A column of height 12 m is fixed at both the ends. Its c/s is a symmetrical I-section. Flanges are (100 mm x 15 mm) and web is (130 mm x 15 mm). Find the factor of safety w.r.t buckling for an axial load of 17 kN. Use Euler’s Theory. E=2x10^5 MPa.

Data:

(L) Length = 12 m

Both ends are fixed. i.e. $Le = \frac{1}{2} \ L$

Axial Load = 17 KN

$E = 2 \times 10^5 \ Mpa$

1] Moment of Inertia:

$I_{xx} = \frac{100 \times 160^3}{12} - \frac{(100- 15) \times 130^3}{12} = 18.57 \times 10^6 \ mm^4 $

$I_{yy} = 2 \times \frac{15 \times 100^3}{12} + \frac{130 \times 15^3}{12} = 2.50 \times 10^6 \ mm^4$

$Lesser \ moment \ of \ Inertia = I = I_{yy} = 2.50 \times 10^6 \ mm^4$

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

$P = \frac{\pi^2 \ E \ I}{L^2e}$

$Le = \frac{1}{2} \ \times L = \frac{1}{2} \times 12000 = 6000 \ mm$ (Both ends are fixed).

$\therefore$ $P = \frac{\pi^2 \times 2 \times 10^5 \times 2.50 \times 10^6}{6000^2} = 137.07 \times 10^3 \ N $

i.e. P = 137.07 KN.

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

$\therefore$ $\ F.O.S = \frac{Crippling \ load}{Safe \ load} = \frac{137.07}{17} = 8.06 $

$\therefore$ Factor of safety is 8.06.