# 08 plasticity 06 strain hardening

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Section 8.6

Solid Mechanics Part II 219 Kelly

Strain Hardening

In the applications discussed in the preceding sections, the material was assumed to be

perfectly plastic. The issue of strain hardening materials is addressed in this section.

8.6.1 Strain Hardening

In the one-dimensional (uniaxial test) case, a specimen will deform up to yield and

then generally harden, Fig. 8.6.1. Also shown in the figure is the perfectly-plastic

idealisation (horizontal line). In the perfectly plastic case, once the stress reaches the

yield point (A), plastic deformation ensues, so long as the stress is maintained at Y. If

the stress is reduced, elastic unloading occurs. In the strain-hardening case, once

yield occurs, the stress needs to be continually increased in order to drive the plastic

deformation. If the stress is held constant, for example at B, no further plastic

deformation will occur; at the same time, no elastic unloading will occur. Note thatthis condition cannot occur in the perfectly-plastic case, where there is one of plastic

Figure 8.6.1: uniaxial stress-strain curve (for a typical metal)

These ideas can be extended to the multiaxial case, where one now has a yield surface

rather than a yield point. In the perfectly plastic case, the yield surface remains the

same size and shape. For plastic deformation, the stress state must be on the yieldsurface and remain on the yield surface. For elastic unloading, the stress state must

move back inside the yield surface.

For the strain-hardening material, the yield surface must change in some way so that

an increase in stress is necessary to induce further plastic deformation. This can be

done in a number of ways. Before looking at how the yield surface might change,

The yield surface is in general described by a function of the form

strain-hardening

0

A

Bstress

strain

Yield point Y perfectly-plastic

elastic

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Section 8.6

Solid Mechanics Part II 220 Kelly

0)( =ijf (8.6.1)

Suppose that the stress state, represented by the vector in stress space, is such that

one is on the yield surface, Fig. 8.6.2. The normal vector to the surface is n. An

increment in stress d now takes place. The notions of (plastic) loading, neutral

As in 8.3.19-20, a normal to the surface is /f , so this scalar product can be

expressed as

33

22

11

df

df

df

d

+

+

= n , (8.6.3)

or, for a general 6-dimensional stress space,

ij

ij

df

(8.6.4)

As mentioned above, neutral loading does not occur for the perfectly plastic material.

0 n d , where fand n now refer to the new loading function.

f

n

dd

d

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Section 8.6

Solid Mechanics Part II 221 Kelly

Figure 8.6.3: A new loading surface, due to stressing to outside the initial yield

surface

Strain Softening

Materials can also strain soften, for example soils. In this case, the stress-straincurve turns down, as in Fig. 8.6.4. The loading function for such a material will in

general decrease in size with further straining.

Figure 8.6.4: uniaxial stress-strain curve for a strain-softening material

Isotropic Hardening

The simplest means by which the loading function (yield surface) can change is

through isotropic hardening. This is where the loading function remains the sameshape but expands with increasing stress, Fig. 8.6.5.

0

stress

strain

surfaceinitial yield

surface

f

nd

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Section 8.6

Solid Mechanics Part II 222 Kelly

Figure 8.6.5: isotropic hardening

Kinematic Hardening

The isotropic model implies that, if the yield strength in tension and compression are

initially the same, i.e. the yield surface is symmetric about the stress axes, they remain

equal as the yield surface develops with plastic strain. In order to model theBauschinger effect, and similar responses, where a strain hardening in tension will

lead to a softening in a subsequent compression, one can use the kinematic

hardening rule. This is where the yield surface remains the same shape and size but

merely translates in stress space, Fig. 8.6.6.

Figure 8.6.6: kinematic hardening

Other Hardening Rules

More complex models can be used, for example the mixed hardening rule, which

combines features of both the isotropic and kinematic hardening models.

initial yield

surface subsequent

yield surface

1

2

stress at

initial yield

elastic

plastic

deformation

(hardening)

elastic

initial yield

surface

subsequent

yield surface

1

2

stress atinitial yield

elastic

elastic

plastic

deformation

(hardening)

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Section 8.6

Solid Mechanics Part II 223 Kelly

8.6.3 The Flow Curve

In order to predict and describe the possible changes to the loading function outlinedin the previous section, one can introduce the concept of the flow curve.

Strain hardening in the uniaxial tension test can be described using a relationship ofthe form

( )ph = (8.6.6)

A typical plot, the flow curve, of this function for a strain-hardening material is shown

in Fig. 8.6.7. The slope of this flow curve is the plastic modulus, Eqn. 8.1.9,

pd

dH

= (8.6.7)

Figure 8.6.7: uniaxial stress plastic strain curve (for a typical metal)

In the multi-axial case, one needs again a flow curve, of the form 8.6.6, but one whichrelates a complex three-dimensional stress state to a corresponding three dimensional

state of plastic strain. This formidable task is usually tackled by defining an effective

stress and an effective strain, which describe in a simple way the amount of stress

and plastic strain, and then by relating these effective parameters using an expressionequivalent to 8.6.6.

