SSC JE Civil Engineering 30 Oct 2020 Shift 2 Paper

Explanation: To find maximum deflection expression. Students are advised to memorise the standard results. This $\delta = \frac{qL^4}{30EI}$ is a standard result for cantilever beam with UVL. For derivation of formula refer this Here $w = \frac{px}{L}$ per unit length at a distance $x$ from $O$ ∴$EI \frac{d^4 y}{dx^4} = -\frac{px}{L}$ $EI \frac{d^3 y}{dx^3} = -\frac{px^2}{2L} + C_1$ S.F. is zero at x = 0, giving $C_1$ = 0 ∴$EI \frac{d^3 y}{dx^3} = -\frac{px^2}{2L}$ ∴ $EI \frac{d^2 y}{dx^2} = -\frac{px^3}{6L} + C_2$ B.M. is zero at x = 0, giving $C_2$ = 0 $EI \frac{d^2 y}{dx^2} = -\frac{px^3}{6L}$ Integrate again $EI \frac{dy}{dx} = -\frac{px^4}{24L} + C_3$ But $\frac{dy}{dx} = 0$ at $x = L$, giving   $0 = -\frac{pL^4}{24L} + C_3$ i. e, $C_3 = \frac{pL^3}{24}$ $y = \frac{p}{24LEI} \left( L^4 x - \frac{x^5}{5} \right) + C_4$ Now y = 0 at x = L, giving $C_4 = -\frac{4pL^5}{5 \times 24EI L} = -\frac{pL^4}{30EI}$ ∴ $y = \frac{p}{24LEI} \left( L^4 x - \frac{x^5}{5} \right) - \frac{pL^4}{30EI}$ $y = -\frac{p}{120EI L} (4L^5 - 5L^4 x + x^5)$ Since the deflection will be maximum at the end O i.e. at x = 0, put this in the above obtained expression to get $\delta = \frac{pL^4}{30EI}$, where negative indicates the deflection is towards negative $y$ axis.