Isotropic plasticity models 61

Based on these comments it is evident that no universal method exists that can be used with the many alternatives which can occur in practice. In the next several sections we illustrate some formulations which employ the alternatives we have discussed above.

3.5 Isotropic plasticity models We consider here some simple cases for isotropic plasticity-type models in which both a yield function and a flow rule are used. For an isotropic material linear elastic response may be expressed by moduli defined with two parameters. Here we shall assume these to be the bulk and shear moduli, as used previously in the viscoelastic section (Sec. 3.2). Accordingly, the stress at any discrete time z,,+~ is computed from elastic strains in matrix form as % + I = P , ~ + I ~ + s=~K+mIm & + l + 2 G ( I - ; m m - c:+

-!&+I -

T

e

)%+I

(3.93)

1)

where the elastic modulus matrix for an isotropic material is given in the simple form

D

-=

and

I is the 9 x

KmmT

+ 2 G (I f m e T ) -

(3.94)

9 identity matrix and m is the nine-component matrix m=[1

1

1

o o o o o 0 1 ~

Using Eqs (3.42) and (3.43) immediately reduces the above to D

= KmmT

+ 2 G ( I o $ mmT) -

(3.95)

The above relation yields the stress at the current time provided we know the current total strain and the current plastic strain values. The total strain is available from the finite element equations using the current value of nodal displacements, and the plastic strain is assumed to be computed with use of one of the algorithms given above. In the discussion to follow we consider relations for various classical yield surfaces.

3.5.1 Isotropic yield surfaces The general procedures outlined in the previous section allow determination of the tangent matrices for almost any yield surface applicable in practice. For an isotropic material all functions can be represented in terms of the three stress invariants:*

ZI 2J2

= Ojf = mTg T

2

=~ q ~= j ;5 5 = 151

353 = S i j S j k S k i

(3.96)

= dets

where we can observe that definition of all the invariants is most easily performed in indicia1 notation. * Appendix A presents a summary of invariants and their derivatives

62

Inelastic and non-linear materials

One useful form of these invariants for use in yield functions is given by43 3a,,

=I,

(3.97)

)

JlJ3 3 d 2a

30=si n-’(-

with

- 6 0 6 - 7T 6 6

- 7l

Using these definitions the surface for several classical yield conditions can be given as:

1. Tresca: F=~@cosO Y (-K ) = O

(3.98)

2. Huber-von Mises: (3.99) Both conditions 1 and 2 are well verified in metal plasticity. For soils, concrete and other ‘frictional’ materials the Mohr-Coulomb or Drucker-Prager surfaces is frequently used.52 3. Mohr-Coulomb: 1

.

F = o,,,sin4 + 5 cos0 - -sin4sin0

L

3

)

-

c cos4 = 0

(3.100)

where C ( K ) and 4 ( ~are ) the cohesion and the angle of friction, respectively, which can depend on an isotropic strain hardening parameter K . 4. Drucker-Prager: F = 3 c y ’ ( ~ )+ o ~@ ~- K ( K )= 0

(3.101)

where I

CY=

and again c and

2 sin4

J5(3

-

sin 4 )

K=

6 cos 4 &(3 - sin 4 )

4 can depend on a strain hardening parameter.

These forms lead to a convenient definition of the gradients F,, or Q,,, irrespective of whether the surface is used as a yield condition or a flow potential. Thus we can always write (3.102) and upon noting that

Isotropic plasticity models 63 Table 3.1 Invariant derivatives for various yield conditions Yield condition

Fc,,,

&F,J?

Tresca

0

2 cos 8( I

Huber-von Mises

0

J5

Mohr-Coulomb

sin @

Drucker-Prager

3n'

J#J,

J5 sin 8

+ tan 8 tan 38)

cos 38

]

I . -cos8 1 = tan 8sin 38 + - sin &(tan38 - tan 0) 2I & [ 1

0

&sinB+sindcosO 2 cos 38 0

Alternatively, we can always write: F,g

= F.oJ,8do,, t F.Jz a J 2 -t ~

80

80

F,J3

(3.104)

dJ3 80

~

which can be put into a matrix form as shown in Appendix A. The values of the three derivatives with respect to the invariants are shown in Table 3.1 for the various yield surfaces mentioned. The form of the various yield surfaces given above is shown with respect to the principal stress space in Fig. 3.8, though many more elaborate ones have been developed, particularly for soil (geomechanics) problems.53P55

Fig. 3.8 Isotropic yield surfaces in principal stress space: (a) Drucker-Prager and von Mises; (b) Mohr-Coulomb and Trexa.

isotropic and kinematic hardening (Prandtl-Reuss equations) ~ ~ ~ - " ~ - ~~-~ 3.5.2

jZmodel with

~

"

~

~

" 1

As noted in Table 3.1 a particularly simple form results if we assume the yield function involves only the second invariant of the deviatoric stresses J 2 . Here we present a more detailed discussion of results obtained by using an associated form and the return map algorithm. Since the yield function involves deviatoric quantities only we can initially make all the calculations in terms of these. Accordingly, the elastic deviatoric stress-strain relation is given as 3 = 2Gge = 2 G ( e

-

eP)

(3.105)

~

.

