Thermo- Elasto-plastic Constitutive Equations for Ductile Material and Its Finite Element Implementation- Hani Aziz Ameen

8/6/2019 Thermo- Elasto-plastic Constitutive Equations for Ductile Material and Its Finite Element Implementation- Hani Aziz A

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THERMO-ELASTO-PLASTIC CONSTITUTIVE EQUATIONS FOR

DUCTILE MATERIAL AND ITS FINITE ELEMENT

IMPLEMENTATION

Asst. Prof. Dr. Hani Aziz AmeenDies and Tools Engineering Department

Technical College,

Baghdad,Iraq.

E-mail: [email protected]

ABSTRACTThermo-elasto-plastic constitutive model for ductile material undergoing thermo

mechanical deformation is proposed. The model is based on the assumption that the total

strain is decomposable into strain components due to elastic deformation, temperature

dependent material properties, thermal strain, and plastic strain. The model is consists of

linear elastic model in series with a plasticity model. This plasticity model adopts the VonMises criterion with associated flow rule of the theory of plasticity and isotropic,

kinematic, and mixed hardening rules. However the proposed model acquires the

advantage of having both the initial and subsequently yield surface to be a function oftemperature. A three dimensional finite element algorithm is developed to implement this

constitutive equation. This algorithm adopts the incremental approach. Two examples are

performed to demonstrate the used of the constitutive relation for thermo-elasto-Plastic.

Results show that essential features in the stress-strain diagram obtained experimentallyare exhibited by the model.

Keywords: Finite element method, ductile material, thermo- elasto-plastic

NOMENCLATUREd Total incremental strain vector

)(ed Incremental strain vector due to elastic deformation

)(dmd Incremental strain vector due to thermal change in material properties

)(Td Incremental thermal stain vector

)(pd Incremental plastic strain vector

G Modulus of rigidityF Yield surface

International Electronic Engineering Mathematical Society IEEMS

http://www.ieems.org

In collaboration withInstitute for Mathematics, Bio-informatics, Information-technology and Computer-science IMBIC

International e-Journal ofEngineering Mathematics: Theory and Application

http://www.ieems.org/iejemta.htm

ISSN 1687-6156

Volume (10), March, 2011, pp. 39-50


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Now, from Hookes law [1])(

d][e

Dd ,where [D] is the elasticity matrix [6],

dTdT

.)( , the flow rule is [6] a.

)(dd

p , where Fa / , and it can be proved

that dT)/][( 1)( TDd dm .Substitute all these relations into Eq.(l), it can be deduced that:

adDdTT

DdTDdDd ][)

][]([][

1

(2)

(i) Isotropic Hardening Rule

For isotropic hardening rule, the yield surface is function of

),,( TKFF (3)

A similar postulation but without thermal effect was proposed by Zienkiewicz et al.[7].

Differentiating F in equation (3) by the chain rule, it can be get:

0d)(

)(

dTT

Fd

k

k

F

FdF

p

t

p

t

(4)

Where ),()( p

kfK

By substituting Eq.(2) into Eq.(4), and rearranged to get:

dTT

FdT

T

DdTdDa

d

t

i)

][]([

1 1

)((5)

Where a

k

k

FaDai

p

t

)(][)(

Substitute equation (5) into equation (2), it can be get:

dTTF

aDdT

TDdTDepdDepd

i

iii

)(

1)()()( ][)][(][][ (6)

Where)()()( ][][][ ipi DDDep (7)

][][1

][)(

)()( DaaD

Dt

i

ip

We will find that equation (6) is identical to what was given in Yamada et al [2] without

the thermal effects. To analyze the constitutive equation (6) with the Von Mises criterion,firstly the gradient vector a must be found as follows:

yields surface for Von Mises is yJF 23

Where

])()()[(6

1 213

2

32

2

212 J

Thus

M

J

J

F

Fa a ].[

12

2

(8)

Where


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300000

030000

003000

00015.05.0

0005.015.0

0005.05.01

][ aM

And the value of GaDat

3][ (9)

Now to find

t

p

k

k

F

)(

We have JF 23

Thus

k

kF

(10)

And the work done = )(. pddkHence

dk

d p 1)( (11)

