Geomechanics and Ground Control · Geomechanics · 1 mark · Numerical answer
The Mohr–Coulomb failure envelope in principal stresses is $\sigma_1=30+3.5\sigma_3$, with stresses in MPa. Find the failure-plane angle with the σ₃ axis in degrees, to one decimal.
The principal-stress multiplier is (1+sinφ)/(1−sinφ) = 3.5. Hence sinφ = (3.5−1)/(3.5+1) = 5/9 and φ = 33.749°. The failure plane makes 45°+φ/2 = 61.8745° with the minor-principal-stress axis, or 61.9°. The complementary angle is measured from the other principal axis.
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