Lecture 25
Example Problem
Modulus of Elasticity
Given:
a 1" x 2" x 10" long aluminum specimen loaded with P = 40,000 pounds deformed a total 0.02 inchesas it was tested within the elastic range.
Determine:
a. the stress in the material
b. the strain in the material
c. the modulus of elasticity.
Solution:
a. Direct stress is the axial load divided by the cross-sectional area. This is
40 kips / 2in2 = 20,000 psi.
b. The strain is the amount of deformation divided by the initial length. This is
(0.02 in / 10 in = 0.002 inches per inch.
c. The Modulus of Elasticity is the stress divided by the strain.
20,000 psi / 0.002 in/in = 10,000,000 psi.
Copyright © 1995, 1996, 1997 by Chris H. Luebkeman and Donald Peting