Chapter 6

 

Analytical techniques and 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, there are several powerful mathematical techniques that can be used to find both exact and approximate solutions.  In this chapter we outline two particularly effective methods: slip-line field theory, which gives exact solutions for plane strain boundary value problems for rigid plastic solids; and bounding theorems, which provide a quick way to estimate collapse loads for plastic solids and structures.

 

 

6.1 Slip-line field theory

 

The largest class of solutions to boundary value problems in plasticity exploits a technique known as slip line field theory.  The theory simplifies the governing equations for plastic solids by making several restrictive assumptions:

1.      Plane strain deformation $–$ i.e. displacement components in the basis shown satisfy $u_3=0$  and $u_1,u_2$ are functions of $x_1$ and $x_2$ only

2.      Quasi-static loading

3.      No temperature changes

4.      No body forces

5.      The solid is idealized as a rigid-perfectly plastic Mises solid. The uniaxial stress-strain curve for this material is illustrated in the figure.  The material properties are characterized by the yield stress in uniaxial tension Y.  Alternatively, the material is sometimes characterized by its yield stress in shear $k=Y/\sqrt{3}$.

 

Otherwise, the technique can be used to solve any arbitrary 2D boundary value problem for a rigid plastic solid.  It is quite difficult to apply in practice, because it is not easy to find the slip-line field that solves a particular problem.  Nevertheless, a wide range of important solutions have been found.  The main intent of this section is to illustrate how to interpret these solutions, and to outline the basis for slip-line field theory.

 

 

 

6.1.1 Interpreting a slip-line field

 

An example of a slip-line field solution is shown in the picture on the right. (This is Hill’s solution to a rigid punch indenting a rigid-plastic half-space).  

 

The slip lines consist of a curvilinear mesh of two families of lines, which always cross each other at right angles.  By convention, one set of lines are named $\alpha$ slip-lines (shown in red); the other are called $\beta$ lines (blue).  The velocity distribution and stress state in the solid can always be determined from the geometry of these lines.

 

 

Stress state at a point in the slip-line field

 

By definition, the slip-lines are always parallel to axes of principal shear stress in the solid.  This means that the stress components in a basis oriented with the $\alpha$, $\beta$ directions have the form

${\sigma }_{\alpha \alpha }=\overline{\sigma }{\sigma }_{\beta \beta }=\overline{\sigma }{\sigma }_{\alpha \beta }=k=Y/\sqrt{3}$

where $\overline{\sigma }=\left({\sigma }_{\alpha \alpha }+{\sigma }_{\beta \beta }\right)/2$ is the hydrostatic stress (determined using the equations given below), k is the yield stress of the material in shear, and Y is its yield stress in uniaxial tension.  This stress state is sketched in the figure.  Since the shear stress is equal to the shear yield stress, the material evidently deforms by shearing parallel to the slip-lines: this is the reason for their name.

 

If $\varphi$ denotes the angle between the $\alpha$ slip-line and the $e_1$ direction, the stress components in the $\left\{e_1,e_2,e_3\right\}$ basis can be calculated as

$\begin{array}{l}{\sigma }_{11}=\overline{\sigma }-k\sin 2\varphi \\ {\sigma }_{22}=\overline{\sigma }+k\sin 2\varphi \\ {\sigma }_{12}=k\cos 2\varphi \end{array}$

The Mohr’s circle construction (shown in the picture to the right) is a convenient way to remember these results.

 

 

Relations governing hydrostatic stress along slip-lines (Hencky equations)

 

The hydrostatic stress can be shown to satisfy the following relations along slip-lines

$\begin{array}{l}\overline{\sigma }-2k\varphi =\text{constant}\alpha \text{ slip line}\\ \overline{\sigma }+2k\varphi =\text{constant}\beta \text{ slip line}\end{array}$

If the hydrostatic stress can be determined at any one point on a slip-line (for example at a boundary), it can be deduced everywhere else. Note that if there is a region in the field where both slip lines are straight, the stress is constant.

 

 

The velocity field (Geiringer equations)

 

The velocity field can be expressed as components in a fixed $\{e_1,e_2\}$ basis, or as components parallel and perpendicular to the slip lines. 

