thickness of pier bottom 1.0 m

Weight of pier bottom ( 8x3x1.0x15) 360.0 kN

Total vertical load in submerged condition P' = 3283.88 kN

(1627.88+351.0+945.0+360.0)

Area of pier bottom A' = ( 8x3) 24.00 m2

- Moments due to horizontal forces

About short axis

Due to wind on super structure 731.55 kN.m

Due to wind load on pier above water (1.13x11.70) 13.22 kN.m

Due to water current under submerged condition 837.78 kN.m

Total moment about short axis M' = (731.55+13.22+837.78) 1582.55 kN.m

P'/A' = (3283.88/25) 136.83 kN/m2

M'x/Zx= (1582.55/32) 49.45 kN/m

Maximum compressive stress at base = P'/A+M'x/Zx = 136.83+49.32 = 186.28 kN/m2

38


39/125

= 18.63 t/m2

SBC of foundation = 100.00 t/m2


40/125

40


41/125


42/125

42


43/125

43


44/125

44


45/125


46/125


47/125

Computations :Height of wall above GL h = 7.44-2.0 = 5.44 m

Cosd = 1 Cosf = 0.866

cosd-SQRT(cos2d-cos2f)Coefficient of active earth pressure (Rankine) Ca = cosd. -----------------------------------

cosd+SQRT(cos2d-cos

2f)

= 0.333

cosd+SQRT(cos2d-cos

2f)

Coefficient of passive earth pressure (Rankine) Cp = cosd. ------------------------------------

cosd-SQRT(cos2d-cos

2f)

= 3.00

17.36 kN/m2/m width

16.64 kN/m2

/m width

43.75 kN/m2/m width

1356.50 kN

Equivalent uniform vertical load after dispersion=1356.50/8.5= 159.59 kN/m width

Horizontal reaction due to breaking:

Horizontal Braking force at bearing level=20% of Class AA load = 0.2*700 = 140.00 kN

Horizontal Braking force on one abutment wall = 140/2 = 70.00 kN

Horizontal Braking force at bearing level per m = 70/8.5 = 8.24 kN/m width

Vertical reaction due to breaking:

3.13 kN/m width

Height of wall above GL (including dirt wall) - 5.44 m

Depth of foundation below GL h1 - 2.00 m

Height of backfill behind wall H - 7.44 m

Height of front counterfort above base slab = h1-ds = 2.00-0.50 = - 1.50 m

Height of rear counterfort above base slab =H-hd-d

s=7.44-1.84-0.5

- 5.10 mHeight of filling at the tail of heel above top of base wall - 5.10 m

Saturated soil height above heel h3-ds=4.96-0.50 = 4.46 m

Pr. due to Saturated soil mass P1 =0.5*h2*p1 =0.5*2.48*17.36= 21.53 kN/m

77.43 kN/m

Pr. due to submerged soil mass P3 =0.5*(h3-ds*)p2 =0.5*(4.96-0.50)*8.08= 37.10 kN/m

Hydrostatic lateral force P4 = 0.5*gw*(h3-ds)^2 =0.5*9.81*(4.96-0.50)^2= 97.57 kN/m

Pr. due to Saturated soil mass on submerged soil P2 =

(h3-ds)*1*p1=(4.96-0.50)*1*17.36

ntens ty o act ve eart pressure o t e su merge so p2

= Ca*gsub*(h3-ds)= (0.33*11.19*(4.96-0.50)) =Intensity of active Hydrostatic pressure p3

= gw*h3 = 9.81*(4.96-0.50) =Neglecting passive earth pressure

Vertical load from deck of bridge acting at centre of face wall

(454.69*2+447.12)=

Vertical reaction at one abutment= 20% of Class AA load*(braking level from road*ht. of road level

from beam rest)/(End span*Width of breaking effect)

= 140*(1.2+1.84)/(16.0*8.5) =

Stability Analysis of Right Abutment

Intensity of active earth pressure of the saturated soil p1

= Ca*gsat*h2 = 0.33*21*2.48 =

47


48/125

2.5

Triangular distance at heel top x=4.46x2.5/5.1 = 2.19 m A A

AA=(2.19+2.50)/2 = 2.34 m

5.1 x

4.46

Taking moments about toe

Moments of Loads/ forces considering one bay:

