Chapter 4
Solutions to simple boundary and initial
value problems
In
this chapter, we derive exact solutions to several problems involving
deformable solids. The examples have
been selected partly because they can easily be solved, partly because they
illustrate clearly the role of the various governing equations and boundary
conditions in controlling the solution, and partly because the solutions
themselves are of some practical interest.
4.1 Axially and spherically symmetric solutions to quasi-static
linear elastic problems
4.1.1 Summary of governing equations of
linear elasticity in Cartesian components
At
last, we have all the basic equations we need to solve problems involving
elastic materials subjected to loading.
Specifically, we are given
the following information
1.
Geometry of the
solid
2. Constitutive law for the material (i.e. the linear
elastic-stress-strain equations)
3.
Body force
density (per unit mass) (if any)
4.
Temperature distribution
(if any)
5.
Prescribed
boundary tractions and/or boundary displacements
In addition, to simplify
the problem, we make the following assumptions
1. All
displacements are small. This means that we can use the infinitesimal
strain tensor to characterize deformation; we do not need to distinguish
between stress measures, and we do not need to distinguish between deformed and
undeformed configurations of the solid when writing equilibrium equations and
boundary conditions.
2. The material is an isotropic, linear elastic solid,
with Young’s modulus E and Poisson’s
ratio ,
and mass density
With
these assumptions, we need to solve for the displacement field ,
the strain field and the stress field satisfying the following equations:
Displacementstrain relation
Stressstrain relation
Equilibrium
Equation (static problems only you need the acceleration terms for dynamic
problems)
Traction
boundary conditions on parts of the boundary where tractions are
known.
Displacement
boundary conditions on parts of the boundary where
displacements are known.
4.1.2 Simplified
equations for spherically symmetric linear elasticity problems

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A representative spherically symmetric problem is illustrated
in the picture. We consider a hollow,
spherical solid, which is subjected to spherically symmetric loading (i.e.
internal body forces, as well as tractions or displacements applied to the
surface, are independent of and ,
and act in the radial direction only).
If the temperature of the sphere is non-uniform, it must also be
spherically symmetric (a function of R
only).
The solution is most conveniently expressed using a
spherical-polar coordinate system, illustrated in the figure. The general procedure for solving problems using
spherical and cylindrical coordinates is complicated, and is discussed in
detail in Appendix E. In this section,
we summarize the special form of these equations for spherically symmetric
problems.
As usual, a point in the solid is identified by its
spherical-polar co-ordinates .
All vectors and tensors are expressed as components in the basis shown in the figure. For a spherically symmetric problem
Position Vector
Displacement
vector
Body
force vector
Here,
and are scalar functions. The stress and strain
tensors (written as components in ) have the form
and
furthermore must satisfy .
The tensor components have exactly the same physical interpretation as they did
when we used a fixed basis, except that the subscripts (1,2,3) have
been replaced by .
For
spherical symmetry, the governing equations of linear elasticity reduce to
Strain Displacement Relations
StressStrain relations
Equilibrium Equations
Boundary Conditions
Prescribed Displacements
Prescribed Tractions
These results can either be derived as a special case of the
general 3D equations of linear elasticity in spherical coordinates, or
alternatively can be obtained directly from the formulas in Cartesian
components. Here, we briefly outline the
the latter.
1. Note that we can find the components
of in the basis as follows. First, note that is radial, and can be written in terms of the
position vector as . Next, note and . Using index notation, the components of the
basis vectors in are therefore
where ,
is the Kronecker delta and is the permutation symbol.
2. The components of the (radial)
displacement vector in the basis are .
3. To proceed with the algebra, it is
helpful to remember that , and
4.
The
components of the strain tensor in the basis therefore follow as
5. The strain components can then be found as ,
and . Substituting for the basis vectors and
simplifying gives the strain-displacement relations. For example
where
we have noted .
The remaining components are left as an exercise.
6. Finally, to derive the equilibrium equation, note that
the stress tensor can be expressed as . Substituting for the basis vectors from
item(1) above gives
7.
Substitute the
preceding result into the equilibrium equation
and
work through a good deal of tedious algebra to see that
This result can also be obtained using the virtual work
principle (see problems for Sect 2.4 for details)
4.1.3 General solution
to the spherically symmetric linear elasticity problem
Our goal is to solve the equations given in Section 4.1.2 for
the displacement, strain and stress in the sphere. To do so,
1.
Substitute
the strain-displacement relations into the stress-strain law to show that
2.
Substitute
this expression for the stress into the equilibrium equation and rearrange the
result to see that
Given the temperature distribution and body force this
equation can easily be integrated to calculate the displacement u.
Two arbitrary constants of integration will appear when you do the integral
these must be determined from the boundary conditions at the inner and
outer surface of the sphere.
Specifically, the constants must be selected so that either the
displacement or the radial stress have prescribed values on the inner and outer
surface of the sphere.
In the following sections, this procedure is used to derive
solutions to various boundary value problems of practical interest.
