ΣMR2=0
12R1+4P=16P+4(3P)
R1=2P
ΣMR1=0
12R2+4P=16P+8(3P)
R2=3P
W=12(12)wo=3P
wo=12P
To draw the Shear Diagram
- VA = -P lb
- VB = VA + Area in load diagram
VB = -P + 0 = -P lb
VB2 = VB + R1 = -P + 2P = P lb
- VC = VB2 + Area in load diagram
VC = P - ½(12)(½P) = -2P lb
VC2 = VC + R2 = -2P + 3P = P lb
- VD = VC2 + Area in load diagram
VD = P + 0 = P
VD2 = VD - P = P - P = 0
- Shear at AB and CD are rectangular.
- Shear at BC is parabolic (2nd degree curve).
- Location of zero shear:
By squared property of parabola
x2 / P = 122 / 3P
x = 6.93 ft
12 - x = 5.07 ft
To draw the Moment Diagram
- MA = 0
- MB = MA + Area in shear diagram
MB = 0 - 4P = -4P lb·ft
- ME = MB + Area in shear diagram
ME = -4P + 2/3 (6.93)(P) = 0.62P lb·ft
- MC = ME + Area in shear diagram
MC = 0.62P - [ 1/3 (12)3P - 2.31P - 5.07P ]
MC = -4P lb·ft
- MD = MC + Area in shear diagram
MD = -4P + 4P = 0
- The moment diagram at AB and CD are straight lines (1st degree curves) while at BC is 3rd degree curve.
Based on allowable bending stress
fb=McI
Where (From Solution 577)
c = 6 in
I = 350.67 in4
Thus,
1200=4P(12)(6)350.67
P=1461.125 lb
Based on allowable shear stress
fv=VQIb
Where (From Solution 577)
Q = 35.5 in3
I = 350.67 in4
b = 0.75 in
Thus,
200=2P(35.5)350.67(0.75)
P=740.85 lb
For safe value of P, use P = 740.85 lb. answer