Elasticity under self-weight: for a uniform bar, the ratio of the total elongation caused by the bar’s own weight to the elongation caused by an external end load equal to the bar’s weight is what?

Introduction / Context:Members often experience both self-weight and external loads. Comparing elongations from these two loading modes sharpens understanding of distributed vs concentrated loading in axial members.

Given Data / Assumptions:

• Uniform prismatic bar: length L, cross-sectional area A, density ρ, modulus E.
• Self-weight acts as a uniformly distributed axial load along the length.
• External end load equals the total weight: W = ρ A g L.

Concept / Approach:Elongation under an axial force P over length L is δ = P L / (E A) when P is constant. Under self-weight, internal axial force varies linearly with position; integrate strain along the length to obtain elongation. The ratio becomes independent of A, ρ, and E.

Step-by-Step Solution:Elongation due to end load W: δ_W = W L / (E A) = (ρ A g L) L /(E A) = ρ g L^2 / E.At section x from the top, internal force from self-weight: P(x) = ρ A g (L - x).Differential elongation: dδ = P(x) dx / (E A) = [ρ g (L - x)/E] dx.Integrate 0→L: δ_sw = (ρ g / E) ∫_0^L (L - x) dx = (ρ g / E) (L^2 / 2) = ρ g L^2 /(2E).Ratio: δ_sw / δ_W = [ρ g L^2 /(2E)] / [ρ g L^2 / E] = 1/2.

Verification / Alternative check:Dimensional consistency and independence from section properties confirm the generality of the 1/2 ratio.

Why Other Options Are Wrong:

• 2 or 1 overstate the effect of self-weight relative to the same total end load.
• 1/4 understates it; correct integration yields 1/2.

Common Pitfalls:

• Treating self-weight as a single lumped load at midspan instead of integrating.

Final Answer:1/2