r. h. plaut thangjuly, s.€¦ · 1t.p~rt.6=zr. h. plaut thangjuly,a s. ia · 1989 blacksburg,...

STRESS ANALYSIS OF ROCKET MOTORS WITH VISCOELASTIC PROPELLANT BY A

MIXED FINITE ELEMENT MODEL

bv

Yung Tun Lin

Dissertation submitted to the Faculty of the

Virginia Polytechnic Institute and State University

in partial fulfillment of the requirements for the degree of

DOCTOR OF PHILOSOPHY

in

Engineering Mechanics

APPROVED:

R. A. Heller, Chairmané

J. N. Redä, —l. l- M. P. Singh

(ein-)

1t.p~rt.6=Z A IA ·R. H. Plaut S.ThangJuly,

1989

Blacksburg, VirginiaU

STRESS ANALYSIS OF ROCKET MOTORS WITH VISCOELASTIC PROPELLANT BY A

MIXED FINITE ELEMENT MODEL

bv

Yung Tun Lin

R. A. Heller, Chairman

Engineering Mechanics .

(ABSTRACT)

A mixed variational statement and corresponding finite element model are de-

veloped for an axisymmetric solid body under external symmetric Ioads using the

updated Lagrangian formulation. The mixed finite element formulation treats the

nodal displacements and stresses as the variables that can be approximated inde-

pendently. The method of static condensation is used to keep some stresses across

interfaces of a solid of revolution discontinuous. The stiffness matrix is transformed

from semi-positive definite to positive definite.

A rocket motor is composed of (1) case (2) propellant and (3) hollow air core

and is modelled as an axisymmetric solid. The propellant of a rocket motor is

treated as a viscoelastic material.

Static and dynamic analyses are performed under (1) two opposite line Ioads (2)

two opposite patch Ioads and (3) one line and one patch load combination. The

modified Newton-Raphson method is used in the solutions of nonlinear algebric

equations. The analysis of free vibration is executed first and then the Newmark

direct integration method is used in a transient analysis. Results of these analyses

are compared with solutions obtained from different methods that are independent

of the finite element method.

Acknowledgements

My sincere appreciation and respect are given to my advisor, Professor R. A.

Heller, for his guidance and support during the course of this study. Special thanks

are extended to Dr. J. N. Reddy for some suggestions on the finite element methods

and to Dr. S. Thangjitham for his ideas and help. I am grateful to Drs. M. P. Singh·

and R. H. Plaut for their encouragements and suggestions during this investigation.

This work has been supported by the U. S. Army Missile Command, Huntsville,

AL under Contract No. DAH-0186-C·D185. The support is greatly acknowledged.

Acknowlcdgcmcnts iii

Table of Contents

_ 1.0 lntroductlon ...................................................1

1.1 General comments ..............................................1

1.2 Literature review ...............................................3

2.0 Development of govemlng equations ..................................7

2.1 Introduction ...................................................7

2.2 Principle of virtual work ...........................................7

2.3 Hellinger-Reissner variational principle ................................9

2.4 Another formulation and boundary conditions ..........................14

2.5 Axi·symmetric tinite element model .................................19

3.0 Description ol vlscoelastic sollds ...................................32

3.1 Introduction ..................................................32

3.2 Derivation of the constitutive law ...................................33

3.3 Finite element model ............................................45

4.0 Algorithms ol solution ...........................................49

4.1 Static analysis ................................................49

4.1.1 Static condensation ..........................................49

Table of Contents iv

4.1.2 Modified Newton—Raphson method ................................51

4.2 Transient analysis ..............................................56

4.2.1 Free vibration ..............................................56

4.2.2 Newmark direct integration method ...............................59

5.0 Numerlcal examples and discussions ................................62

5.1 Static analysis with elastic propellant ................................66

5.2 Static analysis with viscoelastlc propellant ............................70 A

5.3 Dynamic analysis ..............................................73

5.3.1 Analysis with elastic materials ..................................73

5.3.2 Analysis with viscoelastlc propellant ..............................77

5.4 Summary ....................................................82

Appendlx A. .................................................... 176

Appendlx B. .................................................... 178

References ..................................................... 181

Table of Contents V

List of Illustratlons

Figure 1. A physical model of a multi-layered cylinder ................. 2

Figure 2. A finite element model ................................. 4

Figure 3. A solid of revolution ................................. 21

Figure 4. Nevvton-Raphson method .............................. 53

Figure 5. Modified Nevvton-Raphson method ....................... 54

Figure 6. Some vibration modes of a hollow tube .................... 58

Figure 7. Force distribution of the rocket motors .................... 63

Figure 8. Displacements of the cylinder vs. length ................... 68

Figure 9. Deformed shape due to self-weight ....................... 85

Figure 10. Displacement and stress distribution due to self-weight ........ 86

Figure 11. Deformed shape due to two opposite line Ioads .............. 87

Figure 12. Displacement and stress distribution due to two line loads ...... 88

Figure 13. Deformed shape due to two opposite patch loads ............. 89

Figure 14. Displacement and stress distribution due to two patch loads ..... 90

Figure 15. Deformed shape due to one line and one patch combination ..... 91

Figure 16. Displacement and stress distribution (line and patch) .......... 92

Figure 17. Radial displacement vs. time (two line Ioads) ................ 93

Figure 18. Radial stress vs. time (two line Ioads) ..................... 94

Figure 19. Tangential stress vs. time (two line loads) .................. 95

Figure 20. Radial displacement vs. time (two patch loads) .............. 96

List of lllustrations vi

Figure 21. Radial stress vs. time (two patch Ioads) .................... 97

Figure 22. Tangential stress vs. time (two patch Ioads) ................. 98

Figure 23. Radial displacement vs. time (line/patch Ioads) .............. 99

Figure 24. Radial stress vs. time (line/patch Ioads) ...................100

Figure 25. Tangential stress vs. time (line/patch Ioads) .................101

Figure 26. Force histories .....................................102

Figure 27. Radial displacement vs. time (two line loads: Force (A)) ........103

Figure 28. Radial stress vs. time (two line loads: Force (A)) .............104

Figure 29. Tangential stress vs. time (two line loads: Force (A)) ..........105

Figure 30. Radial displacement vs. time (two patch loads: Force (A)) .......106

Figure 31. Radial stress vs. time (two patch loads: Force (A)) ............107

Figure 32. Tangential stress vs. time (two patch loads: Force (A)) .........108

Figure 33. Radial displacement vs. time (line/patch loads: Force (A)) .......109

Figure 34. Radial stress vs. time (line/patch loads: Force (A)) ............110

Figure 35. Tangential stress vs. time (line/patch loads: Force (A)) .........111

Figure 36. Radial displacement vs. time (two line loads: Force (5)) ........112

Figure 37. Radial stress vs. time (two line loads: Force (5)) .............113

Figure 38. Tangential stress vs. time (two line loads: Force (5)) ..........114

Figure 39. Radial displacement vs. time (two line loads: Force (C)) ........115

Figure 40. Radial stress vs. time (two line loads: Force (C)) .............116

Figure 41. Tangential stress vs. time (two line loads: Force (C)) ..........117

Figure 42. Radial displacement vs. time (two line loads: Force (D)) ........118

Figure 43. Radial stress vs. time (two line loads: Force (D)) .............119

Figure 44. Tangential stress vs. time (two line loads: Force (D)) ..........120

Figure 45. Radial displacement vs. time (two line loads: Force (E)) ........121

Figure 46. Radial stress vs. time (two line loads: Force (E)) .............122

List of lllustrations-

vii1

Figure 47. Tangential stress vs. time (two line Ioads: Force (E)) ..........123

Figure 48. Radial displacement vs. time (two patch Ioads: Force (B)) .......124

Figure 49. Radial stress vs. time (two patch loads: Force (B)) ............125

Figure 50. Tangential stress vs. time (two patch loads: Force (B)) .........126

Figure 51. Radial displacement vs. time (two patch loads: Force (C)) .......127

Figure 52. Radial stress vs. time (two patch loads: Force (C)) ............128

Figure 53. Tangential stress vs. time (two patch Ioads: Force (C)) .........129

Figure 54. Radial displacement vs. time (two patch loads: Force (D)) .......130

Figure 55. Radial stress vs. time (two patch loads: Force (D)) ............131

Figure 56. Tangential stress vs. time (two patch Ioads: Force (D)) .........132

Figure 57. Radial displacement vs. time (two patch loads: Force (E)) .......133

Figure 58. Radial stress vs. time (two patch Ioads: Force (E)) ............134

Figure 59. Tangential stress vs. time (two patch loads: Force (E)) .........135

Figure 60. Radial displacement vs. time (line/patch loads: Force (B)) .......136

Figure 61. Radial stress vs. time (line/patch loads: Force (B)) ............137

Figure 62. Tangential stress vs. time (line/patch loads: Force (B)) .........138

Figure 63. Radial displacement vs. time (line/patch Ioads: Force (C)) .......139

Figure 64. Radial stress vs. time (line/patch loads: Force (C)) ............140

Figure 65. Tangential stress vs. time (line/patch loads: Force (C)) .........141

Figure 66. Radial displacement vs. time (line/patch loads: Force (D)) .......142

Figure 67. Radial stress vs. time (line/patch loads: Force (D)) ............143

Figure 68. Tangential stress vs. time (line/patch loads: Force (D)) .........144

Figure 69. Radial displacement vs. time (line/patch loads: Force (E)) .......145

Figure 70. Radial stress vs. time (line/patch loads: Force (E)) ............146

Figure 71. Tangential stress vs. time (line/patch Ioads: Force (E)) .........147

Figure 72. Force histories for viscoelastic propellant ..................148

· List oflllustmtions viii

Figure 73. Radial displacement vs. time (two line loads: Force (F)) ........149

Figure 74. Radial stress vs. time (two line loads: Force (F)) .............150

Figure 75. Tangential stress vs. time (two line loads: Force (F)) ..........151

Figure 76. Radial displacement vs. time (two line loads: Force (G)) ........152

Figure 77. Radial stress vs. time (two line loads: Force (G)) .............153

Figure 78. Tangential stress vs. time (two line loads: Force (G))...........154

Figure 79. Radial displacement vs. time (two line loads: Force (H)) ........155

Figure 80. Radial stress vs. time (two line loads: Force (H)) .............156

Figure 81. Tangential stress vs. time (two line loads: Force (H)) ..........157

Figure 82. Radial displacement vs. time (two patch loads: Force (F)) .......158

Figure 83. Radial stress vs. time (two patch loads: Force (F)) ............159

Figure 84. Tangential stress vs. time (two patch loads: Force (F)) .........160

Figure 85. Radial displacement vs. time (two patch loads: Force (G)) .......161

Figure 86. Radial stress vs. time (two patch loads: Force (G)) ............162

Figure 87. Tangential stress vs. time (two patch loads: Force (G)) .........163

Figure 88. Radial displacement vs. time (two patch loads: Force (H)) .......164

Figure 89. Radial stress vs. time (two patch loads: Force (H)) ............165

Figure 90. Tangential stress vs. time (two patchloads: Force (H)) .........166

Figure 91. Radial displacement vs. time (line/patch loads: Force (F)) .......167

Figure 92. Radial stress vs. time (line/patch loads: Force (F)) ............168

Figure 93. Tangential stress vs. time (line/patch loads: Force (F)) .........169

Figure 94. Radial displacement vs. time (line/patch loads: Force (G)) .......170

Figure 95. Radial stress vs. time (line/patch loads: Force (G)) ............171

Figure 96. Tangential stress vs. time (line/patch loads: Force (G)) .........172

Figure 97. Radial displacement vs. time (line/patch loads: Force (H)) .......173

Figure 98. Radial stress vs. time (line/patch loads: Force (H)) ............174

List or lllustrations ix

Figure 99. Tangential stress vs. time (line/patch Ioads: Force (H)) .........175

List of lllustrations X

List of Tables

Table 1. Mechanical properties of rocket motors .................... 65

Table 2. Comparison of solutions (FEM and exact solutions) ............ 67

Table 3. Comparison between two methods ........................ 69

Table 4. Static responses ..................................... 71

Table 5. Data for the viscoelastic propellant ........................ 72

Table 6. Frequencies of a rocket motor ........................... 74

Table 7. Comparisons between static and dynamic loads (interface) ....... 76

Table 8. Two line loads under force history (B) ..................... 78

Table 9. Two patch loads under force history (B) .................... 79

Table 10. One line and one patch load under force history (B) ........... 80

Table 11. Magnification factors vs. force histories .................... 81

Table 12. Comparison between static and dynamic analyses ............. 83

use of muc;l

xi

1.0 Introduction

1.1 General comments

Solid propellant rocket motors used for tactical missiles are stored in field

storage. They may be subjected to impact damage during handling or may be ac-

cidentally dropped parallel to (or perpendicular to) their axes during the loading

process. The stress distribution due to such causes can result in termination of the

useful service life of rocket motors. The purpose of this study is to obtain the stress

and displacement responses of the propellant in rocket motors (1) to external loads

and (2) to various temperatures at different environments.

A solid rocket motor considered in the present analysis is modelled as a

multi-layered cylinder consisting of (1) case (2) propellant and (3) hollow air core.

The structural diagram is shown in Fig. 1.

The visco·elastic propellant, of course, is the interesting part of the investi-

gation. The case can be treated as either an elastic or a viscoelastic material. The

mechanical properties of a viscoelastic propellant are obtained from experimental

data. The relaxation modulus is expressed in terms of Prony’s series and the time-

temperature shift factor (a,) is derived from the WLF equation. The constitutive law

Introduction 1

A(1) air core

(2) propellant

(3) case

Flgure 1. A physlcal model of a multl-Iayered cyllnder

Introduction 2

is derived from the hereditary integral for relaxation modulus ( Y(t) ) by lengthy

manipulation.

The finite element method has long been established as a versatile and pow-

erful tool of analysis for solids and structural mechanics. The mixed finite element

method is based on the Hellinger-Reissner variational principle that is a stationary

principle. This principle treats all the dependent variables as independent of each

other. The stationary conditions of the principle are the strain-displacement

equations, stress-strain equations, stress-equilibrium equations and both natural

and essential boundary conditions, in short, all of the governing equations of

elasticity.

A muIti·layered cylinder can be modelled as a solid of revolution. An

axisymmetric solid finite element model is used in the present analysis; the finite

element is a ring of constant cross section (Fig.2). Centers of nodal circles lie on

the z axis (axis of revolution).

The external loads can be expressed in terms of Fourier series. By recognizing

the orthogonality properties of trigonometric functions, the coefficients involved in

the calculation of 6 terms can be dropped because of their constancy. The problem

can be reduced to a series of two-dimensional problems. According to the super-

position principle, the original problem is performed by adding the solutions of each

component problem (corresponding to each coefficient of the Fourier series).

1.2 Literature review

The problem has been solved for cylindrical single layer shells [1-4]; the stress

analysis due to thermal loads or external loads for multi-Iayered cylinders is done

Introduction

U3

U

/ \ II \1 \ 11 x 1 \

1 x 1 \I1 J l I1 J \‘I \ I

· ll§lY% 'ix Z /1/\

Z 1/xx 1 V \ 1/

x / \ 1x / \ 1\ / x 1

Flgure 2. A flnite element model“

Introduction 4

by closed form solutions [4-8]. In the above papers [1-8], all material properties

were assumed to be elastic and homogenous.

The textbooks by Flugge [9] and Christensen [10] contain the definitions of the

heredltary integral and relaxation modulus for viscoelastic materials. For

viscoelastic materials the mechanical properties can be obtained from experimental

data [11-12]. The constitutive law of viscoelastic materials has been derived in

[13-19] by the assumption of linear strains between time intervals. Zak [20] solved

the structural analysis of viscoelastic propellant of rocket motors by integral trans-

form techniques.

Chung [21] solved the dynamic stress analysis of viscoelastic shells by a finite

element method. Wilson and Vinson [22] considered the viscoelastic analysis of the

plates by the Rayleigh-Ritz method. The dynamic buckling of viscoelastic cyllndrical

shells by analytical solutions and experimental techniques has been solved by

Bukowskin [23] and FIorence.[24]. They all assumed that the shell is a thin and

single layer.

Wilson [25] and Crose [26] have presented the finite element model for axially

symmetric solids; for more detail, refer to the textbooks by Cook [27], Yang [28] and

Zienkiewicz [29]. Jones [14] has developed displacement finite element methods for

solid rocket motors. This method only considers the nodal displacements as vari-

ables; the stresses are calculated from those known displacements by basic

equations of elasticity [30]. For multi-layered cylinders the tangential stresses

across layers are not necessarily continuous. Apparently the displacement finite el-

ement method can not circumvent this impasse.

Although the mathematical properties of mixed finite element approximations

have seen extensive development in recent years [31-33], the applications of mixed

finite element models to actual physical problems have been rare. In linear, elastic

introduction 5

problems the mixed finite element methods can be developed using the Hellinger-

Reissner variational principle [34-36]. This method treats nodal displacements and

stresses as independent variables. Not all stresses across interfaces of multi-

layered cylinders are continuous; the static condensation technique [28,36-38] is in-

troduced.

The relaxation modulus of viscoelastlc materials is dependent on time and

temperature, as is the constitutive matrix. ln static analysis the modified Newton-

Raphson method is used [27,39]. The equation of motion has been developed for

mixed finite models [40] by using the concept of variations [35]. The Newmark direct

integration method has been described in books [27,29,35] explicitly.

Introduction 6

2.0 Development of govemlng equations

_ 2.1 Introduction

ln this chapter we begin with the statement of virtual work for an arbitrary solid

body under external load and derive the variational functional that will be applied

to the mixed flnite element model of the problem.

