If the principal stresses at a point in an elastic material are 2f tensile, 1.5f tensile and f compressive. What is the value of f at failure using maximum principal stress theory? (Take the elastic limit in simple tension as 210 N/mm² and μ = 0.3)
- A210 N/mm²
- B420 N/mm²
- C170 N/mm²
- D105 N/mm²Correct
Solution
Concept
In Rankine's theory, failure occurs when the maximum principal stress reaches the elastic limit stress in simple tension.
Formula
σ1 = fy
Step-by-step solution
1. Principal stresses: σ1 = 2f (tension), σ2 = 1.5f (tension), σ3 = -f (compression). 2. The maximum principal stress is σ1 = 2f. 3. Failure condition: σ1 = fy, so 2f = 210. 4. f = 210/2 = 105 N/mm².
Explanation
Maximum principal stress theory checks only the largest principal stress. Here the largest stress is 2f, so 2f = 210 N/mm² and f = 105 N/mm². Poisson's ratio is not needed for this theory.
Why the other options are wrong
A: 210 N/mm² is the elastic limit itself, not f. B: 420 N/mm² comes from multiplying by 2 instead of dividing. C: 170 N/mm² does not satisfy 2f = 210 and looks like a value from a different theory.
Shortcut
f = fy / (largest coefficient) = 210/2 = 105 N/mm².
Common mistake
Using μ = 0.3 here. μ is used only in maximum principal strain theory.
Correct answer: D) 105 N/mm²