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Implementation of second-order finite elementsin the GIFTS structural analysis program
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Authors Hunten, Keith Atherton
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Link to Item http://hdl.handle.net/10150/557651
IMPLEMENTATION OF SECOND-ORDER PLANE FINITE ELEMENTS IN THE GIFTS STRUCTURAL ANALYSIS PROGRAM
LyKeith Atherton Hunten
A Thesis Submitted to the Faculty of theDEPARTMENT OF AEROSPACE AND MECHANICAL ENGINEERING
In Partial Fulfillment of the Reguirements / For the Degree of
MASTER OF SCIENCE WITH A MAJOR IN MECHANICAL ENGINEERING
In the Graduate CollegeTHE UNIVERSITY OF ARIZONA
1 9 7 9
STATEMENT BY AUTHOR
This thesis has been submitted in partial fulfillment of requirements for an advanced degree at The University of Arizona and is deposited in the University Library to be made available to borrowers under rules of the Library.
Brief quotations from this thesis are allowable without special permission, provided that accurate acknowledgment of source is made. Requests for permission for extended quotation from or reproduction of this manuscript in whole or in part may be granted by the head of the major department or the Dean of the Graduate College when in his judgment the proposed use of the material is in the interests of scholarship. In all other instances, however, permission must be obtained from the author.
SIGNED:
APPROVAL BY THESIS DIRECTOR This thesis has been approved on the date shown below:
H.A. KAMEL ' DateProfessor of Aerospace and Mechanical Engineering
ACKNOWLEDGMENTS
The opportunity afforded the author to pursue his master's studies is due primarily to Professor Hussein A, Kamel, director of the Interactive Graphics Engineering Laboratory. He has been the source of much inspiration during the last two years while the author was employed as his research assistant, Fellow employees Michael W „McCabe and Patrick G. DeShazo are owed special thanks for their invaluable help with GIFTS and computer programming.
The author gratefully acknowledges the funding support of the United States Coast Guard, without which this study could never have been completed.
Finally, much of the credit for this effort belongs to the r~> ; ■ ■
author's wife, Carol A. Hunten, for her priceless support and understanding during the last three years. Carol deserves special recognition for the excellent typing of this work.
TABLE OF CONTENTS
PageLIST OF ILLUSTRATIONS . . vLIST OF TABLES . . . . . . . . . . . . . . . . . . . . . . . viiABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . viii
1. INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . 12. ELEMENT MATHEMATICAL FORMULATION AND FORTRAN CODING . . . . . 6
. Formulation of Finite Elements Based Upon AssumedDisplacement Functions . . . . . . . . . . . . . . . 6
Element Derivations and FORTRAN Coding . . . . . . . . . 11The Isoparametric Bi-Quadratic Membrane Element QM9 • 12The Linear Strain Triangle Membrane Element TM6 . . . 23The Axisymmetric Isoparametric Bi-Quadratic
Element QA9 . ... . . . . . . . @ . ... . . . . . 33The Axisymmetric Linear Strain Triangle Element TA6 . 43The Linear Strain Rod Element ROD3 . . . . . . . . . 53Kinematically Consistent Force Distribution
on a Second-Order Element's Edge . . . . . . . . 583„ SUBROUTINE DEVELOPMENT, INTEGRATION, AND TESTING . . . . . . 60
Initial Testing o o . o . o o . o . o . o a o o o . o . o 6 0Integration into GIFTS 61Testing the Integrated Subroutines . . . . . . . . . . . 63
4. FINAL TEST PROBLEEB . . . . . . . . . . . . . . . . . . . . . 66Cartesian Problems . . . . . . . . . . .... . . . . . . 66Axisymmetric Problems . . . . . . . . . . . . . . . . . . 73
5 o CONCLUSIONS e o e o o o e o o o o e e o o . o o e e e o o e e 7 8APPENDIX As NOMENCLATURE ............... 78APPENDIX Bs ELEMENT SUBROUTINES . . . . . . . . . . . . . . 79
APPENDIX C: DRIVER PROGRAMS AND OUTPUT . . . . . . . . . . . 145
REFERENCES . . . . . . . . . . . e . . . . . . . . . . . . . l6liv
LIST OF ILLUSTRATIONS
o o e o o o o o o o
1. QM9 Element in Global Space2. QM9 Element in Local Space3« QM9 Parent Square Element4. QH9 Element Stress Evaluation Points 5« TM6 Element in Global Space . . . «6, TM6 Element in Local Space7» TM6 Element Stress Evaluation Points8„ QA9 Element in Local Space . = , .9. QA9 Element in Unit Space .5. « - «10 o The TA6 Element11. The TA6 Integration Points12. TA6 Thermal Force Integration Points 13o R0D3 Element in Global Space . . . . .14. ROD3 Element in Local Space15. ROD3 Element Stress Evaluation Points16. Cartesian Extension Tests17. Axisymmetric Extension Tests18. Cantilever Beam Model19. Built In Beam20. Inertial Loads21. Deflections and Stresses
0 0 0 0 0 8 0 0 0 0 0 0 0 0 0
o o o o o o
0 0 0 6 0 0 0 0 0 0
0 8 0 0 0 0 0 0 0 6 0
0 0 0 0 0 0 6 0 0
0 0 6 0 0 0 8 0 6 0 0 0 0
0 0 0 0 0
6 0 0 0 0 0 0 0 0 8 0 0 0 0 0
0 0 0 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 6 0 0 0 0 0 0 0 0 0 0 6 0
Page
131314 20 24 24 30
3536 44 4752
5354
5764
656566 68 69
v
LIST OF ILLUSTRATIONS — Continued
Figure Page22. Plate Model . . . . . . . . . . . . . . . . . . . . . . . . 7023• Deflected Plate with Stress Contours . . . . . . . . . . . 7124. Deflected Plate with Principal Stress Vectors . . . . . . . 7225- Stresses Around the Hole . . . . . . . . . . . . . . . . . 7326. Pxpe Model . . . . . . . . @ . . . . . . . . . . . . . . . 7 327 - Spherical Pressure Vessel Model . . . . . . . . . . . . . . 75
LIST OF TABLES
Table Page1. The QM9 Interpolation Functions and Their Derivatives . . . 152o Integration Constants o o e o o o o e o e e e o o o o o o o 193c The QA9 Interpolation Functions and Their Derivatives . . . 364. QA9 Integration Constants . . . . . . . . . . . . . . . . . 415. TA6 Integration Constants . . . . . . . . . . . . . . . . . 486. TA6 Thermal Force Integration Constants . . . . . . . . . . 52
vii
ABSTRACT
The present work describes the development and FORTRAN coding of five finite elements Based upon quadratic displacement functions and their integration into the GIFTS structural analysis system. Two of the elements, the cartesian and axisymmetric quadrilaterals, may have curved sides. The cartesian and axisymmetric triangular elements, as well as the rod element, require straight sides with centered midside nodes. All of the elements are intended primarily for two-dimensional analysis and include no "bending effects. None of the elements are original, their authors and histories are. presented.
A detailed mathematical formulation is developed for each of the elements, Each of the elements' subroutines are documented in Appendix B ,
The development of a series of small, simple computer programs for initially testing the subroutines is covered, along with the integration of the subroutines into GIFTS, A series of single element tests of the integrated subroutines are presented. Several problems of 50 to 100 degrees of freedom are compared to their classical solutions as final tests.
\
viii
CHAPTER 1
INTRODUCTION
In solving structural problems using the finite element method $ complex models are generated which need large amounts of data to describe. The process of generating these models is time consuming.The effort is error prone and needs extensive checking.
The term "model generation” is used to denote the total process of generating nodal coordinates and freedoms, element connectivities and properties, as well as loads and boundary conditions. Many attempts have been made to automate this, but they have been invariably fragmented, addressing themselves to only one aspect of the problem and ignoring the interrelationship between the particular effort and other parts of the analysis. GIFTS (Graphics-Oriented Interactive Finite Element Timesharing System) represents an attempt to tie all the loose ends together and provide the user with a unified, coherent approach to model generation, model display, analysis, and result display.
The GIFTS system is a collection of program modules operating on a Unified Data Base (UDB) designed to facilitate the process of finite element analysis, using modern computer systems.
Apart from the lack of a unified approach, finite element program users have been exposed to other inconveniences. Accessability of computers and programs has been inadequate. The turnaround time is
often slow. Analysis programs have different input schemes. If a user decides to use a different package, a considerable amount of work is required to reformat his data. Most available codes can only operate in a large core area, leading to delays in execution of a submitted job. Many small engineering outfits may not have access to a large computer but are in possession of an in-house minicomputer, which may be utilized to generate and display models and solve them.In larger companies, a specialized group may opt for installing a minicomputer linked via a high-speed transmission line to the outfit’s large batch facility. Thus, the mini is utilized for pre- and post-processing of models, and the large computer is used for the solu tion process in a batch mode. Files are exchanged back and forth between the two systems as needed., GIFTS has been designed with the above points in mind. The
following will demonstrate how GIFTS can be used to provide a more user-oriented environment in finite element structural analysis,
' One of the primary design goals for the GIFTS package has been low core requirements. Therefore, it is operational on a number of large timesharing systems based on a multitude of main-frames, as well as on a number of minicomputers. The amount of core required depends on the program module being executed, as well as the particular computer being utilized. To give examples, 2^K to 32K are required on a minicomputer with overlayed program modules. On a timesharing system, 16k to 28k are usually sufficient. The low core requirement does not restrict program capacity.
GIFTS may "be used as a pre- and post-processor for other packageso All data generated by the system, is stored in a number of files using the problem name as a part of the identifier. The format of this Unified Data Base (UDB) is fixed for a particular GIFTS version and is well documented, In order to interface the system to a particular analysis program, it is only necessary to write two simple linking programs,'one to extract the relevant data from the UDB and produce a standard input tape for the chosen analysis package,. and another to insert the results of the computation by the program back into the UDB for future post-processing via GIFTS.
GIFTS is designed to retrieve, for the user, any data file or piece of data contained in a data file and present it in numerical form or, whenever possible, in graphical form.
GIFTS is not a single program but a collection of modules in a program library. Individual modules run independently of each other and communicate only via the UDB, To perform a complete analysis using GIFTS, the user executes a GIFTS procedure using a number of modules in a specified order. A typical static analysis would use modules BULKM and EDITM for model generation. Then the modules BUIKF, BULKB, and EDITLB would be used for applying loads and boundary conditions. To perform the static solution, the modules STIFF, DEGOM,DEFL, and STRESS. would be used. Final post-processing would be performed with the module RESULT.
The GIFTS analysis system supports an element library which, prior to this time, fully supported only first-order elements. These included three- and four-noded membrane elements, three- and
four-noded flat bending elements, three- and four-noded axisymmetric elements, and two-noded rod and beam elements. The pre- and post-processing capabilities in GIFTS have always supported second- and third-order elements.
The second-order elements chosen here to augument the GIFTS element library include six- and nine-noded membrane elements., six- and nine-noded axisymmetric solid elements, and a three-noded rod element, All of these second-order elements are not original. The nine-noded quadrilateral membrane element was authored by Ergatoudis, Irons, and Zienkiewicz (1968). The axisymmetric solid quadrilateral was based on the theory suggested by Wilson (1965) and Doherty, Taylor, and Wilson (1969)0 The six-noded triangles and the three-noded bar were first presented mathematically by DeVeubeke in the collection of papers by Hollister and Zienkiewicz (1965, PP- 145-19?)0 They were later implemented by Argyris (1965, 1968). Because of this, the mathematical formulation required is mostly concerned with manipulation for efficient computation on the digital computer. Thus, the core of the project lies in efficient FORTRAN coding of the mathematics and in large computer program management.
The primary purpose of this study is to provide several more accurate and efficient finite elements for the GIFTS analysis system. It is also intended to provide documentation of the element formulation and coding and of the integration of the subroutines into a large analysis system, A thorough mathematical development of the elements is presented for reference.
Not all of the aspects of the elements presented were coded, The axisymmetric elements' mass calculating subroutines were not written. Also, no facility for generating second-order kinematically consistent loads was added to GIFTS, although this is covered mathematically in the section on element formulation.
The project was broken into three distinct stages. The first stage involved several considerations in initial element formulation and coding. These included compatability with GIFTS (once written, the subroutine should combine with GIFTS with a minimum of change), simplicity and straight-forwardness, and minimum core requirements. The second stage involved initial testing of the subroutines in several small programs, avoiding the extra complications involved with GIFTS, which is comprised of large, complex programs. The initial tests performed in these small programs were carefully selected for simplicity. The final stage involved actually integrating the subroutines into GIFTS. Also, the supporting pre- and post-processing capabilities for second-order elements had not been thoroughly tested so they required additional debugging. Once the subroutines were integrated and the result displays debugged, a series of tests of increasing complexity were run. These ranged from single element tests to problems of 50 to 100 degrees of freedom.
CHAPTER 2
ELEMENT MATHEMATICAL FORMULATION AND FORTRAN CODING
In finite element structural analysis, forces and "boundary conditions are imposed upon the structural model at hand. The structure typically is a collection of finite elements joined together at their nodes, approximating a structure. The solution to the problem involves the computation of the deflections of all the nodes, from which the stresses within each element are computed.
Formulation of Finite Elements Based Upon Assumed Displacement Functions
Since displacements are the primary unknowns, the appropriate functional is that of Total Potential Energy fj (found in most texts on finite element analysis, for example Zienkiewicz 1971)• Initially,
Pip is expressed in terms of an element*s nodal degrees of freedom in displacement. Then, as required by the Principle of Stationary Potential Energy, the nodal degrees of freedom must assume values such that the values of fl are stationary (at a minimum). Applying this will yield a set of simultaneous equations, whose unknowns are the nodal degrees of freedom.
Let the displacements of an arbitrary point within an element be given by
= 1 2 ,(1)
6
where jd is the array of nodal generalized displacements of the element. The matrix f defines the nature of the assumed displacement field. The elements of f are called "displacement interpolation functions" or, more simply, "shape functions". In each element formulation, a set of shape functions will he assumed and the element "built around them.
Element strains, _e, are expressed in terms of displacements, j), "by substituting u in the strain-displacement relations. Two variations of these relationships will "be used, one idealization for cartesian two-dimensional plane stress assumptions, the other for axisymmetric solid assumptions.
The strain-displacement relationships; are:
e a *X
eydx , 0
0B_Sy
Yxy d_ d__dy dx
The axisymmetric strain-displacement relationships are:
8From these relationships, we can derive a strain-displacement
matrix that will allow us to evaluate the strain at any point within the element.
For cartesian elements s
e — ■ df nX ^ 0ey o MByxy 'Bf -MBy Bx _ (n )
For axisymmetric elements:
er = dfdre
z 0e0 f^rz r
BfBz
Mdy
Mdr
u
(5)
More concisely:
e = T p *“■ ~ e s P
(6)The elastic matrix terms are different for each of the above
assumptions. For the cartesian elements
E = E 1 V 0(1-v2) V 1 0
0 0 1-V2 " (7)
9For the axisymmetric elements
E E(l+v)(l-2v)
1-v V V 0
V 1-v V 0
V V 1-v 0
0 0 0 1-v2 (8)
The general formulation for the Potential Energy for one element is developed as follows;
fip = | i t J,y ^ , p 5 S e , p a T i
+ ^ J'r < P (4) 4VS' SY f P dv - £ Ja f H dA (9)
To solve for the nodal degrees of freedom, we set dfl /dp = 0 to obtain the minimum potential energy. The resulting relationship can he expressed by
U v < p 1 Ie,p < ] 5 = -JV4 p (5o> dv
+ J'v f (£) »
+ /. f U dA + P (10)
The above equation has been broken into several separate terms, each having special significance.
10The term
JVTe.pSIe.pdV = k'is commonly called the element stiffness matrix, and defines the physical properties of the element.
The term
s v < P (=y av ■
represents nodal forces that result from a state of initial stress in the element, hut may he easily modified to represent the forces due to thermal loading. To do this, we must first find the amount of thermal stress in the element. This can he represented as
^ 6 = i e 6 = E f a 6
where a is the thermal expansion coefficient and _0 is the matrix of nodal temperatures. Note that this would allow the modeling of initial stress hy applying a temperature field to the element.
Body (inertial) forces are represented hy the term
sv f* (I) av = EF .
To augument this term, we can note that we can derive a mass representation term so we could generate the hody force matrix from BF =M A where M would he the mass matrix and A the acceleration. This mass matrix may he represented in a diagonal form, which does not truly satisfy the kinematic virtual work equations. Though not
11kinematically consistent, this mass representation gives excellent results (Zienkiewicz 1971).
rn = $ £ y ffj dY„
The next term
describes the surface forces applied to the element.The last term, P, is the nodal applied load vector, to which
all of the above force terms will be added before solution for the unknown nodal degrees of freedom (displacements),
For a structure modeled by many elements, the stiffnesses of the elements are summed at common nodes, as are the forces. Boundary conditions are then applied to the equations. The unknown deflections are given by the inverse of this stiffness matrix multiplied by the sum of the forces applied to the structural model.
Element Derivations and FORTRAN Coding The following sections discuss in detail the mathematical
formulation for the five elements under consideration. Each section in turn discusses element geometry, displacement interpolation function assumptions, stiffness matrix formulation, stress computation, and kinematically consistent thermal forces. A diagonalized kinematically consistent mass matrix is presented for the cartesian elements only. The FORTRAN subroutines for the elements are in Appendix B.The code is commented and hopefully serves as its own documentation.
12Following the detailed mathematical presentation is a hrief
discussion of kinematically consistent forces on a second-order edge. The edge is assumed to have a centered midside node as three of the elements discussed, TM6, TA6, and ROD3, assume centered nodes. The other two elements, QM9 and QA9» as is mentioned "below, should have midside nodes as centered as possible.
The Isoparametric Bi-Quadratic Membrane Element QM9This element is based upon quadratic displacement interpola
tion functions, which allow the element to have curved as well as straight sides. The element is sensitive to the location of the midside and center nodes and to excessive boundary curvature. The nodes should be as close to centered as possible to avoid numerical inaccuracies (Fwa, Jofriet, and LeLievre 1978). Checks for element flatness are also made, which will terminate assembly if failed.
Element Geometry. The element is defined as a plane surface embedded in a general three-dimensional space, Fig, 1, For mapping to a two-dimensional space, Fig, 2, the element's local x' axis is defined to pass through nodes 4 and 6. The direction of the local z' axis is defined by the cross-product between the local x" axis and the vector through nodes 2 and 8. The direction of the local y" axis is defined by the cross-product between the local z” and x' axes. The element stiffness is formulated in a unit space, Fig. 3» The transformation between unit space and local two-dimensional space is accomplished through the Jacobian matrix developed below.
13
I
Fig. 1. QM9 Element in Global Space
4 <h
Fig. 2. QM9 Element in Local Space
14
7 8 9
4 o <> 6
1
Fig. 3» QM9 Parent Square Element
The direction cosines of the local axes are calculated from the midside nodes s
-x' = -46 (11)
-z' = -46 X -28 (12)
V = -x' X -z' (13)
- = C-x" -y' -z'] 1 (14)
Element Stiffness. For an arbitrary point within the element in s,t space, assume that its displacement may he interpolated from the nodal displacements by the Lagrangian bi-quadratic interpolants
15listed below in Table 1. Their derivatives are also listed for futurereference.
Table 1. The QM9 Interpolation Functions and Their Derivatives
pt. ! f.f 1 dfl/ds = fsi ! dfl/dt = fti1 j s(l-s)t(l-t)/4" (l-2s)t(l-t)/4 s(l-s)(l-2t)/42 ! -(l-s2)t(l-t)/2 st(l-t) -(1-s2)(l-2t)/23 -s(l+s)t(l-t)/4 ~(l+2s)t(1—t)/4 -s(1+s)(1—2t)/44 -s(l-s)(l-t2)/2 -(l-2s)(l-t2)/2 s(l-s)t5 (l-s2)(l_t2) -2s(l-t2) | -2t(l-s2)6 s(l+s)(l-t2)/2 (l+2s)(l-t2)/2 -s(l+s)t7 -s(l-s)t(l+t)/4 -(l-2s)t(l+t)A -s(l-s)(l+2t)/48 (l-s2)t(l+t)/2 -st(l+t) (l-s2)(l+2t)/29 s(l+s)t(l+t)/4 (l+2s)t(l+t)/4 s(l+s)(l+2t)/4
The x ' and y ' coordinates are related by the matrix ofinterpolation functions f, so that
x' = ft x" (15)
wherex = [x' x' ... x'j
f = ff1 f2 ... f9] .
t „Similarly y" = f jr' where
z = (y y2 •** y9 3 •
16The x,y and s,t coordinate systems are related "by:
dx 3 s
as
where
4 2 ax'at
at
4 s'
4 2
(16, 17)
(18, 19)
fsl fs2 110 fs9) ’
ftl ft2 0 "' ft9.
