Box Abutment

(1)

FRL = 561.936 0.35 Cl bridge

(median edge) 0.3 1 15+16

4.012 ₂ 0.775

varies cap top lvl

0.3 ₃ 0.3 _558.900 0.4 0.45 0.35 FILL 7 12 FILL 17.936 13 14 OGL 4 547.268 16.786 0.85 1.8 0.85 6 5 544.000 Founding level 0.3 4.662 0.3 4.7 0.85 2.3 12.419 1 _10.2 0.85 10 87 ° 2.968 4.172 4.518 8 0.3 4.553 11.225 2.968 9 0.3 4.889 2.3 4.934 13.412 2.984 80 ° 0.85 11.379 0.85 4.662 5.129 11 0.863 11.507 1 2.335 13.959

(2)

(3)

Formation level (median edge) = 561.936

carriageway width = 11.15 m

gradient= 7 %

formation level(outer end) = 561.156

0.749 wearing coat = 56 mm height of superstructure = 1.925 m height of pedestal = 150 mm height of bearing = 125 mm foundation level = 544.000

maximum height of abutment = 17.936 m

0.780 m

minimum height of abutment = 17.156 m

abutment cap level = 558.900

Stem Thickness (top)= 0.450 m

Stem Thickness (bottom) = 0.850 m

maximum height of dirtwall = 3.036 m

minimum height of dirtwall = 2.256 m

DL + SIDL from superstructure = 170 t

LL from superstructure = 80 t

Allowable soil bearing capacity = 45 t/m2 (normal case)

Avg height Avg length

counterfort 1 16.283 m 9.415 m

counterfort 2 16.507 m 9.728 m

side wall 1 16.006 m 9.794 m

side wall 2 16.786 m 9.817 m

Avg height is calculated taking the average of maximum and minimum height of abutment at median and outer edge respectively and then doing the necessary deduction. Average length is the length that average height would go, to result area equivalent to the actual geometry.

(4)

abutment cap maximum dimension = 2.781 m abutment cap minimum dimension = 1.474 m active earth coefficient for normal case = 0.279 active earth coefficient for seismic case = 0.368 unit weight of concrete (substructure) = 2.40 t/m3

unit weight of soil = 1.80 t/m3

braking force = 10 t

maximum centrifugal force(normal) = 18 t maximum centrifugal force(seismic) = 9 t

transverse moment (normal) = 344 t-m

transverse moment (seismic) = 172 t-m

coefficient of friction = 0.5

(5)

STABILITY CALCULATIONS

face of dirtwall

0.574

1

0.441

face of stem wall

2 0.754 0.35 1.881 VERTICAL FORCES W (t) 1 wt of slab 74.969 4.638 2 dirt wall 29.754 9.451 3 abut cap 34.389 10.373 4 stem 313.061 10.519 5 toe 98.095 12.792 6 heel slab 228.772 5.002 7 partition wall 105.3607 5.112 8 counterfort-1 110.379 5.008 9 counterfort-2 115.618 5.164 10 side wall-1 319.796 5.197 11 side wall-2 336.168 5.209 12 end wall 132.512 0.150 13 earth fill -1 1227.29 2.631 14 earth fill-2 1094.909 7.742 15 DL + SIDL 170 10.519 16 LL 80 10.519

For calculating moments about toe for stability and for calculating area and section modulus's of the foundation plan, considering the trapezoid to be an equivalent rectangle for simplicity.

0.85 x 16.006 x 9.794 x 2.4 = 0.85 x 16.786 x 9.817 x 2.4 = 0.3 x 11.225 x 16.396 x 2.4 = 41.585 x 16.396 x 1.8 = 170 = 80 = 4.9595 x 8.92 x 16.786 x 1.8 =

Lever arm from end wall

9.276 x 11.225 x 0.3 x 2.4 = 0.5 x (2.256 + 3.036) x 11.225 x 0.35 x 2.4 = 0.5 x (2.781+1.474) x 11.225 x 0.6 x 2.4 = 0.5 x (0.85 + 0.45) x 17.636 x 11.379 x 2.4 = 0.5 x (1.8+0.853) x 13.412 x 2.3 x2.4 = weight calculations 0.5 x (9.337 + 10.644) x 11.225 x 0.85 x 2.4 = 0.3 x 8.925 x 16.396 x 2.4 = 0.3 x 16.283 x 9.415 x 2.4 = 0.3 x 16.507 x 9.728 x 2.4 =

(6)

11.225

13.225 toe side

13.189

longitudinal section modulus = 383.41 m3

Area = 174.42 m2

transverse section modulus = 384.46 m3

W (t) 1 wt of slab 74.969 8.551 2 dirt wall 29.754 3.738 3 abut cap 34.389 2.816 4 stem 313.061 2.67 5 toe 98.095 0.397 6 heel slab 228.772 8.187 7 partition wall 105.3607 8.077 8 counterfort-1 110.379 8.181 9 counterfort-2 115.618 8.025 10 side wall-1 319.796 7.992 11 side wall-2 336.168 7.980 12 end wall 132.512 13.039 13 earth fill -1 1227.29 10.558 14 earth fill-2 1094.909 5.447 15 DL + SIDL 170 2.67 16 LL 80 2.67 ∑ = 4471.073 903.011 927.834 2555.810 2682.621 96.839 835.873 38.944 1872.956 850.998

Lever arm from Toe side Moment (t-m)

641.060 111.220 1727.824 12957.728 5963.969 453.900 213.600 32834.188

(7)

