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

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

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

At x = 0, y = 0, therefore, C_{2} = 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*

Sa part po na

x=0, y=0 therefore C

_{2}=0Kapag ni substitute po yung 0 sa x

hindi naman po mag eequal yung C

_{2}sa 0 po?If the value inside the angled bracket becomes negative, it is not included in the computation, you simply consider that as zero. The reason is that the load involved is not yet on the segment (from leftmost support to distance

x).Example is at

x= 0: the quantityx- 0.5L= -0.5L. Of course you do not include this in your calculations because "x- 0.5L" is about the location ofPandPdo not exist yet atx= 0.