Solution to Problem 435 | Relationship Between Load, Shear, and Moment

$2w_o (5) = 10(4)(0) + 20(2) + 40(3)$

$w_o = 16 \, \text{kN/m}$
 

$\Sigma M_{midpoint \,\, of \,\, EF} = 0$

$5R_1 = 10(4)(5) + 20(3) + 40(2)$

$R_1 = 68 \, \text{kN}$
 

To draw the Shear Diagram

1. M_A = 0
2. M_B = M_A + Area in load diagram
   M_B = 0 - 10(2) = -20 kN
   M_B2 + M_B + R₁ = -20 + 68 = 48 kN
3. M_C = M_B2 + Area in load diagram
   M_C = 48 - 10(2) = 28 kN
   M_C2 = M_C - 20 = 28 - 20 = 8 kN
4. M_D = M_C2 + Area in load diagram
   M_D = 8 + 0 = 8 kN
   M_D2 = M_D - 40 = 8 - 40 = -32 kN
5. M_E = M_D2 + Area in load diagram
   M_E = -32 + 0 = -32 kN
6. M_F = M_E + Area in load diagram
   M_F = -32 + w_o(2)
   M_F = -32 + 16(2) = 0

To draw the Moment Diagram

1. M_A = 0
2. M_B = M_A + Area in shear diagram
   M_B = 0 - ½ (20)(2) = -20 kN·m
3. M_C = M_B + Area in shear diagram
   M_C = -20 + ½ (48 + 28)(2)
   M_C = 56 kN·m
4. M_D = M_C + Area in shear diagram
   M_D = 56 + 8(1) = 64 kN·m
5. M_E = M_D + Area in shear diagram
   M_E = 64 - 32(1) = 32 kN·m
6. M_F = M_E + Area in shear diagram
   M_F = 32 - ½(32)(2) = 0
7. The location and magnitude of moment at C' are determined from shear diagram. By squared property of parabola, x = 0.44 m from B.