Difficulty: Easy
Correct Answer: Disagree
Explanation:
Introduction / Context:
In column buckling, the concept of effective or equivalent length converts different end restraints into an equivalent pinned–pinned length for use in Euler's critical load formula. The statement claims that for a column with one end fixed and the other end free, the equivalent length is half of the actual length. Understanding end conditions is essential for safe structural design.
Given Data / Assumptions:
Concept / Approach:
Euler buckling uses Pcr = (pi^2 * E * I) / (Le^2), where Le is the effective length. The effective length depends on rotational and translational restraint at the ends. Standard K-factors relate Le to the real length l by Le = K * l.
Step-by-Step Solution:
For pinned–pinned: K = 1.0, Le = l.For fixed–fixed: K = 0.5, Le = 0.5 * l.For fixed–free (cantilever): K = 2.0, Le = 2 * l.For fixed–hinged: K ≈ 0.7, Le ≈ 0.7 * l.Therefore the claim that Le equals half the length for a fixed–free column is incorrect; it should be double the actual length.
Verification / Alternative check:
The buckled mode shape for a cantilever is a quarter sine wave over 2 * l when mapped to the pinned–pinned reference, justifying K = 2.0 and not 0.5.
Why Other Options Are Wrong:
Agree: Incorrect because fixed–free gives K = 2.0, not 0.5. Only true for both ends fixed: Half-length applies to fixed–fixed, not fixed–free. Only true for both ends hinged: Hinged–hinged uses Le = l. Only true for fixed–hinged: That case has K ≈ 0.7, not 0.5.
Common Pitfalls:
Confusing fixed–free with fixed–fixed; memorizing formulas without linking them to end restraints and mode shapes; ignoring that K reflects rotational and translational fixity.
Final Answer:
Disagree
Discussion & Comments