Box Abutment

FRL = 561.936

0.35

(median edge)

Cl bridge

1

0.3

15+16 2

4.012

0.775 varies

0.3

cap top lvl 0.3

3 0.4

0.45

558.900

0.35

7

FILL

12

FILL

17.936

13

14

OGL

4

547.268

16.786

0.85

1.8

6

0.85

5

544.000

Founding level 0.3

0.3

4.662

4.7

0.85

2.3

12.419 1

10.2

10

0.85 87 °

4.172

2.968

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

4.662

0.85

11.379

5.129

11

0.863 11.507

1

2.335

13.959

JAMMU-UDHAMPUR V-6 (A1L + A2L) Formation level (median edge) =

561.936

carriageway width =

11.15 m

gradient= formation level(outer end) =

7% 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

minimum height of abutment =

17.156 m

0.780 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 = LL from superstructure = Allowable soil bearing capacity =

170 t 80 t 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.

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) =

9t

transverse moment (normal) =

344 t-m

transverse moment (seismic) =

172 t-m

coefficient of friction = Allowable soil bearing capacity(seismic)=

0.5 56.25 t/m2 (= 1.25 X 45)

STABILITY CALCULATIONS face of dirtwall

0.574

1 0.441

face of stem wall

2

0.754

0.35

1.881

VERTICAL FORCES

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

wt of slab dirt wall abut cap stem toe heel slab partition wall counterfort-1 counterfort-2 side wall-1 side wall-2 end wall earth fill -1 earth fill-2 DL + SIDL LL

weight calculations 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 = 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 = 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 = 4.9595 x 8.92 x 16.786 x 1.8 = 170 = 80 =

W (t) 74.969 29.754 34.389 313.061 98.095 228.772 105.3607 110.379 115.618 319.796 336.168 132.512 1227.29 1094.909 170 80

Lever arm from end wall 4.638 9.451 10.373 10.519 12.792 5.002 5.112 5.008 5.164 5.197 5.209 0.150 2.631 7.742 10.519 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.

11.225 toe side

13.225

13.189

longitudinal section modulus = Area = transverse section modulus =

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

wt of slab dirt wall abut cap stem toe heel slab partition wall counterfort-1 counterfort-2 side wall-1 side wall-2 end wall earth fill -1 earth fill-2 DL + SIDL LL ∑=

W (t) 74.969 29.754 34.389 313.061 98.095 228.772 105.3607 110.379 115.618 319.796 336.168 132.512 1227.29 1094.909 170 80 4471.073

Lever arm from Toe side 8.551 3.738 2.816 2.67 0.397 8.187 8.077 8.181 8.025 7.992 7.980 13.039 10.558 5.447 2.67 2.67

383.41 m3 174.42 m2 384.46 m3 Moment (t-m) 641.060 111.220 96.839 835.873 38.944 1872.956 850.998 903.011 927.834 2555.810 2682.621 1727.824 12957.728 5963.969 453.900 213.600 32834.188

HORIZONTAL FORCES (without using earth pressure coefficient)

Force calculations 1 earth pressure 2 surcharge 3 braking force

Force F(t) Normal ( Fx0.279) 3110.085 867.714 454.516 126.81 10 10

0.5*1.8*307.853*11.225 = 1.2*1.8*18.746*11.225 = 10

NORMAL CASE 1. WITHOUT SUPERSTRUCTURE Total vertical force = Total resisiting moment = cg of forces from toe= eccentricity from cg of base = Total horizontal force= overturning moment=

4221.07 t 32166.69 t-m 7.62 m 1.03 m

32.68 t/m2 15.72 t/m2

2.2 (B/6)

ok

994.52 t 7582.77 t-m

Fos against Sliding = Fos against Overturning = Max Stress = Min Stress =

<

2.12 4.24 < <

> > 45t/m2 45t/m2

1.5 2

ok ok

ok ok

2. WITH SUPERSTRUCTURE Total vertical Force = Resisting Moment = cg of forces from toe = ecentricity from cg of base = Total horizontal force = overturning moment =

4471.07 t 32834.19 t-m 7.34 m 0.75 m

Max Stress = Min Stress =

38.06 t/m2 13.21 t/m2

2.2 (B/6)

