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AANNIISSOOTTRROOPPIICC AAXXIISSYYMMMMEETTRRIICC FFEEMM

ADVANCED FINITE ELEMENT METHODS ASEN 5367

PROFESSOR CARLOS A. FELIPPA

PETER GRAY AND VIVEK KAILA

May, 2006

CONTENTS

1. Introduction 3 2. Problem Statement 4

3. Anisotropic Materials 5 4. Element Choice and Formulation 8

4.1 Total Potential Energy Formulation 8 4.2 Hellinger Reissner Formulation 10 4.3 Stress Hybrid Formulation 12

5. Benchmark Problems and Results 14 5.1 Comparing different Stress Assumptions 14 5.2 Plate Bending Problem 17

5.2.1 Isotropic Material 18 5.2.2 Orthotropic Material 20 5.2.3 Analysis 24

5.3 Radially Loaded Column 25 5.4 Axisymmetric Column Loading 27

6. Conclusion 33

7. References 34

A. Appendix 35

A.1 Case I Radial Alignment 35 A.2 Case II Transverse Alignment 36 A.3 Case III Circumferential Alignment 37

2

1. INTRODUCTION The use of anisotropic materials has become fairly prevalent and more significant beginning in the latter part of the twentieth century. The need for new composite fiber materials and their development was influenced in part by the rapid growth of the aircraft and aerospace industry. This meant that composite materials were available in which the strength, stiffness and weight of the materials could be modeled around the desired application. This is seen quite nicely in the case of aircrafts where it is desired to orient an increased stiffness in the direction of the load, at the same time however, maintaining a desirable strength to weight ratio. In this work we present a finite element solution for solving these anisotropic problems for problems that can be simplified by applying the symmetric boundary conditions about the axis of symmetry. We have decided to use three types of element formulations to solve the axisymmetric problem using finite elements. These are the Isoparametric displacement element, the stress-displacement HR element, and the stress hybrid element. We have presented three benchmark problems. These include the plate bending problem in which a point load is applied to the center of a circular flat plate, the column bending problem, in which a lateral force is applied to an infinite column, and the column wrap problem, where we simulate an infinite isotropic column wrapped around by an orthotropic composite. For each benchmark problem we experiment with assigning an increased stiffness in one direction, giving the other two directions relatively smaller and equal stiffness values, essentially making the problem orthotropic. By doing this we hope to simulate composites with unidirectional fiber alignment. We repeat the problem (in cylindrical coordinates) for radial alignment (along the r axis), transverse alignment (along the axis), and circumferential alignment (alongz ). A brief theory of anisotropy is presented in the next section. This is followed by the element derivations for the stiffness matrix. A small discussion about the assumption of the stress fields is included in this section. Results for the three benchmark problems are presented followed by a conclusion and appendix.

3

2. PROBLEM STATEMENT This section presents, briefly, the statement of the axisymmetric problem. A three dimensional problem is reduced to two dimensions by applying symmetry about the axis of revolution, provided the load application is symmetric about the axis as well. Using a cylindrical coordinate system is most pertinent to the problem at hand. The displacement field is defined by the translation in the radial and transverse directions,

(1) The kinematic equations govern the strain-displacement relationship and are given as follows,

(2) The constitutive equation governs the stress-strain relationship and is given by,

(3) The equilibrium equations for the reduced two dimensional axisymmetric problem is,

(4) The boundary conditions are the displacement boundary conditions (primary/dirichlet) or the force boundary conditions (flux/naumann).

4

3. ANISOTROPIC MATERIALS A generally anisotropic body is one with different elastic properties for different directions drawn through a point. Each strain component is a linear function of six components of stress [1]. This is show in the stress-strain relationship for a three dimensional material where, the symmetric elasticity matrix includes 36 non-zero entries, 21 of which are independent:

(5)

=

12

31

23

3

2

1

66

5655

464544

36353433

2625242322

161514131211

12

31

23

3

2

1

EEESYMEEEEEEEEEEEEEEEEEE

LLLLM

LLM

LLM

LM

M

Simplifications can be made, however, by assuming symmetry in the material properties. Composite fibers are often aligned to produce a transversely isotropic material where the axial fiber direction is strong and orthogonal directions are weakly bound, as shown in Fig.1. Axis is the direction of fiber alignment and and are the transverse direction with equal material properties.

