RCC Retaining WallS

Density of concrete = 25 kN/m3

Use Grade of concrete = M 25

Steel grade = Fe 500

Angle of repose of earth = 30 °

Density of back fill γ = 20 kN/m3

Safe bearing capacity of foundation soil = 280 kN/m2

δ = 45 °

Fck = 25 N/mm2

Fy = 500 N/mm2

Dia of rod = 32 mm

Clear cover = 40 mm

Consider rod dia = 32 + 40 / 2

Effiective cover = 52 mm

Stem/Wall

Overall depth of wall " D " = 1200 mm

Effiective deprh " d " = 1148 mm

Toe/Heel slab

Overall depth of slab " D " = 900 mm


451.113

0.3 3.55 m

447.563 45 ⁰

h1 7.321 w1

H 12.071

Stem

h 8.521 440.242 w2

0.3

Toe 0.9 1.2 2.4 439.942

Heel

h2 1.2 w3

0.9

w4 0.5 439.042

d c

4.5

Design of RCC Retaining Wall

i. Stability Analysis

Weight of wall w1 = 0.3 x 7.621 x 1 x 25

= 53.03 kN

Leaverarm from point c = 2.55 m

Weight of wall w2 = 0.5 x 0.9 x 7.621 x 1 x 25

= 114.32 kN

Leaverarm from point c = 3 m

Weight of foundation w3 = 4.5 + 1.2 x 0.5 x 0.4 x 1 x 25

= 28.5 kN


Weight of foundation w4 = 4.5 x 0.5 x 1 x 25

= 56.25 kN


Preassure behind the wall

Pa = 1/2 Ca γ H2

Earth pressure coefficent Ca = 0.232

H = 12.071 m

Pa = 338.04 kN/m

Horizontal component Ph = Pa cos δ

= 292.75 kN

Vertical component Pv = Pa sin δ

= 239.03 kN

S.no

1

2

3

4

5

6

Resultant force on the wall meet the base at distance z from the heel end a

z = 1845.74 / 783.88

= 2.35 m

Eccentracity e = z - b/2

= 0.1 m

But e < b/6

0.1 m < 0.75 m

Moment of Ph 292.75 4.02 1176.86

Total 783.88 1845.74

w4 56.25 2.25 126.56

Pv 239.03

w2 114.32 3 342.96

w3 28.5 2.25 64.13

Loads Magnitude of the load in kN Distance fro c in mMoment about c

kN.m

w1 53.03 2.55 135.23

i. Check for bearing pressure

Pressure at the toe and at the heel are given by

P = W/bl(1±6e/l)

Pmax = 197.42 Kn/m2

Pmin = 150.97 Kn/m2

15.097 t/m2

19.742 t/m2

ii. Factor of safety against overturning

z = 2.35 m

Resisting moment about d = 1842.118 kN.m

Overturning moment about d = 1176.86 kN.m

Factor of safety against overturning = 1842.118

1176.86

= 1.57 > 1.5 O.K

iii. Factor of safety against sliding

Forces causing sliding = 544.85 kN

Frictional force (µ W) µ = 0.55 = 299.67 kN

Factor of safety against sliding = 299.67

292.75

= 1.02 > 1 O.K

This is lessthan 1.5. Hence a Key shoild be provided

iv. Design of the stem/wall

Bending moment for the stem per metre run

M = Ca γ h3/6

= 0.232 x 20 x 7.621 x 7.621 x 7.621 / 6

= 1026.89 kN.m

Factroed B.M "M" = 1.5 x 1026.89

= 1540.335 kN.m

= N.mm

Effective depth required " d² " =

d² = 463258.647

d = 680 mm

Hence provide effective depth = 1148 mm

1540335000

M

0.133 x Fck x b

=1540335000

0.133 x 25 x 1000

Main reinforcement

Consider dia of bar = 32 mm

0.87 x Fy x d ( 1 - 0.42 x 0.46)

Ast Req = 3823.12 mm²

For 32 mm dia bars "As" = 803.84 mm²

Spacing of bars " S " = 1000 As/Ast Req = 1000 x 803.84 / 3823.12

= 210.26 mm say 210 mm

Provided steel reinforcement Ast Pro = 1000 As/S

= 1000 x 803.84 / 210

= 3827.81 mm²

HENCE SAFE

Distribution reinforcement


Ast Req = 0.12 % BD

(As per clause 40.1 of IS 456-2000)

