Question:medium
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\(^2\).
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\(^2\).
Assume Young's modulus of steel as \(2 \times 10^5\) N/mm\(^2\).
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Use Euler's formula with effective length equal to actual length since both ends are hinged.
Updated On: Jul 28, 2026
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Correct Answer: 269
Solution and Explanation
Step 1: Convert all data to consistent SI units first.
Diameter $d = 100$ mm $= 0.1$ m, length $L = 6$ m, and $E = 2 \times 10^5$ N/mm$^2$, which in pascals is $E = 2 \times 10^5 \times 10^6 = 2 \times 10^{11}$ Pa.
Step 2: Get the section's moment of inertia in mm to cross-check.
With $d = 100$ mm, $I = \pi d^4/64 = \pi (100)^4/64 = 4.909 \times 10^6$ mm$^4$, which converts to $4.909 \times 10^{-6}$ m$^4$, the same value found from the metre-based calculation.
Step 3: Both ends pinned means no reduction factor is needed.
For a pin-pin column the effective length factor is 1, so $P_{cr} = \pi^2 E I / L^2 = \pi^2 \times 2 \times 10^{11} \times 4.909 \times 10^{-6} / 36$. The numerator is $9.688 \times 10^6$, and dividing by 36 gives $269151$ N.
Final Answer:
$P_{cr} = 269.2$ kN, rounding to 269 kN, which sits in the 268 to 270 kN range. \[ \boxed{P_{cr} \approx 269 \text{ kN}} \]
Diameter $d = 100$ mm $= 0.1$ m, length $L = 6$ m, and $E = 2 \times 10^5$ N/mm$^2$, which in pascals is $E = 2 \times 10^5 \times 10^6 = 2 \times 10^{11}$ Pa.
Step 2: Get the section's moment of inertia in mm to cross-check.
With $d = 100$ mm, $I = \pi d^4/64 = \pi (100)^4/64 = 4.909 \times 10^6$ mm$^4$, which converts to $4.909 \times 10^{-6}$ m$^4$, the same value found from the metre-based calculation.
Step 3: Both ends pinned means no reduction factor is needed.
For a pin-pin column the effective length factor is 1, so $P_{cr} = \pi^2 E I / L^2 = \pi^2 \times 2 \times 10^{11} \times 4.909 \times 10^{-6} / 36$. The numerator is $9.688 \times 10^6$, and dividing by 36 gives $269151$ N.
Final Answer:
$P_{cr} = 269.2$ kN, rounding to 269 kN, which sits in the 268 to 270 kN range. \[ \boxed{P_{cr} \approx 269 \text{ kN}} \]
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