1x 2x 3x

2x

3x

1x Figure 1: Unidirectional Fiber Alignment in Transversely Isotropic Material

5

Application of the symmetry on the 2 and 3 axes reduces the stress-strain relationship by eliminating the shear-extension coupling and shear-shear coupling from the elasticity matrix:

(6)

=

12

31

23

3

2

1

66

55

44

33

2322

131211

12

31

23

3

2

1

000000000000

EESYM

EEEEEEE

LLLLM

LLM

LLM

LM

M

For axisymmetric problems the constitutive equation can be further simplified by applying axisymmetry to a generalized two dimensional case. Expressing the constitutive equation in terms of the compliance matrix ( )1= EC yields [1]:

(7)

=

rz

z

r

rz

zzrr

zzrrz

rzzrr

rz

z

r

GEEE

EEEEEE

/10000/1//0//1/0///1

The fiber direction can be oriented in either the radial, transverse, or circumferential directions, r , z , or , respectively. By using to represent the direction of fiber alignment and and to represent the transverse directions where properties in the direction can be equated to , three similar compliance matrices are constructed for each orientation.

1x

2x 3x

3x 2x

For radial, r , orientation:

(8)

=

rz

z

r

rz

z

r

GEEE

EEEEEE

12

22211

22211

11111

/10000/1//0//1/0///1

6

For transverse, , orientation: z

(9)

=

rz

z

r

rz

z

r

GEEE

EEEEEE

12

11122

22111

11112

/10000/1//0//1/0///1

For circumferential, , orientation:

(10)

=

rz

z

r

rz

z

r

GEEE

EEEEEE

12

11111

11222

11222

/10000/1//0//1/0///1

7

4. ELEMENT CHOICE AND FORMULATION The choice of elements used is usually dependent on more than one factor, and this decision is often not an easy one. This work uses three different element types derived for axisymmetric problems. The three, although derived using the variational principle, are different in their field types. The first of these is the total potential energy (TPE) formulation which employs displacements as the master field. This is also called a single field type variation. The second is the Hellinger-Reissner (HR) formulation which arises from a mixed field type variation and has displacements and stresses as master fields. The third type of element is derived using the hybrid principle where field continuity restrictions between elements are relaxed. In this case the infield (or interior) functional is combined with an interface potential to form the hybrid functional. The finite element equations can then be derived by satisfying the variational equations to obtain the stiffness matrix, force and stress terms. In this work the interior functional uses a stress master field (a TCPE formulation) and the interface potential is derived from boundary displacements of the interfaces. The following subsections provide brief descriptions of each of the three element formulations mentioned above. For the axisymmetric problem we confine this to a four node quadrilateral elements with two translational degrees of freedom ( and ) on each node. ru zu 4.1 Total Potential Energy Formulation The displacement master field is assumed to vary according to the isoparametric formulation of the element. (11)

=

n

zz

rr

z

r

N

NN

uu

uu

uu

n

n

M2

1

.....

......

1

1

The TPE functional from linear elasticity is stated as follows,

=V S

iiiiV

iju

iju

TPE

t

dSutdVubdVe 21 (12)

Assuming body forces to be zero we can rewrite this in matrix-vector form. Also for an axisymmetric element with a one radian ring,

8

rdAdV = (13)

and so,

= t

e S

TuTTPE urdtrde 2

1)(

(14)

Since and are slaves of the displacement field, the strain energy is, e

=)()( 2

121

ee

rdEeerde uuTuT (15)

The strains are obtained from the gradients of the displacement field. And so,

(16) Bueu =where,

DNB = (17) D is the strain-displacement gradient matrix and is the shape function matrix, N

(18) and,

(19)

The stiffness matrix is obtained by numerical integration over the element domain and so,

(20)

9

where l and k denote the gauss point coordinates in the natural coordinates system and and are the gauss point weights. kw lw

4.2 Hellinger-Reissner Formulation We begin the derivation of the element stiffness matrix by first stating the general HR functional from linear elasticity,

=V S

iiklijklijV

iju

ijHR

t

dSutdVCdVe 21 (21)

Note as before we have no body forces and displacements as a master field. In addition we also have stresses as a master field which is no longer a slave of the displacement field. The displacements are allowed to vary according to the isoparametric formulation of the element. The stresses however vary with the geometry as follows, Sa= (22)

where represents a linear field in S r or z or both, and a is a vector of coefficients. In this work we begin by assuming the following in equation (23), (23)

zaraa

raa

raa

razaa

rz

zz

rr

1084

73

62

951

++=

+=

+=

++=

where 0zzz = and 0rrr = . and are the element centroids. 0z 0r These equations are made to satisfy the axisymmetric equilibrium equations and so the matrix is, S

