$\Sigma M_{midpoint\,\,of\,\,CD} = 0$
$4w_1 (11) = 440(8)(5)$
$w_1 = 400 \, \text{lb/ft}$
$\Sigma M_{midpoint\,\,of\,\,AB} = 0$
$2w_2 (11) = 440(8)(6)$
$w_2 = 960 \, \text{lb/ft}$
To draw the Shear Diagram
- VA = 0
- VB = VA + Area in load diagram
VB = 0 + 400(4) = 1600 lb
- VC = VB + Area in load diagram
VC = 1600 - 440(8) = -1920 lb
- VD = VC + Area in load diagram
VD = -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
- MA = 0
- MB = MA + Area in shear diagram
MB = 0 + ½ (1600)(4) = 3200 lb·ft
- Mx = MB + Area in shear diagram
Mx = 3200 + ½ (1600)(40/11)
Mx = 6109.1 lb·ft
- MC = Mx + Area in shear diagram
MC = 6109.1 - ½ (8 - 40/11)(1920)
MC = 1920 lb·ft
- MD = MC + Area in shear diagram
MD = 1920 - ½ (1920)(2) = 0