Thus
si
ti
as
5at
a as a , at aax'" - ax' as dx' at
a _ as a at ady' - 3y' ds dy' dt
(20)
(21)Or, in matrix forms
a ds« dx' dx'■ a ds_5y; _dy'
1 1 "a 1• as
!a: iat
atdx'atdy'
From these relationships, we form a Jacohians
J ,-x y ,st
(22)
‘ds at = "dx' 1dx' dx' ds dsds dt dx' *3L_dy' dy'j _at dt _r- 4- «=14 f f —s; Z -1 = f—s OV] -1 ' —st,x'y'.f, x' 4t * i; Z J .4] (23)
The determinate of the Jacohian is used in the transformation of stiffness matrix formulated in s,t space to the stiffness matrix in local x',y' space as followss
J , s' * -st,x. y ds . dt (24)
For any position within the element defined "by s,t thestrain-displacement matrix T can "be derived as shown "below.■ —e,p
First, the displacement interpolants must "be packed in a way to "be compatible with two degrees of freedom per nodes
• • —2^- u , p t-f i -2 f2 -2 (25)
where
—2 = 1 0
0 1
To obtain the strain-displacement matrix T , the T matrix^e,p -u,pis pre-multiplied by a differential operator matrix derived from the stress-strain relationships s
Further partitioning yields
2e,p = l lSz I9] (-27)
where
3s 25: + ik 3fi 3x' 3s dx' 3t
3s 2fi + 3fi3y" 3s 3y' 3t
3s 25: . il 25:3y^ 3s dy" 3t
3S 25: + dt, 25: dx' 3s dx' 3t
A three-by-three Gauss-Legendre numerical integration scheme is used to evaluate the stiffness matrix„ The element stiffness in local space in terms of x" and y" is
k' = t /. T E T dA— A -~e pP — "-e pP
(28)The element stiffness in local space in terms of s and t is
3 3i' ■ 2 W1 ¥j^e,p K-tj)i=l i=l
J ,-st,x y(29)
where the values for W^, Wj, s^, and t^ are given in Table 2,
19Table 2. Integration Constants
i or j s. or t. 1 J W. or W. 1 J
1 -0.774596669 0.5555555562 0 0.8888888893 0.774596669 0.555555556
— . . . -----------------------------4
Transformation of the Element Stiffness from Local to GlobalCoordinate Systems. Let us define a coordinate transformation matrixT - such that -P »P*
T - „-P ,P* “P0 c-p0
0
0
0-p0 -p
symmetric
0 0 0 0 c-p0 0 0 0 0 C-P0 0 0 0 0 0 “P0 0 0 0 0 0
00 0 0 0 0 0 0 0 C-P (30)
where
-P " •
20Thus, the global three-dimensional stiffness matrix k* is
derived from
is* = 4 ' . P* - V . P * • (31)
Element Stresses. Once the solution of the assembled structural model is performed, the elemental stresses, expressed in local coordinates, are computed from the displacements. A necessary intermediate step is the calculation of the local element strains from the global displacements by
e' = I e',p' V . F * * * • (32)
Four points compatible with GIFTS mesh generation and post-processing were chosen for evaluation of elemental stresses. The strain-displacement matrix is evaluated at each of these four stress points, which are located at the centroid of each of the four sub-quadrilaterals shown in Fig. 4.
321
e — n o d e
# - s t r e s s p oi nt
Fig. 4. QM9 Element Stress Evaluation Points
21The strains at each evaluation point are multiplied "by the
elasticity matrix to obtain the stresses at each points -
S = E e « , . (33)
Mass Calculation. A diagonal form of the full, kinematically consistent mass matrix is used (Zienkiewicz 1971)• The total elemental mass is expressed "by
1 1 .m = f. f t dA = f t f f I J ds dt . (3 )
A -1 -1 I
The diagonalization results in the formv; ■m* = f t f f ff^ f2 ... f^J J | ds dt
-1 -1
= Fm* m* ... m*J . (35)
Finally
S = (§*) 3* (36)
where
9M* = £ m* .
i=l 1
Thermal Forces, The element stresses are related to the strains caused "by thermal and applied forces "by
S = E (e - 6q) (37)
where the thermal strain vector is given "by /
22
e, = cc *1
10
and
= T, ’i32
9,
0 "being the temperature at the elements' nodes. The matrix Tg g was chosen to "be the same as f, the displacement interpolation matrix,
The equivalent nodal forces resulting from constrained thermal expansion are given "by the equation
J’v i;,p 5 a -e,i 8 dV
Y — esP (i-v)
E a t p _t (l-v) JA — e,p
2 e,e S 4V
— 8,9 i “(38)
This form of the equation, since it cannot "be evaluated exactly, must "be numerically integrated. The weighting functions
23and integration constants are the same as in the stiffness matrix integration. The form of the thermal force equation using numerical integration is
P. = E a t p r mt (l-v) x y —e,p
E a t p f t (l-v) s t —e,p ds dt
3 3
(39)
To reduce the number of calculations» we can take advantage of the following manipulations!
P. =3 3
|J|]
3 3 (40)
The Linear Strain Triangle Membrane Element TM6This formulation is based upon linear strain assumptions and
requires the element to have straight sides. In the stiffness matrix computation subroutine, the midside node coordinates are calculated
from the coordinates of the comer nodes, ensuring straight sides.
24Thus, if the boundaries of the mesh are curved, the midside nodes in the mesh will be ignored during stiffness assembly.
Element Geometry. The element is defined in three-dimensional space (Fig. 5)• Before the stiffness is formulated, the element is mapped to a two-dimensional local space with the origin at the centroid of the triangle (Fig. 6).
6
X
zFig. 5. TM6 Element in Global Space
Y
Fig. 6. TM6 Element in Local Space
25The direction cosines for the transformation are calculated
from the first edge 13 and the edge 16s
-x" = -13 '
-z" = -13 X -16
X 5.x*’ ( 3)
£ = [^x- Q y ' 5.2"] » (44)
Thus 9 the-transformation to local coordinates is effected hy
2Eq = V 3 [xj + Z2 + 2 ^ (45)
2£j = [xj - x y (46)
where j = 1,6.The elemental area is calculated "by
A = [H (H - L13)(H - L36)(H - L6l)] ^ (4?).
where
H = I (Liy+ L36 + L6l *
Element Stiffness. For an arbitrary point within the triangular element, assume that its displacement may "be interpolated from the nodal displacements "by a second-order polynomial function as followss
u(x,y) = a^ + agx" + a^y" + a^x"2 + + a^y"2 . (48)
Taking into account all nodes and all deformation modes of theelement, in matrix notations
1> £ y l x t 2 x i y i y ; 2 " - l i
1X 2 y 2 6 % i
1x 3
0 0 - 3 i
1 0 0 • - 4 i
1 • 0 - 5 i
r —
-
y f % iJ
where i = 1,6.Solving for the constants a
= u(x,y)
(49)
_1a = B u, (50)
Thus, to interpolate a displacement for a point in the element, we could write
a = 2u,p £ (51)
where a row of T^ would typically appear as
ali + a21xl + a3iyi + a4ixi2 + a51xlyi + a6iyi2 '
The strains and displacements are related "by the T matrix,6 spwhere
2?
ddx'
d.By'
By'BB x '
T-a,p
Tg ... T j (52)
where
a2i + 2a4ix + a5iy
a3i * a5ix + 2a6iy
a3i * a5ix" + 2a6iy'
a2i + 2a4ix' + a5iy'
For ease of computation, this is further reduced to
Se.p = So + Sxx + Syy (53)
where
- n - Clol — o2 8 8 8 - n / Jlo6- “01 a2i
31
‘31
21
- x " [ - x l —x 2 8 8 8 —x6-l Ixi 2a4i
a5i
a 512 a4i
28
^-yl -y2 0 * ° —y6^ 11
*51 0
0 2a6i
_2a6i : a5i_
From these relationships, we can form the stiffness matrix:
S' - /v (t£ + T*x' + ^y') E (T0 + f^x- + Tyy') dT (54)
Recognizing that the element is of constant thickness and noting constants, we can write
k ' = t T u E f /. dA + t E T /, dA — —o o A —x ---x A
+ t E Ty /A y'2 dA (55)
Integrating the expression, we obtain
k" = t fr" E T A + t"*' E T I — 1— o o —x x yy
+ (T E T + T E T ) I + T E T I 1—x y —y x' xy —y y xx-1 (56)
where
Ixx = H + 4 + y62) -
yy J2 (xi + x3 + xf) -
Ixy = 12 (xiyl + x3y3 + V e )
Transformation of the Stiffness Matrix from Local to GlobalCoordinate Systems. Let us define a coordinate transformation matrixT - . such that-P iP*
-P'.P* -P
-P
-P symmetricC-P0
0-P
-p (57)
where
C = [C , C . —p 1— x —y J
Thus, the global three-dimensional stiffness matrix, k*, is derived from
r = ip'p* is v .p* • (58)
Element Stresses. Once the solution is performed, the element stresses, expressed in local coordinates, are computed from the global displacements. A necessary intermediate step is the calculation of the local element strains from the global displacements by
30The strain-displacement matrix is evaluated at each of the
three stress points, again compatible with GIFTS mesh generation and post-processing, which are located at the centroid of each of the sub-triangles shown in Fig. 7.
_ • — n o d e
# _ s t r e s s p o i n t
Fig. ?. TM6 Element Stress Evaluation Points
The strains at each of the evaluation points are multiplied by the elasticity matrix to obtain the stresses at each point:
S' = E e' . (60)
Mass Calculation. A diagonal form of the full, kinematically consistent mass matrix is used (Zienkiewicz 1971). This is expressed
i>ym = J t A Tf^ f2 ... f^J dA (61)
where
f ± = [l x' y" x'2 x'y' y'2] a
as in the stiffness formulation.The integral of the first term is the area of the element, the
next two are the first moments of area and, as they are evaluated at the centroid, they are zero. The last three terms are the second moments of area and are evaluated as in the stiffness formulation. Thus, we now have
A note of interest is that this matrix reduces to one-third ofthe mass of the plate applied at each of the midside nodes. This was verified "by testing many different geometries in the stand-alone test driver programs mentioned in Chapter 3= Output from this test is shown in Appendix 0. '
Thermal Forces. The stress-strain relationship can "be described as follows
(62)
S = 1 (s " ee) (63)
where the thermal strain vector is given "by
320. "being the temperature at the element nodes and again T „ , thei vtemperature interpolation matrix, was taken to he the same as the displacement interpolation matrix f,
The equivalent nodal forces resulting from constrained thermal expansion are given "by
59 =
4 £=,P S a Te>e e dV
J-V e,p Tl-vJ 6 dV
Eta-j. Tt XI-v7 A -e,p 6 dA .
(64)
Substituting the strain displacement matrix and multiplying throughthe column vector we can separate terms to reduce the
complexity of integrations
33E t a p TitvJ A "a21+2a41x ' 0 a3l+a51x ' "
+a5ly ' +2a6ly '
0 a3l+a5lx ' a21+2a4lx '+2a6iy ' +a51y ' .
I 5
a26+2a^ x ' 0 a36+a56x '+a56y ' +2a66y"
0 a36+a56x' a26+2a46x"+2a66y' +a56y ' _
-e,9 i ^
E t a r (l-v) A 1 H
»..
. .1 + "2a4l" x' +1 y'
a31 a52 2a6l• ; •
"26 2a46 a56a56 . 2a66
T l & y SaBo + l xx ' + V I l e . e i ^ • (65)The temperature interpolation matrix is evaluated along
with the x' and y' terms from T which are factored out:e, p
XT-vJ tv Ctj + —x + :y] [l x' y' x'2 x'y' y' ] a _0 dA .(66)
34
P-0
Performing the integration, we get:
4+ T +o —x r A 0 0 Axx Axy v]
0 A A A A AXX xy XXX xxy xyy0 A A A A A
_ xy yy xxy xyy yyyj(67)
where
A = area of element
Axx = Ja X2 dA = (x2 + x2 + x2) ^
Ayy = /A y2 dA = (yl + y3 + y6) 12
Axy = /a xy dA = (xlyl + x3y3 + x6y6) 12
AXXX = / A x3 ^ = [2 (x^ + Xj + Xg) - 3x 1x3x6] | q
Axxy = xA x2y dA = [2 (x2yi + x2y3 + x2y6)
- + x3xgy^ + x6Xly3)] ^
Axyy = /A xy2 dA = [2 ( x ^ i + x3y2 + x6y^)
- (xiy3y6 + x3y6yi + x6yiy3)]
Ayyy = /a y3 dA = [2 (yj* + y^ + y|) - 35
The Axisymmetric Isoparametric Bi-Quadratic Element QA9This element is "based upon quadratic displacement interpola
tion functions, which allow elements with curved sides. The element is sensitive to the location of the midside and center nodes and to excessive boundary curvature (Fwa, Jofriet, and LeLievre 1978).Because of this, the nodes should be as close to centered as possible.
two-dimensional space, Fig. 8. The element stiffness is formulated in unit space, Fig. 9* The transformation between unit space and two-dimensional space is accomplished through the Jacobian matrix detailed below.
Element Geometry. The element is defined in a general
9
Z
V*-1 2 3
Fig. 8. QA9 Element in Local Space
36
7 8 9
321Fig. 9* QA9 Element in Unit Space
Element Stiffness. For an arbitrary point within the element in s,t space, assume that its displacement may be interpolated from the nodal displacements by the Lagrangian bi-quadratic interpolants listed below in Table 3» Their derivatives are also listed for future reference.
Table 3« The QA9 Interpolation Functions and Their Derivatives
Pt. fi df./ds = f . lz si- ■ ----------— -----dV at = f+.i :
1 s(l-s)t(l-t)/4 (l-2s)t(l-t)/4 s(i-s)(i-2t)A :2 -(l-s^)t(l-t)/2 st(l-t) -d-s2)(l-2t)/2 !
3 -s(l+s)t(l-t)/4 -(l+2s)t(l-t)/4 -s(i+s)(i-2t)/4 :4 -s(l-s)(l-t^)/2 -(l-2s)(l-t2)/2 s(l-s)t ;
5 (i-s^)(l-t^) -2s(i-t2) -2t(i-s2) :
6 s(l+s)(l-t2)/2 (l+2s)(l-t2)/2 -s(l+s)t
7 -s(l-s)t(l+t)/4 -(l-2s)t(l+t)/4 -s(l-s)(l+2t)/4 I
8 (l-s2)t(l+t)/2 -st(l+t) (l-s2)(l+2t)/2 :
9 s(l+s)t(l+t)/4 (l+2s)t(l+t)/4 s(l+s)(l+2t)/4 j
The r and z coordinates are related "by the matrix of interpolation functions f, so that
r = f r (68)
where
r = ^ r2 ... ,
f = ^ f2 ... f93 '
Similarly
z = f 2 (69)
where
Z = ^ Zg ... .
The r,z and s,t coordinate systems are related hys
E = 4 ^ If = 4 s (70.71)
If = 4 S If = 4 s <72' 73)where
38Thus,
d _ ds B , 3t 3dr dr 3 s dr dt
d _ _ds _d_ ' _dt _d_dz - dz ds dz dt
Or, in matrix forms
nd_dr
h
dsdrdsdz
dtdrdtdz
"ddsddt _
J—rz, st
From these relationships 9 we can form a Jacohians
ds " dt_ |] = | dr • . dz | -1drds dz
dt j dr - dzldr ds ds jdt dr ■ dz |dz dt dtj
II III r/—s
1—
^1i_y
A - 4 5 _ A X----
—1 — —st,rz
(74)
(75)
(76)
(77)
The determinate of the Jacohian is used in the transformation of the stiffness matrix formulated in s,t space to the stiffness matrix in r,z space, as shown "belows
st ,rz r ds dt . (78)
For any position within the element defined "by s,t, thestrain-displacement matrix T can "be derived as shown "below,-e,p
First, the displacement interpolants must he packed in a way to he compatible with two degrees of freedom per nodes
-u,p = . f2 -2 0" ° f9 - 2 ^
where
1 0
0 1
To obtain the strain-displacement matrix, the T matrix is-u,ppre-multiplied by a differential operator matrix derived from the stress-strain relationships. Note that this matrix has one more row than the cartesian element QM9 has. This term is used to evaluate hoop stress;
40Further partitioning yields
-e,p = -ZsP (81)where
Si = ds f£i dt ffiBr Bs 3r Bt
Bs 5fi , Bt afi Bz Bs Bz Bt
as 5 + it !!iBz Bs Bz Bt 0
as!!i + at!£idr 5s dr 5t
A three-hy-three Gauss-Legendre numerical integration scheme is used to evaluate the stiffness matrix. The element stiffness in terms of r and z is given "by
k* = 2 rr J\ T E T rdA- A - e , p e,p (82)
The element stiffness in terms of s and t is given "by 1 1 , ,“1 —1 e»p -st ,rz' r ds dt (83)
The form required for numerical integration is
It* = I S 2,rW. W j T ^ s ^ t . ) E T ^ p t s . , ^ ) I1-1 J-l
(84)
where the values for W., ¥ s., and t . are shown in Tahle 4.1 J 1 J
Table 4. QA9 Integration Constants
i or j s^ or tj or ¥ j
1 -0,774596669 0.5555555562 0 O.8888888893 0.774596669 0.555555556
Note that since the integration points are all within the elemental "boundaries, the terms including will not he singular at r = 0, hut numerical inaccuracies may arise.
Element Stresses. The elemental stresses are evaluated in the same manner and positions as in the QM9 element described previously,
Thermal Forces. The element stresses are related to the strains caused by thermal and applied forces by
S = E (e - eQ) (85)
where the thermal strain _e0 is given by:
0. "being the temperature at the element's nodes. The matrix T» A is x -0the same as f , the displacement interpolation matrix.
The equivalent nodal forces resulting from constrained thermal > expansion are given "by the equation
4 = J’t 2e ,p 5 e9 ®
J V2.,pS a T, e.dv
JV -e,p (l-2v) 6 dV
2 rr E a « t (l-2v) A —e, p "0,6 0 r dA
(86)This form of the equation, since it cannot "be evaluated
exactly, must "be numerically integrated. The values for W^, ¥ ^, s^, and tj are the same as in the stiffness matrix integration. The formsuitable for numerical integration is
To reduce the number of calculations, we can take advantage of the following manipulations:
The Axisymmetric Linear Strain Triangle Element TA6This derivation is "based upon linear strain assumptions. It
assumes the element has straight sides. In the stiffness matrix subroutine, the midside node coordinates are calculated from the coordinates of the comer nodes, ensuring straight sides. Thus, if the "boundaries of the mesh are curved, the midside nodes in the mesh will "be ignored during stiffness assembly.
z
Element Geometry. The element is defined and formulated in a two-dimensional space (Fig. 10).
44
6
1 2 3
Fig. 10. The TA6 Element
The elemental area is calculated by
A = [H (H - L13) (H - L^) (H - L6l)]* (89)
"Where
H = i (l13 + + l6i) .
Element Stiffness. Assume that displacements of an arbitrary point within the triangular element may be interpolated from the nodal displacements by a second-order polynomial as follows:
u(r,z) = a^ + a^r + a^z + a^r + a^rz + a^z . (90)
Taking into account all nodes and all deformation modes ofthe element:
where i = 1,6.Solving for the constants a
a = B'1 u „ ‘ (92)
Thus, to interpolate a displacement for a point in the element
a = 2U,P £ (” )
where a row of T would typically appear as u,p
ali + a2iri + a3iZi + V i + a5iVi + a6izi •
Strains and displacements may he related "by the T matrix;6 9P
which, with further partitioning, yields
Te,p = E^I., ... T,,]
where
(95)
h = a2i + 2% i r + a5iz
- (a^ + a2.r + ay.z
+ * a51xy + a6iE
a3i + a5ir + 2a6iK
a3i + a5ir + 2a6iz
a2i + 2alHr + a5iz
where i = 1,6.From these relationships we can form the stiffness matrix:
k* = T E T r de dr dz — —e,p e,p
2 it J\ T E T r dr dz A —e,.p-- e,p (96)
Since this expression cannot he explicitly integrated, numerical integration must he used. The scheme was developed hy Hammer, Marlow, and Stroud (1956)» The form of the stiffness integral suitable for numerical integration is
47The coordinates o f the four integration points are defined by
an area-coordinate scheme:
ri = alrl + a2r3 + a3r6 (98)
zi = + a 2 z 3 + a 3 z 6 (99)
where i = 1,4. Their positions are shown in Fig. 11, and the values for (X and the weighting constant W , are given by Table 5.