+

=

0000000001

00001000001

rrrrz

rrz

S (24)

10

For the axisymmetric element,

=)()(

21

et

e S

TTuTHR urdtrdCrde (25)

The strains are derived from the displacements and so as in the TPE functional,

(26) Bueu = Replacing and in the functional we get, ue

=)()(

21

et

e S

TTTTTHR urdtCSardSaBurdSa (27)

Taking the constant terms out of the integral we may rewrite the above functional as, (28) ufFuaMua TTTHR = 2

1

where the matrices and vector are, FM , f (29)

=)(e

BrdSM T

(30.a)

=)(e

CSrdSF T (30.b) =

St

rdtf

where M is , is and is 87 F 77 f 18 . The variational principal is applied to derive the 77 stiffness matrix K ,

11

(31)

MFMKMuFa

u

a

T

HR

HR

1

1

0

0

=

=

=

=

4.3 Stress-Hybrid Formulation The stress hybrid formulation makes use of the TCPE interior functional and the interface potential based on the interface displacements. As before, we begin with the stress hybrid functional from linear elasticity,

+=V S

iiS

jijiklijklijCd

t

dStddSnddVC 21 (32)

In this case the stresses are chosen to vary with r and , and we use the same stress assumption as is used for the Hellinger Resissner formulation in equation (22),

zSa= .

For axisymmetric element the above functional can be rewritten as, (33)

ufGuaFua TTTCd +=21

where, (34)

=)(

Pre

dTG T

(35)

=)(e

CSrdSF T

12

(36) =

St

dtf Pr where,

(37) where ij is the isoparametric side coordinate that goes from -1 on node i to +1 on node j . This can be written as,

(38)

The matrix T is formed by row stacking each product jiji nt = formed from the in plane traction components and which is associated with the interior stress field on the interface potential.

xit yjt

Since the functional is stationary with respect to the stress and displacements we can derive the elemental stiffness matrix K ,

(39)

(40)

and so,

(41) GFGK T 1=

13

5. BENCHMARK PROBLEMS AND RESULTS 5.1 Comparing different Stress Assumptions In this section we present two choices of assumed stresses for the stress based HR element and the stress hybrid element. It is intended as a demonstration of how this choice dictates the final solution. In the stress field assumed given in the midterm exam, the radial, transverse and hoop stress components had a linear dependency on r and z . The shear component however, is dependent on r only. Therefore we have,

(42) This stress field with 11 coefficients can be reduced to an equivalent field with 7 coefficients by making the above set of four equations satisfy the axisymmetric equilibrium equations. And so we get,

(43) Quite clearly, this matrix choice usually poses computational problems. This is attributed to the r term in the denominator of some of the terms which means we cannot solve problems with such as the solid circular plate with a central load described in 5.2. 0=r To fix this, in this work we chose a different stress distribution in which only the radial and shear term varied with r and z . The transverse and hoop stresses vary with r only. There is no underlying reason for this choice. It is based solely on empirical observation, and seen to work better and do away with the problem of singularity at the axis. Thus the assumption is made as in equation (23),

14

zaraa

raa

raa

razaa

rz

zz

rr

1084

73

62

951

++=

+=

+=

++=

And as before we can reduce this to a 74 system where the matrix S is as in equation (24),

+

=

0000000001

00001000001

rrrrz

rrz

S

In addition to removing the problem at the axis we find, more importantly, that this gives a much better approximation of the displacements compared to the stress field assumed earlier. This is verified by applying it to the pressure in a tube benchmark problem described by the figure below,

6P=10

2

z

Figure 2: Internal Pressure in a Thick Tube

The problem is descritized using four stress HR elements. The results are as follows,

15

5 6 7 8 9 10

2.252.5

2.753

3.253.5

Radial displacements Hblack=exact ,red=FEML

Figure 3: Transverse Displacement Using Assumed Stress Field in Case 1

5 6 7 8 9 10

2.22.42.62.8

33.2

Radial displacements Hblack=exact ,red=FEML

Figure 4: Transverse Displacement Using Assumed Stress Field in Case 2

We see a much better displacement resolution using the second approximation compared to the first one. We can therefore see that the choice of the stress assumption greatly changes the quality of approximation, and in some cases extend beyond that to as in this case by helping us avoiding problems near the axis.

16

5.2 Plate Bending Problem

he first benchmark problem is a fairly classical problem involving the application of a

he cross section of the circular plate is discretized using ten elements. The problem is

10 Elements

10=F 1000

Tpoint load at the center of a circular flat plate. The analytic solution for this is well established. The application of the load at the center of the plate allows the problem to be symmeterized about the axis. We can thus solve it using the axisymmetric finite elements we developed in the last section. The problem description is given in the figure below.

