Solution to Problem 448 | Relationship Between Load, Shear, and Moment

1. A uniformly distributed load in AB is acting downward at a magnitude of 40/2 = 20 kN/m.
2. Upward concentrated force of 40 + 36 = 76 kN acts at B. No load in segment BC.
3. A downward point force acts at C at a magnitude of 36 - 16 = 20 kN.
4. Downward uniformly distributed load in CD has a magnitude of (16 + 24)/4 = 10 kN/m & causes zero shear at point F, 1.6 m from C.
5. Another upward concentrated force acts at D at a magnitude of 20 + 24 = 44 kN.
6. The load in segment DE is uniform and downward at 20/2 = 10 kN/m.

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
   M_AB is downward parabola with vertex at A.
3. M_C = M_B + Area in shear diagram
   M_C = -40 + 36(1) = -4 kN·m
   M_BC is linear and upward
4. M_F = M_C + Area in shear diagram
   M_F = -4 + ½ (16)(1.6) = 8.8 kN·m
5. M_D = M_F + Area in shear diagram
   M_D = 8.8 - ½ (24)(2.4) = -20 kN·m
   M_CD is downward parabola with vertex at F.
6. M_E = M_D + Area in shear diagram
   M_E = -20 + ½ (20)(2) = 0
   M_DE is downward parabola with vertex at E.