Therefore

H

k

k

p

p

1)(

)(

Where

)( pddH

Now the value of

M

HG a

ti ][3)(

, it can be proved that 2][ M at , thus

HG i 3)(

After finding the value of )(i the value of )()( ][ ipD is:

tip SSGH

GD

))3/(1(

3][

2

)()(

(12)

Where

}{ xzyzxyzyx SSSS

mmmxx S zzyy Sand,S,

Hence)()()( ][][][ ipiep DDD

Now, equation (6) will be:)()()()()( FdmTMi

ddddd (13)

Where

dDd iepM )()( ][

dTDd iepT

][ )()(


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dTT

DDd iep

dm ][][

1)()(

dTTF

aDd

iF ][

)()(

(ii) Kinematic Hardening RuleFor kinematics hardening rule the yield surface is:

),,( TFF

The total differentiation of F is

0

dT

T

Fd

Fd

FdF

tt

(14)

And it can be proved that

F

F

F

r

Where

r From Prager shift [8]

adCCddp

)(

Where

),( HC

Thus equation (14) will be

dT

T

FCdadF

t

)ad( (15)

From equations (2) and (15), it can be deduced that

dTT

FdT

T

DdTa

d

t

i)

][[D](d

1 1

)((16)

Where

)]([)( aCaDatk

Substitute equation (16) into equation (2), gives:

dTT

Fa

DdT

T

DdTDdDd

k

kk

ep

k

ep

k

)(

)(1

)()()( ][)][

(][][ (17)

Where)()()( ][][][ kpkep DDD (18)

])[]([1

][)()()()( DaaD

D

ktkkp

And

k

)(

To analyze equation (17), firstly find)(k

a , similar as in equation (8), hence,

)(

)(

)(][

1 kak

kM

a


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Where )(k is the effective stress of)(k

and the terms ( )().( ktaa will be

)(

2

222)((3

12

3

).(

k

zxyzxyt

aa

Hence

)(

2

222

)()((3

12

33

k

zxyzxyk

CG

After finding the value of )(k the value of)()( ][ kpD is

)(

2

)()(

)()( )(31

][ kt

kk

kp SS

G

D

Hence)()()( ][][][ kpkep DDD , Eq.(17) will be

)()()()()( FdmTMkddddd (19)

Where

dDd kepM )()( ][

dTDd kepT

][ )()(

dTT

DDd kep

dm ][][

(k)1

)()(

dTT

F

aDd

k

kF ][

)(

)()(

(iii) Mixed Hardening RuleThe yield surface for mixed hardening rule is

),,,()(TFF

p and the loading criterion may be written:

)(),,()()( pi hTFf

Where, )(ih is a function which governs the isotropic expansion or contraction of the yieldsurface. Hence the rate of plastic strain is now simply split into two components as [3].

))(())(()( ipkppddd (21)

Where)())(( pkp

Mdd (22)

And )())(()1(

pipdMd (23)

Where M


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0)1( )()(

)()()()(

dT

T

Fd

d

dhadaMCdadf p

p

ikktkt

(25)

Substitute equation (2) into equation (25), it can be get:

)(

)(

)()(

1)(

)()

][]([

1 pp

ikkt

md

d

dhdT

T

FdT

T

DdTdDa

d (26)

Where

))1(][)()()()()( kktkktm aaMCaDa

Substitute equation (26) into equation (2) it can be get

)(

)(

)(

)(

)(

)(

)(1

)()()(

][][

)][

(][][

p

p

i

m

k

k

km

ep

m

ep

m

d

d

dhaDdT

T

Fa

D

dTT

DdTDdDd

(27)

Where)()()( ][][][ mpmep DDD (28)

And

])[]([1

][)()(

)(

)()( DaaD

Dktk

m

mp

We will find that equation (27) is identical to what was found in Axelsson [3] without the

thermal effects. Hence, to analyze equation (27), we have:)()( kp

add 5.0

)()()(

.)(3

2

ptpp

ddd

)())((.

pipdMd , equation (26) will be

5.0)()(

))((

)()()()(

1)(

)3

2())1(][

)(][

]([

kkt

ip

kktkkt

kt

aad

dhMaaMCaDa

dTT

FdTk

T

DdTdDa

d

(29)