When describing the motion of a solid body under external Ioads, a Lagrangian

frame of reference is typically adopted. Referring the variables to the initial,

undeformed configuration is known as the "total Lagrangian description”, and refer-

ring the variables to the current configuration is known as the updated Lagrangian

description. The updated Lagrangian description will be used to descrlbe the mo-

tion in this chapter.

2.2 Prlnclple of vlrtual work

The governing equations can be derived from the principles of virtual work

(i.e., virtual displacement, virtual force or mixed virtual displacements and forces).

Considering the principle of virtual displacements, the principle requires that the

sum of the external virtual work done on a body and the internal virtual work stored

Development of governing equations 7

in the body should be zero, i.e., a continuous body, V , should be in equilibrium

under the action of body forces, X, , and surfacetraction(s),Suppose

that on the portion, S, , of the boundary ,S. the displacements are E

and on the portion, S2, the tractions are The boundary portions, S, and S, , are

disjoint, and their sum is the total boundary, S .

Let the displacement vector of the body in equlibrium be u= (u,, u2, ua), and let

6,, and 6,, be the associated stress and strain components, respectively.

The principle of virtual work states that

öW,+öWE=0 (2.1)

öW, = virtual work coming from internal forces

ÖWE = virtual work coming from external forces

For a linear elastic solid, Eq.(2.1) can be written as .

=J.

X,öu,dV

*1**.*.-V

V S2

where 6 denotes the variational symbol ( öu, means the virtual displacement of u, )

and dV and dS2 are the volume and surface elements in configurations over which

the integrations are executed.

The linear strain-stress relations of a (visco·) elastic solid are

{8} = ¤{{8} — {80}) + {G} (23)

Dcvclopmcnt of governing cquations 8

where {a}, {6} are the stress and strain vectors. Assume U, the strain energy, ex-

ists such that

V

and a quantity U', called complementary enel'QY„ such that

ÖU· =IÖ0Ü£ÜdV (2.5)V

With the potential energy of the applied load, V , as a function of the displace·

ments, u,, given by

V = —f XiuidV+‘I· tiuidV (2.6)

the virtual work statement is equivalent to that of minimizing the total potential en-

ergy, H:

ÖH = Ö(U + V) = O (2.7)

2.3 Hellinger-Reissner varlational prlnclple

In deriving the variation functional (Eq. (2.7), we have assumed that strains were

related to the displacements (strain-displacement relationship),

Development of goveming equations 9

1. Bü =

? (Ui,} + Ui,

i)andon the boundary S, (displacement boundary conditions)

U; = E (2-9)

lf we want to relax these constraints (Eqs. (2.8) and (2.9) in the variational

statement Eq. (2.7)), we lntroduce Lagrangian multipliers, 1,,, defined in the domain

V (associated with the strain-displacement equations) and v,, defined only on the

boundary S, (associated with the displacement boundary conditions). Then a new

variational principle is obtained:

1 -H1 = H '* " *5 (Ui,} +—·v[ vi(ui —•ui)dSV

S1

After substltutlng the total energy into the above equation, Eq. (2.10) can be written

as

n1= U

-12

XiüidvVVS2 (2.11)

—( w(¤; — EWSS1

Performing the variation of Eq. (2.11), l.e.,

ÖH, = 0 (2.12)

gives

Dcvclopmcnt of govcming cquations l0I

ÖH1 ölÜ(sÜ —é- (ULL + UL L))dV —

jl}lÜ(6gÜ - öuLj)dV

V V V

vLÖuLdSS2 $1 S1 ·

jöimlmj—-I,

@öuLdS—].

vLÖULdS -16,1„(g„-é (UL} + UL i))dV

S2 S1 V

övi(uiS1

Using integration by parts,

V V S

Substituting Eq. (2.14) into Eq. (2.13), then the Eq. (2.13) becomes

O = fVösU(oÜ — lÜ)dV —- XL)dV + öUL(njÖLmlmj — vL)dS1

— 1+1‘

öUi(njöim'1·mj— GWS * 1;/Ö4y(@y * 5011,;+ E, i)WV (2.15)S2

*1 öV1{E · EWSS1

As the above is true for any variations, we immediately observe that the last

two Euler conditions give us the constraint conditions.- Since the above equation

(2.15) is true for any variations, the coefficients of the variations are zero:

· Development of goveming equations ll

öui Z [öim,lmi]_i + Xi = O in V (2.16a)

öcü : aii = ÄÜ in V (2.16b)

öui : niöimlmi — vi = 0 in Si (2.16c)

6Ui I Djöimztmj— 6= 0 in S2

I Bü = (Ui.] + Uji i) in

Vövi: ui = E in Si (2.160

We can observe that the Lagrangian multiplier can be identified as followsz

Äii=oÜvi

= {Ü (2.17b)

Apparently lii and vi are stress components and boundary tractions, respectively.

Hence Eq. (2.10) can be rewritten in explicit form,

Hi = H -1;/¤Ü(sÜ— é- (ui_i -+- ui_ i))dV — ti(ui — E)dS (2.18)

This expression is called the”Hu

- Washizu" variational principie; however this

formulation is not a practical principle. We recall that the sum of the complementary

and strain energies is written as

Development of goveming equations I2

V

Another variational form can be obtained by replacing U by U' in Eq. (2.18),i.e.,

H2 =

jl

GÜEUÖV — U·(¢7ij)-~|·

uiXidV*j~V

V S2 (2.20)1 -— '[ —- u;)dS

A special form is obtained by assuming that the strains are related to the dis-

placements by _1

1EU = *5 (Ui,} + U},

i)Wecan write Eq. (2.20) as the "HelIinger - Reissner variational princigle" that can

be stated as

H2 = GÜSUÖV — U.(cÜ)—‘I‘

uiXßV —1*

u;@dS—j·

Mu; — @)dS (2.21)V V S2 S1

wherein 6,) and u, are independent variables.

For linear solids we assume zero initial strains ({60} = 0); then Eq. (2.3) can be

written as

{S} = ¤({¤} — {6}) (2-22)


The complementary energy of the system is

· D 2U =g—({¤}-{6}) (2-23)

and the strain — displacement relationship is 6,, =%(u,_,+ ul_,).

Now we differentiate Eq. (2.21) with respect to displacement and stress; this

yields the two approximate equations.

lt is remarked that the prescribed displacements, u, , on the boundary, 8,, are

_ removed (it means no prescribed displacements on boundary) in deriving these two

equations. They are given by

V v

J-öu,(Xi — @)d8 = 0 (2.24b)

V S2

The above two equations are the mixed formulations; stresses are no longer

dependent functions of the displacements.

2.4 Another formulatlon and boundary conditions

Although we derived the mixed finite element model, no explicit statement has

explained the specificatlon of the boundary conditions. In order to determine the

nature and characteristics of the boundary conditions, we use the variational form


of basic equations : equilibrium equation and stress - displacement relationship.

This will give us the classification of the boundary conditions of the mixed finite el-

ement model into natural and essential boundary conditions, and yield explicit for-

mulation of this method.

ln this section we want to derive the explicit equations in cylindrical coordinates

from the equilibrium equations and stress - displacement relationship since a rocket

motor is modelled as a multilayered cylinder.

The equilibrium equations in cylindrical coordinates are

öarr 1 öorß öorz (°rr—"08)——- ——— —— -—l = .25ör+r ö0+öz+ r +X,.O (2a)

öarß 1 ÖG66 ö°z6 2°z8—— ————— ——— —- = 2.25b()

öarz 1 öazß öazz 020——- ··i· ·—·· X = O 2.25

and stress - displacement relationships are

öu,"ET=D11Urr+D12088+D130zzM

1 1 MT + T 6 + Dzadzz (2-26b)

öuzjf = Disdrr + 023*70 6 + Dasdz z (2-26c)

öuü öu

DCVCIOPIDBHÜ of g0V€l’IIiIIg BQIIRÜOIIS

1 öu ÜU6 U6

737 +7— T = D6:662 6 (2-266)

öu, 1 öu,7+TE = DÜGGZÜ (2-260

We will construct the variatlonal form of these nine equations by taking all

terms to one side and multiplying each equation by a different test function

w, ,i= 1, 2, 3,...9 and then integrating the resulting equations over the domain of a

typical finite element.

The test functions are arbitrary. However, in order to achieve some physical

meaning for these equations, the test functions are chosen as the first variations of

the displacements and stresses, i.e.,

— w1 = 6u,—· W2 = 6u0— w3 = 6u,

W4 = ÖGN-

W6 = 6666 (2-27)W6 = ödzzW7 = öarzW8 = öaröwg = 6626

ln this case the signs of the test functions are arbitrary and these result in a

symmetric matrix. ln order to arrive at the physical meaning of boundary conditions,

we perform the calculations by multiplying Eqs. (2.25) with 6u,, 6u0 and 6u, and

multiplying Eqs. (2.26) with 66,,, 6600 , 66,,, 66,,, ÖÖ,6 and ÖÖZO,

Development of g0V£I'I’IiI‘Ig BQUQÜOIIS

86 8 86

V v V

= I x,66,6v 6,,66,61661 + '[ 6, ,66,666-10 (2·288)v s s

v

8 86 86

v v v

X9öu0dV 6, Oöußrdüdz 6, Üöugrdrdz (2-288)v s s

= IVXÜÖUÜÖV6öu

1 66 0 66u

v

=jl

x,66,6v 6, ,66,r666z + I 6, ,66,66616 (2-288)v s s

u,dV + Lt,6, ,dS

6u,

v

DCVCIOPIIIEIK of g0VCI’Ili|'lg CQIISÜOIIS

ÖU6 1 1 ÖU6( 80 T

_T 60

_D12°rr '“ D22"0 0 " D23°z z)ö°00dV = 6 (2286)

V

öuz( 82

— D‘|3Urr_

D23U0 6 " D336z z)ö6zzdV = 0 (2-280V

ÖU8 öuz(-57 + E- — D44a, z)ö0,zdV = 0 (2,28g)

V

1 öu, öuo ug(Y36- + —

Y— D55d,. 8)Öd,.8dV = O (2.2Bh)

V

öuz 1 öuz .(E-· + Y 717— DGSUZ 8)Ö0‘z8dV = 0 (2.28I)

V

The above three Eqs. (2.28a, b, c) are obtained by integrating by parts, and

noting that dV = r dr d8 dz with

t, = 6, rn, + 6, ang + cr, znz (2.29a)

t9 = U,. 8/7,- + G8 Gr/6 + dz8/12·


fz = 0, zh, + 0z gng + 02 znz (2.29c)

where n = (n,, ng, nz) is a unit normal to the boundary, S.

We also note the nature of boundary conditions in the first three equations.

Since only the variations of the displacement components u,, ug and uz appear in

the boundary integral,U

ür = ur(76 =U6Üz

= UZ

constitute the essential boundary conditions of the problem. Prescribing the coeffl-

cients of the variations of the displacements u,, ug and uz, the natural boundary

conditions are constructed:

ir = Ur rnr + Ur znz(6 = O', 6/7, + 0'Z6/7ztz

= Ur znr + dz znz

Apparently t= (t,, tg, tz) is a stress vector on the prescribed boundary, S, in cy-

lindricalcoordinates.2.5

Axl-symmetric flnlte element model

A solid of revolution is axially symmetric if its geometry and material properties

are independent of the circumferential coordinate, 6.

DOVCIOPITICIIY of gOV¢|’l’l(IIg €q\l8(iOI'IS

A rocket motor can be modelled as a multi-layered cylinder, and its material

properties are independent of 6. Hence it can be treated as a solid of revolution

(Fig. 3).

The centers of all nodal circles lie on the z — axis (axis of revolution). Each el-

ement is a solid of revolution about the z- axis. Fig. 3 depicts a solid of revolution

modelled by rectangular flnlte elements, and each finite element is a ring of con-

stant cross section.

If the loading is axially symmetric, there is no displacement in the 6 — direction.

The displacement vector has only u, and u, components.

lf the loading is not axially symmetric, it can be expanded into a Fourier series.

The given series is expressed as the sum of several component Ioadings, so the

analysis can be performed for each component. Applying the principle of superpo- ‘

sition, the original problem is solved by adding the solutions of the component

problems.

Let the loading be expressed as a Fourier series

X, = ZX, „ cos(n 6)

X, = ZX, „ cos(n 6)

F, = EQ, cos(n 6)_ _ (2.32)tz = Zt, ,, cos(n 6)

Xg = 2Xg ,, sin(n 6)

tg = Zig ,, sin(n 6)

where X's and Ps are, respectively, body forces per unit volume and prescribed

surface tractions on the boundary, S, ln the r, 6 and z directions.


Z

Ä"-/il ‘

I°‘

U 11

Z8

Q ‘0000000000000000

' I'

Flgure 3. A Solid of Revolutlon

DCVCIOPITICIÜ of g0VCI’|'Ilhg CQLIQÜOIIS

Similarly, for displacements, we assume that these can also be expressed as

u,. = EIUM cos(n 6)

ug = ZUG „ sin(n 6) (2.33)

uz = ZUZ „ cos(n 6)

where u's are radial, circumferential and axial displacements,respectively.Eqs.

(2.33) represent a state of symmetry with respect to 6 about the plane

6 = 0°. The strain-displacement relationship in cylindrlcal coordinates is expressed

asU

88,, ·ä· 0 0

1 1 896 6 T TE)? O

6 u'szz OO=

• U9 (2.34)

E .Q. 0 L’zöz ör

UzLE. .ä._l. O*9 r 86 8r r

E 0 L LLZ9 8z r 86

From this relationship the strains can be written as

Development of g0VBf|liIIg CQIIBÜOIIS

6, , = ZE, ,6 cos(n6)

66 6 = 266 ,6 cos(n6)

62 2 = 262 ,6 cos(n6)(2.35)

S6: = Em,. ¤¤S(¤9)

6,6 = 26,6 ,, 6:6066)626 = Z6'26 ,6 sln(n6)

where

E =öü,,,’"ör

- _ urn JL -86 n "° r_

r u6n

_ öü2,66= r· = E"

- @¤„„ öüzn (2·36)@6 6 „ =7;+-

_ _Q_ - öu6n Ü6n8r6n“’ r urn+ Ö, " ;-

E ü'z 6 n öz r zn

Furthermore, from the constitutive law (Eq. (2.3)) the stress components should

have the following formulatlonsc

Development ol' goveming equations 23

6, , = Z5, ,, cos(nö)

60 0 = Z50 ,, cos(nö)

62 2 = 252 ,, cos(nö)(2.37)

¤„ = 2822 „ ¤¤S(¤9)

6,0 = 25,0 ,, sin(nö)

UZ9 = 2529 ,, $ll'l(l'IÜ)

where the barred quantities are functions of r and z but not of ö.I

Substituting Eqs. (2.32), (2.33) and (2.37) into the Eq. (2.28) and noting the coef· .

ficients of Eqs. (2.28) (D,,,..., DSG) for isotropic materials

- 1D11 = D22 = Das = ‘g‘ (2-338)

D12 = D1:1 = D23 = —é (2-33b)

2(1 -l— v)D44 = D55 =-' DGS = "

Tthefollowing equations can be obtained:

-ööürn 2 Fön — 2 Ll_- —

- 2{6,,, cos (nö) + -7 öu,,, cos (nö)- , 6,0,,öu,,, sin (nö) +

_ (2.396)

- Döurn 2 _ — — 2 ‘ — 26,2,, -5- cos (nö)}dV - X, ,,öu,,, cos (nö)dV + t,öu,,, cos (nö)dS

s

Development of goveming equations Z4

66- 6-fg {L 58,,6Ü8,, cos2(nö) + 5,8,,( — lßl) sin2(nö) +

ööü (2.39b)5 —QQ- sin2(nö)}dV = Y 65 sin2(nö)dV + [65 sin2 nö dSzön 82 ö n ön ö ön

S

66- 66- .<[‘{52,, sin2(nö) + 5,2,, sin2(nö) — 528,,652,, sin2(nö)}dV

(2.39c)

= ,,652,, cos2(nö)dV + j‘?26ü2,, cos2(nö)dS ‘

s

öürn - 2 1 - - 2 v - - -{T66,,, cos (nö) — —E 6,,,66,,, cos (nö)

(68,,cos2(nö)}dV= 0

Ü - - 1 .. - 269nöUÜf7 °°S (ng)

(2.396)

+ iv (5,,, + 52,,)658,, cos2(nö)}dV = O

65 _ _ _662,, cos2(nö) 52,,652,, (66,, + V;-n)öUzn (2.39,)

cos2(nö)}dV = O


öü öü 2 1 ++ %)öarzn C°S2("6) " Erznöarzn Sin2(n0)}dV (2.3gg)

= O

- öü F _ _ 2 1 +(2 22,2

sin2(n9)}dV = 0

ÖÖÜ _ _ _ 2(1 -l- v) _ _ _uzn)öown own s1n2(n0)öown}dV (2239,)

= O

where dV = rdrdüdz for a typical finite element with a finite volume dV. If

5,,, ,50,,,..., Un, can be interpolated by the following form

5,n = (2.406)

özn = Zöznilvi (2*40c)

Development of goveming equatlons 26

aröln = Zarünilvi (2-40d)

E-f'ZI‘l =EEFZITÜNÜam

= 2öz6„1^'1 (2-400

Ürn = Zlürni/Vi (2-409) V

Üßn = ZÜ6„1^': (2-40h)

üzn = 2ÜzniNi (2-40])

substituting Eq. (2.40) into Eq. (2.39) a typical equation

E1<J„ · (X)., = (1=)„ (2-41)

is obtained where [Kl, is the element stiffness matrix, {X},, is the element nodal ,

variable vector and {F}„ is the the element force vector; these correspond to the

n - th term of the Fourier seriesl

---.... - - - T{X} = {urn

-Ußn ·

uzn ·am

·°r8n • Grzn ·

°”9n • azn • Uzßn} (2-428)

‘For convenience, we drop the subscript n in the subsequent analysis


{F} = {Y1 J2 · fa » °· °- °·O- 0- 0}T (2-42b)

[K] is a 9 by 9 symmetric matrix.Its coefficients are ([K] is a 6 by 6 symmetric

matrix for axisymmetric load, see Appendix A.)