V H RestoringOverturn

ing

2 3 4 5 6 = 2 X 4 7

Base slab=2.63*5.0*0.5*25 = 164.38 2.50 (5.0/2) 411

Vertical/upright Wall=2.63*5.1*1.0*25 = 335.33 2.00 (1.5+1.0/2) 671

Dirt wall=2.63*1.84*0.325*25 = 39.32 2.34 (1.5+1.0-0.325/2) 92Rear Counterfort=0.5*5.1*2.50*25*0.3 = 47.81 3.33 (1.5+1.0+2.5/3) 159

Neglecting wt. of front counterfort 0.00 0.00 0 0

594.08 3.75 (1.5+1.0+2.5/2) 2228

57.78 3.75 (3.0+1+4.5/2) 217

0 0.00 0

287.67 3.75 (1.5+1+2.5/2) 1079

419.72 1.84 1.5+(1.0-0.325)/2) 771

8.23 1.84 1.5+(1.0-0.325)/2) 15

Overturning Moments:

Active earth pressure components:

56.61 5.79 (4.96+2.48/3) 328

203.63 2.48 4.96/2 505

97.57 1.65 4.96/3 161

256.60 1.65 4.96/3 424

21.66 8.64 7.44+1.2 187

TOTAL 1954 636 5642 1605

Check for maximum and minimum pressure at base:Point of application of resultant from toe x = (5609-1605)/1954 = 2.07 m

2.5 m

e= b/2-x = 2.50-2.07 = 0.43 m

b/2 =5.0/2=

Vertical force due to breaking = 3.13*2.63 =

Due to saturated soil mass 'P11' = P11*Lc

= 21.53*2.63 =

Due to saturated soil mass on submerged soil

'P12' = P12*Lc = 77.43*2.63 =

Due to submerged soil mass 'P2 = P2*Lc

= 37.1*2.63 =Due to hydrostatic pressure 'P3' = P3*Lc

= 97.57*2.63 =Due to braking force at bearing level = 8.24*Lc

= 8.24*2.63 =

Restoring Moments:

Soil on heel upto top of dirt wall (between

counterforts) = (2.5*4.46*2.33*11.19) +

(2.5*2.48*2.33*21.0) =

Soil on counterfort= (0.5*2.5*4.46*11.19*0.3) +

(2.5*0.3*2.48*21.0) =

Neglecting passive earth pressure at baseVertical component of hydrostatic pressure on

the heel slab = 2.5*(4.96-0.5)*9.81*2.63 =

Vertical Component of Equivalent load frombridge =(159.59*2.63)

Element

Load/Force in kN

Distance from toe in m

Moment about toe=

Load*dist.

kNm.

1

48


49/125

e 1.5 Hence safe

Net value of active and Passive earth pressure =

(56.61+203.63+97.57+256.60)-0 =

49


50/125

(B

Su

h2=2.48 m

2.24 mH=7.44 m

h3=4.96 m

p3=43.75

kN/m2/m width

bh = 2.5m

p1=17.36 + p2=16.64 b=

kN/m2/m width kN/m

2/m width

bh

=

pmin = 71.18

kN/m2/m

at heel

Diagram showing Retaining Wall, Earth pressures and pr

P1H

P1V

P1 =21.53kN/m width

P2 = 77.43kN/m width

He

Rear

X

Hydrosatic Pr.

P4=97.57

kN/m width

P2 V

P2V+P3V


51/125

Assuming M20 concrete and Fe 415 grade steel,

Permissible stress in concete in bending cbc= 7 N/mm

Permissible stress in steel in tension st= 230 N/mm

Modular ratio m = 13.33

Lever arm z = 0.90 d

Moment of Resistance M.R. = 0.92 bd2

Density of RCC gc = 25 kN/m

Characteristic strength of steel fy = 415 N/mm

Saturated density of backfill soil gsat = 21.00 kN/m

Density of water gw = 9.81 kN/m

Angle of internal friction/Angle of repose of backfill f = 30 o

Coefficient of active earth pressure (Rankine) Ca = 0.333

Coefficient of passive earth pressure (Rankine) Cp = 3.0

C/c spacing of counterforts L = 2.63 m

Thickness of counterfort tc = 2.63-0.60=

Clear spacing of counterforts Lc = 2.03 m

Thickness of base slab ds = 0.50 m

DESIGN OF UPRIGHT WALL :

Dirt wall 325 th.

hd =

Saturated 1840 h2 =

2480

Upright wall H =

Submerged 7440

5100 h3 =

4960

500

1000

Considering 1 m width of slab,

Width of slab b = 1000 mm

Total depth of soil filling behind wall at heel end H= 7.44 mAssuming overall thickness of upright slab tw = 1000 mm

