NUMERICAL IMPLEMENTATION OF PRAGER’S KINEMATIC HARDENING MODEL IN EXACTLY INTEGRATED

FORM FOR ELASTO-PLASTIC ANALYSIS

L,iszu5 SZAB~ and ADS KovAcs

Department of Technical Mechanics, Technical University of Budapest, H-l 111 Mtiegyetcm rkp. 5, Budapest, Hungary

(Receiued 12 November 1986)

Abstract-An exact solution of the Prandtl-Reuss equation for Prager’s kinematic hardening model is described in the paper. As compared with numerical integration methods, the method described permits stress calculations lo be made quickly and accurately in the case of solution of time-independent elastic-plastic problems using the finite element method. The FORTRAN code for the case of plane stress and plane strain as well as for axisymmetrical problems is given.

NOTATION

L 1, 1, 0, 0, 01 Young’s modulus Poisson’s ratio shear modulus, El[2(1 + v)] bulk modulus, E/[3( 1 - 2v)] increment transpose time va%or norm, (rT*r)“2 yield stress radius of the Mises cylinder, fi*, slope of 0 vs L* curve differentiation with respect to time

1. INTRODlJCl7ON

Increasing the accuracy of methods to determine the state of stress in solving time-independent elastic- plastic problems using the !inite element method is still a current task. Among the different methods available, the subincremental techniques [1, 21 are basic.

Several new methods have been recommended in recent years for numerical integration of the consti- tutive equations of small elasto-plastic strain such as the fourth-order RungoKutta method [3], the t: method [4], the ‘return mapping’ algorithm [S], the ‘particular state determination’ algorithm [5], etc. In addition to numerical methods, exact solutions have been described by several authors, e.g. to the Prandtl- Reuss equation for the elastic-ideally plastic case [I,

to constitutive equations in the strain space [8], and to Prager’s kinematic hardening model [9].

This work presents the numerical implementation of an exact solution published by Xucheng and Liangming [9] for the Prager kinematic hardening model. After a brief description of the model the FORTRAN code of plane and axisymmetrical problems is given. This subroutine can be used in the elasto-plastic finite element program given in [2] without any difficulty. Finally some simple numerical examples are presented to show the accuracy and other characteristics of the method.

2. THEORY

In case of the Prandtl-Reuss theory the consti- tutive equation in deviator form can be written as

&=2.G.e- 9-G’

a;.(3G + H’) s’aT*e (1)

where, if the kinematic hardening model is used,

a=s-a. (2)

For the Praeger model, the velocity of ‘back stress’ vector a is defined by

3G *a*+ &(3G + H’) ‘* (3)

With the angle

B

between vectors a and &, the differential equation for @, given by Refs [7-91, takes the following form:

(5)

815

816 L&zL~ Szad and hrju Kovics

Plane strain

In case of plane strain: E, = 0 and As, = 0. Conse- quently the strain and the strain increment vectors are

eT = k,, &y1 Y,/Jz 01

AeT = [A-G, A+ Ar,/fi, 01

W

(8b)

respectively, and the stress vector is

bT = [e,, ov, fi*rxy, a,]

where in plastic state:

(9)

Fig. 1. Plane view of stress states. The ‘back stress’ vector is given as:

The solution of differential eqn (5) can be written as aT = [a,, ayr fi*axy, 41. (10)

j(r) = 2 tan-’ (exp[ - F]tan(/?,/*)), Plane stress

In the case of plane stress: 6, = 0 and Au, = 0. Thus

1, < t B t,+, (6) the stress vector is

where /3, is the angle in the instant t, and is assumed constant.

UT = [e=, by, JLXY, 01 (11)

As shown in Fig. 1, a(t) can be expressed with the and the strain and strain increment vectors are aid of simple trigonometry. Substituting the result in eqn (3) and following the procedure given in Ref. [9], sT = [e,, s,, r,lJX &,I Wa) a(t) can be. calculated. On the basis of Ref. [‘lb-after necessary corrections-the quantities associated with AeT = [As,, AeY, AY,/& A&,1 (Qb)

the instant t,+] can be calculated as follows:

ai+, =p+a,+Z.g*G de

a ,+1=ai+y[(l -p)a,+2+b.GAe]

s I+1 = a,+] + ai+l

8 ,+, = si+) + K.(iT.A6)i + i(i’+u,)i

2 00 .- &P+, =eP+3 3G +H, k

respectively.

