Analytical techniques and solutions for
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
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 and are functions of and only
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 .
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
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
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 slip-lines (shown in red); the other are
called 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
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 ,
directions have the form
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 denotes the angle between the slip-line and the direction, the stress components in the basis can be calculated as
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
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
The velocity field (Geiringer equations)
velocity field can be expressed as components in a fixed basis, or as components parallel and
perpendicular to the slip lines.
Application to the Hill slip-line field
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.
first the state of stress at point a.
Clearly, at this point. The stress state can be
transformed from a basis aligned with the slip-lines to the fixed basis using the Mohr’s circle construction shown
in the figure. Recall (or use the Mohr’s
circle to see) that
is the hydrostatic component of stress. The boundary conditions at a require
that . The first condition is clearly satisfied,
since the slip-lines intersect the boundary at . We can satisfy the second condition by
setting . Finally this gives the stress parallel to the
surface as .
stress must be constant in the triangular region ABC, as the slip lines in this
region are straight.
consider the stress state at b.
Here, we see that . We can use the Hencky equation to determine at b.
so following one of the slip lines we get
Using the basis-change
equation we then get
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.
How to distinguish the and families
of slip lines
slip-line fields are presented without specifying which set of slip-lines
should be taken as the and which should be the set it is up to you to work out which is
which. In fact, the slip-lines are
interchangeable switching and will simply change the sign of all the
can see this clearly using the Hill solution.
The figure on the right shows the solution with and lines switched over. At point a,
and therefore to satisfy we must now choose . To find the stress under the contact, we can
trace a slip line to point b. Here, we see that ,
so the Hencky equation
Using the basis-change
equation we then get
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.
a rigid-perfectly plastic solid, with a von-Mises yield surface characterized
by yield stress in uniaxial tension or its yield stress in shear . Let 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 direction, which is constrained to deform in
plane strain, with and independent of . It is loaded by subjecting part of its
boundary to a prescribed velocity, and the remainder to a prescribed traction, so that
the Greek subscripts can have values of 1 or 2. In practice we will
compute the velocity field rather than the displacement field.
Summary of governing
Strain-rate velocity relation
The plastic flow
strain deformation then requires
the flow rule shows that the remaining components of plastic strain rate
observe that these conditions imply that
where is the shear yield stress of the material, and
we have used the condition that
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
with appropriate boundary conditions.
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 on the right.
of solving for the stress components ,
we will calculate the hydrostatic stress and the angle between the direction and the direction.
Recall that we can relate to ,
and k using Mohr’s circle of stress:
from the picture, we see that
now re-write the governing equations in terms of ,
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
is a quasi-linear hyperbolic system of PDEs, which may be solved by the method
first step is to find eigenvalues and eigenvectors that satisfy
straightforward exercise (set 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
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
the two possible values of the eigenvalue . This shows that
a. There are two sets of characteristic lines (one for
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
the characteristics satisfying are designated slip lines, while the orthogonal set are
designated slip lines
A representative set of
characteristic lines is sketched on the right.
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
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 and simplifying the trig formulas then yields
Hencky Equation: Conditions relating and along slip lines are often expressed as
These are known as the
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
A straightforward algebraic
exercise then yields
These are known as the
6.1.3 Examples of
slip-line field solutions to boundary value problems
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)
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
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
The resultant force on EF is
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 ;
we also know that on CB (because the solid to the left of CB has
no forces acting on it). These conditions can be satisfied by choosing ,
so that the stress state at a is . Tracing a slip-line from a to b, we see that . Finally, the slip lines intersect CD at 45
degrees, so CD is subjected to a pressure acting normal to CD, while the component of
traction tangent to CD is zero.
CD has length H, so the resultant
force acting on CD is
(v) By symmetry, the
resultant force acting on AB is
(vi) Equilibrium then gives
Double-notched plate in
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
can be found by tracing a slip-line from either boundary into the constant
stress region at the center of the solid.
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 designated as shown, and . Following the slip-line to b, we see that ,
so the Hencky equation gives . The state of stress at b follows as
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 .
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
the slip-line, which starts at point A (with
cylindrical-polar coordinates ), and ends at B (with cylindrical-polar
1. At point B, the surface is traction free, which
requires . To satisfy ,
the slip-line must meet the surface at 45 degrees ( ). In
addition, to satisfy the hydrostatic stress .
2. Note that the shear stress component 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, .
4. Integrating this result from to gives - i.e. the slip-lines are logarithmic spirals.
At B, this gives or
6. Note that and apply the Hencky equation from B to A to
basis change equation shows that
At a generic
the same procedure gives
This result can be compared
with the axisymmetric elastic-plastic solution in Section 4.2.
Notched Bar in Bending
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 (radian).
slip-line field can be used to determine the moment M required to deform the bar as a function of the notch angle .
To do so, note that
- The stress acting on the line NO is constant,
since slip-lines are straight.
- You can determine the stress at a point D between
O and N by following the slip-line CD. The stress must satisfy at C, so the slip-lines must meet the
surface at 45 degrees ( ) and we must choose . This gives at D.
- 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 ( ), and the hydrostatic stress must satisfy
. At B, we see that . Using the Hencky equation along the slip-line AB, we find that . Finally from the basis change formulas.
- The height d of point O can be found from the
condition that the axial force applied to the bar must vanish. Integrating along the line NOP and setting the result
to zero shows that
- Finally, taking
moments for the region of the bar to the right of NOP about O shows that
Substituting for d
and simplifying shows that
Overstressing: At first sight, this solution is valid for any notch
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
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
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 slip lines, and so are subjected to a combined
shear stress k and normal stresses as shown.
When the value of 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 slip-line shows that, at collapse ,
and so for the rigid region to remain below yield . Substituting the values of from parts (2) and (3) then gives .
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 and ,
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
- At point A, the
surface of the wedge is traction free. The slip-lines must intersect the
surface at 45 degrees, which shows that and that .
- Tracing the slip-line from A to B and noting gives .
- 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 and .
- The stress is uniform
in the region CDEF, so that .
- The hydrostatic
stresses at B and C must be related by the Hencky equation for a slip-line, which gives .
- Finally, elementary
geometry shows that .
- Hence, solving (5) and
(6) gives .
- Geometry gives .
- 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
where is the hydrostatic stress along the slip-line
BC. The results of (7), (8) and (9) can
be solved for d, R and b
- Finally, taking moments about O gives .
result is valid only if ,
which requires . In addition, the notch angle must satisfy to avoid overstressing the rigid corner at P.