Eccentrically loaded columns – defining an equivalent axial load for design How is the equivalent axial load commonly determined for an eccentrically loaded column to reflect the effect of the produced bending moment?

Civil Engineering Steel Structure Design Difficulty: Medium
Choose an option
  • A
    By adding the axial load and the contribution due to eccentricity, i.e., axial load plus the product of bending moment from eccentricity and an appropriate bending factor
  • B
    By adding the axial load and eccentric load and then subtracting the product of bending moment and the bending factor
  • C
    By dividing the sum of axial and eccentric loads by the product of bending moment and bending factor
  • D
    None of these
  • E

Answer

Correct Answer: By adding the axial load and the contribution due to eccentricity, i.e., axial load plus the product of bending moment from eccentricity and an appropriate bending factor

Explanation

Introduction / Context:Eccentric axial load P acting at an eccentricity e produces a combined effect: direct compression and bending with moment M = P * e. Some design approaches convert this combined action into an “equivalent axial load” using a bending factor to simplify checks against allowable stress.

Given Data / Assumptions:

  • Column subjected to axial load P with eccentricity e.
  • Bending moment M = P * e considered.
  • Working-stress context using equivalent load concepts.

Concept / Approach:The extreme-fibre stress is usually checked as σ = P/A ± M/Z. Equivalently, some handbooks propose a transformed axial load by adding a term proportional to M (via a bending factor) to P to maintain a single check against allowable compressive stress.

Step-by-Step Solution:Compute M = P * e.Introduce a bending factor α reflecting section properties and stress distribution.Define P_eq ≈ P + α * M to represent the combined effect conservatively.

Verification / Alternative check:Direct stress interaction (P/A + M/Z ≤ σ_allow) provides the exact check; the equivalent-load approach with α * M is an expedient that parallels interaction.

Why Other Options Are Wrong:

  • Subtracting or dividing by M and a factor has no rational basis for a conservative equivalent load.
  • Hence option (a) is the only reasonable description.

Common Pitfalls:Relying only on “equivalent load” without verifying combined-stress interaction can miss critical tension on one face.

Final Answer:By adding the axial load and the contribution due to eccentricity, i.e., axial load plus the product of bending moment from eccentricity and an appropriate bending factor

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