PYQs / GATE CE / 2016 / Set 2 / Q51 GATE CE 2016 Set 2 — Question 51 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. NAT +2 / -0 Hard Rankine Theory Earth Pressure Geotechnical Engineering Active Earth Pressure
Geotechnical Engineering → Earth Pressure → Active Earth Pressure
Last updated 5 September 2026
Question The soil profile at a site consists of a 5 m thick sand layer underlain by a
c − ϕ c-\phi c − ϕ soil as shown in figure. The water table is found 1 m below the ground level. The entire soil mass is retained by a concrete retaining wall and is in the active state. The back of the wall is smooth and vertical. The total active earth pressure (expressed in
k N / m 2 kN/m^2 k N / m 2 ) at point A as per Rankine's theory is
_________
Solution 1. Calculate Active Earth Pressure Coefficients (K a K_a K a For Sand layer: ϕ ′ = 32 ∘ ⟹ K a 1 = 1 − sin 32 ∘ 1 + sin 32 ∘ ≈ 0.307 \phi' = 32^\circ \implies K_{a1} = \frac{1 - \sin 32^\circ}{1 + \sin 32^\circ} \approx 0.307 ϕ ′ = 3 2 ∘ ⟹ K a 1 = 1 + s i n 3 2 ∘ 1 − s i n 3 2 ∘ ≈ 0.307 For c − ϕ c-\phi c − ϕ ϕ ′ = 24 ∘ ⟹ K a 2 = 1 − sin 24 ∘ 1 + sin 24 ∘ ≈ 0.422 \phi' = 24^\circ \implies K_{a2} = \frac{1 - \sin 24^\circ}{1 + \sin 24^\circ} \approx 0.422 ϕ ′ = 2 4 ∘ ⟹ K a 2 = 1 + s i n 2 4 ∘ 1 − s i n 2 4 ∘ ≈ 0.422
2.
Calculate Effective Vertical Stress (σ v ′ \sigma'_v σ v ′
Layer 1 (0-1m): σ v 1 ′ = 1 × 16.5 = 16.5 kN/m 2 \sigma'_{v1} = 1 \times 16.5 = 16.5 \text{ kN/m}^2 σ v 1 ′ = 1 × 16.5 = 16.5 kN/m 2 Layer 2 (1-5m): σ v 2 ′ = 4 × ( 19 − 9.81 ) = 4 × 9.19 = 36.76 kN/m 2 \sigma'_{v2} = 4 \times (19 - 9.81) = 4 \times 9.19 = 36.76 \text{ kN/m}^2 σ v 2 ′ = 4 × ( 19 − 9.81 ) = 4 × 9.19 = 36.76 kN/m 2 Layer 3 (5-8m): σ v 3 ′ = 3 × ( 18.5 − 9.81 ) = 3 × 8.69 = 26.07 kN/m 2 \sigma'_{v3} = 3 \times (18.5 - 9.81) = 3 \times 8.69 = 26.07 \text{ kN/m}^2 σ v 3 ′ = 3 × ( 18.5 − 9.81 ) = 3 × 8.69 = 26.07 kN/m 2 Total σ v ′ \sigma'_v σ v ′ 16.5 + 36.76 + 26.07 = 79.33 kN/m 2 16.5 + 36.76 + 26.07 = 79.33 \text{ kN/m}^2 16.5 + 36.76 + 26.07 = 79.33 kN/m 2
3.
Calculate Pore Water Pressure (u u u
Water table is at 1m depth. Depth of water above A = 8 − 1 = 7 m 8 - 1 = 7 \text{ m} 8 − 1 = 7 m u = 7 × 9.81 = 68.67 kN/m 2 u = 7 \times 9.81 = 68.67 \text{ kN/m}^2 u = 7 × 9.81 = 68.67 kN/m 2
4.
Calculate Total Active Earth Pressure (p a p_a p a
Using Rankine's formula for c − ϕ c-\phi c − ϕ p a = K a 2 σ v ′ − 2 c ′ K a 2 + u p_a = K_{a2} \sigma'_v - 2c'\sqrt{K_{a2}} + u p a = K a 2 σ v ′ − 2 c ′ K a 2 + u p a = ( 0.422 × 79.33 ) − ( 2 × 25 × 0.422 ) + 68.67 p_a = (0.422 \times 79.33) - (2 \times 25 \times \sqrt{0.422}) + 68.67 p a = ( 0.422 × 79.33 ) − ( 2 × 25 × 0.422 ) + 68.67 p a = 33.48 − ( 50 × 0.6496 ) + 68.67 p_a = 33.48 - (50 \times 0.6496) + 68.67 p a = 33.48 − ( 50 × 0.6496 ) + 68.67 p a = 33.48 − 32.48 + 68.67 = 69.67 kN/m 2 p_a = 33.48 - 32.48 + 68.67 = 69.67 \text{ kN/m}^2 p a = 33.48 − 32.48 + 68.67 = 69.67 kN/m 2 This value falls within the range 69.0 to 70.5.