1004.52 t 7770.23 t-m

Fos against Sliding = Fos against overturning = Transverse Moment =

<

2.23 4.23

> >

1.5 2

344.45 t-m < <

45t/m2 45t/m2

ok ok

ok ok

ok

Seismic Lever arm (Fx0.368) from base 1144.511 7.369 167.2619 9.373 10 18.746

SEISMIC CASE 1. WITHOUT SUPERSTRUCTURE Total vertical force = resisting moment = cg of forces from toe= eccentricity from cg of base = Total Horizontal force = overturning moment =

4221.07 t 32166.69 t-m 7.62 m 1.03 m

39.44 t/m2 8.96 t/m2

2.2 (B/6)

ok

1311.77 t 10001.65 t-m

Fos against Sliding = Fos against overturning = Max Stress = Min Stress =

<

1.61 3.22 < <

> > 56t/m2 56t/m2

1.25 1.5

ok ok

ok ok

2. WITH SUPERSTRUCTURE Total vertical force = resisting moment = cg of forces from toe = eccentricity from cg of base = Total horizontal force = overturning moment =

4471.07 t 32834.19 t-m 7.34 m 0.75 m

Max stress = Min Stress =

43.92 t/m2 7.35 t/m2

2.2 (B/6)

1321.77 t 10189.11 t-m

Fos against sliding = Fos against overturning= Transverse Moment =

<

1.69 3.22

> >

1.25 1.5

172.22 t-m

< <

56t/m2 56t/m2

ok ok

ok ok

ok

Pressure diagram (Normal) : For abutment + superstructure heel portion 0.3

4.662

0.3

(2) 13.21 t/m2

13.77 t/m2

22.56 t/m2

4.7

0.85

2.3

(1)

stem

toe

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 : overburden pressure for heel: overburden pressure for heel:

6.68 t/m2 32.25 t/m2 (region 1) 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 4.662

Pressure diagram (Seismic) : For abutment + superstructure heel portion 0.3

4.662

0.3

(2) 7.35 t/m2

8.18 t/m2

21.11 t/m2

4.7

0.85

2.3

(1)

stem

toe

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 : overburden pressure for heel: overburden pressure for heel:

6.68 t/m2 32.25 t/m2 (region 1) 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 4.662

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. Grade of steel Dia of bar used Q for concrete grade used

35 500 25 170 t/m2 0.327 0.891 24000 t/m2 1167 t/m2

k value for concrete j value for concrete Permissible stress in steel Permissible stress in concrete 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 = Provide 25 φ @ 200 c/c

=

Distribution steel = 0.25% X 2,454.369 = Provide 12 φ @ 150 c/c = Nominal reinforcement = Provide 8 φ @ 200 c/c

2158.91 mm2 (at bottom) 2454.37 mm2 613.59 mm2 (at bottom) 753.98 mm2

=

250 mm2/m 251.33 mm2

Check for shear : shear force at distance d from face of stem= bending moment at d = effective depth available at d= relief in shear force = (M*tan(beta)/d) =

18.11 t 5.35 t-m 1.005 m 2.42 t

design shear force= shear stress =

24.4 °

1.8

d 2.3

18.11 - 2.42 = 15.69 t 15.61 t/m2 0.16 N/mm2 (τv)

0.85 % steel provided = τc =

0.24 0.24 N/mm2 Hence OK

>

0.16 N/mm2

DESIGN OF TOE(SEISMIC CASE) 2.3 30.65 t/m2

37.24 t/m2

1.713 m

35.6 t/m2


35 500 25 170 t/m2 0.327 0.891 24000 t/m2 1167 t/m2

k value for concrete j value for concrete Permissible stress in steel Permissible stress in concrete Cover in foundation = Seismic relief factor =

75 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 = Provide 25 φ @ 250 c/c

=

Distribution steel = 0.25% X 1,963.495 = Provide 12 φ @ 220 c/c = Nominal reinforcement = Provide 8 φ @ 200 c/c

1687.59 mm2 (at bottom) 1963.50 mm2 490.87 mm2 (at bottom) 514.08 mm2

=

250 mm2/m 251.33 mm2

Check for shear : shear force at distance d from face of stem= bending moment at d = effective depth available at d= relief in shear force = (M*tan(beta)/d) =

21.39 t 6.33 t-m 1.005 m 2.86 t

design shear force= shear stress =

24.4 °

1.8

d 2.3

21.39 - 2.86 = 18.53 t 18.44 t/m2 0.19 N/mm2 (τv)

