**Segment AB:**
$V_{AB} = -2x \, \text{kN}$

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

$M_{AB} = -x^2 \, \text{kN}\cdot\text{m}$

**Segment BC:**

$\dfrac{y}{x - 2} = \dfrac{2}{3}$

$y = \frac{2}{3}(x - 2)$

$F_1 = 2x$

$F_2 = \frac{1}{2}(x - 2)y$

$F_2 = \frac{1}{2}(x - 2) \, [ \, \frac{2}{3}(x - 2) \, ]$

$F_2 = \frac{1}{3} (x - 2)^2$

$V_{BC} = -F_1 - F_2$

$V_{BC} = -2x - \frac{1}{3} (x - 2)^2$

$M_{BC} = -(x/2)F_1 - \frac{1}{3}(x - 2)F_2$

$M_{BC} = -(x/2)(2x) - \frac{1}{3}(x - 2) \, [ \, \frac{1}{3} (x - 2)^2 \, ]$

$M_{BC} = -x^2 - \frac{1}{9}(x - 2)^3$

**To draw the Shear Diagram:**

- V
_{AB} = -2x is linear; at x = 0, V_{AB} = 0; at x = 2 m, V_{AB} = -4 kN.
- V
_{BC} = -2x - 1/3 (x - 2)^{2} is a second degree curve; at x = 2 m, V_{BC} = -4 kN; at x = 5 m; V_{BC} = -13 kN.

**To draw the Moment Diagram:**

- M
_{AB} = -x^{2} is a second degree curve; at x = 0, M_{AB} = 0; at x = 2 m, M_{AB} = -4 kN·m.
- M
_{BC} = -x^{2} -1/9 (x - 2)^{3} is a third degree curve; at x = 2 m, M_{BC} = -4 kN·m; at x = 5 m, M_{BC} = -28 kN·m.