Shear Strength of Soils

In geotechnical engineering, the safety of structures such as foundations, retaining walls, and earth slopes relies on the soil's ability to resist shear failure. Shear strength is the internal resistance per unit area that the soil mass can offer to resist failure and sliding along any plane inside it.

Soil failure is typically not a tensile or compressive yielding, but rather a shear sliding of soil particles past one another.

The Mohr-Coulomb Failure Criterion

The most widely used framework to describe soil shear strength is the Mohr-Coulomb failure criterion. It states that the shear strength of a soil ($\tau_f$) on any failure plane is a linear function of the normal stress ($\sigma$) acting on that plane.

The general equation is:

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

Where:

• $\tau_f$ = shear strength of the soil [kPa or psf]
• $c$ = apparent cohesion of the soil [kPa or psf]
• $\sigma$ = total normal stress on the failure plane [kPa or psf]
• $\phi$ = angle of internal friction [degrees]

Cohesion ($c$) represents the shear strength of the soil when no normal stress is applied (typical for clays and cemented soils). The angle of internal friction ($\phi$) represents the frictional resistance generated between soil particles as normal stress increases (typical for sands and gravels).

Effective Stress Principle in Shear Strength

Because soils are multiphase materials containing water, the shear strength actually depends on the effective stresses transmitted directly between soil particles, not the total stress.

Modifying the Mohr-Coulomb criterion for effective stress gives:

$\tau_f = c' + \sigma' \tan(\phi')$

Where:

• $c'$ = effective cohesion
• $\sigma'$ = effective normal stress ($\sigma - u$, where $u$ is pore water pressure)
• $\phi'$ = effective angle of internal friction

For saturated coarse-grained soils (like clean sand), $c'$ is approximately zero, making the shear strength purely frictional: $\tau_f = \sigma' \tan(\phi')$.

Worked Example

Problem: A saturated sand layer is subjected to a total normal stress of 150 kPa and a pore water pressure of 50 kPa on a specific plane. Laboratory tests determine that the sand has an effective angle of internal friction $\phi'$ of 32 degrees and effective cohesion $c'$ of 0 kPa. Calculate the available shear strength on this plane.

Solution:

1. Determine the effective normal stress $\sigma'$: $\sigma' = \sigma - u = 150 \text{ kPa} - 50 \text{ kPa} = 100 \text{ kPa}$
2. Apply the effective Mohr-Coulomb equation: $\tau_f = c' + \sigma' \tan(\phi')$
3. Substitute the known values: $\tau_f = 0 + 100 \tan(32^\circ)$
4. Calculate the tangent (using degrees): $\tan(32^\circ) \approx 0.6249$
5. Calculate shear strength: $\tau_f = 100 \times 0.6249 = 62.49 \text{ kPa}$

The soil can safely resist up to approximately 62.5 kPa of shear stress on this plane before sliding failure occurs.

Engineering Check

In clayey soils under rapid loading conditions (where water cannot drain), the total stress (undrained) shear strength parameters are used instead of effective parameters. Care must always be taken to select the correct parameters ($c', \phi'$ vs. $c_u, \phi_u$) corresponding to the actual drainage conditions of the engineering project.