$\begin{array}{c}\frac{d{v}_1}{ds}+\tan \varphi \frac{d{v}_2}{ds}=0\\ \frac{d{v}_{\alpha }}{ds}=v_{\beta }\frac{d\varphi }{ds}\end{array}\}\alpha \text{slip line}\begin{array}{c}\frac{d{v}_1}{ds}-\cot \varphi \frac{d{v}_2}{ds}=0\\ \frac{d{v}_{\beta }}{ds}=-v_{\alpha }\frac{d\varphi }{ds}\end{array}\}\beta \text{slip line}$

 

 

 

Application to the Hill slip-line field

 

The stress state throughout a slip-line field can be deduced by working step-by-step along the slip lines.  We illustrate the procedure using Hill’s indentation solution.

 

Consider first the state of stress at point a.  Clearly, ${\varphi }_a=\pi /4$ at this point. The stress state can be transformed from a basis aligned with the slip-lines to the fixed $\{e_1,e_2\}$ basis using the Mohr’s circle construction shown in the figure.  Recall (or use the Mohr’s circle to see) that

${\sigma }_{11}=\overline{\sigma }-k\sin 2\varphi {\sigma }_{22}=\overline{\sigma }+k\sin 2\varphi {\sigma }_{12}=k\cos 2\varphi$

where $\overline{\sigma }$ is the hydrostatic component of stress.  The boundary conditions at a require that ${\sigma }_{12}={\sigma }_{22}=0$.  The first condition is clearly satisfied, since the slip-lines intersect the boundary at $\varphi =\pi /4$.  We can satisfy the second condition by setting $\overline{\sigma }=-k$.  Finally this gives the stress parallel to the surface as ${\sigma }_{11}=-2k$. 

 

The stress must be constant in the triangular region ABC, as the slip lines in this region are straight.

 

Next, consider the stress state at b.  Here, we see that ${\varphi }_b=-\pi /4$.  We can use the Hencky equation to determine $\overline{\sigma }$ at b.  Recall that

$\begin{array}{l}\overline{\sigma }-2k\varphi =\text{constant}\alpha \text{ slip line}\\ \overline{\sigma }+2k\varphi =\text{constant}\beta \text{ slip line}\end{array}$

so following one of the $\alpha$ slip lines we get

$\begin{array}{l}{\overline{\sigma }}_b-2k{\varphi }_b={\overline{\sigma }}_a-2k{\varphi }_a\\ \Rightarrow {\overline{\sigma }}_b=-k-\pi k\end{array}$

Using the basis-change equation we then get

${\sigma }_{11}=-\pi k{\sigma }_{22}=-(\pi +2)k{\sigma }_{12}=0$

The pressure under the punch turns out to be uniform (the stress is constant in the triangular region of the slip-line field below the punch) and so the total force (per unit out of plane length) on the punch can be computed as

$P=w(2+\pi )k$

where w is the width of the punch.

 

 

 

How to distinguish the $\alpha$ and $\beta$ families of slip lines

 

Usually, slip-line fields are presented without specifying which set of slip-lines should be taken as the $\alpha$ and which should be the $\beta$ set $–$ it is up to you to work out which is which.  In fact, the slip-lines are interchangeable $–$ switching $\alpha$ and $\beta$ will simply change the sign of all the stresses.

 

You can see this clearly using the Hill solution.  The figure on the right shows the solution with $\alpha$ and $\beta$ lines switched over.  At point a, $\varphi =3\pi /4$, and therefore to satisfy ${\sigma }_{22}=0$ we must now choose $\overline{\sigma }=k$.  To find the stress under the contact, we can trace a $\beta$ slip line to point b. Here, we see that $\varphi =\pi /4$, so the Hencky equation

$\begin{array}{l}{\overline{\sigma }}_b+2k{\varphi }_b={\overline{\sigma }}_a+2k{\varphi }_a\\ \Rightarrow {\overline{\sigma }}_b=k+\pi k\end{array}$

Using the basis-change equation we then get

${\sigma }_{11}=\pi k{\sigma }_{22}=(\pi +2)k{\sigma }_{12}=0$

at point b.  The normal stress acts upwards on the surface $–$ so that this represents the stress induced by a rigid punch that is bonded to the surface, and pulled upwards.