Height of saturated soil below top at heel end h2 = (7.44)/3 = 2.48 m

Height of submerged soil below saturated soil at heel end h3= (7.44)*2/3 = 4.96 m

Intensity of active earth pressure just above submerged soil p 1 = 0.33*21*2.48 = 17.36 kN/m

Intensity of submerged soil mass p2 = 0.33*11.19*(4.96-0.5) = 16.64 kN/mIntensity of active earth pressure at the base = 17.36+16.64 = 34.00 kN/m

2

Hydrostatic pressure at base p3 = 9.81*(4.96-0.5) = 43.75 kN/m2

Total Horizontal pressure intensity at base p= (34.00+43.75) = 77.75 kN/m2

Since the upright slab will be designed as a continuous slab,

the maximum bending moment for this slab for a one metre deep strip

M= pL^2/12 = 77.75*2.03^2/12 = 26.70 kN-m

71.19 kN-m

Total Bending moment = (26.70+71.19)*1000000 = 97893048 N-mm


DESIGN OF RIGHT ABUTMENT AND COMPONENTS

Bending moment due to braking effect = breaking force per metre*(Overal ht. + Ht. of eye

level of driver)= 8.24*(7.44+1.2) =

17.36 kN/m2

16.64 kN/m2

51


52/125


53/125


54/125

Effective depth provided d = 425 mm

M.R. of balanced section = 0.92bd2

= 0.92*1000*425^2 = = 166175000 N-mm

Since Bending moment is less than the moment of resistance, the section is under reinforced.

Area of steel Ast = M/(st*z) = 73338284/(230*0.90*425) = 834 mm

Minimum reinforcement required = 0.12% if sectional area = 0.12*1000*500 = 600 mm2

Assuming 20 mm dia. Bars,

Area of one bar ast = 314.3 mm2

Spacing S = 1000*ast/Ast = 1000*314.3/834 = 377.0 mm

Provide 20 mm dia. @250 mm c/cActual steel provided per metre length Ast = 22/28*20^2*1000/250 = 1257.1 mm

600.0 mm2

Providing 16 mm dia bars, Spacing S = 1000*ast/Ast = 1000*201.14/600 = 335.2 mm

Provide 16 mm dia bars at 250 mm c/c for both faces of toe slab.

Provide main reinforcement of 20 mm diameter bars @ 250 mm C/CProvide distribution reinforcement of 16 mm diameter bars @ 250 mm C/C

Check for Shear:

Maximum Shear Force V = Net upward pressure * 2.03/2 = 213.56*2.03/2 = 216.8 kN/m2

Nominal shear stress = tv =V/bd = 216.8*1000/(1000*425) = 0.51 N/mm

% age of steel used p = Ast*100/bd = 1257.1*100/(1000*425) = 0.3 %

For M20 and Fe 415 steel and p= 0.30%,Permissible shear stress in concrete as per Table 23 of I.S.456-2000, tc = 0.236 N/mm

tv >tc Hence shear reinforcement is required.

Design shear Vs= (0.50-0.236)*1000*425 = 116463.4 N

Assuming 12 mm dia - 4 legged stirrups in one meter width of toe slab,

Spacing Sv = ssv*Asv*d/Vs = 230*4*22/28*12^2*425/116463.4= 379.9 mm

Since width of toe slab is 1.5m, provide 12 mm dia stirrups in whole width having 4x1.5=6 legged @300 mm c/c

Minimum spacing of shear steel = Asv*fy/0.4b = 6*22/28*12^2*415/(0.4*1500) 313.0 mm

Spacing provided 300 mm is OK.

Provide 12mm dia 6 legged stirrups @ 300 mm c/c. in between counterforts

DESIGN OF HEEL SLAB :

Dirt wall 325 th.

hd =

Saturated 1840 h2 =

2480

Upright wall H =

Submerged 7440

5100 h3 =

4960

500

D C

2500 1000 1500

pmin = 71.18 kN/m2

pD = 148.62 kN/mm2



Assuming

Width of heel slab bh = 2.50 m

Distribution steel in one direction = 0.12% of GCA =0.12*1000*500/100 =

17.36 kN/m2

16.64 kN/m2

54


55/125

and depth of heel slab ds = 0.5 m


Loading on a one metre wide strip of the heel slab will consist of following

Weight of the soil (assuming the soil is totally saturated) = (7.44-0.5)*21.0 = 145.74 kN/m2

