Chapter 6
Solutions for Plastic Solids

 

 

 

Plasticity problems are much more difficult to solve than linear elastic problems.  In general, a numerical method must be used, as discussed in Chapters 7 and 8. Nevertheless, exact solutions can be found for some solids with simple shapes, and special tricks are available to find approximate solutions for plastically deforming solids.  In this chapter we derive solutions for spherically and axially symmetric solids subjected to internal pressure, which illustrate the general behavior of elastic-plastic solids subjected to loading, including the elastic limit, plastic collapse, and shakedown in solids subjected to cyclic loading.   We then discuss bounding theorems, which provide a quick way to estimate collapse loads and shakedown limits for plastic solids and structures.

 

 

 

6.1 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.

 

 

 

6.1.1 Summary of governing equations

 

A representative problem is sketched below

 

 

 We are given the following information

 

1. The geometry of the solid

 

2. The constitutive law for the material (i.e. the elastic-plastic stress-strain equations)

 

3. The body force density $b_i$ (per unit mass) (if any)

 

4. The temperature distribution $\Delta T$ (if any)

 

5. Prescribed boundary tractions $t_i$ and/or boundary displacements $u_i$

 

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 $\nu$,  yield stress Y and mass density ${\rho }_0$

 

3. We will neglect temperature changes.

 

 

With these assumptions, we need to solve for the displacement field $u_i$, the strain field ${\epsilon }_{ij}$ and the stress field ${\sigma }_{ij}$ satisfying the following equations:

· Displacement-strain relation ${\epsilon }_{ij}={\displaystyle \frac{1}{2}\left({\displaystyle \frac{\partial u_i}{\partial x_j}+{\displaystyle \frac{\partial u_j}{\partial x_i}}}\right)}$

 

· Incremental stress-strain relation

$d{\epsilon }_{ij}=d{\epsilon }_{ij}^e+d{\epsilon }_{ij}^p$

$d{\epsilon }_{ij}^e={\displaystyle \frac{1+\nu }{E}\left(d{\sigma }_{ij}-{\displaystyle \frac{\nu }{1+\nu }d{\sigma }_{kk}{\delta }_{ij}}\right)}$

$d{\epsilon }_{ij}^p=\left\{\begin{array}{c}0\sqrt{{\displaystyle \frac{3}{2}{S}_{ij}{S}_{ij}}}<Y\\ d{\overline{\epsilon }}^p{\displaystyle \frac{3}{2}{\displaystyle \frac{S_{ij}}{Y}\sqrt{{\displaystyle \frac{3}{2}{S}_{ij}{S}_{ij}}}=Y}}\end{array}\right.$

where $S_{ij}={\sigma }_{ij}-{\sigma }_{kk}{\delta }_{ij}/3$

 

·         Equilibrium Equation ${\displaystyle \frac{\partial {\sigma }_{ij}}{\partial x_i}+{\rho }_0{b}_j=0}$ (static problems only $–$ you need the acceleration terms for dynamic problems)

 

·         Traction boundary conditions ${\sigma }_{ij}{n}_i=t_j$ on parts of the boundary where tractions are known.

 

·         Displacement boundary conditions $u_i=d_i$ on parts of the boundary where displacements are known.

 

 

 

6.1.2 Simplified equations for spherically symmetric problems

 

A representative spherically symmetric problem is illustrated in the figure. 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 $\theta$ and $\varphi$, 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 Figure 6.2.  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 $(r,\theta ,\varphi )$. All vectors and tensors are expressed as components in the Cartesian basis $\left\{e_r,e_{\theta },e_{\varphi }\right\}$ shown in the figure.  For a spherically symmetric problem we have that

 

· Position Vector       $x=r{e}_r$

 

· Displacement vector $u=u(r)e_r$

 

· Body force vector $b={\rho }_{0}b(r)e_r$

 

 

Here, $u\left(r\right)$ and $b\left(r\right)$ are scalar functions. The stress and strain tensors (written as components in $\{e_r,e_{\theta },e_{\varphi }\}$ ) have the form

$\sigma \equiv \left[\begin{array}{ccc}{\sigma }_{rr}& 0& 0\\ 0& {\sigma }_{\theta \theta }& 0\\ 0& 0& {\sigma }_{\varphi \varphi }\end{array}\right]\epsilon \equiv \left[\begin{array}{ccc}{\epsilon }_{rr}& 0& 0\\ 0& {\epsilon }_{\theta \theta }& 0\\ 0& 0& {\epsilon }_{\varphi \varphi }\end{array}\right]$

and furthermore must satisfy ${\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }$ ${\epsilon }_{\theta \theta }={\epsilon }_{\varphi \varphi }$. The tensor components have exactly the same physical interpretation as they did when we used a fixed $\{e_1,e_2,e_3\}$ basis, except that the subscripts (1,2,3) have been replaced by $(r,\theta ,z)$.

