Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57

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Comput. Methods Appl. Mech. Engrg.

journal homepage: www.elsevier .com/locate /cma

Mixed spectral/hp element formulation for nonlinear elasticity

Yue Yu a, Hyoungsu Baek a,b, Marco L. Bittencourt c, George Em Karniadakis a,⇑a Division of Applied Mathematics, Brown University, Providence, RI 02912, USAb Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139, USAc Departamento de Projeto Mecânico, Faculdade de Engenharia Mecânica, Universidade de Campinas, Campinas, SP 13083, Brazil

a r t i c l e i n f o

Article history:Received 26 January 2011Received in revised form 22 July 2011Accepted 11 November 2011Available online 22 November 2011

Keywords:Ellipticity conditioninf–sup conditionPressure projectionHigh-orderVolume locking

0045-7825/$ - see front matter � 2011 Elsevier B.V. Adoi:10.1016/j.cma.2011.11.007

⇑ Corresponding author.E-mail addresses: Yue_Yu_1@brown.edu (Y. Yu)

mlb@fem.unicamp.br (M.L. Bittencourt), George_Karnbrown.edu, gk@dam.brown.edu (G. Em Karniadakis).

a b s t r a c t

We present a mixed formulation for high-order spectral/hp element methods and investigate its stabilityin large strain analysis of nearly incompressible solids. Specifically, we employ the displacement/pressureformulation and use a pressure projection method along with general Jacobi polynomial bases to derivethe discrete equations. We first study the convergence of the method for smooth and singular solutions ofelastic and hyperelastic problems for various combinations of mixed elements. Subsequently, we inves-tigate numerically the validity of ellipticity and inf–sup conditions for various test problems includingcases with large deformation. We find that the inf–sup condition is satisfied even when the pressureinterpolation order Op is equal to the displacement order Ou, but to resolve the locking-free behavior itis required that Op = Ou � 1, and this choice also gives the best convergence rate. In large deformationanalysis, surprisingly, the results are more stable for Ou P Op P Ou � k because the ellipticity in kernelis better satisfied, with higher k as Ou increases. Compared to the displacement-only formulation, themixed formulation seems to have superior accuracy and it is overall more efficient, especially for prob-lems with high bulk modulus.

� 2011 Elsevier B.V. All rights reserved.

1. Introduction With regards to nearly incompressible problems, there are two

High-order finite elements, especially the p- and hp-versions,lead to exponential convergence rate for smooth solutions andfor singular solutions (hp-version) [1–4]; they also enable flexibil-ity in discretization as their accuracy does not depend strongly onthe aspect ratio of the elements [1]. In the field of solid mechanics,p-FEM and hp-FEM have been studied theoretically [5–10] for lin-ear elasticity, and numerical results have been provided for hyper-elastic [11–13] and plastic materials [14], exhibiting locking-freebehavior for both nearly-incompressible [7,12,13] and plate/shellproblems [8,15–21].

From the numerical standpoint, the spectral/hp element versionbased on general Jacobi polynomials provides further flexibility ingenerating elements (e.g., hexahedral, tetrahedral, triangularprisms and even pyramids). Moreover, the polynomial basis is hier-archical and is well conditioned for enhanced computationalefficiency and scalability. These aspects have led to the implemen-tation of a displacement formulation of the method for linear elas-ticity and geometrically nonlinear problems in [22], where highaccuracy and parallel scalability were demonstrated for severalprototype problems as well as for arterial mechanics problems.

ll rights reserved.

, hbaek@mit.edu (H. Baek),iadakis@brown.edu, gk@cfm.

main strategies for overcoming volumetric locking: the displace-ment-only formulation with higher-order elements and the modi-fied variational forms, including the mixed formulations andreduced integration methods. Both analytical and computationalresults were provided for the first strategy, e.g. [7,12,13], demon-strating that locking-free results can be obtained by using suffi-ciently high polynomial order Ou for displacement. However,when the Poisson ratio is close to 0.5 (or bulk modulus for hyper-elastic materials satisfies k� 1), one has to increase Ou to resolvethe volumetric locking problem, see, e.g. [7,13]. On the other hand,the modified variational form does not have this problem, andhence it exhibits the correct approximability for both h- and p-versions [5]. Among the methods of the second strategy there areseveral interesting modifications, e.g. enhanced strain formula-tions [23–27] and displacement/pressure formulations [5,6,10,15,28–36]. Here we will concentrate on the later. For the large defor-mation problems, these formulations were mainly applied to low-er-order elements (usually Ou 6 3), although combining the mixedformulations and high-order elements was proposed in the analy-sis of Stenberg and Suri [5]. Also, the violation of ellipticitycondition in large deformation problems was found for the dis-placement/pressure formulations but has not been investigatedfor different combinations of high-order elements, e.g. see [36–38].Therefore, for large deformation problems, computational resultsof mixed formulation on high-order elements are needed to inves-tigate both the locking phenomena and ellipticity condition. In

Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57 43

addition, a reduced-integration method (noted also as reduced-constraint method in [10], or pressure projection method in[33,34] and herein) for hyperelastic material was introduced byChen and Pan [33] and Chen et al. [34], hence the displacement/pressure formulation can be reduced to displacement only withoutlosing the similarity to mixed formulations [10].

In the current paper we extend the method presented in [22] tofully nonlinear elastic problems with large deformations followinga mixed formulation and using a pressure projection scheme [33,34]unlike the displacement-only formulation employed in [22].Unfortunately, the mixed formulation approach complicates boththe numerical implementation as well as the theoretical frame-work of the method as compatibility of the spaces for the displace-ment and pressure fields should be established even for linearelasticity problems, e.g. see [28]. In addition, for large deformationsthe issue of ellipticity of the discrete formulation will be addressedfollowing [36,37], where the conditions for lower order elementswere studied. To this end, upon formulating and implementing in3D the mixed formulation, we set up several numerical tests thataddress these issues and highlight the fast convergence, the lock-ing-free behavior, and the stability of the high-order mixed meth-od in large deformation analysis of nearly incompressible solids.Compared to the displacement-only formulation of p-type finiteelements presented in [12,13], which also overcomes volume lock-ing for sufficiently large interpolation order, our numerical resultshere demonstrate that the mixed formulation – despite the afore-mentioned difficulties and the added computational efforts –seems to be more efficient, especially for problems with high val-ues of the bulk modulus.

We consider the problem in Lagrangian coordinates X, both forinfinitesimal and finite deformations. Specifically, we aim to obtainthe displacement field u(X) in a 3D solid domain X � R3, with gi-ven traction and body forces, satisfying [39,40]

divðSÞ þ f ¼ 0 ð1Þ

in some weak sense, where f is the body force and S is the appropri-ate stress tensor. In our implementation, we employ the hierarchi-cal shape functions based on Jacobi polynomials for spatialdiscretization (see Appendix A) to obtain high-order accuracy. Inthe mixed formulation, we introduce the hydrostatic pressurep ¼ trS

3 into the equilibrium equation, and solve for (u,p) [36]. How-ever, by introducing another variable p, the computational cost forinverting the linear system increases. To this end, we employ apressure projection method [33,34], by which we only need to com-pute (once or iteratively via Newton–Raphson in the nonlinear case)the linear system for the u part, while the pressure field is obtainedfrom the displacement field [33].

The paper is organized as follows: in Section 2 we presentbriefly the linear and nonlinear material models we use, and inSection 3 we present the weak formulation and the projectionmethod. In Section 4 we demonstrate the convergence rates ofthe method for linear and nonlinear elasticity problems. In Section5 we present the main results, first focusing on the issue of stabil-ity, by investigating numerically the validity of the ellipticity andinf–sup conditions, then turning to the locking phenomena. We fin-ish in Section 5 with the main conclusions while the appendix con-tains details on the Jacobi polynomial basis and the incrementmethod employed in obtaining the nonlinear solutions.

2. Material models

We mainly consider two models represented by the followingstrain energy density functions:

� Linear elastic model [39,40]:

WðEÞ ¼ k2ðtrEÞ2 þ lE : E; ð2Þ

where E ¼ 12

@u@X

� �T þ @u@X

� �� �is the strain tensor, and l, k are the

shear and Lame modulus, respectively.� Mooney–Rivlin model [33,34,40]:

WðEÞ ¼ A10ðI1J�2=3 � 3Þ þ A01ðI2J�4=3 � 3Þ þ k2ðJ � 1Þ2; ð3Þ

where E ¼ 12

@u@X

� �T þ @u@X

� �þ @u

@X

� �T @u@X

� �� �is the Green Lagrange

strain tensor, A10, A01 are constants describing the behavior ofthe material, and k is the bulk modulus. Also I1, I2, J are relatedwith the deformation gradient F ¼ Iþ @u

@X and the right Greendeformation tensor C = FTF as

I1 ¼ trðCÞ; I2 ¼12ððtrCÞ2 � trðC2ÞÞ; J ¼ detðFÞ:

This model is a simple isotropic approximation for materialswith rubber-like properties [41], in the region of large deforma-tion and large strain. In this expression, we separate the distor-tional and dilatational deformation using Penn’s approach [42]as this facilitates the use of the pressure projection techniquewe will use in the next section; see also [33].

