Difficulty: Easy
Correct Answer: 1/8 (deflection becomes one-eighth)
Explanation:
Introduction / Context:
Deflection control often governs beam sizing. For rectangular sections, the second moment of area is very sensitive to depth, so changing depth can dramatically reduce deflections without changing material or span.
Given Data / Assumptions:
Concept / Approach:
For a central point load, elastic deflection at midspan is δ = (P * L^3) / (48 * E * I). For a rectangular section, I = b * h^3 / 12. If h → 2h, then I_new = b * (2h)^3 / 12 = 8 * (b * h^3 / 12) = 8 I.
Step-by-Step Solution:
Original deflection: δ_1 = (P L^3) / (48 E I).New second moment: I_2 = 8 I.New deflection: δ_2 = (P L^3) / (48 E I_2) = δ_1 / 8.Hence, the deflection becomes one-eighth of the original value.
Verification / Alternative check:
Dimensional reasoning confirms that deflection scales inversely with I, and I scales with h^3; doubling h scales deflection by 1/2^3 = 1/8.
Why Other Options Are Wrong:
1/2 and 1/4 would correspond to linear or quadratic sensitivity; the correct sensitivity is cubic in depth.2, 4, 8 imply deflection increases with depth, which contradicts beam theory.
Common Pitfalls:
Using section modulus instead of second moment for deflection, or changing both breadth and depth accidentally when only depth is varied.
Final Answer:
1/8 (deflection becomes one-eighth).
Discussion & Comments