Difficulty: Medium
Correct Answer: δ = wl^2/(6E)
Explanation:
Introduction:Members with variable cross-section exhibit nonuniform stress under self-weight. A conical bar fixed at the large end and tapering to zero at the free tip has a neat closed-form elongation independent of base diameter.Given Data / Assumptions:
Concept / Approach:At a section x measured from the fixed end, area A(x) decreases linearly, and the axial force equals the weight of the segment below. Strain is σ/E, and total elongation is the integral of strain over the length.Step-by-Step Solution:
Let A0 = πd^2/4; A(x) = A0*(1 − x/l)^2Weight below section: P(x) = w*∫_x^l A(ξ)dξ = wA0l*(1 − x/l)^3/3Stress: σ(x) = P(x)/A(x) = (wl/3)(1 − x/l)Strain: ε(x) = σ(x)/E = (wl/(3E))(1 − x/l)Elongation: δ = ∫_0^l ε(x) dx = (wl/(3E))∫_0^l (1 − x/l) dx = wl^2/(6E)Verification / Alternative check:The result is independent of d because both P(x) and A(x) scale with A0 and cancel. Dimensional check: w has units of force/volume; multiplying by l^2/E gives length.Why Other Options Are Wrong:
Common Pitfalls:Setting up A(x) incorrectly; forgetting that the line of action for self-weight is the weight of the portion below the section; missing the linear taper square in area.
Final Answer:
δ = wl^2/(6E)
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