6
Fig. 11. TA6 Integration Points
48Table 5- TA6 Integration Constants
i al a2 a3 ¥i
1 1. 1 1 • -273 3 3 48.11 2 2 2513 15 15 48
o 2 11 2 25J 15 15 15 48:
4 2 2 11 2|15 15 15 48
Note that since the integration points lie inside the houn-idary, the terms including — will not he singular as r tends to zero,
hut numerical inaccuracies could result.Element Stresses. The elemental stresses are calculated in
the same positions and in the same manner as previously covered in the section on the TM6 element„
Thermal Forces. The applied to thermal stress-strain relationship can he described as followss
S = I (e - ee) (100)
where the thermal strain vector may he written as
e = a ee l 6e = ^6,6 6ii 02i_0_
!>.e6.
0. being the temperature at the element's nodes and T& ,, thei v 9 y
temperature interpolation matrix, is chosen to he the same as the displacement interpolation matrix f=
The equivalent nodal forces resulting from constrained thermal expansion .are given "by
-d _ 2 tr E g P-0 (l-2v) a21+2aM r
+a^z0 r(all+a21r
+a% z + a4 1 ^+a^rz+a^^z2)
a31+a51r+2a61z
0 a31+a51r+2a6lz
0a21+2aw r
+a51z.
6 6 0 *
a26+2aU6r+a56z
0 r(al6+a26r+a26 ^ V 2
+Bigrz+a66z2)
a36+a56r
+2a66z
0 a36+a56r
+2a66a0 a26+2ai)6
+a56z
51
= 2 tt E g (1-2V) (a 1+2a2,1rfa..z) + rxall
+a21r+a31z+a4lr +a51rz+a6lz ^(a31+a51rf2a<, z)'6r
( * r{a16+a26rta36z+V / 2+ayJrz+a6622)
(a36+a56r+2a<^z^
— 6,0 — r dz
(101)
The temperature interpolation matrix is evaluated as follows:
= [l r z r^ rz z ^ J a . (102)The expression for Pg must he numerically integrated. A quintic integration scheme is used, which was originally developed "by Hammer et al. (1956). The location of the integration points is shown in Fig. 12 and the constants of integration in Tahle 6.
-e = fl f J/i tp p1 1 1 1
1 0_
1 1 10 I (103)
2 tr E a 4 rr£ h
<PTt-e,p 9. r. 1 1
1 32
Fig. 12. TA6 Thermal Force Integration Points
Table 6. TA6 Thermal Force Integration Constants
i C1 C2 C3 "l1 1
313
13 0.225
2 al 01 0i 0.132194153 al 0i 0.132394154 Pi 0i ai 0.132394155 a2 a2 02 0.125939186 02 a2 02 0.125939187 02 02 a2 0.12593918
where = 0.0597158?P. = 0.47014206&2 = 0.79742699 P2 = 0.10128651
The Linear Strain Rod Element ROD3This element is based upon linear strain assumptions. The
middle node has been assumed to be centered between the end nodes, simplifying the element. When using this element with the QM9, care must be taken to avoid curved edges, as the ROD3 element must be straight.
Element Geometry. The element is defined in a general three-dimensional space, Fig. 13. The element stiffness is formulated in one-dimensional space along the x" axis along the axis of the rod, with the y" axis perpendicular to the x ' axis at node 1, Fig. 14.
1
Fig. 13. ROD3 Element in Global Space
54
Fig. 14. ROD3 Element in Local Space
The direction cosines for the mapping are calculated from the line connecting nodes 1 and 3> and the origin is moved to node 1,
C = [G , C - C -] . (104)— i— x —y —z
Element Stiffness. For an arbitrary point along the rod, assume that the displacement may be interpolated by the function
u(x') = W^(x') + W2(x*) + ¥^(x#) (105)
where
W1(x") = x'(x' + x p - x"(x* + 3x') + 2x'2( x ' - x')2
¥ (x") = -x'x" + x ' ( x ' + x ' ) - x'2
M (,; - 4 , ' ------
The displacement interpolation matrix T can now he expressed asU 5 p
(X' - L) (x- -|)] (106)
where
L ,= x^ - x^
The strains and displacements are related hy the T matrix.e j p
T-e,P dx0
. d 9y
o
_d,wa-Bx'
-u»p
- C. - 3L); -% (l - 2x‘B) —| (4x - L)] (10?)
From these relationships, we can form the stiffness matrix.
56
AEt < PX E3L 7
-81
-816-8
1187 (108)
Transformation of the Stiffness from Local to GlobalCoordinate System. Let us define a coordinate transformation matrixT - „ such that“P »F*
-P »P* -x0
C --yo
o
c ,—x C »-yc—x V (109)
Thus, the global three-dimensional stiffness matrix k* is derived from
k* = T k T-p'sP* - “P »P* (110)Element Stresses. Once the solution is performed, the element
stresses, expressed in local space, are computed from the global displacements. A necessary intermediate step is the calculation of the local element strains from the global displacements by
s' = s . ' . p ' V . p - ^ v V ( m )The strain-displacement matrix is evaluated at each of the two
stress points, which are located halfway between each of the nodes, Fig, 15.
57
Fig. 15. ROD3 Element Stress Evaluation Points
The strains at each point are then multiplied by the modulus of elasticity to obtain the stresses at each point:
= E(112)
Thermal Forces. The applied-to thermal stress relationship can be described as follows
S = E 61 - e16
e2 ' e26 (U3)
where the thermal strain vector can be written as
where
e = T,
58The temperature interpolation matrix is the same as the
displacement interpolation matrix T—U»PThe equivalent nodal forces resulting from constrained thermal
expansion are given "by
= A E / ^ e ' d ,
A E “ {Xe,pSe,j e ax (114)
Integrating this, we find
P = A E a [-0.5 . 0 0.5]
(115)
Kinematically Consistent Force Distribution on a Second-Order Element’s Edge
The general equation for the discretization of a surface force*applied to an element isP = ^ f Us dA (U6)
where U is the column vector of applied surface forces. If the load —sis a constant, w pounds per inch, the load equation per unit length can be expressed as
P = A f f dA (117)
Using the f matrix from the ROD3 element as a typical second-order edge with center node assumed, the integrated equation has the form
L30 4 2 -1
2 16 2-1 2 4
w
(118)
For unit length and unit constant load, the equivalent distribution is
162 31Z (119)
CHAPTER 3
SUBROUTINE\ eVELOPMENT, INTEGRATION, AND TESTING
Initial Testing Once the element subroutines were written, an initial de-
hugging phase was entered. Rather than integrating the subroutines directly into GIFTS, a series of stand-alone programs were written to drive each class of subroutines. Typical drivers, one for the stiffness and stress subroutines, one for the mass subroutines, and one for the thermal force subroutines, atre shown in Appendix G . Each of the drivers were written using the GIFTS common block structure and library subroutines, minimizing changes in the driven subroutines as they were integrated into GIFTS. The small driver programs also minimized the amount of time spend insuring the framework to drive the subroutine in question was free of errors.
The initial test results produced by the drivers are faster to scan and interpret if round constants are chosen, for example, using 10' for the modulus of elasticity instead of 29A x 10 » The test geometry and boundary conditions were also defined to aid interpretation of results. Square elements ten inches on a side were used for the quadrilaterals, and right triangles ten inches on a side were used for the triangles. A ten inch rod was used for the ROD3 element.
60
61To test the element stiffness subroutines, an eigenvalue test
was used, as discussed by Zienkiewicz (I9?l). For an element with a given number of degrees of freedom, this test produces zero eigenvalues for all rigid body motions. Each element was first tested in local two-dimensional space, then in global three-dimensional.space, testing only one portion of the subroutine at a time. The test for the element stress subroutines consisted of a uniform deflection field, producing a uniform stress field. The element mass subroutine test involved the calculation of the mass matrix for the element. This matrix was then summed to compare With the equivalent solid plate.The input data for' the thermal force subroutine was a uniform temperature field of 100° F , which produced a symmetrical thermal force field.
Integration into GIFTSAfter the initial debugging phase was concluded, the elements
were integrated into GIFTS. The element stiffness and stress modules in GIFTS (STIFF and STRESS) are written very similarly, so the subroutine integrations are discussed together. The element mass and thermal force subroutine integrations are discussed separately.
Prior to this time, the GIFTS program did not have.the capability to analyze second-order elements, though first-, second-, and third-order elements could be generated and stored in the data base,To avoid analysis of models generated with second-order elements,GIFTS tested for second- and third-order elements, and skipped them if any were present during stiffness assembly and stress calculations.
62This test was modified to allow second-order elements» The skip chain that was used to call the appropriate routine, "both in STIFF and STRESS, was modified to call the second- as well as first-order elements „ A new overlay was added to "both STIFF and STRESS containing the new stiffness and stress subroutines.
All element mass calculation is done in module BULKLB. A lumped mass calculation is used for the first-order elements» The error message that GIFTS gave if mass calculations were attempted for second-order elements was changed to a jump to the appropriate element's mass subroutine. There are two of these tests, the test for triangular elements is subroutine TMAPL, the test for quadrilateral elements is in subroutine QMAPL.
The thermal force calculation is done in module DEFL prior to the solution. As the thermal force generation capability was being written at the same time as this project, the problems in integration were minimal. The skip chain to call the appropriate element is identical to the one in STIFF and STRESS, except for subroutine names,
After the analysis subroutines were integrated, the result display capabilities for second-order elements were tested. Once the post-processing errors, such as in the packing of stresses according to stress points interior to the element, and higher order stress contouring, were corrected, the integration of the elements into GIFTS was completed.
Testing the Integrated SubroutinesThe first of the tests involved single element problems for
the quadrilateral elements and two element problems for the triangles. Both element types were tested using a square plate model. The test for the rod element was conducted by placing a bar along the x axis: of the same cross-sectional area as the plate models. All of the tests involved applied and thermal loadings.
The cartesian elements', QM9/TM6/ROD3» stiffness and stress subroutines were first tested in extension (Fig. 16)„ This test was repeated in each of the planes of the cartesian axis system to exercise the direction cosine calculations within the element subroutines. The thermal force subroutines were tested by applying a uniform temperature field to the same models as the extension tests. Once-the initial thermal tests were run, duplicates of the loads that the thermal force subroutines generated were applied to the models with module EDITLB in another loading case. Then the solutions were rerun. The element stress subroutines were then checked for proper thermal strain subtraction by comparing the two loading cases. The thermally located elements showed zero stresses, while the applied force loaded elements showed a constant stress field of the proper value. The mass subroutines were first tested by calculating the mass of the test plates and comparing that with the answer provided by the subroutines. Typical output from one of these tests is shown in Appendix C „
64
Yu
XWNv
Fig. 16. Cartesian Extension Tests
The axisymmetric elements’, QA9/TA6, stiffness and stress subroutines were first tested with a ring problem. The stiffness and stress subroutines were tested with radial and axial displacement tests (Fig. 1?). Since there are no direction cosine calculations within the axisymmetric subroutines, no further orientations were used in these tests. The axisymmetric thermal force subroutines were tested in the same manner as the cartesian elements, by comparing equivalent thermal and applied loading cases.
The ROD3 and the QM9/TM6 elements were then tested together in a cantilever beam problem (Fig. 18). Three loading cases were defined, one an applied end load, another a constant temperature field. The third loading case involved heating of one edge of the beam to make it curl.
65
u
» e • • • •M V V
Fig. 17. Axisymmetric Extension Tests
Yu
Fig. 18. Cantilever Beam Model
CHAPTER 4
FINAL TEST PROBLEMS
Once the initial testing of the integrated elements was concluded, several more complex models were analyzed using the second- order elements to further verify their performance. Two problems
were run with the cartesian elements and two with the axisymmetric elements. The first of the cartesian problems was a beam built in at both ends, the second a plate with a hole in the center. The axisym- metric problems were a uniform pipe and a spherical pressure vessel.
Cartesian Problems The first problem was identical with the one presented in the
GIFTS Modelling Guide (Kamel 1977, pp. 14, 22). The problem was that of a beam built in at both ends. Only one quarter of the beam was modeled, taking advantage of the symmetry in geometry and loading (Fig. 19). The beam was loaded with a twice gravity inertial loading.
Fig. 19 . Built In Beam
66
67The loads for the problem, generated with the help of the
consistent mass routine, are shown in Fig. 20, along with the "boundary conditions. The deflected shape is shown in Fig. 21 with stress contours, documenting the correct plotting of curved second-order elementsides and stress contours. The maximum deflection at the center of
-62.134 x 10 inches compares very well with the theoretical maximum/
deflection of 2.212 x 10 inches (Kamel 1977, p. 17)«The second cartesian problem was a finite plate with a hole in
the middle, uniformly loaded on two edges. The problem was selected because of the curved, non-square elements. This problem has been investigated by many people, for example, Goodier and Timoshenko (1951, p. 248), Peterson (1974, p. 110), and Kamel (1977, p. 31)- Another concern for this test was stress representation, as there is a factor of 3=38 in the magnitude of the stresses from the evenly stressed portion of the plate to the stresses around the hole, making interpretation of the results straight-forward. The plate model is shown in Fig, 22. Only a quarter of the plate was modeled, taking advantage of symmetry.
/
^ 080E-03 LOADING CASE 1
w
JPOEL
LOADS
Fig. 20. Inertial Loads
LOAD PLOT RESULTANTS
ylBTDIR.
mTCoT' L'frifT5
w m
ON00
LOADING CASE 1
WPOEL
DEFL. AND STRESSES 2.00S 5a
STRESS CONTOURS
w i m m M
Fig. 21. Deflections and Stresses ONNO
Fig. 22. Plate Model
The deflected model with stress contours is shown in Fig. 23- A principal stress vector plot is shown in Fig. 24. A graph illustrating the stress variation around the hole in the plate compared to theory is shown in Fig. 25» showing excellent results even with this very coarse mesh. The mesh was not biased to keep midside and center nodes centered.
LOADING CASE 1
d£tf IJlft :
■BEELi— flHB _8IBESS,E8
STRESS CONTOURS
Fig. 23. Deflected Plate with Stress Contours
LOADING CASE 1
TER
DEL
< /OEFL. AND STRESSESFig. 24. Deflected Plate with Principal Stress Vectors
73
1.00modelexact
3.38
Fig. 25• Stresses Around the Hole
Axisymmetric Problems The first problem was a pipe under uniform internal pressure.
The model was constrained in the z (axial) direction at one end only to suppress rigid body motion. This made the problem a free-expanding pipe (Fig. 26).
A
N r
Fig. 26. Pipe Model
74
The radial deflection due to a uniform internal pressure is (GoOdier and Timoshenko 1951> P- 46l)'
uradial = WIT = .i0° --- = 3 = 3898 x 1015 inches (120)radxal E t 29.5 x 10b x 0.1
where E is the modulus of elasticity. The pipe was also subjected toa constant temperature load in another loading case. For a thermal
—6expansion coefficient of 6,5 x 10 , the axial and radial deflectionsare given by
uaxial = 100° x 6,50 x 10 ^ x 5 = 3.25 x 10~^ inches (121)
Uradiai = 100° x 6,50 x 10 ^ x 1 = 6 .50 x 10-^ inches . (122)
The theoretical model for the radial deflection of the pipe assumes constant strain across the pipe thickness, which is assumed to be thin compared to the diameter. The tests showed a radial deflection of 3.380 x 10 ^ inches at the center of the wall for the pressureloading case. Radial deflections of 6.493 x 10 inches and axial
_Zldeflections of 3.25 x 10 inches were produced by the thermal force test, again measured at the center of the wall. All of the results show excellent agreement with the theoretical model. An interesting note is that, when the problem was first run, the loads were applied at the inside of the pipe, giving poorer results. When the loads were moved to the centerline of the pipe wall, much better results were observed. This reenforces that, to get the same results as a theoretical model with a finite element model, the same assumptions must be made.
75The second problem involved a spherical pressure vessel under
uniform internal pressure (Fig. 2?) . The curved, inclined elements made this problem a good exercise for the elements. Taking advantage of symmetry, only the top half of the vessel was modeled.
# * # #
Fig. 2? . Spherical Pressure Vessel Model
This problem has been solved in many texts, for example,Pilkey and Pilkey (1978, p. 165). The radial deflection of I.69I x
_ o10 inches compares very well with the theoretical radial deflection of
■ w ■ r f t - 2 , Z l »... • ‘•695 x 10"3 l na“ ;s
The same assumptions as were made in the pipe model were made here.
CHAPTER 5
CONCLUSIONS
The formulation of the five higher order elements under consideration have "been presented in a form suitable for FORTRAN coding on the digital computer. The functions coded include the stiffness matrix, calculation of mass and thermal loads, and stress calculation. To complement this, a method for calculating kinematically consistent forces along a second-order edge was mathematically developed.
The details of the development.of these elements1 subroutines from conception through final integration and testing were documented. A part of the development involved the writing of a series of small, simple computer programs to initially test the element subroutines.
The debugged pre- and post-processing capabilities of GIFTS for second-order elements were illustrated with many computer plots. These capabilities include plotting of curved second-order edges, stress contouring, and principal stress-vector plots.
Finally, several problems demonstrating the correct performance of these elements were presented. These problems also show that the five elements require careful use. The kinematically consistent loads representation must be carefully followed to obtain accurate solutions. Calculating the loads by hand for the examples in this work quickly pointed out to the author the need for automatic second-order load generation. The need for centered nodes presents
76
difficulties in "biasing GIFTS meshes. This can he resolved hy using the interactive point moving capabilities of GIFTS to center the nodes, but a scheme to do this automatically would be preferable as it would be more accurate.
APPENDIX A
NOMENCLATURE
corner displacements of the element in the local coordinate system.corner displacements of the element in the global coordinate system.comer force vector, of an element in the local coordinate system.corner force vector of an element in the global coordinate system.matrix of displacement interpolation functions.local and global element stiffness matrices respectively.transformation matrix such that p" = T _ p*.-c- —p ,p*stress and strain matrices.generalized Young's modulus such that B_ = E _e. strain-displacement matrix such that _e = T@ p temperature interpolation matrix, the same as f, such that
mass matrix.
APPENDIX B
ELEMENT SUBROUTINES
SUBROUTINE K0M9 C *** REWRITTEN AUGUST 1978 BY KEITH A. HUNTEN.CC STIFFNESS OF BI-QUADRATIC ISOPARAMETRIC GUADRALATERAL MEMBRANE. C
DIMENSION GC (3), G W (3), DC 1(3), DC2 (3), DCS (3), MAPS (3), HAF’4 (3) DIMENSION E(2,2),TC2(3),TC4(3)>TC5(3),TC6(3),TC8(3)COMMON /DAT/SKI 18,18),DC(3,3),TLC(3,9),T E P (3,13),F S I (9),F T I (9)1 ,FU(2,2) ,F J I (2,2),El 1,E 2 1 ,E12,E22COMMON /BKF/MAPI3,2 7 ) ,TC(3,27),SM<6,6)COMMON /MAT/MATPTR,MT,YM,PR,G COMMON /THS/ITYFE,IFTRCS,TH
CEQUIVALENCE (£0(1,1),DC1(1)), (D C(1,2),DC2(D), (DC(l,3),DC3il)) EQUIVALENCE <H A P d , 3),M A P 3 ( 1 )), (MAPI 1,4),MAP4(1)), (Ell,Ei 1,1)) EQUIVALENCE (TC2(1),TC(1,2)),(TC4(1),TC(1,4)),(TC5(1),TC(1,5)) EQUIVALENCE (TC6(1),TC(1,6)),(TC8(1),TC(1,8))
CDATA GC/-.774596669, 0. ,.774596669/DATA GW/ .555555556,.888886889,.555555556/DATA EPSI/l.E-5/
CCC *** COMPUTE DIRECTION COSINES TO THE ELEMENT'S LOCAL AXES IN DC',C COL.Is C(X'), COL.2 =C(Y ), C 0 L . 3 < ( Z ).
D46-0.£(23-0.DO 15 J=l,3D C 1 (J)=TC6(U)-TC4(J)D46-D46t DC1(J)*DC1{J)D C 2 (J )= T C 8 (J )- T C 2 (• J /
15 D2 3-D2 8+D C2(J )* D C 2 (J )D4 6-SG RT(D 4 6 )D2 S-SQ RT(D23 5
CC DEFINE A SMALL LENGTH.
