Practice and answer pack
Rock mechanics and leakoff practice worksheet
Practice effective stress, resolved plane stresses, conductivity normalization and Carter volume accounting with explicit conventions.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Write the convention before the number
Every stress answer should declare compression or tension positive, total or effective stress and the plane-angle definition. Every leakoff answer should declare its area and exposure-time basis.
The worked Mohr example separates stress resolution from the additional assumed cohesion and friction inputs.
Use dimensional checks
Biot's coefficient and dimensionless conductivity should be dimensionless. Carter volume should retain cubic length units after area multiplies its loss thickness.
A modulus ratio must use compatible units, and a leakoff coefficient must match the time unit under the square root.
Explain the model boundary
A resolved stress is not a strength measurement, and a fixed-area leakoff calculation is not a fracture-growth simulation.
Print the worksheet and reveal answers after solving. The synthetic values are designed to expose convention mistakes without implying a field operating limit.
Explore the example
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
Select one of 7 recorded scenarios. Intermediate states are not interpolated.
Plane angle β (degree) = 30. Resolved shear magnitude (psi): 1299.03811; Assumed shear strength (psi): 3050; Effective normal compression (psi): 4250.
- Resolved shear magnitude (psi)
- Assumed shear strength (psi)
- Effective normal compression (psi)
Worked example
Resolved plane stress and an assumed envelope
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.
Relation and unit basis
σn = 6,500 + 1,500 cos 2β; |τ| = |1,500 sin 2β|; strength = 500 + 0.6(σn − 3,000)
Strength is an additional specified envelope. The linked scalar calculator resolves stresses and does not fit c or μ.
1. Define the baseline
Plane angle β (degree) = 30
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.
2. Select the relation
σn = 6,500 + 1,500 cos 2β; |τ| = |1,500 sin 2β|; strength = 500 + 0.6(σn − 3,000)
Strength is an additional specified envelope. The linked scalar calculator resolves stresses and does not fit c or μ.
3. Substitute the values
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
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Resolved shear magnitude (psi) = 1299.03811; Assumed shear strength (psi) = 3050; Effective normal compression (psi) = 4250
This selected plane is below the assumed envelope; that is not a complete all-plane or tensile failure assessment.
Result and interpretation
Resolved shear magnitude (psi) = 1299.03811; Assumed shear strength (psi) = 3050; Effective normal compression (psi) = 4250
This selected plane is below the assumed envelope; that is not a complete all-plane or tensile failure assessment.
Start with the plane-stress example, then work the independent modulus, pressure and fluid-loss exercises.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Plane angle β (degree) | Resolved shear magnitude (psi) | Assumed shear strength (psi) | Effective normal compression (psi) |
|---|---|---|---|
| 0 | 0 | 3,500 | 5,000 |
| 15 | 750 | 3,379.423 | 4,799.038 |
| 30 | 1,299.038 | 3,050 | 4,250 |
| 45 | 1,500 | 2,600 | 3,500 |
| 60 | 1,299.038 | 2,150 | 2,750 |
| 75 | 750 | 1,820.577 | 2,200.962 |
| 90 | 1.8370e-13 | 1,700 | 2,000 |
Scroll the table to see all columns. The highlighted row is the worked baseline. Downloads retain full numerical precision.
Practice problems and answer key
Work through the question before revealing its answer. Use the stated assumptions and include units in every numerical step.
1. What extra information is required to compare a resolved plane stress with a Mohr–Coulomb envelope?
Show the worked answer
A compatible effective-stress definition plus cohesion and friction parameters under the same sign convention.
2. Reproduce the baseline Resolved shear magnitude (psi) without rounding intermediate values.
Show the worked answer
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. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
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. What would be needed to apply this example to a different physical system?
Show the worked answer
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. Calculate α from drained Kd = 10 GPa and grain Ks = 37 GPa.
Show the worked answer
α = 1 − 10/37 = 0.72972973. Use compatible modulus units and the drained frame modulus.
6. How does the Carter component change when fixed-area exposure rises from 60 to 240 min?
Show the worked answer
It doubles because √(240/60) = 2. A fixed spurt component does not double.
Before using the result
- Declare conventions before substitution.
- Check dimensions.
- Keep stress, strength and growth models distinct.
This selected plane is below the assumed envelope; that is not a complete all-plane or tensile failure assessment.
Continue learning
Start with the broader workflow guide
Sources and method
The explanation and synthetic examples are original PetroCalcHub teaching material. The references below support the definitions and model context. They do not supply field validation for the synthetic scenarios.