(7a) The strain component E, can be calculated from the relationship

0)

(7c) e, = - f (u* + UJ - (&,p + EyP). (13)

(7d) Since the strain component Aa, contains the plastic part as well, it cannot be directly calculated from the displacement field. Therefore, in the case of plane

(7e) stress the first step is the calculation of de, using the

where quantities p, g, y, b and k are given in Table 1. Using eqns (7), quantities u,+, , a,+], cP+, can be determined from known quantities u,, a,, sp associ-

Table 1.

l-exp(-2m)+[l-exp(-m)]z~cos~, &?== cm

ated with instant t, and from strain increment As associated with time increment At. 6-1 l+m_2+WexP(-m)lW7,

m [ C 1 c=l+cos~,+(l-cos~,)~exp(-2m)

3. IMPLEMENTATION AND APPLICATION

Now the application of the theory described in the previous section to the plane stress, plane strain and to the axisymmetric case is given.

2*cxp(-m) P” k=m+ln f

0

H’ Y=3C

Implemen~tion of Prager’s kinematic hardening model 817

condition of (o,)~+~ = 0 by iterative solution of the for cyclic shear the other for combined tension and transcendent equation torsion of thin-walled tubes.

F(Ae,) = h),+1+ @Ji+l + K-(i’*Ae)

+:@*a,)=0 (14)

where

iT=[i, l,O, 11. (15)

Note that good results were achieved after six iterations.

Example 1

The first example is the uniaxial tension in case of plane stress state. The exact strain increments were determined from relationships

and

W)

As;= -A (As, + As,) (16)

was assumed as the initial value. Now from eqns (7), the quantities u,+,, a,+, and

EP+, can be calculated from vector As already known.

Axisymmetric case

AsY=(-;-$,~-Aez. (fgb)

For the value Au x = 10 MPa assumed arbitrarily, the strain increment vector with material characteristics specified in Table 2 will be

For the axisymmetric case, the strain, strain increment and stress vectors are

AsT =L: [5.5555, -2.6666, 0.0, 0.01 x IO-‘. (19)

Calculation is made from the limit of plastic yield for 10 steps. The initial values are

(l7a)

(17b)

(174

Four numerical examples were solved to show the accuracy and characteristics of the theory and the FORTRAN subroutine is presented in the Appendix. The first two examples include the calculation of uniaxial tension for the state of plane stress and plane strain selected to test code. The third and the fourth examples contain finite element calculations, one

a: = [200.0, 0.0, 0.0, O.OJ MPa (20)

8; = [t/g, -l/30, 0.0, -i/30] X lo-’ (21)

q=O and sg=O.O. (22)

Values u, calculated by STRINC subroutine as a function of sX are illustrated in Fig. 2. In Fig. 2 the

Table 2.

E = 1.8 x 10’MPa E,==O.l8 x IO-‘MPa H’=2 x IO’MPa

v = 0.3 un = 200 MPa

-2 x IO

0 00a 0 16 024 032 040 048 0 56 064 300- I I I I I I I I

0 006 0.16 024 032 040 0.46 056 064

Epsilon-x x Id’

Fig. 2. Uniaxial stress-strain curve.

818 Lk~6 SZAEU~ and .&DAM KovAcs

x 16‘ 300° 004 008 0 12 0 16 020 024 028 0 32

I I I I I

-L t- +-+ 250

t- t-

t-

o

!k +-+- - 200

0 004 008 0 I2 0 16 020 0 24 0 28 032 (

Equwalent plastic strain x 10e2

36

Fig. 3. Equivalent uniaxial stress vs equivalent uniaxial plastic strain.

analytical values are indicated by the continuous line

while the calculated values are indicated by ’ + ‘, The slope of the straight line along the plastic section is ET. The calculated AE: is equal to -2.666 x lo-‘.

