Question 4 of 4: Choice Question – Rock Mechanics or Block Diagrams

Question D: Choice Question – Rock Mechanics or Block Diagrams (ONE and ONLY ONE of D-I or D-II – 15 total; both solved here)

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.

D-I – Mohr–Coulomb tunnel-stress analysis

Given. Marble surrounding a horizontal circular tunnel: intact-rock Mohr–Coulomb envelope and joint direct-shear strength, isotropic (k=1) in-situ stress, and a stress-concentration factor of 2× at the tunnel roof.

Find. (1) The two labelled strength envelopes; (2) the three Mohr circles describing the in-situ and roof stress states at 400 m; (3) the water pressure P_i needed to open tension fractures; (4) the depth at which NEW fractures form in intact rock at the tunnel wall; (5) the depth at which the EXISTING joint is remobilized at the centre of the roof.

Approach. Take the tunnel's unit weight from its SG to get the in-situ vertical stress P₀=γz; build the roof's three principal stresses from the given 2×/1×/0× stress-concentration ratios; construct each requested Mohr circle from two of those three principals; then use the effective-stress principle (tension fracture) and circle/envelope tangency (new vs. remobilized fractures) to solve Parts 3–5.

1. Part 1 — envelopes. The two strength envelopes are as given: intact τ = 30 MPa + σn·tan40° with a tensile cutoff at σn = −5 MPa (Figure D-I(1), blue); joint τ = 5 MPa + σn·tan25°, with zero tensile capacity so its own cutoff is at σn = 0 (red). No stress data is needed for this part — it is a direct transcription of the two given criteria onto one diagram.
2. Part 2 — roof stresses at 400 m. Unit weight: $\gamma = SG\cdot\rho_w\cdot g = 2.5\times1000\times9.81 = 24{,}525\ \text{N/m}^3 = 0.02453\ \text{MPa/m}$. In-situ vertical stress at z=400 m: $$P_0 = \gamma z = 0.02453\times400 = \boxed{9.81\ \text{MPa}}$$ Since k=1, σh=σv=P₀ — the in-situ state is isotropic, so 2a) plots as a single POINT at (9.81, 0) on the Mohr diagram (radius zero, no unique shear plane). At the roof, the stress-concentration ratios given in the question set the three roof principals: tangential $\sigma_1 = 2P_0 = 19.62\ \text{MPa}$ (perpendicular to the tunnel, across the roof), axial $\sigma_2 = P_0 = 9.81\ \text{MPa}$ (parallel to the tunnel), radial $\sigma_3 = 0$ (normal to the wall, dry/unlined). 2b) The vertical plane striking PARALLEL to the tunnel contains the axial and radial stresses (σ2, σ3): centre $C_b=(9.81+0)/2=4.91$ MPa, radius $R_b=(9.81-0)/2=4.91$ MPa. 2c) The vertical plane striking PERPENDICULAR to the tunnel contains the tangential and radial stresses (σ1, σ3): centre $C_c=(19.62+0)/2=9.81$ MPa, radius $R_c=(19.62-0)/2=9.81$ MPa.

3. Part 3 — water pressure for tension fracturing. Both roof circles from Part 2 share σ3(total) = 0. Injecting water behind the liner raises pore pressure P_i uniformly, translating each circle LEFT by P_i without changing its radius (Terzaghi: $\sigma_n' = \sigma_n - P_i$). Tension fracturing begins the instant the circle's leftmost (minimum-effective-stress) point reaches the intact tensile cutoff, $\sigma_3' = -T$: $$\sigma_3 - P_i = -T \;\Rightarrow\; 0 - P_i = -5 \;\Rightarrow\; \boxed{P_i = 5\ \text{MPa}}$$ This result is independent of which roof circle is used, because σ3=0 is common to both — consistent with the two circles in Figure D-I(2) both resting on the σn-axis at the origin.