4.1.4 Pressurized
hollow sphere
Assume
that
No body forces act on the sphere
The sphere has uniform temperature
The inner surface R=a is subjected to pressure
The outer surface R=b is subjected to pressure
The
displacement, strain and stress fields in the sphere are
Derivation: The solution can be found by applying the procedure outlined
in Sect 4.1.3.
1.
Note
that the governing equation for u (Sect
4.1.3) reduces to
2.
Integrating
twice gives
where A and B are constants of integration to be
determined.
3.
The
radial stress follows by substituting into the stress-displacement formulas
4. To satisfy the boundary conditions, A and B must be chosen so that and (the stress is negative because the pressure
is compressive). This gives two
equations for A and B that are easily solved to find
5.
Finally,
expressions for displacement, strain and stress follow by substituting for A and B in the formula for u
in (2), and using the formulas for strain and stress in terms of u in Section 4.1.2.
4.1.5 Gravitating
sphere
A planet under its own gravitational attraction may be
idealized (rather crudely) as a solid sphere with radius a, with the following loading
A body force per unit mass, where g is the acceleration due to
gravity at the surface of the sphere
A uniform temperature distribution
A traction free surface at R=a
The
displacement, strain and stress in the sphere follow as
Derivation:
1.
Begin
by writing the governing equation for u
given in 4.1.3 as
2.
Integrating
where A and B are constants of integration that must be determined
from boundary conditions.
3.
The
radial stress follows from the formulas in 4.1.3 as
4.
Finally,
the constants A and B can be determined as follows: (i) The
stress must be finite at ,
which is only possible if . (ii) The surface of the sphere is traction
free, which requires at R=a. Substituting the latter condition into the
formula for stress in (3) and solving for A
gives
5.
The
final formulas for stress and strain follow by substituting the result of (4)
back into (2), and using the formulas in Section 4.1.2.
4.1.6 Sphere with
steady state heat flow
The deformation and stress in a sphere that is heated on the
inside (or outside), and has reached its steady state temperature distribution
can be calculated as follows. Assume
that
No body force acts on the sphere
The temperature distribution in the sphere is
where and are the temperatures at the inner and outer
surfaces. The total rate of heat loss
from the sphere is ,
where k is the thermal conductivity.
The surfaces at R=a and R=b are traction free.
The
displacement, strain and stress fields in the sphere follow as
Derivation:
1.
The
differential equation for u given in
4.1.3 reduces to
2.
Integrating
where A and B are constants of integration.
3.
The
radial stress follows from the formulas in 4.1.3 as
4.
The
boundary conditions require that at r=a
and r=b. Substituting these conditions into the result
of step (3) gives two equations for A
and B which can be solved to see that
4.1.7 Simplified
equations for axially symmetric linear elasticity problems

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Two examples of axially symmetric problems are illustrated in
the picture. In both cases the solid is a
circular cylinder, which is subjected to axially symmetric loading (i.e.
internal body forces, as well as tractions or displacements applied to the
surface, are independent of and ,
and act in the radial direction only).
If the temperature of the sphere is non-uniform, it must also be axially
symmetric (a function of r
only). Finally, the solid can spin with
steady angular velocity about the axis.
The two solids have different shapes. In the first case, the length of the cylinder
is substantially greater than any cross-sectional dimension. In the second case, the length of the
cylinder is much less than its outer radius.
The state of stress and strain in the solid depends on the
loads applied to the ends of the cylinder. Specifically
If the cylinder is completely prevented from
stretching in the direction a state of plane strain exists in the solid.
This is an exact solution to the 3D equations of elasticity, is valid
for a cylinder with any length, and is accurate everywhere in the cylinder.
If the top and bottom surface of the short
plate-like cylinder are free of traction, a state of plane stress exists in the solid.
This is an approximate solution to the 3D equations of elasticity, and
is accurate only if the cylinder’s length is much less than its diameter.
If the top and bottom ends of the long cylinder are subjected to a
prescribed force (or the ends are free of force) a state of generalized plane strain exists in the
cylinder. This is an approximate
solution, which is accurate only away from the ends of a long cylinder. As a rule of thumb, the solution is
applicable approximately three cylinder radii away from the ends.
The solution is most conveniently expressed using a
spherical-polar coordinate system, illustrated in the figure. A point in the solid is identified by its
spherical-polar co-ordinates .
All vectors and tensors are expressed as components in the basis shown in the figure. For an axially symmetric problem
Position
Vector
Displacement vector
Body force vector
Acceleration vector
Here, and are scalar functions.
The stress and strain
tensors (written as components in ) have the form
For
axial symmetry, the governing equations of linear elasticity reduce to
Strain Displacement Relations
StressStrain relations
(plane strain and generalized plane strain)
where for plane strain, and constant for generalized
plane strain.
StressStrain relations
(plane stress)
Equation of motion
Boundary Conditions
Prescribed Displacements
Prescribed Tractions
Plane strain solution
Generalized plane strain solution, with
axial force applied to cylinder:
These results can either be derived as a special case of the
general 3D equations of linear elasticity in spherical coordinates, or
alternatively can be obtained directly from the formulas in Cartesian
components. Here, we briefly outline the
the latter.
