24-Bld-A6 Geotechnical Materials and Analysis · May 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, May 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.
Given. Two CU triaxial tests at failure, with total principal stresses $\sigma_1=\sigma_3+(\sigma_1-\sigma_3)$: Test 1: $\sigma_3=150$, $\sigma_1=253$ kPa, $u=82$ kPa. Test 2: $\sigma_3=300$, $\sigma_1=502$ kPa, $u=169$ kPa.
Find. (a) total-stress parameters $c$, $\phi$; (b) whether these test results, as given, can predict long-term (drained) stability, and if so how.
(a) Approach. Use the analytical Mohr-circle failure relation from the formula sheet, $\sigma_1=\sigma_3\tan^2(45°+\phi/2)+2c\tan(45°+\phi/2)$, applied to the *total* stresses from both tests (pore pressure is not needed for the total-stress envelope) — two equations, two unknowns ($c$, $N_\phi=\tan^2(45°+\phi/2)$).
Eliminate c by subtracting the two total-stress equations. $$253=150N_\phi+2c\sqrt{N_\phi},\qquad 502=300N_\phi+2c\sqrt{N_\phi}$$ $$502-253=(300-150)N_\phi\ \Rightarrow\ N_\phi=\frac{249}{150}=1.660$$
Solve for φ. $$\tan(45°+\phi/2)=\sqrt{1.660}=1.288\ \Rightarrow\ 45°+\phi/2=52.19°\ \Rightarrow\ \boxed{\phi\approx14.4°}$$
Back-substitute for c. $$253=150(1.660)+2c(1.288)=249.0+2.576c\ \Rightarrow\ \boxed{c\approx1.6\text{ kPa}}$$ (checked: both test points reproduce $\sigma_1$ to within rounding.)
Mohr circles at failure for both tests: grey circles/envelope use total stresses (part a); blue circles/envelope use effective stresses after subtracting u (part b) — note the effective envelope passes almost through the origin, typical of a normally-consolidated clay.
(b) Answer: YES — and this same Table 1 data is exactly what is needed. Long-term (end-of-life, fully drained) stability of an earthen structure is governed by the soil's effective-stress shear strength, $c'$, $\phi'$, not by the total-stress parameters found in part (a) (those describe only the short-term, end-of-construction/undrained condition). Total-stress parameters from a CU test are only valid for a time frame short enough that no drainage occurs in the field — the opposite of the "long term" condition asked about here.
However, because this was a CU test with pore-water pressure measurement, the effective principal stresses at failure are immediately available: $\sigma'=\sigma-u$. Subtracting the measured $u$ from both $\sigma_1$ and $\sigma_3$ in each test converts the same two data points into an effective-stress Mohr-circle problem, solved by the identical analytical procedure as part (a):
Compute effective principal stresses. Test 1: $\sigma_3'=150-82=68$, $\sigma_1'=253-82=171$ kPa. Test 2: $\sigma_3'=300-169=131$, $\sigma_1'=502-169=333$ kPa.
Solve the same two-equation system for $c'$, $\phi'$. $$N_\phi'=\frac{333-171}{131-68}=\frac{162}{63}=2.571\ \Rightarrow\ \tan(45°+\phi'/2)=1.604\ \Rightarrow\ \boxed{\phi'\approx26.1°}$$ $$171=68(2.571)+2c'(1.604)=174.9+3.207c'\ \Rightarrow\ c'\approx-1.2\text{ kPa}\approx0$$
Interpret. $c'\approx0$ (the small negative value is within the rounding/measurement scatter of a 2-point fit) is exactly what is expected for a normally-consolidated clay, corroborating the analysis. These effective parameters, $c'\approx0$, $\phi'\approx26°$, are the ones that should be used in an effective-stress slope-stability analysis (e.g. Bishop's method) to assess the long-term stability of the earthen structure.