0.85 % steel provided = τc =

0.20 0.22 N/mm2 Hence OK

>

0.19 N/mm2

HEEL DESIGN (NORMAL CASE) (A) box compartments 4.962 m

5.275 m

S4

S2 3.550 m

S1

S3

3.275 m

S4 3.550 m

heel region (2)

conc grade= steel grade= max bar dia= Q value= k value= j value= Fst=

Fc= cover=

heel region (1)

35 500

25 mm 0 t/m2 1.000 0.891 24000 t/m2 1167 t/m2

75 mm

ly = lx =

S1 : One short edge discontinuous 5.275 m 3.275 m

ly/lx =

1.61

<

for continuous slabs l/d: dx = avail dx =

2

(two way slab)

32

102.344 mm 0.763 mm

dx : dy :

0.763 m 0.738 m

5.275 m 9.13 t/m2

0.28 t/m2

Taking the maximum stress Wu : lx : M:

10.00 t/m2 3.275 m α * Wu * lx^2 x+

α M

9.13 t/m2

x-

y+

0.046 4.934

y-

0.061 6.543

0.028 3.003

0.037 3.968

Providing the steel in both dirctions for maximum moment = Steel required = 401.26 mm2 Provide

16 mm

For shear force check Vu = τv =

@ 200 c/c

6.54 t-m/m

1005.31 mm2

0.132 %

7.38 t/m 0.10 N/mm2

HENCE safe in shear

<

τc =

0.18 N/mm2

ly = lx =

S2 : Two adjacent edges discontinuous 5.275 m 3.550 m

ly/lx =

1.49

<


2

(two way slab)

32

110.938 mm 0.763 mm

dx : dy :

0.763 mm 0.738 mm

5.275 m 9.13 t/m2

0.28 t/m2

Taking the maximum stress Wu : lx :

10.00 t/m2 3.550 m x+

α M

9.13 t/m2

M: x-

α * Wu * lx^2

y+

0.056 7.057

y-

0.075 9.452

0.035 4.411

Providing the steel in both dirctions for maximum moment = Steel required = 579.68 mm2 Provide 16 mm @ 200 c/c


0.047 5.923 9.452 t-m/m 1005.31 mm2

0.132 %

8.75 t/m 0.12 N/mm2

HENCE safe in shear

<

τc =

0.18 N/mm2

ly = lx =


ly/lx =

1.52

<


2

(two way slab)

32

102.344 mm 0.763 mm

dx : dy :

0.763 m 0.738 m

17.78 t/m2

9.00 t/m2 4.962 m

Taking the maximum stress Wu : lx : M:

18.00 t/m2 3.275 m α * Wu * lx^2 x+

α M

17.78 t/m2

x-

y+

0.044 8.495

y-

0.057 11.004

0.028 5.406

0.037 7.143


20 mm


@ 200 c/c

11.004 t-m/m

1570.80 mm2

0.21 %

13.28 t/m 0.18 N/mm2

HENCE safe in shear

<

τc =

0.22 N/mm2

ly = lx =


ly/lx =

1.40

<

2


(two way slab) 26

136.538 mm 0.763 m

dx : dy :

cover= max bar dia= Fst= j=

0.763 mm 0.738 mm

75 mm 25 mm 24000 t/m2 0.891

17.78 t/m2

9.00 t/m2 4.962 m


18.00 t/m2 3.550 m x+

α M

17.78 t/m2

x0.053 12.023

M:

α * Wu * lx^2

y+

y-

0.071 16.106

0.035 7.940



0.047 10.662 16.106 t-m/m 1570.80 mm2

0.21 %

15.75 t/m 0.21 N/mm2

HENCE safe in shear

<

τc =

0.22 N/mm2

HEEL DESIGN (NORMAL CASE) (A) box compartments 4.962 m

5.275 m

S4

S2 3.550 m

S1

S3

3.275 m

S4 3.550 m

heel region (2)

conc grade= steel grade= max bar dia= Q value= k value= j value= Fst=

Fc= cover= seismic factor=

heel region (1)

35 500

25 mm 0 t/m2 1.000 0.891 24000 t/m2 1167 t/m2

75 mm 1.5


ly = lx = ly/lx =

1.61

<


2

(two way slab)

32

102.344 mm 0.763 mm

dx : dy :