If we defined the abdomen of equation (29) as Y instead of )(m as in equation (26), it can

be get

5.0)()(

))((

)()()()( )

3

2())1(][ kkt

ip

kktkkt aa

d

dhMaaMCaDaY

Simplified the above equation, it can be get

5.0

)(2

)(222

)(

2

222

)((312

)((31

2

)1(33

k

k

zxyzxy

r

k

zxyzxy

HM

MCGY

Where

Hd

d

d

dhrip

rrip

22))(())((


8/12

46

And)(

k

r

Hence)(

2

)(

)()( ).(31

][ kt

k

mp SS

G

YD

And)()()( ][][][ mpmep DDD

It can be get)()()()()( FdmTMm

ddddd (30)

Where:

dDd mepM )()( ][

dTDdm

ep

T

][)()(

dTT

DDd mep

dm ][][

(k)1

)()(

dTT

F

Y

aDd

kF ][

)()(

3. Finite Element AnalysisA finite element algorithm is developed to implement the constitutive equation derived in

the previous section Adopting, the displacement approach [9], the general equilibriumequation is

v

t BR dv][ (31)

Where R is the rate of the nodal force vectors and [B] is the strain - displacement matrix

[10] , if is the nodal displacement vector then

B ].[ (32)

It should be mentioned that for simplicity, we have applied the integration v

)dv( over the

whole region. In practice the integration is carried out element by element using the

standard assembly rule [11]. Substitute the constitutive equation (13, 19, or 30 for

isotropic, kinematics, or mixed hardening rule respectively).into equation (31), it can be

get:

FKep ][ (33)

Where

v

ep

t

ep dvBDBK ]][[][][ (34)

TFRF (35)

mfR

And

v

ep

ep

t

T dvTT

F

aDDT

T

DTDBF

][][][][][

11


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4. FEM Solution Procedure(1) Select thermal and mechanical load increments.

(2) Perform thermal analysis following Ref[12] and calculate nodal temperature, T(t)

at time t(3) Select material properties based on the average element temperature = (T1+T2+T3+ +T20)/20

(4) Determine H' for isotropic hardening rule and C for Kinematic hardening rule and

H' and, C for mixed hardening ruleWhere

)/(1'

EE

EH

t

t

And

),(' HC

WhereEt from [13] as follow

nnn

k

n

k

EEE

E

E

E

EEEt

EE

Et/)1(

1

))/'(1(1

'))/(1(1

GE 3 and

E

E

EE

)21(3

3

And is determined for each load increment

(5) Compute [B], following [10]

(6) Form the elasto -plastic matrix [Dep] and from equation (7) or (18) or (28) forisotropic, kinematics or mixed hardening rule respectively.

(7) Evaluated the element stiffness matrix [Kep] equation (34)

(8) Evaluated the element force matrix according to equation (35)

(9) Assemble the overall stiffness matrix [Kg] and construct the overall structural

equilibrium equations FKg ][

(10) Modify F for applicable boundary condition

(11) Adjust [Kg] corresponding to step (10)

(12) Solve for by skyline solver and therefore the total displacement components

by (13) Compute from equation (32)

(14) Compute from equation (13), (19) or (30) for isotropic, kinematics or mixed

hardening rule respectively.

(15) Compute the total element stresses and strains by and

(16) Check for convergence, the error in displacement increment is used to check the

accuracy 0001.0.

.

Er

t

t

(17)After checking, updating the stresses, strains, displacements, then go to the next load

increment step (1).


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5. Numerical ExamplesTwo problems are selected for the study of the model, the numerical solution presented

the three dimensional elasto-plastic and thermo- elasto plastic problems using Von Mises

yield criterion and isotropic hardening rule.

