$F_{BH} = 14 \, \text{kN to the right}$
$M_B = 14(2)$
$M_B = 28 \, \text{kN}\cdot\text{m}$ counterclockwise
$F_{CH} = \frac{3}{5} (10)$
$F_{CH} = 6 \, \text{ kN}$ to the right
$F_{CV} = 4/5 (10)$
$F_{CV} = 8 \, \text{ kN}$ upward
$M_C = F_{CH} (2) = 6(2)$
$M_C = 12 \, \text{ kN}\cdot\text{m}$ clockwise
$\Sigma M_D = 0$
$6R_A + 12 + 8(2) = 28$
$R_A = 0$
$\Sigma M_A = 0$
$6R_{DV} + 12 = 28 + 8(4)$
$R_{DV} = 8 \, \text{kN}$
$\Sigma F_H = 0$
$R_{DH} = 14 + 6$
$R_{DH} = 20 \, \text{kN}$
To draw the Shear Diagram
- Shear in segments AB and BC is zero.
- VC = 8
- VD = VC + Area in load diagram
VD = 8 + 0 = 8 kN
VD2 = VD - RDV
VD2 = 8 - 8 = 0
To draw the Moment Diagram
- Moment in segment AB is zero
- MB = -28 kN·m
- MC = MB + Area in shear diagram
MC = -28 + 0 = -28 kN·m
MC2 = MC + 12 = -28 + 12
MC2 = -16 kN·m
- MD = MC2 + Area in shear diagram
MD = -16 + 8(2)
MD = 0
