Question 4 of 5: Mohr–Coulomb Tunnel-Stress Analysis
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
EGBC National Exam — Geological Engineering, 04-Geol-A4 Structural Geology, 2016-May. Open book; any non-communicating calculator permitted; 3 hours; 100 marks. The paper is printed as five lettered mega-questions (A–E): Question A "answer all" 20 T/F items (20 marks), Question B "any and only 8 of the following" (12 term pairs, 24 marks), Question C "any and only 6 of the following" (9 essay topics, 30 marks), Question D a single compulsory 13-mark Mohr–Coulomb/tunnel problem, and Question E a single compulsory 13-mark stereonet-and-deformation problem.
Reference texts: Davis & Reynolds, Structural Geology of Rocks and Regions, 3rd ed. (fold and fault mechanics, stress and strain, Mohr circle analysis); Fossen, Structural Geology, 2nd ed. (rheology, shear zones, fold classification, finite strain, stereographic pi-diagrams); Marshak & Mitra, Basic Methods of Structural Geology (stereonets, block diagrams); Hoek, Practical Rock Engineering; Bieniawski, Engineering Rock Mass Classifications (RQD/RMR, rock mass strength).
Check: the paper names the intact rock "granite" in the envelope-a preamble but then calls the jointed rock "this limestone" one sentence later — a contradiction in the printed paper. Both envelopes are used exactly as the numbers are given, treating the SG=2.7/τ=40+σn·tan45°/T=10 MPa properties as the "intact rock" surrounding the tunnel, regardless of which rock name is attached. The paper also prints "anisotropic isotropic (k=2)" (a self-contradiction); "anisotropic, k=2" is taken as intended since k=2 IS anisotropic.
Given. Intact rock (SG=2.7) surrounding a horizontal circular tunnel: Mohr–Coulomb envelope a and a pre-existing joint's direct-shear envelope b; anisotropic in-situ stress ratio k=2; roof tangential-stress concentration factor (3k−1), given directly by the question (the Kirsch-equation result at the crown of a circular opening); tunnel depth 350 m.
Given data
Quantity
Symbol
Value
Specific gravity of intact rock
SG
2.7
Unit weight
γ = SG·ρw·g
26.49 kN/m³ (0.026487 MPa/m)
Intact-rock cohesion, friction angle
c, φ
40 MPa, 45°
Intact-rock tensile strength
T
10 MPa
Joint cohesion, friction angle
cj, φj
5 MPa, 30°
Joint dip
—
45° N
In-situ stress ratio
k = σh/σv
2 (anisotropic)
Tunnel depth
z
350 m
Roof tangential stress
σ1(roof) = (3k−1)·P0
5×P0 (perpendicular to tunnel); σ3=0 (radial, dry)
Find. (1) The two labelled strength envelopes; (2) the Mohr circles for the in-situ state (c) and the roof's tangential–radial plane at 350 m (d); (3) the depth at which NEW fractures form in intact rock at the tunnel boundary; (4) the depth at which the EXISTING 45°-dipping joint is remobilized at the centre of the roof.
D(1). Both complete Mohr–Coulomb envelopes: intact rock (blue, c=40 MPa, φ=45°) with its tensile cutoff at σn=−10 MPa (dashed), and the pre-existing joint (red, cj=5 MPa, φj=30°, no tensile resistance so its cutoff is at σn=0). The joint envelope sits well below the intact envelope at every σn>0, so any stress state that reactivates the joint does so long before it would fracture intact rock.
Approach. Compute the in-situ vertical stress from the rock's unit weight, build σh from k, form the in-situ Mohr circle (c) and the roof's boundary circle (d) using the given (3k−1) concentration factor, then use tangency between a scaled circle and each envelope to find the two critical depths.
Part 1 — envelopes. The two strength envelopes are as given: intact τ = 40 MPa + σn·tan45° with a tensile cutoff at σn = −10 MPa (Figure D(1), blue); joint τ = 5 MPa + σn·tan30°, with zero tensile capacity so its own cutoff is at σn = 0 (red). This is a direct transcription of the two given criteria — no stress data needed yet.
