$\Sigma M_E = 0$
$6R_1 + 120 = 20(4)(6) + 60(4)$
$R_1 = 100 \, \text{ kN}$
$\Sigma M_B = 0$
$6R_2 = 20(4)(0) + 60(2) + 120$
$R_2 = 40 \, \text{kN}$
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 = 3^{2} / (50 + 40)
3 - x = 2.236
x = 0.764 m from B