{
  "version": 1,
  "topic": "GM09",
  "title": "Vertical versus deviated wellbore stress assumptions",
  "createdAt": "2026-10-24",
  "case": {
    "title": "Shear coupling in a local stress block",
    "premise": "Local normal stresses 7,000 and 5,000 psi. Only local shear changes. No well-trajectory transformation is performed.",
    "formula": "σ± = (7,000 + 5,000 ± √[(7,000 − 5,000)² + 4τ²])/2",
    "variables": "These are eigenvalues of a symmetric two-by-two local stress block, not a complete wellbore stability solve.",
    "steps": [
      {
        "label": "Define the baseline",
        "calculation": "Local shear (psi) = 500",
        "explanation": "Local normal stresses 7,000 and 5,000 psi. Only local shear changes. No well-trajectory transformation is performed."
      },
      {
        "label": "Select the relation",
        "calculation": "σ± = (7,000 + 5,000 ± √[(7,000 − 5,000)² + 4τ²])/2",
        "explanation": "These are eigenvalues of a symmetric two-by-two local stress block, not a complete wellbore stability solve."
      },
      {
        "label": "Substitute the values",
        "calculation": "At τ = 500 psi: 6,000 ± √(4,000,000 + 1,000,000)/2 = 7,118.03399 and 4,881.96601 psi",
        "explanation": "Retain intermediate precision. The displayed values are rounded only after evaluation."
      },
      {
        "label": "Read the result",
        "calculation": "Maximum local principal stress (psi) = 7118.03399; Minimum local principal stress (psi) = 4881.96601",
        "explanation": "Shear widens the eigenvalue separation while the trace remains 12,000 psi."
      }
    ],
    "result": "Maximum local principal stress (psi) = 7118.03399; Minimum local principal stress (psi) = 4881.96601",
    "interpretation": "Shear widens the eigenvalue separation while the trace remains 12,000 psi.",
    "control": "Local shear (psi)",
    "columns": [
      "Local shear (psi)",
      "Maximum local principal stress (psi)",
      "Minimum local principal stress (psi)"
    ],
    "rows": [
      [
        0,
        7000,
        5000
      ],
      [
        250,
        7030.776406404415,
        4969.223593595585
      ],
      [
        500,
        7118.033988749895,
        4881.966011250105
      ],
      [
        750,
        7250,
        4750
      ],
      [
        1000,
        7414.213562373096,
        4585.786437626904
      ]
    ],
    "baseline": 2,
    "rowNotes": [
      "Local shear (psi) = 0. Maximum local principal stress (psi): 7000; Minimum local principal stress (psi): 5000.",
      "Local shear (psi) = 250. Maximum local principal stress (psi): 7030.77641; Minimum local principal stress (psi): 4969.22359.",
      "Local shear (psi) = 500. Maximum local principal stress (psi): 7118.03399; Minimum local principal stress (psi): 4881.96601.",
      "Local shear (psi) = 750. Maximum local principal stress (psi): 7250; Minimum local principal stress (psi): 4750.",
      "Local shear (psi) = 1000. Maximum local principal stress (psi): 7414.21356; Minimum local principal stress (psi): 4585.78644."
    ],
    "provenance": "Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.",
    "questions": [
      {
        "question": "Reproduce the baseline Maximum local principal stress (psi) without rounding intermediate values.",
        "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."
      },
      {
        "question": "Compare the first and last recorded scenarios. Which input changes?",
        "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."
      },
      {
        "question": "What would be needed to apply this example to a different physical system?",
        "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."
      }
    ]
  },
  "recorded": null
}