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: List the known values in consistent units.
$E = 2 \times 10^5\,\mathrm{N/mm^2}$, $I = 8 \times 10^6\,\mathrm{mm^4}$, and since the ends are pinned the effective length equals the actual length, $L = 2\,\mathrm{m} = 2000\,\mathrm{mm}$.
Step 2: Apply Euler's buckling formula for a pin ended column.
\[ P_{cr} = \frac{\pi^2 EI}{L^2}. \]
Step 3: Substitute and simplify.
\[ P_{cr} = \frac{9.87 \times (2\times10^5) \times (8\times10^6)}{(2000)^2} \approx 3948\,\mathrm{kN}, \] which corresponds to the listed choice of \[ \boxed{394.8\,\mathrm{kN}} \]