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retaining wall
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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
Effiective deprh " d " = 848 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
Leaverarm from point c = 2.25 m
Weight of foundation w4 = 4.5 x 0.5 x 1 x 25
= 56.25 kN
Leaverarm from point c = 2.25 m
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
Consider dia of bar = 16 mm
Ast Req = 0.12 % BD
(As per clause 40.1 of IS 456-2000)
= 0.0012 x 1000 x 1200
Ast Req = 1440 mm²
For 16 mm dia bars "As" = 200.96 mm²
Spacing of bars " S " = 1000 As/Ast Req = 1000 x 200.96 / 1440
= 139.56 mm say 135 mm
Provided steel reinforcement Ast Pro = 1000 As/S
= 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
Provided steel reinforcement 32 mm @210 mm c/c
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
Effective depth required " d² " =
d² = 37299.248
d = 193 mm
Hence provide effective depth = 848 mm
Main reinforcement
Consider dia of bar = 12 mm
0.87 x Fy x d ( 1 - 0.42 x 0.46)
Ast Req = 416.72 mm²
For 12 mm dia bars "As" = 113.04 mm²
Spacing of bars " S " = 1000 As/Ast Req = 1000 x 113.04 / 416.72
= 271.26 mm say 245 mm
Provided steel reinforcement Ast Pro = 1000 As/S
= 1000 x 113.04 / 245
= 461.39 mm²
HENCE SAFE
Provided steel reinforcement 12 mm @245 mm c/c
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
Effective depth required " d² " =
d² = 270044.824
d = 519 mm
Hence provide effective depth = 848 mm
Main reinforcement
Consider dia of bar = 25 mm
0.87 x Fy x d ( 1 - 0.42 x 0.46)
Ast Req = 3017.01 mm²
For 25 mm dia bars "As" = 490.625 mm²
Spacing of bars " S " = 1000 As/Ast Req = 1000 x 490.625 / 3017.01
= 162.62 mm say 160 mm
Provided steel reinforcement Ast Pro = 1000 As/S
= 1000 x 490.625 / 160
= 3066.41 mm²
HENCE SAFE
Provided steel reinforcement 25 mm @160 mm c/c
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
Distribution reinforcement
Consider dia of bar = 12 mm
Ast Req = 0.12 % BD
(As per clause 40.1 of IS 456-2000)
= 0.0012 x 1000 x 900
Ast Req = 1080 mm²
For 12 mm dia bars "As" = 113.04 mm²
Spacing of bars " S " = 1000 As/Ast Req = 1000 x 113.04 / 1080
= 104.67 mm say 100 mm
Provided steel reinforcement Ast Pro = 1000 As/S
= 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
Effiective deprh " d " = 248 mm
Provided steel reinforcement 12 mm @100 mm c/c
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
Effective depth required " d² " =
d² = 23593.985
d = 153 mm
Hence provide effective depth = 248 mm
Main reinforcement
Consider dia of bar = 16 mm
0.87 x Fy x d ( 1 - 0.42 x 0.46)
Ast Req = 901.33 mm²
For 16 mm dia bars "As" = 200.96 mm²
Spacing of bars " S " = 1000 As/Ast Req = 1000 x 200.96 / 901.33
= 222.96 mm say 220 mm
Provided steel reinforcement Ast Pro = 1000 As/S
= 1000 x 200.96 / 220
= 913.45 mm²
HENCE SAFE
Distribution reinforcement
Consider dia of bar = 10 mm
Ast Req = 0.12 % BD
(As per clause 40.1 of IS 456-2000)
= 0.0012 x 1000 x 300
Ast Req = 360 mm²
For 10 mm dia bars "As" = 78.5 mm²
Spacing of bars " S " = 1000 As/Ast Req = 1000 x 78.5 / 360
= 218.06 mm say 215 mm
Provided steel reinforcement Ast Pro = 1000 As/S
= 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
Provided steel reinforcement 16 mm @220 mm c/c
Provided steel reinforcement 10 mm @215 mm c/c
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
i. Stability Analysis
Weight of wall
Leaverarm from point c
Weight of wall
Leaverarm from point c
Weight of foundation
Leaverarm from point c
Weight of foundation
Leaverarm from point c
Earth surcharge weight
Leaverarm from point c
Earth pressure P1 due tosurcharge
Leaverarm from point c
Earth pressure P2
Leaverarm from point c
Let x be the distance from c of the point where the resultant force strikes the base
Eccentracity
ii. Factor of safety against overturning
Resultant of vertical forces from c lies at a distance
