simple and accurate four-node axisymmetric element

INTERNATIONAL JOURNAL FOR NUMERICAL METHODS IN ENGINEERINGInt. J. Numer. Meth. Engng 2007; 71:175–200Published online 20 November 2006 in Wiley InterScience (www.interscience.wiley.com). DOI: 10.1002/nme.1939

Simple and accurate four-node axisymmetric element

Magnus Fredriksson and Niels Saabye Ottosen∗,†

Division of Solid Mechanics, Lund University, PO Box 118, Lund SE-221 00, Sweden

SUMMARY

Based on the assumed strain method, a simple four-node axisymmetric solid element is introduced. Theassumed strain field is carefully selected to preserve the correct rank of the element stiffness matrix andto achieve high accuracy. The strain field is developed in conjunction with orthogonal projections andno matrix inversions are needed. The coarse mesh accuracy in bending and in typical axisymmetricload cases is excellent even for nearly incompressible materials. The strain-driven format obtained iswell suited for materials with non-linear stress–strain relations. Several numerical examples are presentedwhere the excellent performance of the proposed simple element is verified. Copyright q 2006 JohnWiley & Sons, Ltd.

Received 7 June 2006; Revised 4 October 2006; Accepted 18 October 2006

KEY WORDS: assumed strain; axisymmetry; Hu-Washizu; quadrilateral; stabilization

1. INTRODUCTION

Low-order finite elements have been the subject for intense research activities for years and severalsuccessful element formulations have been proposed. Originally, many ideas were invented from anad hoc nature. Nowadays, more stringent methods are preferable where the Hu-Washizu variationalprinciple is favourable and extensions into the non-linear regime have become essential. The reasonfor these improvements is that the standard displacement element suffers from drawbacks likevolumetric locking and excessive stiffness in bending. For plane elements, these deficiencies arefairly easy to overcome. Unfortunately, axisymmetry complicates the picture.

Hybrid stress formulations have become essential tools in the development of axisymmetricelements. Based upon the Hellinger–Reissner principle, systematic methods have been proposed andbasic schemes are used to select the assumed stress field, cf. Pian and Chen [1]. The axisymmetricversion of the honoured plane element by Pian and Sumihara [2] was introduced by Tian and Pian[3]. Further improved hybrid stress elements were proposed byWeissman and Taylor [4] introducing

∗Correspondence to: Niels Saabye Ottosen, Division of Solid Mechanics, Lund University, PO Box 118, LundSE-221 00, Sweden.

†E-mail: niels [email protected]

Copyright q 2006 John Wiley & Sons, Ltd.

176 M. FREDRIKSSON AND N. S. OTTOSEN

the FSF and DSF elements and by Wanji and Cheung [5] introducing the refined hybrid stresselement RHAQ6. These elements are all developed in conjunction with an orthogonal projectionwhich was used to make the element stiffness matrix block diagonal.

The three-field Hu-Washizu principle has frequently been used as the variational basis in finiteelement design. Various axisymmetric formulations have been proposed where the element familyby Bachrach and Belytschko [6] which includes the accurate OABI element is representative. Againan orthogonal projection was used both for the stress and the strain fields in order to make thestiffness matrix block diagonal. These elements also take advantage of the �-projection operatorintroduced by Flanagan and Belytschko [7].

In the class of non-conforming elements using incompatible modes, Doherty et al. [8] introducedthe axisymmetric QM5 element which performs very well in bending. Unfortunately, lack of co-ordinate invariance makes this element less attractive, cf. Kavanagh and Key [9]. The axisymmetricversion of the Q6 element was proposed by Wilson et al. [10] and this co-ordinate invariant elementyields excellent results in bending. The fact that Q6 had to be modified to pass the patch test forarbitrary-shaped quadrilaterals did not succeed for the axisymmetric version, cf. Taylor et al. [11].The remedy for incompatible modes elements in axisymmetry did not occur until the enhancedassumed strain method was introduced by Simo and Rifai [12]. Here, the axisymmetric versionof the Qm6 element was motivated from a three-field variational basis. The enhanced assumedstrain method has become popular and numerous element formulations have been proposed. Thetwo enhanced modes element Q1/ME2 by Kasper and Taylor [13] is an element which possessesimproved efficiency with preserved accuracy.

Reduced and selective integration techniques are efficient methods which are suitable for 2D and3D analysis with isotropic materials. For axisymmetry, however, these methods will break down.Improved procedures have been sought for and strain projection methods like the B-bar approachby Hughes [14] has become successful, especially for axisymmetry. Generalization into the non-linear regime has made the B-bar approach even more applicable and variational consistency interms of a three-field Hu-Washizu principle was motivated by Simo et al. [15].

Most efforts made in the development of axisymmetric elements have been done in the fields ofhybrid stress formulations and three-field variational principles. Suppression of volumetric lockingand improved coarse mesh accuracy in bending are properties which have been accomplishedsuccessfully. Extension to inelastic problems may be possible in most cases, but the presence ofthe stress field as an independent field variable makes these formulations slightly inapplicable innon-linear analysis. Hence, a strain-driven format would be more desirable.

The axisymmetric four-node element proposed in this paper is developed with the three-fieldHu-Washizu principle as the variational basis. The interpolated stress field is assumed a priori tobe orthogonal to the difference between the symmetric part of the displacement gradient and theinterpolated strain field. This assumption makes it possible to exclude the interpolated stress fieldfrom the finite element equation, cf. Simo and Hughes [16]. Hence, the proposed element fallswithin the assumed strain method and the strain-driven format sought for is naturally obtained.The interpolated strain field is carefully selected to guarantee rank sufficiency and co-ordinateinvariance of the element stiffness matrix. Related work presented by Fredriksson and Ottosen[17] for plane elements is utilized to improve the coarse mesh accuracy in bending. Volumetriclocking caused by nearly incompressible materials is eliminated always.

