A material may fail if (σ₀ = yield stress in simple tension):
Try it first: pick an option, then read the explanation.
Correct answer: D. all options are correct
Asked in SSC JE 25/01/2018 (Evening)
Detailed Explanation
Theories of failure compare some quantity in the complex stress state with its value at yield in a simple tension test. Different theories pick different quantities.
Why “all options are correct” Is Correct
Each option states a classical criterion: maximum principal stress (Rankine), maximum principal strain (St. Venant), maximum shear stress (Guest–Tresca, τ = σ₀/2 at yield in tension). So a material may fail under any of these conditions — all options are correct.
| Theory of failure | Due to | Failure when | Yield locus (2-D) | Suited to |
|---|---|---|---|---|
| Maximum principal stress | Rankine | σ₁ = σy | Square | Brittle materials |
| Maximum principal strain | St. Venant | σ₁ − μσ₂ = σy | Rhombus | Rarely used |
| Maximum shear stress | Guest / Tresca | σ₁ − σ₃ = σy | Hexagon | Ductile (conservative) |
| Total strain energy | Haigh (Beltrami) | σ₁² + σ₂² − 2μσ₁σ₂ = σy² | Ellipse | Ductile |
| Shear strain (distortion) energy | von Mises (Hencky) | σ₁² + σ₂² − σ₁σ₂ = σy² | Ellipse | Ductile (best fit) |
Why the Other Options Are Wrong
- A. the maximum principal stress exceeds σ₀: Correct, but only one of the criteria.
- B. the maximum strain exceeds σ₀/E: Correct, but only one of the criteria.
- C. the maximum shear stress exceeds σ₀/2: Correct, but only one of the criteria.
Key Point
Failure criteria: σ₁ ≥ σ₀ (Rankine), ε ≥ σ₀/E (St. Venant), τmax ≥ σ₀/2 (Tresca).
Key Concepts for Students
- In simple tension at yield, τmax = σ₀/2.
- Brittle: Rankine; ductile: Tresca or von Mises.
- Energy theories use strain energy instead of stress or strain.
Practice more: this is question 14 in Principal Stress and Principal Planes MCQs – Part 1. Try the whole set and check every answer instantly.