A slender uniform column of length $L$ and flexural rigidity $EI$ is fixed at one end and pinned at the other end. The effective buckling length $L_e$ of this column is approximately:

(A) $L_e \approx 0.7 L$

(B) $L_e = 0.5 L$

(C) $L_e = 1.0 L$

(D) $L_e = 2.0 L$

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles Euler critical buckling load is given by $P_{cr} = \frac{\pi^2 EI}{L_e^2}$, where $L_e = K L$ is the effective length between points of inflection: 1. Pinned-Pinned: $L_e = 1.0 L$ 2. Fixed-Fixed: $L_e = 0.5 L$ 3. Fixed-Free (cantilever): $L_e = 2.0 L$ 4. Fixed-Pinned: From transcendental equation $\tan(\mu L) = \mu L$, the lowest non-zero root gives $\mu L \approx 4.4934$, so $P_{cr} = \frac{20.19 EI}{L^2} = \frac{\pi^2 EI}{(0.699 L)^2} \implies L_e \approx 0.7 L$. ### 2. Distractor Forensics - **Option A ($0.7L$):** Correct effective length for fixed-pinned. - **Option B ($0.5L$):** Fixed-fixed. - **Option C ($1.0L$):** Pinned-pinned. - **Option D ($2.0L$):** Fixed-free. ### 3. Exam Trap & Standard Reference - **Trap Alert:** Fixed-pinned effective length factor is $1/\sqrt{2.046} \approx 0.70$. - **Standard Reference:** Megson, T.H.G., *Aircraft Structures for Engineering Students*, Chapter 9.

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