Design step for cantilever retaining wall
1) Set the dimension of the wall
a) Depth of foundation
h min = 2
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b) Height of stem
H = h + h min
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c) Base width
b0.4H to 0.6H
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d) Length of toe slab = 0.3b to 0.4b
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e) Thickness of base slab = to
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f) Thickness of stem
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- governed by bending moment criteria
- check whether the structure is stable based on the dimension set.
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2) Check
a) Overturning fs 1 = 1.4
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b) Serding fs 2 = 1.4
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otherwise provided a shear key
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c) Failure of sending soil ( base failure )
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= ; e = –
P max = SBC of soil
P min = should be positive (no tension )
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3) Design of stem
Provide main steel, distribution steel, and shear reinforcement for steel of retaining wall.
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4) Design of heel slab
Provide main steel and distribution steel for the heel slab of the retaining wall.
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5) Design of toe slab
Provide main steel and distribution steel for the toe slab of the retaining wall.
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6) Detailing
Detail the reinforced assigned
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Q. A cantilever retaining wall has a 5.5 m tall stem to retain soil level with its top. The soil density is 16
KN/m 3 & has an angle of repose 30safe bearing capacity of soil is 210 KN/m 2. Design retaining wall using M20 and Fe 415. Take frictional coefficient between soil & concrete is 0.45.
Answer
Given
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Height of stem = H = 5.5 m
Density of soil = = 16 KN/m 3
Angle of repose = = 30
Safe bearing capacity = 210 KN/m 2
Coefficient of friction = 0.45
M20 , Fe 415
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1. Dimensioning of wall
a) Depth of foundation
h min = 2 = 1.45
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b) Base width
b = 0.5H
b = 0.5 x 5.5 =2.75 m 2.8 m
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c) Length of toe slab
L = 0.35h
L = 0.35 x 2.8 = 0.98 m 1 m
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d) Thickness of base slab
t = = 0.55 m 550 mm
f) Thickness of stem
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- Moment at base stem = = 147.8 KNm
- Aultimate moment (m u ) = 1.5 x 147.8 = 221.82 KNm
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We know that
m u = 0.138 f ck bd 2
d = = 283.5 mm 290 mm
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- Total depth at base of stem = 290 +60 = 350 mm
- Reducing total depth at top of stem be D’ = 180 mm
- Coefficient depth at top = 180 – 60 = 120 mm
2. Tabulating force and moment
P a = K a rH 2 = 80.67 = 147.895
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Check
a) Overturning
fs 1 = = = 2.20 1.4 safe
b) Serding
fs 2 = = = = 1.01 1.4 not safe
shear key needs to pre provided
c) Failure of undensoil
= = 1.06 m ; e = – = – 1.06 = 0.34 m
P max =
= 125.02 KN/m SBC of soil
P min =
= 19.63 KN/m should be positive (no tension )
3. Design of stem
Effective depth of stem (d)= 350 – 60 = 290 mm
Maximum moment at base of moment (M e ) = 221.83 KNm
M u = 0.87 f y Ast d
M u = 0.87 x 415 x Ast x 290
(main)
using 4-20 mm bars @ 120 mm c/c as main reinforcement
also distribution steel = 0.12% of bd
= x = 318 mm 2
Using 10 mm bars
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Spacing = x 10 3 = 246.9 250 mm
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- Using 10 mm bars @ 250 mm c/c as distribution steel.
- Curtailment of steel ( CL 26.2.1 )
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A ct = =
= 940.23 mm (development length )
Cutting the base at 2 m H from base of stem
H = 5.5 – 2 = 3.5 m
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Moment at (h = 3.5 m) = 38.11 KN m
Total depth at h (3.5m) = 0.18 + = 0.28 m
Effective depth = 0.28 – 0.06 = 0.22 = 220 mm
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We know
M u = 0.87 f y Ast d
57.16 x 10 6 = 0.87 x 415 x Ast x 220
Ast = 776 mm 2
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- Only half the base can be covered at 2 m from the bottom
- Check for shear
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Critical section for shere = h 1 = 5.5 – 0.29 = 5.21 mm
Shear force = K a h 2 = 65.14 KN
Maximum shear force () = 65.14 x 1.5 = 97.71
Thickness at critical section = 0.34 m (similaras)
Effective depth = 0.34 – 0.06 = 0.28 m
= = = 0.3489 0.35 N/ mm 2
max = 2.8 N/mm 2 (for m-20)
Ast = = 1.009 %
= 0.62 N/mm 2 (table – 19)
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Here
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Min shear reinforcement is provided by distribution steel.
4. Design of shear slab
Total depth of heel slab = 500 mm
Effective depth of heel slab = 550 – 60 = 490 mm
Load on heel slab
Total load on heel slab = 107.93 KN/m
Maximum bending moment at B
= – – (0.5 x (74.21 – 19.63) x 1.45 x )
= 113.46 – 20.63 – 19.02 = 73.81 KNm
d = = = 163.53 mm < 490 mm
Ast (min) = = 1003 mm 2
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Now,
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M u = 0.87 f y Ast d
Ast = 424.84 mm 2
Using 16 mm bar
No. of bars = 4.96 5 bar
Spacing = 200 mm
We provide 5- 16 mm bars @ 200 mm c/c as the main reinforcement
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Distribution bars = 0.12% of bd = 660 mm 2
Spacing = 120 mm ( using 10 mm )
We provide 10 mm dia bars @ 120 mm c/c as distribution steel
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5. Design of toe slab
Maximum bending moment = = 56.2 KNm
Factored maximum bending moment = 1.5 x 56.25 = 84.37 KNm
M u = 0.138 f ck bd 2
d = = 174.84 < 490 (safe)
Area of steel
M u = 0.87 f y Ast d
84.37 x 10 6 = 0.87 x 415 x A st x 490
A st = 476.53 mm 2 < A st min (1003 mm 2 )
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We provide 5- 16 mm bars @ 200 mm c/c as the main reinforcement
We provide 10 mm dia bars @ 120 mm c/c as distribution steel
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