By using the hyper-elasticity relations, the second Piola–Kirch-hoff stress tensor S ¼ @W

@E for these two cases can be expressed as

� Linear elastic model:

S ¼ kðtrEÞIþ 2lE: ð4Þ

� Mooney–Rivlin model:

S ¼ 2A10J�2=3Iþ 4A01J�4=3 ðtrEþ 1ÞI� Eð Þ

þ �23

A10I1J�2=3 � 43

A01I2J�4=3 þ kJðJ � 1Þ� �

C�1: ð5Þ

After introducing the hydrostatic pressure p, the models can be re-written as

� Linear elastic model [28]:

WðEÞ ¼ p2

trEþ lE : E; ð6Þ

where the hydrostatic pressure p = ktrE.� Mooney–Rivlin model [40]:

WðEÞ ¼ A10ðI1J�2=3 � 3Þ þ A01ðI2J�4=3 � 3Þ þ p2ðJ � 1Þ; ð7Þ

where the hydrostatic pressure p = k(J � 1), and the last termcorresponds to the ‘‘material constraint’’ [43].

Correspondingly, the second Piola–Kirchhoff stress tensorS ¼ @W

@E related with the hydrostatic pressure can be expressed as

� Linear elastic model:

S ¼ pIþ 2lE: ð8Þ

� Mooney–Rivlin model:

S ¼ 2A10J�2=3Iþ 4A01J�4=3ððtrEþ 1ÞI� EÞ

þ �23

A10I1J�2=3 � 43

A01I2J�4=3 þ Jp� �

C�1: ð9Þ

3. Weak formulation and discretization

Next, we will derive the formulation for solving the displace-ment field u(X) in a three-dimensional object occupying domain

44 Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57

X � R3 with boundary oX = oXD [ oXN. Here Dirichlet type bound-ary conditions are provided on oXD, Neumann type boundary con-ditions (traction) on oXN, X stands for the initial configuration ofthe object, and Xe stands for the initial configuration of eachelement.

3.1. Linear elastostatics

3.1.1. Displacement and mixed formulationsFirst, we consider infinitesimal deformations. The weak form of

the equilibrium equation can be expressed as follows [39]:ZX

S : rvdX�Z@XN

T � vdC�Z

Xqf � vdX ¼ 0; 8v 2 U0; ð10Þ

where the displacement field u(X) 2 U = {w(X) 2 [H1(X)]3jw(X) =uD(X) on oXD}, and U0 = {w(X) 2 [H1(X)]3jw(X) = 0 on oXD}. S, T, fand q are the stress tensor, external traction force (non-followerload) on oXN, external body force, and mass density, respectively.

For the linear elastic model (2), we can rewrite the problem asfollows [28]: find the displacement field u(X) 2 U such that

lZ

X�ðuÞ : �ðvÞdXþ k

ZX

divðuÞdivðvÞdX�Z@XN

T � vdC

�Z

Xqf � vdX ¼ 0; 8v 2 U0; ð11Þ

where the operator � is defined as �ð�Þ ¼ 12

@ð�Þ@X

� �Tþ @ð�Þ

@X

� �� �. We will

refer to the above form as the displacement formulation. For the dis-cretization of (11), one can see [22] for details.

For the mixed formulation, model (6) is considered. Problem (11)should be modified following [28]:

lZ

X�ðuÞ : �ðvÞdXþ

ZX

pdivðvÞdX�Z@XN

T � vdC

�Z

Xqf � vdX ¼ 0; 8v 2 U0; ð12Þ

along with the weak form of the definition for hydrostatic pressurep:Z

XdivðuÞqdX ¼ 1

k

ZX

pqdX; 8q 2 L2ðXÞ: ð13Þ

3.1.2. Local discretizationNext, we consider the discretization of (12). Based on the shape

functions defined in Appendix A, on each element we expand thedisplacement ue ¼ ðue

1;ue2;u

e3Þ and the test function ve ¼ ðve

1;ve2;ve

3Þ:

uei ðXÞ ¼

XNeu

k¼1

ueik/kðXÞ; ve

i ðXÞ ¼XNe

u

k¼1

veik/kðXÞ; i ¼ 1;2;3; ð14Þ

where Neu is the total number of shape functions for displacement

on this element, /k(X) are the shape functions, and ueik; ve

ik are theexpansion coefficients. For the local hydrostatic pressure pe, withthe set of shape functions for pressure f/kðXÞ; k ¼ 1; . . . ;Ne

pg#

f/kðXÞ; k ¼ 1; . . . ;Neug, we can expand the pressure pe:

peðXÞ ¼XNe

p

k¼1

pek/kðXÞ; ð15Þ

where pek are the corresponding expansion coefficients.

3.1.3. Local pressure projectionFollowing the steps in [33], we carry out the projection of the

hydrostatic pressure pe at the element level onto the selected pres-sure function space. For the linear case, this reduced constraintmethod is actually equivalent to the mixed formulation [10]. Here

we consider the problem of approximating pe at the element levelin a least-squares sense, by a linear combination of a sequence ofshape functions f/kðXÞ; k ¼ 1; . . . ;Ne

pg in L2(Xe) as in (15). Thatis, we choose

Pe ¼ ½pe1; . . . ; pe

k; pekþ1; . . . ; pe

Nep�T : ð16Þ

to minimize

kpe � QPek2L2ðXeÞ ð17Þ

where k � kL2ðXeÞ is the L2 norm in the element domain Xe and

Q ðxÞ ¼ /1ðXÞ;/2ðXÞ; . . . ;/NepðXÞ

h i: ð18Þ

So the minimization leads to

MepPe ¼ De; ð19Þ

where the element mass matrix for the pressure is of size Nep � Ne

p asfollows:

Mep ¼

m11 � � � m1Nep

..

. . .. ..

.

mNep1 � � � mNe

pNep

2666437775; mij ¼

ZXe

/i/jdXe ð20Þ

and the right hand side of (19) is a vector of size Nep:

De ¼Z

Xe/1pedXe;

ZXe

/2pedXe; . . . ;

ZXe

/NeppedXe

� ¼ k

ZXe

/1divuedXe; kZ

Xe/2divuedXe; . . . ; k

ZXe

/NepdivuedXe

� ¼ kðKe

pÞT Ue;

ð21Þ

where the vector of unknown expansion coefficients for displace-ment U has length 3Ne

u. In order to have smaller bandwidth forthe coefficient matrix [22], it is organized in a way such that thethree displacement components corresponding to a particular modeare adjacent, as follows:

Ue ¼ ðUe1Þ

T . . . ; ðUekÞ

T; ðUe

kþ1ÞT; . . . ; ðUe

NeuÞT

h iT;

Uek ¼ ½ue

1k; ue2k; u

e3k�:

ð22Þ

The pressure matrix Kep with size 3Ne

u � Nep is composed of Ne

u � Nep

numbers of 3 � 1 submatrices:

Kep ¼

ðKepÞ11 � � � ðKe

pÞ1Nep

..

. . .. ..

.

ðKepÞNe

u1 � � � ðKepÞNe

uNep

266664377775;

ðKepÞij ¼

ZXe

@/i

@x/jdXe;

ZXe

@/i

@y/jdXe;

ZXe

@/i

@z/jdXe

� T

;

i ¼ 1; . . . ;Neu; j ¼ 1; . . . ;Ne

p:

ð23Þ

Therefore, the element coefficients for the projected hydrostaticpressure should be

Pe ¼ kðMepÞ�1ðKe

pÞT Ue: ð24Þ

3.1.4. Global discretizationWe assemble the local displacement expansion coefficients Ue

of all elements into a global form U and upon substitution into(12), we obtain

�ðKu þ KuÞUþ F ¼ 0; ð25Þ

Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57 45

where Ku is the stiffness matrix for lR

X �ðuÞ : �ðvÞdX, and Ku corre-sponds to

RX pdivðvÞdX. Here, the 3Ne

u � 3Neu matrix Ke

u for eachelement is organized into a matrix of Ne

u � Neu blocks, and the

3 � 3 submatrix ðKeuÞij ði; j ¼ 1; . . . ;Ne

uÞ is given by ðKeuÞij ¼R

Xe ðD/iÞT BðD/jÞdXe, where

D/i ¼

@/i@x 0 0 @/i

@y 0 @/i@z

0 @/i@y 0 @/i

@x@/i@z 0

0 0 @/i@z 0 @/i

@y@/i@x

26643775

T

; B ¼ l2I

I

� : ð26Þ

On the other hand, the matrix Ku is assembled based on local matri-ces from the element-wise pressure projections

Keu ¼ kKe

pðMepÞ�1ðKe

pÞT: ð27Þ

On each element, the vector of the external forces Fe has thesame length and ordering of indices as Ue. That is, Fe ¼½ðFe

1ÞT; . . . ; ðFe

kÞT; ðFe

kþ1ÞT; . . . ; ðFe

NeuÞT �T , where Fe

k ¼ ½Fe1k; F

e2k; F

e3k�; Fe

ik ¼R@Xe

NTi/kdCe þ

RXe qfi/kdXe and Ti, fi (i = 1,2,3) are the components

of the external traction and body forces, respectively.In the implementation, the contribution from known expan-

sions (due to Dirichlet boundary conditions oXD) is moved to theright-hand-side before the equation is solved.

Although (25) is equivalent to solving the larger system (U,P),the size of unknowns is by one-fourth smaller. For details on theiterative solution of the linear system, we refer the reader to [22].

