Columns — effective length equivalence: A column of actual length l with both ends fixed is equivalent in buckling strength to a column of what length with one end fixed and the other end free (cantilever)?
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Al/8
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Bl/4
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Cl/2
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Dl
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E2l
Answer
Correct Answer: l/4
Explanation
Introduction / Context:Buckling capacity depends on a column’s effective length, which incorporates end restraints. Converting between end conditions using effective length allows direct comparison of Euler loads for different support cases.
Given Data / Assumptions:
- Euler buckling for long, perfectly straight columns.
- Effective length factor K: fixed–fixed K = 0.5; fixed–free (cantilever) K = 2.0.
- Pcr = π^2 * E * I / (K * l)^2.
Concept / Approach:Set the Euler load for the fixed–fixed column equal to that of a cantilevered column of unknown length Lc. Equality of Pcr implies equality of effective lengths.
Step-by-Step Solution:
Fixed–fixed: Le1 = 0.5 * l.Cantilever: Le2 = 2 * Lc.Equate effective lengths: 0.5 * l = 2 * Lc → Lc = l / 4.Therefore, a fixed–fixed column of length l is equivalent to a fixed–free column of length l/4.Verification / Alternative check:Equate Euler loads explicitly: π^2 E I / (0.5 l)^2 = π^2 E I / (2 Lc)^2 → Lc = l/4, confirming the result.
Why Other Options Are Wrong:
- l/8, l/2, l, 2l do not satisfy equality of effective lengths between the two cases.
Common Pitfalls:Mixing K and C notations; remember C = K^2 if P = π^2 E I / (C l^2).
Final Answer:l/4