Effective Stress

Introduce an effective stress , some function of the stresses, which reduces to the

stress 1 in the uniaxial case. It is to be a measure of the amount of stress in the

plastic flow takes place, the effective stress can be defined throughf.

The yield function can usually be expressed in the form

0)(),( == kFkf ijij (8.6.8)

0

Yperfectly-plastic

p

pd

dH

( )ph =

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Section 8.6

Solid Mechanics Part II 224 Kelly

where kis a material parameter. Consider first the case of isotropic hardening

(kinematic hardening will be considered in a later section). As plastic strain

accumulates, the shape of the yield surface, as described by )( ijF , remains the

same. If one writes

n

ij CF )( = (8.6.9)

then the effective stress is guaranteed to reduce to 1 in the uniaxial case.

For example, for the Von Mises material,

( ) ( ) ( )[ ] nC 6

12/1

2

13

2

32

2

21 =

++ (8.6.10)

With 0, 321 === , one has 3/1,1 == cn and

( ) ( ) ( )

ijijss

J

2

3

2

2

13

2

32

2

21

3

2

1

=

=

++=

(8.6.11)

This is the Von Mises stress 8.3.11, and equals the yield stress in uniaxial tension at

first yield, but it must increase in some way with strain hardening in order to continue

to drive plastic deformation.

Similarly, the effective stress for the Drucker-Prager yield criterion is {Problem 1}

3/1 21

+

+=

JI(8.6.12)

which reduces to 8.6.11 when 0= .

Effective Plastic Strain

The idea now is to introduce an effective plastic strain so a plot of the effective stress

against the effective plastic strain can be used to determine the multi-axial hardening

behaviour. The two most commonly used means of doing this are to define an

effective plastic strain increment:

(i) which is a similar function of the plastic strains as the effective stress is of thedeviatoric stresses

(ii) by equating the plastic work(per unit volume), also known as the plasticdissipation,

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Section 8.6

Solid Mechanics Part II 225 Kelly

p

ijij

p ddW = (8.6.13)

to the plastic work done by the effective stress and effective plastic strain:

pp

ddW =

(8.6.14)

Consider first method (i), which is rather intuitive and non-rigorous. The deviatoric

stress and plastic strain tensors are of a similar character. In particular, their traces are

zero, albeit for different physical reasons; 01 =J because of independence of

hydrostatic pressure, the first invariant of the plastic strain tensor is zero because of

material incompressibility in the plastic range: 0=piid . For this reason, one chooses

the effective plastic strain (increment) pd to be a similar function of pijd as is of

the ijs .

For example, for the Von Mises material one has 8.6.11,ijijss

23 = . Thus one

chooses pij

p

ij

p ddCd = , where the constant Cis to ensures that the expression

reduces to pp dd 1 = in the uniaxial case. Considering this uniaxial case,ppppp ddddd 12

13322111 , === , one finds that

( ) ( ) ( )2132

32

2

21

32

3

2

pppppp

p

ij

p

ij

p

dddddd

ddd

++=

=

(8.6.15)

Consider now method (ii). Consider also the Prandtl-Reuss flow rule, Eqn. 8.4.1,

dsd ip

i = (other flow rules will be examined more generally in 8.7). In that case,

working with principal stresses, the plastic work increment is (see Eqns. 8.2.7-10)

( ) ( ) ( )[ ]

d

ds

ddW

ii

p

ii

p

2

13

2

32

2

213

1++=

=

=

(8.6.16)

Using the effective stress 8.6.11 and 8.6.14 then gives, again with dsdi

p

i = ,

( ) ( ) ( )

( ) ( ) ( )2132

32

2

21

2

13

2

32

2

21

3

2

3

2

pppppp

p

dddddd

dd

++=

++=(8.6.17)

This is the same expression as derived using method (i), Eqn. 8.6.15, but this is so

only for the Von Mises yield condition; it will not be so in general.

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Section 8.6

Solid Mechanics Part II 226 Kelly

Note also that, in this derivation, the Von Mises term 2J conveniently appeared in the

Prandtl-Reuss work expression 8.6.16. It will be shown in the next section that this is

no coincidence, and that the Prandtl-Reuss flow-rule is indeed naturally associated

with the Von-Mises criterion.