First, the standard displacement-based axisymmetric element is summarized where some basicnotations and definitions are presented. Next, this element is reformulated into a convenient formatwhere essential properties and patch test requirements are illustrated. A variational motivation

Copyright q 2006 John Wiley & Sons, Ltd. Int. J. Numer. Meth. Engng 2007; 71:175–200DOI: 10.1002/nme

SIMPLE AND ACCURATE FOUR-NODE AXISYMMETRIC ELEMENT 177

follows and the proposed element is derived explicitly. Finally, several numerical examples are re-ported where the performance of the proposed simple element is compared with other competitiveelements.

2. STANDARD DISPLACEMENT ELEMENT—BASIC NOTATION

The mapping of the four-node quadrilateral is defined by

r =N(�, �)r, z =N(�, �)z (1)

where r and z denote the polar co-ordinates and the nodal co-ordinates are collected into

rT =[r1 r2 r3 r4], zT = [z1 z2 z3 z4] (2)

Moreover, the element shape functions are given by

NI (�, �) = 14 (1 + �I�)(1 + �I�) and N= [N1 N2 N3 N4] (3)

where I = 1, 2, 3, 4 and �, � are the co-ordinates in the parent domain, cf. Figure 1.Similar to (1) and using the isoparametric formulation, the radial displacement u and the axial

displacement v are approximated by

u=[u

v

]=

[N 0

0 N

] [ar

az

]= Na (4)

where aTr =[ur1 ur2 ur3 ur4] and aTz = [vz1 vz2 vz3 vz4] contain the nodal displacements in the r -and z-direction, respectively.

We next introduce the definitions

bT1 = �N(0, 0)

�r= 1

2A[z24 z31 z42 z13]

bT2 = �N(0, 0)

�z= 1

2A[r42 r13 r24 r31]

(5)

Figure 1. Mapping of the bilinear quadrilateral element.



where for instance z24 = z2 − z4, moreover,

sT =[1 1 1 1], hT = [1 − 1 1 − 1], h = �� (6)

where h is obtained by evaluating h = �� at the nodal points. It follows that

bT1 s= 0, bT1r= 1, bT1z= 0, bT1h= 0

bT2 s= 0, bT2r= 0, bT2z= 1, bT2h= 0(7)

Define also the quantity c by

c= 1

A[h − (hTr)b1 − (hTz)b2] (8)

which possesses the following conditions:

cTs= 0, cTr= 0, cTz= 0, cTh= 4

A(9)

Let us now consider a more convenient expression for the displacement field which will be usedin the following sections. According to the procedure described by Belytschko and Bachrach [18]we first write the displacements as

u = �r1 + �r2r + �r3z + �r4h, v = �z1 + �z2r + �z3z + �z4h (10)

where the �-quantities are constant parameters; h is defined by (6)3. Insertion of (1) into (10)yields u = �1 + �2� + �3� + �4�� by which the format (4) is recovered. Evaluation of (10) at thenodal points gives

ar = �r1s + �r2r + �r3z + �r4h, az = �z1s + �z2r + �z3z + �z4h (11)

Premultiply these expressions by sT, bT1 , bT2 and cT, respectively, to obtain

sTar = 4�r1 + �r2sTr + �r3sTz, sTaz = 4�z1 + �z2sTr + �z3sTz

andbT1ar = �r2, bT2ar = �r3, cTar = 4

A�r4

bT1az = �z2, bT2az = �z3, cTaz = 4

A�z4

Determining the �-coefficients from these relations and inserting into (10) give

u =(DT + rbT1 + zbT2 + A

4hcT

)ar , v =

(DT + rbT1 + zbT2 + A

4hcT

)az (12)



where

D= 14 [s − (sTr)b1 − (sTz)b2] (13)

Finally, a comparison of (4) and (12) shows that the N-matrix by (3)2 may be written as

N=DT + rbT1 + zbT2 + A

4hcT (14)

Let us introduce the quantity c defined by

cT =∫AN dA (15)

With this expression as well as (14) we obtain

cT = ADT + VbT1 + WbT2 and V =∫Ar dA, W =

∫Az dA (16)

where use was made of (A6); it appears that V is the element volume per radian. Moreover, withthe use of (7) and (9) we obtain

cTr= V, cTz=W, cTh= 0 (17)

The strain components eu derived from the displacements given by (12) become

eu = ∇u=

⎡⎢⎢⎢⎢⎢⎣

�r

�z

�r z

��

⎤⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(bT1 + A

4

�h�rcT

)ar

(bT2 + A

4

�h�zcT

)az

(bT2 + A

4

�h�zcT

)ar +

(bT1 + A

4

�h�rcT

)az

1

rNar

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(18)

where ∇ is the symmetric part of the gradient operator. The following notation is used:

eu =Ba where B=B0 + B∗ (19)

and

B0 =

⎡⎢⎢⎢⎢⎢⎢⎣

bT1 0T

0T bT2

bT2 bT1

0T 0T

⎤⎥⎥⎥⎥⎥⎥⎦

, B∗ =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

A

4

�h�rcT 0T

0TA

4

�h�zcT

A

4

�h�zcT

A

4

�h�rcT

1

rN 0T

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(20)



The standard displacement finite element method results in

Ka= f (21)

where

K=∫ABTDBr dA, f=

∮LNTtr dL +

∫ANT�br dA (22)

Here, L denotes the boundary curve in the r z-plane, t is the traction vector, b the body forcevector and � is the mass density. Moreover, for isotropic elasticity the constitutive matrix D isgiven by

D= E

(1 + )(1 − 2)

⎡⎢⎢⎢⎢⎢⎢⎢⎣

1 − 0

1 − 0

0 01 − 2

20

0 1 −

⎤⎥⎥⎥⎥⎥⎥⎥⎦

(23)

where E and are Young’s modulus and Poisson ratio, respectively. Clearly, when forming theelement stiffness matrix all quantities involved are expressed in terms of the local co-ordinates(�, �), cf. Appendix A.