0.763 m 0.738 m

5.275 m -10.32 t/m2

2.72 t/m2

2.72 t/m2

Taking the maximum stress

Wu : lx : M:

4.50 t/m2 3.275 m α * Wu * lx^2 x+

α M

x0.046 2.220

y+

y-

0.061 2.944

0.028 1.351

0.037 1.786

Providing the steel in both directions for maximum moment = Steel required = 120.38 mm2 Provide

16 mm

For shear force check Vu = τv = HENCE safe in shear

@ 250 c/c

2.944 t-m/m

804.25 mm2

0.11 %

3.32 t/m 0.04 N/mm2

<

τc =

0.29 N/mm2

ly = lx =


ly/lx =

1.49

<

for continuous slabs l/d: dx =

2

(two way slab)

32

110.938 mm 0.763 mm

avail dx =

dx : dy :

0.763 mm 0.738 mm

5.275 m -10.32 t/m2

2.72 t/m2

2.72 t/m2


Wu : lx :

4.50 t/m2 3.550 m x+

α M

M: x-

0.056 3.176

α * Wu * lx^2

y+

y-

0.075 4.253

0.035 1.985



0.047 2.665 4.253 t-m/m 804.25 mm2

0.11 %

3.94 t/m 0.05 N/mm2

<

τc =

0.29 N/mm2


ly = lx = ly/lx =

1.52

<


2

(two way slab)

32

102.344 mm 0.763 mm

dx : dy :

0.763 m 0.738 m

23.37 t/m2

10.45 t/m2 4.962 m

23.37 t/m2


Wu : lx : M:

36.00 t/m2 3.275 m α * Wu * lx^2 x+

α M

x0.044 16.989

y+ 0.057 22.009

y0.028 10.811

0.037 14.287


25 mm


@ 250 c/c

22.009 t-m/m

1963.50 mm2

0.26 %

26.55 t/m 0.35 N/mm2

<

τc =

0.39 N/mm2

ly = lx =


ly/lx =

1.40

<

2


(two way slab) 26

136.538 mm 0.763 mm

dx : dy :

cover= max bar dia= Fst= j=

0.763 mm 0.738 mm

75 mm 25 mm 24000 t/m2 0.891

17.78 t/m2

9.00 t/m2 4.962 m


27.00 t/m2 3.550 m x+

α M

17.78 t/m2

x0.053 18.034

0.071 24.159

M:

α * Wu * lx^2

y+

y0.035 11.909



0.047 15.993 24.159 t-m/m 1963.50 mm2

0.26 %

23.63 t/m 0.32 N/mm2

<

τc =

0.38 N/mm2

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

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 4.962 m

C 5.275 m

D 3.550 m

E 3.275 m

F 3.550 m

G 5.275 m

H 4.962 m

Considering a 1.0 t/m2 loading throughout this continuous span Moments Calculation by Moment Distribution Method : A B C D member AB BA BC CB CD DC DX DF 0.000 0.450 0.550 0.400 0.600 0.650 0.350 FEM 2.052 -2.052 2.319 -2.319 1.050 -1.050 0.894 Modified 0.000 -3.078 2.319 -2.319 1.050 -1.050 0.894 FEM Balance 0.000 0.342 0.417 0.508 0.761 0.101 0.055 CO 0.000 0.000 0.254 0.209 0.051 0.381 0.000 Balance 0.000 -0.114 -0.140 -0.104 -0.155 -0.247 -0.133 CO 0.000 0.000 -0.052 -0.070 -0.124 -0.078 0.000 Balance 0.000 0.023 0.029 -0.449 0.203 -0.081 0.069 CO 0.000 0.000 -0.225 0.015 -0.041 0.102 0.000 Balance 0.000 0.101 0.123 0.010 0.016 -0.066 -0.036 CO 0.000 0.000 0.005 0.062 -0.033 0.008 0.000

∑

0.000

-2.726

2.731

-2.140

1.729

-0.931

0.849

X XD 1.000 0.447 0.447 0.000 0.028 0.000 -0.067 0.000 0.035 0.000 -0.018 0.425

Shear Force Calculation : Maximum Shear occurs at point B B Due to Load : Due to Moments :

2.481 0.549 3.03

∑

2.6375 0.112 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 = earth pressure coefficient = unit density of soil = Maximum height of soil inside box=

