$\Sigma M_C = 0$

$9R_1 = 5(810)$

$R_1 = 450 \, \text{lb}$

$\Sigma M_A = 0$

$9R_2 = 4(810)$

$R_2 = 360 \, \text{lb}$

**Segment AB:**

$\dfrac{y}{x} = \dfrac{270}{6}$

$y = 45x$

$F = \frac{1}{2}xy = \frac{1}{2}x(45x)$

$F = 22.5x^2$

$V_{AB} = R_1 - F$

$V_{AB} = 450 - 22.5x^2 \, \text{lb}$

$M_{AB} = R_1x - F(\frac{1}{3}x)$

$M_{AB} = 450x - 22.5x^2(\frac{1}{3}x)$

$M_{AB} = 450x - 7.5x^3 \, \text{lb}\cdot\text{ft}$

**Segment BC:**

$V_{BC} = 450 - 810$

$V_{BC} = -360 \, \text{lb}$

$M_{BC} = 450x - 810(x - 4)$

$M_{BC} = 450x - 810x + 3240$

$M_{BC} = 3240 - 360x \, \text{lb}\cdot\text{ft}$

**To draw the Shear Diagram:**

- V
_{AB} = 450 - 22.5x^{2} is a second degree curve; at x = 0, V_{AB} = 450 lb; at x = 6 ft, V_{AB} = -360 lb.
- At x = a, V
_{AB} = 0,
450 - 22.5x^{2} = 0

22.5x^{2} = 450

x^{2} = 20

x = √20

To check, use the squared property of parabola.

a^{2}/450 = 62/(450 + 360)

a^{2} = 20

a = √20

- V
_{BC} = -360 lb is constant.

**To draw the Moment Diagram:**

- M
_{AB} = 450x - 7.5x^{3} for segment AB is third degree curve; at x = 0, M_{AB} = 0; at x = √20, M_{AB} = 1341.64 lb·ft; at x = 6 ft, M_{AB} = 1080 lb·ft.
- M
_{BC} = 3240 - 360x for segment BC is linear; at x = 6 ft, M_{BC} = 1080 lb·ft; at x = 9 ft, M_{BC} = 0.