{
  "version": 1,
  "topic": "GM12",
  "title": "Rock mechanics and leakoff practice worksheet",
  "createdAt": "2026-10-24",
  "case": {
    "title": "Resolved plane stress and an assumed envelope",
    "premise": "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.",
    "formula": "σn = 6,500 + 1,500 cos 2β; |τ| = |1,500 sin 2β|; strength = 500 + 0.6(σn − 3,000)",
    "variables": "Strength is an additional specified envelope. The linked scalar calculator resolves stresses and does not fit c or μ.",
    "steps": [
      {
        "label": "Define the baseline",
        "calculation": "Plane angle β (degree) = 30",
        "explanation": "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."
      },
      {
        "label": "Select the relation",
        "calculation": "σn = 6,500 + 1,500 cos 2β; |τ| = |1,500 sin 2β|; strength = 500 + 0.6(σn − 3,000)",
        "explanation": "Strength is an additional specified envelope. The linked scalar calculator resolves stresses and does not fit c or μ."
      },
      {
        "label": "Substitute the values",
        "calculation": "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",
        "explanation": "Retain intermediate precision. The displayed values are rounded only after evaluation."
      },
      {
        "label": "Read the result",
        "calculation": "Resolved shear magnitude (psi) = 1299.03811; Assumed shear strength (psi) = 3050; Effective normal compression (psi) = 4250",
        "explanation": "This selected plane is below the assumed envelope; that is not a complete all-plane or tensile failure assessment."
      }
    ],
    "result": "Resolved shear magnitude (psi) = 1299.03811; Assumed shear strength (psi) = 3050; Effective normal compression (psi) = 4250",
    "interpretation": "This selected plane is below the assumed envelope; that is not a complete all-plane or tensile failure assessment.",
    "control": "Plane angle β (degree)",
    "columns": [
      "Plane angle β (degree)",
      "Resolved shear magnitude (psi)",
      "Assumed shear strength (psi)",
      "Effective normal compression (psi)"
    ],
    "rows": [
      [
        0,
        0,
        3500,
        5000
      ],
      [
        15,
        749.9999999999999,
        3379.422863405995,
        4799.038105676658
      ],
      [
        30,
        1299.038105676658,
        3050,
        4250
      ],
      [
        45,
        1500,
        2600,
        3500
      ],
      [
        60,
        1299.038105676658,
        2150,
        2750
      ],
      [
        75,
        749.9999999999999,
        1820.577136594005,
        2200.9618943233418
      ],
      [
        90,
        1.8369701987210297e-13,
        1700,
        2000
      ]
    ],
    "baseline": 2,
    "rowNotes": [
      "Plane angle β (degree) = 0. Resolved shear magnitude (psi): 0; Assumed shear strength (psi): 3500; Effective normal compression (psi): 5000.",
      "Plane angle β (degree) = 15. Resolved shear magnitude (psi): 750; Assumed shear strength (psi): 3379.42286; Effective normal compression (psi): 4799.03811.",
      "Plane angle β (degree) = 30. Resolved shear magnitude (psi): 1299.03811; Assumed shear strength (psi): 3050; Effective normal compression (psi): 4250.",
      "Plane angle β (degree) = 45. Resolved shear magnitude (psi): 1500; Assumed shear strength (psi): 2600; Effective normal compression (psi): 3500.",
      "Plane angle β (degree) = 60. Resolved shear magnitude (psi): 1299.03811; Assumed shear strength (psi): 2150; Effective normal compression (psi): 2750.",
      "Plane angle β (degree) = 75. Resolved shear magnitude (psi): 750; Assumed shear strength (psi): 1820.57714; Effective normal compression (psi): 2200.96189.",
      "Plane angle β (degree) = 90. Resolved shear magnitude (psi): 1.8369702e-13; Assumed shear strength (psi): 1700; Effective normal compression (psi): 2000."
    ],
    "provenance": "Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.",
    "questions": [
      {
        "question": "Reproduce the baseline Resolved shear magnitude (psi) without rounding intermediate values.",
        "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."
      },
      {
        "question": "Compare the first and last recorded scenarios. Which input changes?",
        "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."
      },
      {
        "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. 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."
      }
    ]
  },
  "recorded": null
}