30 ° 0.279 1.80 t/m3 16.786 m

Therefor maximum horizontal pressure on wall = Maximum Moment that can occur =

8.43 t/m2 25.3 t-m/m

effective wall thickness required = using maximum bar dia = & clear cover =

0.386 m 25 mm 75 mm

Overall depth required =

0.473 m

Hence provide overall thickness =

0.850 m

effective =87.5mm

take 90 mm

For soil fill inside the box STEEL REQUIREMENT(mm2) Moment Provide D d reqd (m) (t-m/m) (m) Provide Reqd(mm2)

S. No.

Depth (m)

Pressure (t/m2)

1

16.786

8.43

25

0.298

0.800

1665.81

2

13.000

6.53

20

0.252

0.760

1367.11

3

10.000

5.02

15

0.210

0.660

1236.12

4

6.500

3.26

10

0.152

0.560

974.43

5

3.000

1.51

5

0.076

0.450

587.16

20 1745.33 20 1745.33 16 1340.41 16 1005.31 16 1005.31

180 0.22 % 180 0.23 % 150 0.20 % 200 0.18 % 200 0.22 %

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 = earth pressure coefficient = unit density of soil = support thickness = For soil fill inside the box

30 ° 0.279 1.80 t/m3 0.30 m

Depth (m)

Pressure (t/m2)

Vcr (t/m)

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

S. No.

d provided τv τc (N/mm2) (m) (N/mm2)

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 spacing: @ 200 c/c

Design of Partition wall 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

Maximum reaction force = per 1.00 t/m2 load =

3.030 t/m

+ 5.780 t/m

Permissible stress in steel

S. No.

Depth (m)

2.750 t/m

24000 t/m2

Pressure (t/m2)

Tension T(t/m)

Reqd Ast On each face (mm2/m) (mm2/m)

1 16.786

8.43

48.72

2030.03

2 13.000

6.53

37.73

3 10.000

5.02

4

6.500

5

3.000

Ast provided bar dia (mm)

Spacing (mm)

mm2

1015.02

12

90

1256.637

1572.17

786.08

12

90

1256.637

29.02

1209.36

604.68

12

150

753.9822

3.26

18.87

786.08

393.04

12

150

753.9822

1.51

8.71

362.81

181.40

12

150 3

753.9822

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

thickness of end wall =

0.300 m

Minimumreinforcement provided = = Provide

12

0.2% X bd 600.00 mm2 150

753.98 mm2

Design Of COUNTERFORT WALLS 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 = earth pressure coeffiecnt = surcharge height = maximum soil height =

1.80 t/m3 0.279 1.20 m 16.786 m

pressure due to soil at specified depth = pressure due to surcharge at specified depth =

Fst = j=

24000 t/m2 0.891

0.502 X H (t/m2) (varying) 0.603 t/m2 (constant)

pressure diagram will be somewhat like this

0.603 t/m2

2.06 t/m2

X 3.4125 = 16.786 m

9.033 t/m2

D= d= b=

30.83 t/m2

9.662 m 9.596 m 0.300 m

max bar dia = clear cover =

32 φ 50 mm

STEEL REQUIREMENT(mm2) Provide φ nos

S. No.

Depth (m)

Pressure at this depth (t/m2)

Moment (t-m/m)

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

Reqd (mm2)

Two layers of 32φ from the base to height of

6.786 m and lap with 25φ therafter

extending to top Depth (m)


Vu (t)

τv (N/mm2)

τc (N/mm2)

Vs = (τv-τc)bd (t)

16.786

30.83 t/m2

138.02

0.48

0.24

68.93

13.000

24.34 t/m2

85.80

0.30

0.24

16.71

10.000

19.20 t/m2

53.15

0.18

0.22

0

6.500

13.20 t/m2

24.80

0.09

0.22

0

3.000

7.20 t/m2

6.95

0.02

0.22

0

8φ

=

Provide 2 legged stirrup Spacing =

100.531 mm2

335.87868 mm

Provide 8φ 2 legged stirrup at a spacing of 200 c/c (horizontal ties) 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. Average downward pressure =

16.91 t/m2

Total area of reinforcement required to resist this tension = Provide 2 legged 8φ = spacing =

704.59 mm2/m

100.53 mm2

100.53 x 1000 / 704.59

=

142.679

Provide 8φ 2 legged stirrup at a spacing of 120 c/c (vertical ties)

