written 4.9 years ago by | modified 3.9 years ago by |
A column of 6 m effective length is to be made from 3 plates as shown in fig. Using E=210GPa find the factor of safety with respect to buckling for an axial load of 17 kN. Use Euler’s Theory.
written 4.9 years ago by | modified 3.9 years ago by |
A column of 6 m effective length is to be made from 3 plates as shown in fig. Using E=210GPa find the factor of safety with respect to buckling for an axial load of 17 kN. Use Euler’s Theory.
written 3.9 years ago by | • modified 3.9 years ago |
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).