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

$4w_1 (11) = 440(8)(5)$

$w_1 = 400 \, \text{lb/ft}$

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

$2w_2 (11) = 440(8)(6)$

$w_2 = 960 \, \text{lb/ft}$

**To draw the Shear Diagram**

- V
_{A} = 0
- V
_{B} = V_{A} + Area in load diagram

V_{B} = 0 + 400(4) = 1600 lb
- V
_{C} = V_{B} + Area in load diagram

V_{C} = 1600 - 440(8) = -1920 lb
- V
_{D} = V_{C} + Area in load diagram

V_{D} = -1920 + 960(2) = 0
- Location of zero shear:

x / 1600 = (8 - x) / 1920

x = 40/11 ft = 3.636 ft from B

**To draw the Moment Diagram**

- M
_{A} = 0
- M
_{B} = M_{A} + Area in shear diagram

M_{B} = 0 + ½ (1600)(4) = 3200 lb·ft
- M
_{x} = M_{B} + Area in shear diagram

M_{x} = 3200 + ½ (1600)(40/11)

M_{x} = 6109.1 lb·ft
- M
_{C} = M_{x} + Area in shear diagram

M_{C} = 6109.1 - ½ (8 - 40/11)(1920)

M_{C} = 1920 lb·ft
- M
_{D} = M_{C} + Area in shear diagram

M_{D} = 1920 - ½ (1920)(2) = 0