$\Sigma M_E = 0$
$5R_1 + 120 = 6(60) + 40(3)(3.5)$
$R_1 = 132 \, \text{kN}$
$\Sigma M_B = 0$
$5R_2 + 60(1) = 40(3)(1.5) + 120$
$R_2 = 48 \, \text{kN}$
To draw the Shear Diagram
- VA = -60 kN
- VB = VA + Area in load diagram
VB = -60 + 0 = -60 kN
VB2 = VB + R1 = -60 + 132 = 72 kN
- VC = VB2 + Area in load diagram
VC = 72 - 3(40) = -48 kN
- VD = VC + Area in load diagram
VD = -48 + 0 = -48 kN
- VE = VD + Area in load diagram
VE = -48 + 0 = -48 kN
VE2 = VE + R2 = -48 + 48 = 0
- Solving for x:
x / 72 = (3 - x) / 48
48x = 216 - 72x
x = 1.8 m
To draw the Moment Diagram
- MA = 0
- MB = MA + Area in shear diagram
MB = 0 - 60(1) = -60 kN·m
- Mx = MB + Area in shear diagram
MX = -60 + ½ (1.8)(72) = 4.8 kN·m
- MC = MX + Area in shear diagram
MC = 4.8 - ½ (3 - 1.8)(48) = -24 kN·m
- MD = MC + Area in shear diagram
MD = -24 - ½ (24 + 72)(1) = -72 kN·m
MD2 = -72 + 120 = 48 kN·m
- ME = MD2 + Area in shear diagram
ME = 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.
