Chapter 4
Solutions to simple boundary and initial
value problems
4.2 Axially and spherically symmetric solutions to quasi-static
elastic-plastic problems
In
this section, we derive exact solutions to simple boundary value problems
involving elastic-perfectly plastic solids.
The solutions are of interest primarily because they illustrate
important general features of solids that are loaded beyond yield. In particular, they illustrate the concepts
of
1. The elastic limit this is the load required to initiate plastic
flow in the solid.
2. The plastic collapse load at this load the displacements in the solid
become infinite.
3. Residual stress - if a solid is loaded beyond the
elastic limit and then unloaded, a system of self-equilibrated stress is established
in the material.
4. Shakedown - if an elastic-plastic solid is subjected
to cyclic loading, and the maximum
load during the cycle exceeds yield, then some plastic deformation must occur
in the material during the first load cycle.
However, residual stresses are introduced in the solid, which may
prevent plastic flow during subsequent cycles of load. This process is known as `shakedown’ and the
maximum load for which it can occur is known as the `shakedown limit.’ The shakedown limit is often substantially
higher than the elastic limit, so the concept of shakedown can often be used to
reduce the weight of a design.
5. Cyclic plasticity - for cyclic loads exceeding the
shakedown limit, a region in the solid will be repeatedly plastically deformed.
4.2.1 Summary of governing equations
We are given the following
information
1.
Geometry of the
solid
2.
Constitutive law
for the material (i.e. the elastic-plastic 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, elastic-perfectly plastic solid, with Young’s modulus E Poisson’s
ratio ,
yield stress Y and mass density
3.
We will neglect
temperature changes.
With
these assumptions, we need to solve for the displacement field ,
the strain field and the stress field satisfying the following equations:
Displacementstrain relation
Incremental stressstrain relation
where
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.
4.2.2 Simplified
equations for spherically symmetric problems
|
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
with spherical and cylindrical coordinates is complicated, and is discussed in
detail in Chapter 10 in the context of modeling deformation in shells. 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 Cartesian basis shown in the figure. For a spherically symmetric problem we have
that
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 reduce to
Strain Displacement Relations
StressStrain relations
In elastic region(s)
In plastic region(s)
Yield
criterion:
Strain
partition:
Elastic
strain:
Flow
rule:
Equilibrium Equations
Boundary Conditions
Prescribed Displacements
Prescribed Tractions
The equilibrium and strain-displacement equations can be
derived following the procedure outlined in Section 4.1.2. The stress-strain relations are derived by
substituting the strain components into the general constitutive equation and simplifying
the result.
Unlike the elastic solution discussed in Sect 4.1, there is
no clean, direct and general method for integrating these equations. Instead, solutions must be found using a
combination of physical intuition and some algebraic tricks, as illustrated in
the sections below.
4.2.3 Elastic-perfectly
plastic hollow sphere subjected to monotonically increasing internal pressure
Assume
that
The sphere is stress free before it is loaded
No body forces act on the sphere
The sphere has uniform temperature
The inner surface r=a is subjected to (monotonically increasing) pressure
The outer surface r=b is traction free
Strains are infinitesimal
Solution:
(i) Preliminaries:
The sphere first reaches yield (at r=a) at an internal pressure
For pressures in the range the region between and deforms plastically; while the region between remains elastic, where c satisfies the equation
At a pressure the entire cylinder is plastic. At this point the sphere collapses the displacements become infinitely large.
(ii) Solution in the plastic region
(iii) Solution in the elastic region
These
results are plotted in the figures below.
(a)
(b)
(c)
(a) Stress and (b) displacement
distributions for a pressurized elastic-perfectly plastic spherical shell; and
(c) Displacement at r=a as a function
of pressure. Displacements are shown for
Derivation: By substituting the stresses for the
elastic solution given in 4.1.4 into the Von-Mises yield criterion, we see that
a pressurized elastic sphere first reaches yield at r=a. If the pressure is increased beyond yield we anticipate that a
region will deform plastically, while a region remains elastic. We must find separate
solutions in the plastic and elastic regimes.
In the plastic regime
(i) We anticipate that .
The yield criterion then gives .
(ii) Substituting this result into the equilibrium equation
given in Sect 4.2.2 shows that
(iii) Integrating, and using the boundary condition together with the yield condition (i) gives
(iv) Since the pressure is monotonically increasing, the
incremental stress-strain relations for the elastic-plastic region given in
4.2.2 can be integrated. The elastic strains follow as
(v) The plastic strains satisfy . Consequently, using the strain partition
formula, the results of (iv), and the strain-displacement relation shows that
(vi) Integrating gives
where C is a
constant of integration
(vii) The constant of integration can
be found by noting that the radial displacements in the elastic and plastic
regimes must be equal at r=c. Using the expression for the elastic
displacement field below and solving for C gives
This result can be simplified by
noting that from the expression for the location of the
elastic-plastic boundary given below.
