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/m^{3}, a cohesion *C* of 8 kN/m^{3} 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/m^{3}. 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/mm^{2}, f_{y} = 460 N/mm^{2} 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/m^{2}

Z_{0} = 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/m^{2}

**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 O****verturning**

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/m^{2}

87.491 KN/m^{2} < 175 KN/m^{2}^{}Therefore, the bearing capacity check is satisfied.

**Structural Design**

**D****esign 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_{cu}bd^{2}) = (125.693 × 10^{6}) / (30 × 1000 × 302^{2}) = 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 × 10^{6}) / (0.95 × 460 × 0.946 × 302) = 1006 mm^{2}/m

A_{Smin} = (0.13bh)/100 = (0.13 × 1000 × 350) / 100 = 455 mm^{2}

Provide Y16 @ 150mm c/c (A_{sprov} = 1340 mm^{2}/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/mm^{2}^{}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/mm^{2}

For F_{cu} = 30N/mm^{2}, Vc = 0.517 × (30/25)^{1/3} = 0.5494 N/mm^{2}

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/m^{2}

q_{min} = P/B (1 – 6e/B) = 324.87/4 [1 – (6 × 0.0515)/(4)] = 74.927 kN/m^{2}

At ultimate limit state;

q_{max} = 1.4 × 87.491 = 122.487 KN/m^{2}

q_{min} = 1.4 × 74.927 = 104.898 KN/m^{2}^{}q_{C} = 104.898 + [2.85 (122.487 – 104.898)]/4 = 117.430 kN/m^{2}

q_{B} = 104.898 + [3.2 (122.487 – 104.898)]/4 = 118.968 kN/m^{2}

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_{cu}bd^{2}) = (133.553 × 10^{6}) / (30 × 1000 × 342^{2}) = 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 × 10^{6}) / (0.95 × 460 × 0.95 × 342) = 941 mm^{2}/m

A_{Smin} = (0.13bh)/100 = (0.13 × 1000 × 400) / 100 = 520 mm^{2}

Provide Y16 @ 175 mm c/c Top (A_{sprov} = 1148 mm^{2}/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_{cu}bd^{2}) = (34.426 × 10^{6}) / (30 × 1000 × 344^{2}) = 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 × 10^{6}) / (0.95 × 460 × 0.95 × 344) = 242 mm^{2}/m

A_{Smin} = (0.13bh)/100 = (0.13 × 1000 × 400) / 100 = 520 mm^{2}

Provide Y12 @ 175mm c/c (A_{sprov} = 646 mm^{2}/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/mm^{2}

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/mm^{2}

For F_{cu} = 30N/mm^{2}, Vc = 0.456 × (30/25)^{1/3} = 0.484 N/mm^{2}

Since *v* < Vc, no shear reinforcement required.

**DETAILING***I will need my readers to do the detailing sketches. I will accept both manual (hand) and CAD details. But your manual sketches must be very neat and well scaled. You should show the plans and at least one section. If your detailing is good enough, I will publish it on this post with your Facebook, LinkedIn, or G+ URL attached.**So, I will keep updating the post with as many good pictures as I receive. Good knowledge for all to flourish @ Structville*Send your sketches and URL profile to info@structville.com

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A very good and nice article . I appreciate it. Keep it up.

great work!! keep it up

Thank you..

Great work engineer

Muito didático. Obrigada.

Pressure a the bottom should be 28.66

and also Pa =58.43

Hi,

Very good job please can I have it by email.

Thanks for the sharing

Best regards

goods