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Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
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CivilSoil Mechanics
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The angle of failure plane with major principal plane is given by

A

90° + effective angle of shearing resistance

B

90° + half of the angle of shearing resistance

C

45° - half of the angle of shearing resistance

D

45° + half of the angle of shearing resistance

Correct Answer

⚙️ TE • Technical Concept & PrincipleCivilSoil Mechanics
Option D

45° + half of the angle of shearing resistance

Quick Summary:

According to the Mohr-Coulomb failure criterion, the failure plane in a soil mass inclined at an angle α\alphaαto the major principal plane is determined by the internal friction angle (ϕ).\phi).ϕ).The analytical derivation using Mohr's circle shows that the plane of maximum shear stress and the actual failure plane do not coincide unless ϕ=0\phi = 0ϕ=0; for frictional soils, the failure plane occurs at an angle of 45° + ϕ/2\phi/2ϕ/2relative to the major principal plane.

⚙️TETechnical SolutionConcept & Principle
💡 Explanation

According to the Mohr-Coulomb failure criterion, the failure plane in a soil mass inclined at an angle α\alphaαto the major principal plane is determined by the internal friction angle (ϕ).\phi).ϕ).The analytical derivation using Mohr's circle shows that the plane of maximum shear stress and the actual failure plane do not coincide unless ϕ=0\phi = 0ϕ=0; for frictional soils, the failure plane occurs at an angle of 45° + ϕ/2\phi/2ϕ/2relative to the major principal plane.

🔢 Key Formulas

α=45°+ϕ2\alpha = 45°+ \frac{\phi}{2}α=45°+2ϕ​ — Angle of failure plane with major principal plane

τf=c+σntan⁡ϕ\tau_f = c + \sigma_n \tan \phiτf​=c+σn​tanϕ — Mohr-Coulomb failure criterion

⚙️ Working Principle

When a soil sample is subjected to principal stresses σ\sigmaσ_1 and σ\sigmaσ_3, the state of stress is represented by Mohr's circle. The failure envelope is defined by \tau = c + \sigma$$\tan$$\phi.By finding the tangency point of the failure envelope to the Mohr circle, the angle θ(\theta (θ(or α)\alpha)α)of the failure plane with the major principal plane is geometrically derived as 45° + ϕ/2.\phi/2.ϕ/2.

📌 Key Points
  • ▸

    The major principal plane is the plane on which the maximum principal stress (σ\sigmaσ_1) acts.

  • ▸

    The failure plane is the plane of potential sliding within the soil mass.

  • ▸

    If the soil is purely cohesive (ϕ=0),\phi = 0),ϕ=0),the failure plane occurs at 45° to the major principal plane.

  • ▸

    For granular soils, the failure plane angle increases as the internal angle of friction ϕ\phiϕincreases.

✅ Advantages
  • ▸

    Allows prediction of sliding surface orientation in slope stability analysis.

  • ▸

    Essential for calculating active and passive earth pressures in Rankine's theory.

❌ Disadvantages / Limitations
  • ▸

    Assumes the soil is an ideal isotropic, homogeneous material.

  • ▸

    Does not account for non-linear failure envelopes found in some stiff clays.

🛠️ Applications / Uses
  • ▸

    Foundation design (bearing capacity analysis).

  • ▸

    Retaining wall design (Rankine's earth pressure theory).

  • ▸

    Slope stability analysis.

📄 Additional Information
  • ▸

    The major principal plane is the plane on which the normal stress is maximum and shear stress is zero.

  • ▸

    Option C (45° - ϕ/2)\phi/2)ϕ/2)refers to the angle the failure plane makes with the minor principal plane.

📊 Diagram / Illustration
Failure Plane AngleAngle of failure (α) = 45° + φ/2φ = Effective angle of shearing resistanceMajor Principal Plane is at 0°
✅

D is correct — The failure plane in soil makes an angle of 45° + ϕ/2\phi/2ϕ/2with the direction of the major principal stress.

Core Concepts Used
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Mohr-Coulomb Failure Theory Principal Planes Internal Angle of Friction
💡 EXAM TIP

Always remember that the failure plane is inclined at 45° + ϕ/2\phi/2ϕ/2to the major principal plane, but 45° - ϕ/2\phi/2ϕ/2to the minor principal plane. Confusing these two is a common error in exams.

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