16-Civ-A4 Geotechnical Materials and Analysis · December 2017

Question 5 of 6: Pore Pressure and Consolidation State from a CU Test

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Examinations — 16-Civ-A4 Geotechnical Materials and Analysis, December 2017. Closed book, 3 hours, 100 marks. Answer all six questions. Charts (rectangular-area influence chart, Newmark chart) and a formula sheet are provided at the back of the exam.

Reference texts: R.F. Craig & J. Knappett, Craig’s Soil Mechanics (8th ed.); B.M. Das, Principles of Geotechnical Engineering; R.D. Holtz, W.D. Kovacs & T.C. Sheahan, An Introduction to Geotechnical Engineering (2nd ed.); M. Budhu, Soil Mechanics and Foundations.

Question 5: Pore Pressure and Consolidation State from a CU Test (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A CU triaxial test with total stresses σ₃ = 100 kPa and deviator 60 kPa at failure (so σ₁ = 160 kPa), together with the soil’s unique effective envelope (CD): c′ = 0, φ′ = 30°, and the total-stress CU envelope c = 0, φ = 13.3°.

Given data

Confining pressure, σ3	100 kPa (total)
Deviator stress at failure	60 kPa ⇒ σ1 = 160 kPa
Effective (CD) parameters	c′ = 0, φ′ = 30°
Total (CU) parameters	c = 0, φ = 13.3°

Find. The pore-water pressure u_w at failure, the Skempton pore-pressure coefficients A and B, and whether the clay is normally or over-consolidated (with three reasons).

Approach. The effective-stress failure envelope is a property of the soil alone, so the CU test’s effective stresses must lie on it; imposing that condition (with c′ = 0) yields u_w directly. B follows from saturation and A from the measured pore pressure over the deviator stress.

1. Effective-stress failure condition. With c′ = 0, at failure $\sigma_1'=\sigma_3'\tan^{2}\!\left(45^\circ+\tfrac{\phi'}{2}\right)$. Here $\tan^{2}(45^\circ+15^\circ)=\tan^{2}60^\circ=3$.
2. Solve for the pore pressure. Substituting σ′ = σ − u_w: $$(160-u_w)=3\,(100-u_w)\;\Rightarrow\;2u_w=140\;\Rightarrow\;\boxed{u_w=70\ \text{kPa}}.$$ Check: the total-stress angle is $\sin\phi_{cu}=\dfrac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3}=\dfrac{60}{260}=0.231\Rightarrow\phi_{cu}=13.3^\circ$, matching the given CU value.
3. Coefficient B. The specimen is saturated (a CU test is run on a saturated sample), so $$B=1.0.$$
4. Coefficient A. After consolidation under the cell pressure the excess pore pressure is zero, so all of u_w is generated by the deviator stage (Δσ₃ = 0, Δσ_d = 60): $$A_f=\frac{\Delta u}{\Delta\sigma_d}=\frac{70}{60}=\boxed{1.17}.$$
5. Consolidation state. Three consistent indicators show the clay is normally consolidated: (i) the effective envelope passes through the origin (c′ = 0); (ii) large positive pore pressure at failure with A_f = 1.17 > 0.5, i.e. strongly contractive behaviour; (iii) the CU total-stress angle (13.3°) is far below φ′ (30°), the signature of positive shear-induced pore pressure. (An over-consolidated clay would show c′ > 0, A_f near or below zero, and even negative pore pressure.)

Results

Quantity	Value
Pore-water pressure at failure, uw	70 kPa
Skempton B	1.0 (saturated)
Skempton Af	1.17
Consolidation state	Normally consolidated

The result illustrates why the drained (CD) parameters c′, φ′ — not the total-stress CU angle — are the correct inputs for long-term (drained) stability analyses such as the earth-dam problem in Question 1(8): the CU total angle understates the true frictional resistance precisely because of the positive pore pressure quantified here.

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