Design of Cantilever Retaining Wall Supporting Lateritic Earth Fill

Cohesive soil profiles due to their nature are partially self supporting up to an extent called the critical depth (the critical depth is given by 2Z_o, kindly see figure above). As a result, such soils would not exert much pressure on a retaining wall as a granular material would (especially considering active earth pressure). Tropical laterites (especially those found in Nigeria) usually possess angle of internal friction and some cohesion. Their usage for construction purposes is also widespread.

The walls retaining such soils are subjected to active and passive pressure. In this example, we are going to consider active pressure only (all passive pressure neglected), but note that passive pressure could be more critical and hence govern the design (I recommend you read standard geotechnical engineering textbooks for more knowledge on this subject).

Worked Example

The cantilever retaining wall shown below is backfilled with tropical lateritic earthfill, having a unit weight, ρ, of 18 kN/m³, a cohesion C of 8 kN/m³ and an internal angle of friction, φ, of 26°. The allowable bearing pressure of the soil is 150 kNm/², the coefficient of friction is 0.5, and the unit weight of reinforced concrete is 24 kN/m³. Water level is at a great depth and adequate drainage is provided at the back of the wall.

We are to;
1. Determine the factors of safety against sliding and overturning for the active pressure.
2. Calculate ground bearing pressures.
3. Design the wall and base reinforcement assuming f_cu = 30 N/mm², f_y = 460 N/mm² and the cover to reinforcement in the wall and base are, respectively, 40 mm and 50 mm.

Geotechnical Design

Wall Pressure Calculations

Coefficient of active pressure K_A
Using Rankine’s theory
K_A = (1 – sinφ) / (1 + sinφ);
K_A = (1 – sin 26°) / (1 + sin 26°) = 0.39046;

Pressure at the top of wall = 2C√K_A = 2 × 8 × √0.39046 = 9.998 kN/m²
Z₀ = 2 C/γ√K_A = (2 × 8)/(18 × √0.39046) = 1.423m

Pressure at the bottom of retaining wall = H.γ.K_A – 2C√K_A = (5.5 × 18 × 0.39046) – (2 × 8 × √0.39046) = 32.408 kN/m²

Resistance to Sliding

Consider the forces acting on a 1 m length of wall. Horizontal force on wall due to backfill, P_A, is;
P_A = 0.5 × 32.408 × 4.077 = 66.063 kN/m

Weight of wall (W_w) = 0.35 × 5.1 × 24 = 42.84 kN/m
Weight of base (W_b) = 0.4 × 4 × 24 = 38.4 kN/m
Weight of soil (W_s) = 2.85 × 5.1 × 18 = 261.63 kN/m
Total vertical force (W_t) = 324.87 kN/m

Friction force, F_F, is
F_F = μWt = 0.5 × 324.87 = 171.435 kN/m

Neglecting all passive pressure force (F_P) = 0.
Hence factor of safety against sliding is;
F.O.S = F_F/P_A = 171.435/66.063 = 2.595
The factor of safety 2.595 > 1.5. Therefore the wall is very safe from sliding.

Resistance to Overturning

Taking moments about the toe, sum of overturning moments (M_O) is;
M_O = (66.063 kN/m) × (4.077/3) = 89.779 kN.m

Sum of Restoring Moment M_R;
M_R = (W_w × 0.975m) + (W_b × 2m) + (W_s × 2.575m)
M_R = (42.84 × 0.975m) + (38.4 × 2m) + (261.63 × 2.575m) = 792.266 kNm

F.O.S = M_R / M_O = 792.266/89.779 = 8.824
The factor of safety 8.824 > 2.0. Therefore the wall is very safe from overturning.

Bearing Capacity Check

Bending moment about the centre line of the base;
M = [66.064 × (4.077/3)] + (W_w × 1.025m) – (W_s × 1m)
M = [66.064 × (4.077/3)] + (42.84 × 1.025m) – (261.63 × 0.575m) = -16.745 kNm

Total vertical load = 324.87 kN

The eccentricity (e) = M/N = (16.745/324.87) = 0.0515m

Let us check D/6 = 4/6 = 0.666
Since e < D/6, there is no tension in the base

The maximum pressure in the base;
q_max = P/B (1 + 6e/B) = 324.87/4 [1 + (6 × 0.0515)/(4)] = 87.491 KN/m²
87.491 KN/m² < 175 KN/m²

Therefore, the bearing capacity check is satisfied.

Structural Design

Design of the Wall

Conservatively taking the moment at the base of the wall due to the active force;
M = [66.064 × (4.077/3)] = 89.781 kN.m
At ultimate limit state M = 1.4 × 89.781 = 125.693 kN.m

Thickness of wall = 350mm

Effective depth = 350 – 40 – (8) = 302mm (assuming Y16mm bars will be employed for the construction)

k = M/(F_cubd²) = (125.693 × 10⁶) / (30 × 1000 × 302²) = 0.0459
k < 0.156, no compression steel needed
la = 0.5 + (0.25 – k/0.9)^0.5 = 0.5 + (0.25 – 0.0459/0.9)^0.5 = 0.946
A_Sreq = M/(0.95F_y.la.d ) = (125.693 × 10⁶) / (0.95 × 460 × 0.946 × 302) = 1006 mm²/m
A_Smin = (0.13bh)/100 = (0.13 × 1000 × 350) / 100 = 455 mm²
Provide Y16 @ 150mm c/c (A_sprov = 1340 mm²/m)

Check for shear
Design shear force on wall at ultimate limit state V = (1.4 × 66.064) = 92.4896 kN/m
Shear stress v = V/bd = (92.4896 × 1000) / (1000 × 302) = 0.3062 N/mm²

V_c = 0.632 (100A_s/bd)^1/3 (400/d)^1/4
Vc = 0.632 × [(100 × 1340)/(1000 × 302)]^1/3 × (400/302)^1/4
Vc = 0.632 × 0.7627 × 1.0727 = 0.517 N/mm²
For F_cu = 30N/mm², Vc = 0.517 × (30/25)^1/3 = 0.5494 N/mm²
Since v < Vc, no shear reinforcement required.

