Question

As per Euler’s theory, the crippling load for a column of length \( l \) with one end fixed and the other end hinged having Young’s modulus \( E \) and moment of inertia \( I \) is

• A. \( \dfrac{\pi^2 EI}{l^2} \)
• B. \( \dfrac{\pi^2 EI}{4l^2} \)
• C. \( \dfrac{2\pi^2 EI}{l^2} \)
• D. \( \dfrac{4\pi^2 EI}{l^2} \)

Hint:

Always remember effective length factors for Euler buckling problems based on end conditions.

Answer 1

The Correct Option is C

Step 1: For a column fixed at one end and hinged at the other, the buckled shape corresponds to roughly 1.43 half sine-waves fitting between the ends, rather than the single half-wave of a pinned-pinned column.
Step 2: This buckling mode is mathematically equivalent to a pinned-pinned column of a shorter equivalent length \( L_e = \dfrac{l}{\sqrt{2}} \), because that portion of the deflected shape between the effective inflection points matches one full pinned-pinned buckling half-wave.
Step 3: Applying Euler's formula with this equivalent length:
\[ P_{cr} = \dfrac{\pi^2EI}{L_e^2} = \dfrac{\pi^2EI}{l^2/2} = \dfrac{2\pi^2EI}{l^2} \]

Answer 2

A useful check is that the fixed-hinged case should give a crippling load strictly between the weaker pinned-pinned case and the stronger fixed-fixed case, since fixing one end only partially stiffens the column compared to fixing both. Testing the options against this ordering:

1. \( \dfrac{\pi^2EI}{l^2} \): This is the pinned-pinned value, the weakest of the standard cases; the fixed-hinged column, having one end restrained against rotation, must carry more load than this, so this option is too small.
2. \( \dfrac{\pi^2EI}{4l^2} \): This is even smaller than the pinned-pinned case, corresponding to the fixed-free column, the weakest standard case of all; it cannot represent the stiffer fixed-hinged condition.
3. \( \dfrac{2\pi^2EI}{l^2} \): This value is twice the pinned-pinned load and exactly half the fixed-fixed load of \( \dfrac{4\pi^2EI}{l^2} \), correctly placing the fixed-hinged case between the two extremes as physically expected.
4. \( \dfrac{4\pi^2EI}{l^2} \): This is the fixed-fixed value, the strongest of the standard cases; the fixed-hinged column, having only one end restrained, cannot carry this much load, so this option is too large.

Only \( \dfrac{2\pi^2EI}{l^2} \) falls in the physically expected range between the pinned-pinned and fixed-fixed extremes, matching the fixed-hinged end condition.

Therefore, the correct answer is \( \dfrac{2\pi^2EI}{l^2} \).