**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}.