Chapter 6
Analytical techniques and solutions for
plastic solids
6.2 Bounding theorems in plasticity
and their applications
To
set the background for plastic limit analysis, it is helpful to review the
behavior of an elastic-plastic solid or structure subjected to mechanical
loading. The solution to an
internally-pressurized elastic-perfectly plastic sphere given in Section 4.2
provides a representative example. All
elastic-perfectly plastic structures will exhibit similar behavior. In particular
An inelastic solid
will reach yield at some critical value of applied load.
If the load exceeds yield, a plastic region starts to
spread through the solid. As an increasing area of the solid reaches yield, the
displacements in the structure progressively increase.
At a critical load, the plastic region becomes large
enough to allow unconstrained plastic flow in the solid. The load cannot be
increased beyond this point. The solid is said to collapse.
Strain
hardening will influence the results quantitatively, but if the solid has a
limiting yield stress (a stress beyond which it can never harden) its behavior
will be qualitatively similar.
In
a plasticity calculation, often the two most interesting results are (a) the
critical load where the solid starts to yield; and (b) the critical load where
it collapses. Of course, we don’t need
to solve a plasticity problem to find the yield point we only need the elastic fields. In many design problems this is all we need,
since plastic flow must be avoided more often than not. But there are situations where some
plasticity can be tolerated in a structure or component; and there are even
some situations where it’s desirable (e.g. in designing crumple zones in cars). In this situation, we usually would like to
know the collapse load for the solid.
It would be really nice to find some way to get the collapse load
without having to solve the full boundary value problem.
This
is the motivation for plastic limit analysis.
The limit theorems of plasticity provide a quick way to estimate
collapse loads, without needing any fancy calculations. In fact, collapse loads are often much easier
to find than the yield point!
In
this section, we derive several useful theorems of plastic limit analysis and
illustrate their applications.
6.2.1
Definition of the plastic dissipation
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Consider
a rigid perfectly plastic solid, which has mass density ,
and a Von-Mises yield surface with yield stress in uniaxial tension Y. (By definition, the elastic strains are
zero in a rigid plastic material). The solid is subjected to tractions on the its boundary. The solid may also be subjected to a body
force b (per unit mass) acting on
the interior of the solid. Assume that
the loading is sufficient to cause the solid to collapse.
Velocity discontinuities: Note that the velocity and stress fields in a
collapsing rigid plastic solid need not necessarily be continuous. The solution often has shear discontinuities,
as illustrated on the right. In the
picture, the top part of the solid slides relative to the bottom part. We need a way to describe this kind of deformation. To do so,
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1. We assume that the velocity field at collapse may have a finite set of such
shear discontinuities, which occur over a collection of surfaces . Let m be a unit vector normal to the surface at some point , and let denote the limiting values of velocity and stress on the two sides of the surface.
2. To ensure that no holes open up in the material, the
velocity discontinuity must satisfy
3. The solids immediately adjacent to the discontinuity
exert equal and opposite forces on each other.
Therefore
4. We will use the symbol to denote the relative velocity of sliding
across the discontinuity, i.e.
5. The yield criterion and plastic flow rule require that
on any surfaces of velocity
discontinuity.
Kinematically admissible collapse
mechanism: The kinematically
admissible collapse mechanism is analogous to the kinematically admissible
displacement field that was introduced to define the potential energy of an
elastic solid. By definition, a
kinematically admissible collapse mechanism is any velocity field v satisfying (i.e. v
is volume preserving)
Like
u, the virtual velocity v may have a finite set of
discontinuities across surfaces with normal (these are not necessarily the discontinuity
surfaces for the actual collapse mechanism).
We use
to
denote the magnitude of the velocity discontinuity. We also define the virtual
strain rate
(note that ) and the effective virtual plastic strain
rate
Plastic Dissipation: Finally, we define the plastic dissipation associated
with the virtual velocity field v as
The terms in this
expression have the following physical interpretation:
1.
