$\Sigma M_H = 0$
$8R_1 = 4000(4)$
$R_1 = 2000 \, \text{lb}$
$\Sigma M_A = 0$
$8V_H = 4000(4)$
$V_H = 2000 \, \text{lb}$
$\Sigma M_D = 0$
$10R_2 = 2000(14) + 400(10)(5)$
$R_2 = 4800 \, \text{lb}$
$\Sigma M_H = 0$
$14R_3 + 4(4800) = 400(10)(9)$
$R_3 = 1200 \, \text{lb}$
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}