SMALL-EFSI*(D 4 6+D2 8)DO 20 .>1,3 DC1(J)-£C1(J)/D46
20 DC2(J)-DC2(U)/D28
79
80
CALL CROSS (DCliDC2iDC3iXX)CALL CROSS (DC3,DC1,DC2,XX)
CC **# CONFUTE LOCAL COORDINATES, POINT 5 IS THE ORIGIN. ALSO, CHECKC PLANENESS OF ELEMENT.
DO 35 1-1,9 DO 30 J-1,3 8-0.DO 25 K-1,3
25 S-'S+(TC(K, I }-TC5tK))*DC(K,J)SO TLCiJ,I)-S
IF (TLC(3,I) .LT. SHALL) GO TO 35 STOP
35 CONTINUEC? WRITE (12,922) (J , (T L C (I,J ) ,I-i,2),J - l ,9)C?922 FORMAT (' TLC I X Y /,/,(lX,I4,2E13.7))CC INITIALIZE LOCAL STIFFNESS.
EP^YM/(1.-PR**2)CC=PR*EPEll-EPE22=EPE12-CCE21-CC
CC CLEAR UPPER HALF OF MATRIX.
DO 40 J - l,13 DO 40 I-1,J
40 SK(I,J)=0.CC *** (3X3) GAUSSIAN INTEGRATION OF ELEMENT STIFFNESS MATRIX.
DO 70 1-1,3 SS-GC(I)W1-GW(I)SSP-(1.+SS)SSM-(1.“SS)SS2-2.*SS SS2P-1.+SS2 SS2T1-1. -332 SSSQ-1.-SS*SS DO 70 J-1,3 TT=GC(J)TTP-(1.+TT)TTM-(l.-TT)TT2-2.*TTTT2P=1.+TT2TT2M-1.-TT2TTSQ-l.-TTrTTrSI(l)-0.25#SS2M#TT*TTi1FSI(2)-SSvTTM*TT
FSI(3)--0.25*SS2P*TT*TTM F S I (4)^-0.5*SS2M*TTSQ FSI(5)=-SS2*TTSQ FSI(6)-0.5*SS2P*TTSQ FSI(7)=-0.25$SS2M*TT*TTP FSI(8)=-SS*TT*TTP FSI(9)^0.25*SS2P*TT*TTP F T I (1)=0.23*SSM*SS*TT2M F T I (2)^-0.5*SSSG*TT2M F T I (3)=-0.25*SS*SSP*TT2M FTI(4)=SS*SSM*TT F T I ( 5 ) = - S S S Q * n 2 F T I (6)- -SS*SSP*TT FTI(7);-0.23*SS*SSM*TT2P F T I (8)=0,5#SSSQ*TT2P FTI(9)=0.25*SS*SSP*TT2P
C? IF (ISM .GE. 9) GO TO 42c? sio.C? S2-0.C? DO 41 KKK-1»9 C? S l - S l + F S K K K K )C?41 S 2 - S 2 + F T K K K K )C? WRITE (12,900) S3,TT,SI,32 C7900 FORMAT (' 3,T,FSI,FTI ,/2X,4E14.7) C? ISW-ISW+1C?42 CONTINUE CC *** COMPUTE JACOBIAN IN 'FJ'.
81=0.32=0.DO 45 K=l,9 X=TLC(1,K)S1=S1+FSI(K)*X
45 S2=S2+FTI(K)*X FJ(1,1)=S1 FJ(2,1)=S2 31=0.S2=0.DO 50 K=l>9 Y=TLC(2,K)S1=S1+FSI(K)*Y
50 S2=S2+FTI(K)*Y FJ(1,2)=S1 FJ(2,2)=S2
CC H v INVERT JACOB I AN IN 'FJI'.
DET=FJ(1,1)*FJ(2,2)-FJ(1,2)* F J (2,1) REC=1./DETFJI(l,l)-FJ(2,2)tREC FJI(2,2)=FJ(1, D v R E C
82
FJI(1,2)=-FJ(1,2)*REC F J I «2 j 1)- - F J (2 >1)#REC
CC *** COMPUTE T E P ( S S M T J ) .
XX =FJI(li1)XY-FJI(2>1)EXX=FJI(1,2)EY-FJI(2 j2)DO 55 Jl=l,9J 2 F - 2 W 1J2=J2P-1T E P d i J2)-XX*FSI(J1 )+EXX#FTI (J 1 )TEP(l»J2P)-0.TEP(2iJ2)-0.TEP(2,J2P)=XY#FSI(J1) t E Y #FTI (J1)TEP(3»J2)=TEP(2 jJ2P)
55 TEP(3iJ2P)-TEP(1»J2)C? WRITE (10,921) I,J, G w ( I ),GW ( J ) ,DET,TH,E(1,1),E(2,2),0 C?921 FORMAT {' I,J , W I ,WU,DET,TH,El 1,E22,G',214,/,1X ,7 E 1 1.4)
W 2 - W 1* G W (J )wDET*TH DO 60 K=l,2 DO 60 LL-1,2 W3=W2*E(K,LL)DO 60 JJ=1,13 W4=W3*TEP(LL,JJ)DCi oO II— 1, JJ
60 SK (II,J J )- S K(II ,JJ)+ W 4*TE P(K , I I )W3=G#W2DO 65 JJ - 1 , 18W4-W3*TEP(3,JJ)DO 65 11=1,JJ
65 SK (II, JJ)-SK(II ,JJ)+W 4 * T E P (3,11)70 CONTINUEC? WRITE (12,920) ( 3 K ( I , I M = 1 , 1 3 )C7920 FORMAT (' SK LOCAL',/,(IX,6E12.5))CC *** UPPER HALF OF ELEMENT STIFFNESS READY. COPY LOWER HALFC OF DIAGONAL SUBMATRICES.
DO 75 1=1,17,2 1 1= 1+1
75 SK(I1,I)=SK(I,I1)CC *** TRANSFORM AND ASSEMBLE ELEMENT STIFFNESS (UPPER HALF).C ASSEMBLE LOWER SUBMATRICES BY TRANSPOSITION.
DO 95 IP=1,9 K0=2*IP-2 DO 90 IO-IP,9 L0=2*IQ-2 Dil 85 1=1,3 DO 55 J=l,3
83
3-0.DO 30 K=l,2 KGK-KO+K DO 80 L-1,2
SO S - S + D C (Ii K)* S K (K O K ,L O + L )* D C (JiL)85 SH(I»J)-S
CALL SAS (IP?IQ? 3)IF (IP .EQ. IQ) GO TO 90
CC ASSEBLE LOWER SUBMATRIX.
DO 89 1-1,3 DO 89 J-I,3 TMP=SM(I,J)SM(I?J)-3M(J,I)
39 S M U ? I ) = T M PCALL SAS (IQ?IP?3)
90 CONTINUE95 CONTINUE
RETURN END
SUBROUTINE POM?C *** REWRITTEN AUGUST 1978 BY KEITH A. HUNTEN.CC THERMAL FORCES FOR BI-QUADRATIC ISOPARAMETRIC QUADRALATERALC MEMBRANE ELEMENT.C
DIMENSION GC(3),GW(3),DC1(3),DC2(3),DC3(3)DIMENSION 02(9),L3(9),STR( 12),S F K 1 8 )DI MENS I ON TC2 (3), TC4 (3), TC5 (3), TC6 (3), TC3 (3)COMMON /DS09/ BC(3,3),TLC(3,9),T E P (IS,IS),F? (9,9),WF (9),F S I (9)
1 ,F T I (9),FJ(2,2 ) ,FJI(2,2),E l 1,E 2 1 > E12, E22COMMON /MAT/MATPTR,MT,YM,PR,G,STY,DEN,THEX COMMON /THS/ITYPE,IPTRCS,TH COMMON /FAR/LHV(11),L G L (5),L F X (14),MOST COMMON /DAT/TC(3,27),NEL COMMON /TMP/TEMP(27)COMMON /TLD/VTLD(6,27)
CEQUIVALENCE (DC(1,1),D C 1 (1)),(BC(1,2),DC2(1)),(DC(1,3),D C 3 ( 1)) EQUIVALENCE (TC2(1),TC(1, 2)),(TC4(1 ),TC(1, 4)),(TC5(1 ),TC(1,5)) EQUIVALENCE (T C 6(1),TC (1,6)),(TC3(1),T C ( 1,8))
CDATA GO/-.774596669,0.,.774596669/DATA GW /.555555556,.688888889,.555555556/DATA 02/0,2,4,6,8,10,12,14,16/, 03/0,3,6,9,12,15,18,21,24/
CCC *** COMPUTE DIRECTION COSINES OF X',Y',Z' IN DC:C C O L . 1= C(X'), COL.2 -C(Y'), C0L.3= C(Z/).
D46=0.D28=0.DO 15 J=l,3D C 1 (J)-TC6(J)-TC4(J)D4 6=D46+DC1(J )* D C 1(J )DC2(J)-TC8(J)-TC2(J)
15 D2 S-D2 8+E C2(J )* D C 2 (J )D4 6-SQ RT(D 4 6 )D28=SQRT(D23)DO 20 0=1,3 DC1(J)=DC1(J)/B46
20 DC2(J)-DC2(J)/D28CALL CROSS (DC1,DC2,BC3,XX)CALL CROSS (DC3,DC1,BC2,XX)
CC «Hr COMPUTE LOCAL COORDINATES, POINT 5 IS THE ORIGIN.
DO 33 1-1,9 DO 30 .>1,3> 0.CO 25 K M , 3
25 >S -KTC ( K , I ) - T C 5 t K ) )*DC(K,J)30 TLCtJ, I)-S35 CONTINUE
CC-THEXvYHvTH/(1.- P R )CC iHHr ESTABLISH TEP, F9 & wF FOR 9 GAUSSIAN POINTS.
KOIDO 70 > 1 , 3 W>GW(J)TT-GC(J)T T P M 1 . + T T )T T M M 1 . - T T )T T 2 - 2 M TTT2F-1.+TT2TT2M-1.-TT2T T S Q M . - T T * T TF T 1 - 0 . 5 * T T M T MFT2=TTS8FT3-0.5*TT*TTPDO 70 1-1,3SS-GC(I)WI=GW(I)S S P M 1 . + S S )S S M M 1 . - S S )SS2=2*SSSS2P-1.+SS2S S 2 M - 1 .-SS2S3SG-1.-SSvSSFSl-0.5*SSrSSMFS2-SSSQFS3-0.5*SS*SSPFSI(1)4).25*SS2M*TT*TTMF3I ( 2 ) - S S M T M * T TF 3 K 3 ) --0.25*SS2P*TT*TTMFol (4) -~tj. 5*SS2M*TTSQFSI(5)=-SS2*TTSQF S I ( 6 > 0 . 3 * S S 2 P * T T S QF b 1(7)- - 0 .25*SS2M*TT#TTPFSI(S)--SSxTTxTTPF 3 I ( 9 ) - 0 . 2 5 * S S 2 P * T T M T PFTI(1)=0.25*SSM*SS*TT3M
86
F T I (2)1-0.5*SSSG*TT2MF T 1(3)i-0.25*SS*SSP*TT2MFTI(4)=SS*SSM*TTFTI(5)=-S3SQ*TT2FTI(6)=-SS*SSP*TTFTI(7)i-0.25*SS*SSM*TT2PFTI(S)4).5*SSSQ*TT2PFTI(9)4).23*SS*SSP*TT2P31=0.32=0.[Kj 45 K=l,9 X=TLC(1,K)S1=S1+FSI(K)*X
45 S2=S2+FTI(K)*X F J ( M ) = S 1 FJ(2,1)=S2 31=0.32=0.DO 50 K=l,9 Y=TLC(2,K)S1=S1+FSI(K)*Y
50 S2=S2+FTI(K)*YFJ(1,2)=S1 FJ(2,2)=S2
CC INVERT JACOBIAN IN FJI'.
DET=FJ(l,l)*FJ(2,2)-FJ(l,2)rFJ(2,l)X WRITE (10,1230) DET,CC,NELXI230 FORMAT (/,"DET=",F10.3,/,"TCONST=",F10.3,/,"NEL=",I5)
REC=1./E€TFJI(1,1)=FJ(2,2)*REC FJI(2,2)=FJ(1,1)*REC FJI(1,2)=-FJ(1,2)*REC F J I (2,1)=-FJ(2,1)*REC
Cc iHrir COMPUTE TE P( S S L T T J ) .
XX=FJI(1,1)XY=FJI(2,1)EXX=FJI(1,2)EY=FJI(2,2)DO 55 Jl=l,9J2P=2*J1J2=J2P-1TEP(K0,J2)=XXrF3I(J1)+EXX*FTI(Jl)TEP(K0r9,J2P)=XYxFSI(J1)+EY*FTI(J1)TEP(K0,J2P)=0.TEP(K0+9,J2)=0.
55 CONTINUEWF(KO)=CC*WI*WJ*DET F9 (K0,1 )=0,25*SS$SSM*TT*TTM
F ? (KO,2)--0.5*SSSG*TT*TTM F9(KO,3)--0.25*S3*SSP*TT*TTM F 9 (KO ? 4)- - 0 .5*SSM*TTSQ*SS F9(KO,5)=SSSQ*TTSQ F 9 (K O ,6)-0.5*SS#S3P#TTS8 F 9 (KO ,7)--0.25#SS*3SH*TT*7TP F9(K0,8)4).5*SSSQ*TT*TTP F9(K0,9)-0.25*SS*SSP*TT*TTP K0=K0+1
63 CONTINUE 70 CONTINUE CC *** LOOP THROUGH LOADING CASES.
DO 200 LC ASE- i,NDST CALL INTilP (NELiLCASE)
CC ^ OBTAIN LOCAL FORCES.
DO 105 1-1,13 105 SPL(I)-0.CC r++ LOOP THROUGH GAUSSIAN POINTS.
DO 150 N O - 1,9CC +++ OBTAIN TEMPERATURE AT POINT.
THETA -=0.DO 110 1=1,9
110 7H E7A-THETA+F9(NG ,I)*TEhP(I)X WRITE (10,1000) THETA,TEMPXI000 FORMAT (/, " THETA= ",E10.2,/,(3E10.2))CC +++ COMPUTE LOCAL FORCE CONTRIBUTION.
DO 120 1=1,16120 SPL(I)=SPL(I)+ W F (N G )#T HETA *(T E P (NG,I)+TEP(NG+9, 150 CONTINUE CC +++ TRANSFORM TO GLOBAL COORDINATES.
10=0 •JO—0DO 160 N=l,9 UP=SPL(J0+1)VP=SPL(J0+2)DO 155 1=1,3
155 V T LD(I,N ) =UP* DC1(I)+VP*DC2(I)160 00=00+2200 CALL OUTLD (NEL.LCASE)
RETURN
88
SUBROUTINE MOM?C t w LAST UPDATED 2/20/79 BY KAH C *** REWRITTEN AUGUST, 1973 BY KEITH A. HUNTEN.CC DIAGONALIZED MASS OF BI-QUADRATIC ISOPARAMETRICC GUADRALATERAL MEMBRANEC
REAL MASS(9)C
DIMENSION G C (3),G W (3),D C X (3),D C Y (3),D C Z (3),T O (3,9>,F S (3/,FTi3) DIMENSION D C (3,3),TLC(2,9),F S K 9 ) , F T K 9 ) , F J ( 2 , 2)DIMENSION T C 2 (3),T C 4 (3),T C 3 (3),TC6(3),TC8(3)COMMON /CGMMND/L(60),V(20)COMMON /PTS/NU,NSiISTRKT,VC(3)COMMON zELT/NEU,NES,ISTRK2,IT,ICRD,1ST,IAPI,NLDRC2,NCR,NOP 1 ,ISTFTR,N M A T ,N T H S ,F A C (3),WR K P D 2 (3),L C P (27)COMMON /MAT/MATPTR,HI,YM,PR,G,STY,DEM COMMON /TH S/ITYPE,IPTRCS,TH
CEQUIVALENCE (F S (1),F S 1 ),(F S (2),F S 2 ),(F S (3),F S 3 )EGUIVALENCE < FT < i ),F T 1>,(F T (2),F T 2 ),* FT < 3),F T 3 )EQUIVALENCE (DC<1, D . D C X d ) ), ([C(1,2),DCY(I >), (DC(1,3),DCZ{1)) EGUIVALENCE ttC2(1),TC U ,2}),(TC4 U ),TC U ,4)),(T C 5 (1),T C (1,5)i EQUIVALENCE (T C 6 (1),T C (1,6)),(TCS U ),T C (1,8))EQUIVALENCE (V(5),MASS(i))
CDATA GC/-.774596669, 0. ,.774596669/DATA GW/ .555555556,.888888889,.555555556/
CcC *** READ POINT COORDINATES.
DO 10 1-1,9 CALL INPTS ( L C P ( D )DO 10 J-i,3
10 TC(J,I)=VC(J)cc rx# COMPUTE COORDINATE TRANSFORMATION MATRIX TO LOCAL COORDINATES.
D46-0.D2S-0.DO 15 .>1,3
89
S=TC6(J)-TC4(J)DCX(J)=SD46-D46+S*SS=TCS(J)-TC2(J)DCY(J)=S
15 D28=D23+S*SD46-SQRT(D46)D2o-SQRT(D2S;DO 20 J-1,3 DCX(.J)=DCX(J)/D46
20 DCY{J}-DCr{J)/D2SCALL CROSS (DCX,DCY,DCZ,XX)CALL CROSS (BCZ,DCX,DCY,XX)
CC *** COMPUTE LOCAL COORDINATES. POINT 5 IS THE ORIGIN.
DO 35 1-1,9 DO 30 .>1,2 S-0.DO 25 K-1,3
25 S=S+(TC(K,I)-TC5(K))$DC(K,J)30 TLCIJ,I)-S35 CONTINUEX WRITE (10,922) (J,(TLC(I,J), 1 - 1 , 2 ) , > 1 , 9 )X922 FORMAT (' TLC I X Y',/,(1X,I4,2E13.7)5 CC INITIALIZE MASSES.
EM-0.[hj 40 > 1 , 9
40 MASS(J)=0.CC ##* (3X3) GAUSSIAN INTEGRATION OF ELEMENT DIAGONAL MASS MATRIX.
DO 70 1-1,3 S > G C ( I )Wl-GW(I)SSP=(1.+S3)SSM=(1.-S3)SS2=2.*SS SS2P=1.+SS2 3 S 2M-1.-SS2 5350=1.-SS&SS FS1=0.5*SS$SSM FS2-SSSQ FS3=0.5*SS*S5P DO 70 > 1 , 3 TT=GC(J)TTP=(1.+TT)TTM=(1.-TT)TT2=2.*TTTT2F-1.+TT2TT2M-1.-TT2
TT3Q-1.-TT*TTFT1-0.5*TT*TTMFT2-TT3QFT3=0.5*TT$TTPFSI(1)=0.25*SS2M*TT*TTMFSI(2)=SS*TTM*TTFSI(3)--0.25*SS2P*TT*TTMF S I (4)^-0.5*SS2M*TTSQFSI(5)=-SS2*TTSQFSI(6)-0.5*SS2P*TTSQFSI(7)^-0.25*SS2M$TT*TTPFSI(8r=-SS*TT*TTPFSI(9)=0.25*SS2P*TT*TTPFTI(1)=0.25*SSM*SS*TT2MFTI(2)=-0.5*SSSQ*TT2MF T I (3)--0.25*SS*SSF*TT2MFTI(4)%SS*SSM*TTFTI(5/--SSSQ*TT2FTI(6)=-SS*SSP*TTFT I < 7 )--0.25*SS*SSf!*TT2PFTI(3)=0.5*SSSQ*TT2PFTI(9)=0.25*SS*SSP*TT2P
CC *** COMPUTE JACOBIAN IN 'FJ'.
91=4.32=0.DO 45 K=l,9 X=TLC(1,K)S1=S1+FSI(K)*X
45 S2=S2+FTI(K)*X FJ(1,1)=S1 FJ(2il)=S2 31=0.32=0.DO 50 K=l,9 Y = TLC(2 jK)S1=S1+F3I(K)*Y
50 S2=S2+FTI(K)*YFJ(1,2)=S1 FJ(2,2)=S2
CC COMPUTE DETERMINATE OF JACOBIAN.
