Problem Diagram


Recall Dave, a college student, decided
to construct a hanging shelf from the ceiling. He plans to place 12,
1foot square boxes on the shelf. The left boxes will hold heavy books
(45 lb each) and the right boxes will hold clothes (20 lb each). The
shelf itself weighs 35 lb.
Now, Dave needs to determine what size wood rods are needed
to support the shelf and boxes. The rods come in increments of 1/16"
and the wood will be Douglas Fir.
First the actual load in each rod needs to be determined. A
free body diagram will help analyze the shelf and solve for the rod load.
Remember, there are two roads at each end.

Free Body Diagram of Shelf


ΣF_{y} =
0
2 A_{y} + 2 B_{y}  45(6)  20(6)
 35 = 0
A_{y} + B_{y} = 212.5 lb
Remember, the 2 before A_{y} and B_{y} is due to each end having two rods.
ΣM_{A}
= 0
45(6)(3)  20(6)(9) + 2 B_{y} (12)  35
(6) = 0
B_{y} = 87.5 lb
A_{y} = 212.5  87.5 = 125 lb
Thus, the two rods at A will each carry 125 lbs and the two rods at B
will carry 87.5 lbs. Using the worse case, 125 lbs, the design load will
be 4 times the actual load for the required factor of safety,
P = 4(125) = 500 lb
Using the
appendix, it is determined that Douglas Fir wood can withstand
a stress of 7,500 psi before it will fail. Using this failure stress, gives

1/16" Increment Rods 
Diameter (D) 
Area (A) 
1/16" 
0.0.625 in^{2} 
1/8" 
0.125 in^{2} 
3/16" 
0.1875 in^{2} 
1/4" 
0.25 in^{2} 
5/16" 
0.3125 in^{2} 
3/8" 
0.375 in^{2} 


σ =
P/A
(7,500 psi) = (500 lb)/A
A = 0.0667 in^{2}
The diameter of a cicular bar for 0.0667 in^{2} is π (D/2)^{2} =
A = 0.0667 in^{2}
D = 0.2913 in
The smallest rod to be at least 0.2913 inches is 5/16" 