Mohr–Coulomb Soil Strength

The Mohr–Coulomb criterion

Soil has almost no tensile strength and very little stiffness in bending. Its useful strength is its resistance to shear — to one block of grains sliding past another. Charles-Augustin de Coulomb (1773) and Christian Otto Mohr (1900) gave us the relation engineers still use to predict that resistance:

τ_f = c′ + σ′ · tan φ′

where:
  τ_f = shear strength on the failure plane (kPa)
  c′  = effective cohesion intercept       (kPa)
  φ′  = effective angle of internal friction (degrees)
  σ′  = effective normal stress on the plane (kPa)
       σ′ = σ − u  (total stress minus pore water pressure)

Read it as two contributions added together. The frictional term σ′ tan φ′ says soil gets stronger the harder you press it together — squeeze the grains and they interlock and resist sliding. The cohesion term c′ is whatever strength remains when the confining stress drops to zero, supplied by cementation, electrochemical bonding, or capillary suction. A clean dry sand has c′ ≈ 0 and lives entirely off friction; a freshly cut clay bank can stand vertically for a while purely on cohesion before it slumps.

Why it is drawn as a circle touching a line

At a point in the ground the stress state is fully described by the major and minor principal stresses σ₁′ and σ₃′. Mohr's circle plots every possible (σ′, τ) pair on planes through that point as a circle of radius (σ₁′ − σ₃′)/2 centred at (σ₁′ + σ₃′)/2 on the normal-stress axis. The Mohr–Coulomb strength relation is a straight line of slope tan φ′ and intercept c′ in the same axes — the failure envelope.

As you load the element, the circle grows. While the circle sits entirely below the envelope, no plane is overstressed and the soil is stable. The moment the circle becomes tangent to the envelope, exactly one plane has reached its strength, and the soil shears along it. Crucially the failure plane is not the plane of maximum shear (the top of the circle) — it is the tangent point, which sits at an angle

θ_f = 45° + φ′/2   (from the major principal plane)

For φ′ = 30°, slip surfaces form at 60° to the horizontal — exactly the inclined shear bands you see slicing through a failed triaxial sample.

Typical strength parameters

Soil type	Friction angle φ′	Effective cohesion c′	Dominant behaviour	Where it governs
Loose clean sand	28–32°	~0 kPa	Purely frictional	Footing settlement, liquefaction
Dense sand / gravel	36–45°	~0 kPa	Frictional, dilatant	High bearing capacity, fills
Normally consolidated clay (drained)	20–28°	0–5 kPa	Frictional, low c′	Long-term slope stability
Stiff overconsolidated clay (drained)	20–25°	5–25 kPa	Cohesive + frictional	Cut slopes, deep excavations
Saturated clay (undrained, short term)	φ_u ≈ 0°	s_u = 20–200 kPa	Purely cohesive (total stress)	End-of-construction stability
Residual clay on a slip surface	φ_r′ = 8–15°	~0 kPa	Strain-softened residual	Reactivated landslides

Notice the same clay appears twice with wildly different "strength." That is not a contradiction — it reflects whether the pore water has time to drain. A rule of thumb: φ′ rarely exceeds 45° for natural soils, and once a slip surface has formed, the friction angle drops from its peak value to a much lower residual value, which is why old landslides reactivate so easily.

Worked example: shear strength of a sand layer

A clean dense sand has φ′ = 38° and c′ = 0. A foundation imposes an effective normal stress of σ′ = 150 kPa on a horizontal plane 3 m down. What is the shear strength available on that plane?

τ_f = c′ + σ′ · tan φ′
    = 0 + 150 · tan(38°)
    = 150 · 0.781
    = 117 kPa

Now suppose heavy rain raises the water table and adds 40 kPa of pore pressure that was not there before. The total normal stress is unchanged, but the effective stress falls:

σ′ = σ − u  →  σ′ drops from 150 by the new 40 kPa: 150 − 40 = 110 kPa
τ_f = 0 + 110 · tan(38°) = 110 · 0.781 = 86 kPa

The available strength dropped roughly 25% with no change in the applied load — purely because water now carries part of the normal stress. This single effect explains why so many slope failures happen during or just after heavy rainfall.