 

For spherical symmetry, the governing equations reduce to

 

· Strain Displacement Relations ${\epsilon }_{rr}={\displaystyle \frac{du}{dr}{\epsilon }_{\varphi \varphi }={\epsilon }_{\theta \theta }={\displaystyle \frac{u}{r}}}$

 

· Stress$—$Strain relations

 

In elastic region(s)

$\begin{array}{l}{\sigma }_{rr}={\displaystyle \frac{E}{\left(1+\nu \right)\left(1-2\nu \right)}\left\{(1-\nu ){\epsilon }_{rr}+\nu {\epsilon }_{\theta \theta }+\nu {\epsilon }_{\varphi \varphi }\right\}}\\ {\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }={\displaystyle \frac{E}{\left(1+\nu \right)\left(1-2\nu \right)}\left\{{\epsilon }_{\theta \theta }+\nu {\epsilon }_{rr}\right\}}\end{array}$

$\left|{\sigma }_{\theta \theta }-{\sigma }_{rr}\right|<Y$

In plastic region(s)

 

Yield criterion:   $\left|{\sigma }_{\theta \theta }-{\sigma }_{rr}\right|=Y$

Strain partition:

$d{\epsilon }_{rr}=d{\epsilon }_{rr}^p+d{\epsilon }_{rr}^{e}d{\epsilon }_{\varphi \varphi }=d{\epsilon }_{\varphi \varphi }^p+d{\epsilon }_{\varphi \varphi }^{e}d{\epsilon }_{\theta \theta }=d{\epsilon }_{\theta \theta }^p+d{\epsilon }_{\theta \theta }^e$

Elastic strain:

$\begin{array}{l}d{\epsilon }_{rr}^e=d{\sigma }_{rr}/E-\nu (d{\sigma }_{\theta \theta }+d{\sigma }_{\varphi \varphi })/E\\ d{\epsilon }_{\theta \theta }^e=d{\epsilon }_{\varphi \varphi }^e=d{\sigma }_{\theta \theta }(1-\nu )/E-\nu d{\sigma }_{rr}/E\end{array}$

Flow rule:         

$\begin{array}{l}d{\epsilon }_{rr}^p=d{\overline{\epsilon }}^p({\sigma }_{rr}-{\sigma }_{\theta \theta })/Y\\ d{\epsilon }_{\theta \theta }^p=d{\epsilon }_{\varphi \varphi }^p=d{\overline{\epsilon }}^p({\sigma }_{\theta \theta }-{\sigma }_{rr})/(2Y)\end{array}$

 

· Equilibrium Equations

${\displaystyle \frac{d{\sigma }_{rr}}{dr}+{\displaystyle \frac{1}{r}\left(2{\sigma }_{rr}-{\sigma }_{\theta \theta }-{\sigma }_{\varphi \varphi }\right)+{\rho }_0{b}_r=0}}$

 

· Boundary Conditions

 

Prescribed Displacements $u_r(a)=g_a{u}_r(b)=g_b$

Prescribed Tractions ${\sigma }_{rr}(a)=t_a{\sigma }_{rr}(b)=t_b$

 

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.

 

 

 

6.1.3 Elastic-perfectly plastic hollow sphere subjected to monotonically increasing internal pressure

 

A pressurized spherical thick-walled sphere is illustrated in the figure.  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 $p_a$

 

· 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 $p_a/Y=2(1-a^3/b^3)/3$

 

· For pressures in the range $2(1-a^3/b^3)/3<p_a/Y<2\log (b/a)$ the region between $r=a$ and $r=c$ deforms plastically; while the region between $c<r<b$ remains elastic, where c satisfies the equation $p_a/Y=2\log (c/a)+{\displaystyle \frac{2}{3}(1-c^3/b^3)}$

 

 

· At a pressure $p_a/Y=2\log (b/a)$ the entire cylinder is plastic.  At this point the sphere collapses $–$ the displacements become infinitely large.

