Shear Strength of Soil Calculator
Calculate soil shear strength using the Mohr-Coulomb failure criterion. Input cohesion, friction angle, normal stress, and pore pressure for drained.
Shear Strength of Soil Calculator
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Formula: tau = c + (sigma - u) * tan(phi)
Worked example โ Effective shear strength = 56.19 kPa
Formula
tau = c + (sigma - u) * tan(phi)
Shear strength tau equals the cohesion c plus the effective normal stress (total normal stress sigma minus pore water pressure u) multiplied by the tangent of the friction angle phi. This is the Mohr-Coulomb criterion in effective stress form. When excess pore pressure is present, the effective stress is reduced, lowering the available shear strength.
Worked Examples
Example 1: Sandy Soil with Low Cohesion
Problem:Calculate shear strength of soil with c = 10 kPa, phi = 30 degrees, normal stress = 100 kPa, and pore pressure = 20 kPa.
Solution:Effective normal stress = 100 - 20 = 80 kPa tan(30) = 0.5774 Effective shear strength = 10 + 80 * 0.5774 = 56.19 kPa Total shear strength = 10 + 100 * 0.5774 = 67.74 kPa
Result:Effective shear strength = 56.19 kPa
Example 2: Undrained Clay Analysis
Problem:A saturated clay has undrained cohesion cu = 50 kPa and phi-u = 0 degrees under undrained conditions.
Solution:Shear strength = cu + sigma * tan(0) = 50 + 0 = 50 kPa Undrained strength is constant regardless of normal stress. This is the total stress approach for short-term loading.
Result:Undrained shear strength = 50 kPa (constant)
Frequently Asked Questions
What is the Mohr-Coulomb failure criterion?
The Mohr-Coulomb criterion is the most widely used model for predicting shear failure in soil. It states that shear strength on any plane equals the cohesion intercept plus the effective normal stress on that plane multiplied by the tangent of the internal friction angle. The two parameters, cohesion c and friction angle phi, can be determined from direct shear tests or triaxial compression tests. The failure envelope is a straight line on a Mohr diagram, and failure occurs when the Mohr circle of stress touches this line.
What is the difference between drained and undrained shear strength?
Drained shear strength applies when pore water can freely drain during loading, so no excess pore pressure builds up. This condition uses effective stress parameters c-prime and phi-prime and governs long-term stability. Undrained shear strength (Su or cu) applies when loading is too fast for drainage, common in saturated clays under rapid loading. In undrained conditions, the friction angle is effectively zero, and all resistance comes from cohesion. The choice between drained and undrained analysis depends on the loading rate relative to the permeability of the soil.
How are cohesion and friction angle determined from laboratory tests?
The most common methods are the direct shear test and the triaxial compression test. In a direct shear test, a soil sample is sheared along a predetermined plane at different normal stresses, and the peak shear stress is recorded for each. Plotting shear stress versus normal stress gives a straight line whose y-intercept is the cohesion and whose slope is tan(phi). Triaxial tests apply confining pressure and then increase the axial load until failure, producing Mohr circles at different confining pressures. The failure envelope tangent to these circles defines c and phi.
What are typical friction angle values for different soil types?
Loose sand typically has a friction angle of 28 to 32 degrees, medium dense sand ranges from 32 to 36 degrees, and dense sand can reach 36 to 42 degrees. Gravel usually has higher values, from 34 to 48 degrees. Normally consolidated clay has effective friction angles of 20 to 28 degrees, while overconsolidated clay ranges from 25 to 35 degrees. Silt falls between sand and clay at about 26 to 34 degrees. These values depend on grain shape, size distribution, mineralogy, and the density of the deposit.
Why is pore water pressure subtracted from the normal stress?
Terzaghi's principle of effective stress states that water in the pores carries pressure but transmits no shear, so only the grain-to-grain contact stress, sigma minus u, mobilises friction. Raising the pore pressure pushes the grains apart and reduces the frictional term without changing the cohesion intercept. With c = 10 kPa, phi = 30 degrees and sigma = 100 kPa, a pore pressure of 20 kPa cuts strength from 67.7 kPa to 56.2 kPa. This is why rapid drawdown and rainfall infiltration trigger slope failures.
What is the difference between peak, critical state and residual friction angle?
A dense soil must dilate to shear, and that extra work shows up as a peak friction angle several degrees above its true frictional value. At large strain the sample reaches constant volume and mobilises the critical state angle, which depends on mineralogy and grain shape rather than density. On a surface that has already slipped a long distance, plate-shaped clay particles align parallel to the shear plane and the residual angle can fall below 10 degrees. Reactivated landslides must be analysed with residual, not peak, parameters.
What do the bearing capacity factors Nc, Nq and N-gamma represent?
They convert the same friction angle into the three contributions to a shallow foundation's ultimate bearing capacity in Terzaghi's equation: Nc scales the cohesion term, Nq scales the surcharge from soil above founding level, and N-gamma scales the self-weight of the failure wedge below the footing. All three grow steeply with phi. At phi = 30 degrees the Vesic values are roughly Nq = 18.4, Nc = 30.1 and N-gamma = 22.4, so a few degrees of error in phi changes a footing design substantially.
How do the Rankine coefficients Ka and Kp relate to the friction angle?
Rankine's active coefficient is Ka = tan squared of (45 - phi/2) and the passive coefficient is Kp = tan squared of (45 + phi/2); they are reciprocals of each other. Active pressure is the reduced lateral thrust a wall feels once it yields slightly away from the soil, while passive resistance is the much larger force generated when the wall pushes into the soil. At phi = 30 degrees, Ka is 0.333 and Kp is 3.0, a ninefold difference that dominates retaining wall and sheet pile design.
At what angle does the failure plane form in a triaxial specimen?
Mohr-Coulomb geometry places the failure plane at 45 + phi/2 degrees measured from the plane on which the major principal stress acts, so in a standard triaxial test with vertical loading the shear band appears at that angle from the horizontal. For phi = 30 degrees that is 60 degrees. This follows from the tangent point where the Mohr circle touches the envelope, and it is why the observed inclination of the rupture surface in a failed specimen is a useful check on the fitted friction angle.
Where does apparent cohesion in a partly saturated soil come from?
Above the water table, curved water menisci at grain contacts create matric suction that pulls particles together, adding an effective normal stress that behaves like cohesion. A moist sand bank can stand vertically for a while on this apparent cohesion alone, even though its true effective cohesion is nearly zero. The strength is temporary: saturation by rainfall destroys the menisci, and complete drying removes them too. Never rely on it in a permanent design, and use the saturated envelope for long-term stability.
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Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer ยท Editorial policy
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