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Solution 1. Calculate Active Earth Pressure Coefficients (K a K_a K a For Sand layer: ϕ ′ = 32 ∘ ⟹ K a 1 = 1 − sin 32 ∘ 1 + sin 32 ∘ ≈ 0.307 \phi' = 32^\circ \implies K_{a1} = \frac{1 - \sin 32^\circ}{1 + \sin 32^\circ} \approx 0.307 ϕ ′ = 3 2 ∘ ⟹ K a 1 = 1 + s i n 3 2 ∘ 1 − s i n 3 2 ∘ ≈ 0.307 For c − ϕ c-\phi c − ϕ ϕ ′ = 24 ∘ ⟹ K a 2 = 1 − sin 24 ∘ 1 + sin 24 ∘ ≈ 0.422 \phi' = 24^\circ \implies K_{a2} = \frac{1 - \sin 24^\circ}{1 + \sin 24^\circ} \approx 0.422 ϕ ′ = 2 4 ∘ ⟹ K a 2 = 1 + s i n 2 4 ∘ 1 − s i n 2 4 ∘ ≈ 0.422
2.
Calculate Effective Vertical Stress (σ v ′ \sigma'_v σ v ′
Layer 1 (0-1m): σ v 1 ′ = 1 × 16.5 = 16.5 kN/m 2 \sigma'_{v1} = 1 \times 16.5 = 16.5 \text{ kN/m}^2 σ v 1 ′ = 1 × 16.5 = 16.5 kN/m 2 Layer 2 (1-5m): σ v 2 ′ = 4 × ( 19 − 9.81 ) = 4 × 9.19 = 36.76 kN/m 2 \sigma'_{v2} = 4 \times (19 - 9.81) = 4 \times 9.19 = 36.76 \text{ kN/m}^2 σ v 2 ′ = 4 × ( 19 − 9.81 ) = 4 × 9.19 = 36.76 kN/m 2 Layer 3 (5-8m): σ v 3 ′ = 3 × ( 18.5 − 9.81 ) = 3 × 8.69 = 26.07 kN/m 2 \sigma'_{v3} = 3 \times (18.5 - 9.81) = 3 \times 8.69 = 26.07 \text{ kN/m}^2 σ v 3 ′ = 3 × ( 18.5 − 9.81 ) = 3 × 8.69 = 26.07 kN/m 2 Total σ v ′ \sigma'_v σ v ′ 16.5 + 36.76 + 26.07 = 79.33 kN/m 2 16.5 + 36.76 + 26.07 = 79.33 \text{ kN/m}^2 16.5 + 36.76 + 26.07 = 79.33 kN/m 2
3.
Calculate Pore Water Pressure (u u u
Water table is at 1m depth. Depth of water above A = 8 − 1 = 7 m 8 - 1 = 7 \text{ m} 8 − 1 = 7 m u = 7 × 9.81 = 68.67 kN/m 2 u = 7 \times 9.81 = 68.67 \text{ kN/m}^2 u = 7 × 9.81 = 68.67 kN/m 2
4.
Calculate Total Active Earth Pressure (p a p_a p a
Using Rankine's formula for c − ϕ c-\phi c − ϕ p a = K a 2 σ v ′ − 2 c ′ K a 2 + u p_a = K_{a2} \sigma'_v - 2c'\sqrt{K_{a2}} + u p a = K a 2 σ v ′ − 2 c ′ K a 2 + u p a = ( 0.422 × 79.33 ) − ( 2 × 25 × 0.422 ) + 68.67 p_a = (0.422 \times 79.33) - (2 \times 25 \times \sqrt{0.422}) + 68.67 p a = ( 0.422 × 79.33 ) − ( 2 × 25 × 0.422 ) + 68.67 p a = 33.48 − ( 50 × 0.6496 ) + 68.67 p_a = 33.48 - (50 \times 0.6496) + 68.67 p a = 33.48 − ( 50 × 0.6496 ) + 68.67 p a = 33.48 − 32.48 + 68.67 = 69.67 kN/m 2 p_a = 33.48 - 32.48 + 68.67 = 69.67 \text{ kN/m}^2 p a = 33.48 − 32.48 + 68.67 = 69.67 kN/m 2 This value falls within the range 69.0 to 70.5.
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