Question 5 of 5: Circular Drift Stability, Support Pressure and Concrete Liner Design

Question 5: Circular Drift Stability, Support Pressure and Concrete Liner Design (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 ($a=3.05$ m), hydrostatic far-field stress $P_o=55.2$ MPa ($k=1$), rock $S_c=104.8$ MPa, $\phi=30^{\circ}$; concrete liner thickness $t=15$ cm, liner $S_c=34.5$ MPa.

Find. (a) Cohesion $C$. (b) Failure axial stress at $\sigma_3=31.2$ MPa confinement. (c) Factor of safety at the (unlined) drift wall, with a Mohr-Coulomb illustration. (d) Internal support pressure required for stability. (e) Factor of safety at the concrete liner's critical point under the grout pressure from (d).

Approach. $\Psi=45^{\circ}+\phi/2=60^{\circ}$ throughout. Part (a) inverts $S_c=2C\tan\Psi$. Part (b)/(c) apply the Mohr-Coulomb locus directly. Part (c)/(d) use the Kirsch solution for a circular opening in a hydrostatic ($k=1$) field, where the $\cos2\theta$ terms vanish and the wall tangential stress reduces to $\sigma_{\theta\theta}=(1+k)P_o-P_i=2P_o-P_i$ for an internal support pressure $P_i$; setting the available Mohr-Coulomb strength at confinement $\sigma_3=P_i$ equal to $\sigma_{\theta\theta}$ gives the minimum stabilizing $P_i$ (the thick-wall-cylinder formula supplied with the exam). Part (e) treats the 15 cm concrete liner as a thick cylinder loaded only by that external grout pressure, with the critical (maximum compressive) tangential stress at its inner face.

1. Part (a) — cohesion. With $\Psi=45^{\circ}+15^{\circ}=60^{\circ}$ ($\tan60^{\circ}=1.7321$): $$C=\frac{S_c}{2\tan\Psi}=\frac{104.8}{2(1.7321)}=\boxed{30.25\ \text{MPa}}$$
2. Part (b) — failure axial stress at $\sigma_3=31.2$ MPa. Using $\tan^2\Psi=\tan^2(60^{\circ})=3$: $$\sigma_1=\sigma_3\tan^2\Psi+S_c=31.2(3)+104.8=\boxed{198.4\ \text{MPa}}$$
3. Part (c) — stress and safety at the unlined drift wall. In a hydrostatic field ($k=1$) the Kirsch $\cos2\theta$ terms vanish identically, so the tangential (hoop) stress is uniform all around the unlined opening: $$\sigma_{\theta\theta}=(1+k)P_o=(2)(55.2)=\boxed{110.4\ \text{MPa}},\qquad \sigma_{rr}=0\ \text{(free, unlined surface)}$$ At $\sigma_3=0$, the available Mohr-Coulomb strength is just $S_c=104.8$ MPa, so $$FS=\frac{S_c}{\sigma_{\theta\theta}}=\frac{104.8}{110.4}=\boxed{0.949}$$ $FS<1$: the unlined drift wall is theoretically unstable, motivating parts (d)/(e).
4. Part (d) — required internal support pressure. With an internal (support) pressure $P_i$ acting at the wall, the tangential stress there drops to $\sigma_{\theta\theta}=2P_o-P_i$, while the wall's radial stress rises to $\sigma_{rr}=P_i$ (the new confining stress for the strength criterion). Setting available strength equal to demand, $2P_o-P_i=P_i\tan^2\Psi+S_c$, and solving: $$P_i=\frac{2P_o-S_c}{\tan^2\Psi+1}=\frac{2(55.2)-104.8}{3+1}=\frac{5.6}{4}=\boxed{1.4\ \text{MPa}}$$
5. Part (e) — concrete liner critical stress and FoS. The 15 cm liner spans $r_i=3.05-0.15=2.90$ m (inner, exposed face) to $r_o=3.05$ m (outer, grouted face), loaded only by the external grout pressure $P_i=1.4$ MPa found in (d). By the supplied thick-wall-cylinder formula, the tangential stress is largest (most compressive) at the liner's inner face: $$\sigma_t=\frac{2(r_o^2P_i)}{r_o^2-r_i^2}=\frac{2(3.05^2)(1.4)}{3.05^2-2.90^2}=\frac{2(9.3025)(1.4)}{0.8925}=\boxed{29.2\ \text{MPa}}\ \text{(compressive)}$$ $$FS_{\text{liner}}=\frac{S_{c,\text{liner}}}{\sigma_t}=\frac{34.5}{29.2}=\boxed{1.18}$$