$\Sigma M_A = 0$

$6R_C = 5 \, [ \, 6(800) \, ]$

$R_C = 4000 \, \text{lb}$

$\Sigma M_C = 0$

$6R_A = 1 \, [ \, 6(800) \, ]$

$R_A = 800 \, \text{lb}$

**Segment AB:**

$V_{AB} = 800 \, \text{lb}$

$M_{AB} = 800x \, \text{lb}\cdot\text{ft}$

**Segment BC:**

$V_{BC} = 800 - 800(x - 2)$

$V_{BC} = 2400 - 800x \, \text{lb}$

$M_{BC} = 800x - 800(x - 2)(x - 2)/2$

$M_{BC} = 800x - 400(x - 2)^2 \, \text{lb}\cdot\text{ft}$

**Segment CD:**

$V_{CD} = 800 + 4000 - 800(x - 2)$

$V_{CD} = 4800 - 800x + 1600$

$V_{CD} = 6400 - 800x \, \text{lb}$

$M_{CD} = 800x + 4000(x - 6) - 800(x - 2)(x - 2)/2$

$M_{CD} = 800x + 4000(x - 6) - 400(x - 2)^2 \, \text{lb}\cdot\text{ft}$

**To draw the Shear Diagram:**

- 800 lb of shear force is uniformly distributed along segment AB.
- V
_{BC} = 2400 - 800x is linear; at x = 2 ft, V_{BC} = 800 lb; at x = 6 ft, V_{BC} = -2400 lb. When V_{BC} = 0, 2400 - 800x = 0, thus x = 3 ft or V_{BC} = 0 at 1 ft from B.
- V
_{CD} = 6400 - 800x is also linear; at x = 6 ft, V_{CD} = 1600 lb; at x = 8 ft, V_{BC} = 0.

**To draw the Moment Diagram:**

- M
_{AB} = 800x is linear; at x = 0, M_{AB} = 0; at x = 2 ft, M_{AB} = 1600 lb·ft.
- M
_{BC} = 800x - 400(x - 2)^{2} is second degree curve; at x = 2 ft, M_{BC} = 1600 lb·ft; at x = 6 ft, M_{BC} = -1600 lb·ft; at x = 3 ft, M_{BC} = 2000 lb·ft.
- M
_{CD} = 800x + 4000(x - 6) - 400(x - 2)^{2} is also a second degree curve; at x = 6 ft, M_{CD} = -1600 lb·ft; at x = 8 ft, MCD = 0.