 

 

(ii) Solution in the plastic region $a<r<c$

$u={\displaystyle \frac{\left(1-2\nu \right)}{E}r\left\{2Y\log (r/a)-p_a\right\}{e}_r+{\displaystyle \frac{Y\left(1-\nu \right)c^3}{E{r}^2}{e}_r}}$

${\sigma }_{rr}=2Y\log (r/a)-p_a{\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }=2Y\log (r/a)-p_a+Y$    

 

(iii) Solution in the elastic region $c<r<b$

$u={\displaystyle \frac{Y{c}^3}{3E{b}^3{r}^2}\left\{2\left(1-2\nu \right)r^3+\left(1+\nu \right)b^3\right\}{e}_r}$

${\sigma }_{rr}={\displaystyle \frac{2Y{c}^3}{3b^3}\left(1-{\displaystyle \frac{b^3}{r^3}}\right)}$       

   ${\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }={\displaystyle \frac{2Y{c}^3}{3b^3}\left(1+{\displaystyle \frac{b^3}{2r^3}}\right)}$

These results are plotted below. Displacements are shown for $\nu =0.3$.

 

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 $a<r<c$ will deform plastically, while a region $c<r<b$ remains elastic. We must find separate solutions in the plastic and elastic regions.

 

In the plastic region $a<r<c$

 

1. We anticipate that ${\sigma }_{rr}<0{\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }>0$. The yield criterion then gives ${\sigma }_{\theta \theta }-{\sigma }_{rr}=Y$

 

2. Substituting this result into the equilibrium equation given in Sect 4.2.2 shows that

${\displaystyle \frac{d{\sigma }_{rr}}{dr}+{\displaystyle \frac{1}{r}\left(2{\sigma }_{rr}-{\sigma }_{\theta \theta }-{\sigma }_{\varphi \varphi }\right)=0\Rightarrow {\displaystyle \frac{d{\sigma }_{rr}}{dr}-2{\displaystyle \frac{Y}{r}=0}}}}$

 

3. Integrating, and using the boundary condition ${\sigma }_{rr}=-p_{a}r=a$ together with the yield condition 1.  gives

${\sigma }_{rr}=2Y\log (r/a)-p_a{\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }=2Y\log (r/a)-p_a+Y$

 

4. 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

${\epsilon }_{rr}^e=\left({\sigma }_{rr}-2\nu {\sigma }_{\theta \theta }\right)/E{\epsilon }_{\varphi \varphi }^e={\epsilon }_{\theta \theta }^e=\left((1-\nu ){\sigma }_{\theta \theta }-\nu {\sigma }_{rr}\right)/E$

 

5. The plastic strains satisfy ${\epsilon }_{rr}^p+2{\epsilon }_{\theta \theta }^p=0$.  Consequently, using the strain partition formula, the results of (iv), and the strain-displacement relation shows that

$\begin{array}{l}{\epsilon }_{rr}+2{\epsilon }_{\theta \theta }={\epsilon }_{rr}^e+2{\epsilon }_{\theta \theta }^e={\displaystyle \frac{(1-2\nu )}{E}\left({\sigma }_{rr}+2{\sigma }_{\theta \theta }\right)}\\ \Rightarrow {\displaystyle \frac{du}{dr}+{\displaystyle \frac{2u}{r}={\displaystyle \frac{1}{r^2}{\displaystyle \frac{d}{dr}\left(r^{2}u\right)={\displaystyle \frac{(1-2\nu )}{E}\left(6Y\log (r/a)-3p_a+2Y\right)}}}}}\end{array}$

 

6. Integrating gives

$u={\displaystyle \frac{\left(1-2\nu \right)}{E}r\left\{2Y\log (r/a)-p_a\right\}+C/r^2}$

where C is a constant of integration

 

7. The constant of integration can be found by noting that the radial displacements in the elastic and plastic regions must be equal at r=c.  Using the expression for the elastic displacement field below and solving for C gives

$C={\displaystyle \frac{3}{2}{\displaystyle \frac{\left(1-\nu \right)c^3{b}^3}{E(b^3-c^3)}\left\{p_a-2Y\log (c/a)\right\}}}$

This result can be simplified by noting that $p_a-2Y\log (c/a)=2Y(1-c^3/b^3)/3$ from the expression for the location of the elastic-plastic boundary given below.

