---
title: "Shear tests; direct shear calculation"
subject: "Soil Mechanics"
chapter: 10
key: "semester_4/soil_mechanics:shear-tests-direct-shear-strength-calc"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/shear-tests-direct-shear-strength-calc
site: BCT Engineering
---

# Shear tests; direct shear calculation

Soil Mechanics · Chapter 10: Shear Strength of Soil

- Asked in: 2074 Bhadra, 2 marks
- Video: [Direct Shear Test Calculations| Shear Parameters from DST results](https://www.youtube.com/watch?v=L2GM5o_fsd4) — Master 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/shear-tests-direct-shear-strength-calc
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-10

## Question

Write down the names of the shear strength tests that can be performed in the laboratory. How do you calculate shear strength in the direct shear test?

## Short answer

Lab shear tests: direct shear, triaxial, unconfined compression, vane. In direct shear, a normal load N is applied, and horizontal force T_f at failure is measured. Shear strength τ_f = T_f/A. Test at multiple normal stresses yields c (intercept) and φ (slope) from the τ_f vs σ plot.

## Answer

**Laboratory shear strength tests:** direct shear test, triaxial test (UU, CU, CD), unconfined compression test, and laboratory vane shear test.

**Shear strength in the direct shear test**

A soil specimen in a split box (commonly $60\times60$ mm) is loaded with a constant normal load $N$. The upper half is pushed horizontally until failure along the horizontal plane between the two halves. The horizontal force at failure is $T_f$ (read from the proving ring).

$$
\sigma = \frac{N}{A}, \qquad \tau_f = \frac{T_f}{A}
$$

where $A$ is the area of the specimen (corrected for the reduced contact area as the box moves, if required).

The test is repeated at three or more different normal stresses. The points $(\sigma, \tau_f)$ are plotted and a best-fit straight line is drawn. Its intercept on the $\tau$-axis is the cohesion $c$ and its slope is $\phi$:

$$
\tau_f = c + \sigma\tan\phi
$$

For sand, $c = 0$ and $\phi = \tan^{-1}(\tau_f/\sigma)$.

## Check yourself

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

1. How is shear strength computed in direct shear?
   A. tau_f = normal load / area
   B. tau_f = horizontal force at failure / area
   C. tau_f = horizontal force x area
   D. tau_f = normal load x tan(phi)
   Answer: B. Horizontal force at failure over the specimen area gives the shear stress at failure. (Key point: Shear stress)

2. Why are three or more normal stresses used?
   A. To calculate the average of the specimen
   B. To plot (sigma, tau_f) points and draw the best-fit strength line
   C. To measure pore pressure
   D. To find the water content
   Answer: B. A line through at least three points gives c (intercept) and phi (slope). (Key point: Number of tests)

3. What are c and phi for sand in direct shear?
   A. c = tau_f and phi = 0
   B. c = sigma and phi = 45
   C. c = 0 and phi = arctan(sigma/tau_f)
   D. c = 0 and phi = arctan(tau_f/sigma)
   Answer: D. Sand has no cohesion so the line passes through the origin. (Key point: Sand)

## Related in Soil Mechanics

- [Shear strength tests (lab and field)](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Ashear-strength-tests-names) — Soil Mechanics, chapter 10, asked 3×
- [Limitations of direct shear test](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Adirect-shear-limitations) — Soil Mechanics, chapter 10
- [Draw UCS and direct shear results](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aucs-direct-shear-results-plot) — Soil Mechanics, chapter 10
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- [Triaxial tests by drainage; UCS vs cu](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Atriaxial-tests-by-drainage-ucs) — Soil Mechanics, chapter 10

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