In the elastic regime
The solution can be found directly
from the solution to the internally pressurized elastic sphere given in Sect
4.1.4. From step (iii) in the solution
for the plastic regime we see that the radial pressure at r=c is .
We can simplify the solution by noting from the expression for the location of the
elastic-plastic boundary. Substituting
into the expressions for stress and displacement shows that
Location of the elastic-plastic boundary
Finally, the elastic-platsic boundary
is located by the condition that the stress in the elastic region must just
reach yield at r=c (so there is a
smooth transition into the plastic region).
The yield condition is ,
so substituting the expressions for stress in the elastic region and
simplifying yields
If ,
Y, a and b are specified this equation can be solved (numerically) for c.
For graphing purposes it is preferable to choose c and then calculate the corresponding value of
4.2.4 Elastic-perfectly
plastic hollow sphere subjected to cyclic internal pressure
Assume
that
The sphere is stress free before it is loaded
No body forces act on the sphere
The sphere has uniform temperature
The outer surface r=b is traction free
Suppose that the inner surface of the sphere r=a is repeatedly subjected to pressure and then unloaded to zero pressure.
Solution:
(i) Preliminaries:
If the maximum pressure applied to the sphere does not exceed the
elastic limit (i.e. ) the solid remains elastic throughout the
loading cycle. In this case, the sphere
is stress free after unloading, and remains elastic throughout all subsequent
load cycles.
For pressures in the range the region between and deforms plastically during the first
application of pressure; while the region between remains elastic, where c satisfies the equation . In this case, the solid is permanently
deformed. After unloading, its internal
and external radii are slightly increased, and the sphere is in a state of residual stress.
If the maximum internal pressure satisfies ,
the cylinder deforms plastically during the first application of pressure. It then deforms elastically (no yield) while
the pressure is removed. During
subsequent pressure cycles between zero and the maximum pressure, the cylinder
deforms elastically. Residual
stresses introduced during the first loading cycle are protective, and prevent
further plasticity. This behavior is
known as `shakedown’ and the maximum load for which it can occur ( ) is known as the `shakedown limit’
If the maximum internal pressure reaches the
shakedown limit ,
the residual stress just reaches yield at r=a
when the pressure is reduced to zero after the first loading cycle.
For internal pressures ,
a plastic zone forms between as the pressure is reduced to zero, where d satisfies the equation . During subsequent cycles of loading, the
region is repeatedly plastically deformed, stretching
in the hoop direction during increasing pressure, and compressing as the
pressure is reduced to zero. The region
between deforms plastically during the first cycle of
pressure, but remains elastic for all subsequent cycles. This is a `shakedown region.’ The
remainder of the sphere experiences elastic cycles of strain.
In the preceding discussion, we have assumed that the
cylinder is thick enough to support an arbitrarily large pressure. The internal pressure cannot exceed the
collapse load ,
so some regimes are inaccessible for thinner walled spheres.
The stress fields at maximum and minimum load for these
various ranges of applied load are listed below. The displacements can be computed, but the
formulas are too long to record here.
The residual stress distributions (after unloading to zero
pressure) are shown in the figure on the right, for a sphere with b/a=3. The solution for c/a=1.25 is below the shakedown limit;
the other two solutions are for pressures exceeding the shakedown limit. The region of cyclic plasticity can be seen
from the discontinuity in the hoop stress curves. Note that the residual
stresses are predominantly compressive for this reason, bolt holes, pressure vessels
and gun barrels are often purposely pressurized above the elastic limit so as
to introduce a compressive stress near the loaded surface. This protects the component against fatigue,
since fatigue cracks do not propagate under compressive loading.
Solution for pressures below the
elastic limit
The displacement, strain and stress
field at maximum load are given by the elastic solution in Section 4.1.4
Solution for pressures between the
elastic and shakedown limits
At maximum pressure, the displacement and
stress fields are given by the elastic-plastic solution in Section 4.2.3.
At zero pressure, the solution is
(i) Solution for
(ii) Solution for
Solution for pressures exceeding the
shakedown limit
At maximum pressure, the displacement, strain
and stress fields are given in Section 4.2.3.
At zero pressure, the solution is
(i) Solution for cyclic
plastic region
(ii) Solution for
shakedown region
(iii) Solution for the
elastic region
Derivation of stress after unloading in the cyclic plastic regime
(i) We anticipate that .
The yield criterion then gives .
(ii) Substituting this result into the equilibrium equation
shows that
(iii) Integrating, and using the boundary condition together with the yield condition in step (i)
gives
Derivation of stress after unloading in the shakedown regime
(i) In this region, the stress at
maximum load are given by the expressions for in 4.2.3, i.e.
The solid then unloads elastically
while the pressure is removed.