3.2. Nonlinear elastostatics

3.2.1. Displacement and mixed formulationsNext we consider a solid object with finite deformation. Similar

to the linear elastostatics, X stands for the position vector of amaterial point in the initial configuration X, and x(X) denotesthe position vector in the deformed configuration. The weak formof displacement formulation for the equilibrium is: find the dis-placement field u(X) 2 U = {w(X) 2 [H1(X)]3jw(X) = uD(X) on oXD}(where XD denotes the Dirichlet boundary in the initial configura-tion), which satisfies

PðuÞ ¼Z

XS :

12

@v@X

� �T

FðuÞ þ FðuÞT @v@X

� � !dX�

Z@XN

T � vdC

�Z

Xqf � vdX ¼ 0; 8v 2 U0; ð28Þ

where U0, T, f and q have the same meanings as for linear elasto-statics case, but with respect to the initial configuration. Note herethat we assume the external T to be deformation-independent (i.e.,non-follower load). Also, F(u) is the deformation gradient tensor de-fined as FðuÞ ¼ @x

@X ¼ Iþ @u@X.

To rewrite (28) into the mixed formulation framework, we con-sider S related with the hydrostatic pressure as in (9), and divide itinto two parts: with and without the pressure term p. Although itcan be generalized to other cases, here we take S as in (9), i.e., S canbe divided aseS ¼ 2A10J�2=3Iþ 4A01J�4=3ððtrEþ 1ÞI� EÞ

þ �23

A10I1J�2=3 � 43

A01I2J�4=3� �

C�1; ð29Þ

S ¼ JpC�1: ð30Þ

Correspondingly, the elasticity tensor C ¼ @S@E can also be divided into

two parts (see [35,43] for the explicit expressions for these twofourth-order tensors of Mooney–Rivlin model):

eC ¼ @eS@E

; C ¼ @S@E

; ð31Þ

where E is the Green Lagrange strain tensor, with EðuÞ ¼ 12 ðF

TðuÞFðuÞ � IÞ. Also, C can be further divided into two parts

C ¼ C1 þ C2; C1 ¼ p@ðJC�1Þ@E

; C2 ¼ JC�1 @p@E

: ð32Þ

Following [22], we use Pintðu; pÞ to denote the virtual work dueto the internal stress, which is the first term in (28), and we usePextðuÞ to denote the last two terms, representing the virtual workdue to the external forces. Separating the stress tensor and theelasticity tensor into volumetric and distortional parts, we can re-write (28) as

Pðu;pÞ ¼Z

X

eSðuÞ þ SðuÞ� �

:12

@v@X

� �T

FðuÞ þ FðuÞT @v@X

� � !dX

�Z@XN

T � vdC�Z

Xqf � vdX ¼ 0; ð33Þ

which, along with the weak form of the definition for hydrostaticpressure p:Z

XðJðuÞ � 1ÞqdX ¼ 1

k

ZX

pqdX; 8q 2 L2ðXÞ ð34Þ

forms a closed system for the mixed formulation [40].

3.2.2. Newton–Raphson procedureAssuming that p can be approximated by a function of u as in

the linear case, in (33) P is nonlinear with respect to u, so we needto solve for u iteratively. In our implementation we employed theNewton–Raphson iterative procedure. Let uk denote the solution atthe kth iteration, this procedure can be written as follows.

Loop over k until convergence is reached for a pre-specifiedtolerance:

1. Based on the known information uk, update the derivative of Pwith respect to u, which is denoted as DP½u�ðukÞ.

2. Solve the following system of equations for the increment Duwith a linear solver:

DP½u�ðukÞDu ¼ �PðukÞ; 8v 2 U0: ð35Þ

3. Update the solution

ukþ1 ¼ uk þ Du: ð36Þ

Before spatial discretization, we write out the explicit form for thederivative:

DP½u�ðuÞðDuÞ ¼ ðDPint ½u�ðuÞ þDPext½u�ðuÞÞðDuÞ

¼Z

X

12

@v@X

� �T

FðuÞ þ FðuÞT @v@X

� � !: ðeCðuÞ þ CðuÞÞ

:12

@ðDuÞ@X

� �T

FðuÞ þ FðuÞT @ðDuÞ@X

� � !dX

þZ

XðeSðuÞ þ SðuÞÞ :

@ðDuÞ@X

� �T@v@X

� � !dX

¼Z

X

12

@v@X

� �T

FðuÞ þ FðuÞT @v@X

� � !: ðeCðuÞ þ C1ðuÞÞ

:12

@ðDuÞ@X

� �T

FðuÞ þ FðuÞT @ðDuÞ@X

� � !dX

þZ

X

12

@v@X

� �T

FðuÞ þ FðuÞT @v@X

� � !: ðJC�1dpÞdX

þZ

XðeSðuÞ þ SðuÞÞ :

@ðDuÞ@X

� �T@v@X

� � !dX; ð37Þ

46 Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57

where

dp ¼ 12

kJC�1 :@ðDuÞ@X

� �T

FðuÞ þ FðuÞT @ðDuÞ@X

� � !; ð38Þ

with C�1 denoting the inverse of the right Green deformation tensorC.

Note that the external loads are deformation-independent un-der our assumption, thus DPext½u�ðuÞ ¼ 0.

3.2.3. Local discretizationSimilar to the linear elastostatics case, the discretization for the

displacement formulation was detailed in [22]. For the mixed for-mulation, to discretize (35) in one element we expand Due in termsof the shape functions as in (14):

ðDueÞiðXÞ ¼XNe

u

k¼1

ðDueÞik/kðXÞ; i ¼ 1;2;3: ð39Þ

Also, the local hydrostatic pressure pe is expanded as in (15), whilethe increment of hydrostatic pressure dpe is expanded as

dpeðXÞ ¼XNe

p

k¼1

dpek/kðXÞ; ð40Þ

where ðDueÞik and dpek are the corresponding expansion coefficients.

3.2.4. Local pressure projectionSimilarly as in Section 3.1.3, we carry out the projections on

each element for both hydrostatic pressure pe and its incrementdpe. For the pe part, we write the coefficients as in (16), and weminimize the function as in (17). So the minimization leads to(19) with the element mass matrix for pressure the same as in(20) and the right hand side of size Ne

p:

De ¼Z

Xe/1pedXe;

ZXe

/2pedXe; . . . ;

ZXe

/NeppedXe

� ¼ k

ZXe

/1ðJðueÞ � 1ÞdXe; kZ

Xe/2ðJðueÞ � 1ÞdXe; . . . ;

�kZ

Xe/Ne

pðJðueÞ � 1ÞdXe

: ð41Þ

Therefore, the projected hydrostatic pressure pe⁄ can be approxi-mated as

pe ¼ Q ðMepÞ�1De: ð42Þ

For the increment of hydrostatic pressure, a similar procedurecan be followed. We organize the coefficients dPe for dpe as in(16) and we minimize

kdpe � QdPek2L2ðXeÞ; ð43Þ

to obtain

MepdPe ¼ dDe

; ð44Þ

with

dDe ¼Z

Xe/1dpedXe;

ZXe

/2dpedXe; . . . ;

ZXe

/NepdpedXe

� T

¼ k2

ZXe

/1JC�1 :@ðDuÞe

@X

� �T

FðuÞ þ FðuÞT @ðDuÞe

@X

� � !dXe; . . . ;

"Z

Xe/Ne

pJC�1 :

@ðDuÞe

@X

� �T

FðuÞ þ FðuÞT @ðDuÞe

@X

� � !dXe

#T

¼ kðKepÞ

TDUe; ð45Þ

where DUe are the coefficients for Due organized in the same way asin (22), and the matrix Ke

p is a 3Neu � Ne

p matrix organized into

Neu � Ne

p matrix of blocks, with each block being a 3 � 1 submatrixðKe

pÞij ði ¼ 1; . . . ;Neu; j ¼ 1; . . . ;Ne

pÞ:

ðKepÞij;m ¼

X3

k;l¼1

ZXe

J/jC�1kl

12

@/i

@XkFml þ Fmk

@/i

@Xl

� �dXe; m ¼ 1;2;3;

ð46Þ

where Fij; C�1ij and Xi are the components for the deformation tensor

F, the inverse of the right Green deformation tensor C and the posi-tion vector in initial configuration X, respectively. Therefore, theprojected increment for the hydrostatic pressure dpe⁄ can beapproximated as

dpe ¼ kQ ðMepÞ�1ðKe

pÞTðDUÞe: ð47Þ

3.2.5. Global discretizationSimilarly as in (25), we assemble the local displacement expan-

sion coefficients (DU)e of all elements into the global form DU andsubstitute it into (35), to obtain

ðKu þ Ku þ Ku ÞDU ¼ Ru; ð48Þ

where Ku; Ku and Ku are the stiffness matrices for the discretiza-tion ofZ

X

12

@v@X

� �T

FðuÞ þ FðuÞT @v@X

� � !: eCðuÞ

:12

@ðDuÞ@X

� �T

FðuÞ þ FðuÞT @ðDuÞ@X

� � !dX

þZ

X

eSðuÞ :@ðDuÞ@X

� �T@v@X

� � !dX; ð49Þ

ZX

12

@v@X

� �T

FðuÞ þ FðuÞT @v@X

� � !: C1ðuÞ

:12

@ðDuÞ@X

� �T

FðuÞ þ FðuÞT @ðDuÞ@X

� � !dX

þZ

XSðuÞ :

