Design of Br. No. 141.1

DESIGN OF BRIDGE NO. 141/1

(A) Forces & moments about abutment 1. Earth pressure Calculation

Back fill soil properties-

C = 0, θ = 300, δ = 200, γd = 1.8 T/m3, γs = 1.05 T/m3

= 0.297

KAH = 0.297. cos 200 = 0.28

a. LWL Condition

F = ½ ×6.15×3.90×12.0

= 114.02 T

M = 114.02 ×0.42×6.15

= 294.51 Tm.

1.2 HFL condition

F = [(½ ×1.512×3.0) + (3.15 ×1.512) + (½ ×3.15×0.92)] × 12

= [2.268 + 4.76 +1.44] × 12

= 27.21 + 57.12 + 17.28

= 101.61 T

M = 27.21 (0.42 × 3.0 + 3.15) + 57.12 × 3.15/2 + 17.28 × 0.42 × 3.15

= 232.81 Tm.

1.3 Live load Surcharge

F = 0.28×1.8×1.2×6.15×12

= 44.63 T

M = 44.63 × 6.15/2

= 137.23 Tm

...Cos2Φ

Ka =Cos θ cos(δ+θ)[1 +√ sin (δ+θ) sin (δ+β) ]2

cos (δ+θ) cos (θ+β)

0.28×1.8×3.0 = 1.512

Live Load Calculation 1. Class A Double lane

RB×9.4 = 6.8 (2.8+5.8+8.8) – 11.4 ×0.2

... RB = 12.34 T for double lane 24.68 TRA = 19.46 T for Double lane 38.92 T

2. Class 70R Tracked Load

RB × 9.4 = 70×2.085

... RB = 15.12 T

... RA = 54.47 T

Braking force = 0.2 × 70 = 14 TChange in reaction = (14×2.02)÷9.4 = 3.00 T

... RB = 15.12 – 3.00 = 12.12 T

... RA = 54.47 + 3.00 = 57.47 T3. Class 70 R wheeled load

RB×9.4 = 17× (1.17+4.22+5.59) +12(7.72+9.24) - 17× 0.2 = 41.14

... RB = 41.14 T ... RA = 50.86 TFrom above it is seen that case 70R Tracked load gives highest reaction.

... Max LL = 57.47 T

Live load from Super Structure

Type Load Normal SeismicL.A. Moment H.F. L.A. Moment

D.L. 117.65 0.03 3.52 16.94 5.74 97.23

D.L.+L.L. 175.12 0.03 5.25

Load from Dirt Wall, Abutment Cap, Abutment & Retaining Wall ――

(i) Dirt Wall = 0.25 × 0.82 × 12 × 2.4 = 5.90 T

Mn = 5.9 × 0.375 = - 2.21 TmFs = 0.85 TMs = 0.85 × 5.74 = 4.879 Tm

(ii) Abutment Cap = 0.225 × 0.67 × 12 × 2.4 = 4.34 TMn = 0.165 × 4.34 = - 0.716 TmFs = 0.624 TMs = 3.25 Tm

(iii) Abutment (LWL)W = [(0.67+1.0) / 2] × 5.105 × 12 × 2.4

= 122.76 TMn = - 9.82 TmFs = 17.67 TMs = 42.39 Tm

C.G. Calculation Area X Y Ax Ay

0.67 × 5.105 = 3.42 0.335 2.55 1.14 8.72½ × 0.33 × 5.105 = 0.84 0.78 3.403 0.655 2.85

4.26 1.795 11.57

... X = 0.42 m from left

... Y = 2.71 m from top(HFL)W = 1.955×0.67×12×2.4+½ ×1.955×0.12×12×2.4+3.15×12×0.81×1.4+0.5×3.15×

0.19×12×1.4= 37.72+3.37+42.86+5.02= 88.97 T

Mn = - 37.72×0.165+3.37×0.21-42.86×0.095+5.02×0.35= - 7.83 Tm

Fs = 5.43+0.48+6.17+0.72= 12.80 T

Ms = 4.12×5.43+0.48×3.8+6.17×1.58+0.72×1.05= 34.69 Tm

Return Wall ――W = 2× [(0.25×0.3×3.6×2.4) + (1.75×3.6×0.25×2.4)]

= 2(0.648 + 1.89) = 5.076 TMn = 2[(0.684(3.6/2 + 0.5)) + 1.89 (3.6/3+0.5)]

= 2[1.43 + 3.213]= 9.416 Tm

Fs = 2×0.114× (0.648 + 1.89)= 2×(0.093 + 0.272) = 0.73 T

Ms = 2×(0.093×6+0.272×5.26) = 3.976 Tm.