ÜN-14 1 36

K55 = I- % 1v,1~g6A = 1<5° (2.43b)

ÖN-16 1 38 29KU dA = KU = KU (2.43C)

N.dA (2.43d)

ÜN. N·K55 )dA (2.436)

K57 = Nil\QdA (2.43f)


44 -1Kr = ITNlNfdA = K6? = Kgs (2·439)

Kf - I-E- ~,:~g6A = Kf - ;<,§° (2.43h)

1 +:<§5 = -24%~,~j6A

= :<,§?6 - :<,§?° (2.43i)

Other coefficients of [K] are zero, and

:, Y, „~,dA + IZ, „~,ds (2.448)

:2 - 79 „~,dA + fi), „~,6:s (2.44b)

:2 272 „:v,dA + fil „~,ds (2.446)

From Eq. (2.39), we find, for each coefficient n , integrands of the element

stiffness and load vector containing cos2(n6) and sin2(n6) in every term. Integrating

with respect to 6 leaves the problem two - dimensional (r, z).


p Furthermore, from Eq. (2.39) we can observe that the n—th circumferential

component of loading is associated with the n—th circumferential component of

stresses and displacements. This means that the coefficients of the Fourier series

are decoupled.

The resulting constant of TE (and 21: for n = 0 ) can be canceled from every term

of the equations. The final element stiffness matrix is of two - dimensional case

(r, z) . The element stiffness matrix depends upon n (the n — th term of the Fourier

series); the different values of n represent different problems that do not interact

with one another. Therefore the solution is the superposition of n different prob-

lems.The

axisymmetric model can reduce a three - dimensional problem physically

to a two - dimensional problem mathematically. lt saves computational storage and

decreases the CPU time so that it adds to the availability of the program.

The preceeding discussion involved only loadings and displacements that have

6 = 0° as a plane of symmetry. ln fact, anti - symmetric loadings can be considered.

Eqs. (2.32) can be augmented:

X, = ZX,„ cos(n0) + X,„ sin(n6)

Xz = ZX, „ cos(n6) + Xlzn sin(n6)

- = 2- ~ _t, t, ,, cos(n0) + t, ,, sm(n0)

(2.45)tz = Zt, „ cos(n6) + tz „ sin(n6)

X6 = 2)% ,, sin(nl9) — X9 ,7 cos(n8)

tg = Zig „ sin(n6) -— Z, „ cos(n9)I


where the anti - symmetric modes are represented by ~ . Displacements, stresses,

and strains can be modified similarlyß

The n=0 term in Eqs. (2.45) represent the axially symmetric problems, and

even terms n = 2, 4, 6... correspond to loadings and deformations that have 0 = O°

as a plane of symmetry with respect to 0. Similarly, odd terms correspond to anti -

symmetric loadings.

2 the tilde quantitiles are functions of r and z but not of 9

DQVCIOPIDCDI of g0VC|‘I'liIIg BQUIIÜOIIS

3.0 Description of viscoelastic solids

3.1 lntroductlon

The formulation of the isothermal viscoelastic stress - strain relation states that -

the current values of the stress tensor depend upon the complete past history of the

components of the strain tensor. Usually this relation is described by a ”hereditary

integral" for relaxation, Y(t), that ls a characteristic of viscoelastic materials.

The basic assumptions of the viscoelastic solids are :

(1). isotropic thermal expansion

(2). linear, small deformation

(3). thermorheologically simple temperature response

(4). Poisson’s ratio is a constant

The viscoelastic material accepts (1) material property data (Prony series) ob-

tainable from standard laboratory stress relaxation tests from which the shear and

bulk relaxation moduli are derived and (2) a temperature - time shift factor : a,, a

function of temperature. The latter is the key element in a thermorheologically sim-

ple concept in which a reference temperature (usually room temperature) relaxation

function can be utilized for elevated temperature responses by elongation of the

time scale. This is done by replacing the real time with a reduced time with the

Description of viscoelastic solids 32

magnitude determined by the WLF (Wi//iams, Lande! and Ferry) time- temperature

shift function.

The following sections deal with the construction of the constitutive law of

viscoelastic materials and derivation of the flnite element model.

3.2 Derlvatlon of the constitutive law

Following a procedure parallel to those in elasticity whereby the deviatoric

components of stress Sy and strain ey are introduced, then,

t Ö HQ

ISylt)= G1(C(t) — C (t)) -:9;- dt (3-1)*00

and

' I öe(r)vkklt) = G2(€(t) " C (Ü) 98 (3-2)*00

where

1SÜ= ÜÜ‘°§öÜÜkk1

9/7 = EU — T3'Öüökk9

= (811 + 822 + 862) (35)


and 6,, is the Kronecker symbol.

To be consistent with common notations in elasticity, the symbols for the

isotropic relaxation function in simple shear and dilatation are frequently taken as

G(t) and K(t), respectively, where —

G lG(l)=3.6@20) * 1

KU)

=Theviscoelastic stress functions in terms of G(t) and K(t) relaxation functions

are now given by

t

2 G(C — f )-jajdr

f (3.7)

+ {K(€ — — €')}Ö··@d‘l'3 V ÖT0

where { and {' are the reduced times at t and 1 and are defined as

g: (6.63)a¢(V)0

5- = 1-*1%- (6.6:6)ar(T)0


Here a, is the time- temperature shift factor based on an assumption of

thermorheological simplicity for converting stress relaxation data gathered at a se-

ries of temperatures to a single curve. It is assumed that stress curves obtained

from relaxation tests performed at different temperatures are appropriately shifted

along the Log time axis.

An empirical function relating temperature to the shift factorTa,

, which is appli-

cable to many viscoelastic materials, was formulated by Williams, Landell and

Ferry, and is known as the WLF equation. It has the form

C1(T* TR)l =i— 3.9¤¤ a„(T) C2+(T_ TR) ( )

where the constants C, and C2 are determined from experimental data, and the ref-

erence temperature, TR, is the absolute temperature for the base curve.

The stress relaxation test generates a curve which can be approximated nu-

merically by a Prony series. lt has the form

8 t/1,,,,E(t)=EO+ ZETT, exp(—-T) (3.10)

m=1

where EO, E,,,, and 1,,, are moduli and relaxation times of parallel Maxwell elements;

EO is the long-term modulus (t—> oo).

For the special case where Poisson's ratio is a constant with time, the shear

modulus can be written as

E(t)G t

=_.l3.11O 2(1 + v)

( )

and the bulk modulus as


2 EU)2 t "' i 'l"K(t) A()+ 3 G(t)

3(1_2v)(3.12)

Denoting the first term on the right hand side of Eq. (3.7) as I,,

2'/1=2 G({ - { )Tdt (3.13)

0

and combining the Eqs. (3.10) and (3.11), then G(t) has the following exponential

form:

8 {/rm‘

G(t) = go + gm exp( — -T-) (3.14a)m=1

where

2g<> = im;(3.14b)

2gm (1 + V)

Substituting Eq. (3.14a) into Eq. (3.13),

'8

I 62„(V)/1 =2 [220+ 29,,, @><¤—C„„(é—€)]—g;··dr (3-15)

0 lT7=1

where é=é and§’

= äare reduced time parameters, and §,„ =%- .


Simplifying the Eq. (3.15),

I 8 gö€1j(T) ö81j(T)/1 = 2 911Tdr + 2 9m sxpl — C„,(C — C')1Tdr (3-16)

0 m=‘I 0

The second integration term on the right hand side of Eq. (3.16) is now sepa-

_ rated into two parts : the first part has limits from "zero” to”t- At " and the second

has limits from " t- At”

to”

t (current time step). Hence,

' I Ö8Ü(T)9„, @><i¤[ — C,,,(C — C )1Td1 =0 (3.17)"^' I 6,;,,(1) ' I ösÜ(·r)

9,,, @><p[ — C,,,(C — C )1Tdr + 9„, 6><p[ - C„,(C — C )1Tdr0 1-81

The first term on the right hand side of Eq. (3.17) can be written as

'—^'I9,,,6><p[ — C„,(C — C )1Tdr0

*·^* I 6,„(-)= 9,,16><P{—·C„,EC—(€—AC)+(C—AC)—C]}Td¢

O (3.18)‘·^'

I ösÜ(·c)= 9,,, 6><¤( - C,,,AC) @><r>{ — C„,[(C — AC) — C 1}Tdr

0

= 9,,1 6><¤( — C,„AC)pi,Z°°


where Aé = %-, ande

2-Ar_ Ö6··(t)Pfnijét

=.['@><P{ — Cm[(C — AC)- C']}# dr (3-19)

0

The second part of Eq. (3.17) now is integrated by parts:

Qm @><p[ — Cm(C — C')] -5;- dr C¢—Ar

=‘—- GX -— _ÜEÜÜ)

[ C (C C’)1 a'T 6 20Ö? gm p m “

Cm 2-Ar( )

— ————exp — — 1

t özsü 8* L C rc mdöt2 Cm mr—A1

ln the current time step [t—At, t], we have assumed 6,] to be a linear function

of time:

ÖT At

Hence its second derivative is zero.

Substituting Eq. (3.21) into Eq. (3.20) yields,

Description of viscoelastic solids 38 .

' Ö8Ü{T)9m expt — Cm(C -

C’)] —g— dr:—A:

8 -·(t) ö ;·(t — At)Cmté —(C

A:)A

(3-22)— —————A, 9m Cm t — ¤><p( — Cm C)]

m

= (81j(t) " 8U(t ' Atngmßin

where —

, _ 1 — @><1>(—CmAC)

Then substituting Eqs. (3.18) and (3.22) back into Eq. (3.15) gives

6:-A:/1 = 290@g(*) + 2 gm @><P( — Cm^§)Pmy

m=16

+ 22am(¤„)(¢) — @q(¢ — A1))ßi„m=1 (3.24)

6

= (290 + 2 9m://::1)¢y(()m=1

8:-A: :+ 2 amt @><¤( — CmAC):¤m) — ¤,y(¢ — A1)/fm]

m=1

Description of viscoclastic solids 39

I We can write Eq. (3.24) in an abbreviated form,

I1 = G)sU(l) + GU (3.25)

where

8

GI = 2QOm=1

and

8

GU = 2 2 9mi: @><r>( — C„„Aé)i¤Ä„ÜU°‘ — ¤U(¢ — A0ßi„] (3-27)m=1

The coefficient, G,, is called the instantaneous shear modulus, while G,) is the

result of hereditary shear stresses. It is noted that the term p§,;f' in Eq. (3.27) is the

m — th component of the hereditary stress 6,) at the last time step (t- At) and it can

be derived from a recurrence formula as shown below. By definition (Eq. (3.19))


Ipmü = <6><p[ — C,„(é — é')]Tdr0"^‘

6¤,~<1>'”

6 -1 >=j‘

@><1¤[ — C,„(C — §’)]Tdr+I[ 6><p[ — C„,(~§ - é')] äldr0 r-Ar

=f GXPE-C,„(€-C+€-C’)]·%;Ldr0 I ÖSÜITI (3.28)

+ <6><¤[ — C,„(~f — é’)]Tdr1-A:

"^' _I ösÜ(r)

= exp( — C„,A€) <6><p[ — Cmlé — é )]Tdr ·0

+ (81y(Ü — SUU — Amßin= exp( — c„.Ac>p£,;;" + <¤„<0 -

6„(1where{= { —A{ =%

Consldering the second term of Eq. (3.7) and noting that from Eq. (3.12)

l(t) = K(t) —%G(t), the l(t) will have an exponentlal form:

8 t/·r,l(t) = ho + hm exp( — 7%) (3.29a)

m=1

h_ Eov

°‘

(1+ v)(1—2v) 3 2%_ Emv ( - )hm —

(1 + 0)(1 -20)

and substituting Eq. (3.29a) into Eq. (3.7), it gives

{

Description of viscoelastic solids 4l

2 8

/2 = 2 [//0 + hm exp — C„„(é — é )]7; de (330)0 m=1

Using the same procedure derived in the above context yields,

I2 = Hle + H,, + H22 + H33 (3.31)

where

8

m=1

and

8A— t IHkk exp( (3-33)

m=1

Combining the results of I, and I2, we have the following formuiation in matrix

form:


6, ,(t) G, + H, H, G, 0 0 0 6,,(1)06 6(t) H, G, + H, H, 0 0 0 66 6(t)0ZZ(t)

=H, H, H, + G, O 0 0 • 6zz(t)

0, 6(I) 0 0 0 G, 0 0 6, 6(t)0,z(t) 0 0 0 0 G, O 6,z(t)6z 6(t) 0 0 O 0 0 G, 6Z6(t)

(3.34)G11 +H11+H22+H33G22 + H11 + H22 + Has‘

+ Gaa+H11+H22+HaaG12G1aG2a

Then Eq. (3.34) can reduced to a compact form _4

— 1611)) = E,N16<1)) + ~{8<1>}A (3.36)

= c{6(1)} + N{0(t)}

where C=E,N and {0}, {6} and {8} are the vectors of stress, strain and hereditary

stress, respectively. Hence

{6} = {6, ,(t), 66 6(t), 6,Z(t), 6, 6(t), 6, z(t), 6z 6(t)}T (3.37)

{6}= {6,,(t), 66 6(t), 6„(t), 6,6(t), 6,Z(t), 6z6(t)}T (3.38)

{G} = {$1 r(* — A!) 86 6(F — N)» GZ ZU — N)· (3 39)8,6(1— A1), 6,,(1 — A1), 8Z6(1— At)}T

Description of viscoclsstic solids 43

8

E, = Eo + 2:*5,,,/*1,,1 (3-40)m=1

8

8,%* — 4*) = 2E„,[ @><P( — C,„A€)PÄ,]8t — 8%* — A*)ß‘„,] (3-41)m=1

uand

(1 — v) v v 0 O O .

v (1 — v) v 0 O 0

1 v v (1 — v) 0 0 ON = (1 + ")(1 -2V) Q

‘Q Q Q Q

(3.42)

20 0 0 0 00 0 0 0 0 lf!

Eq. (3.36) gives a general viscoelastic constitutive relation that applies to the

problem. Rewriting Eq. (3.36) yields,

(,;(:)} = ¤(¤(¢)} - {Su — 4:)} (3.43)

where D = C", and {€(t)} are the hereditary strains that depend upon the last time

step, and

A t- At{€(t — At)} = (3.44)


3.3 Flnlte element model

The procedures used to derive a finite element model of viscoelastic materials

are the same as those in Chap. 2 for elastic materials. The only difference is that

the element force, right hand side of Eq. (2.43), must include the force due to the

effect of hereditary stresses.

To derive this finite element model, we assume that surface traction(s), heredi-

tary - stresses, stresses,...and displacements can be expanded into Fourier series:

622(0 = 262 „(¢) ¤¤S(¤9)

66 6(0 = 266 „(¢) ¤¤S(~9)

6220) = 262 „(#) <=¤S(¤9)

Ur z(t) = Zar: n(t) CO$(n9)

ar 9(t) = 2520 „(t) sin(n0)(3.45)

62 9(t) = 252 9 „(t) sin(n8)

(......)

¤„(¢) = Z6„„(t) ¤¤S(¤9)

u0(t) = ZU9„(t) sin(n8)

u2(t) = 2Ü2„(t) cos(n9) 2

Then we assume again that 5,„(t), 5„„(t),....17„,(t) can be interpolated by shape

functions, N,, i =1,...r in a typical finite element with a finite volume:


Fm = Zörnilvi

(.............) (3.46)

Üzn = Züzni/Vi

Invoking the Hellinger-Reissner principle and substituting Eqs. (3.46) into Eq.

(2.24), the following compact formulation is obtained:

[Kto]?. · tx}: = im — Ao}: (3-4v>

It is remarked that superscript indicates "visco-" and subscript indicates”

n—th" _

coefficient of the Fourier series.

[K(t)]; depends upon the time; it is the same as the [K]„ of Eq. (2.41) except

E is replaced by E,, the element force {F(t—At)} is the force of the last time step,

and

{F} = {f1, f2,....fg} (3.48)

where

f1 = „N,dA + „N,dS (3.49a)

f2 = fi-9 „N;dA + fte „N,dS (3.49b)


A

{6 = „N;dA + [{2 „NidS (3.49c)

{4 = — fg, „(t — At)NidA (3.49d)

{5 = — [gr 6 „(t — At)NidA (3.496)6

{6 = — I2, Z „(t — At)NidA (3.49{)

{7 = — [S6 „(t — At)N,dA (3.49g)

{6 = —([*22

„(t — At)NidA (3.49h)

fg = —4[QQZ

6 „(t — At)N;dA (3.49i)

Description of viscoelastic solids

l

47

I

From Eq. (3.49) the element force ls calculated from the last time step and it ls

affected by the past time history.


4.0 Algorithms ofsolutlon4.1

Statlc analysis

4.1.1 Static condensatlon .

ln the mixed finite element model, the stress components are given as nodal

variables. Hence, the assembly of the element stiffness matrices into the global

stiffness matrix results in stresses that are continuous across each interface be-

tween elements. For the multi-layered cylinder, the stress components 6,,,,6,,, and

6,,,, are discontinuous across the interface}

To circumvent this difficulty, these stresses are condensed out at the element

level so that the discontinuity of the stresses across the interface no longer exists.

