Direct Shear Test Analyzer
Analyze direct shear test data to determine Mohr-Coulomb shear strength parameters including cohesion and friction angle with linear regression.
Direct Shear Test Analyzer
Calculator
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Formula: tau = c + sigma * tan(phi)
Worked example โ c: 0 kPa | phi: 31.0 degrees | Envelope: tau = 0.600 x sigma_n
Formula
tau = c + sigma * tan(phi)
This Direct Shear Test Analyzer computes results from your provided inputs using the calculator's underlying model.
Worked Examples
Example 1: Clean Sand - Three Point Test
Problem:Three direct shear specimens are sheared at normal stresses of 50, 100 and 200 kPa and fail at shear stresses of 30, 60 and 120 kPa. Fit the Mohr-Coulomb envelope and report cohesion and friction angle.
Solution:The three points are exactly proportional: 30/50 = 60/100 = 120/200 = 0.600 A least-squares line through them therefore passes through the origin, so the cohesion intercept c = 0 kPa Slope = tan(phi) = 0.600, so phi = arctan(0.600) = 30.96 degrees, i.e. 31.0 degrees Envelope: tau = 0 + sigma_n x tan(31.0 deg) Zero cohesion with phi near 31 degrees is the signature of a clean, uncemented medium-dense sand.
Result:c: 0 kPa | phi: 31.0 degrees | Envelope: tau = 0.600 x sigma_n
Example 2: Silty Clay - Four Point Test
Problem:Four specimens sheared at normal stresses of 25, 50, 100 and 150 kPa fail at 28, 40, 62 and 84 kPa. Fit the envelope by least squares.
Solution:Mean normal stress = 325/4 = 81.25 kPa; mean shear stress = 214/4 = 53.50 kPa Sum of (sigma - mean_sigma)(tau - mean_tau) = 1434.4 + 421.9 + 159.4 + 2096.9 = 4112.5 Sum of (sigma - mean_sigma)^2 = 3164.1 + 976.6 + 351.6 + 4726.6 = 9218.8 Slope = tan(phi) = 4112.5 / 9218.8 = 0.4461, so phi = arctan(0.4461) = 24.0 degrees Intercept c = 53.50 - (0.4461 x 81.25) = 53.50 - 36.24 = 17.3 kPa Envelope: tau = 17.3 + sigma_n x tan(24.0 deg). The genuine cohesion intercept of 17 kPa is what distinguishes this silty clay from the sand above.
Result:c: 17.3 kPa | phi: 24.0 degrees | Envelope: tau = 17.3 + 0.446 x sigma_n
Frequently Asked Questions
What is the direct shear test?
The direct shear test is a laboratory geotechnical test used to determine the shear strength parameters of soil, specifically cohesion (c) and the angle of internal friction (phi). A soil specimen is placed in a split box where the upper half is displaced horizontally while a normal load is applied vertically. The test is performed at several different normal stresses, and the peak shear stress at failure is recorded for each. Plotting shear stress versus normal stress and fitting a straight line yields the Mohr-Coulomb failure envelope.
How do you interpret the Mohr-Coulomb failure envelope?
The Mohr-Coulomb failure envelope is a straight line on a plot of shear stress (tau) versus normal stress (sigma). The equation is tau = c + sigma * tan(phi), where c is the y-intercept representing cohesion and phi is the angle the line makes with the horizontal axis representing the friction angle. Cohesion represents the shear strength at zero normal stress and is significant in clay soils. The friction angle reflects the frictional resistance between soil particles and dominates in granular soils.
What are the limitations of the direct shear test?
The direct shear test forces failure along a predetermined horizontal plane, which may not be the weakest plane in the soil. Drainage conditions cannot be precisely controlled, making it difficult to determine truly undrained or drained parameters for fine-grained soils. The stress distribution on the failure plane is non-uniform, with stress concentrations at the edges. Despite these limitations, the test is widely used because it is simple, fast, inexpensive, and provides reasonable shear strength parameters for many practical applications.
What is the difference between peak and residual shear strength?
Peak shear strength is the maximum shear stress a soil can sustain before failure occurs, corresponding to the highest point on the shear stress versus horizontal displacement curve. Residual shear strength is the constant shear stress that the soil maintains at large displacements after the peak has been exceeded. Dense sands and overconsolidated clays show significant peak-to-residual strength reduction due to dilation and particle rearrangement. The residual friction angle is important for analyzing reactivated landslides and pre-existing shear surfaces.
Should I use a direct shear test or a triaxial test?
Use the shear box when you need drained effective-stress parameters quickly for sands and gravels, when the failure plane is genuinely predetermined such as along a bedding surface or a soil-structure interface, or when you need large displacements to reach residual strength. Use the triaxial cell when drainage must be controlled and pore pressure measured, which is the case for undrained strength of clays, for stress paths other than simple shear, and for saturated specimens where back pressure is needed. The triaxial test also keeps the principal stress directions fixed, whereas they rotate in a shear box.
Why must the shearing rate be slow in a consolidated drained test?
The measured envelope is only an effective-stress envelope if the pore water pressure stays at zero throughout shearing. Shearing faster than the specimen can drain generates excess pore pressure that is neither measured nor controlled, so the reported normal stress is not the effective stress and the friction angle is wrong. The rate is therefore set from the consolidation stage of the same specimen: the time to reach failure must be several times the time the specimen needed to consolidate. In practice that means minutes for clean sand but many hours, sometimes days, for a plastic clay.
How does dilatancy inflate the measured friction angle in dense sand?
A dense sand must push grains up and over their neighbours to shear, so the specimen expands vertically against the applied normal load and part of the measured shear resistance is the work done in that expansion rather than true interparticle friction. The result is a peak friction angle several degrees above the constant-volume or critical-state value that the same sand reaches at large strain once it has loosened to a steady void ratio. Because dilatancy is suppressed by high confining stress, the peak envelope for dense sand is genuinely curved and a straight-line fit across a wide stress range will misrepresent both ends.
What does a poor regression fit or a negative cohesion intercept mean?
A low R-squared or an intercept below zero almost always signals a curved failure envelope rather than random scatter, most commonly in dense sand or overconsolidated clay where dilatancy fades as normal stress rises. The remedy is not to force a straight line through the whole set but to quote a secant friction angle over the stress range that matters for the problem, taking the line from the origin to the point at the design stress. A negative intercept has no physical meaning, so this tool clamps cohesion at zero, but the underlying curvature still deserves attention.
How is residual strength measured in a shear box?
Residual strength needs displacement far beyond what a single 60 mm box travel can provide, so the box is sheared, reversed to its starting position, and sheared again through many cycles until the shear stress stops falling. Each reversal disturbs the slip surface slightly, so the ring shear apparatus, which shears an annular specimen continuously in one direction without limit, gives cleaner residual values. Residual friction angles in high-plasticity clay can fall to the low single digits because platy clay minerals become aligned parallel to the slip plane, which is why reactivated landslides must be analysed with residual rather than peak parameters.
Why is the specimen area corrected as shearing proceeds?
As the top half of the box slides over the bottom half, the overlapping area shared by the two halves shrinks, so the same applied forces act over less and less contact. If the original nominal area, 3600 mm2 for a standard 60 mm box, is used all the way to failure, both the normal stress and the shear stress are underestimated at the moment failure occurs. Standard practice corrects the area for the measured horizontal displacement before computing stresses. The correction is small at the few millimetres of travel typical of a peak in sand, but it becomes significant in reversal testing for residual strength.
References
Background & Theory
History
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer ยท Editorial policy
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