Dead load of base slab (0.5 x25) = 12.5 kN/m2

Average Upward Pressure pav = (71.18+148.62)/2 = 109.90 kN/m

Therefore net downward pressure W = 145.74-12.5-71.18= 48.34 kN/m2

Since the rear counterfort is provided, the heel should be designed as

a continuous slab, continous over the rear counterforts.Maximum bending moment for this slab for a one metre deep strip

M= WL^2/12 = 48.34*2.03^2/12 = 16.60 kN-m

= 16600359 N-mm

Equating M.R. = B.M. we have

0.92bd2 =16600359

Computed depth d = sqrt(B.M./0.92b) = sqrt(16600359/(0.92*1000)) 134.3 mm

Assumed overall depth of toe slab D = 500 mm

Effective cover = 75 mm

Effective depth provided d = 425 mm


= 0.92*1000*425^2 = = 166175000 N-mm


Area of steel Ast = M/(st*z) = 16600359/(230*0.90*425) = 189 mmMinimum reinforcement required = 0.12% if sectional area = 0.12*1000*500 = 600 mm

2


Area of one bar ast = 314.3 mm

Spacing S = 1000*ast/Ast = 1000*314.3/600 = 523.8 mm

Provide 20 mm dia. @250 mm c/c

Actual steel provided per metre length Ast = 22/28*20^2*1000/250 = 1257.1 mm

Distribution steel in one direction = 0.12% of Gross cross-sectional area

=0.12*1000*500 = 600 mm2

Providing 16 mm dia bars, Spacing S = 1000*ast/Ast = 1000*(22/28*16^2)/600 = 279.4 mm

Provide 16 mm dia bars at 250 mm c/c for both faces of toe slab.

Provide main reinforcement of 20 mm diameter bars @ 250 mm C/C

Provide distribution reinforcement of 16 mm diameter bars @ 250 mm C/C

Check for Shear:

Maximum Shear Force V = Net vertical pressure * 2.03/2 = 48.34*2.03/2 = 49.1 kN/m2

Nominal shear stress = tv =V/bd = 48.34*1000/(1000*425) = 0.12 N/mm


For M20 and Fe 415 steel and p= 0.30%,

Permissible shear stress in concrete as per Table 23 of I.S.456-2000, tc = 0.236 N/mm

tv


56/125

DESIGN OF MAIN COUNTERFORTS :

Main (Rear) Counterfort

Dirt wall 325 th.

Saturated 1840

5440 b

Submerged 2.1

5100 d 3.6

X X X X

Toe slab d' 1.5

Heel slab

500 2.50

2500 1000 1500


Assuming thickness of counterfort tc = 0.5 m

The critical section for the main counterfort at which the bending moment should be determinedis at section X-X ( just at the level of front counterfort)

Height of critical section from top of upright wall = 3.60-1.50 = 2.10 m

Height of critical section from top of dirt wall =2.10+1.84 = 3.94 m

Let be the inclination of the main reinforcement with the vertical

Therefore tan = 2.50/3.60 = 0.6944

= tan-1(0.6944) = 34.8 degrees

Assuming effective cover for the counterfort = 75 mm

Effective depth provided d = 2.10*1000*Sin(34.8) -75 = 1123.50 mm

Assuming saturated condition,

Bending Moment at Critical Section = wh3/6*(1-Sinf)/(1+sinf)*Lc=wh

3/6*Ca*Lc

= 21*[(3.94^3)/6]*0.33*2.03 = 144.85 kN-m

Effective depth required = sqrt(B.M.*10 6/0.92b) = sqrt[144.85*10 6/0.92*0.5*1000] = 561.16 mm

Effective depth provided > Effective depth required. Hence safe.

M.R. = 0.92bd2

= 0.92*500*1123.50^2 = 580634476.4 Nmm

580.63 kNm


Area of steel Ast = M/(st*z) =144.85*10^6/230*0.90*1123.5 = 622.86 mm2

1150.57 mm

Using 20 mm diameter bars then area of each bar = 314.286 mm2

No. of bars required = 1150.57/314.286 = 3.66 Nos.

Provide 20 mm diameter 4 Nos.

