--- title: "Shear strength on horizontal plane, WT 3.5 m" subject: "Soil Mechanics" chapter: 10 key: "semester_4/soil_mechanics:shear-strength-cohesionless-water-table-3m" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/shear-strength-cohesionless-water-table-3m site: BCT Engineering --- # Shear strength on horizontal plane, WT 3.5 m Soil Mechanics · Chapter 10: Shear Strength of Soil - Asked in: 2078 Poush - Video: [Calculate the potential shear strength on a horizontal plane at a depth of 3 m below the surface in](https://www.youtube.com/watch?v=BX6tYKqDIBY) — Unique Series - Civil Mantraa - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/shear-strength-cohesionless-water-table-3m - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-10 ## Question Calculate the potential shear strength on a horizontal plane at a depth of 3 m below the surface in a formation of cohesionless soil when the water table is at a depth of 3.5 m. The degree of saturation may be taken as 0.5 on the average. Void ratio = 0.50; grain specific gravity = 2.70; angle of internal friction = 30°. What will be the modified value of shear strength if the water table reaches the ground surface? ## Short answer Water table at 3.5 m: γ = 19.29 kN/m³, σ′ = 57.9 kPa, shear strength = 57.9 tan30° = 33.4 kPa. Water table at surface: γsat = 20.93 kN/m³, σ′ = (20.93−9.81)×3 = 33.4 kPa, strength = 33.4 tan30° = 19.3 kPa. ## Answer **Assumptions:** $\gamma_w = 9.81$ kN/m³; the soil above the water table is partly saturated ($S_r = 0.5$); $c=0$ and capillary effects are neglected. Strength $= \sigma'\tan\phi$ on the horizontal plane. **Case 1: Water table at 3.5 m** (below the 3 m depth, so there is no pore pressure at the plane) $$ \gamma = \frac{(G_s + S_r e)\gamma_w}{1+e} = \frac{(2.70 + 0.5\times0.50)(9.81)}{1.50} = 19.29\ \text{kN/m}^3 $$ $$ \sigma' = \sigma = 19.29\times 3 = 57.9\ \text{kPa} $$ $$ s = \sigma'\tan\phi = 57.9\tan30^\circ = 33.4\ \text{kPa} $$ **Case 2: Water table at the ground surface** (fully saturated, $S_r = 1$) $$ \gamma_{sat} = \frac{(G_s + e)\gamma_w}{1+e} = \frac{(2.70+0.50)(9.81)}{1.50} = 20.93\ \text{kN/m}^3 $$ $$ \sigma' = (20.93 - 9.81)\times 3 = 33.4\ \text{kPa} $$ $$ s = 33.4\tan30^\circ = 19.3\ \text{kPa} $$ **Answer: shear strength is about 33.4 kPa with the water table at 3.5 m, and falls to about 19.3 kPa (a reduction of about 42%) when the water table rises to the ground surface.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the shear strength on the horizontal plane for this soil? A. s = c + sigma tan phi with c from capillarity B. s = sigma' tan phi, since c = 0 C. s = sigma / tan phi D. s = gamma z only Answer: B. For cohesionless soil strength equals effective normal stress times tan phi. (Key point: Strength formula) 2. How is the unit weight above the water table calculated? A. gamma = (G_s + S_r e) gamma_w / (1 + e) = 19.29 kN/m³ B. gamma = (G_s + e) gamma_w/(1+e) = 20.93 C. gamma = G_s gamma_w/(1+e) = 17.66 D. gamma = G_s gamma_w = 26.5 Answer: A. S_r = 0.5 gives (2.70 + 0.25)(9.81)/1.5 = 19.29 kN/m³. (Key point: Unit weight partly saturated) 3. Why is sigma' equal to sigma in case 1? A. Pore pressure equals total stress B. Buoyancy is considered C. Water table is at the plane D. The water table (3.5 m) is below the 3 m plane so there is no pore pressure Answer: D. No pore pressure acts at 3 m, so sigma' = 19.29 x 3 = 57.9 kPa, s = 33.4 kPa. (Key point: Water table below plane) 4. What happens to strength when the water table reaches the surface? A. It rises because gamma_sat is larger B. It stays at 33.4 kPa C. It falls to about 19.3 kPa since sigma' = (gamma_sat - gamma_w) x 3 = 33.4 D. It becomes zero Answer: C. Pore pressure reduces effective stress from 57.9 to 33.4 kPa, so strength drops about 42%. 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