To do this, Eqs. (2.41) or (3.46) are rewritten in partitioned form as

K K X F

li 11 12il_[{ 1}]=|¥{1}] (4.1)K21 K22 {X2} {F2}

where {X,} is the vector of continuous variables across the interface,

3 For the general loading (symmetric), as for axisymmetric loading 6,,, and 6,,, are the discontinuousvariables.

Algorithms ol' solution 49

{X1} = {Um· Ü6,, - Üzn - Fm - örßn -F,2r,}T (4-28)

and {X2} is the vector of discontinuous variables across the interface,

{X2} = {?6,, - azn ·<7z6,,}T (4-2b)

Hence the nodal stress vector for a given element becomes

K11{X1} + K12{X2} = {F1} (4-38)

K21{X1} + K22{X2} = {F2} (4-3b)'

From Eq.(4.3b) .{

{X2} = K2;({F2} ‘ K21{X1}) (4-4)

Substituting Eq.(4.4) into Eq.(4.3a),

{F1} (4-5)

results, or

[R]{X1} = {F} (4-6)

where

[X] = K11 K21 (4-78)

{F} = {F1} — K22 {F2}{

(4-7b)

Algorithms of solution S0

Eq. (4.7) is assembled as usual and is solved for the nodal variables after ap-

plying the boundary conditions. The discontinuous nodal stresses that were con-

densed out are then computed by using Eq. (4.4) at the element level. Since these

stresses are no longer nodal variables, they will be discontinuous between inter-

faces.

4.1.2 Modified Newton-Raphson method

If a rocket motor is composed of (1) elastic case and (2) viscoelastlc propellant,

then E, is a function of reduced time, and depends upon stress and temperature at

the current time step. A direct solution cannot be used in this analysis, hence anI

iterative solution technique is sought.

We can rewrite Eq. (4.6)as‘

{R} = [k]{x} — {F} (4.6)

Suppose that we know the solutions of Eq. (4.6) at the i-th iteration and are

interested in the solutions at the (i + 1)- th iteration in the t- th time step.

We expand Eq. (4.8) about the i- th solutions in TayIor’s series:

ö{R} 2 1 ö2{R} 2 2{RI+1}= {Ri} +——;lö{X}l+ ———.l6 {xi}+IÖ{X} 2 ö{x}2 (4.9)

= O

Assuming the second and higher—order terms in ö{X} are negligible, we can

write Eq. (4.9) as

* for convenience, the bar above K and F and subscript 1 are dropped

Algorithms of solution 5l

6{x}§ = — [i<$]"{Rf} (4.10)

where [K;] is the slope (tangent) of the curve {R} at the i—th iteration in the

t- th time step and

1 1 1{X};+; = {X}; + ö{X}; (4.11)

We iterate through Eqs. (4.10) and (4.11) until the ratio of the norms of incre-

mental variables and total variables satisfies

I IA(x)¥ I I< ro:.(4.12)II1x)1+1I I

where TOL is a convergence tolerance, a small value.

The above statements (from Eq. (4.8) to Eq. (4.12)) are called the ”Newton-

Raphson method" [41]. A geometric interpretation of the Newton-Raphson method

is shown in Fig. 4.

The Newton-Raphson method requires that the tangent matrix [KT] be calcu-

Iated at each iteration. This will result in massive expenses when a lot of degrees

of freedom are included; therefore occasional updates of the tangent are desirable.

The modified Newton-Raphson method updates the imbalance force {R} at each

iteration for a fixed time step while it keeps the tangent matrix [KT] fixed

(Fig. 5).

The [K;] is updated only at the beginning of each time step; the modified

Newton-Raphson method may need more iterations to reach an equilibrium point.

The solution is started by assuming a free state. iterations continue until the

convergence tolerance is achieved. The problem is restarted for the next time step,

Algorithms of solution 52

I= i..____II

I II II I

·I II I I .I I II I II I II I I

1 2u u uc

Flgure 4. Newton-Raphson method


I=

' I. I I

I II |I 'II

"II 'II

u' uz Uc U

Flgure 5. Modlfled Newton-Raphson method


assuming a zero state at the beginning of that time step. Thus a series of problems

is solved.

We may note that the main advantage of this method ls that it is not necessary

to update the stiffness matrix (tangent), [KT] , in a time step but requires more it-

erations.

The solution procedures are shown below for each time step.

1. At the beginning of each time step, the stress vector {6} from the

previous time step is given. For the initial time step (starting time

step), the stress vector {6(t—At)} denotes the initial stress state at

t= 0, indicated by {6°} . lt is customary to assume a stress-free state

at the start of the analysis; {6°} is usually set to be zero. ·

2. Assume the time-temperature shift factor a, is an average value,

and ls given by

af = (af + a{_At)/2.0 (4.13)

3. Calculate hereditary integral, using the recursive formula, Eq.

(3.27) and hereditary strains, {€(t— At)}.

4. Evaluate element stiffness matrices [K] and element forces

{F}„,=p{F}„p_, where p is the load factor that corresponds to the

time step under conslderation, and compute [KT] and {R}.

5. Assemble element tangent matrices [KT] and {R}, save [KT] and

its lnverse matrix for use in this time step.

6. Apply the boundary conditions on the assembled equations.

7. Solve the assembled equations.

8. Update the solution vector using Eq. (4.11).

9. Check for convergence.


10. lf the convergence tolerance is satisfied, go to the next time step

and repeat steps 1 - 9 until the final time step is reached; otherwise,

compute {R} in step 4 and go on.

4.2 Translent analysis

The Hellinger-Reissner variational form is rewritten as

I-ID =

I-T2—TwhereT is the kinetic energy, ITD ls the H-R variational form for the dynamic prob-

lems, and H2 is the H-R variational form for the static problems.”

By using the same technique as for the static analysis, the typical equation be-

comes

EM]1S()+E1<J{x)=(1=) (4-15)

where [M] is called the mass matrix, and

mf = '·.p1v,~jdA = M§2 = M,j?° (4.16)

with p the density; other coefficients of [M ] are zero.

4.2.1 Free vibratlon

If the system is subject to zero external loads, assume that the solution vector

to Eq. (4.15) can be expressed as


{X} = {X}6i“°' (4.17)

Substituting Eq. (4.17) into Eq. (4.15) yields

[K]{x)—w2l[M]{x}=o (4.18)

Rearranging

([1<] — A[M]){x} = o (4.19)

where A=w2 is the eigenvalue of Eq. (4.19) and w is called "circular frequency”

(radians per second).

In determining the frequencies (or periods) of the system, the displacement fl- ·

nite element method is usually used.

Using the same technique in the displacement finite element model, the

equation of motion is the same as Eq. (4.19) but {X} ls the displacement vector

{X} = {ur, ug, uz} (4.20)

In free vibration analysis of a solid of revolution, we are interested in the first

three terms, n = 0, n =1 and n = 2 (with m =1,

2,...).5 Fig. 6 depicts the geometric

interpretations.

Expanding Eq. (4.19) for n=0, n=1 and n=2 into a more explicit form, for

n = O

11 12 11 0 Ur( 21 22

_'l22 ){ }=0 (421)

KU 0 UZ

5 n’s correspond to circumferential modes and m's correspond to longitudinal modas


I@ ü ir ‘·'

Ü * 0 Ü Q •2

·„ CIRCUMFERENTIAL

{ Ah moocsfl • 3 n rg • L

LONGITUOINAL_____, __ _ __ MOCESm • 1 m • 2 rn • 3

Figure 6. Some vlbratlon modes ol a hollow tube


for n=1 and n=2

11 12 13 11Kü Kü MU Ü Ü Ur

21 22 23 22( KU KU KU — Ä. 0 MU 0 ){u6 }= 0 (4.22)

31 32 33 33

The coefficients of the stiffness and mass matrices of a displacement finite element

model are presented in Appendix B.

Even though the displacement finite element method is used, the big difference

between moduli of two materials in the rocket motor still incurs difficulty in solving

Eq. (4.19).‘

To circumvent this difficulty, an alternate method is suggested. Rewriting Eq.

(4.19) into another form to avoid the singularity of the stiffness matrix,

[(K + FM) - (A + I”)M]{X} = 0 (4.23)

wherein F is chosen so large that K + FM is not a singular matrix.

From Eq. (4.23) eigenvalues, A+ I", can be found; finally A can be determined

for different n .

By knowing the period of the system, a time increment for transient analyses

can easily be set (usually 2-]% period).

4.2.2 Newmark direct Integration method

There are several approximation schemes available for solving Eq. (4.15). The

most commonly used one is the Newmark direct integration method.


In the Newmark direct integration method, the displacements and velocitles at

the (n +1) -th time step ( At1 = At, = .... = At ) are approximated by

{x}n+1 = {X}-, + [(1 — ¤)lY}„ + °l{X}n+1]At (4-24)

{x}n+1 = {X}-, + {><}„4¢ + [( g — ß){X}„ + ß{X}„+1J(At)2 (4-25)

where a and ß are the parameters that control the accuracy and the stability of the

scheme, and the subscript, n, indicates that the solution is evaiuated at the n-th

time step (i.e., time t=t,,). The choices oz = 1/2 and ß= 1/4 are known to result in

an unconditionally stable scheme, Substituting Eqs. (4.24) and (4.25) into Eq. (4.15), _

A A[K]{x}n+1 = {F}n+1 (4-25)

is obtained, where

[R] = [K] + a0[M] (4.27)

A . „.

{F}n+1 = {F}n+1 + [M](8o{X}„ + 81{X)„ + 82{X}„) (4-28)

and ao =1/(ßAt2), a, = aoAt, and az =1/2ß -1.

Combining Eqs. (4.24) and (4.25), the acceleration can be found:

81 {X}- — 82{Y}„ (4-28)

lf the solution {X} is known at the (n + 1) -th time step, then the first and sec-

ond derivatives of {X} can be calculated from Eqs. (4.24) and (4.29).

The solution procedures are summarized below for the (n + 1) -th time step.


1. Compute the element stiffness matrices [K] and element mass

matrices [M] .2. lnitialize the starting conditions, i.e., {X},, {X}, and {X},.

3. Select the time increment At, parameters ot and ß.

4. Calculate [K] and {lg} by using Eqs. (4.27) and (4.28).

5. Assemble element stiffness matrices and mass matrices and apply

the boundary conditions.

6. Solve the assembled equations.

7. Calculate velocities and accelerations at (n+1)—th time step by

using Eqs. (4.24) and (4.29).

8. Repeat steps 1-7 until the final time step. ·


5.0 Numerlcal examples and discussions

Responses of rocket motors are calculated under the following four loading

conditions : (Fig.7)

(1). Self weight -

(2). Two patch loads centered at 0 = 0° and 0 =180°' on the midspan of the rocket

motors.

(3). One line load ( centered at 6=0°) and one patch load (centered at 0=180°,

on the midspan of the rocket motors ).

(4). Two line loads centered at 0 = 0° and 8 =180°.

The loads can be expressed in terms of Fourier series (symmetric), i.e.,

f(x) = pA/1: + Z(2p sin(nA) cos(n8)/mr) cos(nx) (5.1)

where p= 1 psi. and A=1°. The following boundary conditions are applied : Mo-

tors are supported on fixed end diaphragms, i.e., at z = 0, z=l

'The area of patch loads is viewed as square with dimension of 0.15 in. ( 2° ) by 0.15 in. in currentexamples

Numerical examples and discussions 62

III. IpsiVü

,/ \ Z

I /@ P= 9#

P=9#f 2

#: —— ·W=

äI

Flgure 7. Force dlstrlbutlon ot the rocket motors A

Numericd Examples end discussions 63

U, = OUG = 0 (5-2)UZ = O

at the inner and outer surfaces (r= r, and r= ro ) (except prescribed surface(s))

6,(r„) = the applied load divided by the area of application

Ur(ri) = 0- Ur(ro) = 0

Ur 0(ri) = 0· Ur 6(ro) = 0

Ur z(ri) = 0* of z(ro) = 0

at the interface

u,(prope//ant) = u,(stee/) ·

u9(prope//ant) = u8(stee/) '

uZ(prope//ant) = uZ(stee/)5

6,(prope//ant) = 6,(stee/)( '3)

6, 8(propel/ant) = 6, 0(stee/)6, Z(prope/lant) = 6, Z(stee/)

Initial conditions :

{X}0 = O. (5.4)

{X)0 = 0

where subscript 0 indicates initial time ( t=t„ ). ln practice, one does not know

{X}; it must be calculated from Eq.(4.15):

(Si) = [M]”1({F} — [K]{X}) (5-5)

with the length of a rocket motor = 60 inches. Its mechanical and geometric prop-

erties are listed in Table 1.

Numcrical cxamplcs and discussions 64

Table 1. Mechanical properties of rocket motors

Material Air Propellant Steellayer 1 layer 2 layer 3Outer Radius (in) 1.875 4.3 4.4

Elastic Modulus (psi) 281.84 30,000,000Poisson's Ratio 0.49 0.25Density (lb/cubic in.) .0622. 0.3Weight/unit length (lb/ln.) _ 2.926 .820

Numcrical Examplcs and discussions 65

5.1 Statlc analysls wlth elastlc propcllant

In order to verify the program, comparison is made with the following three

cases.

(1).A hollow cylinder subjected to a uniform pressure (1. psi.) on the outer surface

(see Table 2). The exact solution (closed form) can be found on pages 70-71 [23].

The results of the finite element method converge to the exact solution when using -

a fine mesh.

(2). Isotropic single layer (steel) cylinder under two opposite line loads [3] (Fig.8).

Cederbaum and Heller [3] worked on the responses of a finite length, moderately

thick composite laminated cylinder due to dynamic loads by using the theory ofu

shells. They obtained solutions (displacement at mid-length as a function of cylinder

length) represented by ”triangIes" in Fig. 8; the solutions obtained from FEM are

presented by "stars” in the same figure. The two solutions are almost the same in

a shorter length range, but for the longer length range there exists some discrep-

ancy. The possible reason is that the ratio of width to length of a finite element is

very large.

(3). Responses of finite length cylinder (length = 60 inches) with those of plane-

strain problem (infinite cylinder) [8] under two opposite line loads (Table 3). Results

of the plane-strain problem [8] were obtained by using the semi—series solution (the

solution of stress function represented in a series form [23]). The displacement

solutions of FEM are in agreement with the solutions of Ref. [8], but not stresses.

Apparently the ends of a finite cylinder have a major effect on the stress solutions

in the mixed FEM.7

7 Radial and tangential displacements (inch), radial, tangential and shear stresses (psi.)

Nnnieiieel exeinples and discussions 66

Table 2. Comparlson ot solutlons (FEM and exact solutions)

i' ‘ arr 090

r=4.3 in. Exact solution -.9896 psi. -1.454 psi.FEM solution -1.031 psi. -1.465 psi.

r=4.4 in. Exact solution -1.0 psi. -1.444 psi.FEM solution -1.041 psi. -1.457 psi.


0.00038A A _.,..--A—-—-·—·

0.00030

··s , .- *—· "” *‘ f I-

Z */ /•·—• Ü.ÜOÜ24 /‘·’ /l-< / .Z /L;] /

ä 0.00018/’ ·

u ^ 11/¤•<an17)

0.00012 » 1 .

E 1/

/

0.00008 //

¢

0.000000 30 80 120 180 300 800

LENGTH (INCH)LEGEND : A A A uam vom s0w.(s1u·:u. mr:.)=•= =•= ¤•= FINITEEIEIDIT SOLU.

Flgure 8. Dlsplacements of the cyllnder vs. length: at the mldspan ot the cyllnder

Numerical Examples and discussions 68

Table 3. Comparison between two methods

Ur Us *7n Gro Goa

F.E.M. .00025 in. .00012 in. .00552 psi .00567 psi .0040 psisolution

Closedform .00026 in. .00012 in. .00597 psi .00512 psi .0053 psisolution


Figs.(9), (10), (11), (12), (13), (14), (15) and (16) show the deformed shape and

response distributions of the rocket motors subjected to (1) self-weight (2) two op-

posite line Ioads (3) two opposite patch Ioads and (4) one line load and one patch

load.°

The responses (absolute maximum) under four different Ioadings are listed in

the Table 4.

5.2 Static analysis wlth viscoelastic propellant

lf the rocket propellant is treated as a viscoelastic material, with its mechanical .

properties listed in Table 5, the hereditary integral solution (Sect. 3.2) is used.

Figs. (17), (18), (19), (20), (21), (22), (23), (24) and (25) are the radial displace-

ments, radial stresses and tangential stresses on the interface at midspan (0 =0°)

due to (1) two opposite line loads, (2) two opposite patch Ioads and (3) one line load

and one patch load at different temperatures ( 35°, 45°, 55°, 65°, 75° and 85° F).

From the above Figs. (17-25), it is seen that the long term radial displacements

(t—>oc) will attain the elastic responses (Figs. 12, 14 and 16) no matter what the

temperature is. Further, it can be observed that the radial displacements increase

as the temperature increases; conversely, the stresses decrease.

' L'} ll?}$*2.3planeperpendicular to the longitudinal axis at midspan. .One line load - 1 psi. (9 lbs / (.15 in. x 60 in.)) one patch load

-400 psi. (9 lbs / (.15 in. x 15 in.))

and total weight of a cylinder (60 inches) is 224.77 pounds

Numeiical examples and discussions g 70

Table 4. Statlc responses

Ur U6 arr drä GOG _

Two patch .00043 in. .00014 in. .04467 psi .01077 psi .03758 psiIoads

One line .00040 in. .00012 in. .04613 psi .01055 psi .03908 psiand one patch load

Two line .00014 in. .00006 in. .00346 psi .00214 psi .00351 psi _Ioads

ISelf-weight .00025 in. .00025 in. .18830 psi .02705 psi .18030 psi

Numcrical Examples and discussions 7|

Table 5. Data for the vlscoelastic propellanf

81 1EU) = Eo +,7gEm @><¤( —Q;)

EO = 281.4E, = .19789 X 10+5E2 = .79896 X 10++EO = .25217 X 10++

O‘

E4=.11526><1O+4 _ ‘EO = .73456 X 10+5 (Ps")EO = .20414 X 10+5E, = .27178 X 10+5

EO = .86427 X 10+*5

T, = .333 X 10+**:2 = .333 X 10+5TO = .333 X 10-7

TO = .333 X 10+5·

TO = .333 X 10+*1, = .333 X 10+*TO = .333 X 10+5 1 ·

Numerical Examplcs and discussions 75

5.3 Dynamlc analysls

When the time duration of the applied load is very short, the propellant may be

considered to be an elastic material whose modulus is calculated using the appro-

priate terms of the Prony’s series of Table 5.

