Solution to Problem 410 | Shear and Moment Diagrams

$y = \dfrac{w_o}{L}x$
 

$F_x = \frac{1}{2}xy$

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

$F_x = \dfrac{w_o}{2L}x^2$
 

Shear equation:
$V = -\dfrac{w_o}{2L}x^2$
 

Moment equation:
$M = -\frac{1}{3}x F_x = -\dfrac{1}{3}x \left( \dfrac{w_o}{2L}x^2 \right)$

$M = -\dfrac{w_o}{6L}x^3$
 

To draw the Shear Diagram:

1. V = - w_o x² / 2L is a second degree curve; at x = 0, V = 0; at x = L, V = -½ w_oL.

To draw the Moment Diagram:

1. M = - w_o x³ / 6L is a third degree curve; at x = 0, M = 0; at x = L, M = - 1/6 w_oL².