Figure 5: Plate Bending Benchmark Problem

Trepeated using the four node Isoparametric displacement element, the Hellinger Reissner (HR) element, and the stress hybrid element. The problem is solved first for a purely isotropic material and for orthotropic materials with the constitutive matrices chosen to simulate composites with fiber alignment along the radial, transverse and circumferential directions. The figure below is the relatively course mesh that is used to carry out the computations.

Figure 6: Plate De

=E

F

0=

10

2

Thin SS circular plate

1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

11 13 15

16

17

18

19

20

21

22

Elements : 10

12 14

scritzation using

17

5.2.1 Isotropic Material

for a purely isotropic case. The isotropic problem uses an lasticity modulus of 1000 with

This section presents the resultse 0= . The transverse displacements for the three cases

ed. Th ent is,

under 10=F units are presented. The analytic solution for a thin plate under circular load for an isotropic problem is first present e transverse displacem

(44)

The radial stresses and hoop stresses are,

where the bending moments are,

(45)

2 4 6 8 10

-0.08

-0.06

-0.04

-0.02

TransverseHzL displacements Hblack=exact,red=FEML

2 4 6 8 10

1

2

3

4

5

Radial stresssigrr Hblack=exact,red=FEML

2 4 6 8 10

2

3

4

5

6

Hoopstress sigqq Hblack=exact,red=FEML

Figure 7: Isotropic Plate Bending Using Isoparametric Displacement Elements

18

2 4 6 8 10

-0.6

-0.4

-0.2

TransverseHzL displacements Hblack=exact ,red=FEML

2 4 6 8 10

10

20

30

40

Radial stresssigrr Hblack=exact,red=FEML

2 4 6 8 10

10

20

30

40

Hoopstress sigqq Hblack=exact,red=FEML

Figure 8: Isotropic Plate Bending Using HR Elements The plots show quite clearly that the displacement elements fair quite well for the plate bending and even with 10 elements is able to resolve the transverse displacements fairly well. The stress resolve much better only when the number of elements is increased. The Hellinger-Reissner element on the other hand is not able to track the displacement or the stresses for the plate bending problem. Increasing the number of elements did not significantly improve results. Since these contribute little to the result, these plots have been excluded.

he stress hybrid el ces the exact same

plagues the stress elements for the orthotropic plate bending roblems as well.

T ement as discussed in the earlier section produstiffness matrix as the HR element for the plate bending problem. These too produce transverse displacements about an order higher than the exact solution.

he nature of this discrepancy in the solutions of the stress elements from the Tdisplacement elements is not understood entirely. It is interesting to note that the stress elements behave quite nicely for the radial loading problem that is discussed in the column loading benchmark in the next section. The stress elements in our experience however, seem unsuited for the transverse loading presented above. And as we will see in .2.2, the same problem5

p

19

5.2.2 Orthotropic Material In this problem the three following cases are considered: Case I: Radial fiber alignment:

Case II: Transverse fiber alignment:

100=E1000=rE 100=zE

100=E100=rE 1000=zE 100=rzG

100=rzG

Case III: Circumferential fiber alignment:

he solutions are presented below. Due to the undependable nature of the solutions from g problem (as mentioned in the

revious subsection, 5.2.1) these solutions are omitted in this section but are included in re using exclusively Isoparametric displacement

he elasticity modulus in the radial direction is kept larger in the radial direction and the and circumferential directions.

E 100=r 1000=E100=zE 100=rzG Tthe HR and Stress hybrid element from the beam bendinpthe appendix. Results presented aelements. Also, due to the absence of an analytic solution to this problem, the benchmark roblem was repeated using ANSYS. Results of the two cases are presented side by side. p

Case I Tproperties are kept smaller in the transverse

20

Figure 9: Radial Alignement, Max Transverse Displacement at A: -0.464677

Figure 10: ANSYS Solution, Max Transverse Displacement at A: -0.469069

Displacement component uz

A

21

Figure 11: Radial Alignment, Max Radial Stress at A: 25.619

Figure 12: ANSYS Solution, Max Radial Stress at A: 25.345

Nodal stress sigrr

A

22

Figure 13: Radial Alignment, Max Hoop Stress: 2.5619

Figure 14: ANSYS Solution, Max Hoop Stress: 2..535

A

Nodal stress sig

23

The results from our Mathematica code are in close agreement with the ANSYS solutions.