HORIZONTAL FORCES (without using earth pressure coefficient) Force F(t) Normal ( Fx0.279) Seismic (Fx0.368) Lever arm from base 1 earth pressure 3110.085 867.714 1144.511 7.369 2 surcharge 454.516 126.81 167.2619 9.373 3 braking force 10 10 10 18.746 NORMAL CASE 1. WITHOUT SUPERSTRUCTURE

Total vertical force = 4221.07 t Total resisiting moment = 32166.69 t-m cg of forces from toe= 7.62 m

eccentricity from cg of base = 1.03 m < 2.2 (B/6) ok Total horizontal force= 994.52 t

overturning moment= 7582.77 t-m

Fos against Sliding = 2.12 > 1.5 ok

Fos against Overturning = 4.24 > 2 ok

Max Stress = 32.68 t/m2 < 45t/m2 ok Min Stress = 15.72 t/m2 < 45t/m2 ok

2. WITH SUPERSTRUCTURE

Total vertical Force = 4471.07 t Resisting Moment = 32834.19 t-m cg of forces from toe = 7.34 m

ecentricity from cg of base = 0.75 m < 2.2 (B/6) ok Total horizontal force = 1004.52 t

overturning moment = 7770.23 t-m

Fos against overturning = 4.23 > 2 ok

Transverse Moment = 344.45 t-m Max Stress = 38.06 t/m2 < 45t/m2 ok Min Stress = 13.21 t/m2 < 45t/m2 ok 0.5*1.8*307.853*11.225 = 1.2*1.8*18.746*11.225 = 10 Force calculations

(8)

SEISMIC CASE

1. WITHOUT SUPERSTRUCTURE

Total vertical force = 4221.07 t resisting moment = 32166.69 t-m cg of forces from toe= 7.62 m

eccentricity from cg of base = 1.03 m < 2.2 (B/6) ok Total Horizontal force = 1311.77 t

Fos against overturning = 3.22 > 1.5 ok

Max Stress = 39.44 t/m2 < 56t/m2 ok

Min Stress = 8.96 t/m2 < 56t/m2 ok

2. WITH SUPERSTRUCTURE

Total vertical force = 4471.07 t resisting moment = 32834.19 t-m cg of forces from toe = 7.34 m

eccentricity from cg of base = 0.75 m < 2.2 (B/6) ok Total horizontal force = 1321.77 t

Fos against sliding = 1.69 > 1.25 ok

Fos against overturning= 3.22 > 1.5 ok

Transverse Moment = 172.22 t-m

Max stress = 43.92 t/m2 < 56t/m2 ok

(9)

Pressure diagram (Normal) : For abutment + superstructure

heel portion

0.3 4.662 0.3 4.7 0.85 2.3

(2) (1) stem toe

13.21 t/m2 13.77 t/m2 22.56 t/m2 23.12 t/m2 31.98 t/m2 33.58 t/m2

gross pressure diagram 38.06 t/m2

13.189

overburden pressure for toe : 6.68 t/m2

overburden pressure for heel: 32.25 t/m2 (region 1) overburden pressure for heel: 31.55 t/m2 (region 2)

Net Pressure consideration for Toe

2.3

26.90 t/m2 31.38 t/m2

Net Pressure consideration for heel(region 1) 4.7

9.13 t/m2 0.28 t/m2

Net Pressure consideration for heel (region 2)

17.78 t/m2 9.00 t/m2

(10)

Pressure diagram (Seismic) : For abutment + superstructure

heel portion

0.3 4.662 0.3 4.7 0.85 2.3

(2) (1) stem toe

7.35 t/m2 8.18 t/m2 21.11 t/m2 21.94 t/m2 34.97 t/m2 37.33 t/m2

gross pressure diagram 43.92 t/m2

13.189

overburden pressure for toe : 6.68 t/m2

overburden pressure for heel: 32.25 t/m2 (region 1) overburden pressure for heel: 31.55 t/m2 (region 2)

Net Pressure consideration for Toe

2.3

30.65 t/m2 37.24 t/m2

Net Pressure consideration for heel(region 1) 4.7

-10.32 t/m2 2.72 t/m2

Net Pressure consideration for heel (region 2)

23.37 t/m2 10.45 t/m2

(11)

DESIGN OF TOE(NORMAL CASE) 2.3 26.90 t/m2 31.38 t/m2 1.713 m 30.2 t/m2 Grade of conc. 35 Grade of steel 500

Dia of bar used 25

Q for concrete grade used 170 t/m2

k value for concrete 0.327

j value for concrete 0.891

Permissible stress in steel 24000 t/m2

Permissible stress in concrete 1167 t/m2

Cover in foundation = 75

Maximum moment at the face of stem = 79.05 t-m

effective depth required = 0.682 m

effective depth available = 1.713 m

Main Ast required = 2158.91 mm2 (at bottom)

Provide 25 φ @ 200 c/c = 2454.37 mm2

Distribution steel = 0.25% X 2,454.369 = 613.59 mm2 (at bottom)

Provide 12 φ @ 150 c/c = 753.98 mm2

Nominal reinforcement = 250 mm2/m

Provide 8 φ @ 200 c/c = 251.33 mm2

Check for shear :

shear force at distance d from face of stem= 18.11 t

bending moment at d = 5.35 t-m

effective depth available at d= 1.005 m

relief in shear force = (M*tan(beta)/d) = 2.42 t

design shear force= 18.11 - 2.42 = 15.69 t

24.4 ° shear stress = 15.61 t/m2 0.16 N/mm2(τv)