Check for shear:Shear force due to horizontal thrust at the level of front counterfort F =

wh2/2 * [(1-Sinf)/(1+Sinf)] * Spacing of counterforts = ((21*3.94^2)/2)*0.333*2.03 = 110.30 kN

Net shear force V = F-(M/d)*Sin b = 110.30 - (114.85/1123.50)*Sin (34.8) = 36.71 kN

Effective depth for shear d' = d/Cos b = 1123.5/Cos 34.8 = 1368.20 mmNominal shear stress tv = V/bd' = 36.71*1000/500*1368.2 = 0.054 N/mm

%age of steel provided = Ast*100/bd = 4*22/28*20^2*100/500*1368.2 = 0.184


Permissible shear stress as per Table 23 of I.S.456-2000 tc = 0.18+((0.184-0.15)/0.1)*0.04) = 0.194 N/mm2

tv


57/125


58/125

Design data:

Assumptions :

h2=3950

h=10490

Front

Counterfort

9400

X X X h3=7890

h1=1350

ds=600

Toe slab tw=

bt=3000 1000 bh=5000

9000

Overall height of upright wall including dirt wall and base H =(268.84-257.00 11.84 m

Slope of backfill d = 0 o

SBC of soil (being rocky soil) = 1000 kN/m2

Depth of soil filling at heel end H = 11.84 m

Height of dirt wall hd = 1.84 m

Thickness of dirt wall td = 0.325 m


Total length of face wall of abutment = 8.50 m

Depth of foundation assumed h1 = 1.35 m

Thickness of vertical wall/Upright wall tw = 1.00 m

Thickness of counterforts tc = 0.60 mBase width b = 9.00 m

Adopt width of heel bh = 5.00 m

Width of toe beyond face of wall bt = 3.00 m

C/c Spacing of Counterforts L = 2.63 m

Clear spacing of Counterforts Lc = 2.03 m




Angle of friction between soil and wall surface b = 29 o

Coefficient of friction between soil and base of wall m = 0.55Saturated density of backfill soil gsat = 21 kN/m

3

Density of water gw = 9.81 kN/m3

Submerged density of soilgsub = 21.0-9.81= 11.19 kN/m3


Loading for Class AA tracked vehicle 700 kN

Computations :


Stability analysis of RCC Counterfort Retaining Wall for Left Abutment

hd=1840

H=11840

Heel slab

Assuming that top 1/3rd height is saturated and lower 2/3rd is

submerged at heel end and computing moments about toe

27.63 kN/m2

27.20 kN/m2

58


59/125

Height of wall above GL h = 11.84-1.35 = 10.49 m

Cosd = 1 Cosf = 0.866cosd-SQRT(cos

2d-cos

2f)

Coefficient of active earth pressure (Rankine) Ca = cosd. -----------------------------------

cosd+SQRT(cos2d-cos

2f)

= 0.333

cosd+SQRT(cos2d-cos

2f)

Coefficient of passive earth pressure (Rankine) Cp = cosd. ------------------------------------cosd-SQRT(cos

2d-cos

2f)

= 3.00

27.63 kN/m2/m width

27.20 kN/m2/m width

71.55 kN/m2/m width

1356.50 kN


Horizontal reaction due to breaking:Horizontal Braking force at bearing level=20% of Class AA load = 0.2*700 = 140.00 kN




3.60 kN/m width




Height of front counterfort = h1-ds = 1.35-0.60 = - 0.75 m above base slab

Height of rear counterfort=H-hd-ds =11.84-1.84-0.6 = - 9.40 m above base slab


Pr. due to Saturated soil mass P1 =0.5*h2*p1= 0.5x3.95x27.63= 54.52 kN/m

Pr. due to Saturated soil mass on submerged soil P2 =(h3-ds)*1*p1=(7.89-0.6)*1*27.63= 201.49 kN/m

Pr. due to submerged soil mass P3=0.5*(h3-ds*)p2 =0.5*(7.89-0.6)*27.2= 99.20 kN/m

Hydrostatic lateral force P4= 0.5*gw*(h3-ds)^2 =0.5*9.81(7.89-0.6)^2= 260.91 kN/m