Example 2

In the case of plane strain state, the strain increments are determined from the following relationships:

a,,=(!$ +-$Ae, (23a)

A.sg= -($ +&,)~Ao,. (23b)

As in the first example, the strain increment vector associated with Au, = 10 MPa will be

As* = [4.2555, -3.9666, 0.0, 0.01 x lo-‘. (24)

Initial values used in the calculation are:

uz = [230.9401, 0.0, 0.0, 115.471 MPa (25a)

E; = Il.1376, -0.48754, 0.0, 0.01 x lo-’ (25b)

p~g=O and sg=O. (25c)

The relationship between calculated equivalent stress and equivalent plastic strain (C) is shown in Fig. 3.

x 10-I -008 -006 -004 -002 0 002 004 006 008

Gamma-xy x 16’

Fig. 4. Cyclic shear stress-strain curve.

Implementation of Prager’s kinematic hardening model 819

Z&Y /

_A .

/

\ l B

0. .- C Q,

Q” -

(b)

Fig. 5. Tension-torsion test.

The calculated values are indicated by ‘ + ’ while the slope of the continuous line is identical with H’.

Example 3

This example presents detail on the solution of cyclic shear using the finite element method. Calcu-

lations were made according to the loading program (rA = 145.4734 Mpa, ~a = 135.4735 MPa) shown in Fig. 4. The material characteristics are given in Table 2. One isoparametric element with eight nodes has been used in finite element calculation. The numerical model is shown in Fig. 5b (u, = 0) while the results of analytical and finite element calculations are given in Fig. 4.

Example 4

This example shows the finite element calculation of a thin-walled tube under combined tension and torsion for the non-proportional load path [3,6] in Fig. 6 and Table 3. The numerical model is shown in Fig. 5 while the material characteristics are listed in Table 4. Calculation was made for four load steps along section AB while for three load steps along section BC. Figure 6 shows the calculated u, - E, diagram while Fig. 7 shows the calculated ?XY -3 relationship. Broken lines along section AB and BC are the reference results in [3,6] obtained by 80 load steps.

Table 3.

e,(MPa) r,(MPa)

A 151.2 93.1

: 251.2 53.1 259.2 0.0

Table 4.

E= 1.95 x IO’MPa H’ = 1.949 x IO' MPa

v = 0.3 u, = 181 MPa

x 10-I 3oQ” 000 016 0.24 032 0.40 048 056 064

I I I I I I I I

0 006 016 024 032 0.40 040 0.56 064

Epsilon-x x IO-’

Fig. 6. Axial stress-strain response.

820 Llsz~6 Sue6 and ADAM KovAcs

x 10.’

0 004 008 012 0 16 0 20 0 24 0 28 0 32 120% I I I I I I I I

--- - Reference sol&Ion 2 60-. 0 z I-

40 ‘-

b I I I I I I

0 004 008 0 12 0 16 0 20 0 24 0 28 032 C

Epsilon-xy x IO-’

Fig. 7. Shear stress-strain response.

36

4. CONCLUSION

In this work, adaptation of the results published

in [9] to finite element calculations has been shown. By means of the recommended method the stress increment can be calculated in an exact way in cases of proportional and non-proportional load using the Prandtl-Reuss equations with the Prager kinematic hardening model. The method described needs a shorter running time as compared with the sub- increment techniques.

REFERENCES

I. J. M. M. C. Marques, Stress computation in elasto- plasticity. Engng Cornput. 1, 42-51 (1984).

2. D. R. J. Owen and E. Hinton, Finire Elements in Plusricify. McGraw-Hill, New York (1980).

3.

4.

5.

6.

7.

8.

9.

