$\Sigma M_C = 0$
$4R_1 + 120(2)(1) = 100(2) + 120(2)(3)$
$R_1 = 170 \, \text{lb}$
$\Sigma M_A = 0$
$4R_2 = 120(2)(1) + 100(2) + 120(2)(5)$
$R_2 = 410 \, \text{lb}$
To draw the Shear Diagram
- VA = R1 = 170 lb
- VB = VA + Area in load diagram
VB = 170 - 120(2) = -70 lb
VB2 = -70 - 100 = -170 lb
- VC = VB2 + Area in load diagram
VC = -170 + 0 = -170 lb
VC2 = -170 + R2
VC2 = -170 + 410 = 240 lb
- VD = VC2 + Area in load diagram
VD = 240 - 120(2) = 0
- Solving for x:
x / 170 = (2 - x) / 70
70x = 340 - 170x
x = 17 / 12 ft = 1.42 ft
To draw the Moment Diagram
- MA = 0
- Mx = MA + Area in shear diagram
Mx = 0 + (17/12)(170)
Mx = 1445/12 = 120.42 lb·ft
- MB = Mx + Area in shear diagram
MB = 1445/12 - ½ (2 - 17/12)(70)
MB = 100 lb·ft
- MC = MB + Area in shear diagram
MC = 100 - 170(2) = -240 lb·ft
- MD = MC + Area in shear diagram
MD = -240 + ½ (2)(240) = 0
