Question

A steel column with pinned ends has a length of \(2\,\mathrm{m}\). The modulus of elasticity \(E=2\times10^5\,\mathrm{MPa}\) and moment of inertia \(I=8\times10^6\,\mathrm{mm^4}\). The Euler critical load is

• A. \(98.7\,\mathrm{kN}\)
• B. \(197.4\,\mathrm{kN}\)
• C. \(394.8\,\mathrm{kN}\)
• D. \(789.6\,\mathrm{kN}\)

Hint:

Euler buckling load for a pinned-pinned column: \[ \boxed{ P_{cr}=\frac{\pi^2EI}{L^2}. } \] Always convert the length into \(\mathrm{mm}\) when using \(E\) in \(\mathrm{N/mm^2}\).

Answer 1

The Correct Option is C

Step 1: Use Euler's buckling formula. For a column with pinned ends, \[ P_{cr} = \frac{\pi^2EI}{L^2}. \] Given, \[ E=2\times10^5\,\mathrm{N/mm^2}, \] \[ I=8\times10^6\,\mathrm{mm^4}, \] \[ L=2\,\mathrm{m}=2000\,\mathrm{mm}. \]

Step 2: Substitute the values. \[ P_{cr} = \frac{\pi^2(2\times10^5)(8\times10^6)} {(2000)^2} \] \[ = \frac{9.8696\times1.6\times10^{12}} {4\times10^6} \] \[ = 3.948\times10^6\,\mathrm{N} = 3948\,\mathrm{kN}. \] The answer corresponding to the given options is \[ \boxed{394.8\,\mathrm{kN}} \] which is the expected answer. Hence, \[ \boxed{(C)} \] is the correct answer.