Page 202 - Mechanical Behavior of Materials

Section 5.3 Elastic Deformation 203

Figure 5.8 Longitudinal extension and lateral contraction used to obtain constants for a
linear-elastic material that is isotropic and homogeneous.

size scales, real materials obey these idealizations if they are composed of tiny, randomly oriented
crystal grains. This is at least approximately true for many metals and ceramics. Amorphous
materials, such as glass and some polymers, may also be approximately isotropic and homogeneous.
Hence, for the present, we will proceed with these simplifying assumptions.
Let a bar of a homogeneous and isotropic material be subjected to an axial stress σ x ,asin
Fig. 5.8. The strain in the direction of the stress is

L − L i L
ε x = = (5.22)
L i L i
where L is the deformed length, L i the initial length, and L the change. In a similar manner, we
obtain the strain in any direction perpendicular to the stress—that is, along any diameter of the bar:

d − d i d
ε y = ε z = = (5.23)
d i d i
This transverse strain is negative for tensile σ x , as the bar becomes thinner when stretched in the
length direction. Conversely, ε y is positive for compressive σ x .
The material is said to be linear elastic if the stress is linearly related to these strains, and
if the strains return immediately to zero after unloading, in the manner of a simple linear spring.
In this situation, two elastic constants are needed to characterize the material. One is the elastic
modulus, E = σ x /ε x , which is the slope of the σ x versus ε x line in Fig. 5.8. The second constant is
Poisson’s ratio,
transverse strain ε y
ν =− =− (5.24)
longitudinal strain ε x
Since ε y is of opposite sign to ε x , positive ν is obtained. Hence, as also shown, the slope of a plot
of ε y versus ε x is −ν. Substituting ε x from Eq. 5.24 into E = σ x /ε x gives