**Segment AB:**
$V_{AB} = -20x \, \text{kN}$

$M_{AB} = -20x(x/2)$

$M_{AB} = -10x^2 \, \text{kN}\cdot\text{m}$

**Segment BC:**

$V_{BC} = -20(3)$

$V_{BC} = -60 \, \text{kN}$

$M_{BC} = -20(3)(x - 1.5)$

$M_{BC} = -60(x - 1.5) \, \text{kN}\cdot\text{m}$

**Segment CD:**

$V_{CD} = -20(3) + 40$

$V_{CD} = -20 \, \text{kN}$

$M_{CD} = -20(3)(x - 1.5) + 40(x - 5)$

$M_{CD} = -60(x - 1.5) + 40(x - 5) \, \text{kN}\cdot\text{m}$

**To draw the Shear Diagram**

- V
_{AB} = -20x for segment AB is linear; at x = 0, V = 0; at x = 3 m, V = -60 kN.
- V
_{BC} = -60 kN is uniformly distributed along segment BC.
- Shear is uniform along segment CD at -20 kN.

**To draw the Moment Diagram**

- M
_{AB} = -10x^{2} for segment AB is second degree curve; at x = 0, M_{AB} = 0; at x = 3 m, M_{AB} = -90 kN·m.
- M
_{BC} = -60(x - 1.5) for segment BC is linear; at x = 3 m, MBC = -90 kN·m; at x = 5 m, M_{BC} = -210 kN·m.
- M
_{CD} = -60(x - 1.5) + 40(x - 5) for segment CD is also linear; at x = 5 m, M_{CD} = -210 kN·m, at x = 7 m, M_{CD} = -250 kN·m.