SSC JE Civil Engineering 23-Sep-2019 Shift 1 Solved Paper

The expression for crippling load for a column with both ends hinged
● The load at which the column just buckles (or bends) is called crippling load. Consider a column AB of length l and uniform crosssectional area, hinged at both of its ends A andB.
● Let P be the crippling load at which the column has just buckled. Due to the crippling load, the column wiII deflect into a curved form ACB as shown in Figure below:But moment = $\mathit{E}\mathit{I}\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}$
Equating the two moments, we have
$\mathit{E}\mathit{I}\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}=-\mathit{P}.\mathit{y}$
or $\mathit{E}\mathit{I}\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}+\mathit{P}.\mathit{y}=0$
or $\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}+\frac{\mathit{P}}{\mathit{E}\mathit{I}}.\mathit{y}=0$
The solution of the above differential equation is
$\mathit{y}={\mathit{C}}_1.\cos \left(\mathit{x}\sqrt{\frac{\mathit{P}}{\mathit{E}\mathit{I}}}\right)$ $+{\mathit{C}}_2.\sin \left(\mathit{x}\sqrt{\frac{\mathit{F}}{\mathit{E}\mathit{I}}}\right)$ ...(i)
Where ${\mathit{C}}_1$ and ${\mathit{C}}_2$ are the constants of integration and the valuesare obtained as follows:
At A, x = 0 and y = 0.
Substituting these values in equation (i), we get
$0={\mathit{C}}_1.\cos 0°+{\mathit{C}}_2\sin 0°$
$={\mathit{C}}_1\times 1+{\mathit{C}}_2\times 0$ (∵ cos 0° = 1 and sin 0° = 0)
= ${\mathit{C}}_1$
∴ ${\mathit{C}}_1=0.$ ...(ii)
(ii) At B, x = l and y = 0 (See Fig. above)
Substituting these values in equation (i), we get
$0={\mathit{C}}_1.\cos \left(\mathrm{l}\times \sqrt{\frac{\mathit{P}}{\mathit{E}\mathit{I}}}\right)$ $+{\mathit{C}}_2.\sin \left(\mathrm{l}\times \sqrt{\frac{\mathit{P}}{\mathit{E}\mathit{I}}}\right)$
$0={\mathit{C}}_2.\sin \left(\mathrm{l}\times \sqrt{\frac{\mathit{P}}{\mathit{E}\mathit{I}}}\right)$ [∵ ${\mathit{C}}_1$ = 0 from equation (ii)]
$={\mathit{C}}_2\sin \left(\mathrm{l}\sqrt{\frac{\mathit{P}}{\mathit{E}\mathit{I}}}\right)$ ...(iii)
From equation (iii), it is clear that either ${\mathit{C}}_2=0$
$\sin \left(\mathrm{l}\sqrt{\frac{\mathit{P}}{\mathit{E}\mathit{I}}}\right)=0$
As if ${\mathit{C}}_1$ = 0, then if ${\mathit{C}}_2$ is also equals to zero, then from Eqn no. (i), we will find that y = 0. This means that the bending of the column will be zero or the column will not bend at all, which is not true.
∴ $\sin \left(\mathrm{l}\sqrt{\frac{\mathit{P}}{\mathit{E}\mathit{I}}}\right)=0$
= sin 0 or sin π or sin 2π or sin 3π or...
or $\mathrm{l}\sqrt{\frac{\mathit{P}}{\mathit{E}\mathit{I}}}=0,$ or π or 2π or 3π or....
Taking the least practival value.
$\mathrm{l}\sqrt{\frac{\mathit{P}}{\mathit{E}\mathit{I}}}=\mathit{\pi },$
$\mathit{P}=\frac{{\mathit{\pi }}^2\mathit{E}\mathit{I}}{{\left(\mathrm{l}\right)}^2}$.