Buckling of columns — parameters affecting Euler load: The critical (buckling) load for a given column depends on which set of properties?
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AArea of cross-section only
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BLength and least radius of gyration only
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CModulus of elasticity only
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DAll of the above (area through I, length, least radius of gyration, and E)
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EOnly end load magnitude
Answer
Correct Answer: All of the above (area through I, length, least radius of gyration, and E)
Explanation
Introduction / Context:Euler’s formula for long columns shows how geometry, material stiffness, and end conditions jointly govern buckling strength. Understanding each parameter's role is essential in member sizing.
Given Data / Assumptions:
- Pcr = π^2 * E * I / (K * L)^2 (long, straight columns).
- I = A * k^2, where k is radius of gyration.
- Least radius of gyration controls the weakest axis buckling.
Concept / Approach:Critical load scales linearly with E and I, and inversely with the square of effective length (K * L). Since I involves area A and radius k, area influences Pcr through the second moment of area. Thus multiple properties simultaneously affect buckling.
Step-by-Step Solution:
Write Euler: Pcr = π^2 * E * I / (K * L)^2.Substitute I = A * k^2 → Pcr ∝ E * A * k^2 / (K^2 * L^2).Identify dependencies: E, A, k (least), and L (with end condition factor K).Therefore, all listed geometric and material factors matter.Verification / Alternative check:Comparing two sections with same area but different k (e.g., solid vs thin-walled tube) shows different Pcr, confirming dependence on k in addition to A.
Why Other Options Are Wrong:
- Single-factor answers ignore key contributors.
- “Only end load magnitude” is irrelevant; buckling capacity is a property of the member.
Common Pitfalls:Equating larger area with higher buckling load without considering how area is distributed (k).
Final Answer:All of the above (area through I, length, least radius of gyration, and E)