Question 1 of 5: Kirsch boundary stress — single tunnel and a second, parallel tunnel

Question 1: Kirsch boundary stress — single tunnel and a second, parallel tunnel (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. Depth $z=400$ m; Tunnel A diameter 4 m ($a_A=2$ m); unit weight $\gamma=26\ \text{kN/m}^3$; UCS $\sigma_c=60.0$ MPa; tensile strength $\sigma_t=3.0$ MPa; two stress ratios $k=\sigma_h/\sigma_v \in \{0.3,\,2.57\}$; Tunnel B diameter 6 m ($a_B=3$ m), driven parallel at the same centreline level, 10 m centre-to-centre spacing.

Find. Whether the rock strength on Tunnel A's boundary is exceeded for (a) $k=0.3$ and (b) $k=2.57$; and how Tunnel B's presence changes the stress state in the 5 m pillar between the two openings.

Approach. Compute the pre-mining vertical stress $\sigma_v=\gamma z$, then apply the Kirsch circular-opening boundary solution at the crown ($\theta=90^{\circ}$) and sidewall ($\theta=0^{\circ}$) for each $k$, comparing the extreme compressive value against the UCS and the extreme tensile value against the tensile strength. For the second tunnel, superpose Tunnel B's general (non-boundary) Kirsch field onto the near-boundary point of Tunnel A (and vice-versa), subtracting the once-double-counted far-field term, to check whether the pillar stress changes either conclusion.

1. Vertical in-situ stress. $$\sigma_v=\gamma z = 26\ \text{kN/m}^3 \times 400\ \text{m} = 10{,}400\ \text{kPa} = \boxed{10.4\ \text{MPa}}$$
2. Part (a) — $k=0.3$: boundary stresses at the crown and sidewall. For a circular opening in a biaxial field, the Kirsch solution reduces at the boundary ($r=a$) to $\sigma_{\theta\theta}=\sigma_v[(1+k)+2(1-k)\cos2\theta]$, giving $\sigma_{\theta\theta}=\sigma_v(3k-1)$ at the crown/floor ($\theta=90^{\circ}$) and $\sigma_{\theta\theta}=\sigma_v(3-k)$ at the sidewalls ($\theta=0^{\circ}$). $$\text{crown: }\sigma_v(3(0.3)-1)=10.4(-0.1)=\boxed{-1.04\ \text{MPa (tension)}}\qquad \text{side: }\sigma_v(3-0.3)=10.4(2.7)=\boxed{28.08\ \text{MPa (compression)}}$$ The crown tension (1.04 MPa) is below the 3.0 MPa tensile strength, and the sidewall compression (28.08 MPa) is well below the 60.0 MPa UCS — strength is NOT exceeded anywhere on the boundary for $k=0.3$.
3. Part (b) — $k=2.57$: boundary stresses. $$\text{crown: }10.4(3(2.57)-1)=10.4(6.71)=\boxed{69.78\ \text{MPa (compression)}}\qquad \text{side: }10.4(3-2.57)=10.4(0.43)=\boxed{4.47\ \text{MPa (compression)}}$$ The sidewall value is comfortably below UCS, but the crown compressive stress (69.78 MPa) exceeds the 60.0 MPa UCS — strength IS exceeded at the crown/floor for $k=2.57$.
4. Part (c) — second, 6 m tunnel: pillar interaction by Kirsch superposition. With Tunnel A centred at the origin and Tunnel B 10 m away, the critical near-boundary point is P on A's boundary, 8 m from B's centre ($r=10-a_A=8$, $\theta=0$ relative to B). B's general (non-boundary) Kirsch tangential stress there is $$\sigma_{\theta\theta,B}(r,0)=\frac{\sigma_v}{2}\Big[(1+k)\big(1+\tfrac{a_B^2}{r^2}\big)+(1-k)\big(1+3\tfrac{a_B^4}{r^4}\big)\Big]$$ Superposing (adding B's contribution to A's own boundary value, then subtracting the once-counted far-field $\sigma_v$ to avoid double-counting it): $$\sigma_{\text{total}}(P)=\underbrace{\sigma_v(3-k)}_{\text{A alone}}+\underbrace{\sigma_{\theta\theta,B}(r{=}8,0)}_{\text{B's field at P}}-\sigma_v$$ For $k=0.3$: $28.08+11.57-10.4=\boxed{29.25\ \text{MPa}}$, still well below the 60 MPa UCS. For $k=2.57$: $4.47+12.53-10.4=\boxed{6.60\ \text{MPa}}$ at the pillar-facing point — smaller than the crown value from part (b), so the second tunnel does not change either conclusion: the $k=0.3$ case remains fully within strength everywhere, and the $k=2.57$ case still fails at the crown (not at the pillar), governed by the result already found in part (b).