Question 4 of 5: Choice Question – Rock Mechanics or Yield-Criteria Sketching
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, 2013-Dec. Open book; any non-communicating calculator permitted; 3 hours. The paper is printed as five lettered mega-questions (A–E): Question A instructs "answer 15 of these 20" T/F items, Question B "any and only 8 of the following" (14 term pairs), Question C "any and only 5 of the following" (9 essay topics), Question D "ONE and ONLY ONE of D-I or D-II," and Question E "ONE and ONLY ONE of E-I or E-II."
Check: page 1's NOTES state "FOUR questions constitute a complete exam paper," yet the paper prints FIVE lettered mega-questions (A–E). Read together with "choices in each main question," this is taken to mean a complete SELECTED paper is A + B + C + (D-I or D-II) + (E-I or E-II) — i.e. every lettered question is compulsory, with the internal choice living inside D and E (and inside A/B/C's own "answer N of M" sub-instructions) — not that one of A/B/C/D/E may be skipped outright.
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); Marshak & Mitra, Basic Methods of Structural Geology (stereonets, block diagrams); Hoek, Practical Rock Engineering; Bieniawski, Engineering Rock Mass Classifications (RQD/RMR, rock mass strength).
Question D: Choice Question – Rock Mechanics or Yield-Criteria Sketching (ONE and ONLY ONE of D-I or D-II – 15 total; both solved here)
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=50+σn·tan45°/T=10 MPa properties as the "intact rock" surrounding the tunnel, regardless of which rock name is attached.
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).
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
50 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
400 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 400 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-I(1). Both complete Mohr–Coulomb envelopes: intact rock (blue, c=50 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 = 50 MPa + σn·tan45° with a tensile cutoff at σn = −10 MPa (Figure D-I(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 400 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: $$P_0 = \gamma z = 0.026487\times400 = \boxed{10.59\ \text{MPa}}$$ With k=2, the horizontal in-situ stress is $\sigma_h = kP_0 = 21.19$ MPa. 2a) In-situ circle (c): $\sigma_1=\sigma_h=21.19$, $\sigma_3=P_0=10.59$, so centre $C_c=(21.19+10.59)/2=15.89$ MPa, radius $R_c=(21.19-10.59)/2=\boxed{5.30}$ MPa. 2b) At the roof, the question's own (3k−1) factor gives the tangential stress $\sigma_1(\text{roof})=(3k-1)P_0=(6-1)(10.59)=\boxed{52.97\ \text{MPa}}$, with the radial stress σ3=0 (free, dry, unlined tunnel wall). The vertical plane striking perpendicular to the tunnel (label d) therefore has centre $C_d=52.97/2=26.49$ MPa, radius $R_d=26.49$ MPa — a circle passing through the origin, since σ3=0.
D-I(2)(3)(4). Circle (c), the small in-situ state, and circle (d), the large roof boundary state at the ACTUAL 400 m depth (C=R=26.5 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 $\sigma_1=2c\tan(45^\circ+\phi/2)$: $$\sigma_{1,f}=2(50)\tan(67.5^\circ)=2(50)(2.4142)=241.4\ \text{MPa}$$ Setting $(3k-1)P_0=\sigma_{1,f}$: $$5P_0=241.4 \;\Rightarrow\; P_0=48.28\ \text{MPa} \;\Rightarrow\; z=\frac{P_0}{\gamma}=\frac{48.28}{0.026487}=\boxed{1823\ \text{m}}$$ 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-I(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-I(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 $\sigma_1(\text{roof})=(3k-1)P_0=5P_0$: $$P_0=\frac{23.66}{5}=4.732\ \text{MPa} \;\Rightarrow\; z=\frac{4.732}{0.026487}=\boxed{178.7\ \text{m}}$$ This is the key engineering result: joint remobilization occurs at only 179 m — far LESS than both the tunnel's actual 400 m depth and the 1823 m needed to fracture intact rock. At the tunnel's real depth the roof circle (C=R=26.5, top point at σn=τ=26.5) already lies well past the joint envelope's tangent point (11.8), so the pre-existing roof joint is already reactivated at 400 m — joint-controlled instability, not new intact fracturing, governs this tunnel.
Given. The source's plotted Mohr–Coulomb yield criterion for intact rock (dry) is defined by two points on its upper envelope: (σn,τ)=(0,10) and (70,50) MPa. Separately, a limestone rockmass at 3000 m depth has a N–S horizontal principal stress = 0.5×σvertical, and an E–W principal stress equal to the mean (volumetric) stress at that point.
D-II(a-d). Intact-rock envelope (blue, both signs of τ by symmetry) through the two given points, with cohesion c=10 MPa (τ-intercept) and its extrapolated tensile cutoff at σn=−17.5 MPa; (b) a small uniaxial-tension circle (σ1=0, σ3=−17.5, dashed green) tangent to the cutoff; (c) a larger confined-compression circle (σ3>0, dashed purple) tangent to the positive-σn envelope; (d) the joint envelope (red), same friction angle as the intact rock but zero cohesion, passing through the origin.