Kerb ――W = 2×0.425×0.3×3.6×2.4

= 2.203 TMn = 2.203×2.3 = 5.06 TmFs = 0.317 TMs = 1.99 Tm.Railing ――W = 2[(0.05×3.6) + (3×0.042)]

= 0.641 TMn = 0.641 × 2.3 = 1.47 TmFs = 0.092 TMs = 4.38 Tm

1. Horizontal force due to friction ――Fn = (20/100)×100 = 20 THF=M Rg = 0.06×117.65 = 70.59 TM = 70.59×5.33 = 376.24 TmForces & moment about Pier: - Dead Load = 235.37 TDead load + Live load = 313.80 TFs = 0.144 × 235.3 = 33.88 TMs = 33.88 × 5.74 = 194.47 TmMn = 0.00 Tm2. Load from Pier & Pier Cap ――Pier Cap = 0.8×0.225×12×2.4 = 5.184 TMn = 0Fs = 0.746 TMs = 0.746×5.21 = 3.89 Tm.Pier shift ――(LWL)W = 0.8×5.105×12×2.4 = 117.62 TMn = 0Fs = 16.94 TMs = 43.24 Tm(HFL)W = 0.8×12×3.15×1.4+0.8×12×1.955×2.4

= 42.33+45.04=87.37 T

Mn = 0.00

Fs = 0.144(42.33+45.04) = 6.09+6.48 = 15.58 TMs = 6.09× (3.15+(1.955/2))+6.48×(3.15/2)

= 25.13+10.21= 35.34 Tm

3. Horizontal forces due to friction ――Fn = 20/100×100

= 20 TFn/2 = 10 TH = 0.00 TTotal Horizontal force ――

= 2×10 = 20 TMoment = 20×5.33

= 106.60 TmWater Pressure (HFL)Maximum observed velocity = 0.8m/sec.Manning’s Velocity = 1.8m/sec.Max. Manning’s Velocity

= 1.84√2= 2.6 m/sec

P = 52 KV2

K = 0.9p = 52×0.9×2.62

= 316.4 Kg/m2

Total water pressure F = 316.4×0.8×(3.15/2)

= 398.66 KgFor variation of flow of 20O

F traffic = 398.66 Cos 20O = 374.62 KgF long = 136.35 KgMT = 0.374× 2/3 ×3.15 = 0.785 TmML = 0.136×2/3×3.15 = 0.286 Tm(C ) Design of abutment column (i) DL+EP+LL+HF+LL Surcharge +LWL

Sl. No. Description W FH Mn Fs Ms

1 Earth Pressure   114.02 294.51    2 LL Surcharge   44.63 137.23    3 DL+LL Superstructure 175.12   -5.25 16.94 97.234 Dirt Wall 5.9   -2.21 0.85 4.8795 Abutment Cap 4.34   -0.716 0.624 3.256 Abutment 122.74   -9.82 17.67 42.397 Return Wall 5.076   -9.406 0.73 3.9768 Kerb 2.203   -5.06 0.317 1.999 Railing 0.641   -1.47 0.092 4.38

10 Horizontal force   70.59 376.24    Total   316.02 229.24 774.048 37.223 158.095

Normal Condition Sesmic Condition

W = 316.04 T W =316.04 T

Mn = 774.05 Tm Mn =774.05+158.09= 932.14 Tm

Fn = 229.24 T Fn =229.24+37.22= 266.46 T

Sl. No. Description W FH Mn Fs Ms

1 Earth Pressure   101.61 232.81    2 LL Surcharge   44.63 197.23    3 DL+LL Superstructure 175.12   5.25 16.94 97.234 Dirt Wall 5.9   -2.21 0.85 4.8795 Abutment Cap 4.34   -0.716 0.624 3.256 Abutment 88.97   -7.83 12.8 34.697 Return Wall 5.076   -9.4 0.73 3.9768 Kerb 2.203   -5.06 0.317 1.999 Railing 0.641   -1.47 0.092 4.38