1.8 d 0.85

% steel provided = 0.24

2.3 τc = 0.24 N/mm2 > 0.16 N/mm2

(12)

DESIGN OF TOE(SEISMIC CASE) 2.3 30.65 t/m2 37.24 t/m2 1.713 m 35.6 t/m2 Grade of conc. 35 Grade of steel 500

Dia of bar used 25

Permissible stress in concrete 1167 t/m2

Cover in foundation = 75

Seismic relief factor = 1.5

Maximum moment at the face of stem = 92.69 t-m

effective depth required = 0.738 m

effective depth available = 1.713 m

Main Ast required = 1687.59 mm2 (at bottom)

Provide 25 φ @ 250 c/c = 1963.50 mm2

Distribution steel = 0.25% X 1,963.495 = 490.87 mm2 (at bottom)

Provide 12 φ @ 220 c/c = 514.08 mm2

Nominal reinforcement = 250 mm2/m

Provide 8 φ @ 200 c/c = 251.33 mm2

Check for shear :

shear force at distance d from face of stem= 21.39 t

bending moment at d = 6.33 t-m

effective depth available at d= 1.005 m

relief in shear force = (M*tan(beta)/d) = 2.86 t

design shear force= 21.39 - 2.86 = 18.53 t

24.4 ° shear stress = 18.44 t/m2 0.19 N/mm2(τv)

1.8 d 0.85

% steel provided = 0.20

2.3 τc = 0.22 N/mm2 > 0.19 N/mm2

(13)

HEEL DESIGN (NORMAL CASE)

(A) box compartments

4.962 m 5.275 m 3.550 m 3.275 m 3.550 m conc grade= 35 steel grade= 500 max bar dia= 25 mm

Q value= 0 t/m2 k value= 1.000 j value= 0.891 Fst= 24000 t/m2 Fc= 1167 t/m2 cover= 75 mm

heel region (2) heel region (1)

S4 S2

S3 S1

(14)

ly = 5.275 m

lx = 3.275 m

ly/lx = 1.61 < 2 (two way slab)

for continuous slabs l/d: 32

dx = 102.344 mm avail dx = 0.763 mm dx : 0.763 m dy : 0.738 m 5.275 m 9.13 t/m2 0.28 t/m2

Taking the maximum stress 9.13 t/m2

Wu : 10.00 t/m2 lx : 3.275 m M : α * Wu * lx^2 x+ x- y+ y-α 0.046 0.061 0.028 0.037 M 4.934 6.543 3.003 3.968

Providing the steel in both dirctions for maximum moment = 6.54 t-m/m

Steel required = 401.26 mm2

Provide 16 mm @ 200 c/c 1005.31 mm2 0.132 %

For shear force check Vu = 7.38 t/m

τv = 0.10 N/mm2 < τc = 0.18 N/mm2

HENCE safe in shear

(15)

ly = 5.275 m

lx = 3.550 m

dx = 110.938 mm avail dx = 0.763 mm dx : 0.763 mm dy : 0.738 mm 5.275 m 9.13 t/m2 0.28 t/m2

Wu : 10.00 t/m2

lx : 3.550 m M : α * Wu * lx^2

x+ x- y+

y-α 0.056 0.075 0.035 0.047

M 7.057 9.452 4.411 5.923

Provide 16 mm @ 200 c/c 1005.31 mm2 0.132 %

τv = 0.12 N/mm2 < τc = 0.18 N/mm2

HENCE safe in shear

(16)

ly = 4.962 m

lx = 3.275 m

dx = 102.344 mm avail dx = 0.763 mm dx : 0.763 m dy : 0.738 m 17.78 t/m2 9.00 t/m2 4.962 m

Provide 20 mm @ 200 c/c 1570.80 mm2 0.21 %

τv = 0.18 N/mm2 < τc = 0.22 N/mm2

HENCE safe in shear

(17)

ly = 4.962 m

lx = 3.550 m

dx = 136.538 mm cover= 75 mm

avail dx = 0.763 m max bar dia= 25 mm

Fst= 24000 t/m2

dx : 0.763 mm j= 0.891

dy : 0.738 mm

17.78 t/m2 9.00 t/m2

4.962 m

Wu : 18.00 t/m2

lx : 3.550 m M : α * Wu * lx^2

x+ x- y+

y-α 0.053 0.071 0.035 0.047

M 12.023 16.106 7.940 10.662

Provide 20 mm @ 200 c/c 1570.80 mm2 0.21 %

τv = 0.21 N/mm2 < τc = 0.22 N/mm2

HENCE safe in shear

(18)

HEEL DESIGN (NORMAL CASE)

(A) box compartments

4.962 m 5.275 m 3.550 m 3.275 m 3.550 m conc grade= 35 steel grade= 500

max bar dia= 25 mm

Q value= 0 t/m2 k value= 1.000 j value= 0.891 Fst= 24000 t/m2 Fc= 1167 t/m2 cover= 75 mm seismic factor= 1.5

heel region (2) heel region (1)

S4 S2

S3 S1

(19)

ly = 5.275 m

lx = 3.275 m

dx = 102.344 mm avail dx = 0.763 mm dx : 0.763 m dy : 0.738 m 5.275 m -10.32 t/m2 2.72 t/m2