= 0.0012 x 1000 x 1200

Ast Req = 1440 mm²


Spacing of bars " S " = 1000 As/Ast Req = 1000 x 200.96 / 1440

= 139.56 mm say 135 mm


= 1000 x 200.96 / 135

= 1488.59 mm²

HENCE SAFE

v. Design of the toe and Heel slab

0.9 m 1.2 2.4 m

Toe Heel

15.097 Kn/m2

197.42 Kn/m2

112.34 Kn/m2

160.96 Kn/m2

4.5 m

Provided steel reinforcement 16 mm @135 mm c/c

=

1.5 x 1026.89 x 1000000

0.87 x 500 x 1148 x 0.8068


Ast Req =

1.5 x M

a b c d

e

f g

h

i

j

Bending Moment Calculation of the toe slab

S.no

1

2

3

Factroed B.M "M" = 1.5 x 82.68

= 124.02 kN.m

= N.mm


d² = 37299.248

d = 193 mm


Main reinforcement


0.87 x Fy x d ( 1 - 0.42 x 0.46)

Ast Req = 416.72 mm²



= 271.26 mm say 245 mm


= 1000 x 113.04 / 245

= 461.39 mm²

HENCE SAFE


Ast Req =

1.5 x M

=

1.5 x 82.68 x 1000000

0.87 x 500 x 848 x 0.8068

124020000

M

0.133 x Fck x b

=124020000

0.133 x 25 x 1000

178.27 82.68B.M for toe slab

Upward pressure

efi 16.41 0.60 9.84

Deduct dead load

of the slab 17.00 0.45 7.65

Moment about c

kN.m

Upward pressure

cdif 144.86 0.45 65.19

Loads Magnitude of the load in kN Distance fro c in m

Bending Moment Calculation of the heel slab

S.no

1

2

3

4

5

Factroed B.M "M" = 1.5 x 598.59936

= 897.89904 kN.m

= N.mm


d² = 270044.824

d = 519 mm


Main reinforcement


0.87 x Fy x d ( 1 - 0.42 x 0.46)

Ast Req = 3017.01 mm²



= 162.62 mm say 160 mm


= 1000 x 490.625 / 160

= 3066.41 mm²

HENCE SAFE


Ast Req =

1.5 x M

=

1.5 x 598.59936 x 1000000

0.87 x 500 x 848 x 0.8068

897899040

M

0.133 x Fck x b

=897899040

0.133 x 25 x 1000

Total load 584.46 735.43

Total deduction 152.92 136.83

Deduct for upward

pressure abjh 36.23 1.2 43.48

598.60

Deduct for upward

pressure gih 116.69 0.8 93.35

B.M for heel slab

Weight of backing 457.26 1.2 548.71

Weight of heel slab 42.00 1.2 50.40

Surcharge 85.20 1.6 136.32

Loads Magnitude of the load in kN Distance fro c in mMoment about c

kN.m



Ast Req = 0.12 % BD


= 0.0012 x 1000 x 900

Ast Req = 1080 mm²



= 104.67 mm say 100 mm


= 1000 x 113.04 / 100

= 1130.4 mm²

HENCE SAFE

vi. Design of the key

1.5 Ph = 1.5 x 292.75

= 439.125 kN

Limiting friction force F = 299.67 kN

Unbalanced force = 439.125 - 299.67

= 139.455 kN

Safe horizontal soil pressure = 0.7 x SBC

= 0.7 x 280

= 196 Kn/m2

Let the height of the key

196 x y = 139.455

y = 0.71 m

Say = 0.75 m

Maximum bending moment for the key

M = Wl / 2

= 139.455 x 0.75 / 2

= 52.3 Kn.m

Thickness of shear key proposed = 300 mm

Dia of rod = 32 mm

Clear cover = 40 mm

Consider rod dia = 32 + 40 / 2

Effiective cover = 52 mm

Shear key

Overall width " D " = 300 mm



To provide a FOS of 1.5, the wall should be designed to remain in limiting equilibrium when the horizontal

force on the wall is increased to 1.5 Ph

Factroed B.M "M" = 1.5 x 52.3

= 78.45 kN.m

= N.mm


d² = 23593.985

d = 153 mm


Main reinforcement


0.87 x Fy x d ( 1 - 0.42 x 0.46)