For comparison, the natural frequencies of a motor were calculated using (a)

the rest modulus (E = 281.4 psi.) and (b) the modulus corresponding to a loading

time of zero seconds (E = 32945.2 psi,). _

The calculations were carried out with the aid of the IMSL library routine

"DG2LRG" and the elastic computer program ”AXIF". Results are presented in

Table 6.‘

Based on Table 6, the time increment is chosen to be 0.001 seconds for the

steel case and 0.0018 seconds for the viscoelastic propellant.

5.3.1 Analysis with elastic materials

In order to investigate the responses of rocket motors, different kinds of force

histories are applied (Fig.26), i.e.,

(A) ramp-rectangular load ( with the time interval of ramp

3 -times the basic period )

(B) rectangular pulse with application time greater than the

basic period

(C) rectangular pulse with time 2 0.75 period

(D) triangular pulse with time 2 0.75 period

(E) sine-wave pulse with time 2 0.75 periodA

Force histories (C), (D) and (E) simulate the influence of impact.

Numeiical examples and discussions 73

Table 6. Frequencles of a rocket motor

Frequencles (Hz) (r = ät-)m = 1 m = 2 m = 3

> n = 0 98 108 124 _n = 1 105 110 120n = 2 134 136 145

(A). A cyllnder wlth steel case and propellant (E=281.4 psl) ‘

coFrequen_cles (HZ) (f = E;-)

m = 1 m = 2 m = 3n = 0 678 1015 1240n = 1 244 562 933n = 2 480 588 839

(B). A cyllnder wlth steel case and propellant (E=32945.2 psl)


ln order to investigate the changes between static Ioads and dynamic Ioads, the

force history (A) is applied. Figs. (27), (28), (29), (30), (31), (32) and (33), (34), (35) are

the radial displacements, radial stresses and tangential stresses on the interface at

midspan ( 0 = 0° ) due to two opposite line Ioads, two opposite patch Ioads and one

line load and one patch load, respectively.

From Figs. (27), (30) and (33) the radial displacements° are increased to the

maxima ( i.e., the results due to static Ioads ) in proportion to the gradually in-

creasing ramp force. Beyond these values displacements oscillate sinusoidally with

the constant applied force ( 1 psi. ). Correspondingly, the radial and tangential

stresses‘° due to two opposite patch Ioads or one line load and one patch load (

reach their maxima when the applied forces become constant

( 1 psi. ), see Table 7."

Figs.(36) - (47) are the radial displacements, radial stresses and tangential

stresses on the interface at midspan (0=0°) due to two opposite line Ioads with

four different force histories (B), (C), (D) and (E).

Figs.(48) - (59) are the radial displacements, radial stresses and tangential

stresses on the interface at midspan (0 = 0°) due to two opposite patch Ioads with

four different force histories (B), (C), (D) and (E).

Figs.(60) · (71) are the radial displacements, radial stresses and tangential

stresses on the interface at midspan (0=0°) due to one line load and one patch

load with four different force histories (B), (C), (D) and (E).

°The radial displacements at the interface are almost the same as those at the outer surfaces.

1° The maximum radial and tangential stresses are not located at the loading points, but somewhereelse when two opposite line loads are applied (see Fig.12) .

ll The values in parentheses are the results due to static Ioads

Numerieal examples and discussions 75

A

Table 7. Comparlsons between static and dynamic loads (interface)

Two line .00014 in. - ·loads (.00014 in.)

Two patch .00043 in. .045 psi. .045 psi.loads (.00043 in.) (.04467 psi.) (.04613 psi.)

One line .00040 in. .036 psi. .040 psi.and one patch load (.00040 in.) (.03758 psi.) (.03908 psi.)

Numcrical Examplcs und discussions 76

For application of two opposite line Ioads, Figs.(36) - (47) indicate that the basic

period of the rocket motors is around 0.01 seconds. The "snap" phenomena in

Figs.(37), (40), (43) and (46) occur due to the effect of higher modes (second mode

mainly ). As for (1) two opposite patch Ioads and (2) one line load and one patch

load the basic period can be estimated to be 0.01 seconds from the displacement

vs. time plots ( Figs.(48) - (71) ).u

More interestingly, at the interface the stress responses are opposite to the

displacement responses (tension stress, negative displacement; compression

stress, positive displacement, i.e., the strains 6:,, at the outer surface are opposite

to those at the interface (see Tables 8, 9 and 10))."

From Figs.(36) - (71) the magnification factors ( ratio of maximum dynamic dis-A

placement to static displacement ) are listed in Table 11.

5.3.2 Analysis with viscoelastic propellant

In analyzing the dynamic problem of a rocket motor with viscoelastic propellant,

it is necessary to calculate the modulus of viscoelastic materials from a Prony’s

series at each time (step). Further, the forces due to the hereditary effect must be

updated at each iteration.

Three different kinds of force histories are used in this analysis (Fig. 72). The

total time Iapse of force histories (F), (G) and (H) is 0.18 seconds (100 steps), and

(F) rectangular pulse with a complete application time

(G) rectangular pulse with a shorter time (40 time steps)

(H) triangular pulse with a shorter time (40 time steps)

12 Strains at midspan (0 = 0°)


Table 8. Two llne Ioads under force history (B)

radial strain 6,,

Time (sec.) Outer surface Mid - Interface0.001 .22465-6 -.20005-6 -.81965-50.002 .46905-6 -.35055-6 -.90375-50.003 .63955-6 -.31915-6 -.53485-50.004 .83205-6 -.30915-6 -.30805-50.005 .95225-6 -.36555-6 -.47385-60.006 .92125-6 -.41 145-6 -.23365-50.007 .66985-6 -.29875-6 -.24545-50.008 .43905-6 -.24425-6 -.79165-50.009 .32275-6 -.30545-6 -.10555-4 ·

0.010 .25635-6 -.26415-6 -.10065-4


Table 9. Two patch Ioads under force hlstory (B)

radial strain 6,,

Time (sec.) Outer surface Mid - Interface0.001 .5161 E-5 -.2320E-5 -.7335E-40.002 .6524E—5 -.2813E-5 -.2608E-40.003 .5913E-5 -.2460E-5 -.4882E·40.004 .6724E-5 -.2662E-5 -.4188E-40.005 .66629-5 -.2659E-5 -.4738E-40.006 .7139E-5 -.2864E-5 -.4370E-40.007 .5993E-5 -.2431 E-5 -.2550E-40.008 .5977E-5 -.2546E-5 ·.7066E·40.009 .5725E-5 -.2609E-5 -.3534E-40.010 .5554E-5-.24635-5Numcrical

Examplcs and discussions 79

Table 10. One Ilne and one patch load under force history (B)

radial strain 6,,

Time (sec.) Outer surface Mid - Interface

0.001 .1078E-3 -.4758E-4 -.1538E-20.002 .1372E-3 -.5790E-4 -.5239E-30.003 .1247E-3 -.5165E-4 -.9505E-30.004 .1375E-3 -.5643E-4 -.7901 E-30.005 .12679-3 -.5268E-4 -.1014E·2

0.006 .1351E·3 -.5431 E-4 -.11445-20.007 .1229E-3 -.5038E·4 -.7307E-30.008 .1278E-3 -.5452E-4 -.1367E·20.009 .1207E-3 ·.5305E-4 -.5255E-30.010 .1226E·3 -.5385E-4 -.1430E-2


Table 11. Magnlflcatlon factors vs. force hlstories

(B) (C) (D) (E)

Two patch 1.53 1.53 0.93 1.12loads

One line 1.38 1.38 0.95i

1.15and one patch load

Two line 1.86 1.86 1.29 1.57loads

Numerical Examplcs and discussions 8l

Figs.(73) - (81) are the radial dispiacements, radial stresses and tangential

stresses on the interface at midspan (8=0°) due to two opposite line loads with

three different force histories (F), (H) and (G).


stresses on the interface at midspan (6 =O°) due to two opposite patch loads with

three different force histories (F), (G) and (H).


stresses on the interface at midspan (0=0°) due to one line load and one patch

load combinations with three different force histories (F), (G) and (H).

It is noted that the responses under force history (F) (radial dispiacements,

stresses and tangential stresses) will converge to stable values which are the same .

as those in the static analysis with viscoelastic propellant at a specified time (t =

0.18 sec., see Table 12). Under force histories (G) and (H) the responses (radial

displacement, radial stresses and tangential stresses) of the viscoelastic propellant

will decay to "zero state" after the applied forces are removed at a specified time

step from the cylinder.

5.4 Summary

The analysis of a finite length, multilayered cylinder with viscoelastic propellant

due to static and dynamic Ioads by a mixed finite element model was presented.

This approach was used because the external loads induced boundary condi-

tions on stresses which could not be satisfied by the more frequently applied dis-

placement model. Because loads were expanded into Fourier series, the problem

could be treated as a symmetric one and this simplified the solution. A large

number of Fourier series’ terms were required.


Table 12. Comparison between static and dynamic analyses

two-line Ioads

ur _ arr Ur0 090

dynamic -.00081 inch -.011 psi. .0147 psi. -.0142 psi.responses

static -.00082 inch -.015 psi. .015 psi. -.012 psi.responses

two-patch Ioads

ur arr ar0 G90

dynamic -.00030 inch -.236 psi. .108 psi. -.2175 psi.responses


Iinelpatch Ioads

urdynamic-.00029 inch -.238 psi. .105 psi. -.2205 psi.responses



Motors were supported on end diaphragms that permitted no displacements.

While this assumption is a simplification, it represents, simply supported motors

with end caps and nozzles that prevent radial displacement, reasonabiy well.


.;._ DEFORHED SHRPEUNDEFORMED SHQPE

HRX. • .25E-B HCH

¤III“{Egg!}

1;;:::::}*** FääääääälläIIIIEEEHIIII!!!§”.!EEKIII\!.!!!.!!.!.}.jII

!.!g!.'-==}|II1!.!!.!g==«JIIt!‘!.!=„-}}}IFgääätnl[Ill}

Flgure 9. Deformed shape due to self·wetghtI

Numcrical Examplcs and discussions Y85

I. HDI!. 0l9L. t®'-‘G!£DI ä. MX. • .5ß-3 INZH2. Clälf. OI9\.. IECIVEDI ä. MX. • .52-3 It¤·t3. XIH. STRESB IOISTRIBJTIOJI E. MX. • .Iä£•0 PSI4. SIM STES IDISTRIBJTICNI %. MX. • ZE-I PSI5. TRCETIR STRESS IOISTRIBJ &. MX. • .I&J5•O PSI

Figure 10. Dlsplacement and stress dlstrlbutlon due to self-weight :

along Interface at the mldspan of the cyllnder

Numcrical Exemples and discussions 86

‘——‘—..._.._ BääääääägäffäpgIAX. u . l4l§·3 IPO-!

Il!!!*

I!§=-'!!![/IIII!!.§¤'=!!/III

iII nlEäE§·'=äiIälIIÜIÜÜIHIII

lll_!!!Ll1I|II!!!!!.!„ä§I||I!!.!!!=ägI||l!.!.!!l=§iI||IllllligilllI!.!=‘|='i_!_\i||I!!!l=il{f!|tlgzgztägil__!‘!.

-“-laFigure 11. Deformed shape due to two opposlte Ilne Ioads


I. ÄIR, Ulf,. fE®‘EDI MX. • .l4l§—3 IPO!2. Cl$.H·'. UI?. IEGUEII ä. PRX. • .§l4E·4 IPO-!

3. HDIGI, Sfüä IUISTRIBJTIGII MX. « .34Sä-2 PSI4. S-EM STESS IUISTRIBHIUGI ä. MX. ¤ .2I37E·2 PSI

6. Thiülfl. STRESS fUlS‘|’RlB.l E. MX. n .§U’7E-2 PSI

Flgure 12. Dlsplacement and stress dlstrlbutlon due to two llne loads :

along Interface at the mldspan of the cyllnder


_.........——- OEFORHED SHRPE_____._.,. UNDEFORMED SHQPE

IQX. ¤ Ai-3 l©·|

-l!!|!;!¤*!.!r!;=„llnm-„-.,;I1§§§:=:.!n|!!!?=!.!Ha|

lllnggél/IllIuuuamuI!Il!!!äMI!!!.".:g;\\I|l· I!!!!Eg_g_\\\||I!!!‘I-•g\Y|I§‘¤‘•%ä¥'="·‘="“%4.-nuilil

____¤\

Flgure 13. Deformed shape due to two opposlte patch Ioads

Numcrical Examples and discussions 89

l. ßßll Ol§.. tEFU!€Dl ä. MX. • .43§·3 HCH2. CIE)!. Ol?. lßüüfbl &. MX. • .t377E·3 HCH3. ill NESS IOISTRIBJTIGII ä. MAX. • .4467E-l PSI4. 8·£M Nüä IOISTRIBJTIQU ä. MAX. • .lO77E-I PSIG. TF!~®¢Tll NESS tDlSTRl8.l &. MAX. • .375ä% PSI

C) Ö\4/

Flgure 14. Dlsplacement and stress distrlbutlon due to two patch Ioads

Numcrical Examples and discussions t 90

.........-..-. CEFORHED SHAPE-_-. UNCJEFORMED SHAPE

MAX. • .404§·3 IPO!

ll---- -l--111-!!I !!§§„i11-g= 1;-.2,*.11!!!- f!1;1¤¤·=:=·

— e _!I|- s '

1'ällliäI‘*5ä""'

A IIIIÜHII|!I!!!l¥N||

_ W-- =

I!!!.‘:.ä§§1\i||-1 1 -!!¤•-\\\||-Ü-I Il|!'I=i'i\\”||

I-!!i=iE1|II-Ü--1 1 ¤§1—¤¤¤-E11"-——-—~11---

!.!.!=.*1si--- --;;.1

---

----11

Figure 15. Deformed shape due to one line and one patch combination

Numerical Examples and discussions 9l

I. EINE DI$'L. IEGOEDI &. Mx. • .404I·3 lr¤·I2. CIC)?. DISK. (EIDE) E. MX. • .I2£~3 HCH3. XIII. ST55 IOISTRIHJTIDU &. MX. • .48II-I PSI4. SEM STESS IOISTRIBJTIOII fü. MX. u JGEE-I PSI6. Thßül ST§ IOISTRIBJ ä. MX. • BSE-I PSI

® ®·Flgure16. Dlsplacement and stress dlstrlbutlon (llne and patch)


T•nve•·aIu·6lFl: 6-5-5 86 :1-E1--13 75 Q-9-G 55B-B--9 66

«••-+•—·••46 ·|—l——·l- 35

0.00000

-0.00001-·‘·l-0.00002

:‘Xx-0.0000:1.} Q-0.00004 ¤ \\\

.-0.00006 —‘\ \\\\\\

-̂0.00006.:1 Xca-0.00007 - Öxö

•" -0.00008 \gz" äxN\apnl \

\Q -0.00000s\

\-0.00010 Ö‘*>¥i\-0.00011

‘—}Q\

-0.00012 Qä .-0.0001:1>

F

-0.00014‘-0.00015

3.61-10 3.61-9 3.61-8 3.61-7 3.81-5 3.61-3 3.61-1 36. 3.61+3 3.51+5 3.61+10

Time (sec.)

Flgure 17. Radlal dlsplacement vs. tlme (two Ilne Ioads): vlscoelastlc propellant


Temperal1¤·e<Fl: a-¤—¢ 86 cr-a-13 T6 6-6-G 669-9--9 56 ·••—•••-••· 46 +—•—•- 36

0.00

. ?

-0.01 //

. 6 /, / >'/

-0.02 _ //,6,;//,/.·// .

Ö -0.03 0///

m -0.04 r Ü/m/.2

-0.06 ,' /.6 .. //’M

·"-0.06 "i,;p/

-0.07 ,Ü,’/

lr-0.06_ 6. ' I

-0.00BJI-10 BJI-9 BJI-! BJB—7 3.6I—5 3JI-B BJI-1 36. BJI+B BJI+5 3JI+10

Time (sec.)

Flgure 18. Radlal stress vs. tlme (two llne loads): vlscoelastlc propellant


85 D.a.G 75 €__G_G66

9--9--B 56 «••—+—••• 45 +-1-1- 36

0.00

— 1 VE jr/I

./// i

-0.01 { f /

lg?/

I,

„,5 -0.02 ,7/ 0/gg I/1 7¤- · /7 .\•/

I/1:3 -0.03 , IWU] '1',I[f

gv 7/¤ 1 I¤¤ ,· 1 //

*0-04I,1,

[ /”

I,I

-0.06 _ ,7/. '; {;/

-0.063-II-10 3.6]-9 3.6]-8 3.6K—7 3.6]-5 3.6]-3 3.6]-1 36. 8.6]+3 $$+5 3-6]+10

Time (sec.)

