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: Writing Euler’s crippling load formula.
According to Euler’s theory, the critical (crippling) load for a column is given by: \[ P_{cr} = \frac{\pi^2 EI}{(L_e)^2} \] where \( L_e \) is the effective length of the column.
Step 2: Identifying effective length.
For a column with one end fixed and the other end hinged, the effective length is: \[ L_e = \frac{l}{\sqrt{2}} \]
Step 3: Substituting effective length.
\[ P_{cr} = \frac{\pi^2 EI}{\left(\frac{l}{\sqrt{2}}\right)^2} = \frac{\pi^2 EI}{\frac{l^2}{2}} \] \[ P_{cr} = \frac{2\pi^2 EI}{l^2} \]
Step 4: Conclusion.
The crippling load for the column is \( \dfrac{2\pi^2 EI}{l^2} \).

Answer 2

Euler's crippling load depends on the end-fixity condition through an effective length factor. The four standard cases are: both ends hinged (factor 1), one end fixed and the other free (factor 2), both ends fixed (factor \( \tfrac{1}{2} \)), and one end fixed with the other hinged (factor \( \tfrac{1}{\sqrt{2}} \)). Since the crippling load scales as the inverse square of this factor relative to the pinned-pinned case \( P_{cr} = \dfrac{\pi^2EI}{l^2} \), let's check the options.

1. \( \dfrac{\pi^2EI}{l^2} \): This is the crippling load for a column hinged at both ends (effective length factor 1), not the fixed-hinged case described here, which has a smaller effective length and therefore a larger crippling load.
2. \( \dfrac{\pi^2EI}{4l^2} \): This corresponds to a column fixed at one end and free at the other (effective length factor 2, giving one-quarter the pinned-pinned load), the weakest of the four standard cases, not the fixed-hinged case.
3. \( \dfrac{2\pi^2EI}{l^2} \): For one end fixed and the other hinged, the effective length is \( \dfrac{l}{\sqrt{2}} \), which is shorter than the pinned-pinned length \( l \), so the column is stiffer and the crippling load is scaled up by a factor of \( (\sqrt{2})^2 = 2 \) relative to the pinned-pinned case, giving \( \dfrac{2\pi^2EI}{l^2} \). This matches the fixed-hinged support condition exactly.
4. \( \dfrac{4\pi^2EI}{l^2} \): This is the crippling load for a column fixed at both ends (effective length factor \( \tfrac{1}{2} \), giving four times the pinned-pinned load), the strongest of the four standard cases, not the fixed-hinged case.

Matching the fixed-hinged end condition to its standard effective-length factor confirms the crippling load is \( \dfrac{2\pi^2EI}{l^2} \).

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