DET=FJ(1,1)*FJ(2,2)-FJ(1,2)*FJ(2,1)CC *** COMPUTE MASSES.
w2=Wl*GW(J)*0ET*TH*DENEM=EM+W2K=1DO 55 11=1,3 DO 55 JJ=1,3
91M A S S (K )- M A S S (K )+ W 2 * (F S (J J /* F T (I I ))
55 K-K+l70 CONTINUE
EHS=0.DO 71 1=1,9
71 EMS=EMS>MASS(I)X WRITE (10,930) EM.EMSX930 FORMAT (' EM,EMS',5X,2E15.7)
W3=EM/EMS DO 72 1=1,9
72 M A S S (I )= M A S 3 (I )*N3 RETURNEND
C
92
SUBROUTINE SGM9 C xx* LAST UPDATED DECEMBER 27,1978 BY KAH C *** REWRITTEN AUGUST 1979 BY KEITH A. HUNTEN.CC STRESSES FOR BI-QUADRATIC ISOPARAMETRIC GUABRALATERAL MEMBRANE.C
DIMENSION GC(2),D C 1 (3),DC2(3),D C S (3),E (2,2),12(9),L3(9),S T R(12) DIMENSION TC2(3),TC4(3),TC5(3),TC6(3),TC8(3),SPL(18)DIMENSION TTT(4,9),TSTRN(4)COMMON /DSQ9/DC(3,3), T L C (3,9),T E P {12,13),F S I (9),F T I {9),F J (2,2)
1 ,FJI(2,2),E11,E21,E12,E22COMMON /MAT/MATPTR,MT,YM ,PR,6,STY,DEN,THEXCO COMMON /THS/ITYPE,IPTRCS,TH COMMON /PAR/LHV(11),L G L (5),I P X (14),NDST COMMON /D TS/T C(3,27),S P (54),S P X (6,7),S T (7,10),LCASE,NSTPT,1 ITFLG,NELCOMMON /TMP/TEMP(9)
CEQUIVALENCE (D C (1,1),D C 1 (1)),(D C (1,2),D C 2(1)),(DC(1,3),D C 3 (1)) EQUIVALENCE ( E l l , E d , D )ECUIVALENCE (TC2(1),TC(1,2)),(TC4(1 ),TC(1,4)),(TC5(1),T C ( 1,5)) EQUIVALENCE (TC6( 1),T C d ,6)), (TC3( 1), T C (1,8)), (S P X (1,1),SFL( 1))
CDATA GC /-.5,.5/DATA L2/0,2,4,6,3,10,12,14,16/, L3 /0,6,12,13,24,30,36,42,48/DATA EPSI/l.E-5/
CCC xx* COMPUTE DIRECTION COSINES TO ELEMENT'S LOCAL COORDINATES IN D C . C C0L.1=C(X'), C0L.2-C(Y'), C0L.3<:(Z')
D46-0.D2S-0.DO 15 0=1,3D C 1(J )= T C 6 (J )- T C 4 (J )D4 6=D4 6+D C1(J )* D C 1 (J )DC2(J)=TC3(J)-TC2(J)
15 D28=D28+DC2(J)*DC2(J)D46=SGRT (D46)D2S-SQRT (D23)
C
93
C ^ DEFINE A SMALL LENGTH. bMALL-EPSI*(D 4 6 + D 2 8 )DO 20 .>1,3 DC1(J)=DC1(J)/D46
20 DC2(J)-BC2(.J)/D28CALL CROSS (DC1,BC2,BC3,XX)CALL CROSS (DC3,DC1,CC2,XX)
CC COMPUTE LOCAL COORDINATES, POINT 5 IS THE ORIGIN. ALSOC PLANENESS OF ELEMENT.
DO 35 1=1,9 Du 30 > 1 , 3 3-0.DO 25 K=l,3
25 S-S+(TC(K,I)-TC5(K))#BC(K,J)30 TLC(J,I)=S35 CONTINUECC? WRITE (10,922) (J, (TLC(I,J ) , 1 = 1 , 2 ) , > 1 , 9 )C?922 FORMAT (' TLC I X Y',/,(1X,I4,2E13.7))
C *** COMPUTE ELASTICITY MATRIX E'.EP=YM/(1.-PR**2)VEP=PR*EPCC=PR*EPE11=EPE22-EPE12=CCE 2 1—Lu
CC *** ESTABLISH 'TEP" FOR 4 STRESS POINTS.
NSTPT=4K0=000=1DO 70 > 1 , 2 TT-GC(J)TTP=(1.+TT)TTM=(1.-TT)TT2=2.*TTTT2P-1.+TT2TT2M=1.-TT2TTSQ=1.-TT*TTDO 70 1=1,2SS=GC(I)SSP=(1.+SS)SS M=(1.~SS)332=2.*SS SS2P=1.+SS2 SS2M=1.-SS2 SSSQ-1.-SS*SS
CHECK
94
C » > INTERPOLATION FUNCTIONS (IN TTT'). IF (ITFLG .EG. 0) GO TO 40 TTT-JO,l)=0.25*SSxSSM*TT*TTN T T T (JO,2)--0.5*SSSQ*TT*TTM TTT(JO,3)--0.25*SS*SSP*TT*TTM TTT(JO,4)=-0.5#SSM*TTsQ*SS TTT(JO,5)-SSSQ#TTSQ T T T (JO,6)=0.5*SS*SSP*TTSQ T T T (J O ,7)- - 0 .25*SS*SSM*TT*TTP T T T (JO,8)-0.5*SSSQ*TT*TTP T T T (JO,?)-0.25*SS*SSP*TT*TTP
CC » > AND THEIR DERIVATIVES.40 FSI(i)=0.25*SS2M*TT*TTM
FSI(2)-SS*TTM*TT F S I (3)--0.25*SS2P*TT*TTM FSI(4)=-0.5*SS2M*TTSQ FSI(5)-“ SS2*TTSQ F S I (6)=0.5*SS2F*TTSQ FSI(7)=-0.25*SS2M*TT*TTP FSI(8)=-SS*TT*TTP F S I (9)-0.25#SS2P*TT* TTP FTI(1)=0.25*SSM*SS*TT2M FTI(2)=-0.5*SSSQ*TT2M F T I (3)^-0.25*SS*SSP*TT2M FTI(4)=SS*SSM*TT FTI(5)=-SSSQ*TT2 FTI(6)=-SS*SSP*TT FTI(7)=-0.25*SS*SSM*TT2P FTI(8)=0.5*SSSG*TT2P FTI(9)=0.23*SS*SSP*TT2P
C? IF (ISW .GE. 9) GO TO 42C? 81=0.C? S2=0.C? DO 41 KKK=1,9C ? Si=Sl+FSI(KKK)C?41 S2=S2+FTI(KKK)C? WRITE (10,900) S3,TT,S1,S2 09900 FORMAT (' S,T,FSI,FTI S / 2X,4 E14 .7) C? ISW=ISW+1C9'42 CONTINUE CC w w COMPUTE JACOB I AN IN 'FJ'.
S1=0.32=0.DO 45 K=l,9 X-TLC(1,K)S1=S1+FSI(K)*X
45 S2=S2+FTI(K)*X
o o
95
FJ(2,1)=S291=0.92=0.DO 50 K=l,9 Y=TLC(2,K)S1=S'1+FSI(K)*Y
50 S2=S2+FTI(K)*Y F,I(1,2/=S1 FJ(2,2)=S2
CC *** INVERT JACOBIAN INTO "FJI'.
DET=FJ(1,1)*FJ(2,2)-FJ(1,2)*FJ(2,1) REC=1./DETFJI(1,1)=FJ(2,2)*REC FJI(2,2)=FJ(1,1)*REC FJI(l,2)=-FJ(l,2)rREC F J I (2i1 )=-FJ(2i1 )*REC
*** COMPUTE TEP(SST j TTJ).X X = F J I ( M )XY=FJI(2,1)EXX=FJI(1»2)EY=FJI(2,2)DO 55 Jl=l,9 J2P=2*J1 J2=J2P-1 K1=K0+1 K2--K0+2 K3=K0+3TEP(K1,J2)=XX*FSI(J1)+EXX*FTI•J 1 ) TEP(K1,J2P)=0.TEP(K2,J2)=0.TEP(K2, .J2P)=XYxFSI(Jl )+EY*FTI (J 1 ) TEP(K3,J2)=TEF(K2,J2P)
55 TEP(K3,J2P)=TEP(K1,J2)K0=KG+3 J0=J0+1
70 CONTINUE CC LOOP THROUGH LOADING CASES.
DO 200 LCASE=1,NDST CALL ELTDNS
CC = = OBTAIN LOCAL DEFLECTIONS.
DO 130 N = 1i9 J0=L3(N)I0=L2(N)DO 120 1=1,2 3=0.
96
DO 110 J-1,3 110 3- S + D C (J »I )* S P (J O + J )120 S P L d O + D - S130 CONTINUECC OBTAIN STRAINS.
DO 140 1=1,12 S-0.DO 135 .>1,18
135 S= S + T E P (I ,J )v S P L (J ;140 STR(I)-SC? WRITE (10,930) (STR(I),I=1,12)C?930 FORMAT C STRAINS ',/,(lX,3E15.7))CC » > IF TEMPERATURES ARE PRESENT, CALCULATE AND C SUBTRACT THERMAL STRAINS.
IF (ITFLG .EQ. 0) GO TO 149CC » > GET THIS LCCASES TEMPERATURES AND CALCULATE C TEMPERATURE AND THERMAL STRAINS AT EACH OFC THE STRESS POINTS.
CALL INTMP (NEL,LCASE)X WRITE (10,1111) THEXCO,TEMPXllll FORMAT (/," THEXCO=",E10.3 , /,"TEMP",/,(3E 1 0 . 3 ) )
DO 142 ISP=1,4 SPT=0.ECi 141 1=1,9SP T-SPT+TEMP(I )* T T T (IS P,I)
141 CONTINUETS T R N (IS P)=SPTvTHEXCO
142 CONTINUECC » > SUBTRACT THERMAL STRAINS.
K0=1DO 143 ISP=1,4 STR(K0)=STR(K0)-TSTRN(ISP)STR(K0+1)=STR(K O + 1 )-TSTRN(ISP)K0=K0+3
143 CONTINUE CC = OBTAIN STRESSES.149 J0=0
DO 150 N=1,N3TPTS T 1=E P * S T R (J 0 + 1)+V EPvS TR(J O + 2 )3T 2-vE PvS TR(J 0 + 1)+E P*ST R(J O + 2 )ST3=G*STR(J0+3)ST(1,N)=ST1ST(2,N)=ST2ST(3,N)=ST3ST(4,N)=ST1
97ST(5>N)-ST2ST(6,N)=ST3J0=J0+3
150 S T (7,N )=100.*SGRT (S T 1**2+ST2**2-ST1*3T2+3.*373**2)/STYC? WRITE (10,1000) ((S T (H , N ),H - l ,7),N - l ,N S T F T )C71000 FORMAT (' STRESSES IN SQM9',/,E10.3)200 CALL ELTSTR
RETURN END
C
SUBROUTINE K7M6 C *## REWRITTEN AUGUST, 1973 BY KEITH A. HUNTEN.CC STIFFNESS FOR LINEAR STRAIN TRIANGLE.C
DIMENSION BC(3,5),L P ( 4 ),X L (3,6),X0 (3),L I (3),FL ( 3 > ,Sr1(6,6) DIMENSION D C K 3 ) , O C X (3),D C Y (3),D C Z (3),D (6,6),C T (3,2)COMMON /D AT/TO(3,12),TX(3,12),TY(3,12),SKI 12,12)COMMON /BKF/MAPI3,27),T C (3,27),S M I (6),S M 2 16),S M S (6),S M 4 (6)
1 ,SM5(6),SM6(6)COMMON /MAT/MATPTR,MI,E,PR,G COMMON /THS/ITYPE,IPTRCS,TH
CEQUIVALENCE (D C (1,1),D C 1 (1)),(D C (1,3),D C X (1),C T (1,1)) EQUIVALENCE (DC!1,4),D C Y ( 1)),(DC(1,5),D C Z (1))EQUIVALENCE (S M1(1 ),S M(1, D)
CDATA LP/3,6 ,1,3/, EPS/1.E-5/, LI/5,4,2/
CCC *** COMPUTE DIRECTION COSINES.C SIDE 3-6 IN DC1, 6-1 IN DC2, 1-3 IN DCSC LOCAL X,Y,Z IN DCX,DCY,DCZC H IS HALF THE PERIMETER.
H-O.DO 20 J=l,3 S-0.K K P ( . J )K2-LPIJ+1)DO 10 1=1,3 CD=TC(I,K2)-TC(I,K1)DC(I,J)=CD
10 S=S+CD**2S-SQRT IS)FL(J)=SH-H+SDO 20 1=1,3 DC(I,J)=DC(I,J)/S
20 CONTINUECALL CROSS IDCX,DC1,DCZ,XX)
99CALL CROSS (DCZ,OCX,DCY,XX)H=0.5*H
CC *♦* COMPUTE AREA OF TRIANGLE.
A 136=SGRT (H * (H - F L (!))*(H - F L (2))*(H - F L (3)))CC *** LOCAL ORIGIN, CREATE POINTS 2,4,5.
DO 30 1-1,3TC (1,2) -0. 5v (T C (1,1) + T C (1,3))T C (1, 4 )=0.5*(TC(1, 1 )+TC(1,6))TC(I,5)=0.5*(TC(I,3)+TC(I,6))
30 X0(I}-(TC(I,1)+TC(I,3)+TC(I,6))/3.CC *** OBTAIN LOCAL COORDINATES OF CORNERS.
DO 40 .>1,6 DO 35 1=1,3 3-0.DO 34 K=l,3
34 S=S+DC(K»I+2)#(TC(K,J)- a O ( K ) )XL(I,J)=S
35 CONTINUE40 CONTINUEC? WRITE (10,910) A136, (J,(XL(I,J),1 = 1 , 3 ) , > 1 , 6 )C7910 FORMAT (' A 1 3 6 = y, E 1 0 . 4 , / V LOCAL XYZ ',/,(1X,I3,3E12.4))CC Mir CONSTRUCT 6X6 MATRIX TO OBTAIN COEFFICIENTS BY INVERSION.
DO 70 > 1 , 6 SM1(J)=1.X=XL(1,J)Y=XL(2,J)SM2(J)=X SM3(J)=Y SM4(J)=X*X SM5(J)=X*Y SI16(J)=Y*Y
70 CONTINUECALL INVNM (SM,6,D,6)
C? WRITE (10,930) ((D(I,J),> 1 , 6 ) , 1=1,6)C7930 FORMAT (' 0 •',/,(1X,6E12.5))CC Mir TEP = TO + TX.X + TY.Y C COMPUTE TO,TX,TY
RIXX=A136*(XL(2,1)**2+XL(2,3)**2+XL(2,6)**2)/12.RIXY=A136*(XL(1, 1)»XL(2,1 )+XL(1,3)*XL(2,3)+XL(1,6)»XL(2,6))/12 RIYY=A136* (XL (1,1) M 2 + X L (1,3) **2+XL (1,6) m 2 ) /12.
C? WRITE (10,940) RIXX,RIXY,RIYYC?940 FORMAT (' IXX,IXY,IYY /,/,lX,3E12.4)
> 1DO 30 K - l ,6 Jl-i.j+1
100
T0(2iJ)-0.T0(1,J1)=0.C1=D(2,K)C2-D(3»K)T0(1»J)-C1 T0(2» J D - C 2 T0(3,J)=C2 T0(3>J1)=C1 TX(i,.Jl)^0.TX(2,J)-0.Ci-2.*D(4»K)C2-D(3>K)TX(1,J)=C1TX(2?Ji)-C2TX(3,J)=C2TX(3iJl)-ClC1=2.*D(6,K)TY(1»J)-C2TY(1,J1)=0.TY(2»J)-0.TY(2,J1X1 TY(3iJ)-C1 TY(3,J1)=C2 J-J+2
80 CONTINUE CC? WRITE (10,950) ((T0(I,J),J=1,12),I=1,3),((TX(I,J),J=1,12)C? 1 1-1,3),((TY(I,J),J=l,12),1-1,3)C7950 FORMAT (' TOXY ',/,(lX,6E11.3))CC # « FORM E MATRIX.
EF-E/(1.-PR**2)C? WRITE (10,960) E,PR,G,EP,VEP 07960 FORMAT (' E,PR,G,EP',/1X,4E12.5)CC *** FORM STIFFNESS MATRIX.
DO 90 J-1,12 DO 85 K=J,12 C1=EP*T0(1,K)C2-EF*T0(2,K)SUM-TO (1,J)*(C1 +PRC2) +T0 (2, J) * (P R C 1 +02) +T0 (3, J ) *G*TO (3,K)SuM-SUM*A136C1=€P*TX(1,K)C2-EP*TX(2,K)3 - T X (1,J)*(C1+FR*C2)+TX(2,J)*(PR*C1+C2)+TX(3,J)*G»TX(3,K) SUM-SUM+RIYY*S3-TY (1,0)* (C1+PRC2) +TY (2, J )*( PRC1+C2) +TY (3,0) *G*TX (3, K )SUM-SUM+RIXY*3C1=€P*TY(1,K)C2-EF*TY(2,K)
S^TX(l,J)*(Cl+PR*C2)+TX(2,J)*(PRCl-H::2)+TX(3>J)*G*TY(3iK)Si;i1=SUM+RIXY*SW Y ( C 1 + P R C 2 ) +TY (2,J)*( P R C 1 + C 2 ) +TY (3, J ) *GxTY (3,K)SUM^SUM+R1XX#3 SUM StmTH SK(K,J)=SL!M SK(J,K)=SUM
85 CONTINUE 90 CONTINUEC? WRITE (10,920) ( S K ( I , I ) , M , 12)C7920 FORMAT (y SKP (1X.4E12.4))CC LOCAL STIFFNESS READY. TRANSFORM AND ASSEMBLE.
K0=0DO 150 11-1,6 LO-OECi 140 J J— 1,6
CC — SKIP BOTTOM SUBMATRICES.
IF <JJ .LT. II) GO TO 140CC — FORM SUBMATRIX (II,JJ). (!! CAN BE SPEEDED UP !!)
DO 130 1-1,3 DO 130 U-1,3 S—0.DO 120 K=l,2 C1-CT(I,K)KOK=KO+K DO 120 L-1,2
120 S-S+C: 1 *SK (K O K , L O + L ) C T (J, L )130 SM(I,J)-SC? IF (II .NE. JJ) GO TO 131 C? Kl-KO+l C? K2-K0+2 C? Ll-LO+l C? L2=L0+2C? WRITE (10,9100) II,JJ,K0,L0,((SK(I,J),J=L1,L2),I=K1,K2) C791G0 FORMAT (' SK /,2I4,5X,/ K0,L0 ',215,/,(1X,2E15.7))C7131 CONTINUE
CALL SAS (II,JJ,3)IF (II .EQ. JJ) GO TO 140 DO 135 1=1,3 DO 135 J-i,3 X=SM(I,J)SM(I,J)=SM(J,I)
133 SM(J,I)=XCALL SAS (J J , 11,3)
140 L0=L0+2150 K0=K0+2
RETURN
102
SUBROUTINE MTM6 C *** REWRITTEN AUGUST 1978 BY KEITH A. HUNTEN.CC DIAGONAL MASS MATRIX FOR LINEAR STRAIN TRIANGLE.C
REAL MASS(6)C
DIMENSION BC(3,5),LP(4),XL(3,6>,X0(3),LI(3),FL(3>DIMENSION DC1(3),DCX(3),DCY(3),DCZ(3),D(6,6),CT(3,2)DIMENSION SM(6,6),SMI(6),SM2(6),SMS(6),SM4(6),SM5(6),SMo(6) DIMENSION TC(3,6),TCI(3),TC2(3),TC3(3),TC4(3),TC3(3),TC6(3) COMMON /COMMND/L(60)iV(20)COMMON /PTS/NUiNS,ISTRKT,VC(3)COMMON /ELT/NEUiNESjISTRK2,IT,IORD,1ST,IAPL,NLDRC2,NCR,NGF 1 ,ISTPTR,NMAT,NTHS,FAC(3),WRKPD2(3),LCP(27)COMMON /MAT/MATPTR,MT,E,PR,G,STY,DEN COMMON /THS/ITYFE,IPTRCS,TH
CEQUIVALENCE (DC(1,1),DC1<1)),(DC(1,3),DCX(1),CT(1,1))EQUIVALENCE (DC(1,4),DCY(1)),(DC«1,5),DCZ(1))EQUIVALENCE (SM(1,1),SM1U>),(3M(1,2),SM2(D), (3MU,3),SM3(1)) EQUI VALENCE (SM (1,4), SM4 (1)), (SM (1,5), SM5 (1)),(SM(1,6), SM6 (U) EQUIVALENCE (TC(1,1),TC1(D),(TC(1,2),TC2(1)),(TC(1,3),TC3(1)) EQUIVALENCE (TC(1,4),TC4(1)),(TC(1,5),TC5(1)),(TC(1,6),TC6(1)) EQUIVALENCE (V(11),MASS(D)
CDATA LP/3,6,1,3/, EPS/1.E-5/, LI/5,4,2/
CCC *** READ POINT COORDINATES.