Flgure 19. Tangentlal stress vs. time (two Ilne Ioads): vlscoolastic propellant


T•rm•r•tu•(Fl: 16-a—¢ 5 D-EI--I3 76 G-G-G Ü9-·B--9 56 «••-•••-••• 46 -•-•—•- 5-0.00006-0.00000-0.00010

ö-0.00011 . ~?§\-0.00012 =.\ _-0.00016

‘\-0.00014 -,x-0.00015*0}

-0.00016 *.(\\-0.00017 \\-0.00018 X}

\-0.00019 ¤k\-0.00020 11 \\\-0.00021

‘

,.1 -0.00022 9}.¤ -0.00028 ‘,\Ää -0.00024 *.1

-.-. -0.00025*0Y-0.00026

‘—lS\\ _·-• -0.00027 =v$* -0.00026 ‘

-0.00029 *0}-0.00030 ‘.}\-0.00031 QD-0.00032-0.00033T-0.00064‘-0.00035 }\§ ·-0.00086 gu °-0.00037 . _ Ä _-0.00088T-0.00039

‘~~1 L-0.00040 ·-0.00041 · _-0.00042 · _ 1-0.00046 . .-0.000446.4:-10 6.4:-• 6.4:-6 6.4:-7 6;-6 64:-: 6;-1 64. 64:+6 6;+4 I.I+l|Time (sec.)

Flgure 20. Radlal dlsplacement vs. tlme (two patch Ioads): vlscoolastlcpnopellant


Tenpcrpturs F): 6-6-6 86 ¤·-E1--13 76 G-G-G 65·- 9-9--9 66 «••-•••-••» 46 ·•—•—+ 36

’ ‘/1 2/

;;7/°VI1// / '

rv " [ //nä ·}' /

2 · //Ü

1,//2. ,-1

'P 1 IIl I /r-1 ' 1 /

E'lrl /)‘( /’° Bi 1 /

$2 ·' '//-2 I.;·’;’ I

-3

l

3.8]-10 3.6]-9 3.6l-6 3.6]-7 3.8]—5 3.6]-3 3.8]-1 38. 3.8]+3 3.8]+5 3.8]+lI)

Time (sec.)

Flgure 21. Radlal stress vs. tlme (tvvo patch Ioads): vlscostastic prop•IIant


T•mperaIu·•<F): *-1-* 85 g.-E;..¤ 75 g-g.G 669--B--9 66 —•••—-•••-••~ 46 +-1-1- 36

0.0

-0.1 J, %-0.2 [y 'Ä

- //-0.3

·”6 /

,é’-0.4_*10.6 I 7/-o_g

I/I? 4

ZT -0.7,/,0/ '

U) 1/jl/-• -0.8 [lll //Q -0.0 7/ 7cu '1 7

-1.0 ,1]m -1 1

[ [

C: -1.2 /'/1 pÄ -1.6 ,7// /*/

I I /-1.4 P1 //-]__5

-

"„;’}

E

-1.6

-1.7-1.8-1.9

-2.0 —lßl-10 3.6l-9 66:-6 6.6:-1 6.6:-6 66:-6 6.6:-1 66. 66:+6 6.6:+6 6.6:+10

Time (sec.)

Flgure 22. Tangentlal stress vs. tlme (two patch Ioads): vlscoelasttc propellant

Numerical examplu and discussions 98

T•¤\P•f¤¤¤¤¢Fl= ü-—a—& 86 5-5--43 75 g-g_G 66

-0.00000 -——-0.00010

““—- 2 x-0.00011-0.00012 ”‘Ö§\\-0.00013 E \-0.00014·-0.00015 ‘,x

\-0.00018‘.‘\-0.00017

‘·‘Ö0

-0.00018 Q } .-0.00010

‘\ \\\

-0.00020-0.00021 \‘}\

^ -0.00022‘—x\

rd \ \\0 -0.00023 ~„\-0.00024 MQ

Y -0.00025‘ \\

TL -0.00026 Q}E -0.00027 ·5}\Q -0.00026

‘.}

-0.00029‘~§&

-0.00030‘\-0.00031 T Q

-0.00032-0.00033Q}

-0.00064-0.00035‘

-0.00066 · _-0.00037

“‘.j

-0.00038 .-0.00039_-0.00040 * _-0.00041

3.8]-10 3.8]-9 3.8]-8 3.8]-7 3.8]-5 3.8]-3 3.8]-1 38. 3.8]+3 3.8]+5 3.8]+1.0

Time (sec.)

Flgure 23. Radlal dlsplacement vs. tlme (llnelpatch Ioads): vlscoelastlcpropellant


T•m¤••‘¤u•u=1:6-e-e 86 a-a-6 75 6-9-6 669--9-9 66 •••—•••-••• 46 -•—•—+ 5

O e $

r .,;’ L /

427/°(, /

11[ /

Ö -1 ,1 j/ ·/ /B. . 1 [ jsa l/ .

I

gg1,]s-.

,'/4-)1U1•-1‘ l

d/Iy//'U 1B?.-2 ,1 ß

1 I Q ~.

-3$6]-10 $6]-9 3.6]-8 3.6B-7 3.6]-5 $6]-8 3.6]-1 36. $6]+3 $6]+5 $$+10

Time (sec.)

Flgure 24. Radlal stress vs. tlme (llnelpatch l¤ads): vlscoelastic propellant

Numerical examples end discussions IW

T•tm¢&lu‘•(F): g-g-5 85 5.-5.63 76 6-6.6, 659**9*9 56 ••-*—••• 46 +—•—+ 36

0.0 _•O.1 ‘·-0.2 A:-0.3 I,}1/-0.4 I4}?-0.6 Ié-0.6Ii'? _

T5 —<>·7 :I/1/m 1 I¤-I -0.8

-0.9 "[ //G) H I

5 -1.0 I I //ui -1.1 ,"¢/Q 1l

l / »an -1.2 ,//1 p

-1.4 xn I //

-1.5-1.6

-1.7

-1.6

-1.9

-2.06.6:-10 6.6:-0 6.6:-6 6.6:-1 6.6:-6 6.6:-6 6.6:-1 66. 6.6:+6 6.6:+6 6.6:+16

Time (sec.)

Flgure 25. Tangentlal stress vs. tlme (llnelpatch Ioads): vlscoelastlc pmpollant

Numerical examples and discussions l0l

5cg 43 sQ 2ß 1

00.00 0.01 0.02 0.08 0.04 0.05 0.08

Porc• history (A): time (see.)

A 5-* 4

8•

2 22 1

Ö

0.000 0.005 0.010 0.015 0.020 0.025 0.080

Force history (B): time (sec.)A 5

E- 4**

3

Q 2Q 1

00.000 0.005 0.010 0.015 0.020 0.05 0.080

Force history (C): time (sec.)

A 5

K 4v 8Q 2Q 1 .

00.000 0.005 0.010 0.015 0.020 0.025 0.080

Force history (D): time (sec.)

A 5° 4

Ä 8Q 2£_ 1

° ,0.000 0.005 0.010 0.015 0.020 0.025 0.030

”.

Force history (E): time (sec.)

Figure 26. Force hlstorles

Numerical Example: und discussions mz

0.00000I · A

-0.00001

-0.00002

-0.00003

-0.00004

Q -0.00006 _

-0.00006 ·

4-9cl -0.00007G)

-0.00008O

-0.00009 .~

U)-•-•Q -0.00010

-0.00011

-0.00012

-0.00013“°°l‘ AAA-0.00015

0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0.050 0.055 0.060

Time (sec.)

Flgure 27. Radlal dlsplacement vs. tlme (two llne Ioads: Force (A)) .

Numerical examples and discussions l03

0.0014

0.0012

0.0010

0.0008

0.0006

0.0004

0.0002

0.0000

"T -0.0002‘5Sgt. -0.0004

‘ _

-0.0006U)Q) -0.0008isU) -0.0010

E -0.0012-.-«U -0.0014Q

Q:} -0.0016

-0.0018

-0.0020

-0.0022

-0.0024

-0.0026

-0.0028

-0.0030

-0.00320.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0.050 0.055 0.060

Time (sec.)

Flgure 28. Radlal stress vs. tlme (two llne Ioads: Force (A))

Numerical examples and discussions [O4

0.00160.00150.00140.00130.00120.00110.00100.00090.00080.00070.00060.00050.00040.0003

"T 0.0002·5; 0.0001gg, 0.0000-# -0.0001

·

cn -0.0002Q -0.0003gg •'Ü.ÖOO4Ü; -0.0005

-0.0006

QS-0.0007

un -0.0008gg -0.0009Q -0.0010E -0.0011

-0.0012-0.0013-0.0014-0.0015-0.0016-0.0017-0.0018-0.0019-0.0020-0.0021-0.0022-0.0023-0.0024-0.0025

0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0.050 0.055 0.060

T1me (sec.)

Figure 29. Tangentlal stress vs. time (two line Ioads: Force (A))

Numerical examples and discussions105

0.00001I

-0.00001

-0.00003

-0.00005

-0.00007

-0.00009

-0.00011

-0.00013

-0.00015g -0.00017 _

I -0.00019-0.00021

E -0.00023

ä -0.00025E -0.00027

ä -0.00029-0.00031

-0.00039

-0.00035

-0.00037

-0.00039-0.00041 A-0.00045 '

'0.0000.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0.050 0.055 0.060

Time (sec.)

Figure 30. Radlal dlsplacement vs. tlme (two patch Ioads: Force (A))‘

Numerical examples and discussions |()6

0.00I '

-0.01 °

rx -0.02.,.1 .03Q, .

~.«0)03

2U -0.03U) .•—•G-•-«'Ußns -0.04

-0.05

-0.08

0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0.050 0.055 0.080

Time (sec.)

Figure 31. Radial stress vs. tlme (two patch Ioads: Force (A))

Numerical examples and discussions l07

0.000-0.001-0.002 ,-0.003-0.004-0.005-0.008-0.007-0.008-0.009-0.010-0.011-0.012-0.013

A -0.014. -0.015

-0.016_

gl -0.017ar -0.018m -0.019m -0.020$5 “8·3ä$gf; -0.02:;_ -0.024Q -0.025¤0 -0.026:1 -0.0276 -0.028E- -0.029

-0.030-0.031 r-0.032-0.033-0.034-0.035

’

-0.038-0.037-0.038-0.039-0.040-0.041-0.042-0.043-0.044

0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0.050 0.055 0.080

Txme (sec.)

Flgure 32. Tangentlal stress vs. tlme (two patch Ioads: Force (A))

Numerical examples and discussionsl08

Eäoo:3:0000:8:30300

;g¢gg22;

,3

zgiäoääi

gg

"0'oooov

Ü

:3

:g'oooo6

·,;

·<>Ü333iä

q;

-oÜ00011

E

"0'oomz”0‘ooo1s

°’

"0‘ooo

4

äéäläääää

äjäsxsia

’0°ooo2o

-0.00021'0'ooozz

-0.00023-0.00024-0.00025jg‘00g26'0:ggoä;'0'oooag"0'oooso:363061”0Ü8oggä

-0.00034-0.00035"0'ooose0'ooosv

‘„3„i¤

843

41

Ü

‘o

F

00

Ü9U

or

.0

N

e

0

U

3

5

me

3

O

r

-

ÜO

ic

1

a

O

¤R

Cxa

adl

0.0

mplc

al d

15

5 an

Is

O·O

d

pl

20

d8

is

C

O

cussgo

emen

0;*25

ns

t

iulo

s

Q

30

tim

(560-0

e (|

C

js

In

·

0

‘o

BIP

40

atc

o

.

n

04

Io

5

a

o

d

.0

s

6

Z F

0

00 0])

°°

loo

0.00

-0.01

rx -0.02-.-E .U)¤«Q

m .

3IH -0.0E/J

'U .Q .

nc -0.04

-0.05-

-0.06

0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.085 0.040 0.045 0.050 0.055 0.060

Time (sec.)

Flgure 34. Radial stress vs. tlme (line/patch Ioads: Force (A))

Numerical examples and discussions H0

0.001i

-0.001

-0.003

-0.005

-0.007 _

-0.009

-0.011

-0.013

-0.015g, -0.017

gz .m -0.019 ·U}O -0.021$4ä -0.023 · _

qj -0.025

EF -0.027

g -0.029

-0.031

-0.033

-0.035

-0.037

-0.039

-0.041

-0.043

-0.045

0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0.050 0.055 0.060

Time (sec.)

Flgure 35. Tangentlal stress vs. tlme (Ilnelpatch Ioads: Force (A))

Numerical examples and discussions lll

0.00000-0.00001-0.00002-0.00008-0.00004-0.00005-0.00006-0.00007-0.00006-0.00000

Q -0.00010_

•-•

" -0.00011+9Q -0.00012

_

-0.000184; -0.00014

2 -0.00015.-4$.1 -0.00016

E -0.00017-0.00016-0.00019-0.00020-0.00021-0.00022-0.00028-0.000%-0.000%-0.000%

0.000 0.005 0.010 0.015 0.020 0.0% 0.080

Time (sec.)

Flgure 36. Radlal dlsplacement vs. tlme (two line Ioads: Force (8))

Numcrical Examplcs und discussions H2

0.04

0.03

0.02

"TE 0.01Q-«äf

IDU)

2 .Q 0.00

· U)•—•6••¤|

E -0.01Q

-0.02

-0.08

-0.040.000 0.005 0.010 0.015 0.020 0.025 0.080

Time (sec.)

Flgure 37. Radlal stress vs. tlme (two llne Ioads: Force (8))‘

Numcrical Examplcs und discussions U3

0.04

0.03

0.02

A

'G 0.01Ö1~«WW0 .g 0.ooV U)

aiE°G -0.01 l[·•

-0.02

-0.03

-°o•°‘I

0.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Flgure 38. Tangentlal stress vs. tlme (two llne Ioads: Force (8))

Numcrical Examplcs und discussions II4

0.00022

0.00020

0.00018

0.00016

0.000140.00012

0.00010

0.00008

fg 0.00006ä 0.00004 _

C'- 0.00002

Ei 0.00000 _E -0.000023 -0.000042 -0.00006

E -0.00008Q -0.00010

-0.00012

-0.00014-0.00016

-0.00018

-0.00020

-0.00022

-0.00024

-0.00026 '

0.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Flgure 39. Radlal dlsplacement vs. time (two llne Ioads: Force (C))

Numerical Examples und discussions IIS

0.05

0.04

0.03

0.02

"T-5 0.01¤— .\/

0.00

E.

m -0.01PIG!•!¤|

E -0.02N

-0.03

-0.04

-0.05

-0.000.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Flgure 40. Radlal stress vs. tlme (two Iine Ioads: Force (C))

Numcrical Examples and discussions II6

0.05

0.04

0.08

0.02

A'E 0.01 IßaQ

So.ooQDI-«

ü_ -0.01Q3¤0GQ -0.02E··•

-0.08

-0.04

-0.05

-0.080.000 0.005 0.010 0.015 0.020 0.025 0.080

Time (sec.)

Flgure 41. Tangentlal stress vs. time (two llne loads: Force (C))

Numerical Examples uid discussions ll7

0.000170.000160.000150.000140.000180.000120.000110.000100.000090.000080.00007

A 0,00006A 0110005

ä 0@0004•—• 0.00003" 0.00002E 0.00001 _0 0.00000E -0.00001QD -0.00002

. Q -0.0000:;E -0.00004m —o§oo00sE -01)0006

-0.00007-0.00008-0.00009-0.00010-0.00011-0.00012-0.00018-0.00014-0.00015-0.00016-0.00017-0.00018

0.000 0.005 0.010 0.015 0.020 0.0% 0.080

Time (sec.)

Flgurg 42. Radlal dlsplacement vs. tlme (tvvo·|Ine Ioads: Force (D))

Numcrical Exampics and discussions N8

0.05

0.04

0.03

0.026

-•-iID¤. 0.01\J .tb

2Q 0.00(IJ•—•

.9.QM

-0.02

-0.03

-0.04

-0.050.000 0.005 0.010 0.015 0.020 0.025 0.080

'I‘ime (sec.)

Flgure 43. Radlal stress vs. tlme (two llne Ioads: Force (D))

Numcrical Examples and discussions H9

0.05

0.04

0.08

0.02ex~•-IEfg 0.01U)ID

2 .Q 0.00U)

gl, .¤ -0.01GSE•¤

-0.02

-0.03

-0.04

-0.050.000 0.005 0.010 0.015 0.020 0.025 0.080

Time (sec.)

Flgure 44. Tangentlal stress vs. tlme (two Ilne Ioads: Force (0))

Numerical Example: and discussions l20

0.000200.000190.000180.000170.000160.000150.000140.000180.00012

. 018*%%0.000093.00003

Q 0:00006U 0.00005g 0.00004•-• 0.00003‘·’

0.00002E 0.00001 -

- 0.00000

ä -0.00001Q -0.00002

-0.00008Q -0.00004«-• -0.00005¤« -0.00006.2 -0.00007Q -0.00008

-0.00009-0.00010-0.00011-0.00012-0.00018-0.00014-0.00015-0.00016

- . 11*%%-0.00019-0.00020

I3:%§é0.000 0.005 0.010 0.015 0.020 0.0% 0.080

Time (sec.)

Flgure 45. Radlal dlsplacement vs. time (two line Ioads: Force (E))

Numcrical Examples und discussions l2l

0.05

0.04

0.03

0.02A-.4lbBe 0.01~.«U1In .GJ‘3 0.00U).-4.9,.¤ -0.01G

Dt'.-0.02

-0.03

-0.04

-0.050.000 0.005 0.010 0.015 0.020 0.0% 0.030

Time (sec.)

Flgure 46. Radlal stress vs. time (two Ilne Ioads: Force (E))

Numcrical Examples und discussions 122

0.05

0.04

0.08

0.02»·~-.-Z3_/

0.01

wIn '

0.00U)

6an -0.01dl

[··•-0.02

' -0.08

-0.04

-0.050.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Flgure 47. Tangentlal stress vs. time (two llne loads: Force (E))

Numericsl Examples and discussions l23

0.0000

-0.0001

-0.0002/\•¤Q .GE 00086 °° .G

ä8G -0.0004

•-I

¤«rn••¤l

Q

-0.0005

-0.0008

-0.00070.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Flgure 48. Radlal dlsplacement vs. tlme (two patch loadsz Force (B))

Numerical Examplcs and discussions l24

0.170.180.150.140.180.120.110.100.090.080.070.080.05

8*82.Q 0.02

py 0.01 ,Q 0.00‘-'

-0.012,2 ‘3·3§ V°’

-0.04 V:3 -0.05UJ -0.08 .•-n -0.07

-0.08•¤ -0.09 _q -0.10Q -0.11

-0.12-0.18 .-0.14

"

-0.15-0.18-0.17-0.18-0.19-0.20-0.21 .