The first
integral represents the work dissipated in plastically straining the solid;
2.
The second
integral represents the work dissipated due to plastic shearing on the velocity
discontinuities;
3.
The third
integral is the rate of mechanical work done by body forces
4.
The fourth
integral is the rate of mechanical work done by the prescribed surface
tractions.
6.2.2. The Principle of Minimum Plastic Dissipation
Let
denote the actual velocity field that causes a
rigid plastic solid to collapse under a prescribed loading. Let v be
any kinematically admissible collapse mechanism. Let denote the plastic dissipation, as defined in
the preceding section. Then
1.
2.
Thus,
is an absolute minimum for - in other words, the actual velocity field at
collapse minimizes . Moreover, is zero for the actual collapse mechanism.
Derivation: Begin by summarizing the equations governing the
actual collapse solution. Let denote the actual velocity, strain rate and
stress in the solid at collapse. Let denote the deviatoric stress. The fields must
satisfy governing equations and boundary conditions
Strain-displacement relation
Stress
equilibrium
Plastic
flow rule and yield criterion
On
velocity discontinuities, these conditions require that
Boundary conditions
We
start by showing that .
1. By definition
2. Note that, using (i) the flow rule, (ii) the condition
that and (iii) the yield criterion
3. Note that from the symmetry of .
Hence
4. Note that . Substitute into the expression for ,
combine the two volume integrals and recall (equilibrium) that to see that
5. Apply the divergence theorem to the volume integral in
this result. When doing so, note that we
must include contributions from the velocity discontinuity across S as follows
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6.
Finally, recall
that on the boundary, and note that the outward
normals to the solids adjacent to S
are related to m by . Thus
Since ,
we find that as required.
Next, we show that . To this end,
1. Let be a kinematically admissible velocity field
as defined in the preceding section, with strain rate
2. Let be the stress necessary to drive the
kinematically admissible collapse mechanism, which must satisfy the plastic
flow rule and the yield criterion
3. Recall that the plastic strains and stresses
associated with the kinematically admissible field must satisfy the Principle
of Maximum Plastic Resistance (Section 3.7.10), which in the present context
implies that
To see this, note that is the stress required to cause the plastic
strain rate ,
while the actual stress state at collapse must satisfy .
4. Note that . Substituting into the principle of maximum
plastic resistance and integrating over the volume of the solid shows that
5.
Next, note that
6. The equilibrium equation shows that . Substituting this into the result of (5) and
then substituting into the result of (4) shows that
7. Apply the divergence theorem to the second integral.
When doing so, note that we must include contributions from the velocity
discontinuity across as follows
8. Recall that on the boundary, and note that the outward
normals to the solids adjacent to S
are related to m by . Thus
9.
Finally, note
that on
since
the shear stress acting on any plane in the solid cannot exceed . Thus
proving that as required.
6.2.3
The Upper Bound Plastic Collapse Theorem
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Consider a rigid plastic solid, subjected to some
distribution of tractions and body forces . We will attempt to estimate the factor by which the loading can be increased before
the solid collapses ( is effectively the factor of safety). We suppose that the solid will collapse for
loading ,
.
To estimate ,
we guess the mechanism of collapse.
The collapse mechanism will be an admissible velocity field, which may have a finite set of discontinuities across
surfaces with normal ,
as discussed in 6.2.1.
The
principle of minimum plastic dissipation then states that
for any collapse mechanism,
with equality for the true mechanism of collapse. Therefore
Expressed
in words, this equation states that we can obtain an upper bound to the
collapse loads by postulating a collapse mechanism, and computing the ratio of
the plastic dissipation associated with this mechanism to the work done by the
applied loads.
So,
we can choose any collapse mechanism, and use it to estimate a safety
factor. The actual safety factor is
likely to be lower than our estimate (it will be equal if we guessed
right). This method is evidently
inherently unsafe, since it overestimates the safety factor but it is usually possible guess the collapse
mechanism quite accurately, and so with practice you can get excellent
estimates.