@ðDuÞ@X

� �T@v@X

� � !dX ð50Þ

andZX

12

@v@X

� �T

FðuÞ þ FðuÞT � @v@X

� � !: ðJC�1dpÞdX; ð51Þ

respectively.Therefore, on each element the stiffness matrices Ke

u and Ke⁄u

are 3Neu � 3Ne

u matrix organized into Neu � Ne

u matrix blocks, witheach block being a 3 � 3 submatrix ðKe

uÞij ði; j ¼ 1; . . . ;NeuÞ given by

ðKeuÞij;ms¼

X3

p;q;k;l¼1

ZXe

12

@/i

@XpFmqþFmp

@/i

@Xq

� �eCpqkl12

@/j

@XkFslþFsk

@/j

@Xl

� �dXe

þdms

X3

p;q¼1

ZXe

eSpq@/i

@Xp

@/j

@XqdXe; m;s¼1;2;3 ð52Þ

and

ðKeu Þij;ms¼

X3

p;q;k;l¼1

ZXe

12

@/i

@XpFmqþFmp

@/i

@Xq

� �C1

pqkl12

@/j

@XkFslþFsk

@/j

@Xl

� �dXe

þdms

X3

p;q¼1

ZXe

Spq@/i

@Xp

@/j

@XqdXe; m;s¼1;2;3; ð53Þ

respectively. For the local matrix Keu on each element, upon substi-

tution of (47) into (51), we have

Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57 47

ðKeu Þij;ms ¼ k

X3

p;q¼1

XNep

a¼1

ZXe

12

@/i

@XpFmq þ Fmp

@/i

@Xq

� �� JðC�1Þpq/aððMe

pÞ�1ðKe

pÞTÞaj;sdXe

¼ kX3

p;q¼1

XNep

a¼1

ZXe

12

@/i

@XpFmq þ Fmp

@/i

@Xq

� �� JðC�1Þpq/adXeððMe

pÞ�1ðKe

pÞTÞaj;s

¼ kXNe

p

a¼1

ðKepÞia;mððM

epÞ�1ðKe

pÞTÞaj;s ¼ kðKe

pðMepÞ�1ðKe

pÞTÞij;ms:

ð54Þ

In the above expressions, Fij; eSij; Sij; eC ijkl; Cijkl; ðC�1Þij and Xi are thecomponents for the deformation tensor F, the distortional part ofthe second Piola–Kirchhoff tensor eS, the volumetric part S, the dis-tortional part of elasticity tensor eC , the volumetric part C, the inverseof the right Green deformation tensor C and the position vector ininitial configuration X, respectively. Also, dms stands for the Kroneck-er delta. The ⁄ symbols for C1

pqkl and Spq in (53) means they are depen-dent on the hydrostatic pressure pe, while in these expressions thelocal hydrostatic pressure after projection pe⁄ is substituted in.

Regarding the residual vectors on each element, in (48) theright-hand-side Re

u has length 3Neu, with the same ordering as

DUe. Similarly to what we did for P, we can divide this vector intotwo contributions (here we denote its element as ðRe

uÞij; i ¼ 1;2;3and j ¼ 1; . . . ;Ne

u):

Reu ¼ ðR

euÞ

int þ ðReuÞ

ext; ð55Þ

where ðReuÞ

int represents the contribution from the internal stress

ðReuÞ

intij ¼ �

X3

k;l¼1

ZXe

eSkl þ Skl

� �12

@/j

@XkFil þ Fik

@/j

@Xl

� �dXe ð56Þ

and ðReuÞ

ext represents the contribution from the external loads

ðReuÞ

extij ¼

Z@Xe

N

Ti/jdCe þZ

Xeqfi/jdXe: ð57Þ

Here S is also marked with ⁄ in (56) because the hydrostatic pres-sure after projection pe⁄ is employed.

Correspondingly, the Newton–Raphson iterative procedurewith pressure projection will be modified as follows.

Loop over k until the convergence tolerance is reached:

1. Based on the known information Uk, following Section 3.2.4 toupdate for the projected hydrostatic pressure pe⁄.

2. Update the volumetric stress tensor S and volumetric elasticitytensor C1 based on (30) and (32).

3. Update the stiffness matrices (52)–(54) if needed. Update theinternal right hand side as in (56).

4. Solve (48) for the increment DU with a linear solver.5. Update for displacement

Ukþ1 ¼ Uk þ DU: ð58Þ

Fig. 1. p-convergence test: cubic object discretized with 8 hexahedral elements.

4. Convergence studies

In this section, we investigate the accuracy of the pressure pro-jection method in nearly incompressible problems by comparingsimulation results against analytical solutions.

4.1. Linear elastostatics

We first investigate the convergence of the scheme for linearelastostatic problems and solve the equilibrium equations for

infinitesimal deformation (12) using the pressure projection meth-od. Three kinds of convergence (p,h,hp) are studied for solutionswith and without singularities. For the material properties, we willuse Young’s modulus E and the Poisson ratio m, which are relatedwith the material properties k and l in (2) by

l ¼ E2ð1þ mÞ ; k ¼ mE

ð1þ mÞð1� 2mÞ : ð59Þ

4.1.1. Smooth solution: p-convergenceConsider the deformation of a cubic solid occupying the domain

0 6 x 6 1, 0 6 y 6 1 and 0 6 z 6 1 discretized by 8 hexahedral ele-ments as in Fig. 1, where the thick solid lines mark the edges of theelements. For the test example, we use an analytic solution similarto [22], but with m = 0.49, that is, which corresponds to nearlyincompressible material. The displacements on the face x = 0 are

ux ¼ 0:0; uy ¼ B sin bY ; uz ¼ 0; ð60Þ

where ux, uy and uz are the displacements in x, y and z directions,respectively; also, B, b are prescribed constants. A traction forcefield, T = (Tx,Ty,Tz), is applied on the rest of the faces and is given by

Tx ¼ nxk aA1m� 1

� �cos aX þ bB cos bY

� �;

Ty ¼ nyk bB1m� 1

� �cos bY þ aA cos aX

� �;

Tz ¼ nzkðaA cos aX þ bB cos bYÞ;

ð61Þ

where n = (nx,ny,nz) is the outward-pointing unit vector normal tothe surface, and A, a are also prescribed constants. A body forcefield, qf = (fx, fy, fz) is applied on the solid with components:

fx ¼kð1� mÞ

ma2A sin aX; f y ¼

kð1� mÞm

b2B sin bY ; f z ¼ 0: ð62Þ

This problem has the following analytic solution for the displace-ment of the object:

ux ¼ A sin aX;

uy ¼ B sin bY ;uz ¼ 0;

8><>: ð63Þ

so we see that the object is not totally incompressible (p =k(aAcosaX + bBcosbY)).

48 Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57

In our test, we use the following parameters:

A ¼ B ¼ 0:2; a ¼ b ¼ 5:0; E ¼ 1000; m ¼ 0:49 ð64Þ

to study the convergence behavior. We first vary the element orderOu for displacement (i.e., highest polynomial degree in the expan-sions) from 1 to 10, for various pressure orders Op = 0, 1, 3, 5, asin the upper plots of Fig. 2. Then in the lower plots, for fixed dis-placement orders Ou = 4, 6, 8, 10 we show the results for differentchoices of pressure orders Op. To compare the computational errorswith respect to the exact solution, for each element order pair(Ou,Op) we calculate the relative errors for pressure in the L2 norm,and for displacement in the energy norm as follows

ep ¼kpa � pck2

kpak2; eu ¼

kua � uckE

kukE; ð65Þ

where pa = kdiv (ua) and ua are the analytic solutions for pressureand displacement, respectively, while pc and uc are results fromcomputation. We plot these errors in logarithmic scale, as a functionof Ou (or Op in the lower plots) which is in linear scale.

From the upper plots of Fig. 2, we see that for every fixed inter-polation order Op of pressure, the computational results do notconverge to the analytic solution when Ou increases. On the otherhand, in the lower plots, for each fixed Ou considered approxi-mately straight lines are obtained in the range of Op < Ou, indicatingthat the numerical errors decrease exponentially fast as the ele-ment order Op for pressure increases. So for slightly compressibleproblems, the element order Op for pressure is more importantfor obtaining accurate results. Also, as shown in the upper rightplot, for each fixed Op, the two cases with Ou = Op and Ou = Op + 1give the best results for displacement. Combining these resultswith results of the upper left plot, which shows that Ou = Op givesthe largest pressure error, we can see that in this test Ou = Op + 1is the best choice for accuracy.

4.1.2. Smooth solution: h-convergenceNow we focus on the h-convergence on the problem tested in

Section 4.1.1 using uniform tesselations with i � i � i (i = 1,2, . . . ,8) hexahedral elements, while i = 2 was used for all cases in

0 2 4 6 8 1010−6

10−4

10−2

100

Order for displacement (Ou)

Rel

ativ

e er

rors

for p

Op=0

Op=2

Op=4

Op=6

0 2 4 6 8 1010−10

10−7

10−4

10−1

101

Order for pressure (Op)

Rel

ativ

e er

rors

for p

Ou=10

Ou=8

Ou=6

Ou=4

Fig. 2. p-refinement test for the smooth problem in linear case: upper: relative errorsfunction of the element order Op for pressure; left column: relative errors for pressure;

Section 4.1.1 (seeing Fig. 1). The displacement order Ou = 5 is em-ployed here while runs with four choices of pressure ordersOp = 1, 3, 4, 5 are tested. Similar to the p-convergence tests, inFig. 3 we calculate the relative errors of pressure and displacementin appropriate norms. We plot these errors in logarithmic scale as afunction of element size h ¼ 1

i which is in inverse logarithmic scale.From Fig. 3, we see that in most of cases, a fixed element order

Op for pressure gives a Op + 1 order accuracy for h-convergence, ex-cept when Op = Ou which shows an Op order. The results, especiallythose for the case of Op = Ou � 2, are consistent with the analysis in[5,10]. Also, the right plot shows a 5.5 order accuracy with thechoice Op = Ou � 1 = 4 (green curve), which is the best convergencerate we obtain from h-refinement in this test.

