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A new element for analyzing large deformation of thin Naghdi shellmodel. Part II: Plastic
Aazam Ghassemi ⇑, Alireza Shahidi, Mahmoud FarzinDepartment of Mechanical Engineering, Najafabad Branch, Islamic Azad University, Isfahan, Iran
a r t i c l e i n f o
Article history:Received 17 August 2009Received in revised form 12 October 2010Accepted 15 November 2010Available online 3 December 2010
Keywords:Large deformationThin Cosserat shellPlasticConstrained director
a b s t r a c t
In this paper a new element is developed that is based on Cosserat theory. In the finite ele-ment implementation of Cosserat theory shear locking can occur, especially for very thinshells. In the present investigation the director vector is constrained to remain perpendic-ular to the mid surface during deformation. It will be shown that this constraint yieldsaccurate results in very large deformation of thin shells also the rate of convergency is verygood. For plastic formulation, the model introduced by Simo is used and it has beenreduced for constrained director vector and the consistent elasto-plastic tangent moduliis extracted for finite element solution. This model includes both kinematic and isotropichardening. For numerical investigations an isoparametric nine node element is employedthen by linearization of the principle of virtual work, material and geometric stiffnessmatrices are extracted. The validity and the accuracy of the proposed element is illustratedby the numerical examples and the results are compared with those available in theliterature.
� 2010 Elsevier Inc. All rights reserved.
1. Introduction
Large deformation analysis of shells is usually studied using two different approaches:
– three-dimensional theory;– direct method or Cosserat theory in which a director is assigned to a non-euclidean plane [1–4].
Similarly in numerical analysis of large deformation of shells and plates two methods have also been employed.For three-dimensional theory, a three-dimensional modified element was developed by Ahamd [5]. Other researchers like
Hughes and Liu [6] and Hughes and Carnoy [7] developed this element. This element is recorded in standard finite elementtext books like, Bathe [8], Hughes [9] and Blytschko [10]. Another approach for this theory was developed, namely, co-rota-tional method. Numerical formulation of this approach was firstly presented by Wempner [11] whose article introducesco-rotational finite element in nonlinear analysis of shells. The work of Argyris [12] can also be mentioned along thisapproach. This approach is suitable for large deformation with small strains. Some examples of this approach are worksof Parish [13], Beuchter et al. [14], Sansour et al. [15], Peng et al [16], Jiang et al. [17] and Liu et al. [18].
Another theory is direct method. This method is one of the best theories for modeling large deformation of shells. Thistheory was presented by Cosserat for first time and further elaborated upon by a number of authors such as Naghdi [19],
0307-904X/$ - see front matter � 2010 Elsevier Inc. All rights reserved.doi:10.1016/j.apm.2010.11.029
⇑ Corresponding author.E-mail addresses: [email protected] (A. Ghassemi), [email protected] (A. Shahidi), [email protected] (M. Farzin).
Applied Mathematical Modelling 35 (2011) 2650–2668
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journal homepage: www.elsevier .com/locate /apm
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Antman [20]. The basic assumption of this theory is that the mid surface of the shell is regarded as an inextensible one-direc-tor surface. Typically this approach yields an exact analytical definition of the initial geometry of the shell and the represen-tation of the stress and strain state in curvilinear coordinates and stresses are entirely in term of stress resultants and stresscouples.
Green and Naghdi [21] derived a general form of constitutive equation for an elastic perfectly plastic material with atten-tion to thermo dynamical constraints.
When dealing with stress resultants, it is of course important to be able to identify a yield surface which remarks the lim-iting values of the stress resultants. Ilyushin [22] derived an exact form of the yield surface for a linear elastic, perfectly plas-tic isotropic material which obeys Vonmisses yield criterion. Crisfeild [23] improved this criterion by adding a pseudohardening effect due to progression of yielding across the shell’s thickness. Simo [24–26] extended Ilyushin criterion foran isotropic and kinematic hardening materials also he developed finite element formulation of this theory.
Since the Ilyushin criterion is a multifunction and it’s surface has corner, several researchers such as Mohammed andSkallerud [27] modified Ilyushin ’s criterion to one function.
In the above works, shear deformations in the direction of the thickness are taken into account. In these analyses, as thethickness approaches zero, their numerical analyses mostly experience shear locking.
According to the well known Kirichhoff’s hypothesis, straight lines perpendicular to the mid-surface remain perpendic-ular to the deformed mid-surface. This hypothesis yields satisfactory results only when the thickness approaches zero andthe deformation is not large. This hypothesis can lead to numerical difficulties, if used for large deformations. However it willbe shown in this paper that by employing Cosserat’s surface and constraining the director vector to remain perpendicular tomid surface during deformation, very good results can be obtained for large deformation of thin plates and shells withoutany locking. This constraint is in fact a limiting analysis of the Cosserat theory in which Kirichhoff’s hypothesis is enforcedand hence the shear strains in the direction of the shell’s thickness are ignored. For plasticity solution, the model extended bySimo [26], is used and it is modified for a constrained director surface also the consistent elasto-plastic tangent moduli isextracted for this modified surface.
Using principle of virtual work and linearization process stiffness matrices are extracted. For numerical solution a 9 nodeisoparametric element has been used.
The outline of this paper is as follows. In Section 2 the theory is explained. In this section the algorithm of return mappingfor plastic solution is extracted and the elasto-plastic tangent moduli is derived for numerical solution. In Section 3 finiteelement scheme for solution a constrained Cosserat shell is developed. In this section by using virtual work, the geometricand material stiffness matrices are derived through a linearization process. In Section 4 several numerical examples are pre-sented and the results are compared with literature. Finally conclusions are drawn in Section 5.
2. Theory
In this section the stress and strain vectors have been illustrated and then elasto-plastic constitutive equations are pre-sented. The return mapping algorithm is explained and finally elasto-plastic tangent moduli is derived for presented model.
2.1. Kinamatic relations
Fig. 1 shows geometry of a three-dimensional shell with a mid surface (M). On the mid surface the convective coordinatesystem h1, h2 is considered which has the base vectors a1, a2 and a3 which is orthogonal to a1 and a2. The position vector ofany point with respect to O is [28]:
R ¼ rðh1; h2Þ þ h3a3: ð1Þ
Fig. 2 shows the mid surface of an arbitrary shell in equilibrium states before and after deformation (t = 0, t respectively). Inthis figure x, y, z represent reference Cartesian coordinate system and h1, h2 are the convective coordinate system.
3a
O
3θ
rR
S
M
2/h−
2/h
Fig. 1. Geometry of a three-dimensional shell.