Apart from the reformulation of the shape functions described above, this is the standardformulation used when the stiffness matrix for the isoparametric displacement-based element isestablished. However, let us now reformulate the expression for the B-matrix, so that a morefruitful format is achieved which even opens for various mixed finite element formulations of thefour-node axisymmetric element.

3. REFORMULATION OF ISOPARAMETRIC STIFFNESS MATRIX

In the spirit of Flanagan and Belytschko [7], let us introduce B according to

B= 1

V

∫ABr dA (24)

i.e. the quantity Ba becomes the mean strains in the element. Insertion of (20) gives

B=B0 + 1

V

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

A

4�1c

T 0T

0TA

4�2c

T

A

4�2c

T A

4�1c

T

cT 0T

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

where

�1 =∫A

�h�r

r dA

�2 =∫A

�h�z

r dA

(25)



where use was made of (15). We can write the equation above as

B=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

bT1 0T

0T bT2

bT2 bT1

1

VcT 0T

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

wherebT1 =bT1 + A

4V�1c

T

bT2 =bT2 + A

4V�2c

T

(26)

We then define B∗h according to

B∗h =B − B (27)

Insertion of (20) and (25) gives

B∗h =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

A

4H1c

T 0T

0TA

4H2c

T

A

4H2c

T A

4H1c

T

1

rN − 1

VcT 0T

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

where

H1 = �h�r

− 1

V�1

H2 = �h�z

− 1

V�2

(28)

It follows that ∫AB∗hr dA= 0 (29)

Inserting B= B + B∗h into (22)1 and noting that B is a constant matrix whereas B∗

h possessesproperty (29), we obtain

K=Kb + Kh (30)

where

Kb = V BTDB, Kh =∫AB∗Th DB∗

hr dA (31)

It appears that (30) may be interpreted as the sum of a base matrix and a higher order matrix interms of a stabilization matrix much along the lines originally introduced by Kosloff and Frazier[19] for plane stress and plane strain conditions.

In that spirit, we now investigate the null space of the base matrix Kb. Since D is a positivedefinite matrix, the null space is given by the non-trivial solutions

Ba= 0 (32)



With B and a given by (26) and (11), respectively, and noting properties (7) and (9) as well as(17) we obtain

Ba=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

�r2 + �1V

�r4

�z3 + �2V

�z4

�r3 + �2V

�r4 + �z2 + �1V

�z4

A

V�r1 + �r2 + W

V�r3

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(33)

It is concluded that Ba= 0 when

a=⎡⎣0

s

⎤⎦ rigid body translation (34)

and

a=⎡⎢⎣W

As − z

r

⎤⎥⎦

a=

⎡⎢⎢⎣

1

A

(�1 + �2

W

V

)s − �1

Vr − �2

Vz + h

0

⎤⎥⎥⎦

a=

⎡⎢⎢⎢⎣

�1V

(W

As − z

)

−�2Vz + h

⎤⎥⎥⎥⎦

(35)

whereas (34) is the only possible rigid body movement, (35) corresponds to non-zero strains. Ifthe Kh-matrix in (30) is ignored, then the modes in (35) become false zero-energy modes and themost fundamental issue of Kh is therefore to remove these false zero-energy modes. Apart from(35)3 these modes were identified by Matejovic and Adamık [20].

On this background, it is of interest to investigate the null space of Kh and for this purpose itis advantageous to rewrite B∗

h given by (28). Insertion of (14) and (16) gives

B∗h =E∗

hK∗ (36)



where

E∗h =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

A

4H1 0 0 0

0A

4H2 0 0

A

4H2

A

4H1 0 0

A

4

h

r0

z

r− W

V

1

r− A

V

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

, K∗ =

⎡⎢⎢⎢⎢⎢⎢⎣

cT 0T

0T cT

bT2 0T

DT 0T

⎤⎥⎥⎥⎥⎥⎥⎦

(37)

With this formulation, the higher order matrix Kh given by (31)2 can be written as

Kh =K∗TPK∗ where P=∫AE∗Th DE∗

hr dA (38)

Since the constitutive matrix D is a positive definite matrix and the E∗h-matrix certainly is non-

singular, the null space of Kh coincides with the null space of K∗. We therefore investigate thenon-trivial solutions to

K∗a= 0 (39)

Since all four rows of K∗ are linearly independent, the rank of K∗ is four. Accordingly, we musthave four non-trivial solutions to (39). Written explicitly, (39) reads with (37)2

K∗a=

⎡⎢⎢⎢⎢⎢⎢⎣

cTar

cTaz

bT2ar

DTar

⎤⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎣

0

0

0

0

⎤⎥⎥⎥⎥⎥⎦ (40)

With ar and az given by (11), D by (13) and the properties listed in (7) and (9), we obtain

cTar = 4

A�r4, cTaz = 4

A�z4, bT2ar = �r3, DTar = �r1 (41)

In order that K∗a= 0, we must therefore require �r1 = �r3 = �r4 = �z4 = 0, i.e.

a=[

�r2r

�z1s + �z2r + �z3z

]gives K∗a= 0 (42)

which comprises four linearly independent solutions, as expected. As also expected, (42) containsthe rigid body translation given by (34). Moreover, the null space of B given by (35) is notcontained in the null space of K∗ given by (42); clearly, apart from the rigid body translation,the null spaces of B and K∗ are disjunct. With this information, it is of interest to discuss thepatch test.