Providing the steel in both directions for maximum moment = 2.944 t-m/m

Provide 16 mm @ 250 c/c 804.25 mm2 0.11 %

τv = 0.04 N/mm2 < τc = 0.29 N/mm2

HENCE safe in shear

(20)

ly = 5.275 m

lx = 3.550 m

dx = 110.938 mm avail dx = 0.763 mm dx : 0.763 mm dy : 0.738 mm 5.275 m -10.32 t/m2 2.72 t/m2

Wu : 4.50 t/m2

lx : 3.550 m M : α * Wu * lx^2

x+ x- y+

y-α 0.056 0.075 0.035 0.047

M 3.176 4.253 1.985 2.665

Provide 16 mm @ 250 c/c 804.25 mm2 0.11 %

τv = 0.05 N/mm2 < τc = 0.29 N/mm2

HENCE safe in shear

(21)

ly = 4.962 m

lx = 3.275 m

dx = 102.344 mm avail dx = 0.763 mm dx : 0.763 m dy : 0.738 m 23.37 t/m2 10.45 t/m2 4.962 m

Provide 25 mm @ 250 c/c 1963.50 mm2 0.26 %

τv = 0.35 N/mm2 < τc = 0.39 N/mm2

HENCE safe in shear

(22)

ly = 4.962 m

lx = 3.550 m

dx = 136.538 mm cover= 75 mm

avail dx = 0.763 mm max bar dia= 25 mm

Fst= 24000 t/m2

dx : 0.763 mm j= 0.891

dy : 0.738 mm

17.78 t/m2 9.00 t/m2

4.962 m

Wu : 27.00 t/m2

lx : 3.550 m M : α * Wu * lx^2

x+ x- y+

y-α 0.053 0.071 0.035 0.047

M 18.034 24.159 11.909 15.993

Provide 25 mm @ 250 c/c 1963.50 mm2 0.26 %

τv = 0.32 N/mm2 < τc = 0.38 N/mm2

HENCE safe in shear

(23)

H 4.962 m G 5.275 m F 3.550 m E 3.275 m D 3.550 m A B C X A B C D E F G H 4.962 m 3.550 m 3.275 m 3.550 m 5.275 m 4.962 m 1.0 t/m2

Moments Calculation by Moment Distribution Method :

A X member AB BA BC CB CD DC DX XD DF 0.000 0.450 0.550 0.400 0.600 0.650 0.350 1.000 FEM 2.052 -2.052 2.319 -2.319 1.050 -1.050 0.894 0.447 Balance 0.000 0.342 0.417 0.508 0.761 0.101 0.055 0.000 CO 0.000 0.000 0.254 0.209 0.051 0.381 0.000 0.028 Balance 0.000 -0.114 -0.140 -0.104 -0.155 -0.247 -0.133 0.000 CO 0.000 0.000 -0.052 -0.070 -0.124 -0.078 0.000 -0.067 Balance 0.000 0.023 0.029 -0.449 0.203 -0.081 0.069 0.000 CO 0.000 0.000 -0.225 0.015 -0.041 0.102 0.000 0.035 Balance 0.000 0.101 0.123 0.010 0.016 -0.066 -0.036 0.000 CO 0.000 0.000 0.005 0.062 -0.033 0.008 0.000 -0.018

∑

0.000 -2.726 2.731 -2.140

The side walls along with the stem wall is designed as a continuous seven span.

5.275 m

Considering a loading throughout this continuous span

B C D

Modified

FEM 0.000 -3.078 2.319 -2.319 1.050 -1.050 0.894 0.447

0.425

(24)

Shear Force Calculation :

Maximum Shear occurs at point B

Due to Load : 2.481 2.6375

Due to Moments : 0.549 0.112

∑ 3.03 2.7495

STEEL REQUIREMENT BY MOMENT

Maximum Moment that can occur due to unit load= 3.0 t-m/m Inside the box compartments, the earth is confined and hence pressure acting is static, so "active earth pressure at rest" will act.

for given soil, angle of repose = 30 °

earth pressure coefficient = 0.279

unit density of soil = 1.80 t/m3

Maximum height of soil inside box= 16.786 m

Therefor maximum horizontal pressure on wall = 8.43 t/m2

Maximum Moment that can occur = 25.3 t-m/m

effective wall thickness required = 0.386 m

using maximum bar dia = 25 mm take

& clear cover = 75 mm 90 mm

Overall depth required = 0.473 m

Hence provide overall thickness = 0.850 m For soil fill inside the box

Reqd(mm2) 1 16.786 8.43 25 0.298 0.800 1665.81 20 180 1745.33 0.22 % 2 13.000 6.53 20 0.252 0.760 1367.11 20 180 1745.33 0.23 % 3 10.000 5.02 15 0.210 0.660 1236.12 16 150 1340.41 0.20 % 4 6.500 3.26 10 0.152 0.560 974.43 16 200 1005.31 0.18 % 5 3.000 1.51 5 0.076 0.450 587.16 16 200 1005.31 0.22 % B STEEL REQUIREMENT(mm2) Provide Provide D (m) effective =87.5mm

S. No. Depth (m) Pressure (t/m2)

Moment

(25)

Therefore wall overall thickness can vary from 0.80 m at bottom to 0.45 m at top

STEEL REQUIREMENT BY SHEAR

Maximum Shear that can occur due to unit load= 3.03 t/m Inside the box compartments, the earth is confined and hence pressure acting is static, so "active earth pressure at rest" will act.

