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