 

 

 

 

6.1.2 Derivation of the slip-line field method.

 

Consider a rigid-perfectly plastic solid, with a von-Mises yield surface characterized by yield stress in uniaxial tension $Y$ or its yield stress in shear $k=Y/\sqrt{3}$.   Let $u_i,{\epsilon }_{ij},{\sigma }_{ij}$ denote the components of displacement, strain and stress in the solid. The solid is assumed to be a long cylinder with its axis parallel to the $e_3$ direction, which is constrained to deform in plane strain, with $u_3=0$ and $u_1,u_2$ independent of $x_3$.  It is loaded by subjecting part of its boundary ${\partial }_{1}R$ to a prescribed velocity, and the remainder ${\partial }_{2}R$ to a prescribed traction, so that

$\begin{array}{l}{\dot{u}}_{\alpha }=v_{\alpha }^{\ast }(x_1,x_2)⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄\text{on }{\partial }_{1}R\\ {\sigma }_{\alpha \beta }{n}_{\alpha }=t_{\beta }^{\ast }(x_1,x_2)⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄⥄\text{on }{\partial }_{2}R\end{array}$

where the Greek subscripts $\alpha ,\beta$ can have values of 1 or 2. In practice we will compute the velocity field $v_i={\dot{u}}_i$ rather than the displacement field.

 

 

Summary of governing equations

 

1.      Strain-rate $–$ velocity relation ${\dot{\epsilon }}_{\alpha \beta }=\frac{1}{2}\left(\partial v_{\alpha }/\partial x_{\beta }+\partial v_{\beta }/\partial x_{\alpha }\right)$

2.      The plastic flow rule ${\dot{\epsilon }}_{ij}=3{\dot{\overline{\epsilon }}}^p{S}_{ij}/2Y$

Plane strain deformation then requires

$\begin{array}{l}{\dot{\epsilon }}_{33}=3{\dot{\overline{\epsilon }}}^p[{\sigma }_{33}-({\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33})/3]/2Y=0\\ \Rightarrow {\sigma }_{33}=({\sigma }_{11}+{\sigma }_{22})/2\end{array}$

whereupon the flow rule shows that the remaining components of plastic strain rate satisfy

$\begin{array}{l}{\dot{\epsilon }}_{11}=3{\dot{\overline{\epsilon }}}^p\left({\sigma }_{11}-({\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33})/3\right)/2Y=3{\dot{\overline{\epsilon }}}^p({\sigma }_{11}-{\sigma }_{22})/4Y\\ {\dot{\epsilon }}_{22}=3{\dot{\overline{\epsilon }}}^p\left({\sigma }_{22}-({\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33})/3\right)/2Y=3{\dot{\overline{\epsilon }}}^p({\sigma }_{22}-{\sigma }_{11})/4Y\\ {\dot{\epsilon }}_{12}=3{\dot{\overline{\epsilon }}}^p{\sigma }_{12}/2Y\end{array}$

We observe that these conditions imply that

${\dot{\epsilon }}_{11}+{\dot{\epsilon }}_{22}=0\Rightarrow \partial v_1/\partial x_1+\partial v_2/\partial x_2=0$

$\begin{array}{l}\frac{{\dot{\epsilon }}_{11}-{\dot{\epsilon }}_{22}}{{\sigma }_{11}-{\sigma }_{22}}=\frac{{\dot{\epsilon }}_{12}}{{\sigma }_{12}}\Rightarrow \left({\dot{\epsilon }}_{11}-{\dot{\epsilon }}_{22}\right){\sigma }_{12}={\dot{\epsilon }}_{12}\left({\sigma }_{11}-{\sigma }_{22}\right)\\ \Rightarrow \left(\partial v_1/\partial x_1-\partial v_2/\partial x_2\right){\sigma }_{12}=\frac{1}{2}\left(\partial v_1/\partial x_2+\partial v_2/\partial x_1\right)\left({\sigma }_{11}-{\sigma }_{22}\right)\end{array}$

3.      Yield criterion

$\sqrt{\frac{3}{2}{S}_{ij}{S}_{ij}}-Y=0\Rightarrow \frac{1}{4}{\left({\sigma }_{11}-{\sigma }_{22}\right)}^2+{\sigma }_{12}^2=k^2$

where $k=Y/\sqrt{3}$ is the shear yield stress of the material, and we have used the condition that ${\sigma }_{33}=({\sigma }_{11}+{\sigma }_{22})/2$

4.      Equilibrium conditions

$\begin{array}{l}\partial {\sigma }_{ij}/\partial x_i=0\Rightarrow \partial {\sigma }_{11}/\partial x_1+\partial {\sigma }_{12}/\partial x_2=0\\ \partial {\sigma }_{22}/\partial x_2+\partial {\sigma }_{21}/\partial x_1=0\end{array}$

 

Solution of governing equations by method of characteristics

 

From the preceding section, we observe that we must calculate a velocity field $v_{\alpha }(x_{\alpha })$ and stress field ${\sigma }_{\alpha \beta }(x_{\alpha })$ satisfying governing equations