for given soil, angle of repose = 30 °

earth pressure coefficient = 0.279

unit density of soil = 1.80 t/m3

support thickness = 0.30 m

For soil fill inside the box

S. No. Depth (m) Pressure

(t/m2) Vcr (t/m) d provided (m) τv (N/mm2) τc (N/mm2) 1 16.786 8.43 18 0.800 0.219 0.221 2 13.000 6.53 14 0.760 0.179 0.224 3 10.000 5.02 10 0.660 0.158 0.215 4 6.500 3.26 7 0.560 0.121 0.227 5 3.000 1.51 3 0.450 0.070 0.246

At every section τv < τc , hence nominal shear shear reinforcement will do Provide 2 legged shear stirrup :

bar dia : 10 φ Asv = 78.5 mm2

(26)

(27)

Maximum reaction force = 3.030 t/m + 2.750 t/m per 1.00 t/m2 load =

S. No. bar dia (mm) Spacing (mm) mm2 1 16.786 8.43 48.72 2030.03 1015.02 12 90 1256.637 2 13.000 6.53 37.73 1572.17 786.08 12 90 1256.637 3 10.000 5.02 29.02 1209.36 604.68 12 150 753.9822 4 6.500 3.26 18.87 786.08 393.04 12 150 753.9822 5 3.000 1.51 8.71 362.81 181.40 12 150 753.9822 3

thickness of end wall = 0.300 m

Minimumreinforcement provided = 0.2% X bd

= 600.00 mm2

Provide 12 150 753.98 mm2

In the end wall there is earth pressure acting from both sides hence net pressure is effectively zero. Thus minimum reinforcement would suffice.

Depth (m) Pressure (t/m2) Tension T(t/m) Reqd Ast (mm2/m) On each face (mm2/m) Ast provided This member is designed as a tension member with tensile force equal to the maximum

reaction acting at the juncture of partition wall and the side walls

Design of Partition wall

5.780 t/m

(28)

INNER COUNTERFORTS (T-BEAMS)

influence zone for the counterfort walls= 3.413 m (this is the width from which each counterfort receives the eatrh pressure)

unit weight of soil = 1.80 t/m3 Fst = 24000 t/m2

earth pressure coeffiecnt = 0.279 j = 0.891

surcharge height = 1.20 m

maximum soil height = 16.786 m

pressure due to soil at specified depth = 0.502 X H (t/m2) (varying) pressure due to surcharge at specified depth = 0.603 t/m2 (constant) pressure diagram will be somewhat like this

0.603 t/m2 X 3.4125 = 16.786 m 9.033 t/m2 30.83 t/m2 D = 9.662 m 32 φ d = 9.596 m 50 mm b= 0.300 m φ nos 1 16.786 30.83 1641 7999.044 32 9.00 2 13.000 24.34 1337 6513.915 32 7.00 3 10.000 19.20 1095 5337.109 32 6.00 4 6.500 13.20 813 3964.169 32 4.00 5 3.000 7.20 532 2591.229 32 3.00

Design Of COUNTERFORT WALLS

2.06 t/m2 S. No. Depth (m) Pressure at this depth (t/m2) Moment (t-m/m) STEEL REQUIREMENT(mm2) Reqd (mm2)

max bar dia = clear cover =

(29)

6.786 m and lap with 25φ therafter extending to top 16.786 30.83 t/m2 138.02 0.48 0.24 13.000 24.34 t/m2 85.80 0.30 0.24 10.000 19.20 t/m2 53.15 0.18 0.22 6.500 13.20 t/m2 24.80 0.09 0.22 3.000 7.20 t/m2 6.95 0.02 0.22

Provide 2 legged stirrup 8 φ = 100.531 mm2 Spacing = 335.87868 mm

(horizontal ties)

Average downward pressure = 16.91 t/m2

Total area of reinforcement required to resist this tension = 704.59 mm2/m Provide 2 legged 8φ = 100.53 mm2

spacing = 100.53 x 1000 / 704.59 = 142.679

(vertical ties)

Provide 8φ 2 legged stirrup at a spacing of 120 c/c

Vs = (τv-τc)bd (t) 68.93 16.71 0 0 0 Vu (t) Pressure at this depth (t/m2) τc (N/mm2)

Provide 8φ 2 legged stirrup at a spacing of 200 c/c τv (N/mm2)

To secure connection between counterfort and heel slab, counterfort needs to be designed for tension arising out of the net outward pressure of heel slab.

Depth (m)

(30)

OUTER COUNTERFORTS (L BEAMS)

influence zone for the counterfort walls= 1.775 m (this is the width from which each counterfort receives the eatrh pressure)

unit weight of soil = 1.80 t/m3 Fst = 24000 t/m2

earth pressure coeffiecnt = 0.279 j = 0.891

surcharge height = 1.20 m

maximum soil height = 16.786 m

pressure due to soil at specified depth = 0.502 X H (t/m2) (varying) pressure due to surcharge at specified depth = 0.603 t/m2 (constant) pressure diagram will be somewhat like this

0.603 t/m2 X 1.775 = 16.786 m 9.033 t/m2 16.04 t/m2 D = 9.662 m 32 φ d = 9.596 m 50 mm b= 0.850 m φ nos 1 16.786 16.04 854 4160.876 32 5.00 2 13.000 12.66 695 3388.114 32 5.00 3 10.000 9.99 570 2775.782 32 3.00 4 6.500 6.87 423 2061.395 32 3.00 5 3.000 3.75 276 1347.008 32 3.00 1.07 t/m2

max bar dia = clear cover = S. No. Depth (m) Pressure at this depth (t/m2) Moment (t-m/m) STEEL REQUIREMENT(mm2) Reqd (mm2) Provide