Approach. Read cohesion and friction angle directly off the two given points on the plotted envelope, extrapolate to σn-intercept for the tensile cutoff (parts a–d); separately, build the 3D stress tensor at 3000 m from the given ratios, then apply Terzaghi's principle for part g.
Part (a) — three key strength parameters. The envelope's slope through (0,10) and (70,50) is $\tan\phi=\dfrac{50-10}{70-0}=0.5714 \Rightarrow \phi=\boxed{29.7^\circ}$. The τ-intercept at σn=0 gives cohesion $c=\boxed{10\ \text{MPa}}$. Extrapolating the same line to τ=0 gives the (approximate) tensile-strength cutoff: $$\sigma_t = c/\tan\phi = 10/0.5714 = \boxed{17.5\ \text{MPa}}\ (\sigma_n=-17.5\ \text{MPa})$$ — all three values are marked directly on Figure D-II(a-d).
Part (b) — uniaxial tension. Uniaxial tension has $\sigma_1=0$, $\sigma_3=-T=-17.5$ MPa, giving a small circle of centre $-8.75$ and radius $8.75$ MPa, tangent to the tensile cutoff at exactly one point (dashed green in the figure) — the failure state a rock reaches under pure axial pulling.
Part (c) — confined (triaxial) compression. For an illustrative confining stress $\sigma_3=20$ MPa, the intact criterion in $\sigma_1$–$\sigma_3$ form ($N_\phi=\tan^2(45^\circ+\phi/2)=2.97$) gives $\sigma_{1,f}=2c\sqrt{N_\phi}+\sigma_3 N_\phi = 2(10)(1.723)+20(2.97)=34.5+59.4=\boxed{93.9\ \text{MPa}}$, plotted as the larger dashed-purple circle (centre 56.9, radius 36.9 MPa) tangent to the positive-σn branch of the envelope — the confining pressure both shifts the circle right and lets it grow to a much larger radius before touching the (now higher) envelope, illustrating why confinement raises strength.
Part (d) — joint envelope. A continuous, fully formed planar joint has no cohesion (asperities are already sheared through) but, per the question, the SAME frictional strength as the intact rock: $\tau=\sigma_n\tan29.7^\circ$, a straight line through the origin (red) sitting below the intact envelope everywhere σn>0 — a joint of this kind fails at lower shear stress than intact rock at every confining level, exactly as in D-I.
Part (e)–(g) — 3D stress tensor at 3000 m.
Check: the source gives no unit weight for this limestone (D-I's SG=2.7 belongs to the OTHER rock type in that sub-question); a typical limestone specific gravity of 2.6 (γ=25.51 kN/m³) is assumed here, per Goodman, Engineering Geology: Rock in Engineering Construction.
Part (e) — stress tensor. $\sigma_v=\gamma z = (2.6\times1000\times9.81\times10^{-6})(3000)=\boxed{76.52\ \text{MPa}}$. N–S: $\sigma_{NS}=0.5\sigma_v=\boxed{38.26\ \text{MPa}}$. E–W is the mean stress $p=(\sigma_v+\sigma_{NS}+\sigma_{EW})/3$; solving $\sigma_{EW}=p$ self-consistently gives $\sigma_{EW}=(\sigma_v+\sigma_{NS})/2=\boxed{57.39\ \text{MPa}}$ (down, N–S, and E–W axes as drawn in the source's block-diagram figure).
Part (f) — maximum shear stress and dip. With $\sigma_1=\sigma_v=76.52$ (vertical, max) and $\sigma_3=\sigma_{NS}=38.26$ MPa (N–S horizontal, min): $$\tau_{max}=\frac{\sigma_1-\sigma_3}{2}=\boxed{19.13\ \text{MPa}}$$ occurring on planes at 45° to both σ1 and σ3 — since σ1 is vertical and σ3 is horizontal N–S, these maximum-shear planes strike E–W and DIP $\boxed{45^\circ}$ (either north or south).
Part (g) — effect of Pf=45 MPa. Applying Terzaghi's principle to each total stress: $\sigma_v'=76.52-45=31.52$, $\sigma_{NS}'=38.26-45=\boxed{-6.74\ \text{MPa}}$ (negative — TENSILE), $\sigma_{EW}'=57.39-45=12.39$ MPa. Because the N–S effective stress goes tensile while the other two stay compressive, the rock is driven into extension specifically along N–S — the mechanism most likely to open EXTENSIONAL (hydraulic) fractures/veins striking perpendicular to the tensile direction, i.e. striking E–W and near-vertical (parallel to σv and σEW, both still compressive), matching the general fault-valve mechanism developed in Question C9.
