Question

A column with one end fixed and one end free has a critical buckling load of 100 N. For the same column, if the free end is replaced with a pinned end then the critical buckling load will be \(\underline{\hspace{1cm}}\) N (round off to the nearest integer).

Hint:

For buckling problems, the critical load is affected by the boundary conditions of the column. A column with both ends pinned has a higher critical buckling load than one with one end fixed and the other free.

Answer 1

Correct Answer: 800 - 840

Step 1: Recall Euler's Buckling Formula

The critical buckling load for a column is given by:

$$P_{cr} = \frac{\pi^2 EI}{L_e^2}$$

where:

• $E$ = Modulus of elasticity
• $I$ = Moment of inertia
• $L_e$ = Effective length of the column

Step 2: Determine Effective Lengths

The effective length depends on the end conditions:

For fixed-free end conditions: $$L_e = 2L$$

For fixed-pinned end conditions: $$L_e = \frac{L}{\sqrt{2}} \approx 0.707L$$

Step 3: Write Critical Load Equations

For fixed-free column: $$P_{cr1} = \frac{\pi^2 EI}{(2L)^2} = \frac{\pi^2 EI}{4L^2} = 100 \text{ N}$$

For fixed-pinned column: $$P_{cr2} = \frac{\pi^2 EI}{(0.707L)^2} = \frac{\pi^2 EI}{0.5L^2} = \frac{2\pi^2 EI}{L^2}$$

Step 4: Find the Ratio

Taking the ratio: $$\frac{P_{cr2}}{P_{cr1}} = \frac{\frac{2\pi^2 EI}{L^2}}{\frac{\pi^2 EI}{4L^2}} = \frac{2\pi^2 EI}{L^2} \times \frac{4L^2}{\pi^2 EI} = 8$$

Step 5: Calculate the New Critical Load

$$P_{cr2} = 8 \times P_{cr1} = 8 \times 100 = 800 \text{ N}$$

Answer: The critical buckling load with fixed-pinned end conditions is 800 N.