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The base vectors of convective coordinate system, in initial configuration are denoted by 0ai. Similarly, tai denotes basevectors of convective coordinate system at time t. It should be noted that the director vector is constrained to be perpen-dicular to the mid surface at each time, so ta3 = td. The Position vector of a material point, which is a function of h1 and h2,is:
trðh1; h2Þ ¼txðh1; h2Þtyðh1; h2Þtzðh1; h2Þ
264375: ð2Þ
The base vectors can be written as:
taa ¼ tr;a ¼ @t r
@ha
ta3 ¼ ta3 ¼t a1�t a2kt a1�t a2k
): ð3Þ
Components of the first and the second fundamental tensors of the surface and also components of membrane and bendingstrains are written as [1]:
taab ¼ tx;a � tx;b;tbab ¼ �ta3;a:tab ¼ ta3:taa;b;
t0eab ¼
12
tx;a:tx;b � 0x;a:0x;b� �
;
tjab ¼ ta3:tx;ab ¼1ffiffiffiffiffitap ½tx;ab:tx;1 � tx;2�;
t0qq ¼ tjab � 0jab:
ð4Þ
In the above relations, lower left subscripts in the strain components denote reference configuration. Let computationalstrain vector, 0te, according to
t0e ¼ t0e11; t0e22;2t0e12; t0q11; t0q22;2t0q12
� �T: ð5Þ
Such that 0te contains elastic and plastic parts and it can be decomposed into:
t0 _e ¼ t0 _e
e þ t0 _ep: ð6Þ
In this formulation 0tep is the plastic part of the strain and it will be explained as follow.
2.2. Stress resultants and stress couples
In Cosserat theory, membrane and bending stresses are defined in terms of stress resultants in the direction of thickness[1]. Fig. 3 shows the effective Cauchy stresses at a material point of the deformed configuration and also the effective sym-metric Piola stresses corresponding to the Cauchy stresses.
In the above figure, ttnab, ttmab are membrane stresses and bending moments per unit length in the deformed configura-tion, respectively. The invariant forms of these stresses are:
x
y
z
E1
E2
E3
da 030 =
10a
20a
1θ
2θ
2θ 1at
2at
da tt =3
1θ
Fig. 2. Equilibrium state of a quadrilateral plate at times zero and t.
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ttnn ¼ ttn
ab taa � tab;ttq ¼ ttq
a taa;ttm ¼ ttm
ab taa � tab:
9>=>;: ð7ÞSimilarly 0tnab, 0tmab are membrane stresses and bending moments per unit length in the reference configuration, respec-tively. The invariant forms of these stresses are:
t0n ¼ t0nab 0aa � 0ab;t0m ¼ t0mab 0aa � 0ab:
): ð8Þ
These stresses are related to each other according to the following relations:
abab ¼ Jttnab:abab ¼ Jttmab;
): ð9Þ
where J ¼ dt Sd0S is the transformation Jacobean, namely, the ratio between the element surface after and before deformation.For a convective coordinate the Jacobean term can be written as:
J ¼ detðFÞ where Fji ¼tai � 0aj: ð10Þ
In this formulation F is the deformation gradient tensor.For simplicity computational stress vector, 0tr, has been defined according to:
t0r ¼ t0n
11; t0n
22; t0n
12; t0m
11; t0m
22; t0m
12D E
: ð11Þ
2.3. Constitutive equations
In the previous sections the stress and strain vectors were determined. Because the stress components are in terms ofstress resultants and stress couples, the constitutive equations must be formulated directly according them.
The generalized Ilyushin–Shapiro elasto-plastic model is entirely in terms of stress resultants and stress couples. It shouldbe noted that this yield surface is multifunction.
2.3.1. Elasto-plastic constitutive equationsFor an elasto-plastic material which the elastic region is linear we can write [24]:
t0 _r ¼ C t0 _e � t0 _e
p� �; ð12Þ
where 0tep is the plastic part of the strain and:
C ¼Cm 00 Cb
� �; ð13Þ
Cm ¼Eh
ð1� m2Þ
ð0a11Þ2 mð0a11Þð0a22Þ þ ð1� mÞð0a12Þ2 ð0a11Þð0a12Þð0a11Þ2 ð0a22Þð0a12Þ
sym 1þm2 ð0a12Þ2 þ 1�m2 ð0a11Þð0a22Þ
26643775;
Cb ¼h2
12Cm:
To define the plastic part of the strain, the yield function must be defined. In general for an elasto-plastic material that itsyield surface is one function, the yield function can be defined as follow:
1at2at 12ntt
11ntt
21mtt
120nt
110nt
1
0a2
0a
210mt
220mt
22mttFig. 3. Effective Cauchy and Piola stresses at a material point of the body.
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f = f(0tn, 0tm,Ps,U0) where Ps, s = 1,2, . . .n characterizing the hardening response and U0 is the character of initialconfiguration.
If the associative flow rule is considered then:t0 _e
p ¼ _c @f@t0r
and a general hardening law as _Ps ¼ _chsðt0n; t0m;ps;U0Þ.
In the above equations _c P 0 is the plastic consistency parameter; a function satisfying Kuhn–Tucker complementaryconditions:
_c P 0; f 6 0; _c f ¼ _c _f ¼ 0: ð14ÞThe above condition must be satisfied through plasticity model.
According to Simo [26] the set of Ps, s = 1,2, . . .n is supplemented by a conjugate set of internal variables as, s = 1,2, . . .nthrough the transformation:
P =rH(a) = �Da, where H(a) is hardening potential and for simplicity, it will be assumed that it is strictly quadratic(HðaÞ ¼ 12 aT DaÞ such that D 2 R
n � Rn is constant.According to Simo [26], the hardening relation can be written as:
P ¼p
P
� �¼ �
DaD�a
� �where D ¼ k0
k0and D ¼ 2
3H0I6 ðI6 is the unique matrixÞ: ð15Þ
The constants k0, k0, H0 are yield parameters. k0 is the uniaxial yield stress, k0 is the linear isotropic hardening moduli and H0 isthe kinematic hardening moduli. Variables a 2 R and �a 2 R6 are associative with isotropic and kinematic hardening of theyield surface, respectively.
For a yield surface with multiple functions, the elastic domain can be defined as follow:
Xr ¼ ðr;PÞ 2 R6 � R6jflðr;PÞ < 0n o
for all l 2 f1;2; . . . ;mg; ð16Þ
@Xr ¼ ðr;PÞ 2 R6 � RP jflðr;PÞ ¼ 0n o
for all l 2 f1;2; . . . ;mg; ð17Þ
where oXr is the boundary of the yield surface.The functions fl(r,P) are smooth functions which are assumed to define independent constraints at any (r,P) 2 oXr and
may intersect in a nonsmooth fashion. The closure Xr [ oXr is assumed to be a closed convex set.If the associated flow rule is used, according to Koiter rule, the plastic components of strains and hardening characters can
be written as below:
_ep ¼Xql¼1
_cl@fl@r
; ð18Þ
_a ¼Xql¼1
_cl@Pflðr;P; pÞ; ð19Þ
where q = {l 2mjfl = 0} (active surfaces).
