formula.expert

Mohr–Coulomb Shear Strength

Also known as soil shear strength · c plus sigma tan phi

τf=c′+σ′tan⁡ϕ′\tau_f = c' + \sigma'\tan\phi'

Worked example: c′ 15 kPa, σ′ 100 kPa, φ′ 25° → τ = 61.63 kPa — press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Try different units for next level excitement!

Learning zone

Mohr–Coulomb Shear Strength explained

c′φ′σ′τf

Coulomb wrote the friction half of this in 1773 while working on the retaining walls of Martinique; Mohr supplied the stress-circle framework a century later. The idea is that soil resists shear in two ways — a cohesion intercept that exists at zero normal stress, and a frictional part proportional to the effective normal stress squeezing the grains together. A silty clay with c′ = 15 kPa and φ′ = 25° on a plane carrying σ′ = 100 kPa can resist τ = 15 + 100 × tan 25° = 15 + 46.6 = 61.6 kPa.

The trap is primed: this must be written in effective stress. Raise the pore pressure and σ′ falls, taking the frictional term with it — which is precisely how a stable slope fails after a week of rain without a single grain being removed. The second trap is c′ itself. For a normally consolidated clay or a clean sand the true effective cohesion is essentially zero, and a c′ of 10 or 20 kPa fitted to a scattered triaxial dataset is an artefact of the regression that will quietly carry a large fraction of your slope's factor of safety. Skempton's 1964 work on progressive failure in London Clay showed that long-term strength drifts down to a residual with no cohesion at all.

Mohr–Coulomb Shear Strength formula

τf=c′+σ′tan⁡ϕ′\tau_f = c' + \sigma'\tan\phi'
Where
  • τf\tau_f= Shear strength at failure (kPa)
  • c′c'= Effective cohesion (kPa)
  • σ′\sigma'= Effective normal stress (kPa)
  • ϕ′\phi'= Effective friction angle (°)

Missing one of these? Work it out first, then come back