Solution to Problem 432 | Relationship Between Load, Shear, and Moment

$5R_1 + 120 = 6(60) + 40(3)(3.5)$

$R_1 = 132 \, \text{kN}$
 

$\Sigma M_B = 0$

$5R_2 + 60(1) = 40(3)(1.5) + 120$

$R_2 = 48 \, \text{kN}$
 

To draw the Shear Diagram

1. V_A = -60 kN
2. V_B = V_A + Area in load diagram
   V_B = -60 + 0 = -60 kN
   V_B2 = V_B + R₁ = -60 + 132 = 72 kN
3. V_C = V_B2 + Area in load diagram
   V_C = 72 - 3(40) = -48 kN
4. V_D = V_C + Area in load diagram
   V_D = -48 + 0 = -48 kN
5. V_E = V_D + Area in load diagram
   V_E = -48 + 0 = -48 kN
   V_E2 = V_E + R₂ = -48 + 48 = 0
6. Solving for x:
   x / 72 = (3 - x) / 48
   48x = 216 - 72x
   x = 1.8 m

To draw the Moment Diagram

1. M_A = 0
2. M_B = M_A + Area in shear diagram
   M_B = 0 - 60(1) = -60 kN·m
3. M_x = M_B + Area in shear diagram
   M_X = -60 + ½ (1.8)(72) = 4.8 kN·m
4. M_C = M_X + Area in shear diagram
   M_C = 4.8 - ½ (3 - 1.8)(48) = -24 kN·m
5. M_D = M_C + Area in shear diagram
   M_D = -24 - ½ (24 + 72)(1) = -72 kN·m
   M_D2 = -72 + 120 = 48 kN·m
6. M_E = M_D2 + Area in shear diagram
   M_E = 48 - 48(1) = 0
7. The location of zero moment on segment BC can be determined using the squared property of parabola. See the solution of Problem 434.