Question 5 of 5: Single-Plane-of-Weakness Theory — Two Orthogonal Fracture Sets

Question 5: Single-Plane-of-Weakness Theory — Two Orthogonal Fracture Sets (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. Two orthogonal fracture sets: A ($C_A=0.100$ MPa, $\phi_A=20^\circ$), B ($C_B=0$, $\phi_B=35^\circ$); intact rock $\sigma_1=75+5.29\sigma_3$ (MPa); minor principal stress $\sigma_3=10$ MPa for part (a), $\sigma_3=0$ for part (b).

Find. The strength $\sigma_1(\beta)$ as a function of the angle $\beta$ between $\sigma_1$ and the plane of weakness, for both cases, and the governing lower-bound envelope.

Approach. Apply Jaeger's single-plane-of-weakness formula to set A directly (angle $\beta$ from $\sigma_1$ to A's plane) and to set B using $(90^\circ-\beta)$ since B is orthogonal to A; the intact-rock line is an upper bound (strength can never exceed it); the governing strength at each $\beta$ is the LOWEST of the three.

1. Single-plane-of-weakness formula. For a joint at angle $\beta$ to $\sigma_1$ (valid for $\phi_j<\beta<90^\circ$; outside this range the joint cannot be critically stressed and only the intact rock or the other set governs): $$\sigma_1=\sigma_3+\frac{2(C_j+\sigma_3\tan\phi_j)}{\big(1-\cot\beta\tan\phi_j\big)\sin2\beta}$$ Because B is orthogonal to A, its own controlling angle (measured from ITS plane to $\sigma_1$) is $(90^\circ-\beta)$ when $\beta$ is measured to A's plane, so $\sigma_{1,B}(\beta)=\sigma_{1,B\text{-formula}}(90^\circ-\beta)$.
2. Part (a), $\sigma_3=10$ MPa — evaluate the three candidates and take the lower envelope. Intact: $\sigma_1=75+5.29(10)=\boxed{127.9\ \text{MPa}}$ (a flat ceiling, independent of $\beta$). Joint A alone is minimized at its own critical angle $\beta^*=45^\circ+\phi_A/2=55^\circ$: $$\sigma_{1,A}(55^\circ)=10+\frac{2(0.100+10\tan20^\circ)}{(1-\cot55^\circ\tan20^\circ)\sin110^\circ}\approx\boxed{20.7\ \text{MPa}}$$ Joint B alone (its own axis) is minimized at $\beta^*=45^\circ+\phi_B/2=62.5^\circ$, giving $\sigma_{1,B}\approx36.9$ MPa — higher than A's minimum despite B's larger friction angle, because A's nonzero cohesion is outweighed here by the low confinement making the cohesion term relatively more valuable for A's shallower critical angle. Scanning $\beta=0^\circ$ to $90^\circ$ (Figure Q5(a)): near $\beta\approx0^\circ$ and $\beta\approx90^\circ$ neither joint can be critically stressed, so intact rock (127.9 MPa) governs; joint B governs a low-to-mid $\beta$ band (dropping to its own 36.9 MPa minimum near $\beta\approx27$–35$^\circ$ in this orthogonal framing); joint A takes over and governs from about $\beta\approx30^\circ$ through the global minimum at $\beta=55^\circ$ and symmetrically back up to about $\beta\approx80^\circ$. $$\boxed{\text{Governing (lowest) strength: }\sigma_1\approx20.7\ \text{MPa at }\beta=55^\circ\text{ (joint A critical)}}$$
3. Part (b), $\sigma_3=0$. Intact strength drops to the bare UCS: $\sigma_1=75+5.29(0)=\boxed{75\ \text{MPa}}$. For joint A, the cohesion term alone survives ($\sigma_3\tan\phi_A$ vanishes), so A's minimum strength drops sharply but stays positive (≈0.29 MPa at its own critical angle, since $C_A=0.1$ MPa is still nonzero). For joint B ($C_B=0$), the numerator $2(C_B+\sigma_3\tan\phi_B)=2(0+0)=0$ identically — a cohesionless joint has exactly zero shear strength once unconfined, for every $\beta$ at which it is kinematically free to slip ($\phi_B<\beta<90^\circ$ on its own axis, translating to roughly $5^\circ$–$55^\circ$ in the orthogonal-$\beta$ framing used for the plot). So the lower-bound envelope collapses to essentially $\sigma_1\approx0$ across almost the whole orientation range, rising to the intact UCS (75 MPa) only in the narrow bands near $\beta\approx0^\circ$/$90^\circ$ where NEITHER joint can be critically oriented. $$\boxed{\text{At }\sigma_3=0:\ \sigma_1\approx0\text{ for most }\beta\text{ (set B governs, cohesionless)},\ \text{rising to }75\ \text{MPa only near }\beta\approx0^\circ,90^\circ}$$
   Check: the qualitative conclusion — an unconfined specimen containing a cohesionless joint at (or near) its critical orientation has essentially zero strength — is a genuine, well-known consequence of the theory, not a computational artifact; it is the reason field rock masses with open, cohesionless discontinuities are treated as having negligible tensile/unconfined strength in that orientation band.