$\frac{1}{4}{\left({\sigma }_{11}-{\sigma }_{22}\right)}^2+{\sigma }_{12}^2=k^2$

$\partial v_1/\partial x_1+\partial v_2/\partial x_2=0$

$\left(\partial v_1/\partial x_1-\partial v_2/\partial x_2\right){\sigma }_{12}=\frac{1}{2}\left(\partial v_1/\partial x_2+\partial v_2/\partial x_1\right)\left({\sigma }_{11}-{\sigma }_{22}\right)$

$\begin{array}{l}\partial {\sigma }_{11}/\partial x_1+\partial {\sigma }_{12}/\partial x_2=0\\ \partial {\sigma }_{22}/\partial x_2+\partial {\sigma }_{21}/\partial x_1=0\end{array}$

together with appropriate boundary conditions.

 

We focus first on a general solution to the governing equations.  It is convenient to start by eliminating some of the stress components using the yield condition.  Since the material is at yield, we note that at each point in the solid we could find a basis in which the stress state consists of a shear stress of magnitude k (the shear yield stress), together with an unknown component of hydrostatic stress $\overline{\sigma }$.  The stress state is sketched on the right.

 

Instead of solving for the stress components ${\sigma }_{\alpha \beta }$, we will calculate the hydrostatic stress $\overline{\sigma }$ and the angle $\varphi$ between the $e_1$ direction and the $m_1$ direction.   Recall that we can relate ${\sigma }_{\alpha \beta }$ to $\overline{\sigma }$, $\varphi$ and k using Mohr’s circle of stress: from the picture, we see that

${\sigma }_{11}=\overline{\sigma }-k\sin 2\varphi {\sigma }_{22}=\overline{\sigma }+k\sin 2\varphi {\sigma }_{12}=k\cos 2\varphi$

 

We now re-write the governing equations in terms of $\overline{\sigma }$, $\varphi$ and k.  The yield criterion is satisfied automatically.  The remaining four equations are most conveniently expressed in matrix form

$A_{ij}\frac{\partial q_j}{\partial x_1}+B_{ij}\frac{\partial q_j}{\partial x_2}=0$

where A and B are 4-dimensional symmetric matrices and q is a 1x4 vector, defined as

$q=\left[\begin{array}{c}\varphi \\ v_1\\ v_2\\ \overline{\sigma }\end{array}\right]A=\left[\begin{array}{cccc}0& -2k\cos 2\varphi & -2k\sin 2\varphi & 0\\ -2k\cos 2\varphi & 0& 0& 1\\ -2k\sin 2\varphi & 0& 0& 0\\ 0& 1& 0& 0\end{array}\right]B=\left[\begin{array}{cccc}0& -2k\sin 2\varphi & 2k\cos 2\varphi & 0\\ -2k\sin 2\varphi & 0& 0& 0\\ 2k\cos 2\varphi & 0& 0& 1\\ 0& 0& 1& 0\end{array}\right]$

 

This is a quasi-linear hyperbolic system of PDEs, which may be solved by the method of characteristics. 

 

The first step is to find eigenvalues $\mu$ and eigenvectors $r_i$ that satisfy

$r_i{A}_{ij}=\mu r_i{B}_{ij}$

A straightforward exercise (set $\det (A-\mu B)=0$ to find the eigenvalues, and substitute back to get eigenvectors, or if you’re lazy use a symbolic manipulation program…) shows that there are two repeated eigenvalues, with corresponding eigenvectors

$\mu =\cot \varphi \{\begin{array}{c}r=[1,0,0,-2k]\\ r=[0,1,\tan \varphi ,0]\end{array}\mu =-\tan \varphi \{\begin{array}{c}r=[1,0,0,2k]\\ r=[0,1,-\cot \varphi ,0]\end{array}$

We can now eliminate A from the governing matrix equation

$r_i{B}_{ij}\left(\mu \frac{\partial q_j}{\partial x_1}+\frac{\partial q_j}{\partial x_2}\right)=0\Rightarrow \sqrt{1+{\mu }^2}{r}_i{B}_{ij}\left(\frac{\mu }{\sqrt{1+{\mu }^2}}\frac{\partial q_j}{\partial x_1}+\frac{1}{\sqrt{1+{\mu }^2}}\frac{\partial q_j}{\partial x_2}\right)=0$

