Solution to Problem 416 | Shear and Moment Diagrams

$LR_1 = \frac{1}{3} LF$

$R_1 = \frac{1}{3}\,( \, \frac{1}{2}Lw_o \, )$

$R_1 = \frac{1}{6} Lw_o$
 

$\Sigma M_{R_1} = 0$

$LR_2 = \frac{2}{3} LF$

$R_2 = \frac{2}{3} \, ( \, \frac{1}{2} Lw_o \, )$

$R_2 = \frac{1}{3} Lw_o$
 

$\dfrac{y}{x} = \dfrac{w_o}{L}$

$y = \dfrac{w_o}{L}x$

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

$F_x = \dfrac{w_o}{2L}x^2$
 

$V = R_1 - F_x$

$V = \frac{1}{6} Lw_o - \dfrac{w_o}{2L}x^2$
 

$M = R_1x - F_x\,(\frac{1}{3}x)$

$M = \frac{1}{6}Lw_ox - \dfrac{w_o}{2L}x^2\,(\frac{1}{3}x)$

$M = \frac{1}{6}Lw_ox - \dfrac{w_o}{6L}x^3$
 

To draw the Shear Diagram:
V = 1/6 Lw_o - w_ox²/2L is a second degree curve; at x = 0, V = 1/6 Lw_o = R₁; at x = L, V = -1/3 Lw^o = -R₂; If a is the location of zero shear from left end, 0 = 1/6 Lw_o - w_ox²/2L, x = 0.5774L = a; to check, use the squared property of parabola:
 

To draw the Moment Diagram:
M = 1/6 Lw_ox - w_ox³/6L is a third degree curve; at x = 0, M = 0; at x = L, M = 0; at x = a = 0.5774L, M = M_max.