Problem Diagram
|
|
Recall Dave, a college student, decided
to construct a hanging shelf from the ceiling. He plans to place 12,
1-foot 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
|
|
ΣFy =
0
2 Ay + 2 By - 45(6) - 20(6)
- 35 = 0
Ay + By = 212.5 lb
Remember, the 2 before Ay and By is due to each end having two rods.
ΣMA
= 0
-45(6)(3) - 20(6)(9) + 2 By (12) - 35
(6) = 0
By = 87.5 lb
Ay = 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 in2 |
1/8" |
0.125 in2 |
3/16" |
0.1875 in2 |
1/4" |
0.25 in2 |
5/16" |
0.3125 in2 |
3/8" |
0.375 in2 |
|
|
σ =
P/A
(7,500 psi) = (500 lb)/A
A = 0.0667 in2
The diameter of a cicular bar for 0.0667 in2 is π (D/2)2 =
A = 0.0667 in2
D = 0.2913 in
The smallest rod to be at least 0.2913 inches is 5/16" |