(1) Thick Circular Aluminum RingThe first example is the thick circular aluminum ring. Six element with 20-node

hexahedral is used. Because of the symmetry of the problem numerical computation is

confined to one quarter (Fig(1)). Fig(2) shows the variation of radial displacement ofinternal surface with increasing load. A good agreement between the present finite

element method and the experimental results obtained by Ref[14] are evident. The

material properties are as follows:

E = 85917.232 MPa

323.096.188

y MPa

H=3158.6 MPainternal pressure (P) = 3.4, 6.8, 10.3, 13.7, 20.6, 27.5, 31 MPa

(2) A gear Tooth

Fig(3) shows a gear tooth that is subjected to a line load bx acting in the x-direction at itsupper edge and to thermal load h1 and h2 (convective heat transfer coefficient). This

problem is symmetric with respect to the x-z plane, and the tooth is assumed to be fixed

at its base Half of the problem is discretized into a network of four hexahedral 20-node

element ,as shown in figure (4), [9],[12].

Figure (3) Gear tooth subjected to Figure (4) FE-mesh for the half

Thermo-mechanical load gear shown in Figure (3)


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Values of structural parameters are as follows:

k= 51.9 W/(m.K), Cp= 468.89 J/(Kg.K), 3/35.7887 mKg

Table (1): Properties of material [4]T (K) E (MPa) E' (MPa) )(k (MPa) n

294 199955 4137 310275 10

477 180649 2183.64 258562.5 10

588 177891 1637.56 241325 10

Figure (5) show the elasto- plastic zone for a gear tooth due to the applying of the thermal

and mechanical load.

Figure (5): Elasto-plastic zone in a gear tooth due to thermo-mechanical load

CONCLUSIONThis paper has demonstrated an efficient computational model for thermo-elasto-plastic

analysis of three dimensional problems. This model introduced the general constitutive

relation which can be applied to a particular real material, and it sensitive to thetemperature history. A finite element concept for thermo-elasto plastic analysis has been

suggested and used to study the three dimensional problems of a Von Mises material and

obeying the present stress-strain relations. A computer program has been written to test

the theory, the efficiency of the program could be improved by adoption of the skylinesolution and published results give a reasonable agreement with the obtained results.

REFERENCE(1) S. P. Timoshenko and J. N. Goodier, Theory of elasticity", 3

rdedn, McGraw-Hill,

1970.

(2) Y. Yamada, N. Yoshimura & T. Sakurai "Plastic stress -strain matrix and its

application for the solution of elastic plastic problem by FEM", Int. J. Mech. Sci. (10),

343-354,1968(3) K. Axelsson and A. Samnelsson, Finite element analysis of elastic-plastic

material displaying mixed hardening", Int. J. Num. Eng., Vol (14), 211-225, 1979

(4) T. R. Hsu, "The finite element method in thermo-mechanics, Boston, Allen &

Unwin, 1986(5) C. S. Desai & J.F. Abel, Introduction to the finite element method", Van

Nostrand, Reinhold, NewYork, 1972

(6) O. C. Zienkiewicz, "The finite element method in engineering science, Mc

Graw-Hill, New York, lst

edn,1967(7) O. C. Zienkiewicz, S. Valliappan and I .P. King, Elasto-plastic solutions of

engineering problems 'initial stress', finite element approach", Int. J. for Num. Meth. Eng.


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Vol.(1),75-100,1969

(8) H. Ziegler, A modification of Pragers hardening rule", Quar. of App. Math.

Vol.(17) ,55-65,1959

(9) H. A. Ameen, FEM for elasto-plastic stress analysis of two dimensionalproblems. M. Sc. thesis, Al-Nahreen university, 1994(10) H. A. Ameen Derivation of lagrangian shape functions for hexahedral element.

IeJAAEM , Vol 1, Jan, 2011. (http://www.ieems.org)

(11) S. S. Rao, The finite element method in engineering, 2 edn, pergamon press,

1989(12) H.Tawfiq and H.A.Ameen, Three dimensional heat transfer finite element

analysis with steady state, transient and phase change, Engineering and Technology

Journal, Vol.19, No.5, 2000

(13) R. M. Richard & J. R. Blacklock, " Finite element analysis of inelastic structuresAIAA. J. (7) (3), pp.432-438,1969

(14) D.R.J. Owen and E.S.Salonen "Three-dimensional elasto-plastic finite elementAnalysis", Int. J. for .Num. Meth. Eng., Vol (9), 209-218, 1975.

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Thermo- Elasto-plastic Constitutive Equations for Ductile Material and Its Finite Element Implementation- Hani Aziz Ameen