Difficulty: Medium
Correct Answer: 1.67
Explanation:
Introduction / Context:
Engineering design often addresses uncertainties in both actions (loads) and resistances (strength). A practical way to represent this in an allowable-stress context is to select a factor of safety (FOS) that maintains a margin when both the applied load is higher than computed and the available strength is lower than computed by specified percentages.
Given Data / Assumptions:
Concept / Approach:
Define factor of safety F such that allowable load = E / F. Under worst deviations, the demand becomes 1.25 P and the capacity becomes 0.75 E. Safety requires 1.25 P ≤ 0.75 E / F, which implies F ≤ 0.75 E / (1.25 P) = 0.6 * (E / P). To guarantee safety irrespective of E/P in the original computation (i.e., when E/P ≈ F in the base design), Rearrangement shows the minimum F consistent with these deviations is about 1 / 0.6 ≈ 1.67.
Step-by-Step Solution:
Worst applied load = 1.25 P.Worst available strength = 0.75 E.Require: 1.25 P ≤ (0.75 E) / F.Solve for F: F ≥ (0.75 E) / (1.25 P) = 0.6 * (E / P).For a base design where E/P ≈ F, we get F ≈ 1 / 0.6 = 1.67 (rounded).
Verification / Alternative check:
If F = 1.67, allowable = E / 1.67 ≈ 0.60 E. Under worst deviations, demand = 1.25 P, capacity = 0.75 E / 1.67 ≈ 0.449 E. Ensuring initial E/P comfortably exceeds F provides adequate margin; the 1.67 value is a widely taught minimum in classic problems for ±25% deviations.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:
1.67
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