Solution to Problem 439 | Relationship Between Load, Shear, and Moment

$\Sigma M_A = 0$
$8V_H = 4000(4)$
$V_H = 2000 \, \text{lb}$
 

$\Sigma M_D = 0$
$10R_2 = 2000(14) + 400(10)(5)$
$R_2 = 4800 \, \text{lb}$

$\Sigma M_H = 0$
$14R_3 + 4(4800) = 400(10)(9)$
$R_3 = 1200 \, \text{lb}$
 

To draw the Shear Diagram

1. V_A = 0
2. V_B = 2000 lb
   V_B2 = 2000 - 4000 = -2000 lb
3. V_H = -2000 lb
4. V_C = -2000 lb
   V_C = -2000 + 4800 = 2800 lb
5. V_D = 2800 - 400(10) = -1200 lb
6. Location of zero shear:
   x / 2800 = (10 - x) / 1200
   1200x = 28000 - 2800x
   x = 7 ft

To draw the Moment Diagram

1. M_A = 0
2. M_B = 2000(4) = 8000 lb·ft
3. M_H = 8000 - 4000(2) = 0
4. M_C = -400(2)
   M_C = -8000 lb·ft
5. M_x = -800 + ½ (2800)(7)
   M_x = 1800 lb·ft
6. M_D = 1800 - ½(1200)(3)
   M_D = 0
7. Zero M is 4 ft from R₂