24-Bld-A6 Geotechnical Materials and Analysis · December 2018

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, December 2018. 3 hours, closed book, 100 marks. Section A (Q1–Q3) is compulsory; Section B directs "answer any three of Q4–Q7," 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.

Table 1 — CU triaxial results at failure

Confining stress, $\sigma_3$ (kPa)	Deviator stress, $\sigma_1-\sigma_3$ (kPa)	Pore-water pressure, $u$ (kPa)
150	103	82
300	202	169

Find. (a) Total-stress parameters $c,\phi$ analytically. (b) Whether Table 1 supports a short-term stability analysis.

Approach. At failure $\sigma_1=\sigma_3\tan^2(45°+\phi/2)+2c\tan(45°+\phi/2)$; two tests give two equations in the two envelope unknowns, solved directly (no Mohr-circle plotting needed).

1. Major principal stress at failure. Test 1: $\sigma_1=150+103=253\text{ kPa}$. Test 2: $\sigma_1=300+202=502\text{ kPa}$.
2. Solve for $N_\phi=\tan^2(45°+\phi/2)$. Subtracting the two failure equations eliminates $c$: $502-253=(300-150)N_\phi\Rightarrow N_\phi=1.660$, so $\phi=2(\arctan\sqrt{1.660}-45°)=\boxed{14.4°}$.
3. Solve for $c$. $c=\dfrac{253-150(1.660)}{2\sqrt{1.660}}=\boxed{1.55\text{ kPa}}$. Check: at $\sigma_3=300$, $300(1.660)+2(1.55)(1.288)=502.0$ ✓.

(b) Answer: yes, but only for the short-term (undrained) case, and with a qualification. Each CU test's deviator stress at failure directly gives the mobilized undrained shear strength at that specimen's own consolidation pressure, $s_u=(\sigma_1-\sigma_3)/2$ (51.5 kPa at $\sigma_3=150$; 101 kPa at $\sigma_3=300$) — exactly the quantity a $\phi=0$ (total-stress) short-term stability analysis needs, provided the field effective overburden/loading at each point of the earthen structure is matched against the CU test consolidated to a comparable stress. Used this way — as an $s_u$ vs. consolidation-stress relationship, or via the $c,\phi$ envelope evaluated at the field's own confining stress, rather than one constant pair applied everywhere — the data are genuinely useful for short-term stability. They are not, however, sufficient on their own for long-term (drained) stability, which instead needs the effective-stress envelope. The same pore-pressure measurements let us recover it as a check: $\sigma_3'=\sigma_3-u$, $\sigma_1'=\sigma_1-u$ gives $c'\approx-1.2\approx0\text{ kPa}$, $\phi'=26.1°$ — consistent with a normally consolidated clay ($c'\approx0$) and confirming the two stress states are governed by genuinely different envelopes.

Final Results — Question 6

Quantity	Value
Total-stress: $c$ / $\phi$	1.55 kPa / 14.4°
$s_u$ at $\sigma_3=150$ / $300$ kPa	51.5 kPa / 101.0 kPa
Effective-stress (check): $c'$ / $\phi'$	≈0 kPa / 26.1°
Short-term stability from Table 1?	Yes, via $s_u$ matched to field consolidation stress

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