**To draw the Load Diagram**
- Downward 4000 lb force is concentrated at A and no load in segment AB.
- 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.
- Another point force acting downward with 3700 - 1700 = 1200 lb at D and no load in segment DE.
- The shear in EF is uniformly decreasing, thus a uniform downward force is acting with magnitude of (1700 + 3100)/8 = 600 lb/ft.
- Upward force of 3100 lb is concentrated at end of span F.

**To draw the Moment Diagram**

- The locations of zero shear (points G and H) can be easily determined by ratio and proportion of triangle.
- M
_{A} = 0; M_{B} = M_{A} + Area in shear diagram

M_{B} = 0 - 4000(3) = -12,000 lb·ft
- M
_{G} = M_{B} + Area in shear diagram

M_{G} = -12,000 - ½ (80/77)(4000)

M_{G} = -14,077.92 lb·ft
- M
_{C} = M_{G} + Area in shear diagram

M_{C} = -14,077.92 + ½ (74/77)(3700)

M_{C} = -12,300 lb·ft
- M
_{D} = M_{C} + Area in shear diagram

M_{D} = -12,300 + 3700(3) = -1200 lb·ft
- M
_{E} = M_{D} + Area in shear diagram

M_{E} = -1200 + 1700(4) = 5600 lb·ft
- M
_{H} = M_{E} + Area in shear diagram

M_{H} = 5600 + ½ (17/6)(1700)

M_{H} = 8,008.33 lb·ft
- M
_{F} = M_{H} + Area in shear diagram

M_{F} = 8,008.33 - ½ (31/6)(3100) = 0