Solving for the propped reaction R

$EI \, t_{A/B} = 0$

$(Area_{AB}) \cdot \bar{X}_A = 0$

$\frac{1}{2}L(RL)(\frac{2}{3}L) - \frac{1}{2}(\frac{1}{2}L)(\frac{1}{2}PL)(\frac{5}{6}L) = 0$

$\frac{1}{3}RL^3 - \frac{5}{48}PL^3 = 0$

$\frac{1}{3}R - \frac{5}{48}P = 0$

$R = \frac{5}{16}P$ *answer*

Solving for the midspan deflection δ_{mid}

$\dfrac{y}{\frac{1}{2}L} = \dfrac{RL}{L}$

$y = \frac{1}{2}RL$

$EI \, t_{C/B} = (Area_{BC}) \cdot \bar{X}_C$

$EI \, t_{C/B} = \frac{1}{2}(\frac{1}{2}L)(RL)(\frac{1}{3}L) + \frac{1}{2}(\frac{1}{2}L)(\frac{1}{2}RL)(\frac{1}{6}L) - \frac{1}{2}(\frac{1}{2}L)(\frac{1}{2}PL)(\frac{1}{3}L)$

$EI \, t_{C/B} = \frac{1}{12}RL^3 + \frac{1}{48}RL^3 - \frac{1}{24}PL^3$

$EI \, t_{C/B} = \frac{5}{48}RL^3 - \frac{1}{24}PL^3$

$EI \, t_{C/B} = \frac{5}{48}(\frac{5}{16}P)L^3 - \frac{1}{24}PL^3$

$EI \, t_{C/B} = \frac{25}{768}PL^3 - \frac{1}{24}PL^3$

$EI \, t_{C/B} = -\frac{7}{768}PL^3$

Thus,

$EI \, \delta_{mid} = \frac{7}{768}PL^3$ *answer*

Gusto ko po sana itanong kung hindi po ba pwede hatiin into a rectangle and triangle both RL/2 ang height nung trapezoid sa pag kuha ng midspan deflection. Nung tinatry ko po kasi iba lumalabas na sagot. Thank you.

Pwede naman po, pero dapat pareho lang sagot.