Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

pRoBLEmS
p13.25–p13.30 The strain rosette shown in Figures P13.25–
P13.30 was used to obtain normal strain data at a point on the free
surface of a machine part.
(a) Determine the strain components e x , e y , and γ xy at the point.
(b) Determine the principal strains and the maximum in-plane
shear strain at the point.
(c) Draw a sketch showing the angle θ p , the principal strain
deformations, and the maximum in-plane shear strain
distortions.
(d) Determine the magnitude of the absolute maximum shear
strain.
y
45° 45°
a
b
FIGURE p13.27
c
x
c
y
60°
60°
FIGURE p13.28
b
a
x
Problem ε a ε b ε c ν
P13.25 410 µε −540 µε −330 µε 0.30
P13.26 215 µε −710 µε −760 µε 0.12
P13.27 510 µε 415 µε 430 µε 0.33
P13.28 −960 µε −815 µε −505 µε 0.33
P13.29 −360 µε −230 µε 815 µε 0.15
P13.30 775 µε −515 µε 415 µε 0.30
y
b
120°
120° x
a
120°
c
y
FIGURE p13.29
c
y
b
45°
c
b
x
c
60°
y
b
45°
a
45°
FIGURE p13.25
x
a
FIGURE p13.26
60°
60°
a
FIGURE p13.30
x
13.8 Generalized Hooke’s Law
for Isotropic materials
σ
x
τ
xy
y
τ
yx
σ
FIGURE 13.10
y
σ
τ
y
yx
τ
xy
σ
x
x
Hooke’s law [see Equation (3.4)] can be extended to include the two-dimensional
(Figure 13.10) and three-dimensional (Figure 13.11) states of stress often encountered in
engineering practice. We will consider isotropic materials, which are materials with
properties (such as the elastic modulus E and Poisson’s ratio ν) that are independent of
orientation. In other words, E and ν are the same in every direction for isotropic materials.
Figures 13.12a–c show a differential element of material subjected to three different
normal stresses: σ x , σ y , and σ z . In Figure 13.12a, a positive normal stress σ x produces a
positive normal strain (i.e., elongation) in the x direction:
σ x
ε x =
E
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