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Shear tests; direct shear calculation

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.

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?

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 60×6060\times60 mm) is loaded with a constant normal load NN. The upper half is pushed horizontally until failure along the horizontal plane between the two halves. The horizontal force at failure is TfT_f (read from the proving ring).

σ=NA,τf=TfA\sigma = \frac{N}{A}, \qquad \tau_f = \frac{T_f}{A}

where AA 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 (σ,τf)(\sigma, \tau_f) are plotted and a best-fit straight line is drawn. Its intercept on the τ\tau-axis is the cohesion cc and its slope is ϕ\phi:

τf=c+σtan⁡ϕ\tau_f = c + \sigma\tan\phi

For sand, c=0c = 0 and ϕ=tan⁡−1(τf/σ)\phi = \tan^{-1}(\tau_f/\sigma).

✓ Answer checked by Claude Opus 5.5 (high effort)

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