Prandtl-Reuss Relations in terms of Effective Parameters

With the definitions 8.6.11, 8.6.15 for effective stress and effective plastic strain, one

can now write {Problem 2}

2

3 pdd = (8.6.18)

and the Prandtl-Reuss (Levy-Mises) plastic strain increments can be expressed as

( ) ( )[ ]( ) ( )[ ]( ) ( )[ ]

( )( )( )

zx

pp

zx

yz

pp

yz

xy

pp

xy

yyxxzz

pp

zz

xxzzyy

pp

yy

zzyyxx

pp

xx

dd

dd

dd

dd

dd

dd

/

/

/

/

/

/

23

23

23

21

21

21

=

=

=

+=

+=+=

(8.6.19)

or

ij

pp

ijs

dd

2

3= . (8.6.20)

A relation between the effective stress and the effective plastic strain will now make

equations 8.6.19 complete.

The Flow Curve

The flow curve can now be plotted for any test and any conditions, by plotting the

effective stress against the effective plastic strain. The idea (hope) is that such a curvewill coincide with the uniaxial flow curve. If so, the strain hardening behaviour for

new conditions can be predicted by using the uniaxial flow curve, that is, it is taken

that the effective stress and effective plastic strain for any conditions are related

through 8.6.6,

ph = (8.6.21)

and the effective plastic modulus is given by

( ) pdd

H

= (8.6.22)

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Section 8.6

Solid Mechanics Part II 227 Kelly

The total accumulated effective plastic strain is

== )(

H

dd pp (8.6.23)

which is a function of effective stress only. The inverse of this expression will be

( ) ==pp d (8.6.24)

Work Hardening

The hardening rule 8.6.21 describes how the yield surface evolves. It is a functionof the effective plastic strain, hence the term strain hardening. An alternative

procedure to describe the hardening process is to plot stress, not against plastic strain,

but against the plastic work. Directly from Fig. 8.6.1, by evaluating the area beneaththe stress plastic strain curve, one can obtain the plot shown in Fig. 8.6.8. Here, the

stress is expressed in the form

( ) ( )pp dwWw == (8.6.25)

The flow curve for arbitrary loading conditions is then ( ) pp dwWw == . Eqn.8.6.25 is called a work hardening rule.

Figure 8.6.8: uniaxial stress plastic work curve (for a typical metal)

.When the effective stress and effective plastic strain are defined using Eqns. 8.6.13-

8.6.14, then pp ddW = and the strain hardening and work hardening rules areequivalent. In that case the plastic modulus is

( )pp

p

ppdW

d

d

dW

dW

d

d

dH

=== (8.6.26)

0

Y

ppdW =

pdW

d

( )pWw=

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Section 8.6

Solid Mechanics Part II 228 Kelly

8.6.4 Application: Combined Tension/Torsion of a thinwalled tube with Isotropic Hardening

Consider again the thin-walled tube, now brought to the point of yield through tension

and then subjected to a twist whilst maintaining the axial stress constant, at the initialtensile yield stress. The Prandtl-Reuss equations in terms of effective stress andeffective plastic strain, 8.6.19, reduce to

xy

p

xyxy

xx

p

xxzzyy

xx

p

xxxx

dd

E

d

dd

Edd

dd

Ed

2

31

2

1

1

++

=

==

+=

(8.6.27)

Maintaining xx at a value 0Y and introducing the plastic modulus 8.6.22,

xyxyxy

zzyy

xx

d

Hd

Ed

Yd

Hdd

Yd

Hd

1

2

31

1

2

1

1

0

0

++

=

==

=

(8.6.28)

Using the terminology of Eqn. 8.6.8, the Von Mises condition is

3,3

3

1,0),(

22

0

YkYFkFf =+=== (8.6.29)

and the effective stress is 220 33 +== YF . The expansion of the yield surface

is shown in Fig. 8.6.9 (see Fig. 8.3.2).

Figure 8.6.9: expansion of the yield locus for a thin-walled tube under isotropic

hardening

3/0Y

0Y

plastic

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Section 8.6

Solid Mechanics Part II 229 Kelly

Thus

3/

1

2

31

3/21

3/

2

0

2

2

2

0

20

2

0

2

0

Y

d

Hd

Ed

Yd

HYdd

Y

d

H

Yd

xy

zzyy

xx

++

+=

+==

+=

(8.6.30)

These equations can now be integrated. If the material is linear hardening, so Hisconstant, then they can be integrated exactly using

( )

=

++=

+ ax

axdxax

xaxdx

ax

xarctan,ln

2

122

222

22(8.6.31)

+

+=

+==

++=

0000

2

0

2

00

2

0

2

0

3arctan

3

1

2

3)1(

31ln4

31ln2

11

YYH

E

YY

E

YH

E

Y

E

Y

E

YH

E

Y

E

xy

zzyy

xx

(8.6.32)

Results are presented in Fig. 8.6.10 for the case of 10/,3.0 == HE . The axial

strain grows logarithmically and is eventually dominated by the faster-growing shear

strain.

