**Data:**

E = 210 Gpa

$E = 210 \times 10^3 \ Mpa$

Axial Load = 17 KN

$L_e = 6m = 6000 \ mm$

**1] Moment of Inertia:**

$I_{xx} = \frac{80 \times (120 + 10 + 10)^3}{12} - \frac{(80 – 20) \times 120^3}{12} = 9.65 \times 10^6 \ mm^4$

$I_{yy} = \frac{ 2 \times 10 \times 80^3}{12} + \frac{120 \times 20^3}{12} = 9.33 \times 10^5 \ mm^4$

$Lesser \ moment \ of \ Inertia = I = I_{yy} = 9.33 \times 10^5 \ mm^4$

**2] (P) Crippling load by Euler’s formula:**

$P = \frac{\pi^2 \ E \ I}{L^2e} = \frac{\pi^2 \times 210 \times 10^3 \times 9.33 \times 10^5} {6000^2} = 53.71 \ KN$

**3] $Safe \ load = \frac{Crippling \ Load}{F.O.S}$**

$\therefore \ F.O.S = \frac{53.71}{17} = 3.15$ **(Ans).**