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 = earth pressure coeffiecnt = surcharge height = maximum soil height =

1.80 t/m3 0.279 1.20 m 16.786 m

pressure due to soil at specified depth = pressure due to surcharge at specified depth =

Fst = j=

24000 t/m2 0.891

0.502 X H (t/m2) (varying) 0.603 t/m2 (constant)

pressure diagram will be somewhat like this

0.603 t/m2

1.07 t/m2

X 1.775 = 16.786 m

9.033 t/m2

D= d= b=

16.04 t/m2

9.662 m 9.596 m 0.850 m

max bar dia = clear cover =

32 φ 50 mm

STEEL REQUIREMENT(mm2) Provide φ nos

S. No.

Depth (m)


Moment (t-m/m)

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

Reqd (mm2)

Two layers of 32φ from the base to height of extending to top

6.786 m and lap with 28φ therafter

Depth (m)


Vu (t)

τv (N/mm2)

τc (N/mm2)

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 : Provide 2 legged stirrup Spacing =

8φ 200 mm

=

100.531 mm2

(horizontal ties)

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. Average downward pressure =

16.91 t/m2

Total area of reinforcement required to resist this tension = Provide 2 legged 8φ = spacing =

704.59 mm2/m

100.53 mm2

100.53 x 1000 / 704.59

=

Provide 8φ 2 legged stirrup at a spacing of 120 c/c (vertical ties)

142.679

3413 850

9728

300 modular ratio m =

280/(3σcbc) for M35, σcbc = = 8 assuming effective cover = 100 mm d=

11.67 10478 mm

finding the neautral axis (X) 1) assuming the Neutral axis lies in the flange Using Ast=

9 nos 32 Φ 7238.23 mm2

b*X^2/2= m*Ast(d-X) ====> X^2 + (34)X - 356252 Solving we get X =

580 mm

for moment M = 1641 Stress in concrete (Fc) =

∴ Fc = Stress in steel (Fs) =

∴ Fs =

X^2 + (2mAst/b)X - (2mAst/b)d a= b= c=

1 34 -356252

M/(0.5*b*X*(d-X))

1.68 Mpa < mFc*(d-X)/X 228.64 Mpa

11.67 Mpa

<

240 MPa

1775 850

9817

850 modular ratio m =

280/(3σcbc) for M35, σcbc = = 8 assuming effective cover = 100 mm d=

11.67 10567 mm

finding the neautral axis (X) 1) assuming the Neutral axis lies in the flange Using Ast=

3 nos 32 Φ 2412.74 mm2

b*X^2/2= m*Ast(d-X) ====> X^2 + (22)X - 232474 Solving we get X =

471 mm

for moment M = 423 Stress in concrete (Fc) =

∴ Fc = Stress in steel (Fs) =

∴ Fs =

X^2 + (2mAst/b)X - (2mAst/b)d a= b= c=

1 22 -232474

M/(0.5*b*X*(d-X))

1.00 Mpa < mFc*(d-X)/X 171.66 Mpa

11.67 Mpa

<

240 MPa

Design of ROOF SLAB by Pigeud's curve

Thickness of slab = 0.30 m wearing course = 56.00 mm span in transverse direction = span in longitudinal direction =

B= L=

3.550 m (short) 5.275 m (long)

Maximum bending moment due to dead load = weight of deck slab = weight of wearing course = total weight =

0.720 t/m2 0.123 t/m2 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

poisson's ratio µ =

0.15

for reinforced concrete bridges

Total dead weight W = moment along short span = moment along long span =

15.79 t

(m1+ µ*m2)*W = (m2+ µ*m1)*W =

0.78 t-m 0.40 t-m

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

ONE TRACK

0.85

3.6 SIZE OF PANEL of deck slab= impact factor=

3.550 m

X

5.275 m

25 %

u= v=

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

K= u/B= v/L=

0.67 0.271 0.704

using pigeud's curve= m1= m2=

0.962 m 3.712 m limited to

0.12 0.048

5.275 m

load per track of AA=

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

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

2.6375

X

2.6375

4 3.75 t

1 6.25 t

2 6.25 t

3 3.75 t

8 3.75 t

5 6.25 t

6 6.25 t

7 3.75 t

1.775

Y

X

1.775

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

B= L=

3.550 m 5.275 m

Bending Moment due to wheel 1 = tyre contact dimensions=

6.25 t 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

m1 = m2 =

0.221 0.190

Total load allowing for 25 % impact =

Moment along short span = Moment along long span =

K=

0.70 7.813 t

1.950 t-m 1.744 t-m

Bending Moment due to wheel 2 =

6.25 t

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. Y

X

6.25 t

6.25 t

0.3

0.85

0.85

X

0.3

Y 2.3 Intensity of loading = (6.25*1.25)/(0.5097*0.3984) = Consider the loaded area of u= v=

