Mohr–Coulomb Yield Criteria – Strength Parameters and 3D Stress Tensor: 18-Geol-A4 Structural Geology, December 2015, Q4

Question D: Mohr–Coulomb Yield Criteria – Strength Parameters and 3D Stress Tensor (18 total)

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.

Check: the source's printed axes only carry numeric gridlines (no coordinates typed on the envelope line itself), so the three strength parameters in part (a) are read directly off the plotted envelope by digitizing the drawn line against the printed axis gridlines (vertex/tension-cutoff at σn≈−10 MPa, τ-intercept at σn=0 ≈15.5 MPa, envelope reaching ≈(60,49) MPa at its drawn end) — consistent with the question's own instruction to "Estimate or Calculate." The source gives no unit weight for this limestone; a typical limestone specific gravity of 2.6 (γ=25.51 kN/m³) is assumed, per Goodman, Engineering Geology: Rock in Engineering Construction's equivalent sub-question.

Given. The plotted Mohr–Coulomb yield criterion for intact rock (dry), read off the figure. Separately, a limestone rockmass at 2000 m depth has a N–S horizontal principal stress = 0.5×σ_vertical, and an E–W principal stress equal to the mean (volumetric) stress at that point; pore pressure in part (g) is 45 MPa.

Given data
Quantity	Symbol	Value
Envelope τ-intercept (read from plot)	c	≈15.5 MPa
Envelope slope (read from plot)	tanφ	≈0.556 (φ≈29.1°)
Tension cutoff (read from plot)	σt	≈−10 MPa
Limestone specific gravity (assumed)	SG	2.6
Depth	z	2000 m
N–S horizontal stress ratio	σ_NS/σv	0.5
Pore pressure (part g)	P_f	45 MPa

Find. (a) cohesion, friction angle, tensile strength; (b)–(d) the tension, confined-compression and joint-envelope Mohr circles; (e) the 3D principal stress tensor at 2000 m; (f) maximum shear stress and the dip of the planes it acts on; (g) what forms if P_f=45 MPa.

D(a-d). Intact-rock envelope (blue) read off the plot: cohesion c≈15.5 MPa (τ-intercept), friction angle φ≈29.1° (slope), tensile cutoff at σn≈−10 MPa (dashed). The joint envelope (red, dashed) shares the same friction angle but has zero cohesion, so it sits below the intact envelope at every σn>0.

Approach. Read cohesion, friction angle and tensile cutoff directly off the plotted envelope (part a), sketch the tension and confined-compression circles that are tangent to it (b, c) and the zero-cohesion joint line (d); separately build the 3D in-situ stress tensor at 2000 m from the given ratios (e), find the absolute maximum shear stress and its plane orientation (f), then apply Terzaghi's principle to see which principal direction goes tensile under P_f=45 MPa (g).

Part (a) — three key strength parameters. Reading the envelope's τ-intercept at σn=0 gives cohesion $c\approx\boxed{15.5\ \text{MPa}}$. Its slope gives the friction angle: $\tan\phi\approx0.556\Rightarrow\phi\approx\boxed{29.1^\circ}$. The envelope's vertical tension-cutoff segment sits at $\sigma_n\approx\boxed{-10\ \text{MPa}}$, i.e. a tensile strength $T_0\approx10$ MPa — all three are marked directly on Figure D(a-d). (Internal check: $c+T_0\tan\phi=15.5-10(0.556)=9.9\ \text{MPa}$, matching the plotted vertex height of ≈10 MPa, confirming the three readings are mutually consistent.)
Part (b) — uniaxial tension. Uniaxial tension has $\sigma_1=0$, $\sigma_3=-T_0=-10$ MPa, giving a small Mohr circle of centre $-5.0$ and radius $\boxed{5.0\ \text{MPa}}$, tangent to the tensile cutoff — 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 ($N_\phi=\tan^2(45^\circ+\phi/2)=2.894$) gives $$\sigma_{1,f}=2c\sqrt{N_\phi}+\sigma_3 N_\phi = 2(15.5)(1.701)+20(2.894)=52.7+57.9=\boxed{110.6\ \text{MPa}}$$ plotted as a much larger circle (centre 65.3, radius 45.3 MPa) tangent to the positive-σn branch of the envelope — the confining pressure both shifts the circle right and lets it grow far larger 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: $\tau=\sigma_n\tan29.1^\circ$, a straight line through the origin (red, dashed) sitting below the intact envelope everywhere σn>0 — a joint of this kind slips at lower shear stress than intact rock fractures, at every confining level.

D(e). Principal stress tensor at 2000 m: σv (vertical, largest), σEW (intermediate), σNS (smallest, horizontal N–S) — axes as drawn in the source's N/E/down triad.

