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:
- VAB = -wox for segment AB is linear; at x = 0, VAB = 0; at x = L/2, VAB = -½woL.
- At BC, the shear is uniformly distributed by -½woL.
To draw the Moment Diagram:
- MAB = -½wox2 is a second degree curve; at x = 0, MAB = 0; at x = L/2, MAB = -1/8 woL2.
- MBC = -½woLx + 1/8 woL2 is a second degree; at x = L/2, MBC = -1/8 woL2; at x = L, MBC = -3/8 woL2.