A circular bar of diameter d and length l elongates by e under a gradually applied axial load W. If the same bar is simply supported as a beam of span l and carries a central point load W, what is the mid-span deflection in terms of e, l, and d?

Introduction / Context:
Relating axial deformation to bending deflection through material and geometric properties is a powerful technique. Here, the same load W produces an axial extension in a bar and a mid-span deflection in a simply supported beam made from that bar. Eliminating E and geometric constants links the two responses.

Given Data / Assumptions:

• Uniform circular bar: diameter d, length l.
• Under axial load W (gradually applied) the extension is e.
• The same member, simply supported with central load W, deflects at mid-span by δ.
• Linear elasticity holds.

Concept / Approach:
Axial extension under load W:
e = W * l / (A * E)with A = (π d^2) / 4. Mid-span deflection for a simply supported beam with central load W:
δ = W * l^3 / (48 * E * I)with I = (π d^4) / 64. Eliminate W/E using the axial relation to express δ in terms of e, l, d.

Step-by-Step Solution:
1) From axial case: W/E = e * A / l.2) Substitute into beam deflection: δ = (e * A / l) * l^3 / (48 * I) = e * A * l^2 / (48 * I).3) Compute A/I for circular section: A/I = [(π d^2 / 4)] / [π d^4 / 64] = 16 / d^2.4) Therefore δ = e * (16 / d^2) * l^2 / 48 = e * l^2 / (3 d^2).

Verification / Alternative check:
Dimensional check: e has dimension of length; multiplying by (l^2 / d^2) yields length. Trend check: larger diameter reduces deflection, larger span increases deflection – both physically consistent.

Why Other Options Are Wrong:
Other coefficients (1/6, 2/3, etc.) result from algebraic mistakes in A/I or the beam formula (48EI).

Common Pitfalls:

• Using I for a rectangle instead of a circle.
• Forgetting that the axial load is gradually applied only affects strain energy, not the static e relation.
• Mixing up span l and half-span in the beam formula.

Final Answer:
e * l^2 / (3 d^2)