Design of the Base

The pressure distribution diagram on the base at serviceability limit state is as shown below;
q_max = P/B (1 + 6e/B) = 324.87/4 [1 + (6 × 0.0515)/(4)] = 87.491 kN/m²
q_min = P/B (1 – 6e/B) = 324.87/4 [1 – (6 × 0.0515)/(4)] = 74.927 kN/m²

At ultimate limit state;
q_max = 1.4 × 87.491 = 122.487 KN/m²
q_min = 1.4 × 74.927 = 104.898 KN/m²

q_C = 104.898 + [2.85 (122.487 – 104.898)]/4 = 117.430 kN/m²
q_B = 104.898 + [3.2 (122.487 – 104.898)]/4 = 118.968 kN/m²

On investigating the maximum design moment about point C;

Back fill = 1.4 (18 × 5.1 × 2.85 × 2.85/2) = -521.952 kNm
Heel Slab = 1.4 (24 × 0.4 × 2.85 × 2.85/2) = -54.583 kNm
Earth Pressure = (104.898 × 2.85 × 2.85/2) + [(117.430 – 104.898) × 0.5 × 2.85 × 2.85/3] = 426.017 + 16.965 = 442.982 KNm
Net moment = -521.952 – 54.583 + 442.982 = -133.553 kNm

On investigating the maximum design moment about point B;

Back fill = (conservatively neglected)
Heel Slab = 1.4 (24 × 0.4 × 0.8 × 0.8/2) = – 4.3008 kNm
Earth Pressure = (118.968 × 0.8 × 0.8/2) + [(122.487 – 118.968) × 0.5 × 0.8 × (2 × 0.8/3)] = 37.9776 + 0.751 = 38.727 kNm
Net moment = -4.3008 + 38.727 = 34.426 kNm

Design of the toe
M_B = 133.553 kN.m

Thickness of base = 400 mm

Effective depth = 400 – 50 – (8) = 342 mm (assuming Y16mm bars will be employed for the construction)

k = M/(F_cubd²) = (133.553 × 10⁶) / (30 × 1000 × 342²) = 0.0385
k < 0.156, no compression steel needed
la = 0.5 + (0.25 – k/0.9)^0.5 = 0.5 + (0.25 – 0.0385/0.9)^0.5 = 0.95
A_Sreq = M/(0.95F_y.la.d ) = (133.553 × 10⁶) / (0.95 × 460 × 0.95 × 342) = 941 mm²/m
A_Smin = (0.13bh)/100 = (0.13 × 1000 × 400) / 100 = 520 mm²
Provide Y16 @ 175 mm c/c Top (A_sprov = 1148  mm²/m)

Design of the heel
M_C = 34.426 KNm

Thickness of base = 400 mm
Effective depth = 400 – 50 – (6) = 344 mm (assuming Y12mm bars will be employed for the construction)

k = M/(F_cubd²) = (34.426 × 10⁶) / (30 × 1000 × 344²) = 0.00981
k < 0.156, no compression steel needed
la = 0.5 + (0.25 – k/0.9)^0.5 = 0.5 + (0.25 – 0.00981/0.9)^0.5 = 0.95
A_Sreq = M/(0.95F_y.la.d ) = (34.426 × 10⁶) / (0.95 × 460 × 0.95 × 344) = 242 mm²/m
A_Smin = (0.13bh)/100 = (0.13 × 1000 × 400) / 100 = 520 mm²
Provide Y12 @ 175mm c/c (A_sprov = 646 mm²/m)

Check for shear at the base
The maximum shear force at ‘d’ from the face of the wall is investigated at the either side;

Maximum shear force at point B
Earth fill = 1.4 × 18 × 5.1 × 2.85 = -366.282 kN/m
Wall base = 1.4 × 24 × 0.4 × 2.85 = -38.304 kN/m
Base Earth Pressure = 0.5 (104.898 + 117.430) × 2.85 = 316.817 kN/m
Net shear force = -366.282 – 38.304 + 316.817 = -87.769 kN/m

A little investigation will show that this is the maximum shear force at any section on the base.

Taking the maximum shear stress at the base  v = (87.769 × 1000) / (1000 × 342) = 0.2566 N/mm²

V_c = 0.632 (100As/bd)^1/3 (400/d)^1/4
Vc = 0.632 × [(100 × 1148) / (1000 × 342)]^1/3 × (400/342)^1/4
Vc = 0.632 × 0.694 × 1.0399 = 0.456 N/mm²
For F_cu = 30N/mm², Vc = 0.456 × (30/25)^1/3 = 0.484 N/mm²
Since v < Vc, no shear reinforcement required.

DETAILING
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