DESIGN OF RCC ABUTMENT UNDER SLRB

2.3001.400 0.000

0.900 + 338.715

0.275 + 338.440

0.300 A + 338.140

20.929

0.000

2.000 E B 3.200 D C 317.786

0.600+ 316.986

2.3002.000 7.500

Using M-20 grade concrete, the design parameters to be adopted are the following

= 200

= 6.67m = 13.993k = 0.318j = 0.894Q = 0.948

DIMENSIONS OF THE BASE:-Length of the toe = 2.000 mLength of the heel = 2.000 mThickness at the top = 1.4 mThickness of the bed block = 0.9 mThickness at the bottom = 2.3 mHeight of the stem = 20.929 mThickness of the base slab = 2 m

= 15.329 m

Unit weight of soil considered = 21

Unit weight of concrete considered = 25

W1

W2

W3

W6

W4

W5

sst N/mm2

scbc N/mm2

Height of the wall H

KN/m3

KN/m3

BforceFforce

Rreaction

Angle of internal friction = 32

= 0.307Value of Kp = 3.257Reaction coming on to the abutment = 35.069 KNBreaking force = 0 KNFrictional force = 0 KNSafe bearing capacity of the

soil considered = 600Coefficint of friction between soil and concrete = 0.6Clear cover provided = 0.075 m

804 Main reinfortcement provided(Toe) = 32 mm804 Main reinfortcement provided(Heel) = 32 mm201 Distribution steel provided(Heel) = 16 mm201 Distribution steel provided (stem) = 16 mm804 Main reinfortcement provided (stem) = 32 mm

0 Consider a live load surcharge of = 0 m

DESIGN

Minimum depth of foundation

Ka =

Ka = 0.307

d = 600 / 21 x 0.307²= 2.693 m

However provide d = 2.25 m below ground level

The ratio of the length of the toe slab to the base width b is given by

a =

Where H = Height of the wall including thickness of the base

a = 1 - 600 / ( 2.2 x 21 x 22.929)= 0.153

Width of the baseb =

Value of Ka

KN/m2

d=q0

γ×( 1−sinφ

1+sinφ )2

( 1−sinφ1+sinφ )

1−q0

2 . 2×γH

0 .95×H×2√ K a

(1−α )×(1+3α )

