Difficulty: Medium
Correct Answer: Increasing the length of the rod
Explanation:
Introduction / Context:When a weight drops onto a member, part of the potential energy W * h is converted into strain energy in the rod. The resulting peak stress is higher than for a static load and depends on the member's stiffness and energy absorption capacity. Designers must recognize which parameters reduce the maximum impact stress to improve safety and durability.
Given Data / Assumptions:
Concept / Approach:
For elastic impact, peak stress approximately follows from equating strain energy to drop energy, giving a peak stress that increases with stiffness and decreases with greater compliance. A representative relation shows peak stress increasing with E and decreasing with L and A. Thus, making the member more compliant (larger L and/or larger A for energy capacity per unit stress) reduces the maximum stress developed.
Step-by-Step Solution:
1) Drop energy: U = W * h (neglecting local losses).2) Elastic strain energy in an axially loaded bar: U = ∫(σ^2 / (2E)) * (A dz) = (σ_max^2 * A * L) / (2E) for a uniform bar (conceptual form).3) Higher L or A raises the energy capacity for a given peak stress, thereby lowering σ_max required to absorb U.4) Conversely, larger E raises stiffness, reducing deflection and increasing σ_max for the same U.5) Therefore, to reduce peak stress, increase L (and/or A), or choose a lower E material; among the options, increasing L is correct.Verification / Alternative check:
Energy methods or exact elastic-impact formulas with static extension terms both show σ_max decreasing with increasing L and A, and increasing with E.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing static and impact loading; ignoring that energy balance, not just force equilibrium, governs peak stress under impact.
Final Answer:
Increasing the length of the rod
Discussion & Comments