6.2.4
Examples of applications of the upper bound theorem
Example 1: collapse load for a uniaxial
bar. We will illustrate the bounding
theorems using a few examples. First, we
will compute bounds to the collapse load
for a uniaxial bar. Assume the bar has
unit out of plane thickness, for simplicity.
To
get an upper bound, we guess a collapse mechanism as shown below. The top and bottom half of the bar slide past
each other as rigid blocks, as shown, with a velocity discontinuity across the
line shown in red.
The upper bound theorem gives
In
this problem the strain rate vanishes, since we assume the two halves of the
bar are rigid. The plastic dissipation
is
The body force vanishes,
and
where is the vertical component of the velocity of
the top block. Thus
The best upper bound occurs
for ,
giving for the collapse load.
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Example 2: Collapse load for a bar containing a hole. For a slightly more interesting problem, consider the
effect of inserting a hole with radius a in the center of the
column. This time we apply a force
to the top of the column, rather than specify the traction distribution in
detail. We will accept any solution that
has traction acting on the top surface that is statically equivalent to the
applied force.
A possible collapse
mechanism is shown. The plastic
dissipation is
The rate
of work done by applied loading is
Our upper bound follows as
and the best upper bound
solution is
Example 3: Force
required to indent a rigid platic surface. For our next example, we attempt to find upper and
lower bounds to the force required to push a flat plane punch into a rigid
plastic solid. This problem is
interesting because we have an exact slip-line field solution, so we can assess
the accuracy of the bounding calculations.
A
possible collapse mechanism is shown above.
In each semicircular region we assume a constant circumferential
velocity . To compute the plastic dissipation in one of
the regions, adopt a cylindrical-polar coordinate system with origin at the
edge of the contact. The strain
distribution follows as
Thus the
plastic dissipation is
(note that there’s a
velocity discontinuity at r=a). The work done by applied loading is just
giving the upper bound
This
should be compared to the exact slip-line field solution
computed
in section 6.1. The error is 17% - close
enough for government work.
Example 4:
Orthogonal metal cutting. The picture shows a simple model of machining. The objective is to determine the horizontal
force P acting on the tool (or
workpiece) in terms of the depth of cut h,
the tool rake angle and
the shear yield stress of the material
To
perform the calculation, we adopt a reference frame that moves with the
tool. Thus, the tool appears stationary,
while the workpiece moves at speed to the right.
The collapse mechanism consists of shear across the red line shown in
the picture.
Elementary geometry gives the
chip thickness d as
Mass
conservation (material flowing into slip discontinuity = material flowing out
of slip discontinuity) gives the velocity of material in the chip as
The velocity discontinuity
across the shear band is
The plastic dissipation
follows as
The
upper bound theorem gives
To
obtain the best estimate for P, we
need to minimize the right hand side of this expression with respect to . This gives
The
resulting upper bound to the machining force is plotted on the figure to the
right.
6.2.5
The Lower Bound Plastic Collapse Theorem
The lower bound theorem provides a safe estimate of the
collapse loads for a rigid plastic solid.
Consider a rigid plastic solid, subjected to some
distribution of tractions and body forces . We will attempt to estimate the factor by which the loading can be increased before
the solid collapses ( is effectively the factor of safety). We suppose that the solid will collapse for
loading ,
.
To estimate ,
we guess the distribution of stress in the solid at collapse.
We will denote the guess for the stress distribution by . The stress distribution must
1. Satisfy the boundary conditions ,
where is a lower bound to
2. Satisfy the equations of equilibrium within the solid,
3. Must not violate the yield criterion
anywhere within the solid,
The
lower bound theorem states that if any
such stress distribution can be found, the solid
will not collapse, i.e. .
Derivation
1. Let denote the actual velocity field in the solid
at collapse. These must satisfy the
field equations and constitutive equations listed in Section 6.4.4.
