Euler's column theory: For columns of equal material and length l, the crippling load for a column with one end fixed and the other end free is __________ compared to a similar column hinged at both ends. Choose the correct comparison.
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AEqual to
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BLess than
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CMore than
Answer
Correct Answer: Less than
Explanation
Introduction / Context:Euler's buckling theory relates the critical (crippling) load to the boundary conditions via the effective length. Knowing how end restraint changes buckling capacity is essential for safe column design.
Given Data / Assumptions:
- Same column length l, material (E), and section (I).
- Case 1: One end fixed and the other free.
- Case 2: Hinged (pinned) at both ends.
- Elastic buckling (Euler region).
Concept / Approach:Euler load P_cr = (pi^2 * E * I) / (l_e^2), where l_e is the effective length. For pinned–pinned: l_e = l. For fixed–free: l_e = 2l. Larger l_e reduces P_cr quadratically.
Step-by-Step Solution:
Pinned–pinned: P_cr,pp = (pi^2 * E * I) / (l^2).Fixed–free: P_cr,ff = (pi^2 * E * I) / (2l)^2 = (pi^2 * E * I) / (4l^2).Ratio: P_cr,ff / P_cr,pp = 1/4 → fixed–free carries one-quarter of the pinned–pinned capacity.Therefore, P_cr for fixed–free is less than that for pinned–pinned.Verification / Alternative check:Effective length factors K: K = 1.0 (pinned–pinned), K = 2.0 (fixed–free). Since P_cr ∝ 1/K^2, doubling K quarters the load, consistent with the result.
Why Other Options Are Wrong:
- Equal to/More than: contradict the effective-length dependence where a cantilevered column is the weakest restraint case.
Common Pitfalls:Confusing fixed–free with fixed–fixed (the latter has higher capacity, K = 0.5).
Final Answer:Less than