38.47 t/m2

0.150 X 2.300

sqrt((2.3+2*0.056)^2+0.3^2) = sqrt((0.15+2*0.056)^2+0.3^2) =

u/B =

0.685

m1 = m2 =

0.112 0.119

2.4306 0.3983

v/L =

0.076


4.578 t-m 4.788 t-m

K=

0.70

Now consider the area beyond the actual loading = u= v=

1.7 X 0.15

sqrt((1.7+2*0.056)^2+0.3^2) = sqrt((0.15+2*0.056)^2+0.3^2) =

u/B =

0.517

m1 = m2 =

0.132 0.14

1.8367 0.3983

v/L =

0.076


3.517 t-m 3.674 t-m

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

K=

0.70

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


3.75 t



6.25 t

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 B.M along short span = B.M along long span = Bending Moment due to wheel 6 =

0.823 t-m 0.195 t-m 6.25 t

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. B.M along short span = B.M along long span =

0.823 t-m 0.195 t-m


3.75 t



3.75 t


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= along long span=

5.16 t-m 3.24 t-m


k value for concrete j value for concrete Permissible stress in steel Permissible stress in concrete cover

effective depth required= provided deff =

35 500 16 170 t/m2 0.327 0.891 24000 t/m2 1167 t/m2

50 mm

0.174 m 0.242 m OK

main reinforcement required= So provide ϕ 16 @ 150 c/c

998 mm2 1340 mm2 OK

longitudinal reinforcement required = So provide ϕ 12 @ 150 c/c

625 mm2 754 mm2 OK

Design of Abutment Cap Taking thickness equal to 225 mm for calculation of steel requirement. Y

X

Along X direction

0.225 m

2.781 m

Steel reqd = 1% of cross section

=

6257.25 mm2

steel on both top and bottom face = 1/2 of steel reqd= Provide

ϕ 25 @ 150 c/c

=

3128.625 mm2

3272.492 mm2

Along Y direction

0.225 m

1.000 m Steel reqd = 1% of cross section

2250 mm2

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

ϕ 12 @ 150 c/c

= ==>

1507.964 mm2 2 legged stirrup

1125 mm2

DESIGN OF DIRTWALL

varies

Design values : 0.35 m

g=

1.80 t/m2

ka =

0.279

3.036 2 1.525 t/m2

1 0.603 t/m2

Earth Pressure diagram

1)Earth Pressure due to surcharge equivalent to 1.2m of earthfill

=

=

0.603 t/m2

1.525 t/m2

2.777 t-m/m ka *g*h3/6

= =

Total bending moment at the base of dirtwall due to earth pressure

3.036 m

ka *g*1.2*h2/2

= =

Bending moment at the base of dirtwall due to earth pressure (1)

height of dirtwall,h =

ka *g*h =

Bending moment at the base of dirtwall due to earth pressure (1)

0.350 m

ka *g*1.2 =

2)Earth Pressure due to backfill of earth

width of dirtwall =

=

2.342 t-m/m 5.120 t-m/m

Calculation of force and moment due to the effect of braking :(cosidering 40t bogie loading) 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. 1.7

10t

10.85

10t

2.79m

Dirt Wall

3.036

1.750

2.79

Effective width

=

7.576 m

Braking force, 0.2*20 Braking force = Braking force per metre width

=

4 t 4 t 0.53 t

=

3.036

(Only two wheels can come on dirt wall at a time.) (Impact factor can't be included with braking force)

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

Basic Design Data: Grade of conc. Grade of steel Dia of bar used Permissible stress in concrete Permissible stress in steel m , Modulur ratio K value for concrete j value for concrete Q for concrete

35 500 16 1167 t/m2 24000 t/m2 10 0.327 0.891 170.08

Max. moment in dirtwall (t-m)

7.36

Effective depth required (mm)

208

Effective depth provided (mm)

292

Ast required (mm2)

SAFE

1178

Provide Vertical longitudinal reinforcement: Use

f16

Ast provided (mm2)

@ 150 c/c

(On Approach Side face)

Min. Reinforcent in vertical direction (On outer face) Use

f16

=

1340

=

525

Thus OK For 1m length

@ 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 .