2. Let denote the guess for the stress field.
3. The Principle of Maximum Plastic
Resistance (see Section 3.7.10) shows that ,
since is at or below yield.
4. Integrating this equation over the
volume of the solid, and using the principle of virtual work on the two terms
shows that
This proves the
theorem.
6.2.6 Examples of
applications of the lower bound plastic collapse theorem
Example 1: Collapse load for a plate
containing a hole. A plate with width L contains a hole of radius a
at its center. The plate is
subjected to a tensile force P as
shown (the traction distribution is not specified in detail we will accept any
solution that has traction acting on the top surface that is statically
equivalent to the applied force).
For
a statically admissible stress distribution, we consider the stress field shown
in the figure, with ,
and all other stress components zero.
The
estimate for the applied load at collapse follows as
Example 2: Rigid indenter in contact
with a half-space. We consider a flat indenter with width a that is pushed into the surface of a
half-space by a force P. The stress state illustrated in the figure
will be used to obtain a lower bound to the collapse load in the solid. Note that
1. Regions C, E, F are stress free
2. The stress in regions A and D consists of a state of
uniaxial stress, with direction parallel to the boundaries between AC (or AE)
and CD (or DF) respectively. We will
denote this stress by ,
where m is a unit vector parallel to
the direction of the uniaxial stress.
3. The stress state in the triangular region B has
principal directions of stress parallel to .
We will write this stress state as
The
stresses in each region must be chosen to satisfy equilibrium, and to ensure
that the stress is below yield everywhere.
The stress is constant in each region, so equilibrium is satisfied
locally. However, the stresses are
discontinuous across AC, AB, etc. To
satisfy equilibrium, equal and opposite tractions must act on the material
surfaces adjacent to the discontinuity, which requires, e.g. that ,
where n is a unit vector normal to
the boundary between A and B as indicated in the figure. We enforce this condition as follows:
1. Note that
2. Equilibrium across the boundary between A and B
requires
3. We must now choose and to maximize the collapse load, while ensuring
that the stresses do not exceed yield in regions A or B. Clearly, this requires ;
while must be chosen to ensure that . This requires . The largest value for maximizes the bound.
4. Finally, substituting for gives . We see that the lower bound is .
6.2.7 The
lower bound shakedown theorem
In
this and the next section we derive two important theorems that can be used to
estimate the maximum cyclic loads
that can be imposed on a component without exceeding yield. The concept of shakedown in a solid subjected to cyclic loads was introduced in
Section 4.2.4, which discusses the behavior of a spherical shell subjected to
cyclic internal pressure. It was shown
that, if the first cycle of pressure exceeds yield, residual stresses are
introduced into the shell, which may prevent further plastic deformation under
subsequent load cycles. This process is
known as shakedown, and the maximum
load for which it can occur is known as the shakedown
limit.
We
proceed to derive a theorem that can be used to obtain a safe estimate to the
maximum cyclic load that can be applied to a structure without inducing cyclic
plastic deformation.
We
consider an elastic-perfectly plastic solid with Von-Mises yield surface,
associated flow law, and uniaxial tensile yield stress Y. Assume that
1. The displacement on part of the boundary of the solid
2. The remainder of the boundary is subjected to a prescribed cycle of traction
. The history of traction is periodic, with a
period T.
Define
the following quantities:
1. Let denote the actual history of displacement,
strain and stress induced in the solid by the applied loading. The strain is partitioned into elastic and
plastic parts as
2. Let denote the history of displacement, strain and
stress induced by the prescribed traction in a perfectly elastic solid with identical geometry.
3. We introduce (time dependent) residual stress and residual strain fields, which (by definition) satisfy
Note
that, (i) because on ,
it follows that on ;
and (ii) because it follows that
The
lower
bound shakedown theorem can be stated as follows: The solid is
guaranteed to shake down if any
time independent residual stress field can be found which satisfies:
The equilibrium equation ;
The
boundary condition on ;
When
the residual stress is combined with the elastic
solution, the combined stress does not exceed yield at any time during the cycle of load.