DO 5 1=1,NCR CALL INPTS (LCP(D)DO 5 J-1,3
5 TC(J,I)=VC(J)CC w * COMPUTE DIRECTION COSINES.C SIDE 3-o IN DC1, 6-1 IN DC2, 1-3 IN DCSC LOCAL X,Y,Z IN DCX,DCY,DCZC H IS HALF THE PERIMETER.
H=0.
DO 20 .J=l,3 3=0.K1=LP(J)K2-LP(J+1)DO 10 1=1,3 CB=TC(I,K2)-TC(I,K1)DC(I,J)=CD
10 5-S+CD**2S=SQRT (S)FL(v)=SH=H+SDO 20 1=1,3 DC(I,J)=DC(I,J)/S
20 CONTINUECALL CROSS (DCX,DC1,DCZ,XX)CALL CROSS (DCZ,DCX,DCY,XX)H=0.5*H
CC irr* COMPUTE AREA OF TRIANGLE.
A136=SQRT ( ( H-FL(1))*(H-FL(2))*(H-FL(3)))CC LOCAL ORIGIN, CREATE POINTS 2,4,5.
DO 30 1=1,3TC2(I)=0. 5*(T C1( I)+ TC3( D)TC4(I)=0. 5*(T C1( I)+ TC6( D)TC5(I)=0.5*(TC3(I)+TC6(I))
30 X0(I)=(TC1(I)+TC3(I)+TC6(I))/3.CC *** LOCAL COORDINATES.
Du 4v u— 1,6 DO 35 1=1,3 IP2-I+2 S-0,DO 34 K=l,3
34 S=S+DC(K,IF2)*(TC(K,J)-X0(K))XL(I,J)=S
35 CONTINUE40 CONTINUEC? WRITE (10,900) (J,(XL(I»J)»I=1»2),J=1,6)C7900 FORMAT ( LOCAL COORDINATES',/,(IX,14,5X,2E15.7))Cc CONSTRUCT 6X6 MATRIX TO OBTAIN COEFFICIENTS BY INVERSION.
DO 70 J=l,6 SM(J>=1.X=XL(1,J)Y=XL(2,J)SM2(J)=XSM3(J)=YSM4(J)=X*XSM5(J)=X*Y
104
S M 6 W ) = Y * Y 70 CONTINUE C
CALL INVNM (SH,6,0,6)C? WRITE (10,910) ((D(I,J),J-i,6),1-1,6)C7910 FORMAT (' D ',/,(1X,6E13.7))C
R I XX=A136*(XL(2,1)** 2 + X L (2,3)**2+XL(2,6)*$2)/12.RI XY-A 136 *(X L (1,1)* X L(2,1 )+XL(1,3)* X L (2,3)+ X L (1,6)*XL(2,6))/12.0 R I YY-A136*(X L (1,1)* * 2 + X L (1,3)**2+XL(1,6)**2)/12.
C? WRITE (10,940) A136,RIXX,RIXY,RIYY C7940 FORMAT (' A,IXX,IXY,IYY ',/,1X,4E12.4)CC *** COMPUTE ELEMENT MASS MATRIX.
DO 100 1,6S = A 136*0(1,N )+ R IYY* D(4,N )+R I X Y * D (5,N )+RIX X * D (6,N )
100 MASS(N)=A63 (S*DEN*TH)RETURNEND
105
SUBROUTINE PTM6 C UPDATED 11/10/78 BY KAHC REWRITTEN AUGUST 1978 BY KEITH A. HUNTEN.CC THERMAL FORCES FOR LINEAR STRAIN TRIANGLEC
DIMENSION DC1(3),DCX(3),BCY(3),DCZ(3),D(6,6),CT(3,2) DIMENSION T1(12,3),T2(3,6),DT(6),TDT(3),FL(3),SM(6,6) DIMENSION PL(12),SPL(12),L2(6),L3(6),LP(5),LI(3) COMMON /DAT/TC(3,27),NEL COMMON /TMP/TEMP(27)COMMON /TLD/VTLD(6,27)COMMON /LM6/DC(3,5),XL(3,6),X0(3)COMMON /MAT/MATPTR,MI,YM,PR,G,STY,DEN,THEXO COMMON /PAR/LHV(11),L G L (5),L P X (14),NDST COMMON /THS/ITYPE,IPTRC3,TH
CEQUIVALENCE (DC(1,1),D C 1 (1)),(DC(1,3),DCX(1),CT(1,1)) EQUI VALENCE (DC (1,4), DC: Y (1)), (DC (1,5), DCZ (1))
CDATA L2/0,2,4,6,8,10/, L3/0,3,6,9,12,15/DATA LP /3,6 ,1,3,6/, EPS/1.E-5/, LI/5,4,2/
CCC *#♦ COMPUTE DIRECTION COSINES.C SIDE 3-6 IN BC1, 6-1 IN DC2, 1-3 IN DCSC LOCAL X ,Y ,Z IN DCX,DCY,DCZC H IS HALF THE PERIMETER.
H=0.DO 20 J-1,3 S=0.K 1 < P ( J )K2--LP(.J+1)DO 10 1=1,3 CD=TC(I,K2)-TC(I,K1)DC(I,J)=CD
10 S=S+CD**2S=SGRT(S)FL(J)=SH=H+S
106
DO 20 1=1,3 DC(I,J)=DC(I,J)/S
20 CONTINUECALL CROSS (DCX,BC1,BCZ,XX)CALL CROSS (DCZ,DCX,DCY,XX)H=0.5*H
CC FIND AREA OF TRIANGLE.
A!36 = SQRT{N»(H-FL(1))*(H-FL(2))*(H-FL(3>))CC *** LOCAL ORIGIN, CREATE POINTS 2,4,5.
DO 30 1=1,3TC(I,2)=0.5*(TC(I,1)+TC(I,3)) TC(I,4)=0.5*(TC(I,1)+TC(I,6)) TC(I,5)=0.5*(TC(I,3)+TC(I,6))
30 X0(I)=(TC(I,l)+TC(I,3)+TC(I,6))/3.CC COMPUTE LOCAL COORDINATES OF CORNERS OF ELEMENT.
DO 4u J— 1,o DO 35 1=1,3 S=0.DO 34 K=l,3
34 S= S + D C (K , I + 2 )*(T C (K , J )- X O (K ))XL(I,J)=S
3 5 CONTINUE40 CONTINLECC CONSTRUCT 6X6 MATRIX TO OBTAIN COEFFICIENTS BY INVERSION.
DO 70 J = 1,6 SM(J,1)=1.X=XL(1,J)Y=XL(2,J)SM(J»2)=X SM(J,3)=Y SM(J,4)=X*X SM(J,5)=X*Y SM(J,6)=Y*Y
70 CONTINUECALL INVNM (SM,6,D,6)
CC *** PG=(THEX0$E*TH/(1-V))*T1*T2*D*THETA
CC=THEX0*YM*TH/(1.-PR)CC = = COMPUTE Tl.
J=1DO SO K=l,6Ji-J+iC1=D(2,K)C2=D(3,K)T K J , 1)=C1
10?T K J i i D - C 2 Cl=2.*D(4iK)C2-D(5iK)T1U,2}=C1T1 (J1 j2)-C2C1=2.*D(6,K)T1(J,3)--C2T1(J1,3)=C1J-J+2CONTINUE
FORM T2.A12-A136/12.Xi-XL(1»1)Yl-XL(2i1)X2=XL(1 j3)Y2-XL(2»3)X3=XL(1,6)Y3-XL(2»6)AXX=A12*(X1**2+X2**2+X3**2)AX Y-A1 2v(X i*Y1+X2*Y2+X3*Y3)AY Y = A 12*(Y 1* * 2+Y2**2+Y3**2)A30=A136/30.AX XX=A30*(2.#(X1*$3+X2#*3+X3*#3)-3.*Xl*X2*X3) AX XY=A 30*(2.*(X 1** 2 $ Y 1+X2**2*Y2+X3**2$Y3)
1 -(X1*X2*Y3+X2*X3*Y1+X3*X1$Y2))AXYY-A30*(2.*(X1*Y1**2+X2*Y2**2+X3*Y3**2)
1 -(XI*Y2*Y3+X2*Y1*Y 3+X3*Y1*Y2))AYYY=A30*(2.*(Yl**3+Y2$*3+Y3**3)-3.*Y1*Y2*Y3) T2(1,1)=A136 T2(2,l)=0.7 2 ( 3 , 1 ) 0 .7 2 ( 1 , 2 ) 0 .72(2,2)-AXX72(3»2)-AXY7 2 ( 1 , 3 ) 0 .72(2,3)=AXYT2(3,3)-AYY72(1,4)-AXX72(2,4)=AXXXT2(3,4)=AXXYT2(1,5)=AXY72(2,5)=AXXYT2(3,5)=AXYYT2(1,6)=AYY72(2,6) O X YYT2(3,6)=AYYY
LOOP THROUGH LODINO CASES.DO 200 LCASE = l.NDST
108CALL INTMP (NEL?LCA3E)
CC --- D*THET m .
DO 90 1=1,65=0 •DO 85 0=1,6
85 S=S+D(I,J)*TEMP(J)90 DT(I)=SCC T2*D$THETA.
DO 100 1=1,3 3=0.DO 95 J=l,6
95 S=S+T2(I,J)*DT(J)100 TDT(I)=SCC = = FL=CC*T1*T2*D*THETA.
DO 110 1=1,12 3=0.DO 105 J=l,3
105 S=S+T1(I,J)*TDT(J)110 PL(I)=3*CCCC === GLOBAL FORCES.
10=0J0=0DO 120 N=l,6 UP=PL(.J0+1)VP=PL(.J0+2)DO 115 1=1,3
115 VTLDiI.N)=UP*DCX(I )+ V P*DC Y(I ) 120 00=00+2200 CALL OUTLD (NEL,LCASE)
RETURNEND
C
109
SUBROUTINE STM6 C *** LAST UPDATED DECEMBER 27,1978 BY KAH C *** REWRITTEN AUGUST 1973 BY KEITH A. HUNTEH.CC STRESSES FOR: LINEAR STRAIN TRIANGLE.C
DIMENSION X Y EP(2,3),T T T (18),E F T (3)DI MENS IGN DC1 (3), BCX (3), BCYI3), DCZ (3) ,0(6,6), C T O , 2), SM( 6,6/ DIMENSION TEPI9,12),S F L (12),L2<6),L3 <6),STRi9),LP(5),L I (3)COMMON /D AT/TO(3,12),TX(3,12),TY(3,12),FL(3),SM1(6),SM2(6),SM3(6)
1 ,SM4(6),SM5(6),SM6(6)COMMON /LM6/DC(3,5),XL(3,6),X 0 ( 3 )COMMON /MAT/MATPTR,MT,E,PR,G,STY,DEN,THEXCO COMMON /'PAR/LHV (11), LGL (5), LPX (14), NDST COMMON /D TS/TC(3,27),S P (36),S F X (6,10),ST(7,10),LCASE,NSTPT
1 ,ITFLG,NELCOMMON /TMP/TEMP(27)
CEQUIVALENCE (BC(1,1),D C 1 (1)),(BC(1,3),DCX(1 )»CT(1,1))EQUIVALENCE (D C ( 1,4),D C Y I 1) ),(DC(1,5),D C Z (1))EQUIVALENCE (S M1(1 ),S M(1, D)EQUIVALENCE (S P X (1,1),S F L (1)),(S P X (1,3),S T R (1))
CDATA 12/0,2,4,6,8,10/, 13/0,6,12,13,24,30/DATA LP/3,6,1,3,6/, EPS/l.E-5/, LI/5,4,2/
CCC COMPUTE DIRECTION COSINES.C SIDE 3-6 IN DC1, 6-1 IN DC2, 1-3 IN DCSC LOCAL X,Y,Z IN DCX,DCY,DCZC H IS HALF THE PERIMETER.
H-0.Du 20 J— 1,3 S-0.Kl-LP(J)K2-LP(J+1)DO 10 1-1,3 CB--TC(I,K2)-TC(I,K1)DC(I,U)=CD
10 S-SrCD**2
110
3=SQRT (S)FL(J)=3H-H+SDO 20 1=1,3 DC(I,J)=DC(I,J)/3
20 CONTINUECALL CROSS (DCX,DC1,DCZ,XX)CALL CROSS (DCZ,DCX,DCY,XX)H=0.5*H
CC *** COMPUTE AREA OF TRIANGLE.
A 13&=SGRT (H*(H-FL(1))*(H-FL(2))*(H-FL(3)))CC LOCAL ORIGIN, CREATE POINTS 2,4,5.
DO 30 1=1,3TC(I,2)=0.5*( TC( I,1 )+TC U,3 ))TC(I,4)=0.5*(TC(I,1)+TC(I,6))TC(I,5)=0.5*(TC(I,3)+TC(I,6))
30 X0(I)=(TC(I,1)+TC(I,3)+TC(I,6))/3.C COMPUTE LOCAL COORDINATES OF ELEMENTS CORNERS.
DO 40 J=l,6 DO 35 1=1,3 S=0.DO 34 K=l,3
34 S=S+DC(K,I+2)*(TC(K,J)-X0(K})XL(I,J)=S
35 CONTINUE40 CONTINUEC? WRITE (10,910) (J,<XL(I,J),I=1,3),J=1,6>C?910 FORMAT (y LOCAL XYZ ',/,(lX,I3,3E12.4))Cij v** COMPUTE TH COORDINATES OF THE STRESS POINTS,
DO 60 0=1,2X YEP (J, 1 )=(XL (J, 1)+XL (J, 2)+XL (J, 4)) /3.X Y EP(J,2) =(X L (J,2)+ X L (J ,3)+ X L (J,5))/ 3 .X Y EP(J,3)=(X L (J,4 )+XL(J,5)+XL(J,6))/3.
60 CONTINUECC w * CONSTRUCT 6X6 MATRIX TO OBTAIN COEFFICIENTS BY INVERSION.
DO /0 J— 1,6 SM1(J)=1.X=XL(1,J)Y=XL(2,.j)SM2(J)=X 3M3(J)=Y SM4(J)=X*X SM5(J)=X$Y SM6(J)=Y*Y
70 CONTINUECALL INVNM (SM,6,D,6)
Ill
C? WRITE (10,930) ( ( D ( I , I = l , o )C?930 FORMAT (' 0 ',/,(lX,6E12.5))CC TEP = TO + TX.X + TY.YC COMPUTE TO,TX,TY
J— 1DO 30 K-1,6
C1=D(2,K)C2-D(3,K)T 0 ( 1 , J ) C 1T 0 (1 jJ1)=0.T0(2iJ/-0.T0(2, J D - C 2 70(3,J ) C 2 T0(3,J1)=C1 C1=2.*D(4,K)C2-D(5,K)TX(1,J)-C1 TX(1,J1)=0.TX(2,J)=0.TX(2,J1)-C2TX(3,J)=C2TX(3,J1)=C1C1-2.*D(6,K)TY(1,J)-C2TY(1,J1)=0.TY(2,J)-0.TY(2,J1)<:1 TY(3,J)-C1 TY(3, J D - C 2 J-J+2
30 CONTINUEC? WRITE (10,950) ((T0(I,J),J=1,12),I=1,3),((TX(I,J),J=1,12) C? 1 ,1=1,3),((TY(I,J),J-1,12),1-1,3)C7950 FORMAT (' TOXY ',/,(1X ,6 E 1 1.3))CC *** FORM E MATRIX COEFFICIENTS.
EF-E/(1.-FR**2)VEP=PR#EP
CC *rif ESTABLISH TEP FOR 3 STRESS POINTS.
NSTPT=3I0=-3DO 100 NS P = 1 ,N3TPT X=XYEP(l,iNSP)Y=XYEP(2,NSP)10=10+3
C? WRITE (10,941) NSP,K,IO,X»YC7941 FORMAT (/,' NSP,K,IO,X,Y ,3X,3I3,3X,2E12.6)
112
DO 95 1=1,3 DO 95 J = i , 12
95 T E P (IO+I,J ) = T O (I,J )+X*TX U ,J )+Y*TY(I,J)C? WRITE (10,942) ((TEP(I0+I,J),J=1,12),1=1,3)C7942 FORMAT (' TEP',/,(1X,6E12.5))100 CONTINUE CC *** i f TEMPERATURES ARE PRESENT, ESTABLISH TEMPERATURE C INTERPOLATION MATRIX FOR THE STRESS POINTS.
IF (ITFLG .E3. 0) GO TO 105 K0=1DO 106 NS r-1,N3TPT X=XYEP(1,NSP)Y=XYEF(2,NSr)TTT(K0)=1.TTT(K0+1)=XTTT(K0+2)=YTTT(K0+3)=X*XTTT(K0+4)=X*YTTT(K0+5)=Y*YK0=K0+6
106 CONTINUE K0=-6DO 108 K=l,3 K0=K0+6 DO 108 1=1,6 TTTT=0.DO 107 J=l,6
107 TTTT=TTTT+TTT(KO+I)*0(I,J)103 TTT(KO+I)=TTTTCCC m * LCCiP THROUGH LODI NO CASES 105 DO 200 LCASE= 1,HOST
CALL ELTBNS C? WRITE (10,900) (SP(I),1=1,36)C?900 FORMAT (' SP ',/,(1X,3E15. 8,3F 5.D )CC = = OBTAIN LOCAL DEFLECTIONS.
DO 130 N=l,6 J0=L3(N)I0=L2(N)DO 120 1=1,2 3=0.DO 110 0=1,3
110 S=S+CT(J,I)*SP(JO+J)120 SPL(IOrI)=S 130 CONTINUEC? WRITE (10,944) ( S P L d ),!=!, 12)C7944 FORMAT C SPL ',/,(1X,2E13.8))
113
c — OBTAIN STRAINS.BO 140 I-li9 S^O.DO 135 J - 1 j 12
135 S^S+TEP(IiJ)*SPL(J)140 S T R I D E SC? WRITE (10,920) ( S T R ( I ) , M , 9 )C7920 FORMAT C ST RAINS',/,(IX,3E15.7))CC v** IF TEMPERATURES ARE PRESENT,CALCULATE AND C SUBTRACT THE THERMAL STRAINS.
IF (ITFLG ,EQ. 0) GO TO 143CC > » GET TEMPERATURES FOR THIS LBCA3E.
CALL INTMP (NELiLCASE)CC > » GET THE TEMPERATURES AT THE STRESS POINTS.
KG--6DO 143 1-1,3KO=KO+6TEMPT=0.DO 142 J-1,6TEMPT-TEMPT+TTT(K O + J )* T E M P (J )
142 CONTINUE EPT(I)=TEMPT
143 CONTINUE CC > » C A L C U L A T E STRAINS AND SUBTRACT THEM.
K0=1DO 144 1=1,3 TSTF:N=THEXCO*EPT(I)S T R (K O )= S T R (K O )-TSTRN S I R (K O + 1 )= S T R (K O + 1 )-TSTRN K0=K0+3
144 CONTINUE CC = = OBTAIN STRESSES.148 10=0
•JO—0DO 150 N=1,NSTPTS T 1= E P*ST R(J 0 + 1)+V EP*S TR(J O + 2 )ST2=VE P*S TR(J O + 1 )+E P * S T R (J O + 2 )ST 3=G* STR(J O + 3 )ST(1,N)=ST1ST(2,N)=ST2ST(3,N)=ST3ST(4,N)=ST1ST(5,N)=ST2ST(6,N)=ST3
114
uO-vO+3150 S T ( 7 , N )-100.w'SQRT (ST 1**2+ST2**2-ST 1 *312+3.*573**2)/STYC? WRITE (10,1010) ((ST( 11, J J ) , 11-1,7), J J - l , nSTF'T)C7 1010 FORMAT (/," STRESSES",/,(3E10.2))200 CALL ELTSTR
RETURN END
115
SUBROUTINE K0A9 C WRITTEN NOVEMBER, 1978 BY KEITH A. HIMTENCC STIFFNESS OF AN AXISYMMETRIC BI-QUADRATIC ISOPARAMETRICC QUAORALATERAL MEMBRANE.C
DIMENSION GC(3),GW(3),F(9>DIMENSION E(3,3),TC2(3),TC4(3),TC5(3),TC6(3),TCS(3)COMMON /DAT/SK(13,18),DC(3,3),TLC(3,?),TEP(4,18),FSI(9),FTI(9) 1 ,F.J(2,2),FJI(2,2)COMMON /ORDER/E11,E21,E31,E12,E22,E32,E13,E23,£33 1 ,F1,F2,F3,F4,F5,F6,F7,F3,F9COMMON /BKF/MAP(3,27),TCI3,27),SMI6,6)COMMON /MAT/HATPTR,MT,YM,FR,G COMMON /TH3/ITYFE,IPTRCS,TH
CEQUIVALENCE (FI,FID)EQUIVALENCE (TC2(1),TC(1,2)),(TC4(1),TC(1,4)),(TC5(1),TC(1,5)) EQUIVALENCE ITC611),TCI 1,6)),ITC311),TCI 1,3)),(El1,E(1,1))
C DATA GC/-.774596669, 0. ,.774596669/DAT A GW/ .555555556,.888888889,.555555556/DATA EP3I/1.E-5/, TWOPI/6.283135307/
CCCC *** GENERATE NORMAL PORTION OF ELASTICITY MATRIX.
