$\Sigma M_H = 0$

$8R_1 = 4000(4)$

$R_1 = 2000 \, \text{lb}$

$\Sigma M_A = 0$

$8V_H = 4000(4)$

$V_H = 2000 \, \text{lb}$

$\Sigma M_D = 0$

$10R_2 = 2000(14) + 400(10)(5)$

$R_2 = 4800 \, \text{lb}$

$\Sigma M_H = 0$

$14R_3 + 4(4800) = 400(10)(9)$

$R_3 = 1200 \, \text{lb}$

**To draw the Shear Diagram**

- V
_{A} = 0
- V
_{B} = 2000 lb

V_{B2} = 2000 - 4000 = -2000 lb
- V
_{H} = -2000 lb
- V
_{C} = -2000 lb

V_{C} = -2000 + 4800 = 2800 lb
- V
_{D} = 2800 - 400(10) = -1200 lb
- Location of zero shear:

x / 2800 = (10 - x) / 1200

1200x = 28000 - 2800x

x = 7 ft

**To draw the Moment Diagram**

- M
_{A} = 0
- M
_{B} = 2000(4) = 8000 lb·ft
- M
_{H} = 8000 - 4000(2) = 0
- M
_{C} = -400(2)

M_{C} = -8000 lb·ft
- M
_{x} = -800 + ½ (2800)(7)

M_{x} = 1800 lb·ft
- M
_{D} = 1800 - ½(1200)(3)

M_{D} = 0
- Zero M is 4 ft from R
_{2}