4.1.3. Solution with singularities: p-convergenceAs in the previous two sections, we consider the deformation of

a cubic solid occupying the domain 0 6 x 6 1, 0 6 y 6 1 and0 6 z 6 1, but discretized by i (i = 1,2, . . . ,7) layers of prismatic ele-ments. In Fig. 4 we show the 1- and 3-layer meshes, with the thicksolid lines marking the edges of the elements.

We use a similar analytic solution as in [2,3], which is singularat the line x = y = 0. The displacements on the faces x = 0, y = 0 andz = 0, 1 are known

ux ¼ AffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiX2 þ Y2

q; uy ¼ B

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiX2 þ Y2

q; uz ¼ 0; ð66Þ

where A, B are prescribed constants. On face x = 1, a traction forcefield T = (Tx,Ty,Tz) is applied as

Tx ¼ �k AX � BY � AX

m

� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiX2 þ Y2

p ; Ty ¼lðBX þ AYÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

X2 þ Y2p ; Tz ¼ 0 ð67Þ

and on face y = 1

Tx ¼lðBX þ AYÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

X2 þ Y2p ; Ty ¼

k AX � BY þ BYm

� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiX2 þ Y2

p ; Tz ¼ 0: ð68Þ

The body force field qf = (fx, fy, fz) applied on the solid can be de-scribed as:

0 2 4 6 8 10

10−6

10−4

10−2

100

102

Order for displacement (Ou)

Rel

ativ

e er

rors

for u

Op=0

Op=2

Op=4

Op=6

0 2 4 6 8 1010−10

10−7

10−4

10−1

102

Order for pressure (Op)

Rel

ativ

e er

rors

for u

Ou=10

Ou=8

Ou=6

Ou=4

as a function of the element order Ou for displacement; lower: relative errors as aright column: relative errors for displacement.

0.1250.250.5110−6

10−4

10−2

100

Element size (h)

Rel

ativ

e er

rors

for p

Op=1

Op=3

Op=4

Op=5

1

14

2

15

0.1250.250.5110−6

10−4

10−2

100

102

Element size (h)

Rel

ativ

e er

rors

for u

Op=1

Op=3

Op=4

Op=5

1

2

51

1

1

4

5.5

Fig. 3. h-refinement test for the smooth problem in linear case: relative errors as a function of the element size. Left: relative errors for pressure. Right: relative errors fordisplacement.

Fig. 4. Multi-layer meshes refined near the singularities of the solution. Left: 1-layer mesh. Right: 3-layer mesh.

Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57 49

fx ¼kðAX2ð2m� 1Þ þ 2AY2ðm� 1Þ þ BXYÞ

2mðX2 þ Y2Þ3=2 ;

f y ¼kð2BX2ðm� 1Þ þ BY2ð2m� 1Þ þ AXYÞ

2mðX2 þ Y2Þ3=2 ; f z ¼ 0: ð69Þ

In this problem the displacement of the object has the followinganalytic solution:

ux ¼ AffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiX2 þ Y2

p;

uy ¼ BffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiX2 þ Y2

p;

uz ¼ 0:

8>><>>: ð70Þ

We test this problem on 1-, 2- and 3-layer meshes, with the fol-lowing parameters:

A ¼ B ¼ 1:0; E ¼ 1000; m ¼ 0:49 ð71Þ

to study the convergence behavior. On each mesh, we systemati-cally vary the element order Ou for displacement from 1 to 16. Based

1 2 4 8 1610−4

10−3

10−2

10−1

100

Order for displacement (Ou)

Rel

ativ

e er

rors

for p

1−layer mesh2−layer mesh3−layer mesh

12

Fig. 5. p-refinement test for the singular problem in linear case: elative errors as a functirelative errors for displacement.

on our previous results of smooth solutions, all errors presentedcorrespond to the (best) choice Op = Ou � 1. In Fig. 5, we give the rel-ative errors for pressure and displacement in logarithmic scale as afunction of Ou which is also in logarithmic scale. We see that all re-sults from the 1-layer mesh are approximately on a straight linewith Ou increasing. Instead of the exponential convergence forsmooth solutions as in Section 4.1.1, a second-order convergencerate is obtained. Although 2- and 3-layer meshes are refined nearthe singularity leading to smaller errors compared with what fromthe 1-layer mesh, when Ou becomes sufficiently large (say, largerthan 3 in the right plot), they also show a second-order conver-gence. These results correspond to slightly higher convergence ratescompared to the estimates in [5,10], because the ra type of singular-ity is at an endpoint of the element, and this type of mesh gives a 2-� order accuracy for this case according to [3].

4.1.4. Solution with singularities: hp-convergenceWe now study the hp-convergence on the same problem intro-

duced in Section 4.1.3. Similar as in [2–4], we refine the meshes

1 2 4 8 16

10−4

10−3

10−2

10−1

100

Order for displacement (Ou)

Rel

ativ

e er

rors

for u

1−layer mesh2−layer mesh3−layer mesh

21

on of the element order Ou for displacement. Left: relative errors for pressure. Right:

50 Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57

gradually as the element order increases. Namely, for increasingdisplacement order Ou, we use i = Ou layers mesh in the test. InFig. 6 we calculate the relative errors of pressure and displacementin appropriate norms, for Op = Ou � 2, Op = Ou � 1 and Op = Ou. Weplot these errors in logarithmic scale, as a function of element or-der Ou for displacement (also the number of layers) which is in lin-ear scale. We see that approximately straight lines are obtained forall cases, which indicates the exponential convergence, as expectedfrom the analysis in [2,3]. Also, consistent with the previous sec-tions, we obtain larger errors when Op = Ou � 2, while the resultsfrom Op = Ou � 1 and Op = Ou are similar with each other.

4.2. Nonlinear elastostatics

Next we investigate the convergence of the projection spectral/hp element method for nonlinear elastostatic problems. We em-ploy the Mooney–Rivlin model (3), and solve the equilibrium Eq.(33). We consider the same domain as in the linear case shownin Fig. 1, and both h- and p-convergence tests are performed forproblems with smooth solutions.

4.2.1. Smooth solution: p-convergenceIn this test example, we choose the material property parame-

ters as

0 2 4 6 810−4

10−3

10−2

10−1

100

Number of layers (=Ou)

Rel

ativ

e er

rors

for p

Op=Ou−2Op=Ou−1Op=Ou

Fig. 6. hp-refinement test for the singular problem in linear case: relative errors as a funclayers). Left: relative errors for pressure. Right: relative errors for displacement.

0 2 4 6 810−5

10−4

10−3

10−2

10−1

100

Order for displacements (Ou)

Rel

ativ

e er

rors

for p

Op=0

Op=1

Op=2

Op=3

0 2 4 6 810−8

10−6

10−4

10−2

100

Order for displacements (Ou)

Rel

ativ

e er

rors

for p

Ou=7

Ou=6

Ou=5

Ou=4

Fig. 7. p-refinement test for the smooth problem in nonlinear case: upper: relative errorfunction of the element order Op for pressure; left column: relative errors for pressure;

A10 ¼ A01 ¼ 1; k ¼ 100: ð72Þ

The face X = 0 is clamped, and the displacements on the rest of thefaces are given as follows:

ux ¼ A sin aX; uy ¼ 0; uz ¼ 0; ð73Þ

where A, a are prescribed constants. A body force field, qf = (fx, fy, fz)is also applied on the object:

fx ¼4Aa2 sin aX

9ðaA cos aX þ 1Þ10=3 ðð225ðaA cos aXÞ3 þ 625ðaA cos aXÞ2

þ 675aA cos aX þ 225ÞðaA cos aX þ 1Þ1=3

þ aA cos aXððaA cos aX þ 1Þ2=3 � 1ÞðaA cos aX þ 2Þ þ 6

þ 6ðaA cos aX þ 1Þ2=3Þ;fy ¼ fz ¼ 0:

ð74Þ

The displacements of the solid for this problem can be expressed bythe following analytic functions:

ux ¼ A sin aX;

uy ¼ 0;uz ¼ 0

8><>: ð75Þ

0 2 4 6 810−5

10−3

10−1

101

Number of layers (=Ou)

Rel

ativ

e er

rors

for u

Op=Ou−2

Op=Ou−1

Op=Ou

tion of the element order Ou for displacement (which is equal to the number of mesh

0 2 4 6 8

10−5

10−3

10−1

101

Order for displacements (Ou)

Rel

ativ

e er

rors

for u

Op=0

Op=1

Op=2

Op=3

0 2 4 6 8

10−7

10−5

10−3

10−1

101

Order for displacements (Ou)

Rel

ativ

e er

rors

for u

Ou=7

Ou=6

Ou=5

Ou=4

s as a function of the element order Ou for displacement; lower: relative errors as aright column: relative errors for displacement.

0.1250.250.51

10−6

10−4

10−2

100

Element size (h)

Rel

ativ

e er

rors

for p

Op=0

Op=2

Op=3

Op=41

1

1

1

3

4

0.1250.250.5110−7

10−5

10−3

10−1

101

Element size (h)

Rel

ativ

e er

rors

for u

Op=0

Op=2

Op=3

Op=4

1

1

11

3

4

Fig. 8. h-refinement test for the smooth problem in nonlinear case: relative errors as a function of the element size. Left: relative errors for pressure. Right: relative errors fordisplacement.