3.1. Patch test

The patch test requires that an arbitrary element patch is able to reproduce an arbitrary constantstrain state. In that case, convergence is achieved since for a decreasing element size the stress/strainstate approaches a constant one. It follows that it is acceptable that the patch test is fulfilled forinfinitesimally small elements only. In the most general case with the nodal displacements givenby (11), we obtain the strains eu from (27), (33) and (36), i.e.

eu =Ba= Ba + B∗ha

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

�r2 + �1V

�r4

�z3 + �2V

�z4

�r3 + �2V

�r4 + �z2 + �1V

�z4

A

V�r1 + �r2 + W

V�r3

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

+

⎡⎢⎢⎢⎢⎢⎢⎢⎣

H1�r4

H2�z4

H2�r4 + H1�z4

h

r�r4 +

(z

r− W

V

)�r3 +

(1

r− A

V

)�r1

⎤⎥⎥⎥⎥⎥⎥⎥⎦

(43)

where the expressions in (41) were used.If a constant strain state is aimed at, we must require �r1 = �r3 = �r4 = �z4 = 0 and we then

obtain the restriction �r = ��. Consequently, an arbitrary constant strain state cannot be achievedfor an arbitrary size of the element. However, if the element becomes infinitesimally small, thenwe clearly have (z/r −W/V ) → 0 and (1/r − A/V ) → 0. A glance at (43) shows that a constantstrain state now requires �r4 = �z4 = 0, i.e. the contribution from B∗

ha vanishes. Consequently,when the element size approaches zero, the nodal displacements

a∗ =[

�r1s + �r2r + �r3z

�z1s + �z2r + �z3z

](44)

implies

eu = Ba∗ =

⎡⎢⎢⎢⎢⎢⎢⎢⎣

�r2

�z3

�r3 + �z2

A

V�r1 + �r2 + W

V�r3

⎤⎥⎥⎥⎥⎥⎥⎥⎦

(45)

It is concluded that the most general linear nodal displacement field creates an arbitrary constantstrain field when the element size approaches zero, cf. also the discussion given by Taylor et al. [21].In that case only the B-matrix is of importance whereas the B∗

h-matrix does not contribute.Therefore, in a patch test with infinitely small elements only the B-matrix matters.



4. VARIATIONAL FORMULATION

The proposed element is based on the assumed strain method introduced by Simo and Hughes [16].Following, for instance, Belytschko et al. [22] the variation of the Hu-Washizu functional reads

(1

2

∫AeTDer dA

)+

(∫ArT(eu − e)r dA

)=

∮L

uTtr dL +∫A

uT�br dA (46)

where eu denotes the strains derived from the strain–displacement relations, whereas e is theassumed strains; moreover, r denotes the assumed stresses. It is presupposed that these stressesare orthogonal to the difference between the strains eu and the assumed strains e, i.e.∫

ArT(eu − e)r dA= 0 (47)

The assumed strains are approximated according to

e= Ba where B= B + Bh (48)

where the quantity Bh is, as yet, not specified. Since the displacements are approximated accordingto (4), then (46) reduces with (47) to

Ka= f where K=∫ABTDBr dA (49)

and the external forces f are again given by (22)2.Since the stress field does not enter the finite element equation, we do not need to specify it, and

in principle, it is therefore always possible to select a stress field which fulfils the orthogonalitycondition (47). With (48), (19) and (27) we obtain∫

ArT(B∗

h − Bh)r dA= 0 (50)

Noting property (29), the expression above reduces for a constant stress state to∫ABhr dA= 0 (51)

Finally, with (51) and insertion of (48)2 into (49)2 we obtain

K=Kb + Kh (52)

where

Kb = V BTDB, Kh =∫ABThDBhr dA (53)

It appears that the stiffness matrix consists of a base matrix Kb and a higher order matrix Kh.Since Kb is necessary and sufficient to fulfil the patch test for infinitesimally small elements,the fundamental property of Kh is to act as a stabilization matrix and thereby ensuring that thetotal element stiffness matrix K possesses the correct rank. Since there exists only one rigid bodymotion in terms of a translation in the axial direction, the rank of K is required to be seven. In



relation to (32), we have already observed that the rank of Kb is four so we require the rank ofKh to be three. This requirement as well as (51) are the only conditions related to Bh.

With these remarks, we could choose Bh in a form similar to (36). Since only a rank of three isneeded we may, for simplicity, cancel one row in K∗. The first two rows correspond to the planecase investigated by Fredriksson and Ottosen [17] so the question is whether to cancel the thirdor fourth row. As will be discussed later, the third row containing b2 enables us to exactly modelthe case of a narrow circular ring exposed to axisymmetric bending. Consequently, the fourth rowof K∗ is excluded and we choose

Bh =EhK where K=

⎡⎢⎢⎢⎣cT 0T

0T cT

bT2 0T

⎤⎥⎥⎥⎦ (54)

and then the higher order stiffness matrix Kh given by (53)2 becomes

Kh =KTPK where P=∫AEThDEhr dA (55)

5. PROPOSED AXISYMMETRIC ELEMENT

With the base matrix Kb defined by (53)1 we are now ready to choose an appropriate Eh enteringBh in (54)1. According to (48), the assumed strains are defined by the sum

e= eb + eh where eb = Ba, eh =Bha (56)

and whee B is given by (26). In order to motivate the choice of Eh, we first define a local xy-systemwith axes aligned with the principal axes of inertia, cf. Figure 2. The principal axis, around whichthe moment of inertia is maximum defines the local x-axis and the co-ordinates (r0, z0) specifythe position of the element centroid in the global r z-system. The angle � is defined positive inthe counterclockwise direction and it is measured from the global r -axis to the local x-axis; it ischosen such that 0��<180◦. Moreover, by definition and using (16)2,3, we have

V = r0A, W = z0A (57)

Figure 2. Local element co-ordinate system.



