Rock mechanics and leakoff practice worksheet
PetroCalcHub · Original synthetic teaching exercise · 2026-10-24

Practice effective stress, resolved plane stresses, conductivity normalization and Carter volume accounting with explicit conventions.

Compression-positive principal stresses 8,000 and 5,000 psi; pore-pressure contribution 3,000 psi; assumed c = 500 psi and μ = 0.6. β follows the calculator's angle convention.

1. What extra information is required to compare a resolved plane stress with a Mohr–Coulomb envelope?

Working: ________________________________________________


2. Reproduce the baseline Resolved shear magnitude (psi) without rounding intermediate values.

Working: ________________________________________________


3. Compare the first and last recorded scenarios. Which input changes?

Working: ________________________________________________


4. What would be needed to apply this example to a different physical system?

Working: ________________________________________________


5. Calculate α from drained Kd = 10 GPa and grain Ks = 37 GPa.

Working: ________________________________________________


6. How does the Carter component change when fixed-area exposure rises from 60 to 240 min?

Working: ________________________________________________


ANSWER KEY

1. A compatible effective-stress definition plus cohesion and friction parameters under the same sign convention.
2. At β = 30°: σn = 7,250 psi; σn′ = 4,250 psi; |τ| = 1,299.03811 psi; strength = 500 + 0.6 × 4,250 = 3,050 psi. Resolved shear magnitude (psi) = 1299.03811. This selected plane is below the assumed envelope; that is not a complete all-plane or tensile failure assessment.
3. Plane angle β (degree) changes from 0 to 90. Other controls remain as stated in the premise. Resolved shear magnitude (psi) changes from 0 to 1.8369702e-13.
4. Replace the synthetic controls with compatible documented inputs and check the relation's assumptions. Strength is an additional specified envelope. The linked scalar calculator resolves stresses and does not fit c or μ. This selected plane is below the assumed envelope; that is not a complete all-plane or tensile failure assessment.
5. α = 1 − 10/37 = 0.72972973. Use compatible modulus units and the drained frame modulus.
6. It doubles because √(240/60) = 2. A fixed spurt component does not double.

Method: Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
Sources:
Espinoza: rock yield and failure: https://dnicolasespinoza.github.io/ch4_RockFailure.html
Espinoza: fracture mechanics, Carter leakoff and fluid balance: https://dnicolasespinoza.github.io/ch7_HydroFrac.html