Solution to Problem 428 | Relationship Between Load, Shear, and Moment

$5R_1 = 50(0.5) + 25$

$R_1 = 10 \, \text{kN}$
 

$\Sigma M_A = 0$

$5R_2 + 25 = 50(4.5)$

$R_2 = 40 \, \text{kN}$
 

To draw the Shear Diagram

1. V_A = R₁ = 10 kN
2. V_B = V_A + Area in load diagram
   V_B = 10 + 0 = 10 kN
3. V_C = V_B + Area in load diagram
   V_C = 10 + 0 = 10 kN
4. V_D = V_C + Area in load diagram
   V_D = 10 - 10(3) = -20 kN
   V_D2 = -20 + R₂ = 20 kN
5. V_E = V_D2 + Area in load diagram
   V_E = 20 - 10(2) = 0
6. Solving for x:
   x / 10 = (3 - x) / 20
   20x = 30 - 10x
   x = 1 m

To draw the Moment Diagram

1. M_A = 0
2. M_B = M_A + Area in shear diagram
   M_B = 0 + 1(10) = 10 kN·m
   M_B2 = 10 - 25 = -15 kN·m
3. M_C = M_B2 + Area in shear diagram
   M_C = -15 + 1(10) = -5 kN·m
4. M_x = M_C + Area in shear diagram
   M_x = -5 + ½(1)(10) = 0
5. M_D = M_x + Area in shear diagram
   M_D = 0 - ½(2)(20) = -20 kN·m
6. M_E = M_D + Area in shear diagram
   M_E = -20 + ½ (2)(20) = 0