10 Horizontal force   70.59 376.24    Total   282.25 216.83 784.84 32.353 150.395

Normal Condition Seismic Condition

W = 282.25 T W =282.25 T

Mn = 216.83 Tm FH = 216.83+32.29= 249.12 T

Fn = 724.84 T Mn=724.84+150.39= 875.23 Tm

From Above two tables it is seen that LWL non Seismic is guiding for design.Design load = 316.04 TDesign Moment = 774.05 Tme = 774.05/316.04 = 2.44t = 1000mm b = 12000mmt/e = 1.6/2.44 = 0.409 d’/t = 60/1000 = 0.06 ≈ 0.05Assuming 1% Steelp = 1%n = 10mp = 1/100×10 = 0.10Referring Turneaure & Maurer ChartsCZ = 7.8 K = 0.275

... fe = M/bt2 ×CZ= (774.05×105)÷(1200×(100)2)×7.8 = 50.31 Kg/cm2 <115 Kg/cm2

... ft = n fe (d/kt – 1)= 10×50.31×((97.6/0.275×100) – 1)= 1282.44 Kg/cm2 <2000 Kg/cm2

Ast = 1/100 × 1200 × 100 = 1200 cm2 on each face 600 cm2

Using 32 Φ – 75 Nos. on each face

Summary of forces & moment at pier

LWL Condition Sl. No. Description W FHT FHL FS MNL Mnt Ms

1 DL+LL Superstructure 313.80     33.88     194.47

2 Pier Cap 5.184     0.746     3.89

3 Pier 117.62     16.94     43.24

4 Horizontal force     20   106.60    

5 Total 436.60   20 51.56 106.60   241.60

Normal Condition:-W = 436.60 THF = 20TM = 106.60 TmSeismic Condition: -W = 436.60 THF = 20+51.56 = 71.56 TM = 106.60+241.60 = 348.20 Tm

HFL Condition Sl. No. Description W FHL FHT MNL MNT FS Ms

1 DL+LL Superstructure 313.80         33.88 194.47

2 Pier Cap 5.184         0.746 3.893 Pier 87.37         12.58 35.344 Horizontal force   20   106.60      5 Water Current   0.136 0.374 0.286 0.785    6 Total 406.35 20.136 0.374 106.886 0.785 47.206 233.70

Normal Condition: - W = 406.35 TML = 106.9 TmMT = 0.785 TmHFL = 20.13 THFT = 0.374 TSeismic Condition: -W = 406.35 TML = 106.9+233.70 = 340.60 TmMT = 0.785 TmHFL = 20.13+47.20 = 67.33 THFT = 0.374 TFrom Above two tables it is observed that LWL non Seismic is guiding for design.Design Moment = 348.20 Tm Design load = 436.60 Te = 348.20/436.60 = 0.797mt = 800mm b = 12000mmd’/t = 60/800 = 0.075 ≈ 0.10t/e = 0.8/0.797 = 1.003Assuming P = 0.9%np = 10×0.009 = 0.09

Referring Turneaure & Maurer ChartsCZ = 9.3 K = 0.325

... fc = M/bt2 ×CZ

= (348.20×105)× 9.3 ÷(1200×(80)2) = 42.16 Kg/cm2 <1.5×115 Kg/cm2

... ft = n fc (d/Kt – 1)= 10×42.16× [77.6/ (0.325×80) – 1]= 836.71 Kg/cm2 <1.5×2000 Kg/cm2 (SAFE)

Ast = 0.9/100 × 1200 × 80 = 864 cm2 Provide 55 Nos. 32 Φ on each face.

Design of Raft Slab: - Dimensional Properties: -A = 29.98×12.9 = 386.74 m2

Ix = (29.98×12.93) ÷12 = 5363.19 m4

Iy = (12.9×29.983) ÷12 = 28966.99 m4

Zx = (29.98×12.92) ÷6 = 831.49 m3 Zy = (12.9×29.982) ÷6 = 1932.42 m3 Load directly coming over raft slab: -(i) LWL Condition Self Wt: - 1.29.98×12.9×2.4 = 928.18 TEarth : - 2× (½×12×1.8×3.08×4.62) = 307.36 T

(T) 1170.77 T(ii) HFL Condition Self Wt: - 1×29.98×12×1.4 = 503.66 TEarth : - 2× (½×12×1.05×3.05×4.62) = 179.29 TWater : - 12×3.15(29.98 – 2.0 – 2.4) = 997.16 T

