Shear Strength of Soil (Mohr-Coulomb, Triaxial and Direct Shear)

NCEES FE Civil spec: Geotechnical Engineering, F. Shear strength. FE Reference Handbook: Civil Engineering chapter, Geotechnical section (Mohr-Coulomb failure, shear stress at failure).

What the FE Civil Spec Covers

Geotechnical section F asks you to compute and interpret the shear strength of soil. Expect problems that give strength parameters and ask for the strength on a plane, problems that give triaxial or direct shear results and ask for the friction angle or cohesion, and conceptual questions about drained versus undrained behavior. Shear strength feeds directly into bearing capacity, earth pressure and slope stability, so it is worth getting fluent.

Key Concepts

Soil fails in shear. Strength comes from friction between particles (proportional to the effective normal stress) plus any cohesion or cementation.
Effective stress controls strength. Water pressure carries no shear, so the frictional part depends on $σ^{'} = σ - u$ . Raising pore pressure lowers strength.
Drained vs undrained. Clean sands drain quickly and are analyzed with $c^{'} \approx 0$ and $ϕ^{'}$ . Saturated clays loaded quickly behave undrained and are analyzed with $ϕ = 0, c = s_{u}$ .
Tests. Direct shear measures normal and shear stress on a forced horizontal plane. The triaxial test applies a confining pressure $σ_{3}$ and then a deviator stress until failure; types are UU (unconsolidated undrained), CU (consolidated undrained) and CD (consolidated drained). The unconfined compression test is a UU test with $σ_{3} = 0$ .

Which Test Gives What

Test	Result	Typical use
Unconfined compression	$s_{u} = q_{u} /2$	Quick strength of intact saturated clay
UU triaxial	$s_{u}$ (total stress, phi = 0)	End-of-construction checks on clay
CU triaxial with pore pressure	Total and effective parameters	Staged loading, rapid drawdown
CD triaxial or direct shear	$c^{'}$ , $ϕ^{'}$	Sands, long-term clay stability

Typical drained friction angles: loose sand about 28 to 32 degrees, dense sand about 35 to 40 degrees, gravel higher. Dense sands dilate and show a peak strength above their large-strain (critical state) value; loose sands contract. In an undrained test on saturated soil, a contracting sample builds positive pore pressure and a dilating sample builds negative pore pressure, which is why the same soil can show very different total stress strengths depending on drainage.

Core Formulas

Mohr-Coulomb failure envelope (effective stress form):

τ_{f} = c^{'} + σ_{n}^{'} tan ϕ^{'}

Principal stresses at failure, written in terms of the strength parameters:

σ_{1}^{'} = σ_{3}^{'} tan^{2} (4 5^{\circ} + \frac{ϕ ^{'}}{2}) + 2 c^{'} tan (4 5^{\circ} + \frac{ϕ ^{'}}{2})

For a cohesionless soil, the friction angle follows directly from one Mohr circle:

sin ϕ^{'} = \frac{σ _{1}^{'} - σ _{3}^{'}}{σ _{1}^{'} + σ _{3}^{'}}

Stresses on a plane at angle $θ$ from the major principal plane:

σ_{n} = \frac{σ _{1} + σ _{3}}{2} + \frac{σ _{1} - σ _{3}}{2} cos 2 θ, τ = \frac{σ _{1} - σ _{3}}{2} sin 2 θ

The failure plane is at $θ_{f} = 4 5^{\circ} + ϕ /2$ . For undrained clay, $s_{u} = (σ_{1} - σ_{3})_{f} /2$ , and from an unconfined test $s_{u} = q_{u} /2$ .

Where It Is in the FE Reference Handbook

Search the Civil Engineering chapter, Geotechnical section. It shows the Mohr-Coulomb envelope with a Mohr circle and gives the shear stress at failure equation. The stress transformation equations above also appear in the Mechanics of Materials chapter under Mohr's circle.

Worked Example: Drained Triaxial Test on Sand

A consolidated drained triaxial test on a clean sand ( $c^{'} = 0$ ) uses an effective confining pressure of 100 kPa. The sample fails at a deviator stress of 200 kPa. Find (a) the friction angle, (b) the normal and shear stress on the failure plane, and (c) the shear strength of the same sand at an effective normal stress of 250 kPa.

Step 1: principal stresses at failure.

σ_{3}^{'} = 100 kPa, σ_{1}^{'} = 100 + 200 = 300 kPa

Step 2: friction angle.

sin ϕ^{'} = \frac{300 - 100}{300 + 100} = 0.5 \Rightarrow ϕ^{'} = 3 0^{\circ}

Step 3: failure plane. $θ_{f} = 4 5^{\circ} + 3 0^{\circ} /2 = 6 0^{\circ}$ , so $2 θ_{f} = 12 0^{\circ}$ .

σ_{n}^{'} = 200 + 100 cos 12 0^{\circ} = 150 kPa, τ_{f} = 100 sin 12 0^{\circ} = 86.6 kPa

Check with the envelope: $150 tan 3 0^{\circ} = 86.6$ kPa. The point lies on the envelope, as it should.

Step 4: strength at 250 kPa.

τ_{f} = 0 + 250 tan 3 0^{\circ} = 144.3 kPa

You can sketch the same problem in the Mohr's circle calculator with $σ_{x} = 300$ and $σ_{y} = 100$ to see the circle touch the envelope.

Common Mistakes

Using total stress in the Mohr-Coulomb equation when the parameters are effective stress parameters.
Taking the maximum shear stress (top of the circle) as the stress on the failure plane. The failure plane is at 45 + phi/2, not 45 degrees.
Forgetting to add the confining pressure to the deviator stress to get the major principal stress.
Using $s_{u} = q_{u}$ instead of $q_{u} /2$ for the unconfined compression test.
Doubling angles incorrectly: Mohr circle angles are twice the physical angles.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A soil has effective-stress strength parameters c′ = 10 kPa and φ′ = 30°. At effective normal stress σ′ = 100 kPa, what is its Mohr-Coulomb shear strength? See the worked solution or practice the Soils & Foundations deck
At a point, σx = 80 MPa, σy = 20 MPa and τxy = 30 MPa. The principal stresses are most nearly: See the worked solution or practice the Stress & Strain deck
At a point, σx = 80 MPa, σy = 20 MPa and τxy = 30 MPa. The maximum in-plane shear stress is most nearly: See the worked solution or practice the Stress & Strain deck

Related Calculators and Examples

Frequently Asked Questions

Should I use total or effective stress shear strength parameters on the FE exam?

Use effective stress parameters (c prime and phi prime) for drained, long-term problems such as sands and slopes after pore pressures dissipate. Use the undrained shear strength (phi = 0, c = su) for short-term loading of saturated clay, such as end-of-construction checks.

What angle does the failure plane make in a triaxial test?

The failure plane is inclined at 45 degrees plus phi/2 from the plane on which the major principal stress acts. For phi = 30 degrees that is 60 degrees.

How do I get undrained shear strength from an unconfined compression test?

The undrained shear strength is half the unconfined compressive strength, su = qu / 2, because the Mohr circle passes through the origin and the envelope is horizontal for phi = 0.

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