From the FBD of the section to the left of hinge
$\Sigma M_H = 0$
$4R_1 = 200(6)(3)$
$R_1 = 900 \, \text{lb}$
To draw the Shear Diagram
- VA = 0
- VB = VA + Area in load diagram
VB = 0 - 200(2) = -400 lb
VB2 = VB + R1 = -400 + 900 = 500 lb
- VH = VB2 + Area in load diagram
VH = 500 - 200(4) = -300 lb
- VC = VH + Area in load diagram
VC = -300 - 200(2) = -700 lb
- Location of zero shear:
x / 500 = (4 - x) / 300
300x = 2000 - 500x
x = 2.5 ft
To draw the Moment Diagram
- MA = 0
- MB = MA + Area in shear diagram
MB = 0 - ½ (400)(2) = -400 lb·ft
- Mx = MB + Area in load diagram
Mx = -400 + ½ (500)(2.5)
Mx = 225 lb·ft
- MH = Mx + Area in load diagram
MH = 225 - ½(300)(4 - 2.5) = 0 ok!
- MC = MH + Area in load diagram
MC = 0 - ½ (300 + 700)(2)
MC = -1000 lb·ft
- The location of zero moment in segment BH can easily be found by symmetry.