Case II In this case the elasticity modulus in the transverse direction is kept higher than the other two directions in order to increase the stiffness in the transverse direction. Since including all the Mathematica and ANSYS plots would make this report too lengthy we sim list the results for this case and the next. The solutions are listed as follows: Max. Transverse D

ax. Radial Stress: 4.28416 vs 3.989 (ANSYS)

Min. Transverse Stress: -8.49313 vs -8.493 (ANSYS) Max. Shear Stress: 3.51803 vs 0.942 (ANSYS) Case III Here the elasticity modulus is kept higher in the circumferential direction and equal in the radial and transverse directions. The results are listed as follows: Max. Transverse Displacement: -0.1807 vs -0.1898 (ANSYS)

Max. Radial Stress: 0.299906 vs 0.26517 (ANSYS)

Min. Transverse Stress: -3.25208 vs -3.252 (ANSYS) Max. Shear Stress: 4.8366 vs 3.045 (ANSYS) 5.2.3 Analysis The results presented for the three cases indicate that the maximum transverse displacements in the three cases is least for Case III, i.e. when the stiffness in increased in the circumferential direction. This is quite an interesting observation since the force is applied in the transverse directions would lead one to believe that maximum resistancewould be offered when the maximum stiffn ss is parallel to the direction of the appliedforce.

Also, the maximum radial stress in Case III is measured at 0.299906 compared to 4.28416 in Case II, and compared to 25.619 in Case I. This makes perfect sense when we look at these results in context to the maximum shear stress values for the three cases. Although the shear stress values of our code do not correlate perfectly with ANSYS, we do observe that these are significantly larger for the Case III compared to cases I and II.

ply

isplacement: -0.918983 vs -0.942101 (ANSYS) M

e

24

This implies that the increased stiffness in the circumferential direction allow the stresses to propagate in the circumferential directions and reduce the loads on the radial axis. This

distribute the load along the circumferential direction is probably the reason hy Case III is able to produce the least maximum transverse displacement of the three

problem described by ig. 15, a column with an inward radially applied load. This condition represents an

ability towcases. 5.3 Radially Loaded Column The three elements were tested under the loading conditions for the Fexternal pressure, for example.

Figure 15: Radial Column Loading B

enchmark Problem

he stiffness matrices for the Hybrid4 element approaches infinity as rT approaches zero. d to the center of the column. The

uad4 and HR elements do not contain a actor in the stiffness matrix formulation al comparison, nevertheless, the hole, of

dius is applied to all three elements.

of the axisymmetric elements with the given loading conditions were ssessed with a transversely isotropic material with the three orientations and elasticity

To avoid computation errors, a small hole is introducer/1 fQ

and therefore, do not require a hole. For equ 1.0=rra

The capabilitiesamatrices shown in Section 3. Ten elements were used to form the column. They are constrained in the z - direction to simulate an infinitely long column. Case I: Radially Aligned Fibers. The displacement and HR elements produce nearly exact displacements with the stress hybrid element slightly less stiff. Because of the small r value near the axis of revolution the stress hybrid also produces large errors near the axis.

25

2000=rE , 100=zE , 100=E , 100=G

Radial DisplacementRadial Orientation

-1.2

-1

-0.8

0

-0.6

-0.4

-0.2

lace

men

t

0 2 4 6 8 10

Node Distance

Dis

p Quad4HRHybrid4

Figure 16: Element Comparison for Radial Orientation

Case II: Transversely Aligned Fibers. With the exception of the stress hybrid nodes nearest the axis, all elements produce nearly equivalent displacements. 100= , 2000rE =zE , 100=E , 100=G

Radial DisplacementTransverse ientation

00 2 4

Or

-0.26 8 10

D -0.8

-0.6

-0.4

ispl

acem

ent

Quad4

HR

Hybrid4

-1.2

-1

Node Distance

Figure 17: Element Comparison for Transverse Orientation

26

Case III: Circumferential Aligned Fibers. All elements produce accurate displacements

its..