Finally, if we set

$\frac{\partial x_1}{\partial s}=\frac{\mu }{\sqrt{1+{\mu }^2}}\frac{\partial x_2}{\partial s}=\frac{1}{\sqrt{1+{\mu }^2}}$

and note that

$\left(\frac{\partial q_j}{\partial x_1}\frac{\partial x_1}{\partial s}+\frac{\partial q_j}{\partial x_2}\frac{\partial x_2}{\partial s}\right)=\frac{d{q}_j}{ds}$

we find that

$r_i{B}_{ij}\frac{d{q}_j}{ds}=0$

along characteristic lines in the solid that satisfy

$\frac{\partial x_1}{\partial s}=\frac{\mu }{\sqrt{1+{\mu }^2}}\frac{\partial x_2}{\partial s}=\frac{1}{\sqrt{1+{\mu }^2}}$

The special characteristic lines in the solid can be identified more easily if we note that

$\frac{d{x}_2}{d{x}_1}=\frac{1}{\mu }$

which shows that the slope of the characteristic lines satisfies

$\frac{d{x}_2}{d{x}_1}=\tan \varphi \frac{d{x}_2}{d{x}_1}=-\cot \varphi$

for the two possible values of the eigenvalue $\mu$.  This shows that

a.       There are two sets of characteristic lines (one for each eigenvalue)

b.      The two sets of characteristics are orthogonal (they therefore define a set of orthogonal curvilinear coordinates in the solid)

c.       The characteristic lines are trajectories of maximum shear (to see this, recall the definition of $\varphi$ ).  For this reason, the characteristics are termed slip lines $–$ the material slips (deforms in shear) along these lines.

 

Conventionally the characteristics satisfying $d{x}_2/d{x}_1=\tan \varphi$ are designated $\alpha$ slip lines, while the orthogonal set are designated $\beta$ slip lines

 

A representative set of characteristic lines is sketched on the right.

 

When solving a particular boundary value problem, the central issue will be to identify a set of characteristic lines that will satisfy the boundary conditions.  Field equations reduce to simple ODEs that govern variations of hydrostatic pressure and velocity along each slip line.

 

 

Relations along slip-lines

 

To complete the theory, we need to find equations relating the field variables $q=[\varphi ,\partial v_1/\partial x_1,\partial v_2/\partial x_2,\overline{\sigma }]$ along the slip-lines.  To do so we return to the governing equation

$r_i{B}_{ij}\frac{d{q}_j}{ds}=0$

and substitute for B and r.  For the four separate eigenvectors, we find that $r_i{B}_{ij}$ reduce to

$\begin{array}{l}\begin{array}{c}[0,-2k\sin 2\varphi ,2k(\cos 2\varphi -1),0]\\ [-2k\sin 2\varphi +2k\tan \varphi \cos 2\varphi ,0,0,\tan \varphi ]\end{array}\}\mu =\cot \varphi (\alpha \text{ slip-line)}\\ \begin{array}{c}[0,-2k\sin \varphi ,2k(\cos \varphi +1),0]\\ [-2k\sin 2\varphi -2k\cot \varphi \cos 2\varphi ,0,0,\tan \varphi ]\end{array}\}\mu =-\tan \varphi (\beta \text{ slip-line)}\end{array}$

 

Computing $r_i{B}_{ij}d{q}_j/ds$ and simplifying the trig formulas then yields

$\begin{array}{c}\frac{d{v}_1}{ds}+\tan \varphi \frac{d{v}_2}{ds}=0\\ -2k\frac{d\varphi }{ds}+\frac{d\overline{\sigma }}{ds}=0\end{array}\}\alpha \text{slip line}\begin{array}{c}\frac{d{v}_1}{ds}-\cot \varphi \frac{d{v}_2}{ds}=0\\ 2k\frac{d\varphi }{ds}+\frac{d\overline{\sigma }}{ds}=0\end{array}\}\beta \text{slip line}$

 

Hencky Equation: Conditions relating $\overline{\sigma }$ and $\varphi$ along slip lines are often expressed as

$\begin{array}{l}\overline{\sigma }-2k\varphi =\text{constant}\alpha \text{ slip line}\\ \overline{\sigma }+2k\varphi =\text{constant}\beta \text{ slip line}\end{array}$

These are known as the Hencky equations

 

Geiringer equations: One can also obtain simpler expressions relating velocity components along slip-lines.  It is convenient to express the velocity vector as components in a basis oriented with the slip-lines

 

The necessary basis-change is

$\begin{array}{l}{v}_{\alpha }=v_1\cos \varphi +v_2\sin \varphi \\ v_{\beta }=-v_1\sin \varphi +v_2\cos \varphi \end{array}$

A straightforward algebraic exercise then yields

$\frac{d{v}_{\alpha }}{ds}=v_{\beta }\frac{d\varphi }{ds}\frac{d{v}_{\beta }}{ds}=-v_{\alpha }\frac{d\varphi }{ds}$

These are known as the Geiringer equations.

