Solution to Problem 451 | Relationship Between Load, Shear, and Moment

1. Upward concentrated load at A is 10 kN.
2. The shear in AB is a 2nd-degree curve, thus the load in AB is uniformly varying. In this case, it is zero at A to 2(10 + 2)/3 = 8 kN at B. No load in segment BC.
3. A downward point force is acting at C in a magnitude of 8 - 2 = 6 kN.
4. The shear in DE is uniformly increasing, thus the load in DE is uniformly distributed and upward. This load is spread over DE at a magnitude of 8/2 = 4 kN/m.

To draw the Moment Diagram

1. To find the location of zero shear, F:
   x²/10 = 3²/(10 + 2)
   x = 2.74 m
2. M_A = 0
3. M_F = M_A + Area in shear diagram
   M_F = 0 + 2/3 (2.74)(10) = 18.26 kN·m
4. M_B = M_F + Area in shear diagram
   M_B = 18.26 - [1/3 (10 + 2)(3) - 1/3 (2.74)(10) - 10(3 - 2.74)]
   M_B = 18 kN·m
5. M_C = M_B + Area in shear diagram
   M_C = 18 - 2(1) = 16 kN·m
6. M_D = M_C + Area in shear diagram
   M_D = 16 - 8(1) = 8 kN·m
7. M_E = M_D + Area in shear diagram
   M_E = 8 - ½ (2)(8) = 0
8. The moment diagram in AB is a second degree curve, at BC and CD are linear and downward. For segment DE, the moment diagram is parabola open upward with vertex at E.