Enter the normal stress and the shear stress at failure of each trial of a direct shear test. The calculator fits the Mohr-Coulomb envelope to them by least squares and gives the friction angle, the cohesion and how well the line fits the trials. It follows ASTM D3080, and it’s free and needs no account.

Calculator · ASTM D3080

Trials

The standard asks for at least three different normal stresses: with two, the line always goes through both points and the angle doesn't come from the data.

Trial	Normal stress(kg/cm²)	Shear stress(kg/cm²)
#1
#2
#3

Mohr-Coulomb envelope

With 3 trials at normal stresses from 0.50 to 2.00 kg/cm², the fitted line is τf = 0.332 + σ·0.796. It gives a friction angle of 38.5° and a cohesion of 0.332 kg/cm², with a fit of R² = 0.9983

Shear strength parameters

Quantity	Value
Cohesion · cthe line's intercept	0.332kg/cm²
Slope · tan φleast squares over the trials	0.7957
Friction angle · φatan of the slope	38.51°
The line's fit · R²1 − SSres / SStot	0.9983

This calculator is offered for information and technical support. It doesn't replace professional judgment or the reviews due, and its use is the user's responsibility.

How it calculates

The Mohr-Coulomb criterion takes a soil’s shear strength on a plane as a straight line in the normal stress on that plane:

\tau_f = c + \sigma \tan\phi

With $n$ trials, each a pair $(\sigma_i, \tau_i)$ , the calculator fits that line by least squares. The slope is $\tan\phi$ and the intercept is the cohesion:

\tan\phi = \frac{n \sum \sigma_i \tau_i - \sum \sigma_i \sum \tau_i}{n \sum \sigma_i^2 - \left(\sum \sigma_i\right)^2}, \qquad c = \frac{\sum \tau_i - \tan\phi \sum \sigma_i}{n}

$\phi$ is the arctangent of the slope, and $R^2 = 1 - SS_{res}/SS_{tot}$ says how close the trials fall to the line. ASTM D3080 asks for at least three normal stresses: with two, the line goes through both points whatever they are, so the calculator warns. It also warns about two trials at the same normal stress, an $R^2$ under 0.95, and a friction angle below 0° or above 50°.

The line is fitted to whatever you enter. Peak stresses give the peak envelope, and stresses at large displacement the residual one. The fields are in kg/cm² (1 kg/cm² ≈ 98.1 kPa); $\phi$ doesn’t depend on the unit, and $c$ comes out in the unit of the stresses.

The full procedure, equipment and report are in the Academy lesson: Direct shear test.

How to use it

For each trial, enter the normal stress on the specimen and the shear stress at failure: each force divided by the specimen’s area.
Use three different normal stresses, over the range the soil will carry in the field.
Read $\phi$ in the result, and the cohesion, the slope $\tan\phi$ and $R^2$ in the table, with the trials and the fitted envelope on the plot.

Restore the example brings back the three trials of the worked example. To try $c$ and $\phi$ under a footing, the bearing capacity calculator takes the cohesion in kPa.

Frequently asked questions

What is the direct shear test used for?

To measure a soil’s shear strength parameters, its friction angle and cohesion, on a horizontal failure plane. It’s quick and simple, and it’s most used for sands and for the drained strength of clays, sheared slowly enough for the water to drain.

What is the ASTM standard for direct shear testing?

ASTM D3080, the direct shear test of soils under consolidated drained conditions. It asks for at least three specimens, each sheared under a different normal stress, to draw the envelope.

What are the drawbacks of a direct shear test?

The failure plane is forced along the gap between the two halves of the box, and the stresses on it aren’t uniform. Drainage can’t be controlled and pore pressure isn’t measured, so the test doesn’t give undrained strength reliably. The contact area also shrinks as the halves slide. A triaxial test avoids most of these limits.

How do you find c and φ from a direct shear test?

Plot each trial’s shear stress at failure against its normal stress and draw the best straight line through the points. The line’s slope is $\tan\phi$ and its intercept on the shear axis is the cohesion $c$ . With drained tests these are the effective parameters, $c'$ and $\phi'$ .

Shear strength calculator: direct shear and Mohr-Coulomb