DESIGN OF DIRTWALL Normal Case with Live load

I

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 = b = f = d = Kah = = g

0.35 3.036 1

2

1.525 t/m2

0.603 t/m2

90.0 0.0 30.0 20.0 0.279

degree degree degree degree t/m

1.8

3

Earth Pressure diagram =

3.036 m

1 Earth Pressure due to surcharge equivalent to 1.2m of earthfill

=

2 0.603 t/m

2 Earth Pressure due to backfill of earth

=

2 1.525 t/m

Height of dirtwall

Bending moment at the base of dirtwall due to earth pressure (1) Bending moment at the base of dirtwall due to earth pressure (2)

= =

2.78 t-m/m 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

3.04

10t

2.79m

1.700

2.79

3.036

Effective width Braking force, 0.2*20 Braking force per metre width Bending moment at the base of dirtwall due to effect of braking

= = = =

Therefore total bending moment at the base of the dirtwall

=

7.526 4.0 0.53 1.61

m t t t-m/m

7.34 t-m/m

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. 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. For Backfill Soil : a = b = f = d = Cah = = g

0.30 3.036 1

2

0.730 t/m2

0.288 t/m2

90.0 0.0 33.0 22.0 0.134 1.8

degree degree degree degree t/m

3

Earth Pressure diagram =

3.036 m

1 Earth Pressure due to surcharge equivalent to 1.2m of earthfill

=

2 0.288 t/m

2 Earth Pressure due to backfill of earth 3 Horizondal force due to seismic

= =

2 0.730 t/m 0.231 t/m

Height of dirtwall

Bending moment at the base of dirtwall due to earth pressure (1) Bending moment at the base of dirtwall due to earth pressure (2) Bending moment at the base of dirtwall due to seismic horizondal force (3)

= = =

1.76 t-m/m 1.41 t-m/m 0.35 t-m/m

Total bending moment at the base of dirtwall Therefore Normal case govern for the design

=

3.52 t-m/m

CALCULATION OF DESIGN PARAMETERS Grade of concrete = Grade of steel =

M 35 Fe 500

Permissible stresses: sst =

2 24480 t/m

sbc =

2 1190 t/m

Basic Design Parameters: k j

= =

0.327 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)

=

2 17.30 cm /m

50% of additional reinforcement should be provided as per note of transport ministry Provide Min reinforcement

16 f

@ 100 c/c

at earth face

= 0.06% of cross sectional area

2 11.53 cm /m

=

2 20.11 cm /m Thus OK

=

2 2.1 cm /m

Provide

10 f

@ 150 c/c

outer face

Provide

10 f

@ 150 c/c

Horz both face

=

2

5.24 cm /m Thus OK 2

5.24 cm /m

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. Assuming thickness of abutment cap

=

225.0 mm

Width of abutment cap Thickness of abutment cap Length of abutment cap

= = =

1420 mm 400 mm 11379 mm

Area of steel required (min 1%)

=

36355905 mm

Quantity of steel to be provided at top

=

3 18177953 mm

Quantity of steel to be provided at bottom

=

3 18177953 mm

Quantity of steel to be provided in longitudnal direction ( 0.5*total steel at top ) Assuming a clear cover of

= =

3 9088976 mm 50.0 mm

Length of bar

=

11279.0 mm

Area of steel required in longitudnal direction

=

3

Top Face (a) Longitudnal steel

Provide Provided steel

8 bar of

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

2 805.8 mm

904.3 mm2

(b) Transverse steel Quantity of steel to be provided in transverse direction

=

Quantity of steel required

=

Adopting

12 mm dia bar and clear cover

Length of each stirrups

=

1420

-100

= =

No. of stirrups required in per m length Say

Required spacing 12mm

3 798750 mm /m

50 mm

Volume of each stirrup

Provide Provided steel

3 9088976 mm

3 149212.8 mm

= =

5.4 Nos 6.0 Nos

=

166.7 mm

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

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

1320 mm

753.6 mm2 /m

Box Abutment

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