Question 5 of 6: Direct Shear Test — Shear Strength Parameters & Method Comparison

Question 5: Direct Shear Test — Shear Strength Parameters & Method Comparison (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) cohesion $c$ and angle of internal friction $\phi$; (b) a discussion of direct shear against alternative shear-strength test methods.

Approach. Convert each run's normal and shear FORCE to normal and shear STRESS using the fixed sample area, then fit the Mohr–Coulomb failure line $\tau=c+\sigma'\tan\phi$ to the four $(\sigma,\tau)$ pairs by least squares.

1. Convert forces to stresses. With $A=0.0025\ \text{m}^2$, $\sigma=N/A$ and $\tau=T/A$:

   Run	$\sigma$ (kPa)	$\tau$ (kPa)
   1	32.0	48.8
   2	64.0	58.8
   3	96.0	67.2
   4	128.0	78.0

2. Fit the Mohr–Coulomb envelope. Least-squares regression of $\tau$ on $\sigma$ through the four points gives $$\tau=39.2+0.300\,\sigma\quad(\text{kPa}),\qquad \tan\phi=0.300,$$ $$c=\boxed{39.2\ \text{kPa}},\qquad \phi=\arctan(0.300)=\boxed{16.7^\circ}.$$
   Figure 3. Mohr–Coulomb failure envelope fitted to the four direct shear runs.

Part (b) — direct shear vs. alternative methods. The direct shear test is simple, fast and inexpensive: it forces failure onto a single, known horizontal plane, so a series of tests at different normal loads traces the Mohr–Coulomb envelope directly, without constructing full Mohr circles as in triaxial testing, and repeated reversal along the same plane readily gives residual (post-peak) strength. Its main limitations are that the forced horizontal failure plane is not necessarily the weakest plane in the soil; stress and strain are distributed non-uniformly across the specimen because of the rigid box edges and progressive (rather than simultaneous) failure across the shear surface; drainage cannot be controlled or verified, and pore pressure cannot be measured during the test, so results are only strictly valid for free-draining, drained conditions (which is appropriate here for a sandy soil, but not for a clay tested rapidly); and only one point on the failure surface is measured per specimen, with no access to the intermediate principal stress or the full stress path. The triaxial test is the natural alternative in the laboratory: a cylindrical specimen fails on whatever plane is weakest, stress and strain are far more uniform, and pore pressure and volume change can be measured and drainage controlled (CD, CU or UU tests), at the cost of more expensive equipment, longer test times, and more skilled operation. The unconfined compression test is faster and cheaper than either, but is limited to saturated cohesive soils and yields only the undrained strength ($\phi_u=0$ assumption) — it would not be appropriate for this sandy soil at all. In the field, the vane shear test gives a quick, minimally-disturbed estimate of undrained shear strength in soft clays but not the drained parameters a sand requires, while plate-load and standard/cone penetration tests give field-scale, in-situ estimates of strength and bearing capacity (accounting for real fabric, cementation and layering) but are comparatively slow and expensive to run and interpret. For this sandy soil — free-draining, tested under modest confining stress — the direct shear test is a reasonable and economical choice, provided the practitioner recognizes its plane-forcing and non-uniform-stress limitations when interpreting the fitted $c$ and $\phi$.