Axial deformation under self-weight vs. equivalent end load:\nFor a uniform vertical bar, how does the elongation caused by its own weight compare with the elongation produced if the same total weight is applied as a single axial load at the free end?

Introduction / Context:
Designers often compare axial deformation due to self-weight with the deformation caused by a lumped end load. Recognizing the correct relationship avoids overestimating or underestimating elongation in hanging members such as tie rods, cables, and tall columns.

Given Data / Assumptions:

• Uniform prismatic bar of length L, cross-sectional area A, density rho, and modulus E.
• Self-weight W = rho * g * A * L acts continuously along the length.
• Comparison is with a concentrated end load equal to W.
• Linear elasticity, small deformation, axial loading only.

Concept / Approach:
Deformation is the integral of strain along the bar. For a concentrated end load W, the axial force is constant everywhere, giving a simple expression. For self-weight, axial force varies linearly from zero at the free end to W at the top, so the average internal force is W/2, leading to half the elongation of the same bar under end load W.

Step-by-Step Solution:

Verification / Alternative check:
The self-weight produces a linearly varying force with an average of W/2, so elongation equals that from a constant force of W/2 applied along the length, confirming the half relation.

Why Other Options Are Wrong:

• Equal to/double/quadruple/one-third contradict the integral result and average-force reasoning.

Common Pitfalls:
Using W instead of the varying internal force in the integral; forgetting that the free end carries zero axial force in the self-weight case.

Final Answer:
half