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1.15 Show that Hooke's law for isotropic materials can be written as (Eq. 1.35)
σ
ij
=
λδ
ij
kk
+
2
µ
ij
Show that the compatibility relations can be expressed in terms of stresses as (Eq. 1.69)
1
ν
2
σ
+
+
ν
σ
=
+
ν
δ
ij
p
Vk,k
−
(
p
Vi, j
+
p
Vj,i
)
ij
kk,i j
1
1
1.16 Frequently in the literature the material relationship for an isotropic material is written
as
λ
1
2
=
µ
σ
−
λ)
δ
σ
ij
ij
ij
kk
µ(
µ
+
2
2
3
Is this expression correct?
1.17 Show that, for linearly elastic isotropic materials, the following properties of the
material constants apply
E
>
0
,G
>
0
,
λ>
0
,
0
<ν<
1
/
2
.
Hint:
The conditions
E
>
0 and
ν>
0 follow from Hooke's law for uniaxial
tension in the absence of body forces
=
σ
/
E,
=
=−
νσ
/
E
and the normal
x
0
y
z
0
stress in the
x
direction given a constant positive value
Under this loading
condition, the solid must undergo extension (positive) in the axial direction
x
and
contraction (negative) in the transverse direction
y
and
z
. In order for this to be
true,
E
σ
.
0
>
0 and
ν>
0.
The condition 0
<ν<
1
/
2 can be established by imposing hydrostatic compres-
sion so that
0, while all shear stresses are zero. Then,
the change in volume per unit volume (the dilitation)
e
σ
=
σ
=
σ
=−
p
,
p
>
x
y
z
=
+
+
z
will be
x
y
e
2 follows because
the change in hydrostatic compression is negative. Also, it should be recognized
that a negative value implies the unlikely situation where an elongational strain in
a tension bar corresponds to an expansion in a perpendicular direction.
=−
3
p
/(
3
λ
+
2
G
)
=−
3
p
(
1
−
2
ν)/
E
.
The condition 0
<ν<
1
/
1.18 Show that a material with Poisson's ratio equal to one half is incompressible.
1.19 Show that for an isotropic material the stress-strain relationship
σ
ij
=
E
ijk
k
is
defined by
K
3
G
2
E
E
ijk
=
G
(δ
ik
δ
j
+
δ
i
δ
jk
)
+
−
δ
ij
δ
k
,
K
=
3
(
1
−
2
υ)
or
E
ijk
=
λδ
ij
δ
m
+
µδ
i
δ
jm
+
µδ
im
δ
j
1.20 Show that in the absence of body forces, the equilibrium equations of Problem 1.10
are identically satisfied by an Airy stress function
ψ(
r,
φ)
defined as
2
2
2
1
r
∂ψ
r
2
∂
1
ψ
∂φ
σ
φ
=
∂
ψ
r
2
∂ψ
1
1
r
∂
ψ
σ
r
=
r
+
τ
r
φ
=
∂φ
−
∂
2
∂
r
2
∂
r
∂φ
1.21 Verify that the Airy stress functions of Eq. (1.55) satisfy the equations of equilibrium.
1.22 Show that
Ax
2
Ay
2
2
x
3
6
xy
2
B
is a permissible Airy stress function for
a two-dimensional elastic solid with no body forces.
1.23 Suppose the triangular element of Fig. P1.23 is in a state of plane stress with no body
forces. If the Airy stress function is given by
ψ
=
+
−
+
+
x
4
x
2
y
2
y
4
,
where
C
is a constant, find the stress distributions in the element. Also find the normal stress
distribution along the edges.
ψ(
x, y
)
=
C
(
/
12
+
/
4
−
)
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