Solution to Problem 434 | Relationship Between Load, Shear, and Moment

$6R_1 + 120 = 20(4)(6) + 60(4)$

$R_1 = 100 \, \text{ kN}$
 

$\Sigma M_B = 0$

$6R_2 = 20(4)(0) + 60(2) + 120$

$R_2 = 40 \, \text{kN}$
 

To draw the Shear Diagram

1. V_A = 0
2. V_B = V_A + Area in load diagram
   V_B = 0 - 20(2) = -40 kN
   V_B2 = V_B + R₁ = -40 + 100 = 60 kN]
3. V_C = V_B2 + Area in load diagram
   V_C = 60 - 20(2) = 20 kN
   V_C2 = V_C - 60 = 20 - 60 = -40 kN
4. V_D = V_C2 + Area in load diagram
   V_D = -40 + 0 = -40 kN
5. V_E = V_D + Area in load diagram
   V_E = -40 + 0 = -40 kN
   V_E2 = V_E + R₂ = -40 + 40 = 0

To draw the Moment Diagram

1. M_A = 0
2. M_B = M_A + Area in shear diagram
   M_B = 0 - ½ (40)(2) = -40 kN·m
3. M_C = M_B + Area in shear diagram
   M_C = -40 + ½ (60 + 20)(2) = 40 kN·m
4. M_D = M_C + Area in shear diagram
   M_D = 40 - 40(2) = -40 kN·m
   M_D2 = M_D + M = -40 + 120 = 80 kN·m
5. M_E = M_D2 + Area in shear diagram
   M_E = 80 - 40(2) = 0
6. Moment curve BC is a downward parabola with vertex at C'. C' is the location of zero shear for segment BC.
7. Location of zero moment at segment BC:
   By squared property of parabola:
   (3 - x)² / 50 = 3² / (50 + 40)
   3 - x = 2.236
   x = 0.764 m from B