Question

Consider an initially perfectly straight elastic column with pinned supports at both ends. If \( E \) is the Young's modulus of the material, \( L \) is the length of the column between the supports, and \( I \) is the least moment of inertia of the constant cross-sectional area of the column, then the Euler load is given by:

• A. \( \frac{\pi^2 EI}{L^2} \)
• B. \( \frac{\pi^2 EI}{4L^2} \)
• C. \( \frac{\pi^2 EI}{\sqrt{2}L^2} \)
• D. \( \frac{2\pi^2 EI}{L^2} \)

Hint:

For Euler's buckling problems, always check the end conditions of the column to determine the effective length factor \( K \). This is crucial for calculating the critical load.

Answer 1

The Correct Option is A

Step 1: Recall the formula for Euler's critical load. 
The Euler load (critical buckling load) for a column with pinned supports at both ends is given by: \[ P_{\text{cr}} = \frac{\pi^2 EI}{(KL)^2}, \] where: - \( E \) is the Young's modulus, - \( I \) is the least moment of inertia of the cross-section, - \( L \) is the length of the column, - \( K \) is the effective length factor. 

Step 2: Determine the effective length factor for pinned ends. 
For a column with pinned supports at both ends, the effective length factor \( K \) is \( 1 \). Substituting \( K = 1 \) into the formula: \[ P_{\text{cr}} = \frac{\pi^2 EI}{L^2}. \] 

Step 3: Verify the answer. 
The expression matches the given Option (A). Conclusion: The Euler load for the column is \( \frac{\pi^2 EI}{L^2} \).