Question

A circular rod of length \( l = 2 \, \text{m} \) is subjected to a compressive load \( P \), as shown in the figure. The bending (flexural) rigidity of the rod is \( 2000 \, \text{Nm}^2 \). If both ends are pinned, then the critical load \( P_{cr} \) in N (rounded to the nearest integer) at which the rod buckles elastically is 

• A. 4935
• B. 2000
• C. 5167
• D. 1238

Hint:

For a rod with pinned ends, the critical load can be calculated using the formula \( P_{cr} = \frac{\pi^2 EI}{l^2} \), where \( EI \) is the flexural rigidity.

Answer 1

The Correct Option is A

The critical buckling load \( P_{cr} \) for a rod with both ends pinned can be calculated using the formula for the buckling of a column: \[ P_{cr} = \frac{\pi^2 EI}{l^2} \] Where:
- \( E \) is the Young's Modulus of the material (not provided directly but inferred from the flexural rigidity),
- \( I \) is the area moment of inertia of the cross-section of the rod (again, inferred from flexural rigidity),
- \( l \) is the length of the rod.
The flexural rigidity \( EI \) is given as 2000 Nm². The formula for the critical load is then: \[ P_{cr} = \frac{\pi^2 \times 2000}{2^2} \] Substituting the values: \[ P_{cr} = \frac{\pi^2 \times 2000}{4} = 4935 \, \text{N}. \] Thus, the critical load at which the rod will buckle elastically is 4935 N, and the correct answer is (A).