EN222: Mechanics of Solids
Division of Engineering
Brown University
There is a vast catalog of solutions to linear elastic boundary and initial value problems. In contrast, there are very few exact solutions to boundary value problems involving plastically deforming solids. Such solutions as exist are usually either for highly simplified geometries (e.g. spherical or axial symmetry – an example can be found in the EN175 lecture notes) or simplified material models (such as rigid plastic solids).
5.1 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 method is not widely used these days (because it’s a lot harder to use than an FEM package!) but it does provide analytical solutions to a number of very difficult problems (e.g. many metal forming processes). Many of these solutions would be extremely difficult to obtain numerically, because they involve huge deformations, and contain velocity discontinuities.
Slip-line theory makes three restrictive assumptions
1. Plane strain deformation
2. Quasi-static loading
3. The solid is idealized as a rigid-perfectly plastic Mises solid.

The solid of interest must therefore be a long cylindrical region, subject to boundary conditions
The boundary conditions on the ends of the cylindrical region must satisfy
Note that rather than prescribe displacements on the boundary, it is more convenient to prescribe velocities.
In view of the boundary conditions we assume that the solution must be a plane strain deformation, satisfying
In practice we will not solve for the displacement field, but instead will compute the velocity field
Summary of governing equations
1. Strain-rate – velocity relation
2. The plastic flow rule
where is some constant. Plane strain deformation then requires
whereupon the flow rule shows that the remaining components of plastic strain rate satisfy
We observe that these conditions imply that
3. Yield criterion
where is the shear yield stress of the material, and we have used the condition that
4. Equilibrium conditions
Solution of governing equations by method of characteristics
From the preceding section, we observe that we must calculate a velocity field and stress field satisfying governing equations
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 . The stress state is sketched below.

Instead of solving for the stress components , we will calculate the hydrostatic stress and the angle between the i direction and the direction. Recall that we can relate to , and k using Mohr’s circle of stress

From the picture, we see that
These relations will be essential in interpreting slip-line fields. Remember that you can quickly re-derive them with Mohr’s circle.
We now re-write the governing equations in terms of , and k. The yield criterion is satisfied automatically. The remaining four equations are most conveniently expressed in matrix form
where A and B are 4-dimensional symmetric matrices and q is a 1x4 vector, defined as
This is a quasi-linear hyperbolic system of PDEs, which may be solved by the method of characteristics.
We follow standard procedure. The first step is to find eigenvalues and eigenvectors that satisfy
A straightforward exercise (set to find the eigenvalues, and plug back to get eigenvectors, or if you’re lazy use MAPLE…) shows that there are two repeated eigenvalues, with corresponding eigenvectors
We can now eliminate A from the governing matrix equation
Finally, if we set
and note that
we find that
along characteristic lines in the solid that satisfy
The special characteristic lines in the solid can be identified more easily if we note that
which shows that the slope of the characteristic lines satisfies
for the two possible values of the eigenvalue . 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 ). For this reason, the characteristics are termed slip lines – the material slips (deforms in shear) along these lines.
Conventionally the characteristics satisfying are designated slip lines, while the orthogonal set are designated slip lines
A representative set of characteristic lines is sketched below.

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 along the slip-lines. To do so we return to the governing equation
and substitute for B and r. For the four separate eigenvectors, we find that reduce to
Computing then yields
Hencky Equations
Conditions relating and along slip lines are often expressed as
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
A straightforward algebraic exercise then yields
These are known as the Geiringer equations.
Examples of slip-line field solutions to boundary value problems
If this is your first exposure to slip-line field theory you are probably totally confused at this point. It is not easy to see how all these results are related to an actual boundary value problem!
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.
Some simple examples are helpful.
Rigid punch indenting a half-space (Hill)

Hill’s slip line field solution to a rigid punch indenting a rigid-plastic half-space is shown above. The slip-lines are shown in red; the lines are blue. We now proceed to interpret the solution.

Consider first the state of stress at point a. Clearly, at this point, and the figure shows the Mohr’s circle construction that is helpful in transforming the stress state from a basis aligned with the slip-lines to the fixed {i,j} basis. Recall (or use the Mohr’s circle to see) that
where is the hydrostatic component of stress. The boundary conditions at a require that . The first condition is clearly satisfied if the slip-lines intersect the boundary at 45 degrees. We can satisfy the second condition by setting . Finally this gives the stress parallel to the surface as .
The stress must evidently 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 . We can use the Hencky equation to determine at b. Recall that
so following one of the slip lines we get
Using the basis-change equation we then get
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
where w is the width of the punch.
Plane Strain Extrusion (Hill)
Many slip-line field solutions approximate some important metal-forming process. An example is shown below

It is left as an exercise to show that for this solution the force on the punch is given by