2.3.2. Generalized Ilyushin–Shapiro elastoplastic modelLet the back stress ‘‘�P’’, then the yield function is determined as:
flðrþ P;pÞ ¼ ulðrþ PÞ �j2ðpÞj20
6 0; l 2 f1;2g: ð20Þ
In the above formulation:
ulðrþ PÞ :¼ ðrþ PÞT Alðrþ PÞ; ð21Þ
Such that:
Al ¼1
n20p signðlÞ
2ffiffi3p
n0m0p
signðlÞ2ffiffi3p
n0m0p 1
m20p
24 35; ð22Þwhere signðlÞ :¼ þ1; if l ¼ 1�1; if l ¼ 2
and p :¼
1 � 12 0� 12 1 00 0 3
24 35.In above relations m0 and n0 are the yield parameters associated with membrane and bending response respectively.
These yield parameters are typically related to the uniaxial yield parameter j0 through the relations:n0 = hj0 and m0 ¼ h
2
4 k0 where h is the shell thickness.Also
jðpÞ ¼ j0 þ j0p; ð23Þ
which defines the radius of the yield surface.
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If the associated flow rule is used, according to Koiter rule, the plastic components of strains can be written as below:
_ep ¼ _cX
l2f1;2g
@fl@r
: ð24Þ
So from Eq. (20) we have:
_ep ¼X
l2f1;2g
_cl2Alðrþ PÞ: ð25Þ
The parameters p and P according to the generalized Ilyushin–Shapiro yield function are determined as below:
_a ¼Xml¼1
_cl@pflðr;P; pÞ ! _a ¼X
l2f1;2g
_cl�2j0jðpÞ
j20; ð26Þ
_�a ¼Xml¼1
_cl@�pflðr;P; pÞ ! _�a ¼X
l2f1;2g
_cl2Alðrþ PÞ: ð27Þ
Then we have:
p ¼ �Da ¼ �j0j0
a and P ¼ �D0�a ¼ �23
H0I6�a ðI6 is the unique matrixÞ: ð28Þ
For solving the illustrated yield function and finding stresses and plastic strains an iteration step must be done. This iterationprocess is discussed in return mapping algorithm.
2.3.3. Return mapping algorithmThe algorithm has the standard geometric interpretation of a closest-point-projection, in the energy norm, of a trial state
onto the elastic domain. Because the illustrated surface has corner, the active surface must be defined during return to theyield surface.
2.3.3.1. Discerete algorithmic problem. Let current mid-surface of the shell is known. Consider a time disceretization of theinterval [0,T] � R of interest. We assume that the variables en; epn;an; �an are known. Let DUn+1, Dtn+1 be a given incrementin the displacement and time on the interval t 2 [tn, tn+1]. Then the variables en; epn;an; �an must be updated to enþ1; epnþ1;anþ1; �anþ1 at tn+1 2 [tn,T]. To this end, application of an implicit, backward Euler difference scheme leads to the following non-linear coupled system:
epnþ1 ¼ epn þXql¼1
clnþ1@rflðr;P; pÞnþ1 ¼ epn þXql¼1
clnþ12Alðrþ PÞnþ1;
�anþ1 ¼ �an þXql¼1
clnþ1@Pflðr;P;pÞnþ1;
rnþ1 ¼ Cðenþ1 � epnþ1Þ;
Pnþ1 ¼ �D0�anþ1 ¼ �23
H0�anþ1;
anþ1 ¼ an þXql¼1
clnþ1@pflðr;P;pÞnþ1;
p ¼ �Danþ1 ¼ �j0j0
anþ1;
ð29Þ
where we have set clnþ1 ¼ Dt _clnþ1.
Because the use of convective coordinates, it does not need to objective rates. It is considerable that the discrete coun-terpart of the Kuhn–Tucker loading/unloading takes the following form:
clnþ1 P 0; f lðr;P;pÞnþ1 6 0 and clnþ1flðr;P; pÞnþ1 ¼ 0 ðno sum on lÞ l 2 f1;2g: ð30Þ
Convexity of the yield surface is guaranteed by a positive definite Al(l 2 {1,2}) and consequently the solution is unique.The trial state in the interval t 2 [tn, tn+1] can be determined as below:At the first stage set:
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eptrial
nþ1 :¼ epn; �atrialnþ1 :¼ �an; atrialnþ1 ¼ an;
eetrial
nþ1 :¼ enþ1 � epn;
rtrialnþ1 :¼ C enþ1 � epn� �
;
Ptrialnþ1 :¼ �23
H0atrialnþ1; ptrialnþ1 ¼ �
j0j0
atrialnþ1;
f triall;nþ1 :¼ fl rtrialnþ1;Ptrialnþ1; p� �
:
ð31Þ
As the notion of Simo [26], the yield function, flðr;P; pÞnþ1, can be expressed in terms of the consistency parametersc1nþ1; c2nþ1. By this, the return mapping reduces to solution of the following nonlinear system:
flðrnþ1 þ Pnþ1; pnþ1Þ ¼ �f lðc1nþ1; c2nþ1Þ ¼ 0l 2 f1; 2g as follow:At the first stage it is obvious that:
rtrialnþ1 ¼ rnþ1 þ CDepnþ1 ¼ rnþ1 þ C
Xl2f1;2g
clnþ12Alðrþ PÞnþ1; ð32Þ
Ptrialnþ1 ¼ Pnþ1 þ23
H0D�anþ1 ¼ Pnþ1 þ23
H0X
l2f1;2gclnþ12Alðrþ PÞnþ1: ð33Þ
Define parameter g as:
gnþ1 ¼ rnþ1 þ Pnþ1 ¼gngm
nþ1
; ð34Þ
where rnþ1 ¼nm
nþ1
and Pnþ1 ¼ PnPm
nþ1
and gtrialnþ1 ¼ rtrialnþ1 þ Ptrialnþ1.