-0.22-0.28-0.24-0.25-0.28

0.000 0.005 0.010 0.015 0.020 0.025 0.080

Txme (sec.)

Flgure 49. Radlal stress vs. time (two~ patch Ioads: Force (8))

Numcrical Examples and discussions 125

0.160.150.140.130.120.110.100.090.080.070.060.050.040.03

Q 0.02m 0.01

0.00-0.01 .

$ -0.02-0.04

ä -0.06, -0.06 r0:-0.07¤¤

-0.00 .Q -0.001E- -0.10 ‘

-0.11-0.12 .-0.13-0.14-0.15-0.16-0.17-0.18-0.19-0.20-0.21-0.22-0.23-0.24

0.000 0.005 0.010 0.015 0.020 0.025 0.030

T1me (sec.)

Flgure 50. Tangentlal stress vs. time (two patch loads: Force (B))

Numerical Examples and discussions IZ6

0.0004

0.0008

0.0002

0.0001

6

fa 0.0000¤•-4

W.?

4 ¤0.0001

äCD ..Q

0.0002

NE.m -0.0008•-•G

-0.0004

-0.0005

-0.0008

-_

700000.000 0.005 0.010 0.015 0.020 0.025 0-030

'1‘ime (sec.)

Flgure 51. Radlal dlsplacement vs. tlme (two patch loads: Force (C))

Numerlcal Example; and discussion;U

|Z7

0.8

0.2

0.1

0.0—

-0.1A.é _ _

-0.2~..«IDU9 -0.8

r E'

VJ -0.4•—•G••¤I

'¤ -0.6GM

-0.8

-0.7

-0.8

-0.0

-1.0

0.000 0.005 0.010 0.015 0.020 0.025 0.080

Time (sec.)

Flgure 52. Radlal stress vs. time (two patch loads: Force (C))'


0.3

0.2

0.1

0.0‘

-0.1A

I

~..¢

3Q -0.3 U;„I••-JV) -0.4Q)¤D¤ -0.5Q .r Ä-0.8

-0.7

-0.8

-0.9

-1.0

0.000 0.005 0.010 0.015 0.020 0.025 0.030

'I‘ime (sec.)

Figure 53. Tangentlal stress vs. time (two patch Ioads: Force (C))

Numcrical Examples and discussions I29

0.0004

0.0008

0.0002

E 0.0001U

.

¤•-«Q0 0.0000 ·G

äS -0.0001

E

6•-IG4ID¤•d

Q -0.0002

-0.0008

-0.0004

-0.00050.000 0.005 0.010 0.015 0.020 0.025 0.080

Time (sec.)

Flgure 54. Radlal dlsplacement vs. tlme (two patch loads: Force (D))

Numerical Example: and discussion: Z I30[

0.140.130.120.110.100.090.080.070.06

A 0.050.04 ·

g, 0.08S-!

2 SIZ?Q; .3 0.00 _V U3 -0.01

*3 -0.02$8 -0.0s ’Q -0.04

-0.05-0.06-0.07-0.08 .-0.09-0.10-0.11-0.12-0.13

-0.14

0.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Flgure 55. Radlal stress vs. time (two patch loads: Force (0))

Numcrica! Examplcs and discussions I3!

0.130.12

0.11

0.100.09

0.080.07

0.080.05

fg 0.04‘•’

0.02. 3 0.01

0 00i

t

V? -0.01ga -0.02

-0.03·‘ ‘

.2 -„..„l-0.05-0.08-0.07

-0.08-0.09

-0.10-0.11

-0.12-0.13

0.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Flgure 56. Tangentlal stress vs. time (two patch Ioads: Force (D))

Numcricsl Examples and discussions Ü:

0.0004

0.0008

0.0002

E o.ooo1OG•-xr

— -•-P 0.0000G

ä .-0.0001

6•·-•G-U3••-Q -0.0002 *

-0.0008

-0.0004

-0.00050.000 0.005 0.010 0.015 0.020 0.025 0.080

Time (sec.)

Flgure 57. Radlal dlsplacement vs. time (two— patch Ioads: Force (E))

Numerical Example; and discussion; Y l33

0.140.130.12

0.110.10

0.090.080.070.06

"? 0.060.04

"’0.03

IDcn 0.02

0 01V3 0.00•-•G -0.01

0 O2-0.03

-0.04-0.05-0.06-0.07-0.08-0.09-0.10-0.11-0.12

0.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Flgure 58. Radlal stress vs. tlme (two patch Ioads: Force (E))

Numcrical Examplcs und discussions I34

0.18

0.12

0.11

0.100.090.080.070.06

A 0.05

ä Ziä;szm 0.02 ~g 0.013 0 00. U) ·:;.:3*2;ä

-0:08E-

-0.04-0.05-0.06 '

-0.07-0.08-0.09-0.10-0.11 ~ .

-0.120.000 0.005 0.010 0.015 0.020 0.025 0.080

Time (sec.)

Figure 59. Tangentlal stress vs. time (two patch Ioads: Force (E))

Numcrical Examplcs and discussions ÜS

0.0000

-0.0001

Q -0.0002UQI-l~.«-6->

ElE -0.0005GJOGl-IG-•tn~•-l

Q ~·0.0004

-0.0005

-0.00060.000 0.005 0.010 0.015 0.020 0.025 0.080

Time (sec.)

Flgure 60. Radlal dlsplacement vs. tlme (Ilnelpatch loads: Force (8))”

Numerical Example: end discussion: I36

0.170.180.150.140.130.120.110.100.090.080.070.080.050.04

Q 0.03m 0.02¤« 0.01T; 0.00

-0.01-0.02· N -0.03

ä -0.04 ~Q -0.05

9 ‘8‘8°E -0.08 ·Q -0.09

-0.10 -]\-0.11 j-0.12-0.13 „

-0.14-0.15-0.16-0.17-0.18-0.19-0.20-0.21-0.22-0.23

0.000 0.005 0.010 0.015 0.020 0.025 0.030

Txme (sec.)

Flgure 61. Radlal stress vs. tlme (llne/patch l0ads: Force (8))

Numerical Examplcs and discussions l37

0.170.160.150.14i 0.130.120.110.100.090.080.070.060.05

*7 0.04*5; 0.02C4 0.02‘;)’

0.010.00

ä -0.01· -0.02

-0.06 „Q; -0.04 ' -

-0.05‘

gf -0.06 Q . eQ -0.07E·• -0.08 .

-0.09-0.10-0.11-0.12-0.13

·‘

-0.14-0.15-0.16 ~-0.17-0.18-0.19-0.20-0.21

i

0.000 0.005 0.010 0.015 0.020 0.0% 0.030

Time (sec.)

Flgure 62. Tangentlal stress vs. time (llnelpatch Ioads: Force (8))

Numerical Examplcs and discussions |38

0.0008‘

0.0002

0.0001

E ogooooO ;

_

tl 1h-I .-# .8 -0.0001Cl ;

E E8 -0i0002G :

pä .

¤« :ID I••-• .Q -010008

-0.0004

-0.0005

-0.00060.000 0.005 0.010 0.015 0.020 0.0% 0.080

Time (sec.)

Flgure 63. Radlal dlsplacement vs. tlme (llnelpatch loads: Force (C))

Numerical Examples und discussions [39

0.8

0.2

0.1

0.0

-0.1»·«.$ .nl -0.2

W -0.32.HW -0.4•—•6za 0 5,

Cd-0.8

-0.7

-0.8

-0.9

-1.0

0.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Flgure 64. Radlal stress vs. time (Ilnelpatch Ioads: Force (C))

Numcrical Examples und discussions l40

0.8

0.2

0.1

0.0W

-0.12

al -0.2\•/

ä -0.8 l·•->VJ -0.4¤$¤D .G -0.5GE-•

-0.8

-0.7

-0.8

-0.9

-1.0

0.000 0.005 5 0.010 0.015 0.020 0.025 0.080

Time (sec.)

Figure 65. Tangential stress vs. time (linelpatch Ioads: Force (C))I

Numcrical Examples and discussions Ni

0.0008

0.0002

0.0001AEÜ .Cl

CL8 0.0000Cl

EGJQ -0.0001

I-1¤-•cn••¤l

Q

-0.0002

-0.0008

-0.00040.000 0.005 0.010 0.015 0.020 0.025 0.080

Time (sec.)

Flgure 66. Radlal dlsplacement vs. tlme (llnelpatch Ioads: Force (D))

Numericai Exampies uid discussions 142

0.13

0.12

0.11

0.10

0.090.08

0.07

0.08

Ö 0.050.04

V 0.08g 0.02

0.01·

-•-> 0.00::3 -0.01 iG -0.02 _N -0.02. ‘

-0.04

-0.05

-0.08

-0.07

-0.08 '

-0.09

-0.10

-0.11

0.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Flgure 67. Radlal stress vs. tlme (Ilne/patch Ioads: Force (D))

Numcrical Examplcs and discussions _ N3

0.12

0.11

0.10

0.09

0.08

0.07

0.08

0.05

fg 0.040.03

g) 0.02g 0.01

· $-45*; 0.00

es -0.01ED -0.02 5 .

[2 -0.03

-0.04

-0.05

-0.08

-0.07

-0.08

-0.09

-0.10

-0.11

0.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Figure 68. Tangentlal stress vs. tlme (llnelpatch loads: Force (0))

Numcrical Examples and discussions I44

0.0003

0.0002

0.0001

A€¤

0.0000~••|\-/

QC1

E -0.0001- GJ

OGS•-•

Q -0.0002-•-«Q

-0.0003

-0.0004

-0.0005

0.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Flgure 69. Radial displacement vs. time (Iinelpatch Ioads: Force (E))

Numcrical Examples and discussions NS

0.11

0.10

0.09

0.08

0.07

0.08

0.05

0.042

-5}-· 0.03·E 0.02

Z} 0.01EU 0.00m -0.01•—(

E -0.02E -0.03E -0.04 ‘ °

-0.05 ··»

-0.08

-0.07

-0.08 ·]-0.09

-0.10

-0.11

0.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Flgure 70. Radlal stress vs. tlme (Ilnelpatch Ioads: Force (E))

Numerical Example: und discussions N6

0.11

0.10

0.09

0.08

0.07

0.08

0.05

A 0.04

0.03 .\-/ .

m 0.02mQ) 0.01

0.00

ci -0.01Cm¤ -0.02 1.2 x \-0.03 |

-0.04

-0.05 _

-0.06

-0.07

-0.08

-0.09

-0.10

0.000 0.005 0.010 0.015 0.020 0.025 0.030

Time (sec.)

Flgure 71. Tangentlal stress vs. tlme (line/patch Ioads: Force (E))

Numcrical Examplcs and discussions M7

5

6 4'S3 3u2 2c

"· 100.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.162 0.180

Force history (F): time (sec.)

5

6 4

3 3

8H 2·

"· 100.000 0.018 0.036 0.054 0.0720.090 0.108 0.126 0.144 0.162 0.180

Force history (G): time (sec.)

5

6 4'53 3gg

-no"· 1

00.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.162 0.180

Force history (H): time (sec.)

Figure 72. Force hlstorles for vlscoelastlc propellant

Numerical Example; and discussions _ 148

U

0.00000

-0.00001

-0.00002

-0.00003

{Ä„¤O -0.00004C7••-I

§-/*6*.m -0.00005

-•¤l¤ 1-0.00006]-0.00007 -·

-0.00008

-0.00009

0.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.162 0.180

Time (sec.)

Flgure 73. Radlal dlsplacement vs. tlme (two Ilne Ioads: Force (F))

Numcrical Example: and discussion: 149

-0.01 A

rvvvv-_

•

-0.02

-0.03

ax'ES -0.04gz

CD0)Q)Xn -0.05

4-*U}

•—•QEtg -0.08

QS

-0.07

-0.08

-0.09

0.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.162 0.180

Time (sec.)

Flgure 74. Radlal stress vs. time (two Ilne Ioads: Force (F))

Numcrical Examplcs und discussions l50

-0.005

-0.007

-0.009

-0.011

-0.013 _

-0.015

-0.017

-0.019

-0.021 _*7 -0.023-•¤l

gf -0.025V -0.027Ä -0.029

. -0.031«•-J

°'* -0.0:1:1gl) -0.0:161G -0.0:17

-0.039

-0.041

-0.043

-0.045

-0.047

-0.049-0.051

-0.053

-0.055

-0.057

0.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.14-4 0.162 0.180

Time (sec.)

Flgure 75. Tangentlal stress vs. time (two Ilne Ioads: Force (F))

Numerical Examples and discussions ISI

0.00002

l

0.00001

0.00000

-0.00001

-0.00002zs

|

¤ -0.00003•¤1

ga

iim -0.00004•¤¤I

Q -0.00005 ·-0.00006

-0.00007

-0.00008

-0.00009

0.000 0.018 0.038 0.054 0.072 0.090 0.108 0.128 0.144 0.182 0.180

Time (sec.)

Flgure 76. Radlal dlsplacement vs. tlme (two llne Ioads: Force (G))

Numerical Examplcs und discussions 152

0.03

0.02

0.01

0.00

}-.-2

E-/ -0.02CD03Q)S-4 -0.03+9Ui

•-I

E -0.04'UCU

-0.05

-0.06

-0.07

-0.08

-0.09

0.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.14-4 0.162 0.180

Time (sec.)

Flgure 77. Radial stress vs. time (two Ilne Ioads: Force (G))

Numerical Examples and discussions . iss

0.02

0.01

0.00

2

'5 -0.01 .¤«\-/

mcn2__, -0.02m¤$ä"Q -0.03

E-•

-0.04

-0.05

-0.080.000 0.018 0.038 0.054 0.072 0.090 0.108 0.128 0.144 0.182 0.180

Time (sec.)

Figure 78. Tangentlal stress vs. tlme (two Ilne Ioads: Force (G))

Numerical Examples und discussions [S4

0.00000

-0.00001

-0.00002

-0.00003

.¤OC1

-0.00004· •-n

¤«U)-•-«Q

-0.00005

-0.00006

-0.00007

-0.00008

0.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.182 0.180

Time (sec.)

Figure 79. Radlal dlsplacement vs. tlme (two ilne Ioads: Force (H))

Numerical Examples end discussions 155

0.004

0.003

0.002

0.001

0.000

-0.001

-0.002

-0.003

'°> -0.004EQ! ·0.005 .\-/gg) -0.006

Q; -0.007.5m -0.008

gg -0.009••-4

'O -0.010as

0:: -0.011-0.012

-0.013

-0.014

-0.015

-0.016

-0.017

-0.018

-0.019

0.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.162 0.180

Time (sec.)

Figure 80. Radlal stress vs. tlme (two line loads: Force (H))

Numerical Examples and discussions *56

0.002

0.001

0.000

-0.001

-0.002

-0.003

-0.004"?

-0.005O-4"“

-0.006mä -0.0075m -0.008

¤5 -EB 0.009Q -0.010 I ·E- I

-0.011 ·+

-0.012

-0.013 ·

-0.014

-0.015

-0.016

-0.017

0.000 0.018 0.036 0.054 0.072 0.090 0.106 0.126 0.144 0.162 0.160

.Time (sec.)

Figure 81. Tangentlal stress vs. tlme (two Ilne Ioads: Force (H))

Numerical Example: und discussions IS7

-0.00008

-0.00009

-0.00010

-0.00011

-0.00012

-0.00013

-0.00014

-0.00015

-0.00016

-0.00017

E -0.00016ä -0.00019 .:7 -0.00020

& -0.00021 ·E -0.00022

-0.00023

-0.00024

-0.00025

-0.00026

-0.00027

-0.00028

-0.00029

-0.00030

-0.00031

-0.00032

0.000 0.018 0.038 0.054 0.072 0.090 0.108 0.128 0.144 0.182 0.180

Time (sec.)

Flgurc 82. Radlal dlsplacement vs. tlme (two patch loads: Force (F))

Numcrical Examplcs and discussions _ [S8

-

*1.,.2

_

U)Q2~./U)U')CDS4

4-JU)

T6-•-«'UQ

Q: -2

-8 U

0.000 0.018 0.038 0.054 0.072 0.090 0.108 0.128 0.144 0.182 0.180

Time (sec.)

Flgure 83. Radlal stress vs. tlme (two patch Ioads: Force (F))

Numcrical Examplcs and discussions |59

"°·2 "{UÄVYVVI!1!1VVVV""*‘

-0.3‘ -

‘

-0.4

-0.5

-0.6

-0.7

xx -0.8.„-E

Q -0.9x/

VJ -1.0- m

E0 -1.1m· -1.2

GJ¤I>Q -1.3:5E" -1.4

-1.5

-1.6

-1.7

-1.8

-1.9

-2.00.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.182 0.180

Time (sec.)

Figure 84. Tangentlal stress vs. tlme (two patch Ioads: Force (F))

Numerical Example: and discussions 160

0.000030.00002

‘

0.000010.00000

-0.00001-0.00002-0.00003-0.00004-0.00005-0.00008 (

-0.0000*7-0.00008-0.00009 ·-0.00010

A -0.00011·g -0.00012¤ -0.00013

L'; -0.00014. .-2 -0.00015

gk) -0.00018-•-• -0.00017Q -0.00018

-0.00019-0.00020-0.00021-0.00022-0.00023-0.00024-0.00025-0.00028-0.00027-0.00028-0.00029-0.00030-0.00031-0.00032

0.000 0.018 0.038 0.054 0.072 0.090 0.108 0.128 0.144 0.182 0.180

Time (sec.)

