# Design step for contiliver retaining wall - Civil Engineering

## Design step for cantilever retaining wall

1) Set the dimension of the wall

a) Depth of foundation
h min = 2

b) Height of stem
H = h + h min

c) Base width
b0.4H to 0.6H

d) Length of toe slab = 0.3b to 0.4b

e) Thickness of base slab = to

f) Thickness of stem

1. governed by bending moment criteria
2. check weather structure is stable based on the dimension set.

2) Check

a) Overturning fs 1 = 1.4

b) Serding fs 2 = 1.4

otherwise provided shear key

c) Failure of sending soil ( base failure )

= ; e = –
P max = SBC of soil
P min = should be positive (no tension )

3) Design of stem
Provide main steel, distribution steel, and shear reinforcement for steel of retaining wall.

4) Design of heel slab
Provide main steel, distribution steel for heel slab of retaining wall.

5) Design of toe slab
Provide main steel, distribution steel for toe slab of retaining wall.

6) Detailing
Detail the reinforced assigned

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 &amp; 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 &amp; concrete is 0.45.

Given

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

1. Dimensioning of wall

a) Depth of foundation
h min = 2 = 1.45

b) Base width
b = 0.5H
b = 0.5 x 5.5 =2.75 m 2.8 m

c) Length of toe slab
L = 0.35h
L = 0.35 x 2.8 = 0.98 m 1 m

d) Thickness of base slab
t = = 0.55 m 550 mm
f) Thickness of stem

1. Moment at base stem = = 147.8 KNm
2. Aultimate moment (m u ) = 1.5 x 147.8 = 221.82 KNm
We know that
m u = 0.138 f ck bd 2
d = = 283.5 mm 290 mm

1. Total depth at base of stem = 290 +60 = 350 mm
2. Reducing total depth at top of stem be D’ = 180 mm
3. Coefficient depth at top = 180 – 60 = 120 mm

### 2. Tabulating force and moment

P a = K a rH 2 = 80.67 = 147.895

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

Spacing = x 10 3 = 246.9 250 mm

1. Using 10 mm bars @ 250 mm c/c as distribution steel.
2. Curtailment of steel ( CL 26.2.1 )
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

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

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

1. Only half the base can be covered at 2 m from the bottom
2. Check for shear
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)

Here

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 &lt; 490 mm
Ast (min) = = 1003 mm 2

Now,

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

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

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 &lt; 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 &lt; A st min (1003 mm 2 )

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