5

Width of top of triangle of submerged soil x=7.29*5.0/9.4 = 3.88 m A A

AA=(3.88+5.0)/2 = 4.44 mx

9.4

7.29


= Ca*gsat*h2 = 0.33*21*3.95 =n ens y o ac ve ear pressure o e su merge so p2


= gw*h3 = 9.81*(7.89-0.6) =


(454.69*2+447.12)=

Vertical reaction at one abutment

= 20% of Class AA load*(braking level from road*ht. of road level


= 140*(1.2+2.3)/(16*8.5) =

Neglecting passive earth pressure

Stability Analysis of Left Abutment

59


60/125



V H RestoringOverturn-

ing

Base slab=2.63*9.0*0.6*25 = 355.05 4.50 (9+0/2) 1598

Vertical/upright Wall=2.63*9.4*1.0*25 = 618.05 3.50 (3.0+1/2*1.0) 2163

Dirt wall=2.63*1.84*0.325*25 = 39.32 3.84 (3.0+1.0-0.325/2) 151

Rear Counterfort=0.5*9.4*5.0*25*0.6 = 352.50 5.67 (3.0+1+5.0/3) 1998

Neglecting wt. of front counterfort 0.00 0.00 0 0

1669.60 6.50 (3.0+1+5.0/2) 10852

328.75 6.50 (3.0+1+5.0/2) 2137

940.85 6.50 (3.0+1+5.0/2) 6116

419.72 3.34 3.0+(1.0-0.325)/2) 1401

9.48 3.34 3.0+(1.0-0.325)/2 32

Overturning Moments:

Active earth pressure components:

143.38 9.21 (7.89+3.95/3) 1320

529.92 3.95 7.89/2 2091

260.91 2.63 7.89/3 686

686.19 2.63 7.98/3 1805

21.66 13.04 11.84+1.2 282

TOTAL 4733 1642 26447 6186

Check for maximum and minimum pressure at base:Point of application of resultant from toe x = (26447-6186)/4733 = 4.28 m

4.5 m

e= b/2-x = 4.50-4.28 = 0.22 m

e


61/125

Check against sliding:

1642 kN

Coefficient of friction = 0.55

Total vertical load = 4733 kN

Frictional resistance at base =4733*0.55= 2622 kN

Factor of safety = 2622/1642 = 1.60

> 1.5 Hence safe

Net value of active and Passive earth pressure =

(143.38+529.92+260.91+686.19+21.66)-254.27=

61


62/125

Design data:

Assumptions :

h2=4200

h=10490

Front

Counterfort

10150

X X X h3=8390

h1=2100

ds=600

Toe slab tw=

bt=3000 1000 bh=5000

9000

Overall height of upright wall including dirt wall and base H =(268.84-256.25) = 12.59 m

Slope of backfill d = 0 o

SBC of soil (being rocky soil) = 1000 kN/m2

Depth of soil filling at heel end H = 12.59 m

Height of dirt wall hd = 1.84 m

Thickness of dirt wall td = 0.325 m


Total length of face wall of abutment = 8.50 m

Depth of foundation assumed h1 = 2.10 m

Thickness of vertical wall/Upright wall tw = 1.00 m

Thickness of counterforts tc = 0.60 m

Base width b = 9.00 m

Adopt width of heel bh = 5.00 m

Width of toe beyond face of wall bt = 3.00 m

C/c Spacing of Counterforts L = 2.63 m

Clear spacing of Counterforts Lc = 2.03 m




Angle of friction between soil and wall surface b = 29 o

Coefficient of friction between soil and base of wall m = 0.55

Saturated density of backfill soil gsat = 21 kN/m3

Density of water gw = 9.81 kN/m3

Submerged density of soilgsub = 21.0-9.81= 11.19 kN/m3


Loading for Class AA tracked vehicle 700 kN

Computations :Height of wall above GL h = 12.59-2.10 = 10.49 m

Cosd = 1 Cosf = 0.866cosd-SQRT(cos

2d-cos

2f)


Stability analysis of RCC Counterfort Retaining Wall for Left Abutment

hd=1840

H=12590

Heel slab

Assuming that top 1/3rd height is saturated and lower 2/3rd is

submerged at heel end and computing moments about toe

29.38 kN/m2

29.07 kN/m2


63/125

Coefficient of active earth pressure (Rankine) Ca = cosd. -----------------------------------

cosd+SQRT(cos2d-cos

2f)

= 0.333

cosd+SQRT(cos2d-cos

2f)

Coefficient of passive earth pressure (Rankine) Cp = cosd. ------------------------------------

cosd-SQRT(cos2d-cos

2f)

= 3.00

29.38 kN/m2/m width

29.07 kN/m2/m width

76.45 kN/m2/m width

1356.50 kN


Horizontal reaction due to breaking:

Horizontal Braking force at bearing level=20% of Class AA load = 0.2*700 = 140.00 kN