Part (e) — stress tensor. Vertical (overburden) stress: $$\sigma_v=\gamma z = (2.6\times1000\times9.81\times10^{-6})(2000)=\boxed{51.01\ \text{MPa}}$$ N–S horizontal stress: $\sigma_{NS}=0.5\,\sigma_v=\boxed{25.51\ \text{MPa}}$. E–W is stated to equal the mean (volumetric) stress $p=(\sigma_v+\sigma_{NS}+\sigma_{EW})/3$ at the SAME point; solving $\sigma_{EW}=p$ self–consistently: $$\sigma_{EW}=\frac{\sigma_v+\sigma_{NS}}{2}=\frac{51.01+25.51}{2}=\boxed{38.26\ \text{MPa}}$$ (check: $p=(51.01+25.51+38.26)/3=38.26=\sigma_{EW}$ ✓).
Part (f) — maximum shear stress and dip. Ordering the three principal stresses, $\sigma_1=\sigma_v=51.01$ MPa (largest, vertical) and $\sigma_3=\sigma_{NS}=25.51$ MPa (smallest, horizontal N–S), with $\sigma_2=\sigma_{EW}=38.26$ MPa (intermediate): $$\tau_{max}=\frac{\sigma_1-\sigma_3}{2}=\boxed{12.75\ \text{MPa}}$$ These maximum-shear planes bisect the σ1 and σ3 directions — since σ1 is vertical and σ3 is horizontal N–S, the planes of maximum shear contain the E–W (σ2) direction and dip at exactly $\boxed{45^\circ}$ (either north or south), i.e. they strike E–W.
Part (g) — effect of P_f=45 MPa. Applying Terzaghi's principle to each total principal stress: $$\sigma_v'=51.01-45=6.01,\quad \sigma_{EW}'=38.26-45=-6.74,\quad \sigma_{NS}'=25.51-45=\boxed{-19.49\ \text{MPa}}$$ Only $\sigma_{NS}'$ drops below the intact rock's tensile cutoff found in part (a) ($-19.49<-10$ MPa) — $\sigma_{EW}'$ is negative but not yet past the cutoff, and $\sigma_v'$ stays compressive. With the effective minimum principal stress exceeding the tensile strength specifically along N–S, the rock is driven into HYDRAULIC EXTENSION FRACTURING (tension veins) oriented perpendicular to σNS: near-vertical fractures striking E–W (parallel to the σv–σEW plane, both of which remain safely compressive), matching the fault-valve mechanism developed in Question C9.

Question D — Final results
Part	Quantity	Result
a	c, φ, T₀	c≈15.5 MPa, φ≈29.1°, T₀≈10 MPa
b	Uniaxial tension circle	Centre −5.0 MPa, radius 5.0 MPa
c	Confined compression circle (σ3=20)	σ1,f=110.6 MPa; centre 65.3, radius 45.3 MPa
d	Joint envelope	τ=σn·tan29.1° (c=0)
e	Stress tensor at 2000 m	σv=51.01, σEW=38.26, σNS=25.51 MPa
f	τmax, dip	12.75 MPa, 45°
g	P_f=45 MPa effect	σNS′=−19.49 MPa (past tensile cutoff) → E–W-striking, near-vertical extension fractures

Topic. Reading strength parameters off a plotted Mohr–Coulomb envelope; constructing 3D in-situ stress tensors; effective-stress control on extension fracturing.

Key relations. τ=c+σn·tanφ; $N_\phi=\tan^2(45^\circ+\phi/2)$, $\sigma_1=2c\sqrt{N_\phi}+\sigma_3 N_\phi$; mean stress $p=(\sigma_1+\sigma_2+\sigma_3)/3$; Terzaghi $\sigma'=\sigma-P_f$.

Why this works. A linear Mohr–Coulomb envelope is fully defined by exactly two of its three "key" parameters (any two of c, φ, tensile cutoff fix the third via σt=c/tanφ, which is why the internal check in part (a) works); and the E–W stress in part (e) is forced to a unique value once "average volumetric pressure" is read as the true 3D mean stress, since a stress that equals the mean of itself and two knowns has only one self-consistent solution.

Common pitfall. Confusing the linear-extrapolation tensile cutoff (read directly off the plotted vertex here) with a true Griffith/Brazilian tensile strength, which is typically somewhat lower; and assuming pore pressure must weaken ALL three effective stresses equally into failure, when in an anisotropic tensor it can flip only the SMALLEST principal stress negative (here σNS) while the other two remain safely compressive or only mildly tensile.

Source. Hoek, Practical Rock Engineering, ch.4 (Mohr–Coulomb parameters, Nφ form); Goodman, Engineering Geology: Rock in Engineering Construction (unit weight, in-situ stress); Davis & Reynolds, Structural Geology of Rocks and Regions, 3rd ed., ch.5 (effective stress, extension fracturing).

Question 4 of 5: Mohr–Coulomb Yield Criteria – Strength Parameters and 3D Stress Tensor

Question D: Mohr–Coulomb Yield Criteria – Strength Parameters and 3D Stress Tensor (18 total)