Page 265 - Mechanical Behavior of Materials

Section 6.7 Summary 267

To completely characterize the in-plane strains at a point, ε x , ε y , and γ xy are needed. There is no
direct way of measuring shear strain, but γ xy can be calculated if longitudinal strains are measured
in three different directions. Transformation equations analogous to those of Section 6.2.1 for stress
are needed to calculate the shear strain, and sometimes also ε x and/or ε y , if the measured strains are
not aligned with the desired x-y directions.

Example 6.9
Consider a strain gage rosette mounted on the unloaded free surface of an engineering
component, as shown in Fig. 6.16(a). Note that longitudinal strains ε x , ε y , and ε 45 are measured
in x- and y-directions and in a third direction 45 from the other two. Develop an equation
◦
for calculating the shear strain γ xy from the three measurements that are available, so that the
in-plane state of strain, ε x , ε y , and γ xy , is completely known.

Solution Convert the stress transformation relationship of Eq. 6.4 to the corresponding
equation for normal strain by substitutions from Eq. 6.37:

ε x + ε y ε x − ε y γ xy
ε θ = + cos 2θ + sin 2θ
2 2 2
◦
Substituting θ = 45 gives
ε x + ε y γ xy
ε 45 = +
2 2
Solving for γ xy then provides the desired result:

γ xy = 2ε 45 − ε x − ε y Ans.

6.7 SUMMARY

For a general state of stress, given by components σ x , σ y , σ z , τ xy , τ yz , and τ zx , there is one choice of
a new coordinate system where shear stresses are absent and where the maximum and minimum
normal stresses occur along with an intermediate normal stress. These special stresses are the
principal normal stresses, σ 1 , σ 2 , and σ 3 , and they may be obtained by solving the cubic equation
given by the determinant

(σ x − σ)
τ xy τ zx

τ xy τ yz
(σ y − σ) = 0 (6.44)

τ zx τ yz
(σ z − σ)
If there is only one nonzero component of shear stress, such as τ xy , the principal normal stresses are

2

σ x + σ y σ x − σ y
2
σ 1 ,σ 2 = ± + τ , σ 3 = σ z (a, b) (6.45)
xy
2 2