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CU tests on overconsolidated clay

Short answer

Compute σ3′ = σ3 − u, σ1′ = σ3′ + σd. From p′–q plot, Kf line gives tanα=0.4421, a=18.7 kPa. Then sinφ′ = tanα → φ′ = 26.2°, c′ = a/cosφ′ = 20.8 ≈ 21 kN/m². Effective parameters: c′ ≈ 21 kN/m², φ′ ≈ 26°.

The series of consolidated undrained tests on undisturbed samples of an overconsolidated clay were as below. Determine the shear parameters in terms of effective stresses.

Cell pressure (kN/m²)100200400600
Deviator stress at failure (kN/m²)300410610850
Pore water pressure (kN/m²)-45-1550110

Step 1: Effective principal stresses at failure

σ3′=σ3−u\sigma_3' = \sigma_3 - u and σ1′=σ3′+σd\sigma_1' = \sigma_3' + \sigma_d.

σ3\sigma_3σd\sigma_duuσ3′\sigma_3'σ1′\sigma_1'p′=σ1′+σ3′2p' = \frac{\sigma_1'+\sigma_3'}{2}q=σd2q = \frac{\sigma_d}{2}
100300-45145445295150
200410-15215625420205
40061050350960655305
6008501104901340915425

Step 2: Envelope. Draw the four effective-stress circles and the common tangent. Alternatively, fit a straight line through the points (p′,q)(p', q) (the KfK_f line), q=a+p′tan⁡αq = a + p'\tan\alpha. A least-squares fit gives:

tan⁡α=0.4421,a=18.7 kPa\tan\alpha = 0.4421, \qquad a = 18.7\ \text{kPa}

Step 3: Parameters

sin⁡ϕ′=tan⁡α=0.4421  ⇒  ϕ′=26.2∘\sin\phi' = \tan\alpha = 0.4421 \;\Rightarrow\; \phi' = 26.2^\circ c′=acos⁡ϕ′=18.70.8975=20.8 kPac' = \frac{a}{\cos\phi'} = \frac{18.7}{0.8975} = 20.8\ \text{kPa}

Answer: c′≈21c' \approx 21 kN/m² and ϕ′≈26∘\phi' \approx 26^\circ (in terms of effective stress).

✓ Answer checked by Claude Opus 5.5 (high effort)

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