(31)

6.786 m and lap with 28φ therafter extending to top 16.786 16.04 t/m2 71.80 0.09 0.2 13.000 12.66 t/m2 44.63 0.05 0.2 10.000 9.99 t/m2 27.65 0.03 0.2 6.500 6.87 t/m2 12.90 0.02 0.2 3.000 3.75 t/m2 3.61 0.00 0.2

τv < τc at all positions, so provide nominal shear reinforcement :

8 φ = 100.531 mm2

Spacing = 200 mm (horizontal ties)

Average downward pressure = 16.91 t/m2

Total area of reinforcement required to resist this tension = 704.59 mm2/m Provide 2 legged 8φ = 100.53 mm2

spacing = 100.53 x 1000 / 704.59 = 142.679

(vertical ties)

Two layers of 32φ from the base to height of

Depth (m) Pressure at this depth (t/m2) Vu (t) τv (N/mm2) τc (N/mm2)

To secure connection between counterfort and heel slab, counterfort needs to be designed for tension arising out of the net outward pressure of heel slab.

Provide 2 legged stirrup

(32)

3413

9728

300

modular ratio m = 280/(3σcbc) for M35, σcbc = 11.67

= 8

assuming effective cover = 100 mm d = 10478 mm

finding the neautral axis (X)

1) assuming the Neutral axis lies in the flange Using Ast= 9 nos 32 Φ

b*X^2/2= m*Ast(d-X) ====> X^2 + (2mAst/b)X - (2mAst/b)d

X^2 + (34)X - 356252 a = 1 Solving we get X = 580 mm b = 34 c = -356252 for moment M = 1641 Stress in concrete (Fc) = M/(0.5*b*X*(d-X)) ∴ Fc = 1.68 Mpa < 11.67 Mpa Stress in steel (Fs) = mFc*(d-X)/X ∴ Fs = 228.64 Mpa < 240 MPa 850 7238.23 mm2

(33)

(34)

9817

850

modular ratio m = 280/(3σcbc) for M35, σcbc = 11.67

= 8

assuming effective cover = 100 mm d = 10567 mm

finding the neautral axis (X)

1) assuming the Neutral axis lies in the flange Using Ast= 3 nos 32 Φ

b*X^2/2= m*Ast(d-X) ====> X^2 + (2mAst/b)X - (2mAst/b)d

X^2 + (22)X - 232474 a = 1 Solving we get X = 471 mm b = 22 c = -232474 for moment M = 423 Stress in concrete (Fc) = M/(0.5*b*X*(d-X)) ∴ Fc = 1.00 Mpa < 11.67 Mpa Stress in steel (Fs) = mFc*(d-X)/X ∴ Fs = 171.66 Mpa < 240 MPa 2412.74 mm2 1775 850

(35)

Thickness of slab = 0.30 m wearing course = 56.00 mm

B= span in transverse direction = 3.550 m (short) L= span in longitudinal direction = 5.275 m (long)

weight of deck slab = 0.720 t/m2 weight of wearing course = 0.123 t/m2 total weight = 0.843 t/m2

along short span K = short span/long span = 0.67 ===>> m1 = 0.047

from Pigeud's curve

1/K = 1.49 ===>> m2 = 0.018

along long span 0.15 for reinforced concrete bridges

15.79 t

moment along short span = (m1+ µ*m2)*W = 0.78 t-m moment along long span = (m2+ µ*m1)*W = 0.40 t-m

0.85 3.6 3.550 m X 5.275 m impact factor= 25 % u= 0.962 m v= 3.712 m limited to 5.275 m

K= 0.67 using pigeud's curve=

u/B= 0.271 m1= 0.12

v/L= 0.704 m2= 0.048

SIZE OF PANEL of deck slab=

width of load spread along short span= width of load spread along long span=

Design of ROOF SLAB by Pigeud's curve

Live load bending moment due to IRC Class AA tracked vehicle = Maximum bending moment due to dead load =

poisson's ratio µ =

Total dead weight W =

(36)

35.00 t total load per track including impact = 1.25*35 = 43.75 t effective load on span= 43.75*3.712/3.712 = 43.75 t moment along short span= (m1+µ*m2)*43.75 = 5.57 t-m moment along long span= (m2+µ*m1)*43.75 = 2.89 t-m

Y 2.6375 4 1 2 3 X 3.75 t 6.25 t 3.75 t X 2.6375 8 5 6 7 3.75 t 6.25 t 3.75 t 1.775 Y 1.775 B = 3.550 m L = 5.275 m

load per track of AA=

Live load bending moment due to IRC Class AA wheeled vehicle =

6.25 t

The Class AA wheeled vehicle is placed as shown to produce the severest moments. The front axle is placed along the centre line with 6.25t wheel at centre of panel. Maximum moments in the short span and long span directions are computed for individual wheel loads taken in the order shown

(37)

6.25 t tyre contact dimensions= 300 X 150 mm u= sqrt((0.3+2*0.056)^2+0.3^2)= 0.5097 v= sqrt((0.15+2*0.056)^2+0.3^2)= 0.3984

u/B = 0.144 v/L = 0.076 K = 0.70

m1 = 0.221 Total load allowing for 25 % impact = 7.813 t

m2 = 0.190

Moment along short span = 1.950 t-m Moment along long span = 1.744 t-m

6.25 t Y X 6.25 t 6.25 t X 0.3 0.3 Y Intensity of loading = (6.25*1.25)/(0.5097*0.3984) = 38.47 t/m2

Consider the loaded area of 0.150 X 2.300

u = sqrt((2.3+2*0.056)^2+0.3^2) = 2.4306 v = sqrt((0.15+2*0.056)^2+0.3^2) = 0.3983

u/B = 0.685 v/L = 0.076 K = 0.70

m1 = 0.112

m2 = 0.119

Bending Moment due to wheel 1 =

Here wheel load is placed unsymmetrical to YY axis. But Pigeuds Curves are derived for symmetrical loading. Hence we place an equal dummy load symmetrical about the YY axis and consider the whole loading area. Then we deduct the area beyond the actual loaded area. Half of the resulting value is taken as the moment due to actual loading.

