Short hollow circular column under eccentric load: for a column with external diameter D, internal diameter d, and load W applied with eccentricity e that causes zero stress at one extreme fibre, what is the required eccentricity e?

Introduction / Context:
For short columns under combined axial load and bending, the extreme fibre stress is σ = W/A ± M y / I. Setting the minimum stress to zero yields the e that places the neutral axis at the far edge (no tension), a common design condition for masonry or concrete columns with limited tensile capacity.

Given Data / Assumptions:

• Hollow circular section: external diameter D, internal diameter d.
• Area A = (π/4) (D^2 - d^2).
• Second moment of area I = (π/64) (D^4 - d^4).
• Extreme fibre distance y = D/2, bending moment M = W * e.
• Condition for zero minimum stress: W/A = M y / I.

Concept / Approach:
Impose σ_min = 0 → W/A = (W e) y / I → e = I / (A y). Substitute A, I, and y for a circular annulus and simplify algebraically to obtain e in terms of D and d only.

Step-by-Step Solution:
Start: e = I / (A y).Substitute: I = (π/64)(D^4 - d^4), A = (π/4)(D^2 - d^2), y = D/2.Cancel π and simplify constants: e = (1/8) * (D^4 - d^4) / [ D (D^2 - d^2) ].Factor: D^4 - d^4 = (D^2 - d^2)(D^2 + d^2).Cancel (D^2 - d^2): e = (1/8) * (D^2 + d^2) / D.

Verification / Alternative check:
For a solid circular section (d = 0), e reduces to (D/8), the classic result for zero tension at the far edge under eccentric axial load.

Why Other Options Are Wrong:

• Using (D^2 - d^2) in the numerator fails to account for the D^4 - d^4 factorization.
• Denominator 4D doubles the correct eccentricity.

Common Pitfalls:

• Mixing up I and A expressions for an annulus.
• Forgetting the extreme fibre distance y = D/2.

Final Answer:
e = (D^2 + d^2) / (8 D)