Vertical bar under self-weight and end load: A long bar is held vertically from its upper end and carries a load at its lower end. Considering the additional stress due to the bar's own weight, where will the maximum normal stress occur?

Introduction / Context:Members oriented vertically experience a stress gradient due to self-weight. When an additional load is applied at the free (lower) end and the bar is supported at the upper end, the support section carries the greatest force.

Given Data / Assumptions:

• Bar is long, prismatic, and vertical.
• Supported (held) at the top; load applied at the bottom end.
• Self-weight acts uniformly along the length.

Concept / Approach:Internal axial force at a cross-section equals the sum of the applied end load plus the weight of the portion of bar below that section. The portion of bar contributing to the force is greatest at the top section, hence the maximum stress occurs at the upper support.

Step-by-Step Solution:Let w be weight per unit length and P the end load.Axial force at a section located y below the top: N(y) = P + w * y.This is maximum at y = L (top section reaction), giving N_max = P + w * L and σ_max = N_max / A at the top.

Verification / Alternative check:Plot N(y) versus y: a straight line increasing toward the top; the maximum occurs at the support.

Why Other Options Are Wrong:Lower/mid sections carry less cumulative self-weight, hence lower stress.Uniform stress is not possible with self-weight.

Common Pitfalls:Confusing a hanging bar (top support) with a bar supported from below (then the gradient reverses).

Final Answer:At the built-in upper cross-section.