By this definition, from Eqs. (32)–(34) we have:
gtrialnþ1 ¼ I6 þXcl
nþ1
43
H0Al þ 2CAl� �0@ 1Agnþ1: ð35Þ
From Eqs. (13) and (22)1:
CAl ¼1
n20Cnp
signðlÞ2ffiffi3p
n0m0Cnp
signðlÞ2ffiffi3p
n0m0Cmp 1m20
Cmp
24 35: ð36ÞFor simplicity consider orthogonal and diagonal matrix Q as:
Q ¼1 1 0�1 1 00 0
ffiffiffi2p
264375:
Matrices p and C can be rewritten as:
p ¼ QKpQ T and C ¼ QKCQT : ð37Þ
where Kp and KC can be computed as below:
Kp :¼
32 0 00 12 00 0 3
264375 and KC :¼
E1þm 0 0
0 E1�m 00 0 E2ð1þmÞ
264375: ð38Þ
Such that p and C are commute; i.e., pC ¼ CpLet nn :¼ QTgn and nm :¼ QTgm, so from Eq. (35) we have:
C1 C2C3 C4
� �nn
nm
� �nþ1¼
ntrialn
ntrialm
" #nþ1
; ð39Þ
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where:
C1ðc1; c2Þnþ1 :¼ I3 þ ðc1 þ c2Þnþ12n20
23
H0Kp þ hKCKp� �
¼ I3 þ cpnþ1 w1;
C2ðc1; c2Þnþ1 :¼ ðc1 � c2Þnþ12ffiffiffi
3p
n0m0
23
H0Kp þ hKCKp� �
¼ cmnþ1 w2;
C3ðc1; c2Þnþ1 :¼ ðc1 � c2Þnþ12ffiffiffi
3p
n0m0
23
H0Kp þh3
12KCKp
" #¼ cmnþ1 w3;
C4ðc1; c2Þnþ1 :¼ I3 þ ðc1 þ c2Þnþ12
m20
23
H0Kp þh3
12KCKp
" #¼ I3 þ cpnþ1 w4;
ð40Þ
where cpnþ1 ¼ ðc1 þ c2Þnþ1 and cmnþ1 ¼ ðc
1 � c2Þnþ1From Eq. (39) we have:
nn
nm
� �nþ1¼
C1 C2C3 C4
� ��1ntrialn
ntrialm
" #nþ1
: ð41Þ
With these results, first part of Eq. (20) can be rewritten as:
ul;nþ1 ¼nn
nm
� �nþ1
Xlnn
nm
� �nþ1
where Xl :¼1
n20Kp
signðlÞ2ffiffi3p
n0m0Kp
signðlÞ2ffiffi3p
n0m0Kp 1m20
Kp
24 35: ð42ÞBy replacing Eq. (41) into (42) we have:
ul;nþ1 :¼ntrialn
ntrialm
" #Tnþ1
C1 C2C3 C4
� ��TXl
C1 C2C3 C4
� ��1ntrialn
ntrialm
" #nþ1
: ð43Þ
By defining Hðc1; c2Þ :¼ C1 C2C3 C4
� ��1we have:
ul;nþ1 :¼ntrialn
ntrialm
" #Tnþ1
HTXlHntrialn
ntrialm
" #nþ1
; ð44Þ
where the above equation is only function of c1nþ1 and c2nþ1.Similarly, the second term of the yield function can be written in terms of c1nþ1 and c2nþ1 that is discussed as follow.According to the given yield function we have:
jðpÞnþ1 ¼ j0ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiulðrþ PÞnþ1
qso ĵðc1; c2Þnþ1 ¼ j0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiûlðc1; c2Þnþ1
qð45Þ
And from Eq. (29)5 we have:
anþ1 ¼ an þX
l2f1;2gcl�2j0jðpÞ
j20
� �so pnþ1 ¼ pn þ
Xl2f1;2g
clnþ12jðpÞj0
: ð46Þ
So from Eq. (45) we can write:
p̂ðc1; c2Þnþ1 ¼ pn þX
l2f1;2gclnþ12
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiûlðc1; c2Þnþ1
qand ĵðc1; c2Þnþ1 ¼ j0 þ j0p̂ðc1; c2Þnþ1: ð47Þ
By Eqs. (44) and (47) the yield function can be written completely as a function of c1nþ1 and c2nþ1:
f̂ lðc1; c2Þnþ1 ¼ ûlðc1; c2Þnþ1 �ĵðc1; c2Þnþ1
j20¼ 0; l ¼ 1;2: ð48Þ
Because f̂ l;nþ1 monotonically decrease with clnþ1, for increasing hardening laws, Eq. (48) has a unique solutionclnþ1 P 0; c
lnþ1 P 0.
For determination of c1nþ1 and c2nþ1 the Eq. (48) must be solved. These equations are nonlinear in terms of c1nþ1 and c2nþ1.For solving Eq. (48) and for finding c1nþ1 and c2nþ1 Newton Raphson algorithm is used. So the term
@ f̂l@cb b 2 f1;2g must be
computed.By differentiating of Eq. (46) with respect to cb we have:
@ul@cb¼ 2
ntrialn
ntrialm
" #Tnþ1
HTXl@H@cb
ntrialn
ntrialm
" #nþ1
; ð49Þ
where Hðc1; c2Þ :¼ C1 C2C3 C4
� ��1. Computation of @H
@cb has been discussed in Appendix A.
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Also for Newton Raphson iteration to find c1nþ1 and c2nþ1;@k̂ðc1 ;c2Þnþ1
@cl must be determined that is computed at follow.If active surface is single (l = 1 or l = 2) rename k = {ljfl,n+1 > 0} then
p̂ðc1; c2Þnþ1 ¼ pn þ 2cknþ1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiûkðc1; c2Þnþ1
qand
@ĵ@ck¼ j0 2
@ûk;nþ1@ckffiffiffiffiffiffiffiffiffiffiffiffiffiûk;nþ1
p þ 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiûk;nþ1q0@ 1A: ð50Þ
And if two surfaces are active, then
p̂ðc1; c2Þnþ1 ¼ pn þXl¼1;2
clnþ12ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiûlðc1; c2Þnþ1
q: ð51Þ
So@ĵ@cb¼ j0
Xl¼1;2
clnþ12@ûl;nþ1@cbffiffiffiffiffiffiffiffiffiffiffiffiffiffiûl;nþ1
p þ 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiûb;nþ1q0@ 1A: ð52Þ
By this for one active surface:
@ f̂ l@ck¼ @ûl@ck� 2ĵ
j20
@ĵ@ck
l ¼ k; ð53Þ
Dck ¼ �fk;k@fk;k@ck
�;
Dcv ¼ 0 v – k;
(ð54Þ
And for two active surfaces we have:
@ f̂ l@cb¼ @ûl@cb� 2ĵ
j20
@ĵ@cb
; ð55Þ
Dc1
Dc2
( )¼�f1k�f2k
� � @f1@c1
@f1@c2
@f2@c1
@f2@c2
24 35�1: ð56ÞNow, the algorithm of return mapping can be written as below:
1. Compute rtrialnþ1 ¼ Cðenþ1 � epnÞ; ptrialnþ1 ¼ pn; Ptrialnþ1 ¼ Pnþ1 and f triall;nþ1 for l = 1, 2 from Eq. (20).
2. If f triall;nþ1 6 0 (for l = 1,2) then we are in elastic phase and set ð. . . Þnþ1 ¼ ð� � � Þtrialnþ1 and Exit else go to step 3.
