Question

A column fixed at one end and free at the other end buckles at a load P. Now, both the ends of the column are fixed. What is the buckling load for these end conditions?

• A. 2P
• B. 4P
• C. 8P
• D. 16P

Hint:

Remember the effective length factors for different end conditions of columns: - Fixed-free: \( L_e = 2L \) - Both ends pinned: \( L_e = L \) - One end fixed, one end pinned: \( L_e = 0.7L \) - Both ends fixed: \( L_e = 0.5L \)

Answer 1

The Correct Option is D

Step 1: Recall Euler's buckling formula.
The critical buckling load \( P_{cr} \) for a column is given by Euler's formula: \[ P_{cr} = \frac{\pi^2 EI}{L_e^2}, \] where \( E \) is the modulus of elasticity of the column material, \( I \) is the minimum area moment of inertia of the column's cross-section, and \( L_e \) is the effective length of the column, which depends on the end conditions. Step 2: Determine the effective length for the first end condition (fixed-free).
For a column fixed at one end and free at the other end, the effective length \( L_{e1} = 2L \), where \( L \) is the actual length of the column. The buckling load for this case is given as \( P \): \[ P = \frac{\pi^2 EI}{(2L)^2} = \frac{\pi^2 EI}{4L^2}. \] Step 3: Determine the effective length for the second end condition (both ends fixed).
For a column with both ends fixed, the effective length \( L_{e2} = \frac{L}{2} \), where \( L \) is the actual length of the column. Let the buckling load for this case be \( P' \). \[ P' = \frac{\pi^2 EI}{(L/2)^2} = \frac{\pi^2 EI}{L^2/4} = \frac{4\pi^2 EI}{L^2}. \] Step 4: Relate the buckling load for the second case (\( P' \)) to the buckling load for the first case (\( P \)).
We have \( P = \frac{\pi^2 EI}{4L^2} \) and \( P' = \frac{4\pi^2 EI}{L^2} \). We can express \( \frac{\pi^2 EI}{L^2} \) in terms of \( P \): \[ \frac{\pi^2 EI}{L^2} = 4P. \] Now substitute this into the expression for \( P' \): \[ P' = 4 \times \left( \frac{\pi^2 EI}{L^2} \right) = 4 \times (4P) = 16P. \] Step 5: Select the correct answer.
The buckling load when both ends of the column are fixed is 16P, which corresponds to option 4.