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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

    deformation or elastic unloading.

    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,

    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

    strain-hardening

    0

    A

    Bstress

    strain

    Yield point Y perfectly-plastic

    elastic

    unload

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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

    loading and (elastic) unloading are then defined through:

    0 n d loading

    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.

    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 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

    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

    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

    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 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

    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

    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

    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 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

    loading

    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