J. W. Wissmann and C. Hauck, Efficient elastic-plastic finite element analysis with higher order stress point algorithms. Compu;. Strut. 17,89-95 (1983). _ S. E. Groves. D. H. Allen and W. E. Haisler. An efficient and accurate alternative to subincremenmtion for elastic-plastic analysis by the finite element method. Compur. S~rucf. 20, 1021-1031 (1985). J. C. Simo and R. L. Taylor, A return mapping algorithm for plane stress elastoplasticity. Inl. J. Numer. Meth. Engng 22, 649-670 (1986). C. Nyssen, An efficient and accurate iterative method, allowing large incremental steps, to solve elasto-plastic problems. Compuf. Strucf. 13, 63-71 (1981). R. D. Krieg and D. B. Krieg, Accuracies of numerical solution methods for elastic-perfectly plastic model. Trans. ASME. J. Pressure Vessel Technol. 99. 510-515 (1977). P. J. Yoder and R. G. Whirley, On the numerical implementation of elastoplastic models. Trans. ASME, J. Appl. Mech. 51, 283-288 (1984). W. Xucheng and C. Liangming, Exact integration of constitutive equations of kinematic hardening material and its extended applications. SMiRT-8 Proc. Paper L 2/3, BNS&S (1985).

Implementation of Prager’s kinematic hardening model 821

APPENDIX

On the basis of relationships (7), the STRING subroutine calculates quantities u,+,, E,+, and sr+, associated with known strain increment As from the known values of u,, a, and E:, respectively.

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C4 4

c4 SUBROUTINE STRING 4

c4 ~str~~~,strrin,~lfr,~p~tn,YOUN6,POlSS,HARDS,FO,ntyp~I 4

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C 4

C THIS SUBROUTINE EVALUATES STRESS ,BACKSTRESS RND EDUIVALENT 4

C 4

C PLASTIC STRAIN IN PRMER'S KINENATIC HARDENING RULE 4

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Parrmctrrst

mtrcsslnl

strain(n)

______ strrrm vector (n-4) input 8 stress I th stmp outputt stress i+l th step

stress(l)- siqmr -x strrss(ZI= siqm -y stress(S)- fl4 tru 'XY

strcsr(4)= siqar -t

------ total strain incrcmnt wctor l n = 4 1 input

rlfr(nI

l pstn ----

YOUNG ----

POISS ----

HARDS ----

FS ____

strain(l)= delta epsilon -I strain(Z)= delta epsilon -y strain(l)= (dwltr qrmr -ny)/ fi strrinO)= delta epsilon -t

------ backstress vector (n=4I input I rlfr i th step output1 rlfr i+l th step

rIfr(11= rlfr -I rIfa(Z)= rIfr(5)= I/? 4 :::: -tl,

alfrl4)- rlfr -7.

equivalent plastic strain input : rqv.plrst.strrin 1 th step output8 cqv.plrst.fitrrin itI th step

Younq modulus

Poisson ratio

Plastic hrrdrninq modulus

Yield strrss

ntype ---- Problem typr prrrrrtmr n 1 plane stress n 2 plane strain =5 axisyrrrtrlc

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822 Lisz~6 Sud and .&AM Kovics

/(C*DtmZ))/H

el=mrp(-Hl cZ=mrp(-2.tHl C=l.+comfc D=l.-cosfc HS=(l.*H-2.m(C-rltcomfc) P=2.trl/(C+t2tD) BA=(1.-m2+~1.-el)t~l.-rl if(ntypr.nm.1) go to 26 DBE=B+mtrnd(4) CK=strrin(4) mtrrin~4~=-3m~~P*6Rt~l,-P~l+mvmct~4lt~P4B~tB~tHBl4DBEtE~/BK if1rbm(CK-strr1n~4t1.gt.6.60066611 60 to I6

26 do 36 i=l,4 rIfr~il=rlfr~il*BNt~~l.-Pl+mvmct~i~tHBmBtmtrnd~ill l vrct~it=Pt~rvrct~i~~BRmBmstrnd~il~~mlfm~i~

50 continur if(ntypm.rq.2) rlfr~4~=6.6 DR=BK+DE*DS rtremm~l)=mvmct(ll*DA mtrrma(2)=svrct(2)+DA mtrrmm(5)=svrct~31 if(ntypm.mq.1) l trmmm(4)-0.6 if(ntypm.eq.2) rtrram~4l=B.5tlmtrrmm~l~tmtrmss~2~~ if(ntypr.mq.31 mtrmmmOl=mvrctl4l*DA rpstn=rpstn*mqrt~l.SlmRBm~H*rloq~~Ctm2mD~/2.~~/~l.SmBtH~RDB~ return l nd