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Although the stress
σ
z
is zero by definition of plane stress, the corresponding strain
z
(and
u
z
) does not vanish.
For plane strain, it is assumed that
0. An example where the plane
strain assumption is often imposed is the case of a long cylinder under internal or external
pressure. It is then assumed that the strains along the long axis of the cylinder are zero.
A plane strain section can be relatively thick in the direction of
t
. Another example of the
occurrence of plane strain would be a slice of thickness
t
taken from a dam between rigid
end walls. Use
=
γ
=
γ
=
z
xz
yz
=
γ
=
γ
=
τ
=
τ
=
σ
=
ν(σ
+
σ
)
0 in Eq. (1.32) to obtain
0 and
.
z
xz
yz
yz
xz
z
x
y
Thus, in the plane strain situation, the stress in the
z
direction is not zero.
Then
1
−
ν
−
ν
0
σ
x
x
1
+
ν
E
=
−
ν
1
−
ν
0
σ
(1.41a)
y
y
γ
0
0
2
τ
xy
xy
E
−
1
=
σ
or
x
y
γ
xy
1
−
ν
−
ν
0
n
x
n
y
n
xy
1
+
ν
Et
=
−
ν
1
−
ν
0
(1.41b)
0
0
2
E
−
1
=
s
Also
−
ν
ν
σ
1
0
x
x
E
=
ν
1
−
ν
0
σ
(1.42a)
y
y
(
1
+
ν)(
1
−
2
ν)
1
−
2
ν
τ
0
0
γ
xy
xy
2
σ
=
E
or
1
−
ν
ν
0
n
x
n
y
n
xy
x
Et
=
ν
1
−
ν
0
(1.42b)
y
(
1
+
ν)(
1
−
2
ν)
1
−
2
ν
γ
0
0
xy
2
s
=
E
2
Replace
E
by
(
1
−
ν
)
E
and then
ν
by
ν/(
1
+
ν)
to obtain the plane stress formulation
from the plane strain equations.
1.3.2
Thermal and Initial Strains
The stress-strain relations of Eqs. (1.32b) and (1.34b) can be generalized to include thermal
effects and other types of initial strains that may occur, for example, in lack of fit problems
when a structure is assembled from parts. Denote the initial strains by
0
. Then
should be
replaced by
−
0
and Eqs. (1.32b) and (1.34b) become
E
−
1
σ
0
=
+
(1.43)
E
0
σ
=
E
−
(1.44)
In the case of thermal strains, consider an element of an elastic solid subjected to a
temperature change
T
. If an element of length
dx
is not constrained, it expands to a new
+
α
α
length of
dx
dx
T
, where
is the
coefficient of thermal expansion
which may depend on
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