Question

A column of height 20 m is fixed at both ends. If Young’s modulus is \(E = 17\times 10^{9}\ \text{N/m}^2\) and moment of inertia \(I = 3.255\times 10^{-4}\ \text{m}^4\), then the first critical buckling load lies between ________________ kN.

• A. 501 and 520
• B. 521 and 540
• C. 541 and 560
• D. 561 and 580

Hint:

A fixed–fixed column has four times the buckling capacity of a pinned–pinned column.

Answer 1

The Correct Option is C

Step 1: Effective length for fixed–fixed column.
\[ L_{\text{eff}} = \frac{L}{2} = \frac{20}{2} = 10\ \text{m} \]
Step 2: Use Euler’s buckling formula.
\[ P_{cr} = \frac{\pi^{2}EI}{(L_{\text{eff}})^2} \] \[ = \frac{\pi^{2}(17\times10^{9})(3.255\times10^{-4})}{10^{2}} \]
Step 3: Compute numerator.
\[ 17\times10^{9} \times 3.255\times10^{-4} = 5.5335\times10^{6} \] \[ \pi^{2} \times 5.5335\times10^{6} \approx 54.7\times10^{6} \]
Step 4: Divide by 100.
\[ P_{cr} = 5.47\times10^{5}\ \text{N} = 547\ \text{kN} \] Conclusion:
Hence, the first critical buckling load lies between \(541\) and \(560\) kN.