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  

 

 

 

 

 

 

 

 

(c) A.F. Bower, 2008
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