← STEM Studio
Interactive Explainer · Geotechnical

Soil Mechanics

Soil shear strength comes from two genuinely different sources — cohesion (particles sticking together regardless of load) and internal friction (resistance that grows with normal stress). The Mohr-Coulomb line combines both.

15 kPa
30°
80 kPa
Mohr-Coulomb Failure Envelope
τ = c + σ·tan(φ) = 15 + 80·tan(30°) = 61.2 kPa shear strength

About Soil Mechanics

Soil mechanics quantifies how soil deforms and fails under stress. The Mohr-Coulomb failure criterion — shear strength equals cohesion plus normal stress times the tangent of the friction angle — is the standard model for predicting soil shear strength, combining two physically distinct sources of strength into a single, usable design equation.

Cohesion: Strength Independent of Confining Stress

Cohesion represents soil's inherent shear strength that exists even with zero normal (confining) stress applied — arising from physical particle bonding, cementation, or electrochemical attraction between fine clay particles. Cohesive soils (typically fine-grained clays) retain some shear strength even unconfined, which is why a clay sample can stand as a vertical-sided block, at least temporarily, while a cohesionless sand cannot.

Friction Angle: Strength That Grows with Confining Stress

The friction angle term represents strength arising from actual physical friction and interlocking between soil particles — this component grows directly with normal stress, since higher confining stress presses particles together harder, increasing the frictional resistance to sliding between them. Cohesionless soils (sands, gravels) derive essentially all their shear strength from this frictional term, which is why loose dry sand has no strength at all when unconfined (poured into a pile, it simply flows to its natural angle of repose).

Why the Combined Envelope Matters for Design

As shown above, the Mohr-Coulomb line predicts the maximum shear stress a soil can sustain at a given normal stress before failing — any combination of shear and normal stress falling above this line represents a failure condition. This is the direct basis for slope stability analysis, foundation bearing capacity calculations, and retaining wall design, all of which ultimately check whether the actual stress state at critical points stays safely below this failure envelope.

Frequently asked questions

Why does dry sand have essentially zero cohesion?

Sand particles are relatively large, smooth, and lack the physical bonding or electrochemical attraction mechanisms that give fine clay particles cohesive strength — sand's shear strength comes almost entirely from interparticle friction and interlocking, which is why it needs confining stress (or a stable slope angle at or below its angle of repose) to have any shear strength at all.

Can a soil have both cohesion and a meaningful friction angle at the same time?

Yes — many real soils, particularly mixed or partially cemented soils, exhibit both cohesive and frictional strength components simultaneously, which is exactly why the general Mohr-Coulomb equation includes both terms rather than assuming a soil is purely cohesive or purely frictional.

What does it mean if a stress state plots above the Mohr-Coulomb failure line?

It means the combination of shear and normal stress at that point exceeds what the soil can actually sustain — physically, this predicts shear failure will occur, which is exactly the condition geotechnical stability analyses (slopes, foundations, retaining structures) are designed to avoid with an adequate safety margin below the failure envelope.

🎓

Try our STEM Learning Studio

More calculators, simulators, and guides for this discipline.

Related tools & guides

Geotechnical EngineeringFoundation EngineeringSTEM Studio