Solution to Problem 609 | Double Integration Method

By symmetry
$R_1 = R_2 = P$
 

$EI \, y'' = Px - P \langle x - a \rangle - P \langle x - L + a \rangle$

$EI \, y' = \frac{1}{2} Px^2 - \frac{1}{2}P \langle x - a \rangle^2 - \frac{1}{2}P \langle x - L + a \rangle^2 + C_1$

$EI \, y = \frac{1}{6} Px^3 - \frac{1}{6} P \langle x - a \rangle^3 - \frac{1}{6}P \langle x - L + a \rangle^3 + C_1x + C_2$
 

 

At x = 0, y = 0, therefore C₂ = 0

At x = L, y = 0
$0 = \frac{1}{6} PL^3 - \frac{1}{6} P \langle L - a \rangle^3 + C_1L$

$0 = PL^3 - P(L^3 - 3L^2a + 3La^2 - a^3) - Pa^3 + 6C_1L$

$0 = PL^3 - PL^3 + 3PL^2a - 3PLa^2 + Pa^3 - Pa^3 + 6C_1L$

$0 = 3PL^2a - 3PLa^2 + 6C_1L$

$0 = 3PLa (L - a) + 6C_1L$

$C_1 = -\frac{1}{2}Pa (L - a)$
 

Therefore,
$EI \, y = \frac{1}{6} Px^3 - \frac{1}{6} P \langle x - a \rangle^3 - \frac{1}{6}P \langle x - L + a \rangle^3 - \frac{1}{2}Pa (L - a)x$
 

Maximum deflection will occur at x = ½ L (midspan)
$EI \, y_{max} = \frac{1}{6} P(\frac{1}{2}L)^3 - \frac{1}{6} P(\frac{1}{2}L - a)^3 - \frac{1}{2}Pa (L - a)(\frac{1}{2}L)$

$EI \, y_{max} = \frac{1}{48} PL^3 - \frac{1}{6} P\, [ \, \frac{1}{2}(L - 2a) \, ]^3 - \frac{1}{4}PL^2a + \frac{1}{4}PLa^2$

$EI \, y_{max} = \frac{1}{48} PL^3 - \frac{1}{48}P \, [ \, L^3 - 3L^2(2a) + 3L(2a)^2 - (2a)^3 \, ] - \frac{1}{4}PL^2a + \frac{1}{4}PLa^2$

$EI \, y_{max} = \frac{1}{48} PL^3 - \frac{1}{48} PL^3 + \frac{1}{8}PL^2a - \frac{1}{4}PLa^2 + \frac{1}{6}Pa^3 - \frac{1}{4}PL^2a + \frac{1}{4}PLa^2$

$EI \, y_{max} = -\frac{1}{8}PL^2a + \frac{1}{6}Pa^3$

$EI \, y_{max} = -\frac{1}{24}Pa \, (3L^2 - 4a^2)$

$y_{max} = -\dfrac{Pa}{24EI}(3L^2 - 4a^2)$
 

$\delta_{max} = \dfrac{Pa}{24EI}(3L^2 - 4a^2)$           answer