0.85 0.85

(38)

Now consider the area beyond the actual loading = 1.7 X 0.15 u = sqrt((1.7+2*0.056)^2+0.3^2) = 1.8367 v = sqrt((0.15+2*0.056)^2+0.3^2) = 0.3983 u/B = 0.517 v/L = 0.076 K = 0.70 m1 = 0.132 m2 = 0.14

0.531 t-m 0.557 t-m

3.75 t By similar procedure as for previous case, we get B.M along short span = 0.052 t-m B.M along long span = 0.121 t-m

6.25 t

B.M along short span = 0.823 t-m B.M along long span = 0.195 t-m 6.25 t

B.M along short span = 0.823 t-m B.M along long span = 0.195 t-m

In this case loading is eccentric w.r.t XX axis. By similar procedure as for previous case but with load area extended w.r.t. XX axis, we get

Bending Moment due to wheel 6 = Bending Moment due to wheel 3 =

Net moment along short span = Net moment along long span =

In this case loading is eccentric with respect to both XX and YY axes. A strict simulation would be very complicated and laborious. For A reasonable approximation, eccentricity w.r.t. only XX axis is considered and calculations made as for previous case.

(39)

total bending moment along short span = 5.671 t-m total bending moment along long span = 3.649 t-m

To allow for continuity, the computed momnts are multiplied by a factor of 0.8

Design Bending Moment=

along short span= 5.16 t-m

along long span= 3.24 t-m

Grade of conc. 35

Grade of steel 500

Dia of bar used 16

Permissible stress in steel 24000 t/m2 Permissible stress in concrete 1167 t/m2

cover 50 mm

effective depth required= 0.174 m

provided deff = 0.242 m OK

998 mm2 ϕ 16 @ 150 c/c 1340 mm2 OK

625 mm2 ϕ 12 @ 150 c/c 754 mm2 OK longitudinal reinforcement required =

main reinforcement required=

So provide

(40)

Y X 0.225 m 2.781 m = 6257.25 mm2 3128.625 mm2 Provide ϕ 25 @ 150 c/c = 3272.492 mm2 0.225 m 1.000 m 2250 mm2

steel on both top and bottom face = 1/2 of steel red= 1125 mm2

Provide ϕ 12 @ 150 c/c = 1507.964 mm2

==> 2 legged stirrup Along Y direction

Steel reqd = 1% of cross section

Design of Abutment Cap

Taking thickness equal to 225 mm for calculation of steel requirement.

Steel reqd = 1% of cross section

steel on both top and bottom face = 1/2 of steel reqd= Along X direction

(41)

varies Design values : 0.35 m g = 1.80 t/m2 3.036 2 1 ka = 0.279 1.525 t/m2 0.603 t/m2

1)Earth Pressure due to surcharge equivalent to 1.2m of earthfill = ka *g*1.2

= 0.603 t/m2 2)Earth Pressure due to backfill of earth = ka *g*h

= 1.525 t/m2

Bending moment at the base of dirtwall due to earth pressure (1) = ka *g*1.2*h2/2

= 2.777 t-m/m

Bending moment at the base of dirtwall due to earth pressure (1) = ka *g*h3/6

= 2.342 t-m/m

Total bending moment at the base of dirtwall = 5.120 t-m/m due to earth pressure

Calculation of force and moment due to the effect of braking :(cosidering 40t bogie loading)

1.7 10t 10.85 10t

2.79m 3.036

1.750 2.79 3.036

Effective width = 7.576 m

Braking force, 0.2*20 = 4t (Only two wheels can come on dirt wall at a time.) Braking force = 4 t (Impact factor can't be included with braking force)

Braking force per metre width = 0.53 t

Bending moment at the base of dirtwall due to effect of braking = 2.24 t-m/m Therefore total bending moment at the base of the dirtwall = 7.36 t-m/m

DESIGN OF DIRTWALL

Earth Pressure diagram

width of

dirtwall = 0.350 m

Dirt Wall Max. wheel load is from 40t boggie load. So, considering this case only. Wheel Loads shown below are placed on dirt

wall (not on RCC Solid Slab) along carriage-way. So, dispersion is taken directly at 45 degrees. height of

(42)

Basic Design Data:

Grade of conc. 35

Grade of steel 500

Dia of bar used 16

Permissible stress in concrete 1167 t/m2 Permissible stress in steel 24000 t/m2

m , Modulur ratio 10

K value for concrete 0.327

Q for concrete 170.08

Max. moment in dirtwall (t-m) 7.36

Effective depth required (mm) 208

Effective depth provided (mm) 292 SAFE

Ast required (mm2₎ ₁₁₇₈

Provide Vertical longitudinal reinforcement:

Use f16 @ 150 c/c

Ast provided (mm2₎ _{(On Approach Side face)} ₌ ₁₃₄₀ _{Thus OK}

Min. Reinforcent in vertical direction (On outer face) = 525 For 1m length

Use f16 @ 200 c/c

1005 Thus OK

Therefore providing 20 f @ 130 c/c on the approach side and 12 f @ 200 c/c (Min. % reinforcement) on the outer side in the vertical direction . Also providing 12 f @ 200 c/c on both faces in the horizontal direction .