3. This step is start of Newton Raphson iteration for finding c1nþ1 and c2nþ1. At the first set k = 0 and k = {ljfl,n+1 > 0}4. cknþ1;k ¼ 0 and Dcknþ1 ¼ 0.5. Compute fk,n+1 from Eq. (20) and
@fk;nþ1@ca from Eqs. (53) or (55), according its condition.
6. Compute Dc1
Dc2
� �from Eqs. (54) or (56),according its condition.
7. Let �clnþ1 ¼ clnþ1;k þ Dcl where l = 1,2.
8. If �clnþ1 < 0 for l = 1, 2 then reset k ¼ flj�clnþ1 > 0g and go to step 4 else c
lnþ1;k ¼ �c
lnþ1 and set k = k + 1.
9. Check convergency, if (Dc1 + Dc1) 6 tolerance exit, else go to step 5.
By this the algorithm of the return mapping is completed and the parameters c1nþ1 and c2nþ1 are determined.
2.3.4. Elastoplastic tangent moduliFor linearizing the weak form of equilibrium equations the expression drnþ1denþ1 or elasto-plastic tangent moduli, is needed.
This moduli is computed for an isotropic and kinematic hardening rule and is given at follow. The process of extracting elas-to-plastic tangent moduli is discussed in Appendix B.
drnþ1denþ1
¼ Br � BrXb2k
Xa2k
@fa;nþ1@r
z�1ab ybBr; ð57Þ
where k = {ljfl,n+1 = 0} l = 1, 2. And
B�1r ¼ C�1 þ
Xa2k
canþ1@2fa;nþ1@r2
: ð58Þ
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Also:
yk ¼@fk@r
T
� @fk@r
T
D0Xa2k
canþ1@2fa;nþ1@r2
and zak ¼ ykBe@fa;nþ1@r
þ @fk@r
T
D0@fa;nþ1@r
þ @fk@p
Bp@fa;nþ1@p
; ð59Þ
where B�1p ¼ D�1 þ canþ1
@2 fa;nþ1@p2 .
3. Finite element implementation
In this section the numerical solution is discussed. For the numerical solution the principle of virtual work is used to ob-tain the weak form of the governing differential equations and material and geometric stiffness matrices are derived througha linearization process.
3.1. Variatonal form of the virtual work
After defining stress and strain components, by using the principle of virtual work, at time t we have:Zt Sðttnabdt0eab þ ttjabdabjabÞdtS ¼ tRext: ð60Þ
Or it can be written as:Z0Sðt0nabdt0nab þ t0mabdt0jabÞd0S ¼ tRext ; ð61Þ
where Rext is virtual work of the external forces and can be written in terms of boundary tractions according to the followingrelation:Z
t Sðt �n:dtUþ t �m � dtdÞdtSþ
Z@t Sðt ��n:dtUþ t ��m:dtdÞd@tS ¼ tRext : ð62Þ
In the above relation, t �n; t �m are distributed force and moment vectors at time t per unit area of the deformed surface, respec-tively, and t ��n; t ��m are distributed moment and force vectors at time t per unit length of boundary, @t S, respectively.
Computational stress and strain vectors, 0tr and 0te, are defined as:
t0e ¼ t0e11; t0e22;2t0e12;
t0q11;
t0q22;2
t0q12
� �T; ð63Þ
t0r ¼ t0n
11; t0n
22; t0n
12; t0m
11; t0m
22; t0m
12D E
: ð64Þ
So the formulation (61) can be written as:Z0S
t0r
Tdt0ed
0S ¼ tRext: ð65Þ
At follow, the variational form of the strain vector is computed.An arbitrary element with liner boundaries in Cartesian coordinates can be mapped into a standard isoparametric 9 nodes
element. For simplicity natural coordinates f, g are considered to be convective coordinate by the simple boundary equationsof: ha = ±1 (see Fig. 4).
0x ¼ N10x1 þ N20x2 þ � � � þ N90x9; ð66Þ0y ¼ N10y1 þ N20y2 þ � � � þ N90y9; ð67Þ
where Ni is ith shape function of the isoparametric 9 nodes element.
x0
y0
ξ
η
1
2
3 4
5
6
7
8 9
Fig. 4. Isoparametric nine node element.
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So the base vectors in reference configuration are:
0a1 ¼N1;10x1 þ N2;10x2 þ � � � þ N9;10x9N1;10y1 þ N2;10y2 þ � � � þ N9;10y9
0
264375 and 0a2 ¼ N1;2
0x1 þ N2;20x2 þ � � � þ N9;20x9N1;20y1 þ N2;20y2 þ � � � þ N9;20y9
0
264375: ð68Þ
Thus the first fundamental form of the surface in reference configuration is:
0aab� �
¼0a1 � 0a1 0a1 � 0a20a2 � 0a1 0a2 � 0a2
" #: ð69Þ
Let U be the displacement vector, then the position of any point can be written as:
trðh1; h2Þ ¼ 0rðh1; h2Þ þ tUðh1; h2Þ; ð70Þ
tU ¼u
vw
264375 ð71Þ
In-plane displacements are interpolated as follow:
u ¼ N1u1 þ N2u2 þ N3u3 þ � � � þ N9u9;v ¼ N1v1 þ N2v2 þ N3v3 þ � � � þ N9v9;
ð72Þ
where Ni is ith shape function of the isoparametric 9 nodes element.For out of plane displacements the Hermitian shape functions are employed for the 4 corners as follow:
w ¼X4i¼1
N1i wi þ N2i@w@giþ N3i
@w@fiþ N4i
@2w@fi@gi
!; ð73Þ
where N1i to N4i are the Hermitian shape functions.
