A steel column of rectangular section (15 mm x 10 mm)  and length 1.5 m is simply supported at both ends. Assuming modulus of elasticity, E 200 GPa for steel, the critical axial load (in kN) is _______ (correct to two decimal places)

The correct answer is 1.10 (or 1.10 kN).

Step-by-Step Explanation:

1. Identify the given parameters:
- Cross-section dimensions of the rectangular steel column: width $b=15\text{mm}$ and depth $d=10\text{mm}$.
- Length of the column: $L=1.5\text{m}=1500\text{mm}$.
- Modulus of elasticity of steel: $E=200\text{GPa}=200\times {10}^3\text{N/mm}²$ (or $\text{MPa}$).
- Support conditions: Simply supported at both ends, which means the effective length factor is $k=1.0$, so the effective length is $L_e=L=1500\text{mm}$.

2. Calculate the minimum area moment of inertia ($I_{\min }$):
For a rectangular cross-section, the column can buckle about either of its principal axes. Buckling will occur about the axis with the minimum area moment of inertia to minimize the critical load.
The moments of inertia are given by:
$I_x=\frac{b·d^3}{12}$ and $I_y=\frac{d·b^3}{12}$
Thus, the minimum moment of inertia is obtained by using the smaller dimension as the depth (cubed term):
$I_{\min }=\frac{15\times {10}^3}{12}{\text{mm}}^4$
$I_{\min }=\frac{15\times 1000}{12}=1250{\text{mm}}^4$

3. Calculate the Euler critical axial buckling load ($P_{\mathrm{cr}}$):
Euler's buckling load formula for a column pinned (simply supported) at both ends is:
$P_{\mathrm{cr}}=\frac{{\pi }^{2}E{I}_{\min }}{L^2}$
Substituting the values in terms of Newtons and millimeters:
$P_{\mathrm{cr}}=\frac{{\pi }^2\times (200\times {10}^3)\times 1250}{{1500}^2}$
$P_{\mathrm{cr}}=\frac{{\pi }^2\times 2.5\times {10}^8}{2.25\times {10}^6}$
$P_{\mathrm{cr}}=\frac{250{\pi }^2}{2.25}\approx 1096.62\text{N}$

4. Convert the load to kilonewtons (kN):
$P_{\mathrm{cr}}=\frac{1096.62}{1000}\approx 1.0966\text{kN}$
Rounding to two decimal places, we get:
$P_{\mathrm{cr}}\approx 1.10\text{kN}$