Flgure 85. Radlal dlsplacement vs. time (tvvo·patch loads: Force (6))

Numcrical Exampies and discussions I6!

1

0

2 1-•-ZO'}

D-4\-/

U)

3· s-. -1-‘

0m

•-4

C5n•-4’¤ .GS

QS

-2

-3

0.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.162 0.180

Time (sec.)

Flgure 86. Radial stress vs. time (two patch Ioads: Force (G))

Numcrical Examples and discussions l62

0.2

0.1

0.0

-0.1

-0.2

-0.3

-0.4

-0.5A-0.6

V EL -0.7$ -0.82Q -0.9CD_ -1.0QJCD -1.1CIQ -1.2E-

-1.3

-1.4

-1.5

-1.8

-1.7

-1.8

-1.9

-2.00.000 0.018 0.038 0.054 0.072 0.090 0.108 0.128 0.144 0.162 0.180

Time (sec.)

Figure 87. Tangentiai stress vs. time (two patch Ioads: Force (G))

Numcrical Examplcs und discussions I63

0.00000-0.00001-0.00002-0.00003-0.00004-0.00005-0.00006-0.00007-0.00008-0.00009-0.00010

,„ -0.00011

-5 -0.00012_ :1 -0.00013

C? -0.00014E -0.00015ä -0.00018

-0.00017-0.00018-0.00019-0.00020

-0.00021-0.00022-0.00023-0.00024-0.00025-0.00026-0.00027-0.00028

0.000 0.018 0.038 0.054 0.072 0.090 0.108 0.128 0.144 0.182 0.180

Time (sec.)

Flgure 88. Radial dlsplacement vs. tlme (two patch Ioads: Force (H))

Numcrical Examplcs end discussions 164

0.080.050.040.030.020.010.00

-0.01-0.02-0.03-0.04-0.05-0.08-0.07

Q -0.08In -0.09

C-4 -0.10”

-0.11-0.12

Q} -0.133 -0.14m -0.15__‘

-0.18‘

-.1-0.19

Q: -0.20-0.21-0.221-0.23-0.24-0.25-0.28-0.27-0.28-0.29-0.30-0.31-0.32-0.33-0.34

0.000 0.018 0.038 0.054 0.072 0.090 0.108 0.128 0.144 0.182 0.180

T1me (sec.)

Figure 89. Radlal stress vs. time (two· patch Ioads: Force (H))


0.030.020.010.00

-0.01-0.02-0.03-0.04-0.05-0.08

A -0.07-0.08

¤_‘ -0.09‘·’-0.10

g -0.11Q3

-1 1: KIS, -0.14

¤> -tw 0.15c: -0.18 ·QE -0.17

-0.18-0.19-0.20-0.21-0.22-0.23-0.24-0.25-0.28-0.27-0.28

0.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.182 0.180

Time (sec.)

Flgure 90. Tangentlal stress vs. time (two patch Ioads: Force (H))

Numcrical Examplcs und discussions [66

-0.00008

-0.00009

-0.00010

-0.00011

-0.00012

-0.00013

-0.00014

-0.00015

-0.00016

-0.00017-0.00018

·‘f-0.00019

-0.00020Q -0.00021

-0.00022

-0.00023

-0.00024

-0.00025

-0.00026

-0.00027

-0.00028

-0.00029

-0.00030

0.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.162 0.180

Time (sec.)

Figure 91. Radlal dlsplacement vs. time (linelpatch loads: Force (F))

Numcrical Examples and discussions f 167

-0.2

-0.3-0.4

-0.5-0.8

-0.7

-0.8-0.9

-1.0-1.1

/'\-•-Z -1.2 _{ -1.3“"

-1.4IDm -1.5CD$4 -1.8•->m -1.7T; -1.8••¤l

'¤ -1.9‘

G5 -2.0-2.1-2.2

-2.3-2.4

-2.5-2.8-2.7

-2.8-2.9

-3.0

0.000 0.018 0.038 0.054 0.072 0.090 0.108 0.128 0.144 0.182 0.180

Time (sec.)

Flgure 92. Radlal stress vs. time (line/patch Ioads: Force (F))

Numerical Exsmplcs und discussions 168

-0*2-0.3

‘

-0.4

-0.5

-0.8

-0.7

A -0.8..-5{ -0.99

g -1.01 EU -1.1U1

Q3 -1.2¤0gi -1.3CUE" -1.4

-1.5

-1.8

-1.7

-1.8

-1.9

-2.00.000 0.018 0.038 0.054 0.072 0.090 0.108 0.128 0.144 0.182 0.180

Time (sec.)

Flgure 93. Tangentlal stress vs. tlme (llne/patch Ioads: Force (F))

Numcrical Example: und discussions 169

0.000030.000020.000010.00000 _

-0.00001u

-0.00002-0.00003-0.00004-0.00005 ·-0.00006-0.00007-0.00008-0.00009 '

,.„ -0.00010.¤ -0.00011Q -0.00012

Q; -0.00013..: -0.00014

Q -0.00015 J-0.00016 _-0.00017 j-0.00018-0.00019 J

-0.00020 J

-0.00021-0.00022-0.00023-0.00024-0.00025-0.00026-0.00027-0.00028-0.00029-0.00030

0.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.162 0.180

Time (sec.)

Figure 94. Radial displacement vs. time (llnelpatch Ioads: Force (G))


1

0

rs-.-ZU)

Cu

U1CDQ)La -1+9U1

•-<N-•-4'UNM

-2

-3

0.000 0.018 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.162 0.180

Time (sec.)

/ _„~ 7

Flgure 95. Radlal stress vs. time (llnelpatch Ioads: Force (G))

Numcrical Example: and discussions 17I

0.2

0.1

0.0

-0.1

-0.2

-0.3‘

-0.4

A -0.5

E-0.8

„, -0.7

3 -0.8‘

-0.9 -_ -1.0

-1.1 äL5 “1„2AI-1.3

-1.4

-1.5

-1.8

-1.7

-1.8

-1.9

-2.0 I0.000 0.018 0.036 0.054 0.072 0.090 0.108 0.128 0.144 0.162 0.180

Time (sec.)

Flgure 96. Tangenttal stress vs. time (Ilne/patch Ioads: Force (G))

Numerical Examplcs and discussians . 172

0.00000-0.00001

-0.00002-0.00003

-0.00004-0.00005

-0.00008

-0.00007

-0.00008

-0.00009-0.00010

E -0.00011

ä -0.00012-0.00013

.51* -0.000141

.52 -0.00015Q -0.00018

-0.00017

-0.00018

-0.00019

-0.00020

-0.00021-0.00022

-0.00023

-0.00024

-0.00025

-0.00028

-0.00027

0.000 0.018 0.038 0.054 0.072 0.090 0.108 0.126 0.144 0.182 0.180

Time (sec.)

Figure 97. Radlal dlsplacement vs. tlme (llnelpatch loads: Force (H))

Numcrical Examples and discussions l73

0.050.040.03 .0.020.010.00

-0.01-0.02-0.03-0.04-0.05-0.06-0.07-0.08

Q -0.09m -0.10

CL. -0.11T; -0.12

-0.13Z} -0.14S-• -0.15

*5*, -0.18,_, -0.17

E -0.18' -0.19-0.2031Z -0.21

·-0.22 :1-0.23-0.241-0.25-0.28-0.27-0.28-0.29-0.30-0.31-0.32 _-0.33-0.34-0.35

0.000 0.018 0.038 0.054 0.072 0.090 0.108 0.128 0.144 0.182 0.180

T1me (sec.)

Figure 98. Radial stress vs. time (llnelpatch Ioads: Force (H))

Numcrical Examplcs and discussions U4

0.03

0.010.00

-0.01-0.02-0.03-0.04-0.05-0.06

A -0.07.,.1 -0.08

E -0.09‘-“

-0.10

Q -0.11Q -0.12ja -0.13

, -0.14

QB -0.15Cl -0.16[2 -0.17

-0.18-0.19-0.20-0.21-0.22-0.23-0.24-0.25-0.26-0.27-0.26

0.000 0.018 0.036 0.054 0.072 0.090 .0.108 0.126 0.144 0.162 0.180

Time (sec.)

Flgure 99. Tangential stress vs. tlme (llnelpatch Ioads: Force (H))

Numcrical Examplcs and discussions ns

Appendlx A.{X} = {Um- Um- Um- Erzm ö6n· Um} (7-*)Kf = ‘[~j%¥ dA = K§" (1-2) '

K,}" = fr-/j% dA = )<,§" (1-3)

Kf = j·~,~jdA (1-4)

;<,§° = I- é ~,~jdA = Kj?5 = r<§6 (1-5)

K§" = ~,~jdA (1 -6)

Appendix A. *76

35 _ _g_ _ 36 _ 56Kü —E NiNjdA — Kü — Kü (1.7)

The other coefficients of [K] are zero.

Appcndix A. _ I77

Appendlx B.

Forn=O

V ör ör ör V ör V(21)

N; ^’;1-2v ÜN; Ö";

+(‘””)7"7+T' öl 62 WA

K;2_ E ÜN; WAÜ_

(1+v)(1—2v)V ör öz V öz 2 öz ör (2.2)

=K§‘

z2_ E _ ÜN; ö"Ü 1-2;; ÜN; ÜN; dA 23Ku ' (1+v)(1—2v) fw V) öz öz + 2 ör ör } (°)

Appcndix B. H8

For n=1, 2

11 _ E ÜN; ÜN! ÜN; N! ÜN! N;KU ‘

(1+1)(1-21) fw"! öf ÜF +" ÜF7“+”

61 '?'

N. N- _ ÜN. ÖN-2v 1 1 (2.5)V V 2 Öz Öz

_ N. N-

12_ E ÜN; N! N; N!K'! _

(1 +v)(1 —2v) ör

V+·————2n’.(l,-ö,_)}dA

21

ÜN. ÜN. N. ÖN- _ ÜN. ÖN-E 1 /+1 2v 1 /}dA( +v)(1—2v) Ör Öz Ö; 2 Öz Ör (2.7)=K$‘

E N. N- _2 ÜN. N. ÖN- N-

(28)

- 61v- öN·'

Appcndix B. [79

N. öN· _ 5N. N-“?°=vm%m){””+·ä··l%~6ä+}d^(2.9)

32

K;s= E ÜN!+ 1-2v ÜN! ÜN!

U (1 + v)(1 -2v) öz öz 2 ör ör (210)z 1-2v N· N!

I+f7 T*+—F·}dA

mf = mf = mf = jp~,~/6A (2.11)

Appcndix B. ISOI

References

1. Vlasov, N.Z., General Theory of Shells and lts Applications in Engineering,NASA TT-99, 1964

2. Kraus, H., The Elastic Shells, John Wiley, New York, 1967

3. Cederbaun, G. and Heller, R.A., “Dynamic Deformation of Orthotropic Cyl-inders", J. of Pressure Vessel Tech. Vol. 111, 97-101, 1989

4. Gatewood, B.E., "Note on the Thermal Stresses in a Long Circular Cylindersof (m+ 1) Concentric Material", Q. Appl. Math., Vol. 53, 84-86 1948

5. Singh, M.P., Heller, R.A., and Thangjitham, S., “ThermaI Stresses in Con-centric Cylinders due to Asymmetric and Time Dependent Temperatures"J. of Thermal Stresses, Vol. 7,183-195, 1984

6. Thangjitham, S., Heller, R.A., and Singh, M.P., “Frequency Response forThermal Stresses in Multilayered Cylindrical Structures", J. of ThermalStresses, Vol. 9, 133-150,1986

7. Strub, R.A., "Distribution of Mechanical and Thermal Stresses in MultilayerCylinder", Trans. Journal ASME, Vol. 75, 73-82, 1953

8. Heller, R.A., Thangjitham S. and Lin, Y.T., “Stress and Displacements inLine·Loaded Solid Propellant Rocket Motors", Contract No.DAAH—01-86-C-D185, U.S.Army Missile Command, Huntsville, AL., Oct. 1987

9. Flugge, W., Viscoelasticity, 2nd Edition, Springer-Verlag, New York, 1975

10. Christensen, R. M., Theory of Viscoelasticity, 2nd Edition, Academic Press,New York, 1982

11. Singh, M.P. and Heller, R.A., "Random Probability Techniques for RocketMotor Service Life Predictions", Technical Report RK-CR-82-7, U.S.ArmyMissile Command, Huntsville, AL., Sept. 1981

12. Taylor, S. et al., "Thermomechanical Analysis of Viscoelastic Solids", Uni-versity of California, Berkeley, June, 1968

”References l8l

13. Henriksen, M., “Nonlinear Viscoelastic Stress Analysis - A Finite ElementApproach", lnt. J. Compu. & Struc. Vol. 18, 133-139, 1984

14. Jones, l.W. and Pierre, L.E.,“A

Linear Thermoelastic Material Model forSolid Rocket Motor Structural Analysis", Compu. & Struc. Vol. 21, 235-243,1985

15. Yang, T-W et al., “Endochronic Viscoelastic Creep and Analysis of 2-DStructure", Compu. & Struc. Vol. 25, 425-429, 1987

16. Roy, S. and Reddy, J.N.,“A

Finite Element Analysis of Adhesively BondedComposite Joints including Geometric Nonlinearity, GeometricViscoelasticity, Moisture Diffusion, and Delay FaiIure", VPI-E-82-28, VirginiaPolytechnic Institute and State University, Blacksburg, Dec. 1987

17. Roy, S. and Reddy, J.N. "Finite Element Methods of Viscoelasticity and Dif-fusion in Adhesively Bonded Joints", Int. J. of Num. Meth. in Eng. Vol. 26,2531-2546, 1988

18. Roy, S. and Reddy, J.N., "Nonlinear Viscoelastic Finite Element Analysis ofAdhesive Joints", Tire Science and Technology, Vol. 16, No. 3, 146-170, 1988

19. Reddy, J.M. and Roy, S. "Nonlinear° Analysis of Adhesively BondedJoints", Int. J. of Non-Linear Mechanics, Vol.23, No.2, 97-112, 1988

20. Zak, A. R., “Structural Analysis of Realistic Solid Propellant Materials", J.

of Spacecraft, Vol. 5, 270-275, 1968

21. Chung, T.J. and Eidson, R.L. "Dynamic Analysis of ViscoelastoplasticAnisotropic Shells" Compu. & Struc. Vol. 3, 483-496, 1973

22. Wilson, D.W. and Vinson, J.R. "Viscoelastic Analysis of Lamlnated Plate

BuckIing'°, AIAA J. Vol. 22, No. 2, 982-988, 1984

23. Bukowski, R. and Wojewodzki, W. "Dynamic Buckling of ViscoplasticSpherical ShelI" Int. J. Solids & Struc. Vol. 20, No. 8, 761-776, 1984

24. Florence, A.L. and Abrahamson, G.R."Critical Velocity for Coppapse ofViscoplastic Cylindrical Shells without BuckIing", J. of Applied Mechanics,89-94, March, 1977

25. Wilson, E.L., "Structural Analysis of Axisymmetric Solids", AIAA J., Vol. 3,

No. 12, 2269-2274. 1965

26. Crose, J.G., “Stress Analysis of Axisymmetric Solids with AsymmetricPropertles", AIAA J., Vol. 10, No. 7, 866-871,1972

27. Cook, R.D., Concepts and Applications of Finite Element Methods, JohnWiley, New York,1981

References . 182

28. Yang, T.Y., Finite Element Structural Analysis, Prentice-Hall, New Jersey,1986

29. Zienkiewicz O.C., The Finite Element Method , 3rd Edition, McGraw-Hill,London, 1977

30. Timoshenko, S.P., Theory of Elasticity, 3rd Edition, McGraw-Hill, NewYork,1970

31. Atluri, S.N. et al., (eds.) Hybrid and Mixed Finite Element Methods, JohnWiley, New York,1982

32. Reddy, J.N. and Oden, J.T., “Mixed Finite Element Approximations of Linear

Boundary-Value ProbIems", Q. Appl. Math. Vol. 33, 255-280, 1975

33. Oden, J.T. and Reddy J.N., “On Mixed Finite Element Approximations",SIAM J. Num. Ana. Vol. 13, 393-404, 1976

34. Washizu, K., Variational Methods in Elasticity and Plasticity, 2nd Edition,

Pergamon Press Inc., New York, 1975

35. Reddy, J.N., Energy and Variational Methods in Applied Mechanics, JohnWiley, New York,1984

36. Heyliger, P.R., A Mixed Computational Algorithm based on UpdatedLagrangian Description for Plane Elastic Contact Problems , Ph.D Thesis,Virginia Polytechnic Institute and State University, Blacksburg, VA, 1986

37. Heyliger, P.R. and Reddy, J.N. "A Mixed Computational Algorithm for PlaneElastic Contact Problems - I. Formulation", Compu. & Struc. Vol. 26, No. 4,

621-634, 1987

38. Heyliger, P.R. and Reddy, J.N.“A

Mixed Computational Algorithm for PlaneElastic Contact Problems - I. Numerical Examples", Compu. & Struc. Vol.26, No. 4, 635-653, 1987

39. Reddy, J.N., An Introduction to the Finite Element Method, McGraw-Hill,

New York, 1984

40. Akay, H.U., “Dynamic Large Deflection Analysis of Plates using Mixed FiniteElements", Compu. & Struc. Vol.11, 1-11, 1980

41. Reddy, J.N., “CIass Notes on Finite Element Methods II", Virginia

Polytechnic Institute and State University, Blacksburg, VA, 1988

References l83

r. h. plaut thangjuly, s.€¦ · 1t.p~rt.6=zr. h. plaut thangjuly,a s. ia · 1989 blacksburg,...

Documents