For example,
N11 ¼ H01ðfÞH01ðgÞ;N21 ¼ H01ðfÞH11ðgÞ H01ðfÞ ¼ 1=4ð2� 3fþ f
3Þ;N31 ¼ H11ðfÞH01ðgÞ H11ðfÞ ¼ 1=4ð1� f� f
2 þ f3Þ;N41 ¼ H11ðfÞH11ðgÞ:
ð74Þ
Let’s assume
tU ¼ NbU: ð75ÞIn the above relation, N is the shape function matrix and can be written as follow:
N ¼Nu 0 00 Nv 00 0 Nw
264375; ð76Þ
where
Nu ¼ N1 N2 N3 N4 N5 N6 N7 N9½ �; ð77ÞNv ¼ N1 N2 N3 N4 N5 N6 N7 N9½ �; ð78Þ
Nw ¼ H01f1 H01g1
H01f1 H11g1
H11f1 H01g1
H11f1 H11g1
. . . . . . H01f4 H02g4
H01f4 H12g4
H11f4 H02g4
H11f4 H12g4
h i: ð79Þ
By this definition from Eq. (4) we have:
dteab ¼ ðrT;aN;b þ rT;bN;aÞdbU ¼ EabdbU; ð80Þdtjab ¼
1ffiffiffiffiffitap Q ab �
qab2ffiffiffiffiffiffiffita3p A
� �dbU ¼ KabdbU; ð81Þ
where:
Q ab ¼ tr;1 � tr;2� �T N;ab þ ðtr;2 � tr;abÞT N;1 þ ðtr;ab � tr;1ÞT N;2 ð82Þ
qab ¼ tr;ab:tr;1 � tr;2 ð83Þ
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The symbol ‘‘T’’ is the transpose symbol.Also from Eq. (3)2 we have:
dtd ¼ dta3 ¼1ffiffiffiffiffitap bT12 �
1
2ffiffiffiffiffiffiffita3p ATðr;1 � r;2ÞT
� �dbU ¼ YdbU: ð84Þ
Such that
b12 ¼0 tz;2N
2;1 � tz;1N
2;2
ty;1N3;2 � ty;2N
3;1
tz;1N1;2 � tz;2N
1;1 0
tx;2N3;1 � tx;1N
3;2
ty;2N1;1 � ty;1N
1;2
tx;1N2;2 � tx;2N
2;1 0
26643775 ð85Þ
And
A ¼ 2ta11trT;2:N;2 þ 2ta22trT;1:N;1 � 2ta12trT;2:N;1 � 2ta12trT;1:N;2; ð86Þ
where ‘‘a’’ is determinant of aab.So by substituting (80), (81) and (84) in Eq. (65) we have:Z
0S
t0n
abEab þ t0mabKab
�d0S ¼
Z@t S
t �nT Nþ t �mT Y� �
dtSþZ@t Sðt ��nT Nþ t ��mYÞd@tS; ð87Þ
where N, Eab, Kab and Y are determined through Eqs. (80), (81) and (84) respectively.
3.2. Local cartesian system
For simplicity, the curvilinear convective coordinate system can be mapped to a local Cartesian system [25]. Let’s define alocal Cartesian system {xa,x3} with base vectors {t n1, t n2, t n3} by means of the orthogonal transformation:
Kt ¼ ðE3 � tnÞI3 þ ½E3 � tn� þ1
1þ E3 � tnðE3 � tnÞ � ðE3 � tnÞ; ð88Þ
where tn ¼ ta3 ¼t r;1�t r;2kt r;1�t r;2k
is the normal to the mid surface and E1, E2 and E3 are the base vectors of reference Cartesiancoordinate:
E1 ¼100
264375 E2 ¼ 01
0
264375 E3 ¼ 00
1
264375: ð89Þ
Also [E3 � tn] is skew-symmetric tensor corresponding to E3 � tn vector.Observe that Kt maps E3 ? tn = KtE3 without drilling and tr,a.t n = 0 and tna = KtEa such that tna.tnb = dabAlso it can be seen that at time t:
@xa
@ha¼ tna:tr;a and tr;a ¼
@xa
@hatna; ð90Þ
where ha is a curvilinear system.So in the local Cartesian system 0aab = dab
3.3. Geometry and material stiffness matrices
In this section stiffness matrices are extracted by linearization the virtual work, Eq. (87). It is obvious that Eq. (87) at timet, is nonlinear in term of bU and should be linearized for the numerical analyses.
For linearization, the Newton Raphson method is employed as follow:It is assumed that bUk is known where ‘‘k’’ is iteration number, thenbUkþ1 ¼ bUk þ DbUkþ1: ð91Þ
Rename:Z0S
t0n
abEab þ t0mabKabd0S�
Zt S
tt
��mT Nþ �nT DdtS�Z@t S
tt��nT Nþ tt ��m
T Dd@tS ¼ fðbUÞ: ð92ÞThen
fðbUkþ1Þ ¼ fðbUkÞ þ @fðbUÞ@ bU
�����bUk DbU ¼ 0) DbU ¼@fðbUÞ@ bU
�����bUk!�1ð�fðbUkÞÞ: ð93Þ
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The stiffness matrix @fðbUÞ
@bU� ����bUk
�is computed as follows:
@fðbUÞ@ bU
�����bUk ¼ KM þ KG; ð94Þwhere KM can be computed as below:
KM ¼Z
0S
@t0r
@ bU Bd0S� �
; ð95Þ
where B is defined as:
B ¼
E11E22E12K11K22K12
2666666664
3777777775: ð96Þ
For elastic deformation, KM ¼R
0S BT CBd0S and for plastic deformation, KM ¼
R0S B
T CepBd0S, where Cep is determined from Eq.(57).
And KG ¼R
0St0r
@B
@bU d0S, this term is computed as below:KG ¼
Z0Sðt0n
abEab þ t0m
abKabÞd0S; ð97Þ
Eab ¼12
NT;aN;b þ NT;bN;a
�; ð98Þ
Kab ¼3tqab
4ffiffiffiffiffiffiffiffiffiffiðtaÞ5
q0B@
1CAAT :A� tqab2
ffiffiffiffiffiffiffiffiffiffiðtaÞ3
q0B@
1CAA � 12
ffiffiffiffiffiffiffiffiffiffiðtaÞ3
q0B@
1CAAT :Q ab � 12
ffiffiffiffiffiffiffiffiffiffiðtaÞ3
q0B@
1CAQ Tab:Aþ 1ffiffiffiffiffiffiffiffiðtaÞp Qab; ð99ÞA ¼ 4NT;2trT;2tr;1N;1 þ 4N
T;1
trT;1tr;2N;2 � 2NT;1trT;2tr;1N;2 � 2N
T;1
trT;2tr;2N;1 � 2NT;2trT;1tr;1N;2 � 2N
T;2
trT;1tr;2N;1; ð100Þ
Qab ¼ BT12N;ab þBT2abN;1 þB
Tab1N;2; ð101Þ
Bab1 ¼0 tz;1N
2;ab � tz;abN
2;1
ty;abN3;1 � ty;1N
3;ab
tz;abN1;1 � tz;1N
1;ab 0
tx;1N3;ab � tx;abN
3;1
ty;1N1;ab � ty;abN
1;1
tx;abN2;1 � tx;1N
2;ab 0
26643775; ð102Þ
B2ab ¼0 tz;abN
2;2 � tz;2N
2;ab
ty;2N3;ab � ty;abN
3;2
tz;2N1;ab � tz;abN
1;2 0
tx;abN3;2 � tx;2N
3;ab
ty;abN1;2 � ty;2N
1;ab
tx;2N2;ab � tx;abN
2;2 0
26643775: ð103Þ
Rename:
Fk ¼Z
0SðnabEab þmabKabÞd0S
� �����bUk ð104Þand
Fext ¼Z
t Sðtt �n
T Nþ tt �mT DÞdtSþ
Z@t Sðtt ��n
T Nþ tt ��mT DÞd@tS
� �Þ����bUk : ð105Þ
So from Eq. (93):
DbU ¼ ðKM þ KGÞ�1ðFext � FkÞ: ð106ÞTherefore the algorithm of solution can be summarized as below:
1. Consider interpolation matrix, N, from Eq. (76).2. For n = 0, let bU ¼ 0 and DbU ¼ 0.
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3. Compute KM and KG and Fext and Fk.4. Compute DbU from Eq. (106).5. check for convergence, if norm ðDbUÞ < tolerance exit else, let bUn ¼ bUn�1 þ DbU and n = n + 1 and go to 3.4. Numerical examples
In this section the presented method is tested with some numerical examples. The most advantage of this method is thatthe convergency rate is very high and shear locking is eliminated.