to three significant dig

100=rE , 100=zE , 2000=E , 100=G

Radial DisplacementCircumferential Orientation

-0.25

-0.20

-0.15

-0.10

-0.05

0.00

0.05

0 2 4 6 8 10

Node Position

Dis

plac

emen

t

Quad4HRHybrid4

Figure 18: Element Comparison for Circumferential Orientation

5.4 Axisymmetric Column Loading As an example of the potential applications of a transversely isotropic material, a cylindrical column is wrapped with a thin outer layer consisting of fiber composite. The column is intended to represent the structural support of building and the outer wrap is a material designed to protect the support column from a force resulting from an explosive detonation. It is theorized that an outer wrap with the correct material properties can reduce the stress inside the column and prevent the detonation from destroying the structure. The axisymmetry condition is a large deviation from actual loading conditions an explosion creates on a cylindrical structural column. Fig.19 illustrates this by showing the cross-sections of a column under the simulated and actual load. Fig.19.b shows the loading effect due to the overpressure from an explosion located to the right of the column. This loading condition creates a much more complicated problem; bending along the column will need accounting. Fig.19.a is the simplified, axisymmetric approximation to this load used throughout the scenario. Although it does not simulate an actual explosi rap can still be demonstrated. The outer wrap is intended to protect the inner column from failure. To

easure the effectiveness, the stresses within the column are used for comparison.

on, the benefits of applying an outer, composite w

m

27

Figure 19: Axisymmetric Loading and Unidirectional Overpressure Loading Fig. 20 shows the inner column and outer wrap, which is revolved around the z-axis. Elements 1 through 10 are an isotropic material to simulate concrete. Elements 11 and 12 are the transversely isotropic material with varying properties and fiber orientation. As a control, the sim lements to best ompare the benefits of the composites. The stresses at nodes 1 and 2, which are the enter of the column, are used to measure the effectiveness of the various wrap materials

the material properties used in the simulations. In able 2 E is the modulus in the direction of fiber alignment and E is the modulus in

ulation was also conducted using concrete in the last two eccand orientation. Tables 1 and 2 show T 1 2directions 2x and 3x . It can be seen that the composite materials chosen for the simulation are very strong in the longitudinal direction, even compared to concrete, but comparatively weak in the transverse directions.

Figure 20: Element Descritization

Table 1: Isotropic Material Properties for Concrete

Material E psi (10^6)

Concrete 3.9 0.22

a b

Z

28

Table 2: Orthotropic Material Properties

Material E1 psi (10^6) E2

psi (10^6) G12

psi (10^6) 1T3000/934

Graphite Epoxy 19.0 1.5 1.0 0.22

AS/3501 Graphite Epoxy 20.0 1.3 1.0 0.3

p-100/ERL 1962 Pitch Graphite/Epoxy 68.0 0.9 0.81 0.31

Kevlar 49/934 Aramid/Epoxy 11.0 0.8 0.33 0.34

Scotchply 1002 E-gla 0.26 s/Epoxy 5.6 1.2 0.6

Boron/5505 Boron/Epoxy 29.6 2.68 0.81 0.23

Spectra 900/826 Polyethylene/Epoxy 4.45 0.51 0.21 0.32

E-glass 470-36 E-glass/Vinylester 3.54 1.0 0.42 0.32

The three fiber orientations tested in the outer wrap: radial, transverse, and circumferential, are shown in Fig.21.

Figure 21: Fiber alignment of the three cases

o maintain axisymmetry, r-displacement on nodes 1 and 2 are defined as zero. The z-isplacements on all nodes are defined as zero to simulate an infinite column in the z-irection and avoid th n is subject to t the floor and ceilin n inward radial force t nodes 25 and 26 as ig.20.

Tdd e displacement boundary conditions a normal colum

g. Node by node lumping was used to apply aaa shown in F

29

From the previous results using the three elements for a radially loaded column it is know that that all three elements perform similarly with the given boundary conditions. Accurate column modeling requires nodes 1 and 2 of element 1 to rest on the axis of revolution. To avoid the s the displacement lement was utilized.

For all m , th entia nmen uccessful f r reduction of radial displacements, radial stress, hoop stress, and tr erse stress icent radial an nsverse fiber alignments, however, managed to intensif acement and st . For stress reduction at the center nodes the p-100/ERL 1962 pitch graphite/epoxy yielded the greatest results due to its very high modulus in the direction and weak transverse prop ows the inn sses on nodes 1 and 2 and the outer stress on nodes 21 and 22 for the control case using an isotropic outer wrap and the case using a p-100/ raphite/epo uter wrap. T uter wrap wa le to significreduce the stresses within the cen n.

