Question

A long Column with fixed ends can carry load as compared to one end fixed and the other free is

• A. 2 times
• B. 4 times
• C. 8 times
• D. 16 times

Hint:

The buckling load is inversely proportional to the square of the effective length ($P_{cr} \propto 1/L_{eff}^2$).
- Fixed-Fixed: $L_{eff} = L/2$
- Fixed-Free: $L_{eff} = 2L$
The effective length is 4 times smaller for fixed-fixed. Therefore, the load capacity is $4^2 = 16$ times larger.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks to compare the critical buckling load of a long column with fixed-fixed ends to that of a column with fixed-free ends.

Step 2: Key Formula or Approach:
The critical buckling load ($P_{cr}$) for a long column is given by Euler's formula:
\[ P_{cr} = \frac{\pi^2 EI}{(L_{eff})^2} = \frac{\pi^2 EI}{(KL)^2} \] where $L_{eff}$ is the effective length, $K$ is the effective length factor, and $L$ is the actual length.

Step 3: Detailed Explanation:
We need to find the buckling load for each case. Let's assume both columns have the same actual length $L$, and the same material and cross-section (same $EI$).

Case 1: Fixed-Fixed Ends
- Effective length factor, $K_1 = 0.5$
- Critical load, $P_{cr1} = \frac{\pi^2 EI}{(0.5L)^2} = \frac{\pi^2 EI}{0.25L^2} = 4 \frac{\pi^2 EI}{L^2}$

Case 2: Fixed-Free Ends
- Effective length factor, $K_2 = 2.0$
- Critical load, $P_{cr2} = \frac{\pi^2 EI}{(2.0L)^2} = \frac{\pi^2 EI}{4L^2} = 0.25 \frac{\pi^2 EI}{L^2}$
Now, find the ratio of the loads:
\[ \frac{\text{Load of Fixed-Fixed}}{\text{Load of Fixed-Free}} = \frac{P_{cr1}}{P_{cr2}} = \frac{4 (\pi^2 EI / L^2)}{0.25 (\pi^2 EI / L^2)} \] \[ \text{Ratio} = \frac{4}{0.25} = \frac{4}{1/4} = 4 \times 4 = 16 \]

Step 4: Final Answer:
A long column with fixed ends can carry 16 times the load of a column with one end fixed and the other free.