---
title: "Shear strength of sand, WT rises"
subject: "Soil Mechanics"
chapter: 10
key: "semester_4/soil_mechanics:sand-shear-strength-water-rise"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/sand-shear-strength-water-rise
site: BCT Engineering
---

# Shear strength of sand, WT rises

Soil Mechanics · Chapter 10: Shear Strength of Soil

- Asked in: 2077 Chaitra, 7 marks
- Video: [Total and Effective Stress in Soil](https://www.youtube.com/watch?v=bYsneserYWU) — The Geotechnical Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/sand-shear-strength-water-rise
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-10

## Question

What is the shear strength of soil along a horizontal plane at a depth of 4 m in a deposit of sand having $\phi$ = 35°, $\gamma_d$ = 17 kN/m³, $G_s$ = 2.7? Assume the ground water table is at a depth of 2.5 m from the ground surface. Also find the change in shear strength when the water table rises to the ground surface.

## Short answer

Compute effective stress using dry unit weight above water table and submerged unit weight below. With water table at 2.5 m, effective stress is 58.6 kPa and shear strength is 41.0 kPa. When it rises to the surface, effective stress is 42.8 kPa and strength is 30.0 kPa, a decrease of 11.0 kPa.

## Answer

**Assumptions:** $\gamma_w = 9.81$ kN/m³; sand above the water table is dry ($\gamma = \gamma_d = 17$); $c = 0$.

**Void ratio**

$$
\gamma_d = \frac{G_s\gamma_w}{1+e} \;\Rightarrow\; e = \frac{2.7\times9.81}{17} - 1 = 0.558
$$

$$
\gamma_{sat} = \frac{(G_s+e)\gamma_w}{1+e} = \frac{(2.7+0.558)(9.81)}{1.558} = 20.51\ \text{kN/m}^3
$$

**Case 1: Water table at 2.5 m**

$$
\begin{aligned}
\sigma' &= 17(2.5) + (20.51 - 9.81)(1.5) \\
&= 42.5 + 16.05 = 58.6\ \text{kPa}
\end{aligned}
$$

$$
s = \sigma'\tan\phi = 58.6\tan35^\circ = 41.0\ \text{kPa}
$$

**Case 2: Water table at the surface**

$$
\sigma' = (20.51-9.81)(4) = 42.8\ \text{kPa}, \qquad s = 42.8\tan35^\circ = 30.0\ \text{kPa}
$$

**Change in shear strength:** $30.0 - 41.0 = -11.0$ kPa.

**Answer: shear strength is 41.0 kPa with the water table at 2.5 m; it falls by about 11.0 kPa (27%) to 30.0 kPa when the water table rises to the surface.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is e found from gamma_d?
   A. e = gamma_d/(G_s gamma_w) = 0.64
   B. e = G_s gamma_w / gamma_d - 1 = 0.558
   C. e = G_s - 1 = 1.7
   D. e = gamma_w/gamma_d = 0.58
   Answer: B. gamma_d = G_s gamma_w/(1+e), so e = 2.7(9.81)/17 - 1 = 0.558. (Key point: Void ratio)

2. What is sigma' at 4 m depth?
   A. 17(2.5) + (20.51 - 9.81)(1.5) = 58.6 kPa
   B. 17(4) = 68 kPa
   C. 20.51(4) = 82.0 kPa
   D. 17(2.5) + 20.51(1.5) = 73.3 kPa
   Answer: A. Dry unit weight above the water table and submerged unit weight below it. (Key point: Effective stress with water table at 2.5 m)

3. What is the shear strength in case 1?
   A. 58.6 kPa
   B. 30.0 kPa
   C. 68 tan 35°
   D. 41.0 kPa from 58.6 tan 35°
   Answer: D. s = sigma' tan phi = 58.6 x 0.700 = 41.0 kPa. (Key point: Strength)

4. What is the effect of the water table rising to the surface?
   A. Strength increases because gamma_sat is larger
   B. No change as phi is constant
   C. Strength decreases by about 11 kPa to 30.0 kPa since sigma' = 10.7 x 4 = 42.8
   D. Strength becomes zero
   Answer: C. Buoyancy and pore pressure reduce effective stress, hence strength falls about 27%. (Key point: Water rise)

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