EP-YM/III.+FR)v(1.-2*PR))CC=PR*EPEP-11.-PR)vEPE11=EPE22-EPE33-EPE12-CCEl 3 4 %E21-CCE23-CCE31-CCE32-CC
C
116
C *** CLEAR UPPER HALF OF STIFFNESS MATRIX.DO 40 .>1,18 DO 40 1=1,J
40 SK(I,J)=0.CC *** (3X3) GAUSSIAN INTEGRATION OF ELEMENT STIFFNESS MATRIX.
DO 70 1=1,3 SS=GC(I)W1=GW(I)SSP=(l.+SS)SSM-(l.-SS)3S2-2.*SS SS2F-1.+5S2 SS2M=1.-3S2 SSSQ=1.-SS*SS DO 70 >1 ,3 TT=GC(J)TTP=(1.+TT)TTM=(1.-TT)TT2=2.*TTTT2F-1.+TT2TT2M=1.-TT2TTSi>l.-TT*TT
CC » > SHAPE FUNCTIONS.F1=0.25*SS*SSM*TT*TTM
F2=-0.50*SSSQ*TT*TTMF3=-0.25*SS*SSP*TT*TTMF4=-0.5*SSM*TTSQ*SSF5=SSSQ*TTSUF6=0.5*SS*SSP*TTSQF7=-0.25*SS*SSM*TT*TTPF3=0.5*SSS8*TT*TTPF9=0.25*SS*SSP*TT*TTP
CC » > DERIVATIVES OF THE SHAPE FUNCTIONS.
FSI(1)=0.25*SS2M*TT*TTM FSI(2)=SS*TTM*TT FSI(3)=-0.25*SS2P*TT*TTM FSI(4)=-0.5*SS2M*TTSQ FSI(5)=-SS2*TTS9 FSI(6)=0.3$SS2P*TTSQ FSI(7)=-0.25*SS2M*TT*TTP FSI(8)=-SS*TT*TTP FSI(9)=0.25*SS2P*TT*TTP FTI(1)=0.25*SSM*SS*TT2M FTI(2)=-0.3*SSSG*TT2M FTI(3)=~0.25*SS*SSP*TT2M FTI(4)=SS*SSM*TT FTI(5)=-SSSG*TT2
FT1(6)-~S3*S8P*TT FTI(7)^-0.25*SS#SSM*TT2P FTI(8)0.5*SSSG*TT2P FTI(9)=0.25*SS*SSP*TT2P
CC GET THE CURRENT RADIAL COORDINATE.
R-0.DO 44 KR-1,9
44 R-R+F (KR) * TC (I ? KR)CC w# COMPUTE JACOB I AN IN yFJ/.
S14).3 2 0 .DO 45 K=l,9 X-TC(1?K)51=S1t F3I(K)*X
45 S2=S2+FTI(K)*X FJ(1,1)=S1 FJ(2il)-S2 31=0.82=0.DO 50 K=l,9 Y=TC(2,K)Sl=Sl+FSI(K)rY
50 S2=S2+FTI(K)*YFJ(1,2)=S1 FJ(2,2)=S2
CC INVERT JACOBIAN IN TJI'.
DET--FJ (1,1)*FJ(2,2)-FJ(1,2)*FJ(2,1) REC=1./DETFJI(M)=FJ(2,2)*REC FJI(2,2)=FJ(1,1)*REC FJI(1? 2)=-FJ(1»2)*REC F.JI(2,l)=-FJ(2,l)xREC
CC « ♦ COMPUTE TEPISSLTTJ).
XX=FJI(lil)XY=FJI(2il)EXX=FJI(1,2)EY=FJI(2>2)DO 55 01=1,9J2P=2*J1J2=J2P-1TEP(1,J2)=XX*FSI(J1)+EXX*FTI(J1)TEP(1,J2F)=0.TEP(2,J2)=0.TEP(3,J2P)=0.TEP(2,J2P)=XY*FSI (JD+EN'xFTI (J1) TEP(3,u2)=F(.Jl)/R
118
TEP(4 7 J2)-TEP(2»J2P)55 TEP(4 ? J2P)-TEP(i > J2)
W2-W1m (J )*ABS(BET)$TWOPI*R E«3 60 K=l,3 DO 60 LL=1,3
CC » > MULTIPLY ELASTICITY MATRIX BY M2.
M3-W2^E(K»LL)CC » > COMPUTE TEP)T E TEP FOR THE NORMAL PORTION OF E.
DO 60 J J - M 8 W4-W3»TEP(LLjvU)DO 60 11=1,JJ
60 SK (11 , JJ) =SK (1 1, JJ) +W4*TEP (K, 1 1)CC » > MULTIPLY SHEAR PORTION OF THE E MATRIX BY M2,C AND MULTIPLY INTO STIFFNESS MATRIX.
M3=G*M2 DO 65 JJ=1,18 M4=W3*TEP(4,JJ)DO 65 11=1,JJ
65 SK'dl, JJ)=SK( II, JJ)+M4*TEP(4, II)70 CONTINUEC? WRITE (12,920) (SK(1,1),1=1,13)C?920 FORMAT i' SK',/,(1X,6E12.5))CC iHHr UPPER HALF OF ELEMENT STIFFNESS READY. COPY LOWER HALF C OF DIAGONAL SUBMATRICES.
DO 75 1=1,17,2 1 1= 1+1
75 SK(I1,I)=SK(I,I1)CC *** ASSEMBLE ELEMENT STIFFNESS (UPPER HALF) INTO MASTER STIFFNESS,C ASSEMBLING LOWER SUBMATRICES BY TRANSPOSITION.
DO 95 IP=1,9 K0=2*IP-2 DO 90 IQ=IP,9 L0=2*IQ-2 DO SO K=l,2 KOK=KO+K DO SO L=l,2 LOL=LO+L
SO SM(K,L)=SK(KOK,LOL)CALL SAS (IP,IQ,2)IF (IP .EQ. IQ) GO TO 90
Cc = = AoSEBLE lowe r s u b m a t r i x.DO 89 1=1,2 DO 39 J=I,2 THP=SH(I,J)
119Srlu,J)-SH(J,I)
89 SMIJiD-TfFCALL SA3 (IQ,IP,2)
90 CONTINUE 95 CONTINUE
RETURNEND
C
120
SUBROUTINE FQA9 C WRITTEN NOVEMBER, 1973 BY KEITH A. HUNTERCC THERMAL FORCES FOR AN AXISYnETRIC BI-QUADRATICC ISOPARAMETRIC QUADRALATERAL ELEMENT.C
DIMENSION OC(3),GW(3),STR(12),SPL(IS)DIHENSION TC2(3),TC4(3),TC5(3),TC6(3),TC3(3)COMMON /DSQ9/TEP(27,13),F9(9,9),WF(9),FSI(9)1 ,FTI(9),FJ(2,2),FJI(2,2),E11,E21,E12,E22COMMON /MAT/MATFTR,MT,YM,PR,0,STY,DEN,THEX COMMON /THS/1TYFE,IPTRCS,TH COMMON /PAR/LHV(11),LOL(5),LPX(14),NDST COMMON /DAT/TC(3,27),N£L COMMON /TMP/TEMP(27)COMMON /TLD/VTL0(6,27)
CEQUIVALENCE (TC2(1),TC(1,2)),(TC4(1),TC(1,4)),(TC5(1),TC(1,5)) EQUIVALENCE (TC6(1),TC(1,6)),(TCS(1),TC(1,8))
CDATA GC7-.774596669,0.,.774596669/DATA ON /.555555556,.3SSS38S39,.555555556/DATA TWOPI/6.2331853171/
CCCC r** COMPUTE THERMAL CONSTANT.
CC-THEX*YM#TwOPI/(1.-(2.*FR))CC ESTABLISH TEP, F9 & WF FOR 9 GAUSSIAN POINTS.
KO-1DO 70 J-1,3 WJ-GW(J)TT=CC(J)TTP-(1.+TT)TTM-(l.-TT)TT2-2^TT TT2P-1.+TT2 TT2M-1.-TT2 TTSQ-1.-TT*T7
FT1-0.5*TT*TTM FT2-TTSQ FT3=0.5*TT*TTP DO 70 1-1,3 SS-GC(I)WI-GW(I)SSF-d.+SS)SSf^(l.-SS)SS2-2*SS S32P-1.+3S2 SS2M-1.-3S2 SSSQ-1.-SS*S3 FSl-0.5*SS*S311 F32-SSS8 FS3-0.5*SS*SSP
CC » > SHAPE FUNCTIONS.
F9(K0,1) -0.23#SS*SSI1*TT *TTM F9(K0,2)--0.5*SSS0*TT*TTM F9(KO,3)=-0.25*SS*SSF*TT*TTM F9(K0,4)--0.3*SSM*TTSQ*SS F9(K0»5)-SSSQ*TTSQ F9(K0,6)-0.5*SS*SSP^TTSQ F9(KO,7)=-0.25*SS*SSM*TT*TTP F9(KO,8)=0.5*SSSQ*TT*TTP F9(K0,9)=0.25*SS*SSP*TT*TTP
CC » > DERIVATIVES OF THE SHAPE FUNCTIONS.
FSI(1)=0.25*SS2M*TT*TTM FSI(2)=SS*TTM*TT FSI(3)=-0.25*SS2P*TT*TTM FSI(4)=-0.5*SS2MmSQ FSI(5)--SS2*TTSQ FSI(6)4).5*SS2P*TTSQ FSI(7)=-0.25*SS2M*TT*TTP FSI(8)=-SS*TT*TTP FSI(9)^0.25*SS2P*TT*TTP FTI(1)%0.25*SSM*SS*TT2M FTI(2)-0.5*SSS0*TT2H FTI(3)--0.25»SS»SSP*TT2M FTI(4K=SS*SSM*TT FTI(5)=-SSSQ*TT2 FTI(6)=-SS*SSP*TT FTI(7)-~0.25*SS*S3M*TT2P FTI(3I-0.3*SSSQ#TT2P FTI(9}-0.25*SS*SSP*TT2P
CC »> COMPUTE JACOBIAN AND ITS' INVERSEC AND ITS' DETERMINATE.
S1=0.
122
32=0.DO 45 K=l,9 X=TC(hK)S1=S1+FSI(K)*X
45 S2--32+FTI(K)*XFJ(1,1)=S1 FJ(2,1)=S2 31=0.32=0.DO 50 K=1j9 Y=TC(2,K)Sl=Sl+F3I(K)xY
50 S2=S2+FTI(K)*Y FJ(1,2)=S1 FJ(2»2)=S2
CCC xH- GET THE CURRENT RADIAL COORDINATE RCGRD.
RCGRD=0.DO 53 KR-1,9
53 RCGRD=RC0RD+F9(KO> KR)*TC(1,KR)CC INVERT JACOBIAN IN FJI'.
BET=FJ(1,1)*FJ(2,2)-FJ(1,2)$FJ(2,1)REC=1./DETFJI(l,i)=FJ(2,2/*RECFJI(2,2)=FJ(M)*RECFJI(l,2)=-FJ(l,2)xRECFJI(2,1)=-FJ(2,1)*REC
CC *** COMPUTE TEP(33I»TTJ).
XX=FJIUi1/XY=FJI(2,1)EXX=FJI(1,2)EY=FJI(2,2)DO 55 Jl=l,9J2P=2*J1J2=J2P-1TEP(K0,J2)=XX*FSI(J1)+EXX*FTIW1) TEP(K0+9jJ2P)=XY*FSI(J1)+EYxFTI(Ji)TEP(K0+18,J2)=F9(KO,Jl)/RCORB TEP(K0,J2P)=0.TEP(K0+9,J2)=0.TEP(K0+18,J2P)=0.
55 CONTINUEWFiKO)=CO*NI*WJ*ABS(DET)&RCGRD KO-KO+1
70 CONTINUECC *** LOuP THROUGH LOADING CA8ES.
123
DO 200 LCASE-i, MOST CALL INTMP (NELiLCASE)
CC OBTAIN LOCAL FORCES.
DO 105 1-1,18 105 SPLdJ^O.CC +++ LOOP THROUGH GAUSSIAN POINTS.
DO 150 NG-1,9CC t H- obtain t e mp erature at p o i n t.
THETA =0.DO 110 1=1,9
110 THETA=THETA+F?(NG,I)*TEMP<I)CC +++ COMPUTE LOCAL FORCE CONTRIBUTION.
DO 120 1=1,18120 SFKI)=SPL(I)+WF(NG)*THETA*(TEP(NG,I)+TEP(NG+9,1)+TEP(NG+18,1)) 150 CONTINUE CC > » OUTPUT THERMAL LOADS TO DISK.
10=1DO 160 N=l»9 VTLD(1,N)=SPL(I0)VTLD(2,N )=SPL(I0+1)10=10+2
160 CONTINUE200 CALL OUTLB (NEL,LCASE)
RETURNEND
124
SUBROUTINE SQA9 C *** WRITTEN NOVEMBER,1978 BY KEITH ft. HUNTEN.CC STRESSES FOR AN AXISYMMETRIC BI-QUADRATIC ISOPARAMETRICC CuABRALATERAL .C
DIMENSION GC(2),STR(16),I0RDR(13),SPL(13)DIMENSION TEP(16,18),FSI(9),FTI(9),FJ(2,2)DIMENSION FJI(2,2),TTT(4,9),TSTRN(4)
CCOMMON /MAT/MATPTR,NT, YM,PR,G,STY,DEN,THEXCQ COMMON /FAR/LHV(11),LOL(5),IPX(14),NDST COMMON /DTS/TC(3,27),SP(54),SPX(6,7),ST(7,10),LCASE,NSTPT 1 ,ITFLG,NELCOMMON /TMP/TEMP(9)
CEQUIVALENCE (3PXU, 1),SFL(1))
CDATA GC/-.5,.5/DATA IORDR/1,2,7,3,13,14,19,20,25,26,31,32,37,33 1 ,43,44,49,50/
CCCC *** COMPUTE ELASTIC CONSTANTS.
EP=YM/((1.+PR)*(1.-2.*PR))CC=PR*EPE F - d . - P R / v E P
CC w* ESTABLISH TEP' FOR 4 STRESS POINTS.
MSTPT-4KO-ONSP-1DO 70 J-1,2 TT=GC(J)TTP=(1.+TT)TTM-(l.-TT)TT2-2.#TTTT2P-1.+TT2TT2M=1.-TT2
123
TTSQ-1.-TT*T7 DO 70 1=1,2 SS=GC(I)SSP=(1.+SS)SSM=(1.-SS)SS2=2.#SSSS2P=1.+SS2352r1=l.-SS2SSSQ=1.-S3*SS
CC » > FORM INTERPOLATION FUNCTIONS.
TTT(NSP,1)=0.25*SS*S3M*TT*TTM TTT(NSP,2)=-0.5*SSSG*TT*TTM TTT (NSP, 3)=-0.25*SS*SSF'#TT*TTM TTT(NSP,4)=-0.5*SSM*TTSQ*SS TTT(NSP,5)=SSSQ*TTSQ TTT(NSP,6)=0.5*SS*SSP*TTSQ TTT (fiSP, 7)=-0.25*SS*SSM*TT*TTP TTT(NSP,8)=0.5*3SS8*TT*TTP TTT(NSP,9)=0.25*SS*SSP*TT*TTP
CC »> FORM THE DERIVATIVES OF THE INTERPOLATION FUNCTIONS.
FSI(1)=0.25*SS2M*TT*TTM FSI(2)=SS*TTM*TT FSI(3)=-0.25*SS2P*TT*TTM FSI(4)=-0.5*3S2M*TTSQ FSI(5)=-SS2#TTSQ FSI(6)=0.5*SS2P*TTSQ FSI(7)=-0.25*SS2M$TT*TTP FSI(8)=-SS*TT*TTP FSI(9)=0.25*SS2P*TT$TTP FTI(1)=0.25*SSM*SS*TT2M FTI(2)=-0.5*SSSQ*TT2M FTI(3)=-0.25*SS*SSP*TT2M FTI(4)=SS*SSM*TT FTI(5)=-SSSQ*TT2 FTI(6)=-SS*SSP$TT FTI(7)=-0.25*SS*SSM*TT2P FTI(8)=0.5*SSSG*TT2P FTI(9)=0.25*SS*SSP*TT2P
CC w* GET THE CURRENT INTEGRATION POINTS' RADIAL COORDINATE R.
R=0.DO 44 KR=1,9
44 R=R+TTT(NSP,KR)*TC(1,KR)CC x** COMPUTE JACOBIAN IN 'FJ .
S1=0.32=0.DO 43 K-1,9
126
X-TCdjK)S1-S1+FSI(K)*X
45 S2-S2+FTI(K)xXFJ(1,1)=S1 FJ(2il)-S2 SM.S'2-0.DO 50 K-1,9 Y-TC(2»K)S1-S1+FSI\K)*Y
50 S2-S2+FTI(K)*YF J ( 1 j2)-S1 F J ( 2 j2)-S2
CC *** INVERT JACOBIAN INTO 'FJI'.
DET-FJ(1,1)*FJ(2,2)-FJ(1,2)*FJ(2,1) REC-l./DETFJI(1,1)=FJ(2,2)*REC F J I «2 j 2)- F J (1»1)vREC FJI(I,2)=-FJ(1,2)#REC F J I (2 >1)= - F J (2»1)*REC
CC COMPUTE TF.P FOR THE FOUR EVALUATION POINTS.
XX=FJI(1j1)XY=FJI(2,1)EXX=FJI(1,2)EY-FJI(2»2)DO 55 Jl=l,9J2F-2*J1J2=J2P-1Kl-KO+1K24C0+2K3=K0+3K4=K0+4TEP(K1>J2)-XX«FSI (J1 )+EXX->FTI(Jl) TEP(K1,J2P)=0.TEP(K2iJ2)-0.TEP(K3iJ2P)=0.TEP(K2,J2P)-XY*FSI(J1)+EY*FTI(Ji)TEP(K3,J2) TTT(NSP,J1)/RTEP(K4»J2)-TEP(K2> J2P)
55 TEP(K4,J2P)=TEP(K1,J2)K0-K0+4 NSP=NSP+1
65 CONTINUE 70 CONTINUE CC LOOP THROUGH LOADING CASES.
DO 200 LCASE=1iNDST CALL ELTDNS
127
Cc =zu PACK THE DEFLECTIONS.
DO 130 N-M3 NN-IORDR(N)SFL(N)-SP(NN)
130 CONTINUE CC OBTAIN STRAINS.
DO 140 1=1,16s=o.DO 135 0=1,13
135 S=S>TEP(I,.j)*SFL(J)140 STR(I)=SC »> IF TEMPERATURES ARE PRESENT, CALCULATE AND C SUBTRACT THERMAL STRAINS.
IF (ITFLG .EG. 0) GO TO 149CC »> GET THIS LDCASES TEMPERATURES AND CALCULATEC TEMPERATURE AND THERMAL STRAIN AT EACH OFC THE STRESS POINTS.