1. Note that we can find the components
of in the basis as follows. First, note that is radial a radial unit vector can be written in terms
of the position vector as . Next, note and . Using index notation, the components of the
basis vectors in are therefore
where ,
and we use the convention that Greek subscripts range from 1 to 2.
2. The components of the (radial)
displacement vector in the basis are .
3. To proceed with the algebra, it is
helpful to remember that , and
4.
The
components of the strain tensor in the basis therefore follow as
5. The strain components can then be found as ,
and . Substituting for the basis vectors and
simplifying gives the strain-displacement relations. For example
where
we have noted .
The remaining components are left as an exercise.
6. Finally, to derive the equilibrium equation, note that
the stress tensor can be expressed as . Substituting for the basis vectors from
item(1) above gives
7.
Substitute the
preceding result into the equilibrium equation
and
crank through a good deal of tedious algebra to see that
This result can also be obtained using the virtual work
principle (see problems for Sect 2.4 for details)
4.1.8 General solution to the axisymmetric boundary value problem
Our goal
is to solve the equations given in Section 4.1.2 for the displacement, strain
and stress in the sphere. To do so,
1.
Substitute
the strain-displacement relations into the stress-strain law to show that, for
generalized plane strain
where is constant.
The equivalent expression for plane stress is
2.
Substitute
these expressions for the stress into the equilibrium equation and rearrange
the result to see that, for generalized plane strain
while for plane stress
Given the temperature distribution and body force these
equations can be integrated to calculate the displacement u. Two arbitrary constants
of integration will appear when you do the integral these must be determined from the boundary conditions at the inner and
outer surface of the sphere.
Specifically, the constants must be selected so that either the
displacement or the radial stress have prescribed values on the inner and outer
surface of the cylinder. Finally, for
the generalized plane strain solution, the axial strain must be determined, using the equation for the
axial force acting on the ends of the cylinder.
In the following sections, this procedure is used to derive
solutions to various boundary value problems of practical interest.
4.1.9 Long (generalized plane strain) cylinder subjected to internal and
external pressure.
We consider a long hollow
cylinder with internal radius a and
external radius b as shown in the
figure.
Assume that
No body forces act on the cylinder
The cylinder has zero angular velocity
The sphere has uniform temperature
The inner surface r=a is subjected to pressure
The outer surface r=b is subjected to pressure
For the plane strain solution, the cylinder
does not stretch parallel to its axis.
For the generalized plane strain solution, the ends of the cylinder are
subjected to an axial force as shown.
In particular, for a closed ended
cylinder the axial force exerted by the pressure inside the cylinder acting on
the closed ends is
The displacement, strain
and stress fields in the cylinder are
where for plane strain, while
for generalized plane strain.
Derivation: These
results can be derived as follows. The
governing equation reduces to
The equation
can be integrated to see that
The radial stress follows as
The
boundary conditions are (the stresses are negative because the
pressure is compressive). This yields
two equations for A and B that area easily solved to see that
The remaining results follow
by elementary algebraic manipulations.
4.1.10 Spinning
circular plate
We consider a thin solid plate with radius a that spins with angular speed about its axis. Assume that
No body forces act on the disk
The disk has constant angular velocity
The disk has uniform temperature
The outer surface r=a and the top and bottom faces of the disk are free of traction.
The disk is sufficiently thin to ensure a
state of plane stress in the disk.
Derivation: To derive these results, recall that
the governing equation is
The equation
can be integrated to see that
The radial
stress follows as
The radial
stress must be bounded at r=0, which
is only possible if B=0. In addition, the radial stress must be zero
at r=a, which requires that
The
remaining results follow by straightforward algebra.
4.1.11 Stresses induced by an interference fit between two cylinders
Interference
fits are often used to secure a bushing or a bearing housing to a shaft. In this problem we calculate the stress
induced by such an interference fit.
Consider
a hollow cylindrical bushing, with outer radius b and inner radius a. Suppose that a solid shaft with radius ,
with is inserted into the cylinder as shown. (In practice, this is done by heating the
cylinder or cooling the shaft until they fit, and then letting the system
return to thermal equilibrium)
No body forces act on the solids
The angular velocity is zero
The cylinders have uniform temperature
The shaft slides freely inside the bushing
The ends of the cylinder are free of force.
Both the shaft and cylinder have the same
Young’s modulus E and Poisson’s ratio
The cylinder and shaft are sufficiently long
to ensure that a state of generalized plane strain can be developed in each
solid.
The
displacements, strains and stresses in the solid shaft (r<a) are
In the hollow cylinder,
they are
Derivation: These results can be derived using
the solution to a pressurized cylinder given in Section 4.1.9. After the shaft
is inserted into the tube, a pressure p
acts to compress the shaft, and the same pressure pushes outwards to expand the
cylinder. Suppose that this pressure
induces a radial displacement in the solid cylinder, and a radial
displacement in the hollow tube. To accommodate the interference, the
displacements must satisfy
Evaluating
the relevant displacements using the formulas in 4.1.9 gives
Here, we
have assumed that the axial force acting on both the shaft and the tube must
vanish separately, since they slide freely relative to one another. Solving these two equations for p shows that
This
pressure can then be substituted back into the formulas in 4.1.9 to evaluate
the stresses.