The relation between (x, y) and (r, z) is given by[x

y

]=A

[r − r0

z − z0

]where A=

[cos� sin�

− sin� cos�

](58)

and the global displacement vector a transforms according to[ax

ay

]=L

[ar

az

]where L=

[I cos� I sin�

−I sin� I cos�

](59)

where I is the unit matrix of dimension (4× 4).In order to improve the accuracy in bending we will adopt the methodology based upon the

related work by Fredriksson and Ottosen [17] for the plane strain element. In the local xy-systemand with c defined by (8), we calculate the following quantities

qT = [x1y1 x2y2 x3y3 x4y4]mT = [x21 x22 x23 x24 ]nT = [y21 y22 y23 y24 ]

and

�1 = cTq�2 = cTm�3 = cTn

(60)

We can then identify the items

L11 = �1

(�21 + 14�

23)

1/2, L12 =

12�3

(�21 + 14�

23)

1/2

L21 =12�2

(�21 + 14�

22)

1/2, L22 = �1

(�21 + 14�

22)

1/2

(61)

where the L-components above reflect the distortion of the element. For a rectangular element,we have L11 = L22 = 1 and L12 = L21 = 0.

It is next shown that the plane strain element by Fredriksson and Ottosen [17] can be reformulatedinto an assumed strain format. In the local xy-system, the higher order strains e′h for the planestrain case are given by

e′h = E′hK

′a′, e′Th = [�′xx �′yy �′xy], K′ =

[c′T 0

0 c′T

](62)

and

E′h =

⎡⎢⎢⎣

L11y − L21x −L22x + L12y

−L11y + L21x L22x − L12y

0 0

⎤⎥⎥⎦ , =

1 − (63)

Transformation of the local strains e′h into the global r z-system follows from the standard rules oftensor calculus, here given in matrix notation

eh =TTe′h (64)



where the transformation matrix is defined by

T=

⎡⎢⎢⎢⎣

cos2 � sin2 � 2 cos� sin�

sin2 � cos2 � −2 cos� sin�

− cos� sin� cos� sin� (cos2 � − sin2 �)

⎤⎥⎥⎥⎦ (65)

Insertion of (62)1 into (64) and adopting (59) yields

eh =TTE′hK

′La (66)

With the properties K′L=AK′ and K′ = K we obtain

eh = EhKa where Eh =TTE′hA, K=

[cT 0

0 cT

](67)

Finally, insertion of (63)1 and use of (58) yields

Eh =TT

[Q1A

[r − r0

z − z0

]Q2A

[r − r0

z − z0

]]A (68)

where

Q1 =

⎡⎢⎢⎣

−L21 L11

L21 −L11

0 0

⎤⎥⎥⎦ , Q2 =

⎡⎢⎢⎣

−L22 L12

L22 −L12

0 0

⎤⎥⎥⎦ (69)

i.e. the higher order strains eh have now been expressed in the global r z-system. Since the sumof each column of the Q-matrices approaches zero when → 1

2 no locking will occur when thematerial approaches incompressibility. In conclusion, (67)–(69) comprise a reformulation of theplain strain element suggested by Fredriksson and Ottosen [17].

Returning to axisymmetry, Bh must fulfil condition (51). Consider an arbitrary quantity () andthen calculate the quantity () according to

()= () − 1

V

∫A()r dA (70)

It then follows that∫A ()r dA= 0. With this strategy, we determine the quantity Eh by

Eh = Eh − 1

V

∫AEhr dA (71)

With (68) we then obtain

Eh =TT

⎡⎢⎢⎣Q1A

⎡⎢⎢⎣r − 1

V�1

z − 1

V�2

⎤⎥⎥⎦ Q2A

⎡⎢⎢⎣r − 1

V�1

z − 1

V�2

⎤⎥⎥⎦

⎤⎥⎥⎦A (72)



where

�1 =∫Ar2 dA, �2 =

∫Azr dA (73)

It is easily seen that if r → ∞ then (�1/V ) → r0 and (�2/V ) → z0, i.e. Eh approaches Eh. Withthis discussion, Eh entering (54) is chosen as

Eh =

⎡⎢⎢⎢⎢⎢⎢⎢⎣

Eh

0

−z − z0

r

0

0 0z − z0

r

⎤⎥⎥⎥⎥⎥⎥⎥⎦

(74)

Consider the case of a narrow circular ring exposed to axisymmetric bending. As a very goodapproximation, it can be assumed that the cross-section in the r z-plane rotates as a rigid bodyabout its centroid and then the tangential strain becomes proportional to (z − z0)/r as indicatedin (74). In fact, it is easily shown that the base matrix Kb does not contribute and in Eh onlythe third column is activated with the result that the exact strain energy is reflected. Since thesum of each column in Eh approaches zero when → 1

2 , no locking due to incompressibility willoccur. Moreover, it is straightforward to show that when r → ∞ then the proposed axisymmetricformulation approaches the plane strain case; indeed, the plane strain formulation obtained is thevery accurate one proposed by Fredriksson and Ottosen [17]. Finally, it is trivial to see that thesituation of a cylinder with plane end surfaces and exposed to inner and outer pressures onlyactivates the base matrix Kb and that the exact strain energy is obtained. These limiting casessuggest that the proposed element will perform well over a very broad scale of applications and,indeed, this will be demonstrated in the next section.

6. NUMERICAL RESULTS

The performance of the proposed element formulation is investigated and compared with someselected well-known four-node axisymmetric quadrilaterals; the standard displacement element andsix various mixed elements. The examples presented will test sensitivity to volumetric locking fornearly incompressible materials, response in bending dominated problems as well as other typicalaxisymmetric load cases. Since we know, a priori, that the proposed element fulfils the patch test,such load cases will not be reported. All results presented are normalized by an exact solution.The following axisymmetric elements are considered.

1. Q4A: standard displacement-based formulation with 2× 2 quadrature.2. OABI: Hu-Washizu-based element by Bachrach and Belytschko [6].3. FSF and DSF: stress field elements by Weissman and Taylor [4].4. QSR/C: mixed-enhanced element-type C by Simo and Rifai [12].5. Q1/ME2: mixed-enhanced element by Kasper and Taylor [13].6. RHAQ6: refined hybrid element by Wanji and Cheung [5].7. AXINEW: the proposed element.



