Engineering lesson
Vertical versus deviated wellbore stress assumptions
A deviated well requires stresses transformed into its local coordinate system before wall-stress and failure calculations. A vertical-well formula does not automatically account for orientation or shear coupling.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Transform the stress tensor
The local stress components depend on well inclination and azimuth relative to principal stresses. Off-diagonal terms can appear even when the global tensor is diagonal.
The linked tangential-extremes calculator takes already supplied local components and computes principal values of a two-by-two stress block. It does not perform a full geographic-to-well transformation.
Understand the eigenvalue comparison
For a symmetric block, the extrema are half the trace plus or minus half the square root of the squared normal-stress difference plus four times shear squared.
The explorer varies local shear with normal components fixed. This demonstrates coupling, not a sweep of actual well inclination.
Apply strength after calculating stress
The resulting extrema still require an effective-stress convention and an appropriate failure criterion.
A complete stability model may also need pore pressure, mud pressure, thermal terms, anisotropy and boundary conditions. Keep those dependencies separate from this local algebra.
Explore the example
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
Select one of 5 recorded scenarios. Intermediate states are not interpolated.
Local shear (psi) = 500. Maximum local principal stress (psi): 7118.03399; Minimum local principal stress (psi): 4881.96601.
- Maximum local principal stress (psi)
- Minimum local principal stress (psi)
Worked example
Shear coupling in a local stress block
Local normal stresses 7,000 and 5,000 psi. Only local shear changes. No well-trajectory transformation is performed.
Relation and unit basis
σ± = (7,000 + 5,000 ± √[(7,000 − 5,000)² + 4τ²])/2
These are eigenvalues of a symmetric two-by-two local stress block, not a complete wellbore stability solve.
1. Define the baseline
Local shear (psi) = 500
Local normal stresses 7,000 and 5,000 psi. Only local shear changes. No well-trajectory transformation is performed.
2. Select the relation
σ± = (7,000 + 5,000 ± √[(7,000 − 5,000)² + 4τ²])/2
These are eigenvalues of a symmetric two-by-two local stress block, not a complete wellbore stability solve.
3. Substitute the values
At τ = 500 psi: 6,000 ± √(4,000,000 + 1,000,000)/2 = 7,118.03399 and 4,881.96601 psi
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Maximum local principal stress (psi) = 7118.03399; Minimum local principal stress (psi) = 4881.96601
Shear widens the eigenvalue separation while the trace remains 12,000 psi.
Result and interpretation
Maximum local principal stress (psi) = 7118.03399; Minimum local principal stress (psi) = 4881.96601
Shear widens the eigenvalue separation while the trace remains 12,000 psi.
Local shear changes; no unreported trajectory transformation is implied.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Local shear (psi) | Maximum local principal stress (psi) | Minimum local principal stress (psi) |
|---|---|---|
| 0 | 7,000 | 5,000 |
| 250 | 7,030.776 | 4,969.224 |
| 500 | 7,118.034 | 4,881.966 |
| 750 | 7,250 | 4,750 |
| 1,000 | 7,414.214 | 4,585.786 |
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. When local shear is zero, what do the two eigenvalues become?
Show the worked answer
The two supplied diagonal normal stresses, ordered from largest to smallest.
2. Reproduce the baseline Maximum local principal stress (psi) without rounding intermediate values.
Show the worked answer
At τ = 500 psi: 6,000 ± √(4,000,000 + 1,000,000)/2 = 7,118.03399 and 4,881.96601 psi. Maximum local principal stress (psi) = 7118.03399. Shear widens the eigenvalue separation while the trace remains 12,000 psi.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Local shear (psi) changes from 0 to 1000. Other controls remain as stated in the premise. Maximum local principal stress (psi) changes from 7000 to 7414.21356.
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. These are eigenvalues of a symmetric two-by-two local stress block, not a complete wellbore stability solve. Shear widens the eigenvalue separation while the trace remains 12,000 psi.
Before using the result
- Transform stresses before using local components.
- Keep coordinate and sign conventions.
- Do not call a local eigenvalue calculation a full stability model.
Shear widens the eigenvalue separation while the trace remains 12,000 psi.
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.