= 0.95x15.329x(0.307/(1-0.153)x(1+3x 0.153))^0.5= 7.258 m

The base width with respect to slidingb =

= 0.7 x 15.329 x 0.307( 1 - 0.153) x 0.6

= 6.482 m

Provide b = 0.6Hb = 9.197 m

Provide 7.500 m

Width of toe slab == 1.148 m

Provide 2.000 m

Thickness of the base slab= H / 12= 1.277 m

Provide 2 m

Thickness of the stem

= 15.329 m

Considering 1m length of the retaining wall

=

= 3870.329 Kn-m

Effective depth required = 2020.551 mm

Effective depth provided = 2209 mmHENCE SAFE

a x b

Height H1

Max bending moment at B KagH13/6 + KawH1

2/2

0 .7×H×Ka(1−α )×μ

Stability of wall

S.NO Designation Force MagnitudeLever arm

1 1.4 x 0.275 x 25 9.625 3.6 34.65

2 0.9 x 0.3 x 25 6.75 2.45 16.538

3 1 x 2.3 x 20.354 x 25 1170.355 3.15 3686.618

4 0.5 x 0 x 20.354 x 25 0 2 0

5 2 x 7.5 x 25 375 3.75 1406.25

6 3.2 x 20.929 x 21 1406.429 5.9 8297.931

7 3.2 x 0 x 21 0 5.9 0

8 35.069 35.069 2.45 85.919

9 0 0 21.454 0

10 0 0 21.454 0Total 3003.228 KN 13527.91 KN - m

Total resisting moment

= 13527.906 KN - m

Earth pressureP = 757.452 KN

Check for over turningOver turning moment = 3870.327 KN - m

Factor of safety against over turining moment = 3.495 > 2SAFE

Check for slidingFactor of safety against sliding = 2.379 > 1.5

SAFEAs the section fails against sliding provide a shear key

Moment about toe

W1

W2

W3

W4

W5

W6

W7

RREACTION

BForce

FForce

Mr

Pressure distributionNet moment

= 9657.579 KN - mLever arm from toe = 3.216 m

Eccentricity e = 0.534 m

= 1.250 m

= 571.49

= 229.37

Pressure P at the junction of the stem with toe slab

= 480.26

Pressure P' at the junction of the stem with heel slab

= 425.52

Design of toe slab 1.40.9

A

2.3002.000 B 3.200

D C0.400

7.5002.000

229.37571.49

480.26 425.52

SM

emax

Pressure P1 at toe P1 KN/m2

Pressure P2 at heel P2 KN/m2

KN/m2

KN/m2

The weight of the soil above toe slab is neglectedThe forces acting on it are

(i) Up ward soil pressure(ii) Down ward weight of slab

Down ward weight of slab per unit area = 50

Net pressure intensity under D = 521.49

Net pressure intensity under E = 430.26

Total force = shear force at E = 951.754 KN

x from E = 1.032 m

Bending moment at E = 982.21 KN - m

Effective depth required d = 1017.883 mm

Effective depth provided = 1909 mmHENCE SAFE

Area of steel = 2877.603

This reinforcement is provided at the bottom face with a spacing of.= 275 mm c/c

= 2923.055HENCE SAFE

32mm dia bars at a spacing of 275mm C/C

Distribution steel = 0.15% of area

= 2250

Provide 16mm dia bars at a spacing of = 85 mm c/c

KN / m2

KN / m2

KN / m2

Ast mm2

mm2

mm2

Design of heel slabThe forces that are acting on it are

(i) Down ward weight of soil(ii) Down ward weight of heel slab

(iii) Up ward soil pressure

(i) Down ward weight of soil = 1406.429 KN

Acting at = 1.6 m from B

(ii) Down ward weight of heel slab = 160 KN

Acting at = 1.6 m from B

(iii) Up ward soil pressure = 1047.819 KN

Acting at = 1.44 m from B

Total force = Shear force at B = 518.61 KN

Bending moment at B = 997.427 KN - m

Effective depth required d = 1025.738 mm

Effective depth provided = 1909 mmHENCE SAFE

Area of steel = 2922.185

Required 32mm dia bars at a spacing of = 275 mm C/C

Provided 32mm dia bars at a spacing of = 95 mm C/C

Area of steel provided = 8461.474HENCE SAFE

Distribution steel = 0.15% of area

= 2250

provide 16mm dia bars at a spacing of = 85 mm c/c

Area of steel provided = 2364.235

Nominal Shear stress = 0.259Percentage of steel provided = 0.423 %

Permissible Shear stress = 0.2754SAFE

Ast mm2

mm2

mm2

mm2

N/mm2

N/mm2

Design of stem

Max bending moment == 3870.329 KN - m

Effective depth required d = 2020.551 mm

Effective depth provided = 2209 mm

Area of steel = 9799.065

Required 32mm dia bars at a spacing of = 80 mm C/C

Provided 32mm dia bars at a spacing of = 50 mm C/C

Area of steel provided = 16076.8HENCE SAFE

Curtailment of reinforcement in the stem(I)

Considering a section at depth h below the top of the stem, The effective depth at top of stem = 2209 mm

The effective depth d' at that section = 2209 + ( ( 2209 - 2209 ) / 15.329 ) x h= 2209 + 0 x h

=

= Reinforcement at depth h

= Efective depth at depth h

d

If

1

2

Therefore h

KagH13/6

Ast mm2

mm2

Hence h / H1

Ast1

d1

Ast = Reinforcement at depth H1

= Effective depth at depth H1

Ast1 = 1/2 Ast

Ast1

Ast

H1

A st αH3

d

H= (A st×d )13

( A st1×d1

A st×d )13

=

=( 12×d

1

d )13

h = 15.329 x[ ( 2209 + 0 x h) / ( 2 x 2209)] ^(1/3)

12.2 h = 12.167

50% of the reinforcement is to be curtailed at a height of= 15.329 - 12.167 + 12 x 0.032= 3.462 m from the base of the stem

(II)

1

4

Therefore h

h = 15.329 x[(2209 + 0 x h)/(4x2209)]^(1/3)

9.66 h = 9.657

50% of the remaining reinforcement is futher curtailed at a height of= 15.329 - 9.657 + 12 x 0.032

6.056 m from the base of the stem

The remaining reinforcement is to be continued to the top

Check for shear

Shear force p == 757.452 KN

Nominal shear stress = 0.343

Percentage of steel provided = 0.728 %

Permissible shear stress = 0.3456HENCE SAFE

Distribution and temperature reinforcementAverage thickness of the stem = 2.3 m

Distribution reinforcement = 0.15 % of area

= 3450Provide 16 mm at a spacing of = 55 mm C/C

Substituting the values of d, d1and H1

Ast1

Ast

H1

Substituting the values of d, d1and H1

KagH2/2

N/mm2

N/mm2

mm2

=

=( 14×d

1

d )13

2.300

32mm at 200mm C/C

32mm at 100mm C/C

32mm at 50mm C/C16mm at 85mm C/C

15.329 16mm at 55mm C/C

20.929

6.05616mm at 85mm C/C

3.46232 mm AT 95mm C/C 32 mm AT 95mm C/C

2.000 16mm at 85mm C/C

32mm at 50mm C/CREINFORCEMENT AT THE INNER FACE OF THE STEM

0.600 0.600

2.000 16mm at 85mm C/C32mm dia bars at a spacing of 275mm C/C 3.200

7.500

SECTION OF WALL WITH RCC DETAILS

c

V

Design of shear key

The wall is unsafe in sliding, and hence shear key will have to be providedbelow the stem

Let the depth of the key = a

Intensity of pressure Pp developed in front of the key depends up on the soilPressure p in the front of the key.

== 1564.207 KN/m2

== 1564.207 x a

= 0.307 x 21 / 2 x ( 15.329 + a )^2

= 2 x a x 2142 x a

= 3003.228 + 42 x a

For equilibrium of wall, permitting factor of safety as 1.5 against sliding.

1.5 = 0.6 x ( 3003.228 + 42 x a ) + 1564.207 x a0.307 x 21 / 2 x ( 15.329 + a )^2

4.835 -1441.168 a + -665.758 = 0

a = 298.531 mmProvide = 100 mm

Width of the key = 1000 mm

= a x tanθ= a x tan(45+Ø/2)= a√Kp= 0.18 m

Actual available length of the slab= 2 m

Hence safe

pp Kp x p

Pp pp x a

Sliding force at level D1C1

Weight of soil between bottom of base and D1C1

SW

a2

a1

1 .5=υ×∑W+P p

PH

Size of the key = 100 x 1000 mm

= 0.307 x 21 / 2 x ( 15.329 + a )^2= 767.367 KN

= 1564.207 x a= 156.421 KN

= 3003.228 + 42 x a = 3007.428 KN

Actual force to be resisted by the key at a factor of safety of 1.5 is= 1.5 x PH - m x SW= -653.4063

PH

Pp

SW