Page 262 - Mechanical Behavior of Materials

264 Chapter 6 Review of Complex and Principal States of Stress and Strain

plane, no shear strains γ yz or γ zx occur. This creates a situation analogous to that for generalized
plane stress, where σ z is present, but τ yz = τ zx = 0. Hence, one of the principal normal strains is
ε z = ε 3 , and the other two can be obtained from Eq. 6.38. In addition, Mohr’s circle can be used for
the x-y plane.
For isotropic, linear-elastic materials, ε z can be obtained from Hooke’s law in the form of
Eq. 5.26. Taking σ z = 0 and adding Eqs. 5.26(a) and (b) leads to

E
σ x + σ y = (ε x + ε y ) (6.42)
1 − ν
Substituting this into Eq. 5.26(c) with σ z = 0gives ε z in terms of the normal strains in the x-y
plane:

−ν
ε z = (ε x + ε y ) (6.43)
1 − ν

Since τ yz = τ zx = 0, Eq. 5.27 gives γ yz = γ zx = 0, and it is confirmed that ε z is one of the principal
normal strains.
Consider an orthotropic material under x-y plane stress where the x-y plane is a plane of
symmetry of the material. (This is the situation for most sheets and plates of composite materials.)
The strain ε z is still one of the principal normal strains, as γ yz = γ zx = 0 holds in this case also.
Hence, the strains in the x-y plane can still be analyzed with Eqs. 6.38 and 6.39, and Mohr’s circle
for the x-y plane can still be used. However, ε z cannot be obtained from Eq. 6.43, as the more
general form of Hooke’s law for orthotropic materials (Eq. 5.44) is needed.
The principal axes for stress and strain coincide for isotropic materials. Hence, Hooke’s law
can be applied to the principal strains, and the resulting stresses are the principal stresses, and vice
versa. However, this is not the case for orthotropic materials unless the principal normal stress axes
are perpendicular to the planes of material symmetry.

Example 6.8
At a point on a free (unloaded) surface of an engineering component made of an aluminum
alloy, the following strains exist: ε x =−0.0005, ε y = 0.0035, and γ xy = 0.003. Determine the
principal normal and shear strains. Assume that no yielding of the material has occurred.
Solution Since the material is expected to be isotropic, ε z can be obtained from Eq. 6.43, and
this is one of the principal normal strains. We obtain

−0.345
ε 3 = ε z = (−0.0005 + 0.0035) =−0.00158 Ans.
1 − 0.345
where Poisson’s ratio from Table 5.2 is used, and application of this equation based on elastic
behavior is valid due to the absence of yielding. Substituting the given strains into Eqs. 6.38 and
6.39 gives the axis rotations and values for the other two principal normal strains and for one of