= 222.96 mm say 220 mm


= 1000 x 200.96 / 220

= 913.45 mm²

HENCE SAFE



Ast Req = 0.12 % BD


= 0.0012 x 1000 x 300

Ast Req = 360 mm²



= 218.06 mm say 215 mm


= 1000 x 78.5 / 215

= 365.12 mm²

HENCE SAFE

=

1.5 x 52.3 x 1000000

0.87 x 500 x 248 x 0.8068



0.133 x Fck x b

=78450000

0.133 x 25 x 1000

Ast Req =

1.5 x M

78450000

M

0

3

4

Toe

1

d

Density of concrete

Use Grade of concrete

Steel grade

Angle of repose of earth

Density of back fill

Safe bearing capacity of foundation soil


Weight of wall

Leaverarm from point c

Weight of wall


Weight of foundation


Weight of foundation


Earth surcharge weight


Earth pressure P1 due tosurcharge


Earth pressure P2


Let x be the distance from c of the point where the resultant force strikes the base

Eccentracity


Resultant of vertical forces from c lies at a distance

Resisting moment about d

Overturning moment about d

Factor of safety against overturning


Forces causing sliding

Frictional force

Factor of safety against sliding

iv. Check for bearing pressure


11.886

v. Design of stem


Earth pressure coefficent

Horizontal component of pressure

Bending moment at base of the vertical wall

Considering 1 m width of the vertical slab, its thickness is given by,

Using partial factor of 1.5 on the horizontal earth pressure

Effective depth required

Depth of stem at base

Assume dia of bar

Clear cover

Effective depth provided

Area of tension steel is given by

Main Reinforcement

Use 32

32

Provided steel reinforcement

Distribution Reinforcement

Use 10

Check for shear

The critical section for shear strength is taken at a distance d from the bottom stem

Factored shear force

Nominal Shear stress

vi. Design of toe

Dead load of foundation

Overall depth

Clear cover

Assume dia of bar

Effective depth

The shear strength of M

118.86




For 16

Spacing of bars " S " = 1000 As/Ast Req


Check for shear



vii. Design of heel

Dead load of foundation

Earth surcharge load





For 32

Spacing of bars " S " = 1000 As/Ast Req


Check for shear




0.3

10 °

w1

Stem

w2

0.4

1.5 0.45 2.2

Heel

w3

0.6

w4 0.3

c

4.15

= 25 kN/m3

Use Grade of concrete = M 30

= Fe 500

Angle of repose of earth = 22 °

γ = 20 kN/m3

Safe bearing capacity of foundation soil = 300 kN/m2

δ = 10 °


w1 = 0.3 x 3.4 x 1 x 25

= 25.5 kN


w2 = 0.5 x 0.15 x 3.4 x 1 x 25

= 6.375 kN


Weight of foundation w3 = 4.15 + 0.45 x 0.5 x 0.3 x 1 x 25

= 17.25 kN


Weight of foundation w4 = 4.15 x 0.3 x 1 x 25

= 31.125 kN


Earth surcharge weight = 3.4 x 2.2 x 1 x 25

= 187 kN


Earth pressure P1 due tosurcharge = Ca γ h1 h2

= 0.488 x 20 x 4.15 x 4

= 162.02 kN


= 1/2 Ca γ h22

= 0.5 x 0.488 x 20 x 4 x 4

= 78.08 kN


Let x be the distance from c of the point where the resultant force strikes the base

x = 2.660 m

e = x - b/2

= 0.585 m

6e = 6 x 0.585

base length 4.15

= 0.85 m < 1

Hence ok


Resultant of vertical forces from c lies at a distance

= 390.528

267.25

= 1.46 m

Resisting moment about d = 718.903 kN.m

Overturning moment about d = 319.333 kN.m

Factor of safety against overturning = 718.903

319.333

= 2.25


Forces causing sliding = 80.25 kN

(µ W) µ = 0.45 = 120.263 kN

Factor of safety against sliding = 120.263

80.25

= 1.5

iv. Check for bearing pressure


P = W/bl(1±6e/l)

