Solution to Problem 636 | Deflection of Cantilever Beams

$t_{A/B} = \dfrac{1}{EI}(Area_{AB}) \, \bar{X}_A$

$-10 = \dfrac{1}{10\,000\left( \dfrac{50h^3}{12} \right)}\Bigg[ -\frac{1}{2}(2)(4)(\frac{10}{3}) - \frac{1}{2}(4)(16)(\frac{8}{3}) \Bigg] \, (1000^4)$

$-10 = \dfrac{3}{125\,000h^3}\left[ \, -\dfrac{296}{3} \, \right](1000^4)$

$h^3 = \dfrac{-296(1000^4)}{125\,000(-10)}$

$h = 618.67 \, \text{mm}$           answer