Design step for cantilever retaining wall

Design steps for cantilever retaining wall – In this tutorial, we are going to design the cantilever retaining wall using the 6 steps approach.

Step 1 – Set the dimensions 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 = 0.3_b to 0.4_b

e) Thickness of base slab = to

f) Thickness of stem

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

Step 2 – Check

a) Overturning fs₁ = 1.4

b) Serding fs₂ = 1.4

otherwise, provide a shear key

c) Failure of sending soil ( base failure )

= ; e = –

P_max = SBC of soil

P_min = should be positive (no tension )

Step 3 – Design of stem

Provide main steel, distribution steel, and shear reinforcement for steel of retaining wall.

Step 4 – Design of heel slab

Provide main steel and distribution steel for the heel slab of the retaining wall.

Step 5 – Design of the slab

Provide main steel and distribution steel for the toe slab of the retaining wall.

Step 6 – Detailing

Detail the reinforced assigned

Question – A cantilever retaining wall has a 5.5m tall stem to retain soil level with its top. The soil density is 16KN/m³ & has an angle of repose 30. The safe bearing capacity of soil is 210 KN/m². Design retaining wall using M20 and Fe 415. The frictional coefficient between soil & concrete is 0.45.

Answer

Given

Height of stem = H = 5.5 m

Density of soil = 16 KN/m³

Angle of repose = 30

Safe bearing capacity = 210 KN/m²

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 the 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 the top of stem be D’ = 180 mm
3. Coefficient depth at top = 180 – 60 = 120 mm

2. Tabulating force and moment

P_a = K_arH 2 = 80.67 = 147.895

Check

a) Overturning

fs₁ = = = 2.20 1.4 safe

b) Serving

fs 2 = = = = 1.01 1.4 not safe

shear key needs to pre-provided

c) Failure of undersoil

= = 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

The maximum moment at the base of the 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 the base of the 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 (similar)

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

The total depth of the heel slab = 500 mm

Effective depth of heel slab = 550 – 60 = 490 mm

Load on heel slab

The 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 mm²

Now,

M_u = 0.87 f y Ast d

Ast = 424.84 mm²

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 mm²

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 < 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 mm² )

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

Other related tutorials

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• Design a rectangular beam