Resisting moment about d
Overturning moment about d
Factor of safety against overturning
iii. Factor of safety against sliding
Forces causing sliding
Frictional force
Factor of safety against sliding
iv. Check for bearing pressure
Pressure at the toe and at the heel are given by
11.886
v. Design of stem
Preassure behind the wall
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
Using partial factor of 1.5 on the horizontal earth pressure
Effective depth required
Effective depth provided
For 16
Spacing of bars " S " = 1000 As/Ast Req
Provided steel reinforcement
Check for shear
Factored shear force
Nominal Shear stress
vii. Design of heel
Dead load of foundation
Earth surcharge load
Using partial factor of 1.5 on the horizontal earth pressure
The shear strength of M
Effective depth required
Effective depth provided
For 32
Spacing of bars " S " = 1000 As/Ast Req
Provided steel reinforcement
Check for shear
Factored shear force
Nominal Shear stress
The shear strength of M
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 °
i. Stability Analysis
w1 = 0.3 x 3.4 x 1 x 25
= 25.5 kN
Leaverarm from point c = 2.5 m
w2 = 0.5 x 0.15 x 3.4 x 1 x 25
= 6.375 kN
Leaverarm from point c = 2.3 m
Weight of foundation w3 = 4.15 + 0.45 x 0.5 x 0.3 x 1 x 25
= 17.25 kN
Leaverarm from point c = 2.425 m
Weight of foundation w4 = 4.15 x 0.3 x 1 x 25
= 31.125 kN
Leaverarm from point c = 2.075 m
Earth surcharge weight = 3.4 x 2.2 x 1 x 25
= 187 kN
Leaverarm from point c = 1.1 m
Earth pressure P1 due tosurcharge = Ca γ h1 h2
= 0.488 x 20 x 4.15 x 4
= 162.02 kN
Leaverarm from point c = 1.33 m
= 1/2 Ca γ h22
= 0.5 x 0.488 x 20 x 4 x 4
= 78.08 kN
Leaverarm from point c = 1.33 m
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
ii. Factor of safety against overturning
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
iii. Factor of safety against sliding
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
Pressure at the toe and at the heel are given by
P = W/bl(1±6e/l)
Pmax = 118.86 Kn/m2
Pmin = 9.93 Kn/m2
0.993 t/m2
t/m2
Preassure behind the wall
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
Using partial factor of 1.5 on the horizontal earth pressure
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
Provided steel reinforcement
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
The shear strength of M
Kn/m2
92.848 54.316
Vu = 93.35 kN
Mu = 105.84 kN.m
B.M = 0.138 Fck bd2
Using partial factor of 1.5 on the horizontal earth pressure
1.5 x 105.84 x 1000000 = 0.138 x 30 x 1000 d2
d2
= 38347.8
Effective depth required d = 196 mm
Effective depth provided d = 509 mm
Ast Req =
0.87 x Fy x d ( 1 - 0.42 x 0.46)
=
0.87 x 500 x 509 x 0.8068
Ast Req = 888.73 mm²
mm dia bars "As" = 200.96 mm²
Spacing of bars " S " = 1000 As/Ast Req = 1000 x 200.96 / 888.73
= 226.12 mm say 225 mm
Provided steel reinforcement Ast Pro = 1000 As/S
= 1000 x 200.96 / 225
= 893.16 mm²
HENCE SAFE
Provided steel reinforcement 196 mm @225 mm c/c
30 τc = 0.54 N/mm2
Factored shear force Vu = 1.5 x 93.35
= 140.025 kN
Nominal Shear stress τv = Vu/bd
= 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
Using partial factor of 1.5 on the horizontal earth pressure
The shear strength of M
Mu
158.76 x 1000000
1.5 x 410.14 x 1000000 = 0.138 x 30 x 1000 d2
d2
= 148601
Effective depth required d = 385 mm
Effective depth provided d = 509 mm
Ast Req =
0.87 x Fy x d ( 1 - 0.42 x 0.46)
=
0.87 x 500 x 509 x 0.8068
Ast Req = 3443.9 mm²
mm dia bars "As" = 803.84 mm²
Spacing of bars " S " = 1000 As/Ast Req = 1000 x 803.84 / 3443.9
= 233.41 mm say 230 mm
Provided steel reinforcement Ast Pro = 1000 As/S
= 1000 x 803.84 / 230
= 3494.96 mm²
HENCE SAFE
Provided steel reinforcement 385 mm @230 mm c/c
30 τc = 0.54 N/mm2
Factored shear force Vu = 1.5 x 351.69
= 527.535 kN
Nominal Shear stress τv = Vu/bd
= 1.04 N/mm2
τc > τv
UNSAFE
Mu
615.21 x 1000000
The shear strength of M
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