(T) = 1680.91 T

Check for base pressure: - (LWL Normal)W = 1170.77+2×316.04+2×436.60 = 2676.05 T

M = 774.05–774.05+2×106.60 = 213.20 TmP = (2676.05÷359.76) ± (213.20÷1797.60)Pmax = 7.43+0.11 = 7.54 T/m2

Pmin = 7.43 – 0.11 = 7.32 T/m2

(LWL Seismic)W = 1170.77+2×316.04+2×436.60 = 2676.05 TM = 774.05–774.05+2×158.09 + 2 × 348.20 = 1012.58 TmP = (2676.05÷359.76) ± (1012.58÷1797.60)Pmax = 7.43+0.56 = 7.99 T/m2

Pmin = 7.43 – 0.56 = 6.87 T/m2

HFL (Normal)W = 2×282.25+2×406.35+1680.11 = 3057.31 TML = 2×106.9+724.84 – 724.84 = 213.80 TmMT = 2×0.785 = 1.57 TmP = (3057.31÷359.76) ± (213.80÷1797.60) ± (1.57÷719.52)

= 8.49± 0.118 ± 0.002P = 8.61 T/m2

p = 8.606 T/m2

p = 8.37 T/m2

HFL (Seismic)W = 3057.31 TML = 2×150.39+2×340.60 = 981.98 TmMT = 2×0.785 = 1.57 TmP = (3057.31÷359.76) ± (981.98÷1797.60) ± (1.57÷719.52)P = 9.038 T/m2

p = 9.034 T/m2

p = 7.942 T/m2

Design of raft SlabLWL NormalW = 2×316.04+2×436.60 = 1505.28 TM = 213.20 Tm.P = (1505.28÷359.76) ± (213.20÷1747.60)Pmax = 4.305 T/m2

Pmin = 4.063 T/m2

LWL SeismicW = 1505.28 TM = 2×158.09+2×348.20 = 1012.58 TmP = (1505.28÷359.76) ± (1012.58÷1747.60)Pmax = 4.763 T/m2

Pmin = 3.706 T/m2

HFL NormalW = 2×282.25+2×406.35 = 1377.2 TM = 2×106.9 = 213.80 TmP = (1377.2÷359.76) ± (213.80÷1747.60)Pmax = 3.95 T/m2

Pmin = 3.706 T/m2

HFL SeismicW = 1377.2 TM = 981.98 Tm

P = (1377.2÷359.76) ± (981.98÷1747.38)Pmax = 4.309 T/m2

Pmin = 3.267 T/m2 From above it is observed that LWL Normal condition is guiding for design.

Shear force in A1 P1 SFx = -316.04+4.063×x×12+0.0087×x2/2×12At A1 x = 0SF = -316.04 TSFx = 10.08 = -316.04+4.063×10.08×12+0.0087×10.082×6

= +180.72 TSF = 0x = 0.0522x2+48.77×-316.04

... x = 6.43m.

Shear force in P1 P2 SFx = -(436.60-180.72)+12×4.063×x+12×0.0087× x2/2

At P2 x = 9.82mSF = -255.88+478.78+5.03

= +227.93 TSF = 0At - 255.88+48.756x+0.0522x2

x2+934.02x – 4901.91 = 0

... x = 2.616mShear force in P2 A2 SFx = -(436.60-227.93)+12×4.063x+12×0.0087 x2/2

x = 0At P2 = Sf = -208.67TAt x =10.08mSF = 0At -208.67+48.756x+0.0522x2x = (-933.9±√ (933.9)2+4×3997.50) ÷2 = 4.26m

Bending moment in A1 P1 Mx =774.05 – 316.04x + ((4.063x2 ÷ 2) + (0.0087x3 ÷ 2))×12At A1 x = 0M = 774.05 TmAt x = 6.43 mM max = -245.55 TmAt x = 10.08mM = 83.14Portion P1 P2 Mx =774.05 +106.60 – 316.04 (10.08+x) – 436.60x + [(4.063×(10.08+x/2)2+0.0087

×(10.08+x/2×3)3]×12At A1 x = 9.82M = 95.132 TmAt x = 2.616 mM max = -308.88 TmPortion P2 A2 Mx = 774.05 +2×106.60 – 316.04× y – 436.60 (y – 10.08) – 436.60 (y – 19.9) +

