Solution to Problem 608 | Double Integration Method

$V = \frac{1}{2} w_oL$

$M = \frac{1}{2} w_oL (\frac{2}{3} L)$

$M = \frac{1}{3} w_o L^2$
 

By ratio and proportion
$\dfrac{z}{x} = \dfrac{w_o}{L}$

$z = \dfrac{w_o}{L}x$
 

$F = \frac{1}{2} xz$

$F = \frac{1}{2}x \left( \dfrac{w_o}{L}x \right)$

$F = \dfrac{w_o}{2L}x^2$
 

$EI \, y'' = -M + Vx - F(\frac{1}{3} x)$

$EI \, y'' = -\frac{1}{3} w_o L^2 + \frac{1}{2} w_o L x - \frac{1}{3} x \left( \dfrac{w_o}{2L}x^2 \right)$

$EI \, y'' = -\dfrac{w_o L^2}{3} + \dfrac{w_o L}{2}x - \dfrac{w_o}{6L}x^3$

$EI \, y' = -\dfrac{w_o L^2}{3}x + \dfrac{w_o L}{4}x^2 - \dfrac{w_o}{24L}x^4 + C_1$

$EI \, y = -\dfrac{w_o L^2}{6}x^2 + \dfrac{w_o L}{12}x^3 - \dfrac{w_o}{120L}x^5 + C_1 x + C_2$
 

At x = 0, y' = 0, therefore C₁ = 0
At x = 0, y = 0, therefore C₂ = 0
 

Therefore, the equation of the elastic curve is
$EI \, y = -\dfrac{w_o L^2}{6}x^2 + \dfrac{w_o L}{12}x^3 - \dfrac{w_o}{120L}x^5$           answer