 

In the elastic region

 

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 region we see that the radial pressure at r=c is $p_c=-{\sigma }_{rr}=p_a-2Y\log (c/a)$. We can simplify the solution by noting $p_a-2Y\log (c/a)=2Y(1-c^3/b^3)/3$ from the expression for the location of the elastic-plastic boundary.   Substituting into the expressions for stress and displacement shows that

${\sigma }_{rr}={\displaystyle \frac{p_c{c}^3}{\left(b^3-c^3\right)}\left(1-{\displaystyle \frac{b^3}{r^3}}\right)}$ 

          ${\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }={\displaystyle \frac{p_c{c}^3}{\left(b^3-c^3\right)}\left(1+{\displaystyle \frac{b^3}{2r^3}}\right)}$

$u={\displaystyle \frac{p_c{c}^3}{2E\left(b^3-c^3\right)r^2}\left\{2\left(1-2\nu \right)r^3+\left(1+\nu \right)b^3\right\}{e}_r}$

 

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 ${\sigma }_{\theta \theta }-{\sigma }_{rr}=Y$, so substituting the expressions for stress in the elastic region and simplifying yields

$\begin{array}{l}{\sigma }_{\theta \theta }-{\sigma }_{rr}={\displaystyle \frac{3\left(p_a-2Y\log (c/a)\right)b^3}{2\left(b^3-c^3\right)}=Y}\\ \iff {\displaystyle \frac{p_a}{Y}=2\log (c/a)+{\displaystyle \frac{2}{3}(1-c^3/b^3)}}\end{array}$

If $p_a$, 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 $p_a$ 

 

 

 

6.1.4 Elastic-perfectly plastic hollow sphere subjected to cyclic internal pressure

 

The figure illustrates a thick-walled internally pressurized sphere. 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 $p_a$ and then unloaded to zero pressure.

 

Solution:

 

(i) Preliminaries:

 

· If the maximum pressure $p_a$ applied to the sphere does not exceed the elastic limit (i.e. $p_a/Y<2(1-a^3/b^3)/3$ ) 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 $2(1-a^3/b^3)/3<p_a/Y<2\log (b/a)$ the region between $r=a$ and $r=c$ deforms plastically during the first application of pressure; while the region between $c<r<b$ remains elastic, where c satisfies the equation $p_a/Y=2\log (c/a)+{\displaystyle \frac{2}{3}(1-c^3/b^3)}$.  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 $2(1-a^3/b^3)/3<p_a/Y<4(1-a^3/b^3)/3$, 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 ( $p_a/Y=4(1-a^3/b^3)/3$ ) is known as the ‘shakedown limit’

 

· If the maximum internal pressure reaches the shakedown limit $p_a/Y=4(1-a^3/b^3)/3$, the residual stress just reaches yield at r=a when the pressure is reduced to zero after the first loading cycle.

 

· For internal pressures $p_a/Y>4(1-a^3/b^3)/3$, a plastic zone forms between $a<r<d$ as the pressure is reduced to zero, where d satisfies the equation $p_a=4Y(1-d^3/b^3)/3+4Y\log (d/a)$.   During subsequent cycles of loading, the region $a<r<d$ is repeatedly plastically deformed, stretching in the hoop direction during increasing pressure, and compressing as the pressure is reduced to zero.  The region between $d<r<c$ 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 $p_a/Y=2\log (b/a)$, 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 above 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.

 

(ii) Solution for pressures below the elastic limit $p_a/Y<2(1-a^3/b^3)/3$

 

The displacement, strain and stress field at maximum load are given by the elastic solution in Section 4.1.4

 

(iii) Solution for pressures between the elastic and shakedown limits $p_a/Y<4(1-a^3/b^3)/3$

 

· At maximum pressure, the displacement and stress fields are given by the elastic-plastic solution in Section 6.1.3.

 

· At zero pressure, the solution is

 

Solution for  $a<r<c$

$\begin{array}{l}{\sigma }_{rr}=2Y\log \left({\displaystyle \frac{r}{a}}\right)-p_a-{\displaystyle \frac{p_a{a}^3}{\left(b^3-a^3\right)}\left(1-{\displaystyle \frac{b^3}{r^3}}\right)}\\ {\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }=2Y\log \left({\displaystyle \frac{r}{a}}\right)-p_a+Y-{\displaystyle \frac{p_a{a}^3}{\left(b^3-a^3\right)}\left(1+{\displaystyle \frac{b^3}{2r^3}}\right)}\end{array}$    

 

Solution for $c<r<b$

${\sigma }_{rr}={\displaystyle \frac{2Y{c}^3}{3b^3}\left(1-{\displaystyle \frac{b^3}{r^3}}\right)-{\displaystyle \frac{p_a{a}^3}{\left(b^3-a^3\right)}\left(1-{\displaystyle \frac{b^3}{r^3}}\right)}}$

           ${\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }={\displaystyle \frac{2Y{c}^3}{3b^3}\left(1+{\displaystyle \frac{b^3}{2r^3}}\right)-{\displaystyle \frac{p_a{a}^3}{\left(b^3-a^3\right)}\left(1+{\displaystyle \frac{b^3}{2r^3}}\right)}}$

 

(iv) Solution for pressures exceeding the shakedown limit $p_a/Y>4(1-a^3/b^3)/3$

 

· At maximum pressure, the displacement, strain and stress fields are given in Section 6.1.3.

