---
title: "Dry sand failure plane 25°, find σ3"
subject: "Soil Mechanics"
chapter: 10
key: "semester_4/soil_mechanics:fine-dry-sand-failure-plane-25"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/fine-dry-sand-failure-plane-25
site: BCT Engineering
---

# Dry sand failure plane 25°, find σ3

Soil Mechanics · Chapter 10: Shear Strength of Soil

- Asked in: 2075 Baisakh, 7 marks
- Video: [Numerical on Mohrs Circle (shear strength of soil)](https://www.youtube.com/watch?v=guSU0Ddq-l0) — civil engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/fine-dry-sand-failure-plane-25
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-10

## Question

A specimen of fine dry sand, when subjected to a triaxial compression test, failed at a deviator stress of 500 kN/m². It failed with a pronounced failure plane with an angle of 25° to the axis of the sample. Compute the lateral pressure ($\sigma_3$) to which the specimen would have been subjected.

## Short answer

The failure plane is 65 degrees to the major principal plane, so 45 + phi/2 = 65 degrees giving phi = 40 degrees. For c=0, sigma1 = sigma3 tan^2(65), and sigma1 - sigma3 = 500 kPa. Solving gives sigma3 = 500/(4.599 - 1) = 138.9 kN/m^2, about 139 kPa.

## Answer

**Given:** dry sand ($c=0$), deviator stress at failure $\sigma_1-\sigma_3 = 500$ kN/m². The failure plane makes $25^\circ$ with the axis of the sample.

**Step 1: Angle of the failure plane.** The axis of the sample is the direction of the major principal stress $\sigma_1$, so the failure plane makes $25^\circ$ with the direction of $\sigma_1$ and hence $90^\circ - 25^\circ = 65^\circ$ with the major principal plane (the horizontal plane). The theoretical failure angle is

$$
\theta = 45^\circ + \frac{\phi}{2} = 65^\circ \;\Rightarrow\; \phi = 40^\circ
$$

**Step 2: Flow value**

$$
N_\phi = \tan^2\left(45^\circ+\frac{\phi}{2}\right) = \tan^2 65^\circ = 4.599
$$

**Step 3: Lateral pressure.** For $c=0$:

$$
\sigma_1 = N_\phi\sigma_3 \;\Rightarrow\; \sigma_1 - \sigma_3 = \sigma_3(N_\phi - 1)
$$

$$
\sigma_3 = \frac{500}{4.599 - 1} = \frac{500}{3.599} = 138.9\ \text{kN/m}^2
$$

Check: $\sigma_1 = 638.9$, $\sin\phi = \frac{500}{638.9+138.9} = 0.643 = \sin40^\circ$.

**Answer: $\phi = 40^\circ$ and the lateral pressure $\sigma_3 \approx 139\ \text{kN/m}^2$.**

## Check yourself

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

1. The failure plane makes 25° with the sample axis. What angle does it make with the major principal plane?
   A. 25°
   B. 65°
   C. 45°
   D. 115°
   Answer: B. The axis is the sigma_1 direction, so the plane is 90° - 25° = 65° with the major principal plane. (Key point: Failure plane angle)

2. What is phi?
   A. 40°, from 45° + phi/2 = 65°
   B. 20°
   C. 25°
   D. 65°
   Answer: A. phi/2 = 20°, so phi = 40°. (Key point: Phi)

3. What is N_phi?
   A. tan^2 40° = 0.704
   B. tan^2 25° = 0.217
   C. tan 65° = 2.14
   D. tan^2 65° = 4.599
   Answer: D. N = tan^2(45° + phi/2) = tan^2 65°. (Key point: N value)

4. How is sigma_3 found?
   A. sigma_3 = 500 x 4.599
   B. sigma_3 = 500/4.599 = 108.7
   C. sigma_3 = sigma_d/(N - 1) = 500/3.599 = 138.9 kN/m²
   D. sigma_3 = 500 - 138.9
   Answer: C. With c = 0, sigma_1 - sigma_3 = sigma_3(N - 1). (Key point: sigma_3)

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