Pmax = 118.86 Kn/m2

Pmin = 9.93 Kn/m2

0.993 t/m2

t/m2


Pa = 1/2 Ca γ h2

Earth pressure coefficent Ca = 0.488

h = 3.4 m

Pa = 56.41 kN/m

Horizontal component of pressure Ph = Pa cos δ

= 55.55 kN/m

Bending moment at base of the vertical wall = Ph h/3

= 62.96 kN.m/m

Considering 1 m width of the vertical slab, its thickness is given by,

B.M = 0.138 Fck bd2


1.5 x 62.96 x 1000000 = 0.138 x 30 x 1000 d2

d2

= 22811.6

Effective depth required d = 151 mm

Depth of stem at base D = 450 mm

ɸ = 32 mm

= 25 mm

Effective depth provided d = 409 mm

Area of tension steel is given by AtReq =

= 0.36 x 30 x 1000 x 0.48 x 409

= 4874 mm2/m

Main Reinforcement

mm dia steel at 160 mm c/c

mm dia bar area As = 803.84 mm2

AtPro = 1000 As/Ast Req

= 1000 x 803.84 / 4874

0.36 fck b Xm

0.87 Fy

0.87 x 500

= 164.92 mm

Say 160 mm


Ast Pro = 1000 As/S

= 1000 x 803.84 / 160

= 5024 mm2/m

HENCE SAFE

Distribution Reinforcement

mm dia steel at 160 mm c/c

At = 0.785 x 10 x 10 / 160 / 409 x 100

At = 0.12 %

0.12 % > 0.12 %

30 τc = 0.54 N/mm2

The critical section for shear strength is taken at a distance d from the bottom stem

Shear force V = 1/2 Ca γ h2

cosδ

h = 3.4 - 0.409

= 2.991 m

V = 42.99 kN

Factored shear force Vu = 1.5 x 42.99

= 64.485 kN

Nominal Shear stress τv = Vu/bd

= 0.16 N/mm2

τc > τv

HENCE OK

Dead load of foundation = 11.66 kN/m

D = 600 mm

= 75 mm

ɸ = 32 mm

d = 600 - 75 - 16

= 509 mm

0.991 Critical Shear 2.459

509 509

1500 450 2200

9.93 Kn/m2


Kn/m2

92.848 54.316

Vu = 93.35 kN

Mu = 105.84 kN.m

B.M = 0.138 Fck bd2


1.5 x 105.84 x 1000000 = 0.138 x 30 x 1000 d2

d2

= 38347.8



Ast Req =

0.87 x Fy x d ( 1 - 0.42 x 0.46)

=

0.87 x 500 x 509 x 0.8068


mm dia bars "As" = 200.96 mm²


= 226.12 mm say 225 mm


= 1000 x 200.96 / 225

= 893.16 mm²

HENCE SAFE


30 τc = 0.54 N/mm2


= 140.025 kN


= 0.28 N/mm2

τc > τv

HENCE OK

Dead load of foundation = 11.66 kN/m

Earth surcharge load = 187 kN/m

Vu = 351.69 kN

Mu = 410.14 kN.m

B.M = 0.138 Fck bd2



Mu

158.76 x 1000000

1.5 x 410.14 x 1000000 = 0.138 x 30 x 1000 d2

d2

= 148601



Ast Req =

0.87 x Fy x d ( 1 - 0.42 x 0.46)

=

0.87 x 500 x 509 x 0.8068


mm dia bars "As" = 803.84 mm²


= 233.41 mm say 230 mm


= 1000 x 803.84 / 230

= 3494.96 mm²

HENCE SAFE


30 τc = 0.54 N/mm2


= 527.535 kN


= 1.04 N/mm2

τc > τv

UNSAFE

Mu

615.21 x 1000000


cos 10 0.98481

cos 22 0.92718

0 118.86

0.991 92.848

1.5 79.488

1.95 67.676

2.459 54.316

4.15 9.93

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RCC Retaining WallS