(43)

I Normal Case with Live load

In the design of dirtwall the total height cosidered has been calculated taking into account a slope of 2.5% provided in the carriageway.

For Backfill Soil :

a = 90.0 degree 0.35 b = 0.0 degree 3.036 f = 30.0 degree 1 2 d = 20.0 degree Kah = 0.279 1.525 t/m2 0.603 t/m2 g = 1.8 t/m3

Earth Pressure diagram

Height of dirtwall = 3.036m

1 Earth Pressure due to surcharge equivalent to 1.2m of earthfill = 0.603 t/m2 2 Earth Pressure due to backfill of earth = 1.525 t/m2

Bending moment at the base of dirtwall due to earth pressure (1) = 2.78 t-m/m Bending moment at the base of dirtwall due to earth pressure (2) = 2.95 t-m/m

Total bending moment at the base of dirtwall = 5.73 t-m/m

Calculation of force and moment due to the effect of braking :(cosidering 40t bogie loading)

1.7 10t 10t 3.04 2.79m 1.700 2.79 3.036 Effective width = 7.526 m Braking force, 0.2*20 = 4.0t

Braking force per metre width = 0.53 t

Bending moment at the base of dirtwall due to effect of braking = 1.61 t-m/m

Therefore total bending moment at the base of the dirtwall = 7.34 t-m/m DESIGN OF DIRTWALL

(44)

II Seismic Case

In the design of dirtwall the total height cosidered has been calculated taking into account a slope of 2.5% provided in the carriageway.

For Backfill Soil :

a = 90.0 degree 0.30 b = 0.0 degree 3.036 f = 33.0 degree 1 2 d = 22.0 degree Cah = 0.134 0.730 t/m2 0.288 t/m2 g = 1.8 t/m3

Earth Pressure diagram

Height of dirtwall = 3.036m

1 Earth Pressure due to surcharge equivalent to 1.2m of earthfill = 0.288 t/m2 2 Earth Pressure due to backfill of earth = 0.730 t/m2

3 Horizondal force due to seismic = 0.231 t/m

Bending moment at the base of dirtwall due to earth pressure (1) = 1.76 t-m/m Bending moment at the base of dirtwall due to earth pressure (2) = 1.41 t-m/m Bending moment at the base of dirtwall due to seismic horizondal force (3) = 0.35 t-m/m

Total bending moment at the base of dirtwall = 3.52 t-m/m

Therefore Normal case govern for the design CALCULATION OF DESIGN PARAMETERS

Grade of concrete = M 35 Grade of steel = Fe 500

Permissible stresses:

sst = 24480 t/m2 sbc = 1190 t/m2

Basic Design Parameters:

k = 0.327

j = 0.891

Q = 173.41 t/m2

Required effective depth =(7.34/173.41)^0.5 = 0.206 m

Depth provided = 0.35- 0.05 - (16/2000) = 0.292 m

Thus OK

Required Ast =7.34*10000/(24480*0.891*0.292) = 11.53 cm2/m

50% of additional reinforcement should be provided as per note of transport ministry 17.30 cm2/m

Provide 16 f @ 100 c/c at earth face = 20.11 cm2/m

Thus OK

Min reinforcement = 0.06% of cross sectional area = 2.1 cm2/m Dirt wall has been designed for normal case, wind case and seismic case. permisible stresses are increased by 33%/50% and hence it does not governs.

(45)

Provide 10 f @ 150 c/c outer face = 5.24 cm2/m

Thus OK

Provide 10 f @ 150 c/c Horz both face 5.24 cm2/m

Assuming thickness of abutment cap = 225.0 mm

Width of abutment cap = 1420 mm

Thickness of abutment cap = 400 mm

Length of abutment cap = 11379 mm

Area of steel required (min 1%) = 36355905 mm3

Quantity of steel to be provided at top = 18177953 mm3 Quantity of steel to be provided at bottom = 18177953 mm3

Top Face

(a) Longitudnal steel

Quantity of steel to be provided in longitudnal direction ( 0.5*total steel at top ) = 9088976 mm3

Assuming a clear cover of = 50.0 mm

Length of bar = 11279.0 mm

Area of steel required in longitudnal direction = 805.8 mm2

Provide 8 bar of 12 mm dia bar as longitudnal steel on top face of abutment cap.

Provided steel = 904.3 mm2

(b) Transverse steel

Quantity of steel to be provided in transverse direction = 9088976 mm3

Quantity of steel required = 798750 mm3/m

Adopting 12mm dia bar and clear cover 50mm

Length of each stirrups = 1420 -100 = 1320 mm

Volume of each stirrup = 149212.8 mm3

No. of stirrups required in per m length = 5.4 Nos

Say = 6.0 Nos

Required spacing = 166.7 mm

Provide 12mm 150 mm c/c stirrups throughout in length of abutment cap.

Provided steel = 753.6 mm2

/m Same steel will be provided at bottom also on both long and trans direction

DESIGN OF ABUTMENT CAP

As the cap is fully supported on the abutment. Minimum thickness of the cap required as per cl. 716.2.1 of IRC : 78- is 225 mm.