3.60 kN/m width




Height of front counterfort = h1-ds = 2.10-0.60 = - 1.50 m above base slab

Height of rear counterfort=H-hd-ds =12.59-1.84-0.6 = - 10.15 m above base slab


Pr. due to Saturated soil mass P1 =0.5*h2*p1= 0.5x4.20x27.63= 61.64 kN/m

Pr. due to Saturated soil mass on submerged soil P2 =(h3-ds)*1*p1=(7.89-0.6)*1*29.38= 228.94 kN/m

Pr. due to submerged soil mass P3=0.5*(h3-ds)*p2 =0.5*(8.39-0.6)*29.07= 113.27 kN/m

Hydrostatic lateral force P4= 0.5*gw*(h3-ds)^2 =0.5*9.81(8.39-0.6)^2= 297.91 kN/m

5

Width of top of triangle of submerged soil x=7.79*5.0/10.15 = 3.84 m A A

AA=(3.84+5.0)/2 = 4.42 m x

10.15

7.79



V H RestoringOverturn-

ing

Stability Analysis of Left Abutment

ElementLoad/Force in kN Distance from toe in m

Moment about toe=

Load*dist.

kNm.

Restoring Moments:


= Ca*gsat*h2 = 0.33*21*4.20 =n ens y o ac ve ear pressure o e su merge so p2


= gw*h3 = 9.81*(8.39-0.6) =Neglecting passive earth pressure


(454.69*2+447.12)=

Vertical reaction at one abutment= 20% of Class AA load*(braking level from road*ht. of road level


= 140*(1.2+2.3)/(16*8.5) =


64/125


65/125

140 kN

(Braking)

hd= 1.84 m

Surcharge

8.65 m

h1=0.75 m

ds=0.6 bh=5.0 m

bt=3.0 m tw=1m bh = 5.0m

b =9.0 m

p

D

pmin = 259.43 pC = pD =

kN/m2/m 239.63 233.03

at toe kN/m2/m kN/m2/m

Diagram showing Retaining Wall, Earth pressures and p

159.59 kN/m from bridge

LL+DL

Surcharge

p1=27.63

kN/m2/m width

P H

P2H

P3HP4H

P3v

Saturated soil

Submerged

Heel slabToe Slab

Vertical Wall

RearCounterfor

FrontCounterfortX

C

3.88 m

Dirt wall0.325 m width


66/125


67/125


68/125

Assuming M20 concrete and Fe 415 grade steel,

Permissible stress in concete in bending cbc= 7 N/mm

Permissible stress in steel in tension st= 230 N/mm

Modular ratio m = 13.33

Lever arm z = 0.90 d

Moment of Resistance M.R. = 0.92 bd2


Characteristic strength of steel fy = 415 N/mm

Saturated density of backfill soil gsat = 21.00 kN/m

Density of water gw = 9.81 kN/m

Angle of internal friction/Angle of repose of backfill f = 30o

Coefficient of active earth pressure (Rankine) Ca = 0.33

Coefficient of passive earth pressure (Rankine) Cp = 3.0

C/c spacing of counterforts L = 2.63 m

Thickness of counterfort tc = 0.60 m

Clear spacing of counterforts Lc = 2.63-0.60 = 2.03 mThickness of base slab ds = 0.6 m

DESIGN OF UPRIGHT WALL :

h2=3950

H=11840

9400

h3=7890

600

1000



Total depth of soil filling behind wall at heel end H= 11.84 mAssuming overall thickness of upright slab tw = 1000 mm

Height of saturated soil below top at heel end h2 = 11.84/3 = 3.95 m

Height of submerged soil below saturated soil at heel ebd h 3 = 11.84*2/3 = 7.89 mIntensity of active earth pressure just above submerged soil p 1 = 0.33*21*3.95 = 27.63 kN/m

Intensity of submerged soil mass p2 = 0.33*11.19*(7.89-0.6) = 27.20 kN/m

Total intensity of active earth pressure at the base = 27.63+27.20 = 54.83 kN/m2

Hydrostatic pressure at base p3 = 9.81*(7.89-0.6) = 71.55 kN/m

Total Horizontal pressure intensity at base p= p1+p2+p3=(27.63+27.20+71.55) = 126.38 kN/m