Worked example: a triaxial test on clay

A consolidated-drained triaxial test on a stiff clay gives, at failure, a cell (confining) pressure σ₃′ = 100 kPa and a major principal stress σ₁′ = 360 kPa. We can back out φ′ and check c′ using the principal-stress form of Mohr–Coulomb:

σ₁′ = σ₃′ · tan²(45° + φ′/2) + 2c′ · tan(45° + φ′/2)

Center of the Mohr circle: s = (σ₁′ + σ₃′)/2 = (360 + 100)/2 = 230 kPa
Radius of the Mohr circle:  t = (σ₁′ − σ₃′)/2 = (360 − 100)/2 = 130 kPa

For a c′ = 0 fit:  sin φ′ = t / s = 130 / 230 = 0.565
                   φ′ = 34.4°

The radius t is the largest shear the soil could mobilise; the failure plane carries less shear than that, but it is the plane that reaches the envelope first. A second test at a different cell pressure pins down both c′ and φ′ from the slope and intercept of the line drawn tangent to the two circles.

Effective stress: the heart of the matter

Terzaghi's principle — σ′ = σ − u — is what makes Mohr–Coulomb work in real, wet ground. Only the grain skeleton can resist shear; pore water cannot. So strength is controlled by effective stress, not total stress:

• Drained loading (slow, or permeable soil): excess pore pressure dissipates, σ′ tracks σ, and you analyse with c′ and φ′. This is the long-term condition for clays and the only condition for sands.
• Undrained loading (fast, low-permeability clay): water cannot escape, pore pressure spikes, and the short-term strength is set by a single total-stress parameter — the undrained shear strength s_u, with φ_u taken as zero.
• Rising water table or rainfall: u increases, σ′ falls, the envelope effectively drops toward the stress circle, and a previously stable mass can fail with no change in load.
• Seismic shaking of loose saturated sand: cyclic shear pumps pore pressure up until σ′ → 0. With no effective stress, τ_f → 0 and the sand momentarily behaves like a heavy liquid — liquefaction.

Where the criterion does the work

• Bearing capacity. Terzaghi's bearing-capacity factors N_c, N_q and N_γ are derived directly from c′ and φ′ via Mohr–Coulomb plasticity. A footing fails when the slip surface beneath it mobilises full shear strength along its length.
• Slope stability. Limit-equilibrium methods (Bishop, Morgenstern–Price) slice a slope and compare the driving weight to the Mohr–Coulomb shear resistance along a trial slip surface; the factor of safety is the ratio.
• Lateral earth pressure. Rankine's active and passive coefficients K_a = tan²(45° − φ′/2) and K_p = tan²(45° + φ′/2) come straight from the failure-tangent geometry, and set the loads retaining walls must resist.
• Pile capacity. Shaft friction is α·s_u (undrained) or β·σ′_v (drained), with the friction angle controlling end bearing.

Failure modes and trade-offs

• Drained shear (long term). Pore pressures have equalised; strength is the lowest sustained value. Cut slopes in stiff clay often stand for years, then fail decades later as swelling slowly drops the effective stress — the classic delayed failure.
• Undrained shear (short term). Rapid loading of soft clay — an embankment built quickly — can fail at end of construction before consolidation strengthens it. The fix is staged construction with drainage waits between lifts.
• Progressive / residual failure. Peak strength is not mobilised everywhere at once; once a slip band forms, strength on it drops to residual (φ_r′ as low as 8–10°), and failure propagates. Linear Mohr–Coulomb with peak parameters can dangerously overpredict here.
• Liquefaction. Loose saturated sand under cyclic load loses effective stress and therefore all Mohr–Coulomb strength. Densification, drains or ground improvement are the countermeasures.
• Envelope curvature. A straight line fitted to low-stress tests overpredicts strength at high confining stress (dams, deep foundations). Use stress-dependent friction angles or a curved/Hoek–Brown envelope there.

Mohr–Coulomb versus other strength models

	Mohr–Coulomb	Tresca (φ=0)	Drucker–Prager	Cam-Clay	Hoek–Brown
Form	Linear τ–σ envelope	Constant shear	Smooth cone	Critical-state ellipse	Curved (rock)
Parameters	c′, φ′	s_u	α, k	M, λ, κ	σ_ci, m_i, GSI
Intermediate σ₂	Ignored	Ignored	Included	Included	Ignored
Hardening / softening	None (peak only)	None	None	Yes	Limited
Numerics	Non-smooth corners	Simple	Smooth, easy	Complex	Moderate
Best for	Routine soil limit states	Undrained clay	FE plasticity	Soft clay, settlement	Fractured rock

Mohr–Coulomb endures because two physically meaningful parameters, measurable in a routine direct-shear or triaxial test, capture the dominant behaviour of most soils and feed every classical design formula. Its limitations — straight envelope, no intermediate-stress effect, peak-only strength — are well understood, and engineers reach for the more elaborate models only when those limitations bite.