Figure 8.6.10: Stress-strain curves for thin-walled tube with isotropic linear

strain hardening

0

2

4

6

8

0.2 0.4 0.6 0.8 1

0Y

0Y

E

xx

xy

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Section 8.6

Solid Mechanics Part II 230 Kelly

8.6.5 Kinematic and Mixed Hardening

In the above, hardening rules have been discussed and used for the case of isotropichardening. In kinematic hardening, the yield surface translates in stress-space, in

which case Eqn. 8.6.8 take the general form

0)(),,( == kFkf ijijijij (8.6.33)

The stress ij is known as the back-stress; the yield surface is shifted relative to the

stress-space axes by ij , Fig. 8.6.11.

Figure 8.6.11: kinematic hardening; a shift by the back-stress

There are many hardening rules which define how the back stress depends on

development of plastic strain. The simplest is the linear kinematic (orPragers)hardening rule,

p

ijij

p

ijij cddc == or (8.6.34)

where c is a material constant. Thus the yield surface is translated in the same

direction as the plastic strain increment. This is illustrated in Fig. 8.6.12, where the

principal directions of stress and plastic strain are superimposed.

Figure 8.6.12: Linear kinematic hardening rule

pd 11,

p

d 22 ,

pd

pcdd =

1

2

initial yield

surfacesubsequent

surface

ij

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Section 8.6

Solid Mechanics Part II 231 Kelly

Zieglers hardening rule is

( )( )ijijpijij dad = (8.6.35)

where a is some scalar function of the plastic strain, for example

p

dda =

, wherep is the effective plastic strain and is a material constant. Here, then, the loading

function translates in the direction of ijij , Fig. 8.6.13.

Figure 8.6.13: Zieglers kinematic hardening rule

When there is a combination of isotropic and kinematic hardening, then the hardeningrule will be of the form

( ) 0)( == pijij kFf (8.6.36)

8.6.6 The Consistency Condition

It has been seen that the loading function can in general be expressed in the form

0),( =ijf (8.6.37)

where represents one or more hardening parameters, which are zero when there

is no plastic loading. For example, in isotropic hardening, 8.6.37 can be written in the

form 8.6.8 through

( ) 0)(

)(),(

=+=

=

YF

kFkf

ij

ijij(8.6.38)

Alternatively, for kinematic hardening, the hardening parameter is related to the ij in

8.6.33 (see 8.8). There are two hardening parameters in the mixed hardening rule

8.6.36. The hardening parameters themselves depend on other variables, for examplethe plastic strain.

The increment in fcan now be described by

1

2

d

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Section 8.6

Solid Mechanics Part II 232 Kelly

df

df

df ijij

+

= (8.6.39)

The second term here is zero when there is no plastic straining or perfect plasticity.

When there is plastic deformation, then, for the stress to remain on the yield surface,i.e. for the yield criterion to remain satisfied, one must satisfy the followingconsistency condition:

0=

+

=

d

fd

fdf ij

ij

(8.6.40)

Thus the stress state and also the hardening parameters change to ensure the yieldcriterion remains satisfied.

8.6.7 Problems

1. Use the general formula 8.6.9, nij CF )( = , to derive an expression for theDrucker-Prager materials effective stress, Eqn. 8.6.12.

2. Derive Eqns. 8.6.20,

2

3 pdd =

3. Integrate Eqns. 8.6.30 and use the initial (first yield) conditions to get Eqns.8.6.32.

4. Consider the combined tension-torsion of a thin-walled cylindrical tube. The tubeis made of an isotropic hardening Von Mises metal with uniaxial yield stress 0Y .

The strain-hardening is linear with plastic modulus H. The tube is loaded,

keeping the ratio 3/ = at all times throughout the elasto-plastic deformation(i) Show that the stresses and strains at first yield are given by

E

Yv

E

YYY

Y

xy

Y

xx

YY 0000

6

1,

2

1,

6

1,

2

1 +====

(ii) Use the hardening rule 8.6.18 to express the Prandtl-Reuss equations8.6.18 in terms of effective stress and only. Eliminate using3/ = .

(iii) Eliminate the effective stress to obtain

dH

dE

d

dH

dE

d

xy

xx

1

2

31

3

1

11

++

=

+=

(iv) Solve the differential equations and evaluate any constants of integration(v) Hence, show that the strains at the final stress values

0Y= , 3/

0Y=

are given by

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Section 8.6

Solid Mechanics Part II 233 Kelly

+

+=

+=

2

11

2

3

3

1

2

111

0

0

H

E

Y

E

H

E

Y

E

xy

xx