[(4.063y2÷2)+(0.0087y3÷2×3)]×12At A2 M= +774.08 TMM max X = 4.26m

Y = 24.16mM max = -180.57 Tm

Design shear force =At Abutment = 316.04 TAt Pier = 255.88 T

Design Moment =At Abutment = 774.05 Tm At Pier = 201.69 Tm At mid span = 223.02 Tm (t)

Reinforcement at Abutment – M/bt2 = [774.05×107] ÷ [12000×(936)2] = 0.736Pt = 0.58%At = (0.58/100)×93.6×1200 = 651.45 Cm2

Provide 25 Φ – 135 nos. at bottom (662.67 cm2)ح ) =316.04 × 104) ÷ (12000×936(

= 0.281 N/mm2

100As/bd= (100×662.67) ÷ (1200×93.6) = 0.589Corresponding ح c = 0.34 N/mm2 > 0.281 N/mm2

Reinforcement at Pier – M/bt2 = [201.69 ×107] ÷ [12000×(936)2 = 0.191Pt = 0.2%At = (0.2/100)×93.6×1200 = 224.64 Cm2

Provide 25 Φ – 70 nos. at bottom (343.61 cm2)ح ) =255.88 × 104) ÷ (12000×936 = (0.227 N/mm2

100As/bd= (100×343.61) ÷ (1200×93.6) = 0.306Corresponding ح c = 0.25 N/mm2 > 0.227 N/mm2

Reinforcement at Pier Abutment mid span – M/bt2 = [308.88×107] ÷ [12000×(936)2]

= 0.294Pt = 0.24%At = (0.24/100)×93.6×1200

= 330 Cm2

Provide 25 Φ – 70 nos. Provide 20 Φ @ 140 c/c in transverse direction at both top & bottom.Design of Return Wall Earth Pressure at a depth At C = (0.297×1.8×0.6)+(0.297×1.2×1.8)

= 0.321+0.642= 0.963 T

At D = (0.297×1.8×2.35)+(0.297×1.2×1.8)= 1.256+0.642= 1.898 T

M = ((0.963÷2)×0.6×(3.62÷2)) + [(0.963+1.898÷2)×(1.75÷2)×(3.62÷3)]= 7.28 Tm

BM/m = 7.28÷2.35 = 3.1 Tm/mDesign Momentd = √ (3.875×107)÷(0.91×1000) = 206.4mmProvide d = 250mm

= 250 – (25+16/2) = 216mmM/bt2 = [3.875×107] ÷ [1000×2172]

= 0.83From Sp – 16 Pt = 0.67At = (0.67/100)×100×21.7

= 14.039 Cm2

Provide 16 Φ @ 130 c/c (At = 15.47 cm2) on inside face horizontally For outside face horizontally & inside ~ outside face vertically A = (0.2÷100)×100×21.7 = 4.43 m2

Provide 10 Φ @ 150 c/c (At = 5.24 cm2)

Design for vertical bending: -W = (0.648+1.89)+(2.203+0.33)

=5.071 TM = ( 0.648×3.6/2)+(1.89×3.6/3)+(2.203+0.33)×3.6/2

= 7.994 TmM/bt2 = [7.99×107] ÷ [250×23002]

= 0.06Pt = 0.2%At = (0.2/100)×25×230

= 11.5 Cm2

Provide 4 Nos. 20 Φ at topNos. 20 Φ at bottom.

Design of Dirt wall: - A = 0.297×1.8×1.2 = 0.642B = 0.297×1.8×0.82 = 0.438Fa = 0.642×0.82 = 0.526 TFb = ½×0.438×8.82 = 0.179F = Fa+Fb

= 0.526+0.179= 0.705 T

M = 0.526×0.82/2+0.179×0.42×0.82= 0.277 Tm

Required d = √ (0.277×107)÷(0.91×1000) = 55.20 mm= 5.5 cm

Provide d = 250mm= 250 – (25+12/2) = 219mm

M/bt2 = [0.277×107] ÷ [219×10002]= 0.0577

Refer Sp = 16Pt = 0.2%A = (0.2/100)×100×21.9

= 4.38 Cm2 Provide 10 Φ @ 150 c/c vertically & horizontally on both face.Abutment Cap: - Based on square root formula & satisfied provide 6 nos. 10 Φ bars at top & bottom & two legged 10 Φ stirrups @ 150 c/c.

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