Solution to Problem 449 | Relationship Between Load, Shear, and Moment

1. Downward 4000 lb force is concentrated at A and no load in segment AB.
2. The shear in BC is uniformly increasing, thus a uniform upward force is acting at a magnitude of (3700 + 4000)/2 = 3850 lb/ft. No load in segment CD.
3. Another point force acting downward with 3700 - 1700 = 1200 lb at D and no load in segment DE.
4. The shear in EF is uniformly decreasing, thus a uniform downward force is acting with magnitude of (1700 + 3100)/8 = 600 lb/ft.
5. Upward force of 3100 lb is concentrated at end of span F.

To draw the Moment Diagram

1. The locations of zero shear (points G and H) can be easily determined by ratio and proportion of triangle.
2. M_A = 0; M_B = M_A + Area in shear diagram
   M_B = 0 - 4000(3) = -12,000 lb·ft
3. M_G = M_B + Area in shear diagram
   M_G = -12,000 - ½ (80/77)(4000)
   M_G = -14,077.92 lb·ft
4. M_C = M_G + Area in shear diagram
   M_C = -14,077.92 + ½ (74/77)(3700)
   M_C = -12,300 lb·ft
5. M_D = M_C + Area in shear diagram
   M_D = -12,300 + 3700(3) = -1200 lb·ft
6. M_E = M_D + Area in shear diagram
   M_E = -1200 + 1700(4) = 5600 lb·ft
7. M_H = M_E + Area in shear diagram
   M_H = 5600 + ½ (17/6)(1700)
   M_H = 8,008.33 lb·ft
8. M_F = M_H + Area in shear diagram
   M_F = 8,008.33 - ½ (31/6)(3100) = 0