Ch 1. Stress and Strain Multimedia Engineering Mechanics NormalStress Shear andBearing Stress NormalStrain Hooke'sLaw ThermalEffects IndeterminateStructures
 Chapter 1. Stress/Strain 2. Torsion 3. Beam Shr/Moment 4. Beam Stresses 5. Beam Deflections 6. Beam-Advanced 7. Stress Analysis 8. Strain Analysis 9. Columns Appendix Basic Math Units Basic Equations Sections Material Properties Structural Shapes Beam Equations Search eBooks Dynamics Fluids Math Mechanics Statics Thermodynamics Author(s): Kurt Gramoll ©Kurt Gramoll

MECHANICS - CASE STUDY SOLUTION

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"

Practice Homework and Test problems now available in the 'Eng Mechanics' mobile app
Includes over 400 problems with complete detailed solutions.
Available now at the Google Play Store and Apple App Store.