In addition to AXINEW, both the Q4A and OABI elements are implemented whereas the resultsfor the other elements have been taken from the literature.

6.1. Thick-walled cylinder type I

The first example proposed by MacNeal and Harder [23] involves a typical axisymmetric load caseand it tests the element performance when the material approaches the incompressibility limit;a thick-walled cylinder of unit thickness h = 1 is subjected to a unit internal pressure p= 1. Themesh consists of five elements and both regular and skewed mesh configurations are considered, seeFigures 3 and 4, respectively. For the skewed mesh, tan �a = 0.4 and tan �b = 1. Young’s modulusis E = 1000 and Poisson ratio varies from = 0 to 0.4999. According to Timoshenko and Goodier[24], the exact analytical solution for the radial displacement at r is given by

uref = (1 + )pR2i

E(R2o − R2

i )

[R2o

r+ (1 − 2)r

](75)

where Ri and Ro is the inner and outer radius, respectively. The normalized radial displacementsat the inner radius, i.e. r = Ri are summarized in Table I for the regular mesh configuration andin Table II for the skewed mesh configuration.

Figure 3. Thick-walled cylinder type I, regular mesh.

Figure 4. Thick-walled cylinder type I, skewed mesh.

Table I. Thick-walled cylinder type I, regular mesh.

Q4A OABI FSF DSF QSR/C RHAQ6 AXINEW

0.0 0.994 1.000 0.994 1.000 0.994 0.994 1.0000.25 0.990 1.000 — — 0.991 — 1.0000.30 0.988 1.000 0.990 1.000 0.990 0.990 1.0000.49 0.846 1.000 0.986 1.000 0.986 0.986 1.0000.499 0.359 1.000 0.986 1.000 0.986 0.986 1.0000.4999 0.053 1.000 0.986 1.000 0.986 0.986 1.000



Table II. Thick-walled cylinder type I, skewed mesh.


0.0 0.989 0.998 0.985 0.997 0.994 0.989 0.9960.25 0.984 0.998 — — 0.992 — 0.9960.30 0.982 0.998 0.981 0.997 0.991 0.987 0.9960.49 0.816 0.998 0.976 0.997 0.987 0.983 0.9960.499 0.315 0.998 0.976 0.997 0.987 0.983 0.9960.4999 0.044 0.998 0.976 0.997 0.987 0.983 0.996

Figure 5. Thick-walled cylinder type II, regular and skewed meshes.

As expected Q4A suffers from volumetric locking when the material approaches incom-pressibility. All other elements perform very well especially OABI, DSF and AXINEW whichobtain the exact solution for the regular mesh. For the skewed mesh configuration OABI, DSF andAXINEW are still superior and the performance of these elements is independent of the value ofPoisson ratio.

6.2. Thick-walled cylinder type II

Again a thick-walled cylinder subjected to a unit internal pressure p= 1 is considered, but nowwith two layers of elements; this test was taken from Kasper and Taylor [13]. Here, the followingparameters are used: inner radius Ri = 1, outer radius Ro = 2 and height h = 1. Young’s modulusis E = 250 and Poisson ratio varies from = 0 to 0.49999. Only four elements are used and bothregular and skewed meshes are tested, cf. Figure 5. For the skewed mesh, the central node is offsetby {�r, �z} = {0.12, 0.11}. The analytical solution for the radial displacement is again given by(75) and for the given parameters the exact radial displacement at point A becomes

uref = 1750 (5 + 3 − 22) (76)

The normalized radial displacements are summarized in Tables III and IV. For the regularmesh, OABI and AXINEW yield exact results for this problem even if → 1

2 . Both QSR/C andQ1/ME2 perform well, whereas Q4A locks when the incompressibility limit is approached. Theresult becomes quite similar for the skewed mesh where again Q4A locks in the incompressiblelimit. It should be noted that both OABI and AXINEW are insensitive to Poisson ratio even forthe skewed mesh.

6.3. Thin-walled sphere

A thin sphere subjected to a unit internal pressure p= 1 is considered next. The inner radius isRi = 9 and the sphere thickness is h = 1. Young’s modulus is E = 1000 and different values for



Table III. Thick-walled cylinder type II, regular mesh.

Q4A OABI QSR/C Q1/ME2 AXINEW

0.0 0.982 1.000 0.983 0.982 1.0000.1 0.979 1.000 0.981 0.979 1.0000.2 0.974 1.000 0.979 0.975 1.0000.3 0.964 1.000 0.976 0.970 1.0000.4 0.937 1.000 0.972 0.964 1.0000.49 0.627 1.000 0.967 0.956 1.0000.499 0.145 1.000 0.967 0.955 1.0000.4999 0.017 1.000 0.967 0.955 1.0000.49999 0.002 1.000 0.967 0.955 1.000

Table IV. Thick-walled cylinder type II, skewed mesh.

Q4A OABI QSR/C Q1/ME2 AXINEW

0.0 0.974 0.993 0.979 0.978 0.9920.1 0.970 0.992 0.978 0.973 0.9930.2 0.964 0.992 0.976 0.968 0.9930.3 0.953 0.992 0.973 0.962 0.9930.4 0.922 0.992 0.969 0.954 0.9940.49 0.589 0.992 0.964 0.945 0.9950.499 0.137 0.992 0.963 0.944 0.9950.4999 0.018 0.992 0.963 0.944 0.9950.49999 0.002 0.992 0.963 0.944 0.995

Figure 6. Thin-walled sphere.