 

· At zero pressure, the solution is

 

Solution for cyclic plastic region  $a<r<d$

${\sigma }_{rr}=-2Y\log (r/a){\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }=-2Y\log (r/a)-Y$

 

Solution for shakedown region $d<r<c$

$\begin{array}{l}{\sigma }_{rr}=2Y\log (r/a)-p_a-{\displaystyle \frac{4Y{d}^3}{3b^3}\left(1-{\displaystyle \frac{b^3}{r^3}}\right)}\\ {\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }=2Y\log (r/a)-p_a+Y-{\displaystyle \frac{4Y{d}^3}{3b^3}\left(1+{\displaystyle \frac{b^3}{2r^3}}\right)}\end{array}$

 

(iii) Solution for the elastic region $c<r<b$

${\sigma }_{rr}={\displaystyle \frac{2Y{c}^3}{3b^3}\left(1-{\displaystyle \frac{b^3}{r^3}}\right)-{\displaystyle \frac{4Y{d}^3}{3b^3}\left(1-{\displaystyle \frac{b^3}{r^3}}\right)}}$     

     ${\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }={\displaystyle \frac{2Y{c}^3}{3b^3}\left(1+{\displaystyle \frac{b^3}{2r^3}}\right)-{\displaystyle \frac{4Y{d}^3}{3b^3}\left(1+{\displaystyle \frac{b^3}{2r^3}}\right)}}$

 

 

Derivation of stress after unloading in the cyclic  plastic region $a<r<d$

 

1. We anticipate that ${\sigma }_{rr}>0{\sigma }_{\theta \theta }<0$. The yield criterion then gives ${\sigma }_{\theta \theta }-{\sigma }_{rr}=-Y$

 

2. Substituting this result into the equilibrium equation shows that

${\displaystyle \frac{d{\sigma }_{rr}}{dr}+2{\displaystyle \frac{Y}{r}=0}}$

 

3. Integrating, and using the boundary condition ${\sigma }_{rr}=0r=a$ together with the yield condition in step (i) gives

${\sigma }_{rr}=-2Y\log (r/a){\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }=-2Y\log (r/a)-Y$

 

 

Derivation of stress after unloading in the shakedown regime $d<r<c$

 

1. In this region, the stress at maximum load are given by the expressions for $r<c$ in 6.1.3, i.e.

${\sigma }_{rr}=2Y\log (r/a)-p_a{\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }=2Y\log (r/a)-p_a+Y$

The solid then unloads elastically while the pressure is removed. 

 

2. The change in stress during unloading can be calculated quickly by regarding the region $d<r<b$ 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 $p_a-2Y\log (d/a)$; after unloading the pressure follows from the solution for the cyclic plastic region as $2Y\log (d/a)$.  The change in pressure at r=d during unloading is thus $\Delta p=4Y\log (d/a)-p_a$.

 

3. The change in pressure during unloading can also be expressed as $\Delta p=-4Y(1-b^3/d^3)/3$ using the governing equation for d shown below.

 

4. 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

${\sigma }_{\theta \theta }-{\sigma }_{rr}=-Y\Rightarrow p_a=4Y(1-d^3/b^3)/3+4Y\log (d/a)$

 

Derivation of solution in the elastic region $c<r<b$

 

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 $c<r<b$

 

 

 

6.1.5 Simplified equations for plane strain axially symmetric elastic-perfectly plastic solids

 

An axially symmetric solid is illustrated in the figure. 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 $\theta$ and $z$, 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 $e_3$ axis.

 

We will assume that the cylinder is completely prevented from stretching in the $e_3$ direction, so that a state of plane strain exists in the solid. 

