Solution to Problem 438 | Relationship Between Load, Shear, and Moment

$\Sigma M_H = 0$

$4R_1 = 200(6)(3)$

$R_1 = 900 \, \text{lb}$
 

To draw the Shear Diagram

1. V_A = 0
2. V_B = V_A + Area in load diagram
   V_B = 0 - 200(2) = -400 lb
   V_B2 = V_B + R₁ = -400 + 900 = 500 lb
3. V_H = V_B2 + Area in load diagram
   V_H = 500 - 200(4) = -300 lb
4. V_C = V_H + Area in load diagram
   V_C = -300 - 200(2) = -700 lb
5. Location of zero shear:
   x / 500 = (4 - x) / 300
   300x = 2000 - 500x
   x = 2.5 ft

To draw the Moment Diagram

1. M_A = 0
2. M_B = M_A + Area in shear diagram
   M_B = 0 - ½ (400)(2) = -400 lb·ft
3. M_x = M_B + Area in load diagram
   M_x = -400 + ½ (500)(2.5)
   M_x = 225 lb·ft
4. M_H = M_x + Area in load diagram
   M_H = 225 - ½(300)(4 - 2.5) = 0 ok!
5. M_C = M_H + Area in load diagram
   M_C = 0 - ½ (300 + 700)(2)
   M_C = -1000 lb·ft
6. The location of zero moment in segment BH can easily be found by symmetry.