 

 

 

6.1.3 Examples of slip-line field solutions to boundary value problems

 

When using slip-line field theory, the first step is always to find the characteristics (known as the slip line field).  This is usually done by trial and error, and can be exceedingly difficult.  These days, we usually hope that some smart person has already been able to find the slip-line field, and if we can’t find the solution in some ancient book we give up and clobber the problem with an FEM package.  If the slip-line field is known, the stress and velocity everywhere in the solid can be determined using the Hencky and Geiringer equations.

 

In this section we give several examples of slip-line field solutions to boundary value problems.

 

 

Plane Strain Extrusion (Hill)

 

A slip-line field solution to plane strain extrusion through a tapered die is shown in the picture on the right. Friction between the die and workpiece is neglected.

 

It is of particular interest to calculate the force P required to extrude the bar. The easiest way to do this is to consider the forces acting on the region ABCDEF.  Note that

(i) The resultant force on EF is $-P{e}_1$

(ii) The resultant force on CB is zero (you can see this by noting that no external forces act on the material to the left of CB)

 (iii) The stress state at a point b on the line CD can be calculated by tracing a slip-line from a to b. The Mohr’s circle construction for this purpose is shown on the right. At point a, the slip-lines intersect CB at 45 degrees, so that ${\varphi }_a=-{45}^0$; we also know that ${\sigma }_{11}=0$ on CB (because the solid to the left of CB has no forces acting on it). These conditions can be satisfied by choosing $\overline{\sigma }=-k$, so that the stress state at a is ${\sigma }_{11}=0{\sigma }_{12}=0{\sigma }_{22}=-2k$.  Tracing a $\beta$ slip-line from a to b, we see that ${\overline{\sigma }}_b={\overline{\sigma }}_a+2k({\varphi }_a-{\varphi }_b)=-\pi k/3-k$.  Finally, the slip lines intersect CD at 45 degrees, so CD is subjected to a pressure ${\sigma }_{nn}={\overline{\sigma }}_b-k$ acting normal to CD, while the component of traction tangent to CD is zero.

(iv) CD has length H, so the resultant force acting on CD is $H{\sigma }_{nn}\sin 30e_1-H{\sigma }_{nn}\cos 30e_2$

(v) By symmetry, the resultant force acting on AB is $H{\sigma }_{nn}\sin 30e_1+H{\sigma }_{nn}\cos 30e_2$

(vi) Equilibrium then gives

$P=kH\left(\frac{\pi }{3}+2\right)$

 

 

Double-notched plate in tension

 

A slip-line field solution for a double-notched plate under tensile loading is shown in the picture.  The stress state in the neck, and the load P are of particular interest.

 

Both can be found by tracing a slip-line from either boundary into the constant stress region at the center of the solid.

 

Consider the slip-line starting at A and ending at B, for example.  At A the slip-lines meet the free surface at 45 degrees.  With $\alpha ,\beta$ designated as shown, ${\varphi }_A=\alpha -\pi /4$ and ${\overline{\sigma }}_A=k$.  Following the slip-line to b, we see that ${\varphi }_B=\pi /4$, so the Hencky equation gives ${\overline{\sigma }}_B=k\left(\pi -2\alpha \right)$.  The state of stress at b follows as

${\sigma }_{11}=(\pi -2\alpha )k{\sigma }_{22}=(\pi -2\alpha +2)k{\sigma }_{12}=0$

The state of stress is clearly constant in the region ABCD, (and so is constant along the line connecting the two notches).  The force required to deform the solid is therefore $P=ak(\pi -2\alpha +2)$.

 

 

 

Pressurized cylindrical cavity

 

The slip-line field solution to an internally pressurized rigid-plastic cylinder is shown on the right. The goal is to determine the stress state everywhere in the cylinder, and to calculate the internal pressure necessary to drive the deformation.

 

Consider the $\alpha$ slip-line, which starts at point A (with cylindrical-polar coordinates $r=a,\theta =0$ ), and ends at B (with cylindrical-polar coordinates  $r=b,\theta ={\theta }_b$. 

