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 \( \dfrac{2\pi^2 EI}{l^2} \).