$\Sigma M_E = 0$

$6R_1 + 120 = 20(4)(6) + 60(4)$

$R_1 = 100 \, \text{ kN}$

$\Sigma M_B = 0$

$6R_2 = 20(4)(0) + 60(2) + 120$

$R_2 = 40 \, \text{kN}$

**To draw the Shear Diagram**

- V
_{A} = 0
- V
_{B} = V_{A} + Area in load diagram

V_{B} = 0 - 20(2) = -40 kN

V_{B2} = V_{B} + R_{1} = -40 + 100 = 60 kN]
- V
_{C} = V_{B2} + Area in load diagram

V_{C} = 60 - 20(2) = 20 kN

V_{C2} = V_{C} - 60 = 20 - 60 = -40 kN
- V
_{D} = V_{C2} + Area in load diagram

V_{D} = -40 + 0 = -40 kN
- V
_{E} = V_{D} + Area in load diagram

V_{E} = -40 + 0 = -40 kN

V_{E2} = V_{E} + R_{2} = -40 + 40 = 0

**To draw the Moment Diagram**

- M
_{A} = 0
- M
_{B} = M_{A} + Area in shear diagram

M_{B} = 0 - ½ (40)(2) = -40 kN·m
- M
_{C} = M_{B} + Area in shear diagram

M_{C} = -40 + ½ (60 + 20)(2) = 40 kN·m
- M
_{D} = M_{C} + Area in shear diagram

M_{D} = 40 - 40(2) = -40 kN·m

M_{D2} = M_{D} + M = -40 + 120 = 80 kN·m
- M
_{E} = M_{D2} + Area in shear diagram

M_{E} = 80 - 40(2) = 0
- Moment curve BC is a downward parabola with vertex at C'. C' is the location of zero shear for segment BC.
**Location of zero moment at segment BC:**

By squared property of parabola:

3 - x)2 / 50 = 32 / (50 + 40)

3 - x = 2.236

x = 0.764 m from B