**From the load diagram:**
$\Sigma M_B = 0$

$5R_D + 1(30) = 3(50)$

$R_D = 24 \, \text{kN}$

$\Sigma M_D = 0$

$5R_B = 2(50) + 6(30)$

$R_B = 56 \, \text{kN}$

**Segment AB:**

$V_{AB} = -30 \, \text{kN}$

$M_{AB} = -30x \, \text{kN}\cdot\text{m}$

**Segment BC:**

$V_{BC} = -30 + 56$

$V_{BC} = 26 \, \text{kN}$

$M_{BC} = -30x + 56(x - 1)$

$M_{BC} = 26x - 56 \, \text{kN}\cdot\text{m}$

**Segment CD:**

$V_{CD} = -30 + 56 - 50$

$V_{CD} = -24 \, \text{kN}$

$M_{CD} = -30x + 56(x - 1) - 50(x - 4)$

$M_{CD} = -30x + 56x - 56 - 50x + 200$

$M_{CD} = -24x + 144 \, \text{kN}\cdot\text{m}$

**To draw the Shear Diagram:**

- In segment AB, the shear is uniformly distributed over the segment at a magnitude of -30 kN.
- In segment BC, the shear is uniformly distributed at a magnitude of 26 kN.
- In segment CD, the shear is uniformly distributed at a magnitude of -24 kN.

**To draw the Moment Diagram:**

- The equation M
_{AB} = -30x is linear, at x = 0, M_{AB} = 0 and at x = 1 m, M_{AB} = -30 kN·m.
- M
_{BC} = 26x - 56 is also linear. At x = 1 m, M_{BC} = -30 kN·m; at x = 4 m, M_{BC} = 48 kN·m. When M_{BC} = 0, x = 2.154 m, thus the moment is zero at 1.154 m from B.
- M
_{CD} = -24x + 144 is again linear. At x = 4 m, M_{CD} = 48 kN·m; at x = 6 m, M_{CD} = 0.