Since the upright slab will be designed as a continuous slab,


M= pL^2/12 = 126.38*2.03^2/12 = 43.40 kN-m

hd=1840

Upright wall


DESIGN OF LEFT ABUTMENT AND COMPONENTS

27.63 kN/m2

27.20 kN/m2

68


69/125

107.45 kN-m



0.92bd2

= 150848986

Computed depth d = sqrt(150848986/0.92b) = 404.93 mm

Assumed overall depth of slab tw = 1000 mm

Effective cover = 75 mmEffective depth provided d = 1000-75 = 925 mm


= 0.92*1000*925^2 = = 787175000 N-mm



Minimum reinforcement required = 0.12% of sectional area = 0.12*1000*1000 = 1200 mm2


Area of one bar ast = 491.1 mm

Spacing S = 1000*ast/Ast = 409.2 mm

Provide 25 mm diameter bars @ 250 mm C/C on both sides of wall

Hence actual Ast provided per metre length = 22/28*20^2*1000/250 = 1257.14 mm

Distribution steel in one direction = 0.12% of Gross cross-sectional area =0.12*1000*1000 = 1 1200 mm2

Spacing of 20 mm dia bars S= 1000*ast/Ast = 1000*(22/28*20^2)/1200 = 262 mm

Provide 20 mm dia. @ 200 mm c/c on both sides of wall.

Check for Shear:

128.27 kN

Nominal shear stress = tv =V/bd = 128.27*1000/1000*925 = 0.14 N/mm2



Permissible shear stress in concrete as per Table 23 of I.S.456-2000 =tc =0.204 N/mm 0.18 N/mm

tv is less thantc and hence no shear reinforcement is required.

DESIGN OF DIRT WALL :

h2=3950

H=11840

9400

h3=7890

600

1000

Considering 1 m width of slab and designing as a cantilever slab,


Total depth of soil filling behind wall at heel end H= 1.84 m

Assuming overall thickness of dirt wall = 325 mm

Intensity of active earth pressure acting at the bottom of dirt wall p = 0.33*21*1.84 = 12.88 kN/m2

Maximum shear force/m width at bottom of upright slab V= (Pr. due to saturated soil + Pr. due

to submerged soil+ Hydrostatic pr.)*Lc/2 = (p1+p2+p3)*Lc/2 =(27.63+27.20+71.55)*(2.63-0.6)/2

=

hd=1840Dirt wall 325 thick

Bending moment due to braking effect = breaking force per metre*(Overall ht. + Ht. of eye

level of driver )= 8.24*(11.84+1.2) =

27.63 kN/m2

27.20 kN/m2

69


70/125

Since the dirt wall will be designed as a cantilever slab,


M= pL^2/8 = 12.88*2.03^2/8 = 4.42 kN-m

25.05 kN-m



0.92bd2 = 29472699Computed depth d = sqrt(29472699/0.92b) = 178.98 mm

Assumed overall depth of slab tw = 325 mm

Effective cover = 50 mm

Effective depth provided d = 1000-50 = 275 mm


= 0.92*1000*275^2 = = 69575000 N-mm



Minimum reinforcement required = 0.12% if sectional area = 0.12*1000*325 = 390 mm2

Spacing of 16 mm dia bars S= 1000*ast/Ast = 1000*(22/28*16^2)/518 = 388 mm

Provide 16mm diameter bars @ 250 mm C/C

Provide nominal distribution steel of 12mm dia @ 250 mm c/c

DESIGN OF TOE SLAB :

3950

11840

9400

Toe slab

600

3000

kN/m2



Assuming

Width of toe slab bt = 3.0 m

and depth of toe slab ds = 0.6 m


Loading on a one metre wide strip of the toe slab will consist of following

Upward Pressure pmax = 229.25 kN/m

Deduct dead load of slab (0.6 x25) = 15.0 kN/m2

Therefore Net upward pressure W = 229.25-15.0 = 214.25 kN/m2

Since the front counterfort is provided the toe should be designed as

a continuous slab, continous over the front counterforts.

Maximum bending moment for this slab for a one metre deep strip

M= WL^2/12 = 214.25*2.03^2/12 = 73.58 kN-m

Bending moment due to braking effect = breaking force per metre*(Overall ht. + Ht. of eye

level of driver )= 8.24*(1.84+1.2) =

1840

pmax = 229.25

27.63 kN/m2

27.20 kN/m2

70


71/125


72/125


73/125

Minimum spacing of shear steel = Asv*fy/0.4b = 10*22/28*12^2*415/(0.4*5000) = 234.8 mm

Hence provide 12 mm dia 10 legged stirrups @ 200 mm c/c in between counterforts

DESIGN OF MAIN COUNTERFORTS :

Main (Rear) Counterfort

3950

10490

Documents

Man Bridge Night 27-2-2012