The
theorem is valuable because shakedown limits can be estimated using the elastic solution, which is much easier
to calculate than the elastic-plastic solution.
Proof of the lower bound theorem: The proof is
one of the most devious in all of solid mechanics.
1. Consider the strain energy associated with the
difference between the actual residual stress field ,
and the guess for the residual stress field ,
which can be calculated as
where is the elastic compliance tensor. For later reference note that W has to be positive, because strain
energy density is always positive or zero.
2.
The rate of
change of W can be calculated as
(to
see this, recall that )
3.
Note that .
Consequently, we see that
4.
Using the principle
of virtual work, the second integral can be expressed as an integral over the
boundary of the solid
To see this, note that on ,
while on
5.
The remaining
integral in (3) can be re-written as
6. Finally, recall that lies at or below yield, while is at yield and is the stress corresponding to
the plastic strain rate . The principle of maximum plastic resistance
therefore shows that . This inequality and can only be satisfied simultaneously if . We conclude that either the plastic strain
rate vanishes, or . In either case the solid must shake down to
an elastic state.
6.2.8
Examples of applications of the lower bound shakedown theorem
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Example 1: A simple 3 bar problem. It is
traditional to illustrate the concept of shakedown using this problem. Consider a structure made of three parallel
elastic-plastic bars, with Young’s modulus E
and cross sectional are A, as shown
in the figure. The two bars labeled 1
and 2 have yield stress Y; the
central bar (labeled 3) has yield stress 2Y. The structure is subjected to a cyclic load
with mean value and amplitude .
The
elastic limit for the structure is ;
the collapse load is .
To
obtain a lower bound to the shakedown limit, we must
1. Calculate the elastic stresses in the structure the axial stress in each bar is
2. Find a residual stress distribution in the structure,
which satisfies equilibrium and boundary conditions, and which can be added to
the elastic stresses to bring them below yield. A suitable residual stress distribution
consists of an axial stress in bars 1, 2 and 3. To prevent yield at the
maximum and minimum load in all three bars, we require
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The first two equations show that ,
irrespective of . To avoid yield in all bars at the maximum
load, we must choose ,
which gives .
Similarly, to avoid yield in all bars at the minimum load, we must choose ,
showing that .
The
various regimes of behavior are summarized in the figure.
Example 2: Shakedown limit for a
pressurized spherical shell. We consider an elastic-perfectly plastic
thick-walled shell, with inner radius a
and outer radius b. The inner wall of the shell is subjected to a
cyclic pressure, with minimum value zero, and maximum value
To
estimate the shakedown limit we must
1. Calculate the stresses induced by the pressure in an elastic shell. The solution can be found in Section
4.1.4.
2. Find a self-equilibrating residual stress field, which
satisfies traction free boundary conditions on R=a, R=b, and which can
be added to the elastic stresses to prevent yield in the sphere. The equilibrium equation for the residual
stress can be written
We can satisfy this equation by choosing any suitable
distribution for and calculating the corresponding . For example, we can choose ,
which corresponds to . To avoid yield at maximum load, we must
ensure that ,
while to avoid yield at zero load, throughout the shell. The critically stressed material element
lies at R=a at both the maximum and
zero loads, which shows that
Clearly, the best choice of is
The
estimate for the shakedown limit therefore follows as . This is equal to the exact solution derived
(with considerably more effort) in Section 4.1.4.
6.2.9 The
Upper Bound Shakedown Theorem
In
this section we derive a theorem that can be used to obtain an over-estimate to
the maximum cyclic load that can be applied to a structure without inducing
cyclic plastic deformation. Although the
estimate is inherently unsafe, the theorem is easier to use than the lower
bound theorem.
