$\Sigma M_E = 0$

$5R_1 + 120 = 6(60) + 40(3)(3.5)$

$R_1 = 132 \, \text{kN}$

$\Sigma M_B = 0$

$5R_2 + 60(1) = 40(3)(1.5) + 120$

$R_2 = 48 \, \text{kN}$

**To draw the Shear Diagram**

- V
_{A} = -60 kN
- V
_{B} = V_{A} + Area in load diagram

V_{B} = -60 + 0 = -60 kN

V_{B2} = V_{B} + R_{1} = -60 + 132 = 72 kN
- V
_{C} = V_{B2} + Area in load diagram

V_{C} = 72 - 3(40) = -48 kN
- V
_{D} = V_{C} + Area in load diagram

V_{D} = -48 + 0 = -48 kN
- V
_{E} = V_{D} + Area in load diagram

V_{E} = -48 + 0 = -48 kN

V_{E2} = V_{E} + R_{2} = -48 + 48 = 0
- Solving for x:

x / 72 = (3 - x) / 48

48x = 216 - 72x

x = 1.8 m

**To draw the Moment Diagram**

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

M_{B} = 0 - 60(1) = -60 kN·m
- M
_{x} = M_{B} + Area in shear diagram

M_{X} = -60 + ½ (1.8)(72) = 4.8 kN·m
- M
_{C} = M_{X} + Area in shear diagram

M_{C} = 4.8 - ½ (3 - 1.8)(48) = -24 kN·m
- M
_{D} = M_{C} + Area in shear diagram

M_{D} = -24 - ½ (24 + 72)(1) = -72 kN·m

M_{D2} = -72 + 120 = 48 kN·m
- M
_{E} = M_{D2} + Area in shear diagram

M_{E} = 48 - 48(1) = 0
- The location of zero moment on segment BC can be determined using the squared property of parabola. See the solution of Problem 434.