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Chapter 6

Analytical techniques and solutions for plastic solids

Plasticity problems are much more difficult to solve than linear elastic problems. In general, a numericalmethod must be used, as discussed in Chapters 7 and 8. Nevertheless, there are several powerful mathematicaltechniques that can be used to find both exact and approximate solutions. In this chapter we outline twoparticularly effective methods: slip-line field theory, which gives exact solutions for plane strain boundary valueproblems for rigid plastic solids; and bounding theorems, which provide a quick way to estimate collapse loadsfor plastic solids and structures.

Applied Mechanics of Solids (A.F. Bower) Chapter 6: Plasticity... http://solidmechanics.org/text/Chapter6_1/Chapter6_1.htm

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6.1 Slip-line field theory

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

1. Plane strain deformation i.e. displacement components in the basisshown satisfy and are functions of and only

2. Quasi-static loading3. No temperature changes4. No body forces5. 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 arecharacterized by the yield stress in uniaxial tension Y. Alternatively, the material is sometimes

characterized by its yield stress in shear .

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 particularproblem. Nevertheless, a wide range of important solutions have been found. The main intent of this section isto 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 pictureon the right. (This is Hill’s solution to a rigid punch indenting arigid-plastic half-space).

The slip lines consist of a curvilinear mesh of two families oflines, which always cross each other at right angles. Byconvention, one set of lines are named slip-lines (shown inred); 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

By definition, the slip-lines are always parallel to axes of principalshear stress in the solid. This means that the stress components ina basis oriented with the , directions have the form

where is the hydrostatic stress (determined

using the equations given below), k is the yield stress of thematerial in shear, and Y is its yield stress in uniaxial tension. Thisstress state is sketched in the figure. Since the shear stress is equalto the shear yield stress, the material evidently deforms by shearingparallel 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

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)


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

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

The velocity field (Geiringer equations)

The 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

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

Consider 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-linesto the fixed basis using the Mohr’s circle construction shown in the

figure. Recall (or use the Mohr’s circle to see) that

where is the hydrostatic component of stress. The boundary conditionsat 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 .

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 . 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 theslip-line field below the punch) and so the total force (per unit out of plane length) on the punch can becomputed as

where w is the width of the punch.

How to distinguish the and families of slip lines


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Usually, slip-line fields are presented without specifying which set ofslip-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 areinterchangeable switching and will simply change the sign of all the

stresses.

You can see this clearly using the Hill solution. The figure on the rightshows 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

at point b. The normal stress acts upwards on the surface so that this represents the stress induced by a rigidpunch 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 surfacecharacterized by yield stress in uniaxial tension or its yield stress in shear

. Let denote the components of displacement, strain andstress in the solid. The solid is assumed to be a long cylinder with its axisparallel 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

where 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 equations

1. Strain-rate velocity relation

2. The plastic flow rule

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


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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 governingequations. It is convenient to start by eliminating some of thestress components using the yield condition. Since thematerial is at yield, we note that at each point in the solid wecould find a basis in which the stress state consists of a shearstress of magnitude k (the shear yield stress), together with anunknown component of hydrostatic stress . The stress stateis sketched on the right.

Instead 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

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 matrixform

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.

The first step is to find eigenvalues and eigenvectors that satisfy


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A 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, withcorresponding 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 thata. 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 thematerial 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 on theright.

When solving a particular boundary value problem, the central issue will be to identify a set of characteristiclines that will satisfy the boundary conditions. Field equations reduce to simple ODEs that govern variations ofhydrostatic 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


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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 Hencky equations

Geiringer equations: One can also obtain simpler expressionsrelating velocity components along slip-lines. It is convenient toexpress the velocity vector as components in a basis oriented withthe slip-lines

The necessary basis-change is

A straightforward algebraic exercise then yields

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 linefield). This is usually done by trial and error, and can be exceedingly difficult. These days, we usually hopethat some smart person has already been able to find the slip-line field, and if we can’t find the solution in someancient book we give up and clobber the problem with an FEM package. If the slip-line field is known, thestress 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 atapered die is shown in the picture on the right. Frictionbetween the die and workpiece is neglected.

