Lesson 2.5: Theories of Failure

Theories of Failure are critical for design of mechanical components. GATE and PSU exams often test failure criteria under combined stresses, safe design, and factor of safety.

🔹 1. Introduction

• Definition: Theory of failure predicts when a material will fail under a combination of stresses.

• Importance: Ensures safety and reliability of structures and machine components.

• Types of Failure:

  • Ductile Failure: Significant plastic deformation before fracture

  • Brittle Failure: Sudden fracture with little deformation

🔹 2. Principal Stresses

• Principal Stresses (σ1,σ2,σ3\sigma_1, \sigma_2, \sigma_3σ1​,σ2​,σ3​): Maximum & minimum normal stresses at a point.

• Determination: From 3D stress transformation equations or Mohr’s Circle.

• Applications: Basis for failure theories.

🔹 3. Theories of Failure

a) Maximum Normal Stress Theory (Rankine)

• Applicable to: Brittle materials

• Criterion: Failure occurs when max principal stress = ultimate stress

σ1≥σu\sigma_1 \ge \sigma_uσ1​≥σu​

• Factor of Safety:

n=σuσmaxn = \frac{\sigma_u}{\sigma_{\text{max}}}n=σmax​σu​​

b) Maximum Shear Stress Theory (Tresca)

• Applicable to: Ductile materials

• Criterion: Failure occurs when max shear stress = shear yield stress

τmax≥τy=σy2\tau_{\text{max}} \ge \tau_y = \frac{\sigma_y}{2}τmax​≥τy​=2σy​​

• Factor of Safety:

n=σy2τmaxn = \frac{\sigma_y}{2 \tau_{\text{max}}}n=2τmax​σy​​

c) Distortion Energy Theory (von Mises)

• Applicable to: Ductile materials, widely used in design

• Criterion: Failure occurs when distortion energy = energy at yielding

σeq=σ12+σ22−σ1σ2≥σy\sigma_{\text{eq}} = \sqrt{\sigma_1^2 + \sigma_2^2 – \sigma_1\sigma_2} \ge \sigma_yσeq​=σ12​+σ22​−σ1​σ2​​≥σy​

• Factor of Safety:

n=σyσeqn = \frac{\sigma_y}{\sigma_{\text{eq}}}n=σeq​σy​​

d) Mohr’s Theory

• Combines principal stress and shear stress

• Often used for combined bending, torsion, and axial loading

🔹 4. Combined Stress Examples

• Axial + Torsion:

σ1=σ2+(σ2)2+τ2,σ2=σ2−(σ2)2+τ2\sigma_1 = \frac{\sigma}{2} + \sqrt{\left(\frac{\sigma}{2}\right)^2 + \tau^2}, \quad \sigma_2 = \frac{\sigma}{2} – \sqrt{\left(\frac{\sigma}{2}\right)^2 + \tau^2}σ1​=2σ​+(2σ​)2+τ2​,σ2​=2σ​−(2σ​)2+τ2​

• Bending + Axial + Torsion: Use von Mises or Tresca to compute equivalent stress.

🔹 5. Solved Examples (PYQ Style)

Example 1 (GATE ME 2016):
Shaft subjected to axial stress 50 MPa and torsional shear 30 MPa. Check safety using Tresca criterion, σ_y = 250 MPa.

👉 Solution:
τmax=(σ/2)2+τ2=(25)2+302≈39.05\tau_{\text{max}} = \sqrt{(\sigma/2)^2 + \tau^2} = \sqrt{(25)^2 + 30^2} ≈ 39.05τmax​=(σ/2)2+τ2​=(25)2+302​≈39.05 MPa
Factor of Safety: n=σy/(2τmax)=250/(2∗39.05)≈3.2n = \sigma_y / (2 \tau_{\text{max}}) = 250 / (2*39.05) ≈ 3.2n=σy​/(2τmax​)=250/(2∗39.05)≈3.2 → Safe

Example 2 (PSU Exam):
Brittle rod with σ1 = 120 MPa, σ2 = 40 MPa, σ_u = 150 MPa. Check safety using Rankine theory.
👉 σ_max = 120 < 150 → Safe

🔹 6. Practice Exercises

1. Shaft under torsion 60 MPa and axial 80 MPa, σ_y = 300 MPa. Check safety using Tresca & von Mises.

2. Cantilever beam under bending 100 MPa, brittle material σ_u = 180 MPa. Safe or not?

3. Combined stress: σx=50 MPa, σy=30 MPa, τxy=20 MPa. Compute equivalent stress using von Mises.

4. Draw Mohr’s Circle for σx=40 MPa, σy=20 MPa, τxy=15 MPa and find principal stresses.

🔹 7. Summary

• Purpose: Predict failure under combined stresses

• Key Theories: Rankine (max normal), Tresca (max shear), von Mises (distortion energy)

• Applications: Design of shafts, beams, columns, springs

• Factor of Safety: Ensures safe design against material failure