Question 1 of 6: Circular Tunnel — Boundary Stress and Design

Question 1: Circular Tunnel — Boundary Stress and 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.

1.1 — Boundary stress prediction (Kirsch solution)

Given. Depth H = 500 m; unit weight γ = 28 kN/m³; horizontal-to-vertical field stress ratio K = 0.40; Mohr-Coulomb strength c = 15 MPa, φ = 28°; tensile strength σ_t = 0.1 MPa.

Find. Whether the rock at the periphery of the circular opening fails in compression or tension, and by how much margin.

Approach. Compute the vertical field stress from overburden, apply the Kirsch elastic solution for the tangential (hoop) boundary stress of a circular opening at the sidewall and at the roof/floor, then compare each to the Mohr-Coulomb compressive strength at zero confinement (the boundary is a free surface, so σ₃=0 there) and check for tension against σ_t.

1. Field stresses. $p=\gamma H = 28\ \text{kN/m}^3\times 500\ \text{m}=14{,}000\ \text{kPa}=\boxed{14.0\ \text{MPa}}$ (vertical); horizontal $\sigma_h=0.40p=5.6\ \text{MPa}$.
2. Kirsch boundary tangential stresses. For a circular opening in a biaxial field with vertical stress p and ratio K = σ_h/p, the tangential stress is $\sigma_\theta=(3-K)p$ at the sidewall and $\sigma_\theta=(3K-1)p$ at the roof/floor. Substituting K = 0.40, p = 14.0 MPa: $\sigma_{\theta,side}=(3-0.40)(14.0)=\boxed{36.4\ \text{MPa}}$ (compression); $\sigma_{\theta,roof}=(3\times0.40-1)(14.0)=(0.20)(14.0)=\boxed{2.8\ \text{MPa}}$ (compression, since $3K-1=0.20>0$ — with K between 1/3 and 3 the whole boundary stays in compression, so no location reaches the tensile strength σ_t).
3. Compressive strength at the free boundary. At the excavation wall the radial stress is zero (σ₃=0, a free surface), so the Mohr-Coulomb criterion reduces to the uniaxial compressive strength $\sigma_c=\dfrac{2c\cos\phi}{1-\sin\phi}=\dfrac{2(15)(\cos28^\circ)}{1-\sin28^\circ}=\dfrac{26.49}{0.5305}=\boxed{49.9\ \text{MPa}}$.
4. Compare and predict the response. Sidewall factor of safety $FS_{side}=\sigma_c/\sigma_{\theta,side}=49.9/36.4=\boxed{1.37}$; roof/floor $FS_{roof}=49.9/2.8=17.8$. Both locations stay below the compressive strength (no yielding predicted by this linear-elastic, continuum check) and no boundary point is ever in tension for K=0.40, so σ_t is never approached.

The excavation is predicted to remain stable everywhere on a purely elastic-continuum basis, but the sidewall margin (FS ≈ 1.4) is thin for a permanent opening — typical design targets for permanent mine openings are FS ≥ 1.5–2.0 — and does not allow for the strength-reducing effect of the 500 m of overlying rock's discontinuities, blast damage, or long-term (time-dependent) strength loss, none of which this linear-elastic check captures.

1.2 — Alternative design proposal

Given. The sidewall of a circular opening concentrates stress to 2.6× the vertical field stress because the field is markedly anisotropic (K = 0.40 ≠ 1), while the roof/floor is barely stressed at all (0.2×p).

Find. A reshaped excavation profile that removes this stress concentration and raises the sidewall factor of safety.

Approach. A circular profile is the correct shape only for an isotropic (K=1) field; for an anisotropic field the boundary stress can be made exactly uniform by matching the opening's width-to-height ratio to the stress ratio itself, W/H = K, i.e. by elongating the opening along the direction of the larger (here vertical) principal stress.

1. Matching ellipse design rule. For an elliptical opening of width W (horizontal) and height H (vertical) in a biaxial field, the boundary tangential stresses at the two axis ends are $\sigma_{\theta,side}=p\!\left[1-K+2\dfrac{W}{H}\right]$ and $\sigma_{\theta,roof}=p\!\left[K-1+2K\dfrac{H}{W}\right]$. Setting them equal gives $\left(\dfrac{W}{H}\right)^{2}+(1-K)\dfrac{W}{H}-K=0$, whose positive root is $W/H=K=\boxed{0.40}$ — i.e. $H/W=2.5$, an opening two-and-a-half times taller than it is wide, with its long axis parallel to the larger (vertical) field stress.
2. Predicted uniform boundary stress. Under this matched condition the boundary stress everywhere equals $\sigma_\theta=p+\sigma_h=14.0+5.6=\boxed{19.6\ \text{MPa}}$, well below the circular sidewall value of 36.4 MPa.
3. Revised factor of safety. $FS=\sigma_c/\sigma_\theta=49.9/19.6=\boxed{2.55}$, essentially double the circular design's sidewall margin.

A tall, narrow (1:2.5) elliptical or ovaloid profile that mimics this ratio — a high, narrow tunnel elongated in the vertical (major-stress) direction — is the recommended alternative. The rule cannot be inverted: a flattened 2.5:1 opening in this field would raise the sidewall stress to $p[1-K+2(2.5)]=78.4$ MPa, more than double the circular value, and put the roof into $p[K-1+2K(0.4)]=-3.92$ MPa, i.e. 3.92 MPa of tension against a tensile strength of only 0.1 MPa. If the tunnel function does not permit that aspect ratio, an economical fallback is to keep the circular shape but add pattern rock bolting and mesh at the sidewalls (the locations identified in 1.1 as most heavily stressed), sized to the 36.4 MPa boundary stress rather than the near-zero roof/floor stress; either option specifically targets the location the calculation shows is actually at risk, rather than uniformly over-supporting the whole opening.