Poisson ratio are tested. A regular mesh of 10 elements is used, cf. Figure 6. At the inner radius,the exact radial displacement for a hollow thin sphere subjected to an internal pressure is given by

uref = pRi

E

[1 −

2

R3o + 2R3

i

R3o − R3

i

+

](77)



where Ro is the outer radius, cf. Timoshenko and Goodier [24]. The normalized radial deflectionsat point A and B are listed in Tables V and VI, respectively.

Apparently, all elements obtain non-symmetrical solutions for this problem. This characteristicwas already pointed out by Halleux [25] who suggested additional shape functions for the nodalload evaluation; here standard shape functions are used. All elements respond weaker at point A.For point B, both OABI and AXINEW yield almost the exact solution. It should be noted thatQ4A is the most accurate for low values of Poisson ratio, but it locks when → 1

2 .

6.4. Bending of a long thin-walled cylinder

Consider next a long thin-walled cylinder subjected to an end moment. This problem illustratesthe response when high strain gradients are involved, cf. Doherty et al. [8]. The applied momentis M = 2000 per unit of arc length and the shell thickness h = 1. Young’s modulus is E = 11 250and different values for Poisson ratio are tested. The cylinder is meshed with 17 elements withonly one element through the shell thickness, cf. Figure 7.

Table V. Thin-walled sphere, displacement at point A.

Q4A OABI FSF DSF AXINEW

0.0 1.0235 1.0413 1.0612 1.0589 1.05970.25 1.0271 1.0504 — — 1.07530.30 1.0265 1.0530 1.0762 1.0736 1.07910.49 0.8964 1.0705 1.0848 1.0854 1.09790.499 0.4239 1.0717 1.0851 1.0860 1.09920.4999 0.0677 1.0718 1.0851 1.0861 1.0993

Table VI. Thin-walled sphere, displacement at point B.

Q4A OABI FSF DSF AXINEW

0.0 0.9966 0.9952 0.9938 0.9948 0.99550.25 0.9933 0.9940 — — 0.99380.30 0.9917 0.9938 0.9901 0.9919 0.99350.49 0.8758 0.9921 0.9845 0.9880 0.99360.499 0.4185 0.9919 0.9841 0.9877 0.99370.4999 0.0673 0.9919 0.9840 0.9877 0.9937

Figure 7. Long thin-walled cylinder.



According to Timoshenko and Woinowsky-Krieger [26], the analytical solution for the radialdeflection at point A is given by

uref = − M

2�2D, D = Eh3

12(1 − 2), �4 = Eh

4DR2m

(78)

where D is the flexural rigidity and the mean radius is Rm = 167.7051.This problem verifies the deficiencies of Q4A, namely volumetric locking and poor accuracy in

bending, cf. Table VII. FSF, QSR/C, Q1/ME2 and AXINEW perform excellent and they are notinfluenced by the change of , whereas OABI and DSF clearly soften when the incompressibilitylimit is approached.

6.5. Twisting of a thick ring

The next problem is a simple test which examines the response of a section associated with anon-constant hoop strain. A thick ring is subjected to axisymmetric bending; the applied momentis M = 10 per unit of arc length, measured at the radius Rm. The inner radius is R = 10 and thedimensions of the rectangular cross-section are: width b= 2 and height h = 2, cf. Figure 8. Thering is modelled with one element only; Young’s modulus is E = 210 000. Assuming that the ringrotates as a rigid body around the centroid, the analytical solution for the cross-sectional angulartwist is given by, cf. Odqvist [27]

�ref = MR2m

EKw

(79)

Table VII. Bending of a long thin-walled cylinder, displacement at point A.

Q4A OABI FSF DSF QSR/C Q1/ME2 AXINEW

0.0 0.420 0.985 0.985 0.981 0.985 0.975 0.9850.25 0.465 1.006 0.984 1.017 0.985 — 0.9860.30 0.471 1.018 0.984 1.032 0.985 0.976 0.9870.49 0.243 1.127 0.984 1.129 0.985 0.978 0.9880.499 0.083 1.137 0.984 1.135 0.985 0.978 0.9880.4999 0.014 1.138∗ 0.984 1.136 0.985 0.978 0.988

∗The value 1.02 given in [6] could not be reproduced.

+

Figure 8. Thick ring.



The stiffness Kw for a rectangular cross-section takes the form

Kw = Rmh3

12ln

R + b

R, Rm = R + b

2(80)

The normalized angular twist for different values of Poisson ratio are listed in Table VIII. AgainAXINEW is the most accurate and the insensitivity to Poisson ratio is clear. OABI performs wellbut a minor dependence on is observed. For = 0, Q4A also performs well, but it locks at theincompressible limit. In the discussion following (74), it was claimed that the AXINEW elementprovides the exact strain energy for the problem at hand but Table VIII shows that AXINEW isslightly too flexible. However, (79) is based on the assumption that the cross-section rotates asa rigid body. But in reality, since the moment is simulated by means of nodal forces, a certainflexibility exists.

6.6. Uniformly loaded circular plate

This plate bending problem covers simply supported and clamped circular plates. Both regular andskewed meshes are considered, cf. Figures 9 and 10. The plates are subjected to a unit normalpressure p= 1. The plate radius is R = 10 and the thickness is h = 1. Young’s modulus is E = 1875and the plate flexural rigidity D is defined by (78). Poisson ratio varied from = 0 to 0.4999. Forthe skewed mesh tan � = 0.5, cf. Figure 10. The analytical solution for a uniformly loaded circularplate is given by Timoshenko and Woinowsky-Krieger [26]. The central vertical displacement fora simply supported circular plate is given by

uref = pR4

64D

(5 +

1 + + 4

3

3 +

1 − 2h2

R2

)(81)

Table VIII. Twisting a thick ring.

Q4A OABI AXINEW

0.0 1.003 1.003 1.0030.25 0.910 0.989 1.0020.30 0.867 0.985 1.0020.49 0.560 0.994 1.0010.499 0.290 0.999 1.0010.4999 0.054 1.000 1.001

Figure 9. Simply supported circular plate, regular mesh.



