SSC JE Civil Engineering 23 Sep 2019 Shift 1 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 = $EI\frac{d^2y}{dx^2}$ Equating the two moments, we have $EI\frac{d^2y}{dx^2} = -P\cdot y$ or $EI\frac{d^2y}{dx^2} + P\cdot y = 0$ or $\frac{d^2y}{dx^2} + \frac{P}{EI}\cdot y = 0$ The solution of the above differential equation is $y = C_1 \cdot \cos\left(x\sqrt{\frac{P}{EI}}\right)$ $+ C_2 \cdot \sin\left(x\sqrt{\frac{F}{EI}}\right)$ ...(i) Where $C_1$ and $C_2$ 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 = C_1 \cdot \cos 0^{\circ} + C_2 \sin 0^{\circ}$$= C_1 \times 1 + C_2 \times 0$ (∵ cos 0° = 1 and sin 0° = 0)= $C_1$∴ $C_1 = 0.$ ...(ii)(ii) At B, x = l and y = 0 (See Fig. above)Substituting these values in equation (i), we get$0 = C_1 \cdot \cos\left(l \times \sqrt{\frac{P}{EI}}\right)$ $+ C_2 \cdot \sin\left(l \times \sqrt{\frac{P}{EI}}\right)$$0 = C_2 \cdot \sin\left(l \times \sqrt{\frac{P}{EI}}\right)$ [∵ $C_1$ = 0 from equation (ii)]$= C_2 \sin\left(l\sqrt{\frac{P}{EI}}\right)$ ...(iii)From equation (iii), it is clear that either $C_2 = 0$$\sin\left(l\sqrt{\frac{P}{EI}}\right) = 0$As if $C_1$ = 0, then if $C_2$ 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.∴ $\sin\left(l\sqrt{\frac{P}{EI}}\right) = 0$= sin 0 or sin π or sin 2π or sin 3π or...or $l\sqrt{\frac{P}{EI}} = 0,$ or π or 2π or 3π or....Taking the least practival value. $l\sqrt{\frac{P}{EI}} = \pi,$ $P = \frac{\pi^2 EI}{(l)^2}$.