08 plasticity 06 strain hardening
TRANSCRIPT

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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 onedimensional (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 perfectlyplastic
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 strainhardening 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 perfectlyplastic case, where there is one of plastic
deformation or elastic unloading.
Figure 8.6.1: uniaxial stressstrain 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 strainhardening 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,
consider first the related topic of the loading function.
The Loading Function
The yield surface is in general described by a function of the form
strainhardening
0
A
Bstress
strain
Yield point Y perfectlyplastic
elastic
unload

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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
loading and (elastic) unloading are then defined through:
0 n d loading
As in 8.3.1920, 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 6dimensional stress space,
ij
ij
df
(8.6.4)
As mentioned above, neutral loading does not occur for the perfectly plastic material.
In this case, the criteria for loading and unloading can be expressed as
0 n d , where fand n now refer to the new loading function.
f
n
dd
d
loading
unloading
neutral loading

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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 stressstraincurve 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 stressstrain curve for a strainsoftening material
8.6.2 Changes in the Loading Function
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
dnew loading
surfaceinitial yield
surface
f
nd

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Section 8.6
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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
loading
plastic
deformation
(hardening)
elastic
unloading
initial yield
surface
subsequent
yield surface
1
2
stress atinitial yield
elastic
loading
elastic
unloading
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 strainhardening 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 multiaxial case, one needs again a flow curve, of the form 8.6.6, but one whichrelates a complex threedimensional 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
general 3D stress state. Since the loading function determines whether additional
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
Yperfectlyplastic
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 DruckerPrager 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 multiaxial 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
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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 nonrigorous. 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 PrandtlReuss 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. 8.2.710)
( ) ( ) ( )[ ]
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
PrandtlReuss work expression 8.6.16. It will be shown in the next section that this is
no coincidence, and that the PrandtlReuss flowrule is indeed naturally associated
with the VonMises criterion.
PrandtlReuss 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 PrandtlReuss (LevyMises) 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 thinwalled 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 PrandtlReuss 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 thinwalled tube under isotropic
hardening
3/0Y
0Y
plastic
loading

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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)
leading to {Problem 3}
+
+=
+==
++=
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 fastergrowing shear
strain.
Figure 8.6.10: Stressstrain curves for thinwalled 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
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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 stressspace, in
which case Eqn. 8.6.8 take the general form
0)(),,( == kFkf ijijijij (8.6.33)
The stress ij is known as the backstress; the yield surface is shifted relative to the
stressspace axes by ij , Fig. 8.6.11.
Figure 8.6.11: kinematic hardening; a shift by the backstress
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
loading
surface
ij

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Section 8.6
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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 theDruckerPrager 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 tensiontorsion of a thinwalled cylindrical tube. The tubeis made of an isotropic hardening Von Mises metal with uniaxial yield stress 0Y .
The strainhardening is linear with plastic modulus H. The tube is loaded,
keeping the ratio 3/ = at all times throughout the elastoplastic 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 PrandtlReuss 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
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+
+=
+=
2
11
2
3
3
1
2
111
0
0
H
E
Y
E
H
E
Y
E
xy
xx