Axial extension of a circular bar tapering uniformly from diameter d1 to diameter d2 over length L under axial pull P (E = Young's modulus): Which expression gives the total extension?

Introduction / Context:Members with varying cross-section require integration of strain along the length to compute elongation under axial load. A classic result is the extension of a circular bar that tapers linearly from diameter d1 to d2.

Given Data / Assumptions:

• Bar length L; diameters d1 (one end) to d2 (other end), linearly varying.
• Axial tensile load P; Young's modulus E constant.
• Small deformations; axial stress uniform over each section.

Concept / Approach:For an element at distance x with local diameter d(x), area A(x) = π [d(x)]^2 / 4. Strain at x is P / (E * A(x)). Total extension is the integral of strain over length L. With linear taper, d(x) varies linearly → integration yields a simple closed form.

Step-by-Step Solution:

Verification / Alternative check:Check limits: if d1 = d2 = d (uniform bar) → u = (4 P L)/(π E d^2) = (P L)/(A E), since A = π d^2/4. This confirms correctness.

Why Other Options Are Wrong:

• Using average area (A_avg) is an approximation; exact integral gives 1/(d1 d2) dependence, not function of (d1 + d2)^2.
• Forms with d1^2 + d2^2 do not reduce properly for d1 = d2.
• Factor 2 in the numerator (option E) underestimates extension by half.

Common Pitfalls:Arithmetic slips when integrating 1/[d(x)]^2; forgetting to verify the uniform-bar limit; using arithmetic average of areas instead of the correct harmonic-type relation.

Final Answer:(4 * P * L) / (π * E * d1 * d2)