4.1. Pinched cylinder with end rigid diaphragms
A short cylinder with pinching vertical force at the middle section, and two rigid diaphragms at the ends, is studied. Thegeometric and material property of cylinder is as below:
L = 600 Radius = 300, thickness = 3, E = 3000,m = 0.3, k0 = 24.3, k0 = 300 and H0 = 0. Because of symmetry only one octane ofcylinder is modeled. The results are derived for a 30 � 30 mesh. It can be seen (form Fig. 5) that the results are very close tothe results presented in [29, 30].
4.2. A simply supported plate under uniform lateral load
In this example the deformation of a rectangular plate, under uniform lateral load is studied. The material and geometricproperties are as below:
a ¼ b ¼ 407 mm thickness ¼ 7:6 mm E ¼ 2:11� 105N=mm2; k0 ¼ 250N=mm2 and k0 ¼ 50
Because of symmetry, only one quarter of plate was modeled. The results are found for a 20 � 20 mesh. The results areshown in Fig. 6 and have been compared with literature.
4.3. A simply supported trapezoidal plate under uniform lateral load
In this example the deformation of a simply supported trapezoidal plate under uniform lateral load is studied. The geom-etry and material property of plate are as below:
a ¼ 1m; ;E ¼ 2� 105Mpa; k0 ¼ 250Mpa;ha¼ 0:01; k0 ¼ 1000:
This problem experience very large elasto plastic deformation. The results are shown in Fig. 7. In this figure W0 denotes out ofplane displacement of the center of the plate.
4.4. A simply supported skew plate under uniform lateral load
In this example the deformation of a simply supported skew plate under uniform lateral load is studied. The geometryand material property of the plate are as below:
a ¼ 1m; b ¼ 1 m; E ¼ 2� 105Mpa; k0 ¼ 250Mpa;ha¼ 0:01; k0 ¼ 1000:
01000
200030004000
500060007000
80009000
0 100 200 300 400W0
Fpresented by[12]present mthodSimo[5]Brank[13]
Rigid diaphragm support
Rigid diaphragm support
Fig. 5. Vertical deflection at center of the pinched cylinder.
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The problem is solved for a = 30�, a = 60� and a = 45� and the results are obtained for elastic and plastic deformation and theyare shown in Fig. 8. In this figure W0 denotes out of plane displacement of the center of the plate. This problem also expe-rience very large deformation.
5. Conclusion
A new non-linear method based on the Cosserat theory, with constrained director, has been presented for large elasto-plastic deformation with isotropic and kinematic hardening. The most advantage of this new method is that it eliminatesthe shear locking problem during the thin shell analyses and the convergency rate is very good. The material and geometric
0
2
4
6
8
10
12
0 1 2 3 4
normalized central displacement(w/h)
Col
laps
e lo
ad (
0.50
9))
elastic present methodMohammed[9]
a
b
Fig. 6. Vertical displacement at the center of a simply supported plate under uniform lateral load.
0
0.005
0.01
0.015
0.02
0.025
0 20 40 60 80 100w0/h
Fa^4/(Dh)10^(-8)
isotropic hardening (b/a=0.2)elastic (b/a=0.2)isotropic hardening (b/a=0.6)elastic (b/a=0.6)
Fig. 7. Vertical displacement at the center of a simply supported trapezoidal plate under uniform lateral load.
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stiffness matrices, for finite element solution, have been derived through linearization of virtual work equation. For numer-ical solution, a nine node isoparametric element was implemented. Consistent elasto plastic tangent moduli is derived for FEsolution. The method is computationally efficient and the numerical results exhibited very good agreement with the knownvalues in the literature.
Appendix A
According relation (41) rename:
H ¼C1 C2C3 C4
� ��1¼ H
1 H2
H3 H4
" #: ðA:1Þ
By using Eq. (40) we can write:
H1ði; iÞ ¼1þ cpnþ1 w4ði; iÞ
gði; iÞ ; ðA:2Þ
H2ði; iÞ ¼�cmnþ1 w2ði; iÞ
gði; iÞ ; ðA:3Þ
H3ði; iÞ ¼�cmnþ1 w3ði; iÞ
gði; iÞ ; ðA:4Þ
H4ði; iÞ ¼1þ cpnþ1 w1ði; iÞ
gði; iÞ ; ðA:5Þ
Where gði; iÞ ¼ ð1þ cpnþ1 w1ði; iÞÞð1þ cpnþ1 w4ði; iÞÞ � c2mnþ1
w2ði; iÞw3ði; iÞ: ðA:6Þ
So
@H@cpnþ1
¼@H1
@cpnþ1
@H2@cpnþ1
@H3@cpnþ1
@H4@cpnþ1
264375; ðA:7Þ
where
@H1ði; iÞ@cpnþ1
¼�w1ði; iÞ � c2mnþ1 w2ði; iÞw3ði; iÞw4ði; iÞ � 2cpnþ1 w1ði; iÞw4ði; iÞ � c
2pnþ1
w1ði; iÞðw4ði; iÞÞ2
ðgði; iÞÞ2; ðA:8Þ
@H2ði; iÞ@cpnþ1
¼cmnþ1 w2ði; iÞw4ði; iÞ þ cmnþ1 w1ði; iÞw2ði; iÞ þ 2cmnþ1cpnþ1 w1ði; iÞw2ði; iÞw4ði; iÞ
ðgði; iÞÞ2; ðA:9Þ
@H3ði; iÞ@cpnþ1
¼cmnþ1 w3ði; iÞw4ði; iÞ þ cmnþ1 w1ði; iÞw3ði; iÞ þ 2cmnþ1cpnþ1 w1ði; iÞw3ði; iÞw4ði; iÞ
ðgði; iÞÞ2; ðA:10Þ
@H4ði; iÞ@cpnþ1
¼�w4ði; iÞ � c2mnþ1 w1ði; iÞw2ði; iÞw3ði; iÞ � 2cpnþ1 w1ði; iÞw4ði; iÞ � c
2pnþ1
w4ði; iÞðw1ði; iÞÞ2
ðgði; iÞÞ2ðA:11Þ
0
0.02
0.04
0.06
0.08
0.1
0.12
0 20 40 60 80 100
w0/h
Fa^
4/(D
h)10
^(-8
) elastic(Alfa=30deg)
plastic(Alfa=30deg)
elastic(Alfa=60deg)
plastic(Alfa=60deg)
elastic(Alfa=45deg)
plastic(Alfa=45deg)
a
aα
Fig. 8. Vertical displacement at the center of a simply supported skew plate under uniform lateral load.
