On the principles of Strain Energy, in a rod AB of length L placed vertically with end A fixed at top and end B with a flange, a load W is released from a height h (< L) and hits on to the flange. In order to reduce the axial stress in the rod, which of the following should be done? i. Reduce the modulus of elasticity of the material ii. Increase the modulus of elasticity of the material iii. Increase the length of the rod iv. Decrease the length of the rod
- A(i) and (iii)Correct
- B(i) and (iv)
- C(ii) and (iii)
- D(ii) and (iv)
Solution
Concept
Impact stress
Formula
σ = σstatic [1 + √(1 + 2h A E /(W L))]
Step-by-step solution
1. Impact stress increases with √(E/L). 2. Reducing E and increasing L lowers the stress.
Explanation
A more flexible rod absorbs more energy per unit stress.
Why the other options are wrong
Increasing E or decreasing L raises the impact stress.
Shortcut
Impact stress ∝ √(E/L).
Correct answer: A) (i) and (iii)