Fig. 9. Domain for elllipticity and inf–sup condition tests.

Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57 51

and pressure p = k(aAcosaX), with constants A = 0.04, a = 2.0. Similaras in the linear case, this problem is not totally incompressible.

Similar to the linear case, we study the p-convergence by twoways: first vary the element order Ou for displacement up to 7,with fixed orders of pressure Op = 0, 1, 2, 3; then exchange the posi-tions of Ou and Op, i.e., compare the results from different Ou

(Ou = 7,6,5,4) for increasing Op in the range of 0 6 Op 6 Ou. We cal-culate the relative errors of the computed displacement and pres-sure fields for each element order pair as in previous sections, seeFig. 7, these errors are plotted in logarithmic scale as a function ofelement order (in linear scale).

Consistent with the results from Section 4.1.1 for the linearcase, we see that for each fixed Ou, increasing Op gives almost expo-nential convergence, while for each fixed Op and sufficiently highOu, increasing Ou further does not help to decrease the errors. Also,among the cases in the upper plots, the best results are obtained bythe choice Ou = Op + 1; this is also similar to Fig. 2 for the linearcase.

4.2.2. Smooth solution: h-convergenceIn this section, the same problem as in Section 4.2.1 is used, but

with coefficients A = 0.01, a = 2.0 for stability. To study the h-con-vergence, we perform the tests on evenly discretized meshes withi � i � i (i = 1,2, . . . ,8) hexahedral elements as previously for thelinear case. With a fixed displacement order Ou = 4, results fromdifferent pressure orders Op = 0,2,3,4 are studied. In Fig. 8 we showthe relative errors of pressure and displacement, and plot these er-rors in logarithmic scale, as a function of element size h ¼ 1

i whichis in inverse logarithmic scale. We see similar results as in Fig. 3 forthe linear case: when Op < Ou, convergence rate of Op + 1 order isobtained; when Op = Ou, only Op order is given. Here results fromOp = Ou and Op = Ou � 1 have similar convergence rates, while inthe linear case the later gave a 1/2 order higher.

5. Test problems

In this section we perform several numerical tests in order toinvestigate the performance of the high-order projection methodwith respect to ellipticity condition, the inf–sup condition, volu-metric locking, and shear locking.

5.1. Ellipticity condition and large deformation

As mentioned in [36,38], for the large deformation cases, some-times the solver fails even when the element satisfies the inf–supcondition because the ellipticity condition fails in the discreteproblem (i.e., (48)). Here, we will test the ellipticity condition fol-lowing proposition 6 of [37].

With ðu; pÞ denoting the initial conditions, and test functionq 2 L2(X), considering static and nearly incompressible problemsfor which k� 1, (34) can be approximated as

ZXðJ � 1ÞqdX ¼ 0 ð76Þ

and we linearize it, to obtain

bðDu;qÞ¼Z

X

12

@ðDuÞ@X

� �T

FðuÞþFðuÞT @ðDuÞ@X

� � !: ðJC�1ÞðuÞqdX¼0;

ð77Þ

where Du denotes the increment of u for each step in the Newton–Raphson iteration. In the continuous problem, by denoting bU and bPthe admissible displacement and pressure spaces, the ellipticity inkernel is

infv2K

aðv;vÞkvk2 > 0; ð78Þ

where

K ¼ v 2 bU and bðv; qÞ ¼ 0; 8 q 2 bPn o; ð79Þ

aðv;vÞ ¼Z

X

12

@v@X

� �T

FðuÞ þ FðuÞT @v@X

� � !: ðeCðuÞ þ C1ðu; pÞÞ

:12

@v@X

� �T

FðuÞ þ FðuÞT @v@X

� � !dX

þZ

XðeSðuÞ þ Sðu; pÞÞ :

@v@X

� �T@v@X

� � !dX: ð80Þ

On the other hand, after discretization, the ellipticity in kernel is:

infV2Kh

VTðKu þ KuÞVkVk2 > 0; ð81Þ

0 0.2 0.4 0.6 0.8

−800

−600

−400

−200

0

L (displacement on x=1)

smal

lest

eig

enva

lue

e min

pressure order = 0pressure order = 1pressure order = 2

Ou=2

0 0.2 0.4 0.6 0.8

−800

−600

−400

−200

0

L (displacement on x=1)

smal

lest

eig

enva

lue

e min

pressure order = 0pressure order = 1pressure order = 2pressure order = 3

Ou=3

0 0.2 0.4 0.6 0.8

−800

−600

−400

−200

0

L (displacement on x=1)

smal

lest

eig

enva

lue

e min

pressure order = 0pressure order = 1pressure order = 2pressure order = 3pressure order = 4

Ou=4

0 0.2 0.4 0.6 0.8

−800

−600

−400

−200

0

L (displacement on x=1)sm

alle

st e

igen

valu

e e m

in

pressure order = 0pressure order = 1pressure order = 2pressure order = 3pressure order = 4pressure order = 5

Ou=5

Fig. 10. Ellipticity test: minimum eigenvalues for element order pairs (Ou,Op). Upper left: Ou = 2. Upper right: Ou = 3. Lower left: Ou = 4. Lower right: Ou = 5.

1 For interpretation of color in Fig. 10, the reader is referred to the web version ofis article.

52 Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57

where

Kh ¼ fVjKTpV ¼ 0g: ð82Þ

We consider two test examples, and we investigate the influence ofdifferent element orders for pressure on the ellipticity condition.From intuition, we can see that for the same element order of dis-placement, the smaller element order we use for pressure, the lar-ger Kh will be, and therefore the more likely is that the ellipticitycondition will be violated. We will demonstrate the extent of thisviolation by obtaining the minimum eigenvalue of Ku(uh,ph) inKhðuh;phÞ, where (uh,ph) is the projection of an analytic solution(see below) in the space formed by the shape functions. A similarstrategy was employed in [36] for testing the ellipticity condition.

5.1.1. First numerical exampleIn our first example, the Mooney–Rivlin model is considered

with the following material property coefficients:

A10 ¼ 2:0; A01 ¼ 0:2; k ¼ 1000 ð83Þ

and the domain for initial configuration is as in Fig. 9. In this test, aprescribed displacement L is applied on X = 1:

ux ¼ �L; uy ¼ 0; uz ¼ 0 ð84Þ

to describe the extent of compression. For the faces Y = 0, X = 0, weclamp them in the normal direction. That is, on X = 0, we fix ux = 0,and on Y = 0, we fix uy = 0. To balance the system, on Y = 1 a tractionforce field is applied as

Tx ¼ Tz ¼ 0; Ty ¼ 0:4Lð4L2 � 16Lþ 27ÞL� 22

L� 1: ð85Þ

Here we employ a 3D code to obtain a 2D solution and hence weimpose on Z = 0 and Z = 1 periodic boundary conditions, i.e.,ujZ=0 = ujZ=1.

Using a single hexahedral element as the mesh for this domain,we obtain the minimum eigenvalues of the stiffness matrix as in(81) which is evaluated at the projected exact solution (u,p). Nextwe study the minimum eigenvalues of Ku(uh,ph) in Khðuh; phÞ by sys-tematically changing the element orders of displacement and pres-sure. Similar to what we did for the convergence rate tests, fixing theelement order Ou of displacement (Ou = 2,3,4,5), the minimum

eigenvalue versus L for different element orders Op of pressure(Op = 0, . . . ,Ou) are plotted in Fig. 10: The horizontal axis denotesthe displacement L on X = 1, which represents the extent of compres-sion. The different lines correspond to different constant k whichrepresents the element order combinations (Ou,Op = Ou � k).

In Fig. 10, we see that for a fixed element order pair (Ou,Op), as Lincreases, the minimum eigenvalues should decrease. That means,the larger the deformation is, the more severe the ellipticity is vio-lated, which was also demonstrated in [37,38] for low-order ele-ments. We also notice that for a fixed element order Op forpressure, when Ou � Op 6 1, the minimum eigenvalues are muchlarger than for Ou satisfying Ou � Op P 2. On the other hand, it isinteresting that when Ou � Op P 2, the minimum eigenvalues havealmost the same magnitude independent of an increase in Ou. Forexample, for L = 0.9, when Op = 1 the minimum eigenvalue is�753.43 for Ou = 2, �814.11 for Ou = 3, �839.19 for Ou = 4 and�841.80 for Ou = 5. Similarly, when Op = 2 the minimum eigenvalueis �379.38 for Ou = 2, �538.43 for Ou = 3, �691.14 for Ou = 4 and�733.14 for Ou = 5. These results suggest that for a fixed integerk, the pair (Ou,Ou � k) should yield better performance as Ou in-creases. For example, for k = 2 at L = 0.9 as highlighted by thered1 circles, we obtain the minimum eigenvalue for (Ou,Op) = (3,1),(4,2) and (5,3) as �814.11,�691.14 and �511.86, respectively. Theseresults point to a particular trend, namely that for high-order Ou ofdisplacement there are more values of the pressure order Op thatlead to simultaneous validity of the ellipticity condition and of theinf–sup condition.

