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Shear Strength

Soil has no single strength — it has a friction angle, so its strength depends on depth. Which is why a sand slope stands at the same angle whatever its height.

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Soil strength is frictional, so it rises with confining stress rather than being a fixed value — τ = c′ + σ′tan φ′ — which is why a sand slope stands at the same angle whatever its height and why design uses the critical-state rather than the peak strength.

Strength that depends on depth

A steel bar has one yield stress wherever it is. A sand has a different strength at every depth, because its resistance is friction and friction depends on the normal stress. That single difference is what makes soil mechanics its own subject.

Mohr-Coulomb: τ = c′ + σ′tan φ′. The cohesion intercept is strength at zero confining stress, from electrochemical bonding between clay particles — which is why a clay trench face stands vertically and a sand one collapses immediately.

Why slope angle is independent of height

For a purely frictional soil, the driving force down a slope and the frictional resistance both scale with weight — so weight cancels and the stable angle is φ′ regardless of height. A sand pile and a hundred-metre dune stand at the same angle.

Dense against loose

Dense sand must dilate — grains ride over one another — so it shows a peak strength and then falls away. Loose sand contracts and rises to the same value with no peak. Both converge on the same critical state.

Peak strength is mobilised once and lost as soon as a slip surface forms, so design uses the critical state — or, on a pre-existing shear surface, the residual, which in a stiff plastic clay can be under half the peak. That is why old landslips reactivate at angles that look safe.

Measuring it

TestControls drainage?Suits
Direct shear boxPoorly; failure plane forcedQuick sand tests, teaching
TriaxialYes; pore pressure measuredThe standard — can reproduce field conditions
Unconfined compressionUndrained onlyQuick c_u on clay samples
Vane shearUndrained, in situSoft clay that cannot be sampled undisturbed

The test must match the field loading rate, or it answers the wrong question. Only the triaxial lets drained or undrained conditions be chosen deliberately.

The φ = 0 case

In an undrained saturated clay, additional confining stress goes entirely into the pore water, so σ′ is unchanged and the strength does not vary with confinement — a horizontal envelope. Short-term analysis then needs a single parameter c_u, which is why a field vane test is so useful.

What it is used for

Bearing capacity, slope stability and earth pressure — and each is analysed twice, undrained for end of construction and drained for the long term. Whichever gives the lower factor of safety governs.

The numbers you will be asked for

Mohr-Coulomb

τ = c′ + σ′ tan φ′

Undrained clay

τ = c_u, with φ_u = 0

Slope angle, dry sand

stable angle = φ′, independent of height

Skempton's pore pressure

Δu = B[Δσ₃ + A(Δσ₁ − Δσ₃)]

Typical φ′

30–35° loose sand · 35–45° dense sand · 20–30° clay

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One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.

Why does a sand dune stand at the same angle as a small sand pile?
Dense sand shows a peak strength and then falls. Why does design not use the peak?
Why is the undrained strength envelope for saturated clay horizontal?
Why is a triaxial test preferred over a direct shear box?

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