**Segment AB:**
$V_{AB} = -w_ox$

$M_{AB} = -w_ox(x/2)$

$MAB = -\frac{1}{2}w_ox^2$

**Segment BC:**

$V_{BC} = -w_o(L/2)$

$V_{BC} = -\frac{1}{2} w_oL$

$M_{BC} = -w_o(L/2)(x - L/4)$

$M_{BC} = -\frac{1}{2} w_oLx + \frac{1}{8} w_oL^2$

**To draw the Shear Diagram:**

- V
_{AB} = -w_{o}x for segment AB is linear; at x = 0, V_{AB} = 0; at x = L/2, V_{AB} = -½w_{o}L.
- At BC, the shear is uniformly distributed by -½w
_{o}L.

**To draw the Moment Diagram:**

- M
_{AB} = -½w_{o}x^{2} is a second degree curve; at x = 0, M_{AB} = 0; at x = L/2, M_{AB} = -1/8 w_{o}L^{2}.
- M
_{BC} = -½w_{o}Lx + 1/8 w_{o}L^{2} is a second degree; at x = L/2, M_{BC} = -1/8 w_{o}L^{2}; at x = L, M_{BC} = -3/8 w_{o}L^{2}.

## Comments

## Why does the shear diagram

Why does the shear diagram indicate -W

_{0}L / 8 for segment BC? Isn't it -W_{0}L / 2?## You are correct. The figure

You are correct. The figure is scheduled for revision. As soon as the figure get updated, your comment will get deleted because it will no longer reflect the status of the page.