Solution to Problem 605 | Double Integration Method

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

$EI \, y' = \frac{1}{4}Px^2 - \frac{1}{2}P\langle x - \frac{1}{2}L \rangle^2 + C_1$

$EI \, y = \frac{1}{12}Px^3 - \frac{1}{6}P\langle x - \frac{1}{2}L \rangle^3 + C_1x + C_2$
 

At x = 0, y = 0, therefore, C₂ = 0
 

At x = L, y = 0
$0 = \frac{1}{12}PL^3 - \frac{1}{6}P\langle L - \frac{1}{2}L \rangle^3 + C_1L$

$0 = \frac{1}{12}PL^3 - \frac{1}{48}PL^3 + C_1L$

$C_1 = -\frac{1}{16}PL^2$
 

Thus,
$EI \, y = \frac{1}{12}Px^3 - \frac{1}{6}P\langle x - \frac{1}{2}L \rangle^3 - \frac{1}{16}PL^2x$
 

Maximum deflection will occur at x = ½ L (midspan)
$EI \, y_{max} = \frac{1}{12}P(\frac{1}{2}L)^3 - \frac{1}{6}P (\frac{1}{2}L - \frac{1}{2}L)^3 - \frac{1}{16}PL^2 (\frac{1}{2}L)$

$EI \, y_{max} = \frac{1}{96}PL^3 - 0 - \frac{1}{32}PL^3$

$y_{max} = -\dfrac{PL^3}{48EI}$
 

The negative sign indicates that the deflection is below the undeformed neutral axis.
 

Therefore,
$\delta_{max} = \dfrac{PL^3}{48EI}$           answer