Engineering lesson
Mohr Coulomb failure envelopes with worked stress inputs
A Mohr–Coulomb envelope compares shear stress with cohesion plus effective normal stress times friction coefficient. The site's stress-components calculator resolves plane stresses; the strength envelope is an additional specified comparison.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Resolve the plane before applying strength
For the angle convention shown, σn = (σ1 + σ3)/2 + (σ1 − σ3)cos(2β)/2 and τ = (σ1 − σ3)sin(2β)/2. This matches the linked calculator's definition of β.
An angle measured to the plane normal rather than the plane can change the normal-stress expression. Keep the diagram and equation convention together.
Add effective stress and a strength assumption
The example subtracts an explicitly supplied pore-pressure contribution from normal compression and compares |τ| with c + μσn′. Cohesion and μ are synthetic strength inputs, not outputs of the stress calculator.
A positive margin means the selected plane lies below this assumed envelope. It does not establish stability of every plane or address tensile failure.
Audit the physical scope
Principal stresses, pore pressure and strength must refer to compatible conditions and units. A laboratory envelope may depend on rock fabric, saturation and effective-stress definition.
The explorer changes plane angle only. It is a plane-by-plane teaching check rather than a complete wellbore failure assessment.
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.
The plotted shear and assumed strength are compared at each resolved plane angle; no hidden cohesion is inferred.
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.
Test your understanding
Work through the question before revealing its answer. Use the stated assumptions and include units in every numerical step.
1. Does the linked stress-components calculator determine cohesion?
Show the worked answer
No. Cohesion and friction must come from a separately stated strength model or data before an envelope comparison can be made.
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.
Before using the result
- Match the β convention.
- Use effective normal stress consistently.
- Supply strength parameters separately.
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.