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1.4 Do the following strains constitute a possible (compatible) state of strain?
Ay
2
x
2
y
2
=
+
Bxy
=
A
(
+
)
+
By
x
y
γ
=
Bxy
=
γ
=
γ
=
0
xy
z
xz
yz
Answer:
Ye s i f
A
=
B
/
4
.
1.5 Measured strains for a deformed solid are
x
2
z
2
Az
2
=
A
(
+
)
=
0
=
x
y
z
γ
=
0
γ
=
2
Axz
γ
=
0
yz
zx
xy
Find the corresponding displacements.
2
Ax z
leads to
∂
f
1
(
y, z
)
∂
=
−
∂
f
2
(
x, y
)
∂
Hint:
Integrate
x
=
∂
x
u
and
z
=
∂
z
w.
Then
γ
zx
=
=
z
x
g
(
y
)
,
where
f
1
(
y, z
)
and
f
2
(
x, y
)
are functions of integration. Apply
γ
=
0 and
yz
γ
=
0 to show that
g
(
y
)
=
C
1
. It follows that
f
1
=
C
1
z
+
C
2
,f
2
=−
C
1
x
+
C
3
,
xy
where
C
1
,C
2
,
and
C
3
are constants.
Answer:
A
x
3
3
xz
2
+
Az
3
3
u
=
+
C
1
z
+
C
2
,
v
=
C
3
,
w
=
−
C
1
x
+
C
3
.
x
2
10
−
4
,
10
−
3
,
1.6 The components of a displacement field are given by
u
=
(
+
10
)
v
=
(
yz
)
z
2
10
−
3
w
=
(
−
2
xy
)
.
Compute the distance after deformation between the two points
(
3
,
5
,
7
)
and
(
2
,
4
,
6
)
in the undeformed body. Also, find expressions for the strains
ij
and calculate the strains at
(
3
,
−
2
,
5
)
. Do the strains satisfy compatibility?
ds
=
10
−
4
,
The strains do not satisfy compatibility.
Answer:
1
.
738
,
=
6
·
x
1.7 The strain components for the unrestrained thermal expansion of a solid are
x
y
z
xy
yz
xz
0
Does this state of strain lead to unique, continuous displacements regardless of how
=
=
=
α
T,
γ
=
γ
=
γ
=
T
varies with
x, y,
and
z
? What restrictions are placed on
T
as a function of
x, y,
and
z
?
Answer:
T
can be at most a linear function of
x, y,
and
z
.
1.8 Derive constitutive relations for a planar (
x, y
) solid in plane stress and in plane strain
if the material is orthotropic.
1.9 Show that in general the plane stress assumptions for an isotropic solid of elastic
material can lead to a violation of some compatibility conditions.
1.10 Show that the equilibrium equations in polar coordinates for a two-dimensional ele-
ment are
∂σ
r
∂τ
∂φ
+
σ
−
σ
φ
r
r
∂σ
φ
∂φ
+
∂τ
2
τ
1
1
r
r
φ
r
r
φ
r
φ
r
+
+
p
Vr
=
+
+
p
V
φ
=
0
0
∂
∂
r
r
Figure P1.10 shows notation for this problem, but does not provide a complete free-
body diagram for the application of the conditions of equilibrium.
1.11 The equilibrium equations contain more unknowns than equations so that alone they
cannot be solved for the stresses. However, it is possible to test whether a stress
field satisfies the equilibrium equations. Suppose the body forces are zero. Are the
equations of equilibrium satisfied by the following stress field?
σ
x
2
y
2
z
2
z
3
=
+
+
τ
=
τ
=−
xy
+
x
xy
yx
x
2
y
2
2
z
2
y
2
σ
=
+
+
τ
=
τ
=
−
xz
y
xz
zx
2
x
2
y
2
z
2
x
2
σ
=−
+
+
τ
=
τ
=
−
yz
z
yz
zy
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