CALL INTMP (NEL,LCASE)DO 142 ISP=1,4 SPT=0.DO 141 1=1,9SPT=SPT+TEMP(I)*TTT(ISP,I)
141 CONTINUETSTRNIISP)=SPT*THEXCO
142 CONTINUECC »> SUBTRACT THERMAL STRAINS.
K0=1DO 143 ISP=1,4STR(KO)=STR(KO)-TSTRN(ISP)STR(KO+1)=STR(KO+1)-TSTRN(ISP)STR(KO+2)=STR(KO+2)-TSTRN(ISP)KO=KO+4
143 CONTINUE Ccc OBTAIN STRESSES.149 J0=0
DO 150 N=1,NSTPTST(1, N)=EP*STR(JO+1)+CC*(STR(UO+2)+STR(JO+3))S T (2,N )=E P * S T R (J O + 2 )+ C U * (S T R (J O + 3 )+ S T R (J 0 + 1))ST(3,N5 =EP*STR(JO+3)+CC#(STRIJO+1)+STR(JO+2)) ST(4,N)=G*STR(J0+4)J0=J0+4STT=(ST(l,N)-ST(2,N))**2+(ST(2,N)-ST(3,N))**2+2 1 +(ST(1,N)-ST(3,N))**2
150 ST(7,N)=100.*SGRT(.5+STT+3.*ST(4,N)*+2)
CALL ELTSTRRETURNEND
129
SUBROUTINE KTA6 C w * WRITTEN NOVEMBER,1978 BY KEITH A. HUNTEN CC STIFFNESS FOR AN AXIS'/METRIC LINEAR STRAIN TRIANGLE.C
DIMENSION D (B , -A), CT (3,2), TEP (4,12), W (4), A L F A 1 i 4)DIMENSION ALFA2(4),ALFA3(4),EM(3,3),311(6,6)DIMENSION SKI 12,12)COMMON /BKF/MAP(3,27),T O (3,27),S M I (6),S M 2 (6),S M S (6),S M 4 (6)
1 ,SM5(6),SM6(6)COMMON/EM/E11,E 2 1 ,E 3 1 ,E12,E22,E32,E13.E23,E33 COMMON /MAT/MATPTR,MT,E,PR,G COMMON /THS/ITYPE,IPTRCS,TH
CEQUIVALENCE (Snl(l),SM(l,i)), ( E 1 1,E M(1 ,D)
CDATA ALFA1/.333333333,.733333333,.133333333,.133333333/ DATA ALFA2/.333333333,.133333333,.733333333,.133333333/ DATA ALFA3/.333333333,.133333333,.133333333,.733333333/ DATA W/-.5265,3*.520833333/, T W O P 1/6.283185307/
CCCC *** GENERATE NORMAL PORTION OF ELASTICITY MATRIX.
EP-E/ ((1. + P R ) * (1. - 2 * P R ))OC=PR*EPEP-(1.-PR)*EPEll^EPE22-EPE33-EPE12-CCE13-CCE21-iXE23-CCE31-CCE32--CC
CC *## CREATE POINTS 2,4,5.
DO 30 1-1,2T C ( 1 ,2)-0.5*< TCI 1, 1 )+TC(I»3))
130
TC(I»4)-0.5*(TC(Ii1 )+TC(Ii6)) TC(I,5)=0.5*(TC(I,3)+TC(I,6))
30 CONTINUECCC w * CONSTRUCT 6X6 MATRIX TO OBTAIN COEFFICIENTS BY INVERSION.
DO 70 .>1,6 S M h ' J M .R-TC(1,J)Z-TC(2,J)S M 2 U ) = R S M 3 U N SM4(J)-R*R SM5(J)-R#Z SH6(J)-Z^Z
70 CONTINUECALL INVNM (SM,6,D,6)
CC ZEROIZE THE TOP HALF OF THE STIFFNESS MATRIX.
DO 50 1=1,12 DO 50 .J=I,12 SK(I,.J)=0.
50 CONTINUECC LOOP THROUGH INTEGRATION POINTS, GENERATING STIFFNESS.
DO 100 IP=1,4CC » > CALCULATE CURRENT INTEG. FT.'S COORDINATES.
R-ALFA1(IP)*TC(1,1)+ALFA2(IP)*TC(1,3)+ALFA3(IP)*TC(1,6) Z=ALFA1(IP)*TC(2,1)+ALFA2(IP)*TC(2,3)+ALFA3(IP)*TC(2,6)
CC > » FORM THE STRAIN-DISPLACEMENT MATRIX TEP.
J-lDO 80 n - 1 , c*J1=J+1 TEP(2,J)-0.TEP(1,J1)=0.TEP(3,J1)=0.C1=D(2,K)+2.*D(4,K)*R+D(5,K)*ZC2-D (3, K ) +D( 5, K ) i«P+2. #D(6,K)*ZTEP(1,J)=C1TEP(4,J1)=C1TEP(2,J1)=C2TEP(4,J)=C2TEP(3,J)=(D(1,K)+D(2,K)*R+D(3,K)*Z+D(4,K)*R*R
1 + D (5,K)*R*Z+D(6,K)*Z#Z)ZR.J--J+2
30 CONTINUECC > » COMPUTE CONSTANT
131
WW-w (IF"; *TWOP I *RCC START FORMING STIFFNESS MATRIX.
Dfj 85 K-1,3 DO 85 LL-1,3
CC » > MULTIPLY THE NORMAL PORTION OF THE ELASTICITY MATRIX BY C WEIGHTING FNS. AND CONSTANTS.
Wl-EMIKiLDxWWCC » > COMPUTE TEP/T E TEF FOR NORMAL PORTION OF ELASTICITY MATRIX.
DO 85 JJ - l i 12 W2-WlxTEP(LLiJJ)ECi 85 I I - L J-JSK (II, JJ )-SK(IIv.«v) +W2$TEP (K, 11)
85 CONTINUE CC > » MULTIPLY SHEAR PORTION OF ELASTICITY MATRIX BY WEIGHTING FNS. C AND CONSTANTS, THEN MULTIPLY INTO STIFFNESS.
WI-U*WWDO 90 J J = M 2W2=W1*TEP(4,J.J)DO 90 II-l,JJSrall, JJ)-SH(II, J J ) +W2*TEP (4,11)
90 CONTINUE 100 CONTINUE CC *** COPY LOWER HALF OF DIAGONAL SUBMATRICES.
DO 110 1-1,11,2 IM+1
110 SK(I1,I)-SK(I,I1)CC *** ASSEMBLE ELT. STIFFNESS INTO MASTER STIFFNESS, ASSEMBLING C LOWER SUBMATRICES BY TRANSPOSITION.
DO 150 IP=1,6 K0-2*IP-2 DO 140 IQ-IP,6 L0-2*IQ-2 DO 120 K-1,2 KOK-KO+K DO 120 L-1,2 LOL-LO+L
120 S M (K ,L )- S K (K O K ,L O L )CALL SAS(IP,IQi2)IF (IP .EG. IQ) GO TO 140
CC » > ASSEMBLE IN A LOWER SUEMATRIX.
DO 130 1=1,2 DO 130 0=1,2 TMP=SM(I,J)
132
130 SM(J,I)=TMPCALL SAS (10,IF,2)
140 CONTINUE 150 CONTINUE
RETURN END
133
SUBROUTINE PTA6 C w * WRITTEN NOVEMBER, 1?73 BY KEITH A HUNTEN.CC THERMAL FORCES FOR AN AXISYMETRTC LINEAR STRAIN TRIANGLEC
DIMENSION D(6,6),TM(7,12),V(6),WF(7),TTT(7,6)DIMENSION S M (6,6),A L F A 1(7),A L F A 2 (7),A L F A S (75 DIMENSION PL(12),SPL(12)tW(7)COMMON /DAT/TC(3,27),NEL COMMON /TMP/TEMP(27)COMMON /TLD/VTLD(A,27)COMMON /LM6/D C(3,5) ,XL(3,6),XO(3)COMMON /MAT/MATFTR,MT,YM,PR,G,STY,BEN,THEXCO COMMON /PAR/LHV(11),L G L (5),L P X (14), NDST COMMON /THS/ITYPE,IPTRCS,TH
CDATA V/l.,5*0./, TWOPI/6.233185307/DATA ALFA1 /.333333333,.05971587,.47014206,.47014206 1 , .79742699,.10128651,.10128651/DATA ALFA2 /.333333333,.47014206,.05971587,.47014206
1 ,.10128651,.79742699,.10128651/DATA ALFAS /.333333333,.47014206,.47014206,.05971587
1 ,.10128651,.10128651,.79742699/DATA WF /.225,3*.13239415,3*.12593918/
CCcC *** CREATE POINTS 2,4,5.
DO 30 1-1,2TC(I,2)=0.5*(TC(I,1>+TC(I,3))T C (I, 4 )-0.5*(T C (I,1)+ T C (I ,6)) TC(I,5)=0.5*(TC(I,3)+TC(I,6))
30 CONTINUECC iHHr CONSTRUCT 6X6 MATRIX TO OBTAIN COEFFICIENTS BY INVERSION.
DO 70 0-1,6 SM(.J,1)-1.R - T C I 1,J)Z-TC(2,J)SM(J,2)-R
134SM(Ji3)-Z SI1( Ji4)-R*R SM(J,5)=R*Z SM(J,6)=Z*Z
70 CONTINUECALL IfTv'Nrl (SM,6,D,6)
CC *** COMPUTE CONSTANT.
CC-THEXiCO*YM*TWGP 1/(1. - P R )CC *** LOOP THROUGH THE INTEGRATION POINTS,COMPUTING TEP,W,AND TIT.
DO SO NIP-1,7R4%LFA1(NIP)*TC(1,1)+A L F A 2 (N I P )r T C (1,3)+A L F A 3 (N I P )* T C (1,6) Z-ALFA1(NIP)*TC(2,1)+ALFA2(NI P ) * T C (2,3)+A LFA3 (NI P)*TC(2,6)
CC w w FORM TEP * Cl 110] IN TM.
DO 71 IR-1,6 IR6-IR+6TiKNIP, IR)-(D(2, IR)+2.*B(4, IR)*R+D(5, IR)*Z)1 +(D(l,IR)+D(2,IR)vR+D(3,IR)vZ2 +D(4,IR)*R#R+D(5,IR)*R*Z+D(6,IR)#Z#Z)/R TIKNIP, IR6)-D(3, IR)+D(5» IR)*R+D(6, IR)*Z
71 CONTINUE CC *** FORM TEMPERATURE INTERPOLATION MATRIX TTT.
V(2)=RV(3)=ZV(4)=R*RV(5)=R*ZV(6)=Z$ZDO 72 IC-1,6TTT (NIP, 1 0 - 0 .[sj 72 IR-1,6TTT(NIP,IC)=TTT(NIP,IC)+V(IC)*D(IR,IC)
72 CONTINUE CC *** FORM CONSTANT,
W(NIP)=CC*R*WF(NIP)SO CONTINUE CC LOOP THROUGH LOADING CASES.
DO 200 LCASE -= 1,NDST CALL INTMP (NEL,LCASE)
CC > » ZEROIZE VTLD.
[a] 90 1-1,12 90 V T L D (I)-0.CC *Hr LOOP THROUGH THE INT. PI'S,CALCULATING THE THERMAL FORCES.
DO 150 NIP-1,7
135
C > » CALCULATE THE TEMPERATURE AT THE CURRENT INT. FT. THETA-O.DO 100 IR-l.ATHET A-THET A + T T T (N I P »IR)*TEMP(IR)
100 CONTINUECC » > CALCULATE THE FORCE CONTRIBUTION AT CURRENT INT. PT.
MDO 120 IR-1,6 Il^I+lVTLDII)-W(NIP)*TM(NIP,IR)«THETA+VTLD(I) VTLD(I1)=W(NIP)*TM(NIP,IR+6)*THETArVTLD(Il)1=1+2
120 CONTINUE 150 CONTIilE CC » > OUTPUT THE CURRENT LDCASES' LOADS TO DISK.200 CALL OUTLO (NEL.LCASE)
RETURNEND
136
SUBROUTINE STA6 C WRITTEN NOVEMBER, 1973 BY KEITH A. HUN7ENCC STRESSES FOR AN AXISYMETRIC LINEAR STRAIN TRIANGLE.C
DIMENSION D (6,6),C T (3,2),S M (6,6),I0RDR(12),RZEP(2,3)DIMENSION T E P ( 12,12),SFL(12),S T R ( 12)COMMON /DAT/T0«3,12),T X (3,12),TY(3,12),FL(3),SM1(6),SM2(6),SM3(6) 1 ,S M 4 (6),S M 5 (6),S M 6 (6)COMMON /L M6/DC(3,5),X L (3,6)COMMON /MAT/MATF'TR, MT,E,FR, G,STYCOMMON /F'AR/LHV (11), LGL (5), LPX (14), NDSTCOMMON /DTS/TC(3,27),S P (36),S P X (6,10),ST(7,10),LCASE,NSTPT
CEQUIVALENCE (3M1(1),SM(1,1))EQUIVALENCE (S P X (1,1),S P L (1)),(SPX<1,3),S T RI1))
CDATA IORDR/1,2,7,8,13,14,19,20,25,26,31,32/
CCCC CREATE POINTS 2,4,5.
DO 30 1-1,2TC(I,2)4).5*(TC(I,1)+TC(I,3))TC(I,4)-0.5x(TC(1,1)+TC(I,6))TC(I,5)=0.5*(TC(I,3)+TC(I,6))
30 CONTINUECC CONSTRUCT 6X6 MATRIX TO OBTAIN COEFFICIENTS BY INVERSION.
DO 70 0=1,6 SM1(J)=1.R=TC(1,J)Z=TC(2,J)SM2(.J)=R SM3(J)=Z S M 4 (J )=R*R SM5(J)=R*Z SM6(U)=ZxZ
70 CONTINUECALL INVNM (SM,6,D,6)
137
C *** FORM E MATRIX COEFFICIENTS.EP=E/((1.+PR)*(1.-2.*PR))CC=PR*EPEP=(1.-PR)#EP
CC COMPUTE THE LOCATION OF THE EVALUATION POINTS.
NSTPT-3 KO-ODO 75 1-1,2RZEP( 1,1) - (XL (1,1 )+XL (1,2)+XL (1,4)) /S’. RZEP(I,2)-(XL(I,2)+XL(I,3)+XL(I,5))/3. RZEP(I,3)=(XL(I,4)+XL(I,5)+XL(I,6))/3.
75 CONTINUECC wnr LOOP THROUGH THE EVALUATION POINTS, CALCULATING TEP.
DO 100 NEP-1,3 R=RZEF(1,NEP)Z-RZEP(2,NEP)
DO 80 K-1,6Jl-J+1Kl-KO+1K2-K0+2K3=K0+3K4-K0+4TEP(K2,J)-0.TEP(Kl,Jl)-0.TEP(K3, J D - O .C1=D(2,K)+2.*D(4,K)*R+Di5,K)*ZC2=D(3,K)+D<3,K)*R+2.*B(6,K>*ZTEP(K1,U)-C1TEP(K4,J1)-C1TEP(K2, J D - C 2TEP(K4,J)-C2TEP(K3,J)-(D(1,K)+D(2,K)*R+D(3,K)*Z+D(4,K)*R*R
1 +D(5,K)*R*Z+D(6,K)*Z*Z)/RJ=J+2 K0=K0+4
SO CONTINUE100 CONTINUE CC LOOP THROUGH LODING CASES
DO 200 LCASE-1,NDST CALL ELTDNS
CC > » PACK THE DEFLECTIONS.
DO 130 N-1,12 iNN-IORDR(N)
138
SFL (N ) -S'P (M N )130 CONTINUE CC » > OBTAIN STRAINS.
DO 140 M , 1 2 S-0.DO 135 J-lil2
135 S - S + T E P d . J J ^ S P K J )140 STR(I)=S CC » > OBTAIN STRESSES.
JO— 0DO 150 N - I jNSTPTS T (1,N )-EP*ST R(J O + 1 )+ C C * (S T R (J O + 2 )+ S T R (J O + 3 )) S T (2j N )-E P * S T R (J O + 2 )+ C C * (S T R (J O + 3 )+ S T R (J O + 1 )) S T (3, N)-E P«S TR(J O + 3 )+ C C * (S T R (J 0 + 1)+ S T R (J O + 2 )) ST (4 ? N ) =G*STR ( JCh -4 )J0-J0+4S T T - (S T (1,N )- S T (2,N ))**2+(S T (2,N)-S T(3,N ))**2
1 +2+(ST(1>N)-ST(3 jN))**2150 S T (71N )-100.# S Q R T (.5#STT+3.*ST(4,N)^*2)200 CALL ELTSTR
RETURN END
139
SUBROUTINE KROD3 U -irir* WRITTEN BY KEITH A. HUNTEN CC SUBROUTINE TO COMPUTE STIFFNESS OF RODS' ELEMENTS C
DIMENSION SKIS,3)COMMON /BKF/MAP(3,27),T C I (3),T C 2 (3),T C 3 (3),T C (3,2- COMMON /MAT/MATPTR,HT,E,FR,G COMMON /THS/ITYPE,IPTRCS,AREA,VIS)
CC *** CALCULATE DIRECTION COSINES.
FL-O.DO 1 1-1,3 D-TC3II)-TC1(I)FL=FL + B**2 VI D - D
1 CONTINUE FL-=SQRT(FL)DO 2 1-1,3
2 V(I)-V(I)/FL C - E^AREA/(FL*3.)
CC iHM CALCULATE CONSTANTS FOR LOCAL STIFFNESS.
C 1 - 7 . CC 2 - - 3 . CUS-16.*0 SKIl,1)-C1 SK(2,2)-C3 SK(3,3)-C1 SK(1,2)-C2 S n 1 1,3)-C SKI 2 , 3 X 2
CC > » COMPLETE LOWER HALF OF STIFFNESS BY COPYING.
DO SO 1-1,3 DO 30 J-1,3 IF (I .EG. J) GO TO 30 3K(J,I)-SK(1,J)
30 CONTINUE
),SM(6,6)
140
C *** TRANSFORM LOCAL STIFFNESS TO GLOBAL C AND ASSEMBLE.
DO 100 IP-1i3 DO 100 JP-1,3 DO 50 IK-1.3 DO 50 UK-1,3SM(IK,JK)^V(JK)*SK(IP,JP)*V(IK)
50 CONTINUECALL SAS (IP,UP,3)
100 CONTINUE RETURN END
C
141
SUBROUTINE PRODS C *r* WRITTEN JANUARY 197? BY KEITH A. HUNTEN CC THERMAL FORCES FOR A RODS ELEMENT.C
DIMENSION V(3)COMMON /M AT/MATPTR jM T jY M jPR,G,STY,DEN,THEXCO COMMON /PAR/LHV(11),L G L (5),L P X (14),HOST COMMON /THS/ITYPE,IPTRCS,AREA COMMON /DAT/TC(3,27),NEL COMMON /TMP/T1,T2 COMMON /TLD/VTLD(6,27)
CC *** COMPUTE LENGTH AND DIRECTION COSINES.
FL-O.DO 10 1=1,3 D=TC(I,3)-TC(I,1)FL=FL+D*D V(I)=D
10 CONTINUEFL=SQRT(FL)DO 20 1=1,3
20 V(I)=V(I)/FLCC COMPUTE CONSTANTS, THEN LOOP THROUGH LOADING CASES ANDC CALCULATE THE THERMAL FORCES FOR EACH.
T CON=AREA*THE X CO*YM*0.3 DO 200 LCASE=i,NOST
CC » > GET THIS LOADING CASE'S TEMPERATURES.
CALL INTMP (NEL,LCASE)CC » > COMPUTE LOCAL FORCES.
PC1=-T1*TC0NPC2=T2*TC0N
CC > » TRANSFORM TO GLOBAL FORCES AND OUTPUT TO DISK.
DO 150 J=l,3 VTLD(J,1)=PC1*V(J)
150 VTLD1J,2)=PC2*V(J)
142
200 CALL GUILD (NEL jLCASE) RETURN END
0
SUBROUTINE STRD3 C *## WRITTEN BY KEITH A. HUNTEN.CC STRESSES Cf ROBS ELEMENTS,C
DIMENSION CF(3),ST(7)COMMON /PAR/LHV(II),LOLLS),LP(22)COMMON /D TS/T C1(3),TC2(3),TC3(3),TC(3,24),S P 1 (6),SP2(6)1 ,SP3(6),SP(6,13),ST11,STA(5),FC,ST21,STB(5),2 ,STC(7,S),LCASE,N3TPT,ITFLG,NEL CO MMON /MAT/MATPTR,MT,E,PR,0,SY,DEN,THEXCO COMMON /TMP/T1,T2»T3
CEQUIVALENCE (LP(15),NDST)
CC
DO 5 1=2,6 5 ST(I)=0.
FL=0.DO 10 1=1,3 D= TC3(I)-TC1(I)FL= FL+D**2
10 CF(I)= D FL=SGRT(FL)DO 15 1=1,3
15 CF(I)= CF(I)/FLNSTPT=2DO 25 LCASE=1,NDST CALL ELTDNS
CC