Eccentric loading on a column:\nIf the magnitudes of direct (uniform) stress and bending stress are equal, what happens to the stress at one extreme fibre of the column cross-section?

Introduction / Context:
Columns under eccentric axial load experience a combination of uniform direct stress and linear bending stress. Understanding superposition at the extreme fibres is essential for avoiding cracking or loss of compression on one side.

Given Data / Assumptions:

• Direct axial stress σ0 = P / A (uniform).
• Bending stress σb = M * y / I (linear, tension on one side, compression on the other).
• Magnitudes: |σ0| = |σb| at extreme fibre (y = ymax).

Concept / Approach:
Total stress at any fibre equals σ = σ0 ± σb. At the compressed extreme, signs add; at the opposite extreme, they subtract. If magnitudes are equal, one side becomes zero stress (transition from compression to zero), indicating the brink of tension on that side for further eccentricity.

Step-by-Step Solution:

Verification / Alternative check:
This condition corresponds to the load line passing through the core boundary (kern), where no tension just begins to occur. Any additional eccentricity produces tension at the relieved edge.

Why Other Options Are Wrong:

• Doubling is incorrect; superposition may cancel on one side.
• Uniform stress cannot exist with bending.
• Neutral axis location, not maximum stress, occurs where stress is zero across the section; saying it is maximum at the neutral axis is nonsensical.

Common Pitfalls:
Forgetting sign conventions; assuming bending always increases stress magnitude at both edges; ignoring the eccentric load core concept.

Final Answer:
It is zero at one extreme fibre