24-Bld-A6 Geotechnical Materials and Analysis · Undated paper

Question 6 of 7

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

Notes on this paper

07-BLD-A6 Geotechnical Materials and Analysis, National Examinations (printed exam date May 2019). 3 hours, closed book, 100 marks. Section A (Q1–Q3) is compulsory (40 marks); Section B directs "answer any three of Q4–Q7" (60 marks), but for completeness this solution answers all four.

Reference texts: B.M. Das, Principles of Geotechnical Engineering, 9th ed.; B.M. Das, Principles of Foundation Engineering, 9th ed.

Question 6 (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.

(i) Given.

Quantity	Symbol	Value
Effective cohesion	$c'$	20 kN/m²
Effective friction angle	$\phi'$	30°
Deviator stress at failure	$(\sigma_1-\sigma_3)_f$	180 kN/m²
All-round (cell) pressure	$\sigma_3$ (total)	100 kN/m²

Find. The pore-water pressure $u$ at failure.

Approach. The deviator stress is unaffected by pore pressure ($\sigma_1'-\sigma_3'=\sigma_1-\sigma_3$), so combine that fact with the effective-stress Mohr–Coulomb failure criterion (in terms of $\sigma_3'$) to solve for $\sigma_3'$, then recover $u$ from the known total stress.

1. Effective-stress failure criterion. $\sigma_1'=\sigma_3'\tan^2\!\left(45+\frac{\phi'}{2}\right)+2c'\tan\!\left(45+\frac{\phi'}{2}\right)$. With $\phi'=30^\circ$: $\tan(60^\circ)=1.732$, so $K_p=\tan^2(60^\circ)=3.00$.
2. Use the (pore-pressure-independent) deviator stress. $\sigma_1'-\sigma_3'=180$, so $\sigma_1'=\sigma_3'+180$. Substituting into the failure criterion: $\sigma_3'+180=3.00\,\sigma_3'+2(20)(1.732)$, i.e. $\sigma_3'+180=3.00\,\sigma_3'+69.28$.
3. Solve for $\sigma_3'$. $180-69.28=2.00\,\sigma_3' \Rightarrow \sigma_3'=55.36\text{ kN/m}^2$, and $\sigma_1'=55.36+180=235.36\text{ kN/m}^2$.
4. Pore pressure. $u=\sigma_3(\text{total})-\sigma_3'=100-55.36=\boxed{44.6\text{ kN/m}^2}$. Check via $\sigma_1$: $\sigma_1(\text{total})=100+180=280$; $u=280-235.36=44.6\text{ kN/m}^2$ — consistent.

Quantity	Result
$\sigma_3'$	55.4 kN/m²
$\sigma_1'$	235.4 kN/m²
Pore-water pressure, $u$	44.6 kN/m²

(ii) Answer: NO. A UU test is sheared with no drainage permitted at any stage, so the sample's water content — and hence its undrained shear strength $c_u$ (with $\phi_u\approx0$ for a saturated clay) — stays fixed at whatever it was in the field at the time of sampling. That total-stress strength is only representative of short-term, end-of-construction conditions, when field pore pressures have not yet had time to change. An earthen structure's long-term stability, by contrast, is governed by conditions once excess pore pressures have fully dissipated (or built up to a new steady-state seepage condition) — which requires the effective-stress strength parameters $c',\phi'$ from a CU test (with pore-pressure measurement) or a CD test, not the UU test's total-stress $c_u$. Using UU results for a long-term analysis risks a dangerously unconservative answer if the soil's long-term drained strength envelope is actually lower than the short-term undrained one predicts (a real risk for normally consolidated to lightly overconsolidated clays).

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