Question

The critical buckling load of a 6 m long, 10 cm diameter axially loaded solid steel column with hinged support at both ends is ______ kN (rounded off to nearest integer).
Assume Young's modulus of steel as \(2 \times 10^5\) N/mm\(^2\).

Hint:

Use Euler's formula with effective length equal to actual length since both ends are hinged.

Answer 1

Correct Answer: 269

Step 1: Pick the right effective length.
For a column hinged at both ends, the effective length equals the actual length, so \(L_e = L = 6\) m.

Step 2: Find the second moment of area of the circular section.
For a solid circular section of diameter \(d = 0.1\) m, \(I = \dfrac{\pi d^4}{64} = \dfrac{\pi (0.1)^4}{64} = 4.909 \times 10^{-6}\) m\(^4\).

Step 3: Apply Euler's buckling formula.
\(P_{cr} = \dfrac{\pi^2 E I}{L_e^2}\), with \(E = 2 \times 10^5\) N/mm\(^2\) = \(2 \times 10^{11}\) Pa. So \(P_{cr} = \dfrac{\pi^2 \times 2 \times 10^{11} \times 4.909 \times 10^{-6}}{6^2} = \dfrac{9.688 \times 10^{6}}{36} \approx 269151\) N.

Final Answer:
\(P_{cr} \approx 269.15\) kN, which rounds to 269 kN, inside the official 268 to 270 kN band. \[ \boxed{P_{cr} \approx 269 \text{ kN}} \]