5.1.2. Second numerical exampleIn this second example we will investigate the aforementioned

finding by testing the stability of the numerical solver for differentpairs of interpolation order. We consider the cubic object(0 6 x 6 1,0 6 y 6 1,0 6 z 6 1), with the mesh composed by a sin-gle hexahedral element. The Mooney–Rivlin model is specified bythe following material property coefficients:

A10 ¼ A01 ¼ 1:0; k ¼ 1000: ð86Þ

th

Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57 53

For this case, the face X = 0 is clamped, and a traction force field(Tx,Ty,Tz) is applied on the rest of the faces:

At X = 1:

tx ¼ �2x2

25; ty ¼

4x5; tz ¼ 0; ð87Þ

at Y = 0 and Y = 1:

tx ¼ ny4x5þ 2x3

125

� �; ty ¼ �ny

2x2

25; tz ¼ 0; ð88Þ

at Z = 0 and Z = 1:

tx ¼ ty ¼ tz ¼ 0; ð89Þ

where n = (nx,ny,nz) is the outward-pointing unit vector normal tothe surface. The following body force qf = (fx, fy, fz), is also appliedon the object:

fx ¼4x25

; f y ¼ �45; f z ¼ 0: ð90Þ

This problem has the following analytic solution for the displace-ments of the object:

ux ¼ uz ¼ 0; uy ¼ 0:1x2: ð91Þ

For this test, the increment method (see Appendix B) is appliedby separating the external load into N = 10 parts. In Table 1, we listthe results of the stability study for different pairs of element order(Ou,Op) (Op 6 Ou and Ou = 2, . . . ,11); we leave the pairs with Op > Ou

as blank because they were not tested. Also, we mark with ⁄ thepoint where the range of choices for Op, which yield stable results,increases. We can see that for Ou 6 5, we can only use Op P Ou � 1to maintain stability; when 6 6 Ou 6 10, the range is broadened toOp P Ou � 2; and for Ou P 11, all Op P Ou � 3 are appropriate toobtain numerical stability for this test problem.

Table 1Ellipticity test: stability test for the second test problem. A checkmark denotesstability whereas an ‘‘�’’ instability. The star ‘‘*’’ indicates transition in the stabilitybehavior from one regime to another.

OunOp 0 1 2 3 4 5 6 7 8 9 10 11

2 � p p

3 � � p p

4 � � � p p

5 � � � � p p

6 � � � � ⁄p p p

7 � � � � � p p p

8 � � � � � � p p p

9 � � � � � � � p p p

10 � � � � � � � � p p p

11 � � � � � � � � ⁄p p p p

Fig. 11. Meshes for inf–sup condition test with example in 5.1.1: Left: skewed hexahedrand distorted elements (MESH3).

5.2. Inf–sup condition

In all the examples we presented so far (convergence and ellip-ticity tests) we obtained stable numerical results even whenOp = Ou. To further investigate the inf–sup condition, we revisit herethe problems defined in Sections 4.2.1 and 5.1.1 and perform fur-ther numerical tests with many non-hexahedral elements, includ-ing distorted 3D elements, and compare the accuracy of the results.In addition, we compute the inf–sup constant explicitly for differ-ent interpolation orders (Ou,Op).

For both test problems we use three types of meshes with dif-ferent number of elements as in Fig. 11. Among these meshes,the prismatic elements might be especially problematic accordingto [44,45]. For the problem defined in Section 4.2.1 and for anypairs of (Ou,Op) we tested in the range of Ou 2 {1,2,3,4,5,6,7} andOp 6 Ou, we did not observe any instabilities. The relative errorsfor the choices of Op = Ou � 2, Op = Ou � 1 and Op = Ou are shownin Fig. 12. Although the magnitudes of these errors are different,all the lines are parallel to each other, indicating that we obtainedthe same convergence rate with either the (Ou,Op = Ou � 2),(Ou,Op = Ou � 1) or the (Ou,Op = Ou) pairs. For the problem definedin Section 5.1.1 (see Fig. 9) and imposing a load L = 0.1, we foundsimilar results. Also, following [46–48], we obtained the inf–supconstant

b ¼ infq2bP sup

v2bUbðv; qÞjv jH1kqkL2

ð92Þ

for a mesh with one hexahedron, where b(v,q) is defined as in (77).The results are shown in Fig. 13; we see that for (Ou,Op = Ou) cases,the constant b > 0 and is decreasing as Ou increases. However, for(Ou,Op = Ou � 1) the constant is at least an order of magnitude lar-ger. By decreasing further the pressure interpolation order, i.e., forthe pair (Ou,Op = Ou � 2) the results remain essentially the sameas in the case (Ou,Op = Ou � 1).

5.3. Volumetric locking

For low-order elements the displacement-only formulation suf-fers from the volumetric locking behavior for problems with veryhigh bulk modulus. This can be overcome by increasing the dis-placement order, i.e. typically for Ou P 4 [13]. Here, we comparethe mixed formulation with the displacement formulation for athick-walled tube benchmark problem, first proposed in [12], buthere we are using a three-dimensional mesh. We use a neo-Hook-ean material model with A01 = 0.0; A10 = 0.5. Shown in Fig. 14, thethick-walled tube is under pressure load Pinner = 0.5 MPa (appliedalong the radial direction); the tension on the outer surface is0 MPa. The inner and outer radii are 10 mm and 30 mm, respec-tively. Along the Y-direction, the tube has length 30 mm. Therefore,on the inner and outer surfaces the boundary conditions are (herewe use a follower load approach, see [40])

al elements (MESH1); Middle: prismatic elements (MESH2); Right: Many prismatic

0 2 4 6 810−8

10−6

10−4

10−2

100

Order for displacement (Ou)

Rel

ativ

e er

rors

for p

Op=Ou−2

Op=Ou−1

Op=Ou

0 2 4 6 810−7

10−5

10−3

10−1

101

Order for displacement (Ou)

Rel

ativ

e er

rors

for u

Op=Ou−2

Op=Ou−1

Op=Ou

0 2 4 6 810−6

10−4

10−2

100

Order for displacement (Ou)

Rel

ativ

e er

rors

for p

Op=Ou−2

Op=Ou−1

Op=Ou

0 2 4 6 810−6

10−4

10−2

100

Order for displacement (Ou)

Rel

ativ

e er

rors

for u

Op=Ou−2

Op=Ou−1

Op=Ou

0 2 4 6 810−7

10−5

10−3

10−1100

Order for displacement (Ou)

Rel

ativ

e er

rors

for p

Op=Ou−2

Op=Ou−1

Op=Ou

0 2 4 6 810−7

10−5

10−3

10−1

101

Order for displacement (Ou)

Rel

ativ

e er

rors

for u

Op=Ou−2

Op=Ou−1

Op=Ou

Fig. 12. Inf–sup condition test for the problem in 4.2.1., with relative errors as a function of the element order Ou for displacement: upper: results for MESH1; middle: resultsfor MESH2; lower: results for MESH3; left column: relative errors for pressure; right column: relative errors for displacement.

2 4 6 8 1010−4

10−2

100

Order for displacement

inf−

sup

cons

tant

β

Op=Ou−2Op=Ou−1Op=Ou

Fig. 13. Inf–sup test: inf–sup constants for problem defined in Section 5.1.1 withone hexahedron element mesh.

54 Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57

�rrrð10Þ ¼ 0:5; rrrð30Þ ¼ 0: ð93Þ

On surface X = 0 and Z = 0, the displacements vanish along the nor-mal directions. In other words,

ux ¼ 0 on X ¼ 0; uz ¼ 0 on Z ¼ 0: ð94Þ

In order to describe a tube with infinite length, on surfaces Y = 0 andY = 30, the periodic boundary condition is applied: ujY=0 = ujY=30. Allthe displacements are set to 0 as the initial configuration. As inFig. 14, the domain is decomposed into four hexahedral elements.

We test both mixed and displacement-only formulations forthis problem, with displacement order Ou = 2, 3, 4 in the displace-ment formulation, and displacement/pressure order pairs(Ou,Op) = (2,1), (3,2), (4,3) in the mixed formulation; in additionwe perform a simulation with Ou = 2, Op = 2. The bulk modulusk = 1, 10, 100, 1000, 10,000, 100,000 are tested. The results forthe maximum relative error of displacement along the radial rdirection (marked as ur) at the inner boundary r = 10 mm areshowed in Fig. 15. In this test, the increment method (see AppendixB) is applied by separating the external load into 10 parts.

From Fig. 15 we can see that using low-order elements for dis-placement, the displacement-only formulation suffers from volu-metric locking problem, while the mixed formulation does not.When the order Ou is increased to as high as 4, the displacementformulation overcomes the volumetric locking behavior. This re-sult is in agreement with the findings of [13]. When the bulk mod-ulus is larger than 10,000 (MPa), the accuracy with mixed

Fig. 14. Volumetric locking test: mesh for thick-walled tube.

1 10 100 1,000 10,000 100,0000.001

0.01

0.1

1

10

100

bulk modulus k

max

rela

tive

erro

r in

u r (at r

=10

mm

) (%

)

(Ou,Op)=(2,1)

(Ou,Op)=(3,2)

(Ou,Op)=(4,3)

Ou=2

Ou=3

Ou=4

(Ou,Op)=(2,2)

Fig. 15. Volumetric locking test: maximum relative error for ur at inner boundaryr = 10 mm, with two formulations in thick-walled tube problem.

