r
r
θ
2
x
y
=
1
x
x
=
Figure P727
ˆ
z
g
ρ
= −
f
e
L
z
Fig. P7.30
166
SOLUTIONS MANUAL
Solution:
Using the cylindrical coordinate system shown in the figure above, the bound
ary conditions of the problem can be stated as:
At
z
= 0
, L
:
σ
zz
= 0
,
σ
zr
= 0
,
σ
zθ
= 0
,
At
r
=
a
:
u
r
= 0
,
u
θ
= 0
,
u
z
= 0
.
(1)
The body force component is
f
z
=

ρg
.
Neglecting the end effects, we can assume the following form of solution for the antiplane
strain problem:
u
r
= 0
,
u
θ
= 0
,
u
z
=
U
(
r
)
.
(2)
The nonzero strains and stresses are given by
2
ε
rz
=
dU
dr
,
σ
zr
=
μ
dU
dr
.
(3)
Substitution into the third equilibrium equation yields (the other two equilibrium equa
tions are identically satisfied)
dσ
zr
dr
+
1
r
σ
zr

ρg
= 0
→
1
r
d
dr
r
dU
dr
=
ρg
μ
.
(4)
The solution of the equation is given by
r
dU
dr
=
ρg
μ
r
2
2
+
A,
or
dU
dr
=
ρg
μ
r
2
+
A
r
,
U
(
r
) =
ρg
μ
r
2
4
+
A
log
r
+
B,
where
A
and
B
are constants of integration that must be determined using the boundary
conditions. The boundary conditions associated with the antiplane strain problem are
that (a)
U
(
r
) is finite at
r
= 0, and (b)
U
(
a
) = 0.
The first condition gives
A
= 0
and the second one leads to
B
=

(
ρg/
4
μ
)
a
2
.
The first boundary condition can be
replaced by the requirement that
σ
rz
= 0 at
r
= 0. This will lead to the conclusion
that
dU/dr
= 0 at
r
= 0, from which we arrive at the same result (that is,
A
= 0). The
solution becomes
u
z
(
r
) =
U
(
r
) =

ρga
2
4
μ
1

r
2
a
2
.
(5)
The stress field becomes
σ
θz
= 0
,
σ
zr
=
ρg
2
r.
(6)
Note that the boundary conditions (1) of the 3D problem are not satisfied at
z
= 0
, L
.
Hence, it is only an approximate solution.
7.31
An external hydrostatic pressure of magnitude
p
is applied to the surface of a spherical
body of radius
b
with a concentric
rigid
spherical inclusion of radius
a
, as shown in
Fig. P7.31. Determine the displacement and stress fields in the spherical body. Using
the stress field obtained, determine the stresses at the surface of a rigid inclusion in an
infinite elastic medium.
a
b
p
Rigid spherical core
Spherical shell (
)
,
μ λ
Figure P732
Fig. P7.31
CHAPTER 7: LINEARIZED ELASTICITY
167
Solution:
We use the semiinverse method to solve the problem.
Assume that
u
φ
=
u
θ
= 0 and
u
R
=
U
(
R
). The boundary conditions are:
At
r
=
b
:
σ
RR
=

p
;
At
R
=
a
:
u
R
=
U
(
a
) = 0
(1)
The solution of the Navier equations give the displacement [see Eq. (7.3.13)]
U
(
R
) =
AR
+
1
R
2
B,
σ
RR
(
R
) = (2
μ
+ 3
λ
)
A

4
μ
R
3
B,
(2)
where
μ
and
λ
are the Lam´
e constants.
Using the boundary conditions, we obtain
(2
μ
+ 3
λ
= 3
K
)
U
(
a
) = 0
→
B
=

a
3
A
;
σ
RR
(
R
) =
3
K
+
4
a
3
μ
R
3
A,
(3)
σ
RR
(
b
) =

p
→
A
=

p
3
K
+
4
a
3
μ
b
3
.
(4)
Hence the displacement
u
R
and stress field are given by
u
R
(
R
) =

b
3
pR
3
Kb
3
+ 4
μa
3
1

a
3
R
3
(5)
σ
RR
(
R
) =

1 + 2
α
(
a
3
/R
3
)
1 +
β
p,
α
=
2
μ
3
K
,
β
= 2
α
a
3
b
3
(6)
σ
θθ
(
R
) =
σ
φφ
=

1

α
(
a
3
/R
3
)
1 +
β
p.
(7)
To obtain the stresses at the surface of a rigid inclusion in an infinite elastic medium,
we let
b
→ ∞
and obtain
σ
RR
=

1 +
4
μa
3
3
KR
3
p,
σ
θθ
=
σ
φφ
=

1

2
μa
3
3
KR
3
p .
(8)
At the interface of the rigid inclusion and the elastic medium (
R
=
a
), the stresses are
σ
RR
=

1 +
4
μ
3
K
p,
σ
θθ
(
R
) =
σ
φφ
=

1

2
μ
3
K
p .
(9)
7.32
Consider the concentric spheres shown in Fig. P7.32. Suppose that the core is elastic