Eccentrically loaded rectangular column: If a load P has eccentricities e_x and e_y along the X- and Y-axes, what is the normal stress at a general point (x, y) on the cross-section?

Introduction / Context:
Columns and pedestal footings often carry loads that are not applied through the centroid, creating additional bending about principal axes. Determining the combined stress distribution is essential to check tension zones and permissible compressive stress.

Given Data / Assumptions:

• Rectangular cross-section with principal centroidal axes X and Y.
• Load P applied with eccentricities e_x (along X) and e_y (along Y) from the centroid.
• Second moments of area I_x (about X) and I_y (about Y) are known.

Concept / Approach:
An eccentric load creates resultant bending moments M_x and M_y in addition to direct compression P/A. If e_x is measured along the X-axis, it produces a moment about the Y-axis equal to M_y = P e_x; similarly, e_y produces M_x = P e_y about the X-axis. The stress at (x, y) superposes direct and bending components: σ = P/A + M_y * x / I_y + M_x * y / I_x, which becomes σ(x,y) = P/A + (P e_x / I_y) x + (P e_y / I_x) y.

Step-by-Step Solution:
Write direct stress: σ_d = P/A.Take moments: M_y = P e_x and M_x = P e_y.Bending stresses: σ_b,y = M_y x / I_y; σ_b,x = M_x y / I_x.Superpose: σ = σ_d + σ_b,y + σ_b,x = P/A + (P e_x / I_y) x + (P e_y / I_x) y.

Verification / Alternative check:
Check symmetry: if e_x = e_y = 0, the expression reduces to σ = P/A as expected. If only e_x ≠ 0, stress varies linearly with x, consistent with bending about the Y-axis.

Why Other Options Are Wrong:

• Misplaced I_x and I_y or swapping x and y leads to incorrect stress distribution.
• Neglecting bending (P/A only) ignores eccentricity effects.

Common Pitfalls:

• Sign convention: tension/compression signs depend on the chosen positive x–y directions; here a general algebraic form is provided.

Final Answer:
σ(x,y) = P/A + (P e_x / I_y) x + (P e_y / I_x) y.