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**

- V
_{A} = 0
- V
_{B} = V_{A} + Area in load diagram

V_{B} = 0 - 200(2) = -400 lb

V_{B2} = V_{B} + R_{1} = -400 + 900 = 500 lb
- V
_{H} = V_{B2} + Area in load diagram

V_{H} = 500 - 200(4) = -300 lb
- V
_{C} = V_{H} + Area in load diagram

V_{C} = -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**

- M
_{A} = 0
- M
_{B} = M_{A} + Area in shear diagram

M_{B} = 0 - ½ (400)(2) = -400 lb·ft
- M
_{x} = M_{B} + Area in load diagram

M_{x} = -400 + ½ (500)(2.5)

M_{x} = 225 lb·ft
- M
_{H} = M_{x} + Area in load diagram

M_{H} = 225 - ½(300)(4 - 2.5) = 0 ok!
- M
_{C} = M_{H} + Area in load diagram

M_{C} = 0 - ½ (300 + 700)(2)

M_{C} = -1000 lb·ft
- The location of zero moment in segment BH can easily be found by symmetry.