Figure 10. Simply supported circular plate, skewed mesh.

Table IX. Simply supported circular plate, regular mesh.


0.0 0.676 1.061 1.025 1.025 1.026 — 0.9700.25 0.688 1.054 1.025 1.025 1.028 1.037 0.9740.30 0.645 1.049 1.027 1.027 1.028 — 0.9750.49 0.078 1.109 1.026 1.026 1.034 1.043 0.9770.499 0.021 1.130 1.026 1.026 1.031 — 0.9770.4999 0.015 1.132 1.026 1.026 1.031 1.043 0.977

Table X. Clamped circular plate, regular mesh.

Q4A OABI AXINEW

0.0 0.412 1.048 0.9690.25 0.455 1.050 0.9640.30 0.451 1.047 0.9630.49 0.119 1.094 0.9560.499 0.059 1.110 0.9550.4999 0.052 1.112 0.955

whereas for a clamped circular plate we have

uref = pR2

64D

(R2 + 4h2

1 −

)(82)

For regular meshes, AXINEW is slightly better than RHAQ6 and significantly better than OABIand Q4A, cf. Table IX and X. Similar to the other three remaining elements, AXINEW shows noinfluence when Poisson ratio is changing. A comparison of Table IX and XI shows that with themesh configuration being skew, all elements stiffen. Moreover, RHAQ6 and AXINEW turn out tobe less sensitive to changes of Poisson ratio than the other elements.

6.7. Bending of a short thin-walled cylinder

The final example is inspired by the cantilever beam test by Pian and Sumihara [2]. This test willinvestigate mesh distortion sensitivity in bending. For large values of the radius R this problemdegenerates into the 2D plain strain problem and the influence of the hoop strain vanishes; here,the radius is set to R = 10 000. A cylindrical shell is exposed to bending moments at its ends;



Table XI. Simply supported circular plate, skewed mesh.

Q4A OABI FSF DSF RHAQ6 AXINEW

0.0 0.584 0.849 0.853 0.854 — 0.8860.25 0.611 0.898 0.913 0.915 0.928 0.9090.30 0.580 0.904 0.924 0.926 — 0.9120.49 0.087 0.990 0.965 0.968 0.977 0.9270.499 0.024 1.008 0.967 0.970 — 0.9280.4999 0.017 1.011 0.967 0.970 0.979 0.928

+

Figure 11. Short thin-walled cylinder.

symmetry is used and only the upper half of the ring is modelled, cf. Figure 11. The ring issubjected to an edge moment M = 20 per unit of arc length, measured at R. Young’s modulusis E = 210 000 and for simplicity, only the case for Poisson ratio = 0 is considered. The exactdisplacement at point A is easily derived from the beam theory and for the given parameters weobtain

uref = −75M

E(83)

The results are summarized in Figure 12 where the normalized tip deflection is plottedagainst the mesh distortion measure a. Some numerical values for the skewed meshes are listed inTable XII where the distance a is translated into an element angle.

As expected, and discussed in relation to (74) the AXINEW responds exactly as the 2D planestrain case reported in Fredriksson and Ottosen [17]. For rectangular elements, OABI and AXINEWyield the exact solution whereas Q4A is far too stiff.

For small up to moderate angles, i.e. about 0–25◦, the proposed AXINEW element is the mostaccurate. Even for severely distorted elements where the element angles may be 45◦ or more,AXINEW is by far the most superior, cf. Figure 12.



0 1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

AXINEW

Q4A

OABI

Mesh Distortion: a

Nor

mal

ized

Hor

izon

tal D

ispl

acem

ent a

t A

Figure 12. Bending of a short thin-walled cylinder.

Table XII. Short thin-walled cylinder, samples.

a Angle Q4A OABI AXINEW

0.0 0.0 0.242 1.000 1.0000.25 14.036 0.225 0.778 0.9380.50 26.565 0.188 0.487 0.8131.00 45.000 0.127 0.240 0.637

7. CONCLUSIONS

The variationally consistent assumed strain element proposed shows excellent performance forbending dominated problems and for other typical axisymmetric load cases. Coarse mesh accuracyand low distortion sensitivity are characteristic qualities. The element performance is not affectedby the Poisson ratio not even for materials approaching the incompressibility limit. Evaluation ofthe element stiffness matrix is straightforward and no matrix inversions are needed. The element isco-ordinate invariant and the element stiffness matrix possesses correct rank always. Compared withthe standard displacement element no computational efficiency is gained, but the performance interms of accuracy is improved significantly. Compared with other well-established mixed elements,the proposed element is significantly simpler and the accuracy is similar or better. The strain-drivenformat obtained from the assumed strain method is well suited for non-linear analysis in solidmechanics.

APPENDIX A

In this Appendix, some standard relations are recalled. We have

r = 14 (r + ar� + br� + cr��), z = 1

4 (z + az� + bz� + cz��) (A1)



where

r = r1 + r2 + r3 + r4, cx = r1 − r2 + r3 − r4

z = z1 + z2 + z3 + z4, cz = z1 − z2 + z3 − z4(A2)

as well as

ar =−r1 + r2 + r3 − r4, az = −z1 + z2 + z3 − z4

br =−r1 − r2 + r3 + r4, bz = −z1 − z2 + z3 + z4(A3)

It follows that the determinant j of the Jacobian matrix becomes

j = 116 [4A + (arcz + azcr )� + (bzcr − czbr )�] (A4)

With h = ��, we have

�h�r

= 1

j

(�z��

� − �z��

�

),

�h�z

= 1

j

(−�r

�� + �r

��

)(A5)

We also observe that ∫Ah dA=

∫ 1

−1

∫ 1

−1�� j d� d�= 0 (A6)

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simple and accurate four-node axisymmetric element

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