: Stress Values for the 'Column ' benchmark m

r z

0rcomputational limitations for values ae

aterials tested e circumfer l fiber alig t proved s oansv n the

er column. The d tray the displ resses

1x erty. Table 3 sh er stre

ERL 1962 pitch g xy oter colum

he o s ab antly

Table 3 Wrap proble

Isotropic Wrap

Inner nodes 1,2 -0.1818 -0.08 -0.1818 Outer nodes 21,22 -0.1818 -0.08 -0.1818

Graphite/Epoxy Wrap

Inner nodes 1,2 -0.0768 -0.0338 -0.0768 Outer nodes 21,22 -0.0882 -0.0323 -0.0882

Fig.22 and Fig.23 show the comparisons of radial displacement, and stresses for both the isotropic/concrete and graphite/epoxy wrap outer wraps.

30

Figure 22: Displacement and uniform concrete material for a infinite column

Stress plots for a

31

Figure 23: Displacement and Stress plots for a uniform concrete material with a graphite/epoxy wrap

32

6. CONCLUSION

The use of Finite Element Methods for displacement and stress analysis of axisymmetric columns with aniostropic materials was presented. Three elements were analyzed under conditions to represent circular columns and plates:

Isoparametric TPE quad element Stress Hybrid quad element Stress-displacement Hellinger-Reissner quad element

Transversely isotropic composite materials, those with composite fibers oriented in a single direction to provide two axes of symmetry, were employed in the testing. The testing was conduced for three different fiber orientations: radial, transverse, and circumferential. When applicable, the resulting displacements were compared with analytical, benchmark solutions. When used to model bending of a circular plate with a central point load, only one element, the TPE, proved useful. Both the Stress-Hybrid and HR converged to the wrong solution, as verified by the analytical solution to this problem. The elements were inaccurate in handling isotropic materials as well transversely isotropic materials and mesh refinement was ineffective for improving the solution. Radially applied loads proved more predictable. All three elements approached the same solutions for displacement and stress if a sufficiently refined mesh is utilized. The Hybrid4 element produces singularities at

0=r and inaccurate solutions for small values of r but behaves acceptably otherwise. Both isotropic and transversely isotropic (and all fiber orientations) materials can be accurately handled by all three elements. Both the HR and stress hybrid elements include a linear stress field as part of the formulation of the stiffness matrix. The field allows stress to vary linearly in r and according to seven stress amplitude parameters. The definition of the stress field depen the specific application or modeling scenario and affects element performance. In

de s presented here the field selected produced ing conditions nd did not present problems at

z nds

oveloping the HR element two stress fields were tested. For the axisymmetric problem

accurate solutions to radial load0=r .

iber orientation and therefore the orientation of the strong and weak directions of the aterial determine the elasticity matrix. The orientation has significant effect on the

isplacements and stresses. As an example, an isotropic cylindrical column with a thin, uter wrap of transversely isotropic composite material demonstrated the effect of fiber rientation. A circumferential fiber orientation provided considerable stress reductions in e inner column while the radial and transverse orientations slightly raised stresses.

a Fmdooth

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7. REFERENCES

1. Advanced Finite Element Methods, Class Notes, Carlos A. Felippa, 1990-2006. 2. Principles of Composite Material Mechanics, Gibson R. F, McGraw Hill, New

York, 1994. 3. Elastic Characteristics of a Composite with Anisotropic Matrix Fibers,

omposite Materials, Vol. 18, No. 3, May 1982. isotropic Body, Lekhnitskii, S.G., Mir: Moscow,

5. Mechanics of Composite Materials, Jones R.M, Taylor and Francis, 1999.

Golovcham V. T, Mechanics of C4. Theory of Elasticity of an An

1981. 6. A Template Tutorial, Carlos. A. Felippa.

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A. APPENDIX The A.1 Case I Radial alignment Displac

tresses:

following are results of the plate bending problem using the HR formulation.

ements:

S

Node ur uz1 0 3.81457

5 0.362839 3.834096 0.362839 3.871917 0.43338 3.434238 0.43338 3.425769 0.477081 2.95731

10 0.477081 2.9593211 0.500544 2.4577612 0.500544 2.4572113 0.507749 1.9438214 0.507749 1.9441815 0.501365 1.4323316 0.501365 1.4314717 0.483309 0.93184618 0.483309 0.93472919 0.455023 0.46383520 0.455023 0.45373621 0.417633 022 0.417633 0.0356891

2 0 5.202213 0.252134 4.30061

4 0.252134 4.11297

node sigrr sigzz sig sigrz1 45.8425 69.382 45.8425 02 45.8425 69.382 45.8425 03 37.2711 9.38204 37.2711 8.571434 37.2711 9.38204 37.2711 8.571435 24.7946 1.89142 24.7946 2.932336 24.7946 1.89142 24.7946 2.932337 18.3433 0.423308 18.3433 1.792328 18.3433 0.423308 18.3433 1.792329 13.9025 0.100331 13.9025 1.30261