 

The solution is most conveniently expressed using a cylindrical-polar coordinate system, illustrated in Figure 6.7.  A point in the solid is identified by its cylindrical-polar co-ordinates $(r,\theta ,z)$. All vectors and tensors are expressed as components in the basis $\left\{e_r,e_{\theta },e_z\right\}$ shown in the figure.  For an axially symmetric problem

 

· Position Vector       $x=r{e}_r+z{e}_z$

 

· Displacement vector $u=u(r)e_r$

 

· Body force vector $b={\rho }_{0}b(r)e_r$

 

· Acceleration vector $a=-{\omega }^{2}r$

 

 

Here, $u\left(r\right)$ and $b\left(r\right)$ are scalar functions. The stress and strain tensors (written as components in $\{e_r,e_{\theta },e_z\}$ ) have the form

$\sigma \equiv \left[\begin{array}{ccc}{\sigma }_{rr}& 0& 0\\ 0& {\sigma }_{\theta \theta }& 0\\ 0& 0& {\sigma }_{zz}\end{array}\right]\epsilon \equiv \left[\begin{array}{ccc}{\epsilon }_{rr}& 0& 0\\ 0& {\epsilon }_{\theta \theta }& 0\\ 0& 0& {\epsilon }_{zz}\end{array}\right]$

 

For axial symmetry, the governing equations reduce to

 

· Strain Displacement Relations ${\epsilon }_{rr}={\displaystyle \frac{du}{dr}{\epsilon }_{\theta \theta }={\displaystyle \frac{u}{r}{\epsilon }_{zz}=0}}$

 

· Stress$—$Strain relations (plane strain and generalized plane strain)

 

In elastic region(s)

$\left[\begin{array}{c}{\sigma }_{rr}\\ {\sigma }_{\theta \theta }\\ {\sigma }_{zz}\end{array}\right]={\displaystyle \frac{E}{(1+\nu )(1-2\nu )}\left[\begin{array}{ccc}1-\nu & \nu & \nu \\ \nu & 1-\nu & \nu \\ \nu & \nu & 1-\nu \end{array}\right]\left[\begin{array}{c}{\epsilon }_{rr}\\ {\epsilon }_{\theta \theta }\\ {\epsilon }_{zz}\end{array}\right]}$

$\sqrt{\left\{{\left({\sigma }_{\theta \theta }-{\sigma }_{rr}\right)}^2+{\left({\sigma }_{\theta \theta }-{\sigma }_{zz}\right)}^2+{\left({\sigma }_{rr}-{\sigma }_{zz}\right)}^2\right\}/2}<Y$

 

In plastic region(s)

Yield criterion:   $\sqrt{\left\{{\left({\sigma }_{\theta \theta }-{\sigma }_{rr}\right)}^2+{\left({\sigma }_{\theta \theta }-{\sigma }_{zz}\right)}^2+{\left({\sigma }_{rr}-{\sigma }_{zz}\right)}^2\right\}/2}=Y$

Strain partition:

$d{\epsilon }_{rr}=d{\epsilon }_{rr}^p+d{\epsilon }_{rr}^{e}d{\epsilon }_{\varphi \varphi }=d{\epsilon }_{\varphi \varphi }^p+d{\epsilon }_{\varphi \varphi }^{e}d{\epsilon }_{\theta \theta }=d{\epsilon }_{\theta \theta }^p+d{\epsilon }_{\theta \theta }^e$

Elastic strain:

$\begin{array}{l}d{\epsilon }_{rr}^e=d{\sigma }_{rr}/E-\nu (d{\sigma }_{\theta \theta }+d{\sigma }_{zz})/E\\ d{\epsilon }_{\theta \theta }^e=d{\sigma }_{\theta \theta }/E-\nu (d{\sigma }_{rr}+d{\sigma }_{zz})/E\\ d{\epsilon }_{zz}^e=d{\sigma }_{zz}/E-\nu (d{\sigma }_{rr}+d{\sigma }_{\theta \theta })/E\end{array}$

Flow rule:  

$\begin{array}{l}d{\epsilon }_{rr}^p=d{\overline{\epsilon }}^p{\displaystyle \frac{{\sigma }_{rr}-\left({\sigma }_{\theta \theta }+{\sigma }_{zz}\right)/2}{Y}}\\ d{\epsilon }_{\theta \theta }^p=d{\overline{\epsilon }}^p{\displaystyle \frac{{\sigma }_{\theta \theta }-\left({\sigma }_{rr}+{\sigma }_{zz}\right)/2}{Y}}\\ d{\epsilon }_{zz}^p=d{\overline{\epsilon }}^p{\displaystyle \frac{{\sigma }_{zz}-\left({\sigma }_{rr}+{\sigma }_{\theta \theta }\right)/2}{Y}}\end{array}$

 

· Equation of motion

${\displaystyle \frac{d{\sigma }_{rr}}{dr}+{\displaystyle \frac{1}{r}\left({\sigma }_{rr}-{\sigma }_{\theta \theta }\right)+{\rho }_0{b}_r=-{\rho }_0{\omega }^{2}r}}$

 

· Boundary Conditions

 

Prescribed Displacements $u_r(a)=g_a{u}_r(b)=g_b$

Prescribed Tractions ${\sigma }_{rr}(a)=t_a{\sigma }_{rr}(b)=t_b$

 

The equilibrium and strain-displacement equations can be derived following the procedure outlined in Section 4.1.7.  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.