-

-

-

"

64 Inelastic and non-linear materials

Continuum rate form Before constructing the return map solution we first consider the form of the plasticity equations in rate form for this simple model. The plastic deviatoric strain rates are deduced from (3.106) Including the effects of isotropic and kinematic hardening the Huber-von Mises yield function may be expressed as F = I S - K I - V / 5 2 Y(K ) = o

(3.107)

in which 5 are back stresses from kinematic hardening and K is an isotropic hardening parameter. We assume linear isotropic hardening given by* Y ( K )= Yo Here a rate of as

K

+H

L ~

(3.108)

is computed from a norm of the plastic strains, by using Eq. (3.53),

&

i=

(3.109)

in which the factor is introduced to match uniaxial behaviour given by Eq. (3.108). On differentiation of F it will be found that (3.1 10) Using the above, the plastic strains are given by e p = /in

(3.1 11)

and, when substituted into a rate-of-stress relation, yield

S = 2 G [i-);E]

(3.112)

A rate form for the kinematic hardening is taken as

3

(3.113)

k=nT($-&)- $ H ; / i

(3.1 14)

K

= Hk);fi

The rate of the yield function becomes

and when combined with the other rate equations gives the expression for the plastic consistency parameter as (noting that with the nine-component form nTg= 1) (3.115) * More general forms of hardening may be approximated by piecewise linear segments, thus making the present formulation quite general.

Isotropic plasticity models 65

where (3.1 16)

G*=G+i(Hi+Hk)

Substitution of Eq. (3.1 15) into Eq. (3.1 12), and using Eq. (3.42) to reduce to the sixcomponent form, gives the rate form for stress-strain deviators as

:

[

S = 2 G Io--nn

e

(3.1 17)

We note that for perfect plasticity Hi = Hk = 0 leading to 2G/G* = 1 and, thus, the elastic-plastic tangent for this special case is also here obtained. Use of Eq. (3.117) in the rate form of Eq. (3.93) gives the final continuum elasticplastic tangent

D& = KmmT

+ 2G

[ i

Io - -mmT

-

(3.1 18)

This then establishes the well-known Prandtl-Reuss stress-strain relations generalized for linear isotropic and kinematic hardening.

Incremental return map form The return map form for the equations is established by using a backward (Euler implicit) difference form as described previously (see Sec. 3.4.2). Omitting the subscript on the n + 1 quantities the plastic strain equation becomes, using Eqs (3.106) and (3.110), e p = g!

-

+ AAn

(3.119)

and the accumulated (effective) plastic strain = Kn

+ &ax

(3.120)

Thus, now the discrete constitutive equation is = 2G(g

gP)

(3.121)

H~AAIJ

(3.122)

&-iHiax

(3.123)

-

the kinematic hardening is K

=

+3

and the yield function is F = I S - ~ I-

where Y, = Yo + The trial stress, which establishes whether plastic behaviour occurs, is given by

J2/3~,.

gT" = 2G(g - e:)

(3.124)

which for situations where plasticity occurs permits the final stress to be given as s- = -sTR- 2GAXn

(3.125)

66

Inelastic and non-linear materials

Using the definition of relations as

n, we may now combine the stress and kinematic hardening (3.126)

and noting from this that we must have nTR= n

(3.127)

we may solve the yield function directly for the consistency parameter as14,34 AX =

1s'"

-

&I

Y,

-

(3.128)

2 G*

where G* is given by Eq. (3.1 16). We can also easily establish the relations for the consistent tangent matrix for this J2 model. From Eqs (3.121) and (3.1 19) we obtain the incremental expression ds = 2 G [de - dX - AX dn]

(3.129)

The increment of relation (3.127) givesi4

d n = dnTR= 2 G I-nn'] s - KI [- --

I-

de

(3.130)

-

and from Eq. (3.128) we have G dX = -n' de G* -

(3.131)

Substitution into Eq. (3.129) gives the consistent tangent matrix 2GAX

Is'"

-

&I

G*

1s'"

(3.132)

-IC,/

This may now be expressed in terms of the total strains, combined with the elastic volumetric term and reduced to six-components to give D&=KmmT+2G

2GAX ) I o -

Is'" - & I

(52GAX G* IsTR-&I

)nnT]

(3.133)

We here note also that when AX = 0 the tangent for the return map becomes the continuum tangent, thus establishing consistency of form.

3.5.3 J2 plane stress The discussion in the previous part of this section may be applied to solve problems in plane strain, axisymmetry, and general three-dimensional behaviour. In plane strain and axisymmetric problems it is only necessary to note that some strain components are zero. For problems in plane stress, however, it is necessary to modify the algorithm to achieve an efficient solution process. In a plane stress process only the four stresses a,, cry, ry, and T , . ~need be considered. When considering deviatoric

Isotropic plasticity models 67

components, however, there are five components, s, s,., s,, s,~ and sy. The deviators may be expressed in terms of the independent stresses as

(3.134)

The Huber-von Mises yield function may be written as

F

=

[(a- I C ) ~PT P,T(a - IC)]”*-@ Y ( K d ) 0

(3.135)

Expanding Eq. (3.135) gives the plane stress yield function

F

=

1.5(O