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1.30 A solid bar with a rectangular cross-section is subjected to a uniform axial tension of
σ
x
. No other stresses are present. Find the strain in the axial and lateral directions
Answer:
=
[
σ
−
ν(σ
+
σ
)
]
/
E
=
σ
/
E
x
x
y
z
x
=−
νσ
/
E,
=−
νσ
/
E
y
x
z
x
1.31 Suppose a flat plate lies in the
xy
plane. The applied forces shown in Fig. P1.31 are
uniformly distributed along the edges. Also, there is a temperature increase of 100
◦
K.
If this is an aluminum plate with
E
10
6
K
−
1
,
calculate
=
70 GPa,
ν
=
0
.
33
,
α
=
23
×
the strains and the changes in dimensions if
y
x
100 mm
200 mm
FIGURE P1.31
(a) the plate is unrestrained in the
z
direction, and
(b) the plate is placed firmly between lubricated dies which prevent expansion in the
z
direction. Compare the results of (a) and (b).
Hint:
Use plane stress for (a) and plane strain for (b). Also,
τ
=
0
.
xy
10
−
3
,
Answer:
For case (a),
=
[
σ
−
ν(σ
+
σ
)
]
/
E
+
α
T
=
1
.
92
×
=
x
x
y
z
y
10
−
3
,
10
−
3
,
3
.
06
×
=
2
.
11
×
γ
=
γ
=
γ
=
0
,
x
=
200
x
,
y
=
100
y
,
z
=
z
xy
yz
xz
10
.
z
1.32 For an isotropic elastic body with a temperature change
T
, the stress-strain relation-
ship takes the form [from Eq. (1.35a)]
σ
=
λδ
+
2
µ
−
(
3
λ
+
2
µ)
T
δ
ij
ij
kk
ij
ij
Show that these relations for plane stress
(σ
zz
=
τ
xz
=
τ
yz
=
0
)
reduce to
E
E
σ
=
)
(
+
ν
)
−
+
ν
α
T
τ
=
2
µ
xx
xx
yy
xy
xy
(
1
−
ν
2
1
E
E
σ
yy
=
)
(ν
xx
+
yy
)
−
+
ν
α
T
(
1
−
ν
2
1
1.33 Show that for an isotropic material there are only two independent material constants.
Hint:
Show that
G
=
E
/(
2
(
1
+
ν)).
See a basic mechanics of solids text.
σ
=
λδ
+
µ
µ
=
2
ij
,
G
(1)
ij
ik
kk
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