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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 2. The plastic collapse load 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 4. Temperature distribution 5. Prescribed boundary tractions
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 3. We will neglect temperature changes.
With
these assumptions, we need to solve for the displacement field
where
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
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
Here,
and
furthermore must satisfy
For spherical symmetry, the governing equations reduce to
In elastic region(s)
In plastic region(s) Yield
criterion:Â Â Strain
partition: Elastic
strain:Â Â Â Flow
rule:Â Â Â Â Â Â Â Â Â
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
Solution:
(i) Preliminaries:
(ii) Solution in the plastic region
(iii) Solution in the elastic
region
These results are plotted in the figures below.
(a)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (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
In the plastic regime (i) We anticipate that (ii) Substituting this result into the equilibrium equation given in Sect 4.2.2 shows that
(iii) Integrating, and using the boundary condition
(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
(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
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
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
If
4.2.4 Elastic-perfectly plastic hollow sphere subjected to cyclic internal pressure
Assume that
Suppose that the inner surface of the sphere r=a is repeatedly subjected to
pressure
Solution:
(i) Preliminaries:
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
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
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
(i) Solution forÂ
(ii) Solution for
         Â
Solution for pressures exceeding
the shakedown limit
(i) Solution for
cyclic plastic regionÂ
(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 (ii) Substituting this result into the equilibrium equation shows that
(iii) Integrating, and using the boundary condition
Derivation of stress after unloading in the shakedown regime
(i) In this region, the stress at
maximum load are given by the expressions for
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 (iii) The change in pressure during
unloading can also be expressed as (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
We will assume that the cylinder is completely prevented
from stretching in the
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
Here,
For axial symmetry, the governing equations reduce to
In elastic region(s)
In plastic region(s) Yield
criterion:Â Â Strain
partition: Elastic
strain:Â Â Â Flow
rule:Â Â Â Â Â Â Â Â Â
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
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:
(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.Â
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
In the plastic regime (i) To simplify the calculation we
assume that (ii) We anticipate that (iii) Substituting this result into the equilibrium equation shows that
(iv) Integrating, and using the boundary condition
(v) The elastic strains follow as
(vi) With assumption (i), the flow
rule shows that plastic strains satisfy
(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
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
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
If
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(c) A.F. Bower, 2008 |