 

 

 

6.1.6 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 $p_a$

 

· 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 $\sqrt{3}{p}_a/Y\approx (1-a^2/b^2)$

 

· For pressures in the range $(1-a^2/b^2)<\sqrt{3}{p}_a/Y<2\log (b/a)$ the region between $r=a$ and $r=c$ deforms plastically; while the region between $c<r<b$ remains elastic, where c satisfies the equation $\sqrt{3}{p}_a/Y=2\log (c/a)+1-c^2/b^2$

 

· At a pressure $p_a/Y=\left(2/\sqrt{3}\right)\log (b/a)$ the entire cylinder is plastic.  At this point the sphere collapses $–$ the displacements become infinitely large.

 

 

 

(ii) Solution in the plastic region $a<r<c$

$\begin{array}{l}u={\displaystyle \frac{\left(1-2\nu \right)\left(1+\nu \right)}{E}2r\left\{\left(2Y/\sqrt{3}\right)\log (r/a)-p_a\right\}}\\ +{\displaystyle \frac{\left(1+\nu \right)Y{c}^2(2(1-\nu )b^2+(1-2\nu )(b^2-c^2))}{\sqrt{3}E{b}^{2}r}}\end{array}$

$\begin{array}{l}{\sigma }_{rr}=(2Y/\sqrt{3})\log (r/a)-p_a\\ {\sigma }_{\theta \theta }=(2Y/\sqrt{3})\log (r/a)-p_a+2Y/\sqrt{3}\end{array}$    

 

(iii) Solution in the elastic region $c<r<b$

$u_r={\displaystyle \frac{\left(1+\nu \right)c^{2}Y}{E\sqrt{3}}\left\{{\displaystyle \frac{1}{r}+\left(1-2\nu \right){\displaystyle \frac{r}{b^2}}}\right\}}$

${\sigma }_{rr}={\displaystyle \frac{Y{c}^2}{\sqrt{3}{b}^2}\left\{1-{\displaystyle \frac{b^2}{r^2}}\right\}{\sigma }_{\theta \theta }={\displaystyle \frac{Y{c}^2}{\sqrt{3}{b}^2}\left\{1+{\displaystyle \frac{b^2}{r^2}}\right\}}}$

 

The stress and displacement fields are plotted 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 $\nu =0.3$

 

 

Derivation: By substituting the stresses for the elastic solution given in 4.1.9 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 $a<r<c$ will deform plastically, while a region $c<r<b$ remains elastic. We must find separate solutions in the plastic and elastic regions.

 

In the plastic region $a<r<c$

 

1. To simplify the calculation we assume that $d{\epsilon }_{zz}^p=d{\epsilon }_{zz}^e=0$.  This turns out to be exact for $\nu =1/2$, but is approximate for other values of Poisson ratio.  The plastic flow rule shows that

$d{\epsilon }_{zz}^p=d{\overline{\epsilon }}^p{\displaystyle \frac{{\sigma }_{zz}-\left({\sigma }_{rr}+{\sigma }_{\theta \theta }\right)/2}{Y}}$

in which case $d{\epsilon }_{zz}^p=0$ requires that ${\sigma }_{zz}=\left({\sigma }_{rr}+{\sigma }_{\theta \theta }\right)/2$

2. We anticipate that ${\sigma }_{rr}<0{\sigma }_{\theta \theta }>0$. Substituting the result of (i) into the yield criterion then gives ${\sigma }_{\theta \theta }-{\sigma }_{rr}=2Y\sqrt{3}$.

 

3. Substituting this result into the equilibrium equation shows that

${\displaystyle \frac{d{\sigma }_{rr}}{dr}-{\displaystyle \frac{2Y}{\sqrt{3}r}=0}}$

 

4. Integrating, and using the boundary condition ${\sigma }_{rr}=-p_{a}r=a$ together with the yield condition (i) gives

$\begin{array}{l}{\sigma }_{rr}=\left(2Y/\sqrt{3}\right)\log (r/a)-p_a\\ {\sigma }_{\theta \theta }=\left(2Y/\sqrt{3}\right)\log (r/a)-p_a+2Y/\sqrt{3}\end{array}$

5. The elastic strains follow as

$\begin{array}{l}{\epsilon }_{rr}^e=\left({\sigma }_{rr}-\nu {\sigma }_{\theta \theta }-\nu {\sigma }_{zz}\right)/E\\ {\epsilon }_{\theta \theta }^e=\left({\sigma }_{\theta \theta }-\nu {\sigma }_{rr}-\nu {\sigma }_{zz}\right)/E\end{array}$

 

6. With assumption1., the flow rule shows that plastic strains satisfy ${\epsilon }_{rr}^p+{\epsilon }_{\theta \theta }^p=0$.  Consequently, using the strain partition formula and the strain-displacement relation shows that

$\begin{array}{l}{\epsilon }_{rr}+{\epsilon }_{\theta \theta }={\displaystyle \frac{du}{dr}+{\displaystyle \frac{u}{r}={\displaystyle \frac{1}{r}{\displaystyle \frac{d}{dr}\left(ru\right)={\displaystyle \frac{\left(1-2\nu \right)\left(1+\nu \right)}{E}\left\{{\sigma }_{rr}+{\sigma }_{\theta \theta }\right\}}}}}}\\ ={\displaystyle \frac{\left(1-2\nu \right)\left(1+\nu \right)}{E}2\left(2Y[\log (r/a)+1]/\sqrt{3}-p_a\right)}\end{array}$

 

7. Integrating gives

$u={\displaystyle \frac{\left(1-2\nu \right)\left(1+\nu \right)}{E}2r\left\{\left(2Y/\sqrt{3}\right)\log (r/a)-p_a\right\}+C/r}$

where C is a constant of integration

 

8. The constant of integration can be found by noting that the radial displacements in the elastic and plastic regions must be equal at r=c.  Using the expression for the elastic displacement field below and solving for C gives

$C={\displaystyle \frac{\left(1+\nu \right)c^2(2(1-\nu )b^2+(1-2\nu )(b^2-c^2))}{E(b^2-c^2)}\left\{p_a-\left(2Y/\sqrt{3}\right)\log (c/a)\right\}}$

This result can be simplified by noting that $p_a-\left(2Y/\sqrt{3}\right)\log (c/a)=Y(1-c^2/b^2)/\sqrt{3}$ from the expression for the location of the elastic-plastic boundary given below.

 

 

In the elastic region

 

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 region we see that the radial pressure at r=c is $p_c=-{\sigma }_{rr}=p_a-\left(2Y/\sqrt{3}\right)\log (c/a)$. We can simplify the solution by noting $p_a-\left(2Y/\sqrt{3}\right)\log (c/a)=Y(1-c^2/b^2)/\sqrt{3}$ 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

$\begin{array}{l}{\sigma }_{rr}={\displaystyle \frac{p_c{c}^2}{b^2-c^2}\left\{1-{\displaystyle \frac{b^2}{r^2}}\right\}}\\ {\sigma }_{\theta \theta }={\displaystyle \frac{p_c{c}^2}{b^2-c^2}\left\{1+{\displaystyle \frac{b^2}{r^2}}\right\}}\\ {\sigma }_{zz}=2\nu {\displaystyle \frac{p{c}^2}{b^2-c^2}}\end{array}$

$u_r={\displaystyle \frac{\left(1+\nu \right)c^2{b}^2{p}_c}{E\left(b^2-c^2\right)}\left\{{\displaystyle \frac{1}{r}+\left(1-2\nu \right){\displaystyle \frac{r}{b^2}}}\right\}}$

 

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 ${\sigma }_{\theta \theta }-{\sigma }_{rr}=2Y/\sqrt{3}$, so substituting the expressions for stress in the elastic region and simplifying yields

$\begin{array}{l}{\sigma }_{\theta \theta }-{\sigma }_{rr}={\displaystyle \frac{2\left(p_a-\left(2Y/\sqrt{3}\right)\log (c/a)\right)b^2}{\left(b^2-c^2\right)}=2Y/\sqrt{3}}\\ \iff \sqrt{3}{\displaystyle \frac{p_a}{Y}=2\log (c/a)+(1-c^2/b^2)}\end{array}$

If $p_a$, 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 $p_a$