![]() |
|||||||||||||||||||
|
|||||||||||||||||||
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
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
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
Consider
a rigid perfectly plastic solid, which has mass density
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,
1. We assume that the velocity field 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
5. The yield criterion and plastic flow rule require
that
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 Like
u, the virtual velocity v may have a finite set of
discontinuities across surfaces
to denote the magnitude of the velocity discontinuity. We also define the virtual strain rate
 (note that
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
1. 2. Thus,
Derivation: Begin by summarizing the equations governing the
actual collapse solution. Let
On
velocity discontinuities, these conditions require that
We
start by showing that 1. By definition
2. Note that, using (i) the flow rule, (ii) the
condition that
3. Note that
4. Note 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
6. Finally, recall that
Since
Next, we show that 1. Let
2. Let
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 4. Note that
5. Next, note that
6. The equilibrium equation 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
8. Recall that
9. Finally, note that on
since
the shear stress acting on any plane in the solid cannot exceed
 proving that
6.2.3 The Upper Bound Plastic Collapse Theorem
Consider a rigid plastic solid, subjected to some
distribution of tractions
To estimate
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.Â
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
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
The best upper bound
occurs for
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. Â 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.Â
In each semicircular region we assume a constant circumferential
velocity
Thus the plastic dissipation is
(note that there’s a
velocity discontinuity at r=a). The work done by applied loading is
just
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
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
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
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
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
To estimate We will denote the guess for the stress distribution by 1. Satisfy the boundary conditions 2. Satisfy the equations of
equilibrium 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 2. Let 3. The Principle of Maximum Plastic
Resistance (see Section 3.7.10) shows that 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
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 3. The stress state in the triangular region B has
principal directions of stress parallel to The
stresses in each region must be chosen to satisfy equilibrium, and to ensure
that the stress is below yield everywhere. Â
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 1. Note that 2. Equilibrium across the boundary between A and B requires
3. We must now choose 4. Finally, substituting for  Â
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.Â
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 2. The remainder of the boundaryÂ
Define the following quantities: 1. Let 2. Let 3. We introduce (time dependent) residual stress
Note
that, (i) because
The
lower
bound shakedown theorem can be stated as follows: The solid is
guaranteed to shake down if any
time independent residual stress field
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.Â
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
where 2. The rate of change of W can be calculated as
(to
see this, recall that 3. Note 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 5. The remaining integral in (3) can be re-written as
6. Finally, recall that
6.2.8 Examples of applications of the lower bound shakedown theorem
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
The
elastic limit for the structure is
To obtain a lower bound to the shakedown limit, we must 1. Calculate the elastic stresses in the structure 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
The first two equations show 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.Â
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
Clearly, the best choice of
The
estimate for the shakedown limit therefore follows as
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 2. The remainder of the boundaryÂ
Define the following quantities: 1. Let 2. Let 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
To be a kinematically admissible cycle,
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.Â
To
see this, note thatÂ
6.2.10 Examples of applications of the upper bound shakedown theorem
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
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:
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
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.Â
The upper bound theorem states that shakedown will not occur if
Substituting the elastic stress field and the strain rate shows that
This gives
|
|||||||||||||||||||
(c) A.F. Bower, 2008 |