Fig. 16. Shear locking test: Mooney–Rivlin model. Table 3Shear locking test: comparison of relative errors for L between ABAQUS (ABA) with2500 elements and our formulation (NEK) with one element.

eR ABA Order 1 ABA Order 2 NEK Order 2 NEK Order 3 NEK Order 45.11369e�3 0 9.00268e�1 3.79738e�1 3.86840e�2

eR NEK Order 5 NEK Order 6 NEK Order 7 NEK Order 8 NEK Order 94.14217e�3 2.80897e�3 4.11841e�4 6.83762e�4 2.99113e�3

Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57 55

formulation is more than an order of magnitude higher than that ofthe displacement formulation, reaching almost two-orders of mag-nitude higher accuracy for k = 100,000. It is interesting to also notethat the mixed formulation case of Op = Ou = 2 leads to volumetriclocking identical to the displacement-only formulation for Ou = 2.

Table 2Shear locking test: comparison of length L (deformation on the end x = 5) between ABAQU

L ABA Order 1 ABA Order 23.80724 3.78787

L NEK Order 5 NEK Order 63.77218 3.77723

5.4. Shear locking

Next we investigate the performance of the high-order pressureprojection method in shear locking with a numerical example verysimilar to [11]. It is known [11,21,49] that for thin plates, high-or-der elements can overcome the shear locking problem. Here weuse a 5 � 0.02 � 5 test mesh for the Mooney–Rivlin model. Basedon our results for convergence rate tests, here we simply useOp = Ou � 1 and consider polynomial orders Ou from 2 to 9.

In this example, we test the thin mesh by applying a tractionTy = 0.00001 at the end X = 5. We choose the material coefficientsfor the Mooney–Rivlin material according to [49], as follows:A01 = 0.045, A10 = 0.177, and since incompressibility does not effectthe shear locking phenomenon, we use a smaller bulk modulusk = 0.66666. The numerical results are summarized in Fig. 16. Wealso compare the results for deformation length L between our for-mulation and the commercial code ABAQUS [50] in Tables 2 and 3.In ABAQUS, we used 2500 incompatible-mode elements to over-come the shear locking problem. We take the results with qua-dratic elements in ABAQUS as the reference solution Ls for thecomparison of relative errors eR ¼ jL�Ls j

Ls.

From Fig. 16 and Tables 2 and 3, we can see that using low-or-der polynomial interpolation the deformation is much smaller thanthe accurate solution, which indicates that the shear locking phe-nomenon appears. However, by using higher order elements, thedeformation is larger and it finally converges to the accurate solu-tion. Hence, we obtain converged solutions for this problem forOu P 4, which is consistent with the observations in [21]. We notethat for the shear locking test we obtained similar results witheither the (Ou,Op = Ou � 1) or the (Ou,Op = Ou) pairs.

S (ABA) with 2500 elements and our formulation (NEK) with one element.

NEK Order 2 NEK Order 3 NEK Order 40.37777 2.34947 3.64134

NEK Order 7 NEK Order 8 NEK Order 93.78631 3.79046 3.79920

56 Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57

6. Conclusion

We have presented a pressure projection method for nearlyincompressible materials in solid mechanics using a general Jacobipolynomial basis that can be used to generate 3D polymorphichigh-order elements. We investigated the convergence rates of themethod for both smooth and singular solutions of linear and nonlin-ear elasticity problems, and numerically in some depth the issue ofstability of the method for large deformations. We found that for lin-ear problems the mixed formulation is stable even for elements withpressure interpolation order Op equal to the displacement order Ou,while Op = Ou � 1 gives the best solution accuracy. On the otherhand, to resolve the volumetric locking problem for hyperelasticmaterial, it is required that Op = Ou � 1 and in that case we obtainedaccuracies superior to displacement-only formulation, especially athigh values of the bulk modulus, e.g. by almost two orders of magni-tude. For large deformations we examined the stability of the meth-od by considering the discrete ellipticity in kernel as has beensuggested previously for low-order elements [37]. In this case,stability is enhanced by restricting the pressure space, i.e.Ou P Op P Ou � k, where k depends on the specific value of Ou. Forexample, our numerical tests suggest that for low values of Ou 6 5we require that k = 1; for intermediate values of Ou 6 10 we requirethat k = 2, and for Ou P 11 we require that k = 3. These rather sur-prising results deserve further theoretical investigation, especiallyfor understanding the interplay between the discrete ellipticity con-dition and the discrete inf–sup condition in large deformation anal-ysis. From the practical standpoint, the present study suggests thatthe most robust choice of interpolation order in the mixed formula-tion is Op = Ou � 1, leading to stable and accurate results for linearand nonlinear problems with large deformations, including resolu-tion of the volumetric and shear locking phenomena.

Acknowledgments

The authors thank Profs. Johnny Guzman (Brown University)and Zohar Yosibash (Ben Gurion University) for useful discussions.This work was supported by a NSF/OCI grant.

Appendix A. Shape functions

We employ the hierarchical shape functions based on Jacobipolynomials (see [1] for details). Firstly, we introduce three princi-pal functions /a

i ðxÞ; /bijðxÞ and /c

ijkðxÞ (where the integers i, j, k arein the ranges 0 6 i 6 I, 0 6 j 6 J, 0 6 k 6 K, respectively), whichare defined on the domain �1 6 x 6 1:

wai ðxÞ ¼

1�x2 ; i ¼ 0;

1�x2

1þx2 P1;1

i�1ðxÞ; 1 6 i < I;1þx

2 ; i ¼ I;

8><>: ðA:1Þ

wbijðxÞ ¼

waj ðxÞ; i ¼ 0; 0 6 j 6 J;

ð1�x2 Þ

iþ1; 1 6 i < I; j ¼ 0;

ð1�x2 Þ

iþ1 1þx2 P2iþ1;1

j�1 ðxÞ; 1 6 i < I; 1 6 j < J;

waj ðxÞ; i ¼ I; 0 6 j 6 J;

8>>>>><>>>>>:ðA:2Þ

wcijkðxÞ ¼

wbjkðxÞ; i¼ 0; 06 j6 J; 06 k6 K;

wbikðxÞ; 06 i6 I; j¼ 0; 06 k6 K;

1�x2

� �iþjþ1; 16 i< I; 16 j< J; k¼ 0;

1�x2

� �iþjþ1 1þx2 P2iþ2jþ1;1

k�1 ðxÞ; 16 i< I; 16 j< J; 16 k< K;

wbikðxÞ; 06 i6 I; j¼ J; 06 k6 K;

wbjkðxÞ; i¼ I; 06 j6 J; 06 k6 K;

8>>>>>>>>>><>>>>>>>>>>:ðA:3Þ

here Pa;bn ðxÞða;b > �1Þ are the Jacobi polynomials.

Next we define the shape functions in three-dimensional spacebased on these three principal functions. Denote the coordinates ofthe standard domain as n1, n2, n3, respectively. Then the hierarchi-cal shape functions will be

� For a hexahedral element {(n1,n2,n3)j�1 6 n1,n2,n3 6 1}, theshape function is defined by

/pqrðn1; n2; n3Þ ¼ wapðn1Þwa

qðn2Þwar ðn3Þ: ðA:4Þ

� For a prismatic element {(n1,n2,n3)j�1 6 n1,n3;n1 + n3 6 1;�16 n2 6 1}, the shape function is defined by

/pqrðn1; n2; n3Þ ¼ wapð�g1Þwa

qðn2Þwbprðn3Þ; ðA:5Þ

where �g1 ¼ 2ð1þn1Þ1�n3

� 1.

� For a tetrahedral element {(n1,n2,n3)j�1 6 n1,n2,n3;n1 + n2 +n3 6 1}, the shape function is defined by

/pqrðn1; n2; n3Þ ¼ wapðg1Þw

bpqðg2Þw

cpqrðn3Þ; ðA:6Þ

where g1 ¼2ð1þn1Þ�n2�n3

� 1; g2 ¼2ð1þn2Þ

1�n3� 1.

� For a pyramid element {(n1,n2,n3)j�1 6 n1,n2,n3;n1 + n3 6 1;n2 +n3 6 1}, the shape function is defined by

/pqrðn1; n2; n3Þ ¼ wapð�g1Þwa

qðg2Þwcpqrðn3Þ; ðA:7Þ

where �g1 ¼ 2ð1þn1Þ1�n3

� 1; g2 ¼2ð1þn2Þ

1�n3� 1.

Appendix B. Increment method

The increment method is a widely used method in solvingnonlinear elastic problem such as (28). By dividing the externalload

rext ¼Z@XN

T � vdCþZ

Xqf � vdX ðB:1Þ

into N parts, one can enhance the stability in the Newton–Raphsoniterative procedure while applying a large load directly. Take thedisplacement formulation, for example, (the increment methodfor mixed formulation can be derived similarly), this method canbe rewritten as follows:

� With the initial condition of problem (28), solve

Pðw1Þ ¼Z

XSðw1Þ :

12

@v@X

� �T

Fðw1Þ þ Fðw1ÞT@v@X

� � !dX

� 1N

Z@XN

T � vdCþZ

Xqf � vdX

� �¼ 0; 8v 2 U0;

ðB:2Þby Newton–Raphson iterative procedure, and get w1.� For the ith step, take wi�1 and its derivative as the initial condi-

tion, solve

PðwiÞ ¼Z

XSðwiÞ :

12

@v@X

� �T

FðwiÞ þ FðwiÞT@v@X

� � !dX

� iNðZ@XN

T � vdCþZ

Xqf � vdXÞ ¼ 0; 8v 2 U0;

ðB:3Þ

by Newton–Raphson iterative procedure, and get wi, until i = N.� Take wN as the final solution for problem (28).

Y. Yu et al. / Comput. Methods Appl. Mech. Engrg. 213–216 (2012) 42–57 57

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