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

out-of-plane normal strain e z will not be zero. The third principal strain e p3 can be computed
from Equation (13.15):
0.3
ep3
= ez = − ν
( x y)
( 600 400) 85.7
1 − ν e + e = − − + = me
1 − 0.3
Because e p2 < e p3 < e p1 (see Table 13.2), the absolute maximum shear strain will equal the
maximum in-plane shear strain:
γ absmax = εp1 − εp2 = 843 µ ε − ( − 1,043 µ ε) = 1,887 µ rad
y
Sketch the Deformations and Distortions
The principal strains are oriented 29.0° clockwise from the x direction.
Since ε x − ε y < 0, the principal strain corresponding to
this direction is ε p2 = −1,043 µε. The element contracts in this
direction. In the perpendicular direction, the principal strain is
ε p1 = 843 µε, which means that the element elongates.
The distortion caused by the maximum in-plane shear strain
is shown by the diamond that connects the midpoints of each of
the rectangle’s edges.
29.0°
x
843 με
π
2
− 1,887 μrad
1,043 με
mecmovies
ExAmpLE
m13.5 The strain rosette shown was used to obtain normal
strain data at a point on the free surface of a machine part.
Determine
(a) the strain components ε x , ε y , and γ xy at the point.
(b) the principal strains and the maximum shear strain at the
point.
Example 1
ε a = −215 µε
ε b = −130 µε
ε c = 460 µε
c
y
45°
45°
b
a
x
Example 2
ε a = 800 µε
ε b = −200 µε
ε c = 625 µε
c
y
60°
60°
b
x
a
ExERcISE
m13.5 Strain Measurement with Rosettes. A strain rosette
was used to obtain normal strain data at a point on the free surface
of a machine part. Determine the normal strains, the shear strain,
and the principal strains in the x–y plane.
c
60°
y
b
60°
60°
a
FIGURE m13.5
x
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