Combined axial tension and bending: what is the equivalent axial tensile load Pe that would produce the same extreme-fibre tensile stress as the combined action of axial load P and bending moment M on a section of modulus Z?

Introduction / Context:
Members often experience combined axial force and bending. To compare combined stress with an equivalent uniform axial stress, we can define an equivalent axial load that would generate the same extreme-fibre tensile stress as the combined case.

Given Data / Assumptions:

• Axial tension = P (tensile positive).
• Bending moment = M about a principal axis.
• Section area = A, section modulus = Z.

Concept / Approach:
Extreme-fibre stress under combined loading is: sigma_extreme = (P / A) + (M / Z). An equivalent axial load Pe produces a uniform stress sigma_eq = Pe / A. Setting sigma_eq = sigma_extreme gives Pe / A = (P / A) + (M / Z), so Pe = P + A * (M / Z). This provides a convenient scalar comparison for tension checks and detailing decisions.

Step-by-Step Solution:
Write combined stress: sigma = P / A + M / Z. Set Pe / A = sigma. Solve: Pe = P + (A * M / Z).

Verification / Alternative check:
Dimensional check: M / Z has dimensions of stress. Multiplying by A gives force, consistent with Pe. If M = 0, Pe reduces to P, as expected.

Why Other Options Are Wrong:

• P - A M / Z: Would understate stress for positive M.
• P + M / (A Z) or P + Z M / A: Dimensionally inconsistent.
• (P Z) / (A M): Inverts the relationship; not stress equivalence.

Common Pitfalls:

• For compression with bending, sign conventions and interaction formulas may differ and require separate checks.

Final Answer:
Pe = P + (A * M / Z).