Solution to Problem 445 | Relationship Between Load, Shear, and Moment

$\Sigma M_{R1} = 0$
$5R_2 = 80(2) + 90(3)$
$R_2 = 86 \, \text{kN}$
 

Checking
$R_1 + R_2 = F_1 + F_2$   (okay!)
 

To draw the Shear Diagram

1. V_A = R₁ = 84 kN
2. V_B = V_A + Area in load diagram
   V_B = 84 - 20(1) = 64 kN
3. V_C = V_B + Area in load diagram
   V_C = 64 - ½ (20 + 80)(3) = -86 kN
4. V_D = V_C + Area in load diagram
   V_D = -86 + 0 = -86 kN
   V_D2 = V_D + R₂ = -86 + 86 = 0
5. Location of zero shear:
   
   From the load diagram:
   y / (x + 1) = 80 / 4
   y = 20(x + 1)
   
   V_E = V_B + Area in load diagram
   0 = 64 - ½ (20 + y)x
   (20 + y)x = 128
   [20 + 20(x + 1)]x = 128
   20x² + 40x - 128 = 0
   5x² + 10x - 32 = 0
   x = 1.72 and -3.72
   use x = 1.72 m from B




6. By squared property of parabola:
   z / (1 + x)² = (z + 86) / 4²
   16z = 7.3984z + 636.2624
   8.6016z = 254.4224
   z = 73.97 kN

To draw the Moment Diagram

1. M_A = 0
2. M_B = M_A + Area in shear diagram
   M_B = 0 + ½ (84 + 64)(1) = 74 kN·m
3. M_E = M_B + Area in shear diagram
   M_E = 74 + A₁ (see figure for A₁ and A₂)
   
   For A₁:
   A₁ = 2/3 (1 + 1.72)(73.97) - 64(1) - 2/3 (1)(9.97)
   A₁ = 63.5
   
   M_E = 74 + 63.5 = 137.5 kN·m




4. M_C = M_E + Area in shear diagram
   M_C = M_E - A₂
   
   For A2:
   A₂ = 1/3 (4)(73.97 + 86) - 1/3 (1 + 1.72)(73.97) - 1.28(73.97)
   A₂ = 51.5
   
   M_C = 137.5 - 51.5 = 86 kN·m




5. M_D = M_C + Area in shear diagram
   M_D = 86 - 86(1) = 0