Beam under tent-shaped load: A simply supported beam carries a load that varies gradually from zero at both supports to w per metre at midspan (symmetric). Is the statement “the bending moment at the centre is w l / 4” correct?
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AYes
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BNo
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COnly for very small spans
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DOnly if w is total load, not intensity
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EDepends on end fixity
Answer
Correct Answer: No
Explanation
Introduction / Context:Different load distributions produce different shear and bending moment diagrams. Dimensional consistency and known reference cases help validate formulas. The given statement resembles the classic point-load case at midspan rather than a distributed tent-shaped load.
Given Data / Assumptions:
- Simply supported beam of span l.
- Load intensity is zero at both supports and increases linearly to w per metre at midspan, then symmetrically decreases.
- w denotes load intensity (force per unit length).
Concept / Approach:For a central point load W on a simply supported beam, M_max at midspan equals W * l / 4 (units of forcelength). For a uniformly distributed load q, M_max = q * l^2 / 8 (forcelength). For a tent-shaped distributed load peaking at midspan, the correct midspan moment involves w * l^2 multiplied by a coefficient distinct from 1/8 and certainly not w * l / 4, which is dimensionally inconsistent if w is per metre.
Step-by-Step Solution:
Check dimensions: bending moment units should be force * length.Given w is force/length, w * l / 4 has units of force (incorrect). Correct form must include l^2.Therefore, the stated expression cannot be correct for a distributed load described.Verification / Alternative check:Detailed integration of the symmetric triangular intensity gives M_center proportional to w * l^2 with a coefficient less than that for UDL. Regardless, w * l / 4 is invalid dimensionally and numerically.
Why Other Options Are Wrong:
- “Yes” contradicts both dimensional analysis and known beam results.
- “Only for very small spans” or “Depends on end fixity” are irrelevant; the form remains wrong.
- “Only if w is total load” would redefine symbols, but the stem clearly defines w as intensity.
Common Pitfalls:Memorizing W l / 4 (point load) and misapplying it to distributed loads; ignoring unit consistency; mixing up W (total load) with w (load per unit length).
Final Answer:No