Solution to Problem 429 | Relationship Between Load, Shear, and Moment

$4R_1 + 120(2)(1) = 100(2) + 120(2)(3)$

$R_1 = 170 \, \text{lb}$
 

$\Sigma M_A = 0$

$4R_2 = 120(2)(1) + 100(2) + 120(2)(5)$

$R_2 = 410 \, \text{lb}$
 

To draw the Shear Diagram

1. V_A = R₁ = 170 lb
2. V_B = V_A + Area in load diagram
   V_B = 170 - 120(2) = -70 lb
   V_B2 = -70 - 100 = -170 lb
3. V_C = V_B2 + Area in load diagram
   V_C = -170 + 0 = -170 lb
   V_C2 = -170 + R₂
   V_C2 = -170 + 410 = 240 lb
4. V_D = V_C2 + Area in load diagram
   V_D = 240 - 120(2) = 0
5. Solving for x:
   x / 170 = (2 - x) / 70
   70x = 340 - 170x
   x = 17 / 12 ft = 1.42 ft

To draw the Moment Diagram

1. M_A = 0
2. M_x = M_A + Area in shear diagram
   M_x = 0 + (17/12)(170)
   M_x = 1445/12 = 120.42 lb·ft
3. M_B = M_x + Area in shear diagram
   M_B = 1445/12 - ½ (2 - 17/12)(70)
   M_B = 100 lb·ft
4. M_C = M_B + Area in shear diagram
   M_C = 100 - 170(2) = -240 lb·ft
5. M_D = M_C + Area in shear diagram
   M_D = -240 + ½ (2)(240) = 0