Question 4 of 5: Mohr–Coulomb design of a circular drift and its concrete liner

Question 4: Mohr–Coulomb design of a circular drift and its concrete liner (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. Drift diameter $D=6.1$ m; hydrostatic in-situ stress $\sigma_0=35.2$ MPa; rock UCS $\sigma_c=104.8$ MPa; friction angle $\phi=30^{\circ}$; test confining stress $\sigma_3=31.2$ MPa (part b); concrete liner thickness $t=0.15$ m, liner UCS $=34.5$ MPa.

Find. (a) Cohesion $C$. (b) Axial failure stress at $\sigma_3=31.2$ MPa. (c) Factor of safety at the (unsupported) drift wall, with a Mohr–Coulomb diagram. (d) Internal support pressure needed for stability. (e) Resulting pressure and factor of safety at the concrete liner.

Approach. Convert $(C,\phi)$ using the standard Mohr–Coulomb angle $\psi=45^{\circ}+\phi/2$ and $\sigma_c=2C\tan\psi$. For a hydrostatic field ($k=1$), the Kirsch boundary tangential stress at an unsupported circular opening is uniform around the wall, $\sigma_{\theta\theta}=2\sigma_0$; compare this to $\sigma_c$ for the factor of safety. For part (d), set the internal-pressure-reduced boundary stress equal to the failure envelope evaluated at $\sigma_3=P_i$ (the pressure the support itself applies at the boundary) and solve for $P_i$. For part (e), treat the 15 cm liner as a thick-walled cylinder loaded externally by the grouting pressure from (d) and find its own tangential stress and factor of safety.

1. (a) Cohesion. With $\psi=45^{\circ}+\phi/2=45+15=60^{\circ}$ and $\sigma_c=2C\tan\psi$: $$C=\frac{\sigma_c}{2\tan\psi}=\frac{104.8}{2\tan60^{\circ}}=\boxed{30.25\ \text{MPa}}$$
2. (b) Axial stress at failure, $\sigma_3=31.2$ MPa. Using $\sigma_1=\sigma_c+\sigma_3\tan^2\psi$ with $\tan^2 60^{\circ}=3$: $$\sigma_1=104.8+31.2(3)=\boxed{198.4\ \text{MPa}}$$
3. (c) Factor of safety at the drift wall. In a hydrostatic field ($k=1$) the Kirsch crown-term vanishes ($\cos2\theta$ term drops out since $1-k=0$), so the unsupported boundary stress is uniform around the opening: $$\sigma_{\theta\theta}=2\sigma_0=2(35.2)=\boxed{70.4\ \text{MPa}}$$ Since $\sigma_3=0$ at an unsupported boundary, the rock's compressive capacity there is simply $\sigma_c$: $$FoS=\frac{\sigma_c}{\sigma_{\theta\theta}}=\frac{104.8}{70.4}=\boxed{1.49}$$ The Mohr–Coulomb diagram above confirms this graphically: the $(\sigma_3{=}0,\sigma_1{=}70.4)$ stress circle sits entirely below the failure envelope, with its top ($\tau_{\max}=35.2$) well short of the envelope value at that same $\sigma$ ($\tau_{env}\approx50.6$) — the drift is self-supporting at $FoS=1.49>1$, without any support pressure.
4. (d) Internal pressure for “just” stability. With an internal support pressure $P_i$ applied at the boundary, the reduced tangential stress $\sigma_{\theta\theta}=2\sigma_0-P_i$ must be checked against the failure envelope evaluated at the confining stress the support itself provides, $\sigma_3=P_i$: $\sigma_{\theta\theta}=\sigma_c+P_i\tan^2\psi$. Setting these equal and solving for $P_i$: $$P_i=\frac{2\sigma_0-\sigma_c}{\tan^2\psi+1}=\frac{70.4-104.8}{3+1}=\boxed{-8.60\ \text{MPa}}$$ A negative result has a clear physical meaning here, not an error: it confirms part (c)'s finding that the wall is already stable at $P_i=0$ ($FoS=1.49>1$), so the pressure “required” to just reach $FoS=1$ is less than zero — no internal support pressure is mechanically required for stability at this drift wall.
5. (e) Concrete liner pressure and factor of safety. The question specifies the grouting pressure equals the value found in (d); since that value is $\le0$, no positive pressure is actually applied — grout cannot exert a negative (tensile/suction) pressure on the liner, so the physically applied external pressure is $P_i^{+}=\max(P_i,0)=0$ MPa. Treating the 15 cm liner as a thick-walled cylinder ($r_i=3.05-0.15=2.90$ m, $r_o=3.05$ m) loaded externally by $P_i^{+}$, its own tangential stress is $$\sigma_{t,\text{liner}}=\frac{2r_o^2P_i^{+}}{r_o^2-r_i^2}=\frac{2(3.05)^2(0)}{3.05^2-2.90^2}=\boxed{0\ \text{MPa}}$$ so the liner carries essentially no ground-driven stress, and $FoS_{\text{liner}}=34.5/0\to\boxed{\infty}$ (not the limiting design element). This is the direct, physically consistent consequence of part (d)'s result: because the rock is already self-supporting, a liner installed here serves a constructability/durability role (controlling minor spalling, providing a smooth invert, corrosion/erosion protection) rather than a load-bearing one, and its structural design would instead be governed by construction/handling loads not evaluated in this question.