We
consider an elastic-perfectly plastic solid with Von-Mises yield surface,
associated flow law, and uniaxial tensile yield stress Y. Assume that
1. The displacement on part of the boundary of the solid
2. The remainder of the boundary is subjected to a prescribed cycle of traction
. The history of traction is periodic, with a
period T.
Define
the following quantities:
1. Let denote the actual history of displacement,
strain and stress induced in the solid by the applied loading. The strain is partitioned into elastic and
plastic parts as
2. Let denote the history of displacement, strain and
stress induced by the prescribed traction in a perfectly elastic solid with identical geometry.
To
apply the upper bound theorem, we guess a mechanism of cyclic plasticity that
might occur in the structure under the applied loading. We denote the cycle of strain by ,
and define the change in strain per cycle as
To be a kinematically
admissible cycle,
must be compatible, i.e. for
some a displacement field . Note that only the change in strain per cycle needs to be compatible, the plastic
strain rate need not be compatible at
every instant during the cycle.
The compatible displacement field must satisfy
on .
The upper bound shakedown theorem can then be stated as follows.
If there exists any
kinematically admissible cycle of strain that satisfies
the solid will not
shake down to an elastic state.
Proof: The upper bound theorem can be proved by
contradiction.
- Suppose that the solid
does shake down. Then, from the lower bound shakedown
theorem, we know that there exists a time independent residual stress
field ,
which satisfies equilibrium ;
the boundary conditions on
,
and is such that lies below yield throughout the cycle.
- The principle of
maximum plastic resistance then shows that . Integrating this expression over the
volume of the solid, and the cycle of loading gives
- Finally, reversing the order of integration in
the last integral and using the principle of virtual work, we see that
To
see this, note that on while on .
- Substituting this result back into (2) gives a
contradiction, so proving the upper bound theorem.
6.2.10
Examples of applications of the upper bound shakedown theorem
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Example 1: A simple 3 bar problem. We re-visit
the demonstration problem illustrated in Section 6.2.8. Consider a structure made of three parallel
elastic-plastic bars, with Young’s modulus E,
length L, and cross sectional are A,
as shown in the figure. The two bars
labeled 1 and 2 have yield stress Y;
the central bar (labeled 3) has yield stress 2Y. The structure is
subjected to a cyclic load with mean value and amplitude .
To obtain an upper bound to
the shakedown limit, we must devise a suitable mechanism of plastic flow in the
solid. We could consider three possible
mechanisms:
- An increment of
plastic strain in bars (1) and (2) at the instant of
maximum load, followed by in bars (1) and (2) at the instant of
minimum load. Since the strain at
the end of the cycle vanishes, it is automatically compatible.
- An equal increment of plastic strain in all three bars at each instant of
maximum load
- An equal increment of plastic strain at each instant of minimum load.
By finding the combination of
loads for which
we obtain conditions where
shakedown is guaranteed not to occur.
Note that the elastic stresses in all three bars are equal, and are given by .
Thus
- For mechanism (1):
- For mechanism (2):
- For mechanism (3):
These agree with the lower
bound calculated in Section 6.2.8, and are therefore the exact solution.
Example 2: Shakedown limit for a
pressurized spherical shell. We consider an elastic-perfectly plastic
thick-walled shell, with inner radius a
and outer radius b. The inner wall of the shell is subjected to a
cyclic
pressure,
with minimum value zero, and maximum value
To
estimate the shakedown limit we must
1. Calculate the stresses induced by the pressure in an elastic shell. The solution can be found in Section
4.1.4.
- Postulate a mechanism of steady-state plastic
deformation in the shell. For
example, consider a mechanism consisting of a uniform plastic strain
increment which occurs in a spherical shell with
radius a very small thickness dt at the instant of maximum pressure,
followed by a strain at the instant of minimum load.
The
upper bound theorem states that shakedown will not occur if
Substituting
the elastic stress field and the strain rate shows that
This gives for the shakedown limit. Again, this agrees with the lower bound, and
is therefore the exact solution.