(ii) The change in stress during
unloading can be calculated quickly by regarding the region as a spherical shell with internal radius d and external radius b, subjected to radial pressure at r=d.
At maximum load, the pressure at r=d
is ;
after unloading the pressure follows from the solution for the cyclic plastic
regime as .
The change in pressure at r=d during unloading is thus .
(iii) The change in pressure during
unloading can also be expressed as using the governing equation for d shown below.
(iv) We then can simply add the
(elastic) stress and displacement induced by this pressure change to the
displacement and stress at maximum load, to obtain the solution given above.
Boundary of the cyclic plastic zone
The boundary of the cyclic plastic
zone is determined by the condition that the stress in the shakedown regime
must just reach yield at r=d when the
pressure reaches zero. This gives
Derivation of solution in the elastic region
The solution in this region is
derived in the same way as the solution for the shakedown region, except that
the displacement and stress at maximum load are given by solutions for
4.2.5 Simplified
equations for plane strain axially symmetric elastic-perfectly plastic solids
|
An axially symmetric solid is illustrated in the
picture. 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).
Temperature changes will be neglected, to simplify calculations. However, the solid can spin with steady
angular velocity about the axis.
We will assume that the cylinder is completely prevented from
stretching in the direction, so that a state of plane strain exists in the solid.
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 reduce to
Strain Displacement Relations
StressStrain relations
(plane strain and generalized plane strain)
In elastic region(s)
In plastic region(s)
Yield
criterion:
Strain
partition:
Elastic
strain:
Flow
rule:
Equation of motion
Boundary Conditions
Prescribed Displacements
Prescribed Tractions
The equilibrium and strain-displacement equations can be
derived following the procedure outlined in Section 4.1.2. The stress-strain relations are derived by
substituting the strain components into the general constitutive equation and
simplifying the result.
Unlike the elastic solution in Sect 4.1, there is no clean,
direct and general method for integrating these equations. Instead, solutions must be found using a
combination of physical intuition and some algebraic tricks, as illustrated in
the sections below.
4.2.6 Long (plane strain) cylinder subjected to internal 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 free of pressure
The cylinder does not stretch parallel to its
axis
The
solution given below is approximate, because it assumes that both elastic and
plastic axial strains vanish separately (when in fact only the sum of elastic
and plastic strains should be zero).
Solution:
(i) Preliminaries:
The cylinder first reaches yield (at r=a) at an internal pressure
For pressures in the range the region between and deforms plastically; while the region between remains elastic, where c satisfies the equation
At a pressure the entire cylinder is plastic. At this point the sphere collapses the displacements become infinitely large.
(ii) Solution in the plastic region
(iii) Solution in the elastic region
The
stress and displacement fields are plotted in the figure below, for various
positions of the elastic-plastic boundary.
The results are for b/a=3, and
the displacement is shown for a solid with
Stress distribution Radial displacement
Derivation: By substituting the stresses for the
elastic solution given in 4.1.4 into the Von-Mises yield criterion, we see that
a pressurized elastic cylinder first reaches yield at r=a. If the pressure is increased beyond yield, a region will deform plastically, while a region remains elastic. We must find separate solutions
in the plastic and elastic regimes.
In the plastic regime
(i) To simplify the calculation we
assume that . This turns out to be exact for ,
but is approximate for other values of Poisson ratio. The plastic flow rule shows that ,
in which case requires that
(ii) We anticipate that .
Substituting the result of (i) into the yield criterion then gives .
(iii) Substituting this result into the equilibrium equation
shows that
(iv) Integrating, and using the boundary condition together with the yield condition (i) gives
(v) The elastic strains follow as
(vi) With assumption (i), the flow
rule shows that plastic strains satisfy . Consequently, using the strain partition
formula and the strain-displacement relation shows that
(vii) Integrating gives
where C is a
constant of integration
(viii) The constant of integration
can be found by noting that the radial displacements in the elastic and plastic
regimes must be equal at r=c. Using the expression for the elastic
displacement field below and solving for C gives
This result can be simplified by noting that from the expression for the location of the
elastic-plastic boundary given below.
In the elastic regime
The solution can be found directly
from the solution to the internally pressurized elastic cylinder given in Sect
4.1.9. From step (iv) in the solution
for the plastic regime we see that the radial pressure at r=c is .
We can simplify the solution by noting from the expression for the location of the
elastic-plastic boundary. Substituting
into the expressions for stress and displacement in 4.1.9 shows that
Location of the elastic-plastic boundary
Finally, the elastic-plastic boundary
is located by the condition that the stress in the elastic region must just
reach yield at r=c (so there is a
smooth transition into the plastic region).
The yield condition is ,
so substituting the expressions for stress in the elastic region and
simplifying yields
If ,
Y, a and b are specified this equation can be solved (numerically) for c.
For graphing purposes it is preferable to choose c and then calculate the corresponding value of