4. Part 4 — depth for NEW fractures in intact rock. Around a k=1 (isotropic) circular opening, the boundary state is uniform everywhere: tangential $\sigma_1=2P_0$, radial $\sigma_3=0$ (not just at the roof). With σ3=0, the intact Mohr–Coulomb criterion collapses to the standard UCS relation $\sigma_1=2c\tan(45^\circ+\phi/2)$: $$\sigma_{1,f} = 2(30)\tan(65^\circ) = 2(30)(2.145) = 128.7\ \text{MPa}$$ Setting $2P_0 = \sigma_{1,f}$: $$P_0 = 64.34\ \text{MPa} \;\Rightarrow\; z = \frac{P_0}{\gamma} = \frac{64.34}{0.02453} = \boxed{2623\ \text{m}}$$ New intact-rock fracturing therefore requires a very deep tunnel — intact marble is far stronger than the pre-existing joint (Part 5), which is exactly what the two widely separated envelopes in Figure D-I(1) show.
5. Part 5 — depth for existing-joint remobilization at the roof centre. The joint dips 45° N (strikes E–W, parallel to the tunnel), so it lies IN the N–S vertical cross-section — exactly the Part-2(c) plane, where $\sigma_1=2P_0$ (horizontal, N–S) and $\sigma_3=0$ (vertical). A plane dipping 45° has its pole (normal) inclined 45° from horizontal, so the angle from the σ1-direction to the joint's normal is $\theta=45^\circ$ ($2\theta=90^\circ$). At $2\theta=90^\circ$: $$\sigma_n = \frac{\sigma_1+\sigma_3}{2} = P_0, \qquad \tau = \frac{\sigma_1-\sigma_3}{2} = P_0$$ (the orange point in Figure D-I(2), at the top of circle (c) — the maximum-shear plane). Setting the joint slip criterion $\tau=c_j+\sigma_n\tan\phi_j$: $$P_0 = 5 + P_0\tan25^\circ \;\Rightarrow\; P_0(1-0.4663) = 5 \;\Rightarrow\; P_0 = \boxed{9.37\ \text{MPa}}$$ $$z = \frac{9.37}{0.02453} = \boxed{382\ \text{m}}$$ This is the key engineering result: the joint remobilizes at only 382 m, LESS than the tunnel's actual 400 m depth. The existing joint in the roof is therefore already at or past its slip threshold at the tunnel's real depth — long before the 2623 m needed to fracture intact rock (Part 4) — so joint-controlled roof instability, not new intact fracturing, is the governing failure mode for this tunnel.

D-II – Block diagrams of fold and fault geometries

Each block below is drawn as an isometric crustal block with a North arrow for reference; the front face carries the diagnostic cross-section and the top face carries the map-view/plan expression where that is the more diagnostic view.

D-II(1) Recumbent, moderately-plunging box fold

The four coloured layers on the front face show the same box-fold profile repeated at successive structural levels, consistent with the fold train continuing through the multilayer. The axial surface (dashed) runs sub-horizontally, the defining trait of a recumbent fold — it has been rotated roughly 90° from an originally upright orientation, commonly by progressive shear during thrust transport. The hinge line itself (orange, on the top face) plunges moderately rather than lying dead flat, so the fold's outcrop trace on a map would curve as it is traced along strike.

D-II(2) Tight Dome-and-Basin (Type 1) double folding

Because F1 and F2 have comparable wavelength/amplitude and cross at a high angle, every point where an F1 antiformal crest crosses an F2 antiformal crest becomes a closed dome, and every F1-synform/F2-synform crossing becomes a closed basin — contoured, concentric closures rather than the open, curvilinear trend lines of a single fold generation. This is the exam's own "TYPE 1" label (Ramsay's classification), distinguishing it from the Type 3 coaxial pattern built in D-II(4).

D-II(3) Tight, upright, slightly plunging chevron folding (axis to the north), cut by later steep normal faulting with minor dextral offset

The chevron geometry (sharp, planar-limbed, angular hinges, constant limb dip magnitude on either side of each hinge) is diagnostic of buckling in a mechanically layered sequence with a strong competence contrast, and is upright here (axial surface near-vertical) rather than recumbent. Because the fold axis plunges only slightly to the north, the chevron pattern would appear only gently curved in map view before the later fault offsets it. Cross-cutting relationships establish relative age directly: the fault truncates every fold layer and its own trace is not folded, so faulting is unambiguously the younger event.

D-II(4) Type 3 (F2) coaxial refolded isoclinal (F1) fold — both axes trend E–W, plunge = 0; foliation S1=S2 dips 45° south

Because F1 and F2 share the same E–W, non-plunging hinge line, F2 does not generate the crossing "egg-carton" pattern of Type 1 (D-II(2)) — instead it re-tightens/re-orients the SAME hinge trace, producing the hook/crescent outcrop pattern illustrated conceptually in Question C7. The single foliation S1=S2 (rather than two distinct, crosscutting foliations) is diagnostic of coaxial refolding: because F2 reuses F1's kinematic axes, the two fold generations did not imprint two different cleavage orientations, only one, progressively strengthened fabric.