Question 3 of 5: Principal Stress Orientation from a Vertical Shear Fracture
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2013 — 04-Geol-A5 Rock Mechanics. Three-hour, closed-book exam; one of two approved calculators permitted, plus two sheets of the candidate's own rock-mechanics formulae/notes. Five questions of equal value (20 marks each); the paper instructs candidates to answer only the first 4 of 5 questions appearing in the answer book — all five are answered here as a complete study resource. Selected equations, RMR tables (Bieniawski 1989) and the Modified Lauffer stand-up-time chart are supplied at the back of the exam and are reproduced where used.
Reference texts: Bieniawski, Engineering Rock Mass Classifications (Wiley, 1989) — the RMR system, discontinuity-condition guidelines, and excavation/support tables used in Q1; Hoek, Practical Rock Engineering — Mohr-Coulomb strength parameters from triaxial data, Kirsch stress solutions around circular openings, and thick-wall liner design used in Q2/Q3/Q5; Brady & Brown, Rock Mechanics for Underground Mining (3rd ed.) — tributary-area pillar stress analysis and elastic pillar deformation used in Q4; EGBC Geoscience Professional Practice Guidelines for assumption-disclosure conventions.
Question 3: Principal Stress Orientation from a Vertical Shear Fracture (20 marks)
Given. A vertically-oriented planar shear fracture, formed by biaxial compressive ground stresses; Mohr-Coulomb parameters $C=11.5$ MPa, $S_c=73.4$ MPa.
Find. The orientation of the major ($\sigma_1$) and minor ($\sigma_3$) principal stresses relative to the vertical fracture, in the plane perpendicular to the fracture's strike.
Approach. Back out the friction angle from the given $C$ and $S_c$ via $S_c=2C\tan\Psi$, then use the standard Mohr-Coulomb geometry — the failure (shear) plane forms angle $\Psi=45^{\circ}+\phi/2$ with the plane on which $\sigma_3$ acts, equivalently angle $(90^{\circ}-\Psi)$ with the $\sigma_1$ direction — to orient $\sigma_1$ and $\sigma_3$ against the known vertical fracture.
Recover $\Psi$ and $\phi$ from the strength parameters.
$$\tan\Psi=\frac{S_c}{2C}=\frac{73.4}{2(11.5)}=3.191\ \Rightarrow\ \Psi=\arctan(3.191)=\boxed{72.6^{\circ}},\qquad \phi=2(\Psi-45^{\circ})=\boxed{55.2^{\circ}}$$
A friction angle this high is consistent with a strong, well-cemented crystalline rock, and is used as given — it follows directly from the two supplied strength parameters with no other assumption.
Orient the principal stresses against the vertical fracture. By the Mohr-Coulomb geometry, the shear (failure) plane sits at angle $\Psi=72.6^{\circ}$ from the plane on which $\sigma_3$ acts, i.e. at angle $(90^{\circ}-\Psi)=\boxed{17.4^{\circ}}$ from the $\sigma_1$ direction itself. Since the fracture is vertical, $\sigma_1$ (major principal stress) must therefore act at $17.4^{\circ}$ from vertical — steeply inclined, close to but not aligned with the fracture — while $\sigma_3$ (minor principal stress) acts perpendicular to $\sigma_1$, i.e. at $17.4^{\circ}$ from horizontal, both confined to the plane perpendicular to the fracture's strike (the plane the question asks be sketched).
Figure Q3 — the vertical shear fracture and the true directions of the major ($\sigma_1$) and minor ($\sigma_3$) principal ground stresses, both confined to the plane perpendicular to the fracture's strike.
Check: the figure supplied with the exam shows $\sigma_1$ pointing down and to the right at an acute angle to the fracture; the geometry reproduced here (major principal stress $17.4^{\circ}$ from vertical, minor principal stress $17.4^{\circ}$ from horizontal, mutually perpendicular) is the unique orientation consistent with the given $C$/$S_c$ pair and a vertical failure plane, independent of which specific acute-angle side the original sketch drew the arrow on.