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And
@H1ði; iÞ@cmnþ1
¼2cmnþ1 w2ði; iÞw3ði; iÞð1þ cpnþ1 w4ði; iÞÞ
ðgði; iÞÞ2; ðA:12Þ
@H2ði; iÞ@cmnþ1
¼�w2ði; iÞ � c2mnþ1 w3ði; iÞðw2ði; iÞÞ
2 � c2pnþ1 w1ði; iÞw2ði; iÞw4ði; iÞ � cpnþ1 w2ði; iÞðw1ði; iÞ þw4ði; iÞÞðgði; iÞÞ2
@H3ði; iÞ@cmnþ1
¼�w3ði; iÞ � c2mnþ1 w2ði; iÞðw3ði; iÞÞ
2 � c2pnþ1 w1ði; iÞw3ði; iÞw4ði; iÞ � cpnþ1 w3ði; iÞðw1ði; iÞ þw4ði; iÞÞðgði; iÞÞ2
; ðA:13Þ
@H4ði; iÞ@cmnþ1
¼2cmnþ1 w2ði; iÞw3ði; iÞð1þ cpnþ1 w1ði; iÞÞ
ðgði; iÞÞ2ðA:14Þ
And then we can write:
@H@c1¼ @H@cpnþ1
þ @H@cmnþ1
and@H@c2¼ @H@cpnþ1
� @H@cmnþ1
ðA:14Þ
Appendix B
In this appendix we derive explicit expression for elasto-plastic tangent moduli. From Eq. (12) we have:
drnþ1 ¼ Cðdenþ1 � depnþ1Þ: ðB:1Þ
Also from Eq. (29)1 we can write:
depnþ1 ¼Xa2k
dcanþ1@fa;nþ1@r
þ canþ1@2fa;nþ1@r2
dr
!; ðB:2Þ
where k = {ljfl,n+1 = 0} l = 1, 2 (active surfaces).So from (B.1) and (B.2)
C�1drnþ1 ¼ denþ1 �Xa2k
dcanþ1@fa;nþ1@r
þ canþ1@2fa;nþ1@r2
dr
!: ðB:3Þ
So
C�1 þXa2k
canþ1@2fa;nþ1@r2
!drnþ1 ¼ denþ1 �
Xa2k
dcanþ1@fa;nþ1@r
: ðB:4Þ
By renaming:
B�1r ¼ C�1 þ
Xa2k
canþ1@2fa;nþ1@r2
; ðB:5Þ
drnþ1 ¼ Br denþ1 �Xa2k
dcanþ1@fa;nþ1@r
!: ðB:6Þ
For an active surface fk = 0 so dfk,n+1 = 0So
@fk@r
� �Tdrþ @fk
@P
� �TdPþ @fk
@pdp ¼ 0; ðB:7Þ
Also; da ¼Xa2k
dcanþ1@fa;nþ1@p
þ canþ1@2fa;nþ1@p2
dp: ðB:8Þ
Also dp ¼ �Dda so � D�1dp ¼Xa2k
dcanþ1@fa;nþ1@p
þ canþ1@2fa;nþ1@p2
dp:
So
� D�1 þ canþ1@2fa;nþ1@p2
!dp ¼
Xa2k
dcanþ1@fa;nþ1@p
: ðB:9Þ
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By renaming
B�1p ¼ D�1 þ canþ1
@2fa;nþ1@p2
; ðB:10Þ
dp ¼ �BpXa2k
dcanþ1@fa;nþ1@p
ðB:11Þ
presented yield function, from Eq. (29)1, (29)2 and (29)4, it is obvious that
dP ¼ �D0dep: ðB:12Þ
By replacing (B.6), (B.11), (B.12) in (B.7) we have:
@fk@r
� �TBrdenþ1 � Br
Xa2k
dcanþ1@fa;nþ1@r2
!þ @fk
@r
� �T�D0dep� �
þ @fk@p
�BpXa2k
dcanþ1@fa;nþ1@p
!¼ 0: ðB:13Þ
By replacing dep from (B.2) in the above equation we have:
@fk@r
� �Tdrþ @fk
@r
� �T�D0
Xa2k
dcanþ1@fa;nþ1@r
þ canþ1@2fa;nþ1@r2
dr
! !þ @fk@p
�BpXa2k
dcanþ1@fa;nþ1@p
!¼ 0: ðB:14Þ
So
@fk@r
T
� @fk@r
T
D0Xa2k
canþ1@2fa;nþ1@r2
!drþ @fk
@r
� �T�D0
Xa2k
dcanþ1@fa;nþ1@r
� �þ @fk@p
�BpXa2k
dcanþ1@fa;nþ1@p
! ¼ 0: ðB:15Þ
By renaming:
yk ¼@fk@r
T
� @fk@r
T
D0Xa2k
canþ1@2fa;nþ1@r2
: ðB:16Þ
We have:
ykdrþXa2k
dcanþ1 �@fk@r
T
D0@fa;nþ1@r
� @fk@p
Bp@fa;nþ1@p
!¼ 0: ðB:17Þ
And by replacing dr from Eq. (B.6) we have:
ykBrdenþ1 � ykBrXa2k
dcanþ1@fa;nþ1@r
þXa2k
dcanþ1 �@fk@r
T
D0@fa;nþ1@r
� @fk@p
Bp@fa;nþ1@p
!¼ 0: ðB:18Þ
So
ykBrdenþ1 þXa2k
dcanþ1 �ykBr@fa;nþ1@r
� @fk@r
T
D0@fa;nþ1@r
� @fk@p
Bp@fa;nþ1@p
!¼ 0: ðB:19Þ
By renaming
zak ¼ ykBr@fa;nþ1@r
þ @fk@r
T
D0@fa;nþ1@r
þ @fk@p
Bp@fa;nþ1@p
: ðB:20Þ
So
ykBrdenþ1 �Xa2k
dcanþ1zak ¼ 0; ðB:21Þ
when only one surface is active then:
dcanþ1 ¼ z�1aa yaBrdenþ1 no sum on a:
If both of surfaces are active then:
dcknþ1 ¼Xa2k
z�1ka ykBrdenþ1 ðB:22Þ
And then:
drnþ1denþ1
¼ Br � BrXb2k
Xa2k
@fa;nþ1@r
zabybBr: ðB:23Þ
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