Part 2 — in-situ and roof stresses at 350 m. Unit weight: $\gamma = SG\cdot\rho_w\cdot g = 2.7\times1000\times9.81 = 26{,}487\ \text{N/m}^3 = 0.026487\ \text{MPa/m}$. In-situ vertical stress: $$P_0 = \gamma z = 0.026487\times350 = \boxed{9.27\ \text{MPa}}$$ With k=2, the horizontal in-situ stress is $\sigma_h = kP_0 = 18.54$ MPa. 2a) In-situ circle (c): $\sigma_1=\sigma_h=18.54$, $\sigma_3=P_0=9.27$, so centre $C_c=(18.54+9.27)/2=13.91$ MPa, radius $R_c=(18.54-9.27)/2=\boxed{4.64}$ MPa. 2b) At the roof, the question's own (3k−1) factor gives the tangential stress $\sigma_1(\text{roof})=(3k-1)P_0=(6-1)(9.27)=\boxed{46.35\ \text{MPa}}$, with the radial stress σ3=0 (free, dry, unlined tunnel wall). The vertical plane striking perpendicular to the tunnel (label d) therefore has centre $C_d=46.35/2=23.18$ MPa, radius $R_d=23.18$ MPa — a circle passing through the origin, since σ3=0.
D(2)(3)(4). Circle (c), the small in-situ state, and circle (d), the large roof boundary state at the ACTUAL 350 m depth (C=R=23.2 MPa) — its top point already sits well above the joint envelope (grey 45° construction line = locus of σn=τ points for every σ3=0 boundary circle). The dashed orange circle (C=R=11.8) is the smaller, CRITICAL circle at which that same locus first touches the joint envelope, used in Part 4.
Part 3 — depth for NEW fractures in intact rock. With k=2>1, the roof (tangential factor 3k−1=5) is more critical than the sidewalls (factor 3−k=1), so the roof boundary governs where new intact fracturing first occurs. At the tunnel boundary σ3=0, so the intact Mohr–Coulomb criterion collapses to the UCS relation $\sigma_1=2c\tan(45^\circ+\phi/2)$: $$\sigma_{1,f}=2(40)\tan(67.5^\circ)=2(40)(2.4142)=193.1\ \text{MPa}$$ Setting $(3k-1)P_0=\sigma_{1,f}$: $$5P_0=193.1 \;\Rightarrow\; P_0=38.63\ \text{MPa} \;\Rightarrow\; z=\frac{P_0}{\gamma}=\frac{38.63}{0.026487}=\boxed{1459\ \text{m}}$$ New intact fracturing therefore requires a very deep tunnel — the rock's own strength is far higher than the pre-existing joint's (Part 4), matching the wide gap between the two envelopes in Figure D(1).
Part 4 — depth for existing-joint remobilization at the roof centre. A joint dipping 45° has its pole (normal) inclined 45° from horizontal, so the angle from the σ1-direction (horizontal, tangential) to the joint's normal is $\theta=45^\circ$ ($2\theta=90^\circ$). For any roof boundary state ($\sigma_3=0$, $\sigma_1$ variable with depth) at $2\theta=90^\circ$: $$\sigma_n=\frac{\sigma_1}{2},\qquad \tau=\frac{\sigma_1}{2}$$ — i.e. every such state plots on the grey 45° line ($\sigma_n=\tau$) in Figure D(2)(3)(4), always at the TOP of its own circle. Setting the joint slip criterion $\tau=c_j+\sigma_n\tan\phi_j$ with $\sigma_n=\tau=x$: $$x=5+x\tan30^\circ \;\Rightarrow\; x(1-0.5774)=5 \;\Rightarrow\; x=\boxed{11.83\ \text{MPa}}$$ This is the critical (dashed orange) circle in the figure, so $\sigma_1(\text{roof})_{crit}=2x=23.66$ MPa. Since $\sigma_1(\text{roof})=(3k-1)P_0=5P_0$: $$P_0=\frac{23.66}{5}=4.733\ \text{MPa} \;\Rightarrow\; z=\frac{4.733}{0.026487}=\boxed{178.7\ \text{m}}$$ This is the key engineering result: joint remobilization occurs at only 179 m — far LESS than both the tunnel's actual 350 m depth and the 1459 m needed to fracture intact rock. At the tunnel's real depth the roof circle (C=R=23.2, top point at σn=τ=23.2) already lies well past the joint envelope's tangent point (11.8), so the pre-existing roof joint is already reactivated at 350 m — joint-controlled instability, not new intact fracturing, governs this tunnel.
