$\Sigma M_A = 0$

$6R_D = 4[2(30)]$

$R_D = 40 \, \text{kN}$

$\Sigma M_D = 0$

$6R_A = 2[2(30)]$

$R_A = 20 \text{kN}$

**Segment AB:**

$V_{AB} = 20 \, \text{kN}$

$M_{AB} = 20x \, \text{kN}\cdot\text{m}$

**Segment BC:**

$V_{BC} = 20 - 30(x - 3)$

$V_{BC} = 110 - 30x \, \text{kN}$

$M_{BC} = 20x - 30(x - 3)(x - 3)/2$

$M_{BC} = 20x - 15(x - 3)^2 \, \text{kN}\cdot\text{m}$

**Segment CD:**

$V_{CD} = 20 - 30(2)$

$V_{CD} = -40 \, \text{kN}$

$M_{CD} = 20x - 30(2)(x - 4)$

$M_{CD} = 20x - 60(x - 4) \, \text{kN}\cdot\text{m}$

**To draw the Shear Diagram:**

- For segment AB, the shear is uniformly distributed at 20 kN.
- V
_{BC} = 110 - 30x for segment BC; at x = 3 m, V_{BC} = 20 kN; at x = 5 m, V_{BC} = -40 kN. For V_{BC} = 0, x = 3.67 m or 0.67 m from B.
- The shear for segment CD is uniformly distributed at -40 kN.

**To draw the Moment Diagram:**

- For AB, M
_{AB} = 20x; at x = 0, M_{AB} = 0; at x = 3 m, M_{AB} = 60 kN·m.
- M
_{BC} = 20x - 15(x - 3)^{2} for segment BC is second degree curve; at x = 3 m, M_{BC} = 60 kN·m; at x = 5 m, M_{BC} = 40 kN·m. Note that maximum moment occurred at zero shear; at x = 3.67 m, M_{BC} = 66.67 kN·m.
- M
_{CD} = 20x - 60(x - 4) for segment CD is linear; at x = 5 m, M_{CD} = 40 kN·m; at x = 6 m, M_{CD} = 0.