$\Sigma M_A = 0$
$6R_2 = 2(60) + 7(30)$
$R_2 = 55 \, \text{kN}$
$\Sigma M_C = 0$
$6R_1 + 1(30) = 4(60)$
$R_1 = 35 \, \text{kN}$
To draw the Shear Diagram:
- V_{A} = R_{1} = 35 kN
- V_{B} = V_{A} + Area in load diagram - 60 kN
V_{B} = 35 + 0 - 60 = -25 kN
- V_{C} = V_{B} + area in load diagram + R_{2}
V_{C} = -25 + 0 + 55 = 30 kN
- V_{D} = V_{C} + Area in load diagram - 30 kN
V_{D} = 30 + 0 - 30 = 0
To draw the Moment Diagram:
- M_{A} = 0
- M_{B} = M_{A} + Area in shear diagram
M_{B} = 0 + 35(2) = 70 kN·m
- M_{C} = M_{B} + Area in shear diagram
M_{C} = 70 - 25(4) = -30 kN·m
- M_{D} = M_{C} + Area in shear diagram
M_{D} = -30 + 30(1) = 0