---
title: "Dry sand triaxial, φ = 36°, σ3 = 100"
subject: "Soil Mechanics"
chapter: 10
key: "semester_4/soil_mechanics:dry-sand-triaxial-deviator-stress-36"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/dry-sand-triaxial-deviator-stress-36
site: BCT Engineering
---

# Dry sand triaxial, φ = 36°, σ3 = 100

Soil Mechanics · Chapter 10: Shear Strength of Soil

- Asked in: 2075 Bhadra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/dry-sand-triaxial-deviator-stress-36
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-10

## Question

A sample of dry cohesionless soil was tested in a triaxial machine. If the angle of shearing resistance was 36° and the confining pressure 100 kN/m², determine the deviator stress at which the sample failed.

## Short answer

For dry cohesionless sand, use sigma1 = sigma3 tan^2(45 + phi/2). With sigma3 = 100 kPa and phi = 36 degrees, sigma1 = 385.2 kPa. The deviator stress at failure is sigma1 - sigma3 = 285.2 kPa, about 285 kN/m^2.

## Answer

For dry cohesionless soil, $c = 0$ and

$$
\sigma_1 = \sigma_3\tan^2\left(45^\circ + \frac{\phi}{2}\right)
$$

**Given:** $\phi = 36^\circ$, $\sigma_3 = 100$ kN/m².

$$
\tan^2(63^\circ) = (1.9626)^2 = 3.852
$$

$$
\sigma_1 = 100 \times 3.852 = 385.2\ \text{kN/m}^2
$$

$$
\sigma_d = \sigma_1 - \sigma_3 = 385.2 - 100 = 285.2\ \text{kN/m}^2
$$

Check: $\sin\phi = \frac{285.2}{385.2+100} = 0.588 = \sin36^\circ$.

**Answer: deviator stress at failure $\approx 285\ \text{kN/m}^2$.**

## Check yourself

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

1. Which relation gives sigma_1 at failure for a dry cohesionless soil?
   A. sigma_1 = sigma_3 tan(45° + phi/2)
   B. sigma_1 = sigma_3 tan^2(45° - phi/2)
   C. sigma_1 = sigma_3 + 2c
   D. sigma_1 = sigma_3 tan^2(45° + phi/2)
   Answer: D. For c = 0 the Mohr-Coulomb condition reduces to this. (Key point: Formula)

2. What is tan^2(45° + 36°/2)?
   A. tan^2 36° = 0.528
   B. tan^2 27° = 0.260
   C. tan^2 63° = 3.852
   D. tan 63° = 1.96 only
   Answer: C. 45° + 18° = 63°, tan 63° = 1.9626, squared is 3.852. (Key point: Flow value)

3. What is the deviator stress at failure?
   A. 385.2 kN/m²
   B. 285.2 kN/m²
   C. 100 kN/m²
   D. 214.8 kN/m²
   Answer: B. sigma_1 = 385.2 so sigma_d = 385.2 - 100 = 285.2. (Key point: Deviator stress)

4. How can the answer be checked?
   A. sin phi = 285.2/(385.2 + 100) = 0.588 = sin 36°
   B. sin phi = 285.2/385.2
   C. tan phi = 285.2/100
   D. cos phi = 285.2/485.2 = 36°
   Answer: A. (sigma_1 - sigma_3)/(sigma_1 + sigma_3) must equal sin phi. (Key point: Check)

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