Question 4 of 5: Elastic Constant and Mohr-Coulomb Parameters from a Zero-Axial-Strain Triaxial Path

Question 4: Elastic Constant and Mohr-Coulomb Parameters from a Zero-Axial-Strain Triaxial Path (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. Zero-net-axial-strain path throughout ($\varepsilon_a\equiv0$); initial $\sigma_a$–$p$ line through the origin; onset of yield at $p=85$ MPa, $\sigma_a=39.1$ MPa; post-yield slope angle $29^\circ$ (constant).

Find. (a) The elastic constant implied by the initial (elastic) slope. (b) $\sigma_c$ (unconfined compressive strength), $c$ (cohesion) and $\phi$ (friction angle) for the Mohr-Coulomb criterion.

Approach. Because $\varepsilon_a=0$ is enforced throughout by the servo-control, the initial elastic segment's slope $\sigma_a/p$ is a direct algebraic function of Poisson's ratio; because the confining pressure $p$ grows faster than $\sigma_a$ throughout this particular path ($\sigma_a/p<1$ always, as shown below), $p$ is in fact the MAJOR principal stress and the axial direction $\sigma_a$ is the MINOR principal stress once yield begins — the post-yield slope then gives a second Mohr-Coulomb equation, and the two together (with the yield point itself) solve for $\sigma_c$, $c$, $\phi$.

1. Part (a) — elastic constant. For a triaxial element with axial stress $\sigma_a$ and equal radial stress $p$ on both lateral faces, Hooke's law gives $\varepsilon_a=\dfrac{1}{E}\big[\sigma_a-2\nu p\big]$. Enforcing $\varepsilon_a=0$ for every point on the initial straight line (which starts at the origin) requires $\sigma_a=2\nu p$, i.e. the plotted slope IS $2\nu$. Since the line is straight through the origin, its slope equals the ratio at the last point on it — the yield point itself: $$\text{slope}=\frac{\sigma_a}{p}=\frac{39.1}{85}=0.460\ \Rightarrow\ \nu=\frac{0.460}{2}=\boxed{0.23}$$ This is the elastic constant asked for: Poisson's ratio $\nu\approx0.23$, a physically reasonable value for a weak, foliated rock such as soapstone.
2. Identify which stress is major/minor. Because $\sigma_a=2\nu p$ with $2\nu=0.46\lt1$, $\sigma_a\lt p$ at every point on the elastic path, and the post-yield slope ($\tan29^\circ=0.554\lt1$) keeps $\sigma_a\lt p$ afterward too. So throughout the whole test the confining pressure $p$ is the LARGER stress: $\sigma_1=p$ (major principal, radial) and $\sigma_3=\sigma_a$ (minor principal, axial) — this is a triaxial-extension-type stress path, not ordinary triaxial compression.
3. Part (b) — friction angle from the post-yield slope. Once yielding begins, the stress path tracks the Mohr-Coulomb failure envelope itself, so $d\sigma_3/d\sigma_1=\tan(29^\circ)$ along it. From $\sigma_1=\sigma_3\tan^2\Psi+S_c$ (given on the exam formula sheet, $\Psi=45^\circ+\phi/2$): $$\frac{d\sigma_3}{d\sigma_1}=\frac{1}{\tan^2\Psi}=\tan(29^\circ)=0.554\ \Rightarrow\ \tan^2\Psi=1.804\ \Rightarrow\ \Psi=53.3^\circ$$ $$\phi=2(\Psi-45^\circ)=2(8.33^\circ)=\boxed{16.7^\circ}$$
4. $\sigma_c$ and $c$ from the yield point. Substituting the yield point ($\sigma_1=p=85$, $\sigma_3=\sigma_a=39.1$ MPa) into $\sigma_1=\sigma_3\tan^2\Psi+S_c$: $$S_c=\sigma_c=85-39.1(1.804)=85-70.5=\boxed{14.5\ \text{MPa}}$$ $$S_c=2c\tan\Psi\ \Rightarrow\ c=\frac{S_c}{2\tan\Psi}=\frac{14.5}{2(1.343)}=\boxed{5.38\ \text{MPa}}$$
   Check: the exam text asks to "determine $\sigma_a$, c and $\phi$," but $\sigma_a=39.1$ MPa is already given data at the yield point — read as a typo for $\sigma_c$ (the unconfined compressive strength), which is the natural third Mohr-Coulomb constant to pair with $c$ and $\phi$, and is solved for accordingly above.