It is of particular interest to calculate the force P required toextrude the bar. The easiest way to do this is to consider theforces acting on the region ABCDEF. Note that(i) The resultant force on EF is

(ii) The resultant force on CB is zero (you can see this bynoting 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’scircle construction for this purpose is shown on the right. At point a, the slip-lines intersect CB at 45 degrees, so


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that ; we also know that on CB (because the solid to

the left of CB has no forces acting on it). These conditions can besatisfied 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.(iv) 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 tension

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

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

Consider the slip-line starting at A and ending at B, for example. At Athe 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

The state of stress is clearly constant in the region ABCD, (and so is constant along the line connecting the twonotches). The force required to deform the solid is therefore .

Pressurized cylindrical cavity

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

Consider the slip-line, which starts at point A (with cylindrical-polarcoordinates ), and ends at B (with cylindrical-polarcoordinates .

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.

5. At B, this gives or 6. Note that and apply the Hencky equation from B to A to see that

7. Finally, the basis change equation shows that

8. At a generic point , the same procedure gives


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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 barsubjected to a pure bending moment. The solution is valid for

(radian).

The slip-line field can be used to determine the moment M required todeform 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 arestraight.

1.

You can determine the stress at a point D between O and N by following the slip-line CD. The stressmust satisfy at C, so the slip-lines must meet the surface at 45 degrees ( ) and

we must choose . This gives at D.

2.

Similarly, the stress acting on the line OP is constant, since slip-lines are straight. You can calculate thestress at some point B between P and O by following the slip-line AB. At point A, the surface is free oftraction, 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.

3.

The height d of point O can be found from the condition that the axial force applied to the bar mustvanish. Integrating along the line NOP and setting the result to zero shows that

4.

Finally, taking moments for the region of the bar to the right of NOP about O shows that5.

Substituting for d and simplifying shows that

Overstressing: At first sight, this solution is valid for any notch angle , but in fact this is not the case. Aslip-line field is valid only if the rigid regions in the field do not exceed yield. This means that it must bepossible to find a static equilibrium distribution of stress which does not violate the yield criterion anywhere inthe 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 thereis a stress discontinuity (and singularity) at the corner, and it turns out that theregion that was assumed rigid in this solution is over-stressed (the maximumprincipal 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 inthe picture. The lines OE and OF are adjacent to slip lines, and so are subjectedto 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 theHencky 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 .

A solution for a sharp notch is shown in the figure to the right. In the modified field, the region PBNFG isrigid. The left hand part of the bar rotates about point O, shearing along a pair slip lines formed by the circulararcs AB and GF. To calculate the moment, we need first to calculate the angles and , the radius R of the arcBC, the length b of the constant stress regions adjacent to the notch, and the height d of point O above the baseof 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 45degrees, which shows that and that .

1.


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Tracing the slip-line from A to B and noting gives .

2.

At point D at the base of the beam, the surface is traction free, sothe slip-lines must meet the surface at 45 degrees. This gives

and .

3.

The stress is uniform in the region CDEF, so that .4.The hydrostatic stresses at B and C must be related by the Henckyequation for a slip-line, which gives

.

5.

Finally, elementary geometry shows that .6.Hence, solving (5) and (6) gives .7.Geometry gives .8.We obtain two more equations relating the unknown variables from the condition that the resultant forceacting on any surface that extends from the top of the beam to the bottom must vanish. The resultantforce acting on the surface to the right of PBCD can be calculated as

9.

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 . Thus,10.

This result is valid only if , which requires . In addition, the notch angle must satisfy toavoid overstressing the rigid corner at P.

(c) A.F. Bower, 2008This site is made freely available for educational purposes.

You may extract parts of the text for non-commercial purposes provided that the source is cited.Please respect the authors copyright.


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