10 13.9025 0.100331 13.9025 1.3026111 10.498 0.0274918 10.498 1.0268412 10.498 0.0274918 10.498 1.026843 7.7322 0.0178814 7.7322 0.8488364 7.7322 0.0178814 7.7322 0.848836

15 5.40134 0.0427612 5.40134 0.72403716 5.40134 0.0427612 5.40134 0.72403717 3.38633 0.144145 3.38633 0.63152718 3.38633 0.144145 3.38633 0.63152719 1.61134 0.504932 1.61134 0.56013720 1.61134 0.504932 1.61134 0.56013721 0.776292 1.78445 0.776292 0.55350622 0.776292 1.78445 0.776292 0.553506

11

35

A.2 Case II - Transverse alignment

Displacements & Stresses: Node ur

1 0 uz

7.910980 8.04974

3 8425 7.533424 0.458425 7.514665 0.659707 6.907086 0.659707 6.910867 0.787963 6.156878 0.787963 6.156029 0.86742 5.31051

10 0.86742 5.3107111 0.91008 4.4091612 0.91008 4.409113 0.92318 3.4829714 0.92318 3.4830115 0.911573 2.5582516 0.911573 2.5581617 0.878744 1.6569518 0.878744 1.6572419 0.827315 0.7996920 0.827315 0.7986821 0.759333 022 0.759333 0.00356891

node sigrr sigzz sig sigrz1 45.8425 69.382 45.8425 02 45.8425 69.382 45.8425 03 37.2711 9.38204 37.2711 8.57143

37.2711 9.38204 37.2711 8.571435 24.7946 1.89142 24.7946 2.932336 24.7946 1.89142 24.7946 2.932337 18.3433 0.423308 18.3433 1.792328 18.3433 0.423308 18.3433 1.792329 13.9025 0.100331 13.9025 1.30261

10 13.9025 0.100331 13.9025 1.3026111 10.498 0.0274918 10.498 1.0268412 10.498 0.0274918 10.498 1.0268413 7.7322 0.0178814 7.7322 0.84883614 7.7322 0.0178814 7.7322 0.84883615 5.40134 0.0427612 5.40134 0.72403716 5.40134 0.0427612 5.40134 0.72403717 3.38633 0.144145 3.38633 0.63152718 3.38633 0.144145 3.38633 0.63152719 1.61134 0.504932 1.61134 0.56013720 1.61134 0.504932 1.61134 0.56013721 0.776292 1.78445 0.776292 0.55350622 0.776292 1.78445 0.776292 0.553506

20.45

4

36

A.3 Case III - Circumferential alignment

Displacements & Stresses:

Node ur uz1 0 3.814572 0 5.202213 0.252134 4.300614 0.252134 4.112975 0.362839 3.834096 0.362839 3.871917 0.43338 3.434238 0.43338 3.425769 0.477081 2.95731

10 0.477081 2.9593211 0.500544 2.4577612 0.500544 2.4572113 0.507749 1.9438214 0.507749 1.9441815 0.501365 1.4323316 0.501365 1.4314717 0.483309 0.93184618 0.483309 0.93472919 0.455023 0.46383520 0.455023 0.45373621 0.417633 022 0.417633 0.0356891

node sigrr sigzz sig sigrz1 45.8425 69.382 45.8425 02 45.8425 69.382 45.8425 03 37.2711 9.38204 37.2711 8.571434 37.2711 9.38204 37.2711 8.571435 24.7946 1.89142 24.7946 2.932336 24.7946 1.89142 24.7946 2.932337 18.3433 0.423308 18.3433 1.792328 18.3433 0.423308 18.3433 1.792329 13.9025 0.100331 13.9025 1.30261

10 13.9025 0.100331 13.9025 1.3026111 10.498 0.0274918 10.498 1.0268412 10.498 0.0274918 10.498 1.0268413 7.7322 0.0178814 7.7322 0.84883614 7.7322 0.0178814 7.7322 0.84883615 5.40134 0.0427612 5.40134 0.72403716 5.40134 0.0427612 5.40134 0.72403717 3.38633 0.144145 3.38633 0.63152718 3.38633 0.144145 3.38633 0.63152719 1.61134 0.504932 1.61134 0.56013720 1.61134 0.504932 1.61134 0.56013721 0.776292 1.78445 0.776292 0.55350622 0.776292 1.78445 0.776292 0.553506

37