Difficulty: Medium
Correct Answer: It becomes 1/2 of the original deflection
Explanation:
Introduction / Context:Deflection control is a key serviceability requirement. For rectangular sections, the flexural rigidity E * I depends on the geometry, so changing width or depth alters deflection in predictable ways.
Given Data / Assumptions:
Concept / Approach:For a midspan point load on a simply supported beam, the maximum deflection is:delta_max = P * L^3 / (48 * E * I)For a rectangle, the second moment of area is:I = b * h^3 / 12Therefore, delta_max is inversely proportional to b when h, E, L, and P are fixed.
Step-by-Step Solution:
Original I_1 = b * h^3 / 12.After doubling width: I_2 = (2 b) * h^3 / 12 = 2 * I_1.Since delta ∝ 1 / I, delta_2 = delta_1 / 2.Verification / Alternative check:Check dimensional consistency and observe that increasing section size increases stiffness, reducing deflection proportionally.
Why Other Options Are Wrong:
Common Pitfalls:Confusing the stronger influence of depth (h appears as h^3) with width (linear effect). Doubling depth would reduce deflection by a factor of 8; doubling width only halves it.
Final Answer:It becomes 1/2 of the original deflection
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