1.      At point B, the surface is traction free, which requires ${\sigma }_{rr}={\sigma }_{r\theta }=0$.  To satisfy ${\sigma }_{r\theta }=0$, the slip-line must meet the surface at 45 degrees ( ${\varphi }_B={\theta }_b+\pi /4$ ).  In addition, to satisfy ${\sigma }_{rr}=0$ the hydrostatic stress ${\overline{\sigma }}_B=k$.

2.      Note that the shear stress component ${\sigma }_{r\theta }=0$ throughout the cylinder.  This means that the slip-line must cross every radial line at 45 degrees (or, if you prefer, it must cross every circumferential line at 45 degrees).

3.      Consider a small segment ds of the slip-line.  Since the slip-line is at 45 degrees to the radial direction, $dr=rd\theta$.

4.      Integrating this result from $r=a,\theta =0$ to $(r,\theta )$ gives $r=a{e}^{\theta }$ - i.e. the slip-lines are logarithmic spirals.

5.      At B, this gives $b=a{e}^{{\theta }_B}$ or ${\theta }_B=\log (b/a)$

6.      Note that ${\varphi }_A=\pi /4$ and apply the Hencky equation from B to A to see that ${\overline{\sigma }}_A=k-2k{\theta }_B=k-2k\log (b/a)$

7.      Finally, the basis change equation shows that ${\sigma }_{rr}=-p_A={\overline{\sigma }}_A-k=-2k\log (b/a)$

8.      At a generic point $(r,\theta )$, the same procedure gives ${\sigma }_{rr}=-2k\log (b/a){\sigma }_{\theta \theta }=-2k\log (b/a)+2k$

This result can be compared with the axisymmetric elastic-plastic solution in Section 4.2.

 

 

 

Notched Bar in Bending

 

The figure on the right shows a slip-line field solution for a notched bar subjected to a pure bending moment.  The solution is valid for $\omega \ge 1$ (radian).

 

The slip-line field can be used to determine the moment M required to deform the bar as a function of the notch angle $\omega$. To do so, note that

1. The stress acting on the line NO is constant, since slip-lines are straight.
2. You can determine the stress at a point D between O and N by following the slip-line CD. The stress must satisfy ${\sigma }_{22}=0{\sigma }_{12}=0$ at C, so the slip-lines must meet the surface at 45 degrees ( ${\varphi }_C=\pi /4$ ) and we must choose ${\overline{\sigma }}_C=-k$.  This gives ${\sigma }_{11}=-2k{\sigma }_{22}={\sigma }_{12}=0$ at D.
3. Similarly, the stress acting on the line OP is constant, since slip-lines are straight.  You can calculate the stress at some point B between P and O by following the slip-line AB.  At point A, the surface is free of traction, so the slip-line must meet the surface at 45 degrees ( ${\varphi }_A=\pi /4-\omega$ ), and the hydrostatic stress must satisfy ${\overline{\sigma }}_A=k$.  At B, we see that ${\varphi }_B=-\pi /4$.  Using the Hencky equation along the $\beta$ slip-line AB, we find that ${\overline{\sigma }}_B=k(1+\pi -2\omega )$.  Finally ${\sigma }_{11}=k(2+\pi -2\omega ){\sigma }_{22}=k(\pi -2\omega ){\sigma }_{12}=0$ from the basis change formulas.
4. The height d of point O can be found from the condition that the axial force applied to the bar must vanish.  Integrating ${\sigma }_{11}$ along the line NOP and setting the result to zero shows that

$d=\frac{2+\pi -2\omega }{4+\pi -2\omega }(h-a)$

5. Finally, taking moments for the region of the bar to the right of NOP about O shows that

$\frac{d^2}{2}(2k)+\frac{{\left(h-a-d\right)}^2}{2}(2+\pi -2\omega )k-M=0$

Substituting for d and simplifying shows that

$M=k{(h-a)}^2\frac{2+\pi -2\omega }{4+\pi -2\omega }$

 

Overstressing: At first sight, this solution is valid for any notch angle $\omega$, but in fact this is not the case.  A slip-line field is valid only if the rigid regions in the field do not exceed yield.  This means that it must be possible to find a static equilibrium distribution of stress which does not violate the yield criterion anywhere in the rigid part of the solid.  If this cannot be done, the solid is said to be over-stressed.

 

The slip-line field for a notched bar has a peculiar state of stress at point O $–$ there is a stress discontinuity (and singularity) at the corner, and it turns out that the region that was assumed rigid in this solution is over-stressed (the maximum principal shear stress exceeds k) if the notch is too sharp. 

To see this, consider the rigid region of the solid just to the left of O, as shown in the picture.  The lines OE and OF are adjacent to $\alpha$ slip lines, and so are subjected to a combined shear stress k and normal stresses ${\overline{\sigma }}_B,{\overline{\sigma }}_D$ as shown.  When the value of ${\overline{\sigma }}_B-{\overline{\sigma }}_D$ gets too large, the rigid region OEFO collapses plastically $–$ a possible slip-line field at collapse is shown in the figure.  The slip-line field consists of a 90 degree fan, centered at O.  Applying the Hencky relation along a generic $\beta$ slip-line shows that, at collapse ${\overline{\sigma }}_B-{\overline{\sigma }}_D=k\pi$, and so for the rigid region to remain below yield ${\overline{\sigma }}_B-{\overline{\sigma }}_D\le k\pi$.  Substituting the values of ${\overline{\sigma }}_B,{\overline{\sigma }}_D$ from parts (2) and (3) then gives $\omega \ge 1$.

 

A solution for a sharp notch is shown in the figure to the right. In the modified field, the region PBNFG is rigid.   The left hand part of the bar rotates about point O, shearing along a pair slip lines formed by the circular arcs AB and GF.  To calculate the moment, we need first to calculate the angles $\theta$ and $\psi$, the radius R of the arc BC, the length b of the constant stress regions adjacent to the notch, and the height d of point O above the base of the beam.  To this end, note that

1. At point A, the surface of the wedge is traction free. The slip-lines must intersect the surface at 45 degrees, which shows that ${\varphi }_A=\omega -3\pi /4$ and that ${\overline{\sigma }}_A=k$.
2. Tracing the $\alpha$ slip-line from A to B and noting ${\varphi }_B=\omega +\theta -3\pi /4$ gives ${\overline{\sigma }}_B=k+2k\theta$.
3. At point D at the base of the beam, the surface is traction free, so the slip-lines must meet the surface at 45 degrees.  This gives ${\varphi }_D=\pi /4$ and ${\overline{\sigma }}_D=-k$.
4. The stress is uniform in the region CDEF, so that ${\overline{\sigma }}_C=-k$.
5. The hydrostatic stresses at B and C must be related by the Hencky equation for a $\beta$ slip-line, which gives ${\overline{\sigma }}_B={\overline{\sigma }}_C+2k\psi \Rightarrow \psi -\theta =1$.
6. Finally, elementary geometry shows that $\omega +\theta +\psi =\pi$.
7. Hence, solving (5) and (6) gives $\theta =(\pi -\omega -1)/2\psi =(\pi -\omega +1)/2$.
8. Geometry gives $d+R\sin (\psi -\pi /4)+b\cos (\psi -\pi /4)=h-a$.
9. We obtain two more equations relating the unknown variables from the condition that the resultant force acting on any surface that extends from the top of the beam to the bottom must vanish.  The resultant force acting on the surface to the right of PBCD can be calculated as

$\begin{array}{l}{\overline{\sigma }}_{B}b\left(\cos (\psi -\pi /4)e_1+\sin (\psi -\pi /4)e_2\right)-kb\left(\sin (\psi -\pi /4)e_1+\cos (\psi -\pi /4)e_2\right)\\ +{\displaystyle \underset{\pi /4-\psi }{\overset{\pi /4}{\int }}\left[k(\sin \lambda e_1-\cos \lambda e_2)-{\overline{\sigma }}_{BC}(\lambda )(\cos \lambda e_1+\sin \lambda e_2)\right]Rd\lambda }\\ +(d\sqrt{2}-R)\left[k(e_1+e_2)/\sqrt{2}-{\overline{\sigma }}_C(e_1-e_2)/\sqrt{2}\right]=0\end{array}$

where ${\overline{\sigma }}_{BC}={\overline{\sigma }}_C+2k(\pi /4-\lambda )$ is the hydrostatic stress along the slip-line BC.  The results of (7), (8) and (9) can be solved for d, R and b

10. Finally, taking moments about O gives $M=b^2{\overline{\sigma }}_B/2+kRb+k{R}^2\psi -{\overline{\sigma }}_C(2d^2-R^2)/2$. Thus,

$M=k{(h-a)}^2\left\{{(\omega -\pi )}^2-1\right\}/\left\{{(\omega -\pi -1)}^2-4{\cos }^2\left[(\omega -1)/2\right]\right\}$

This result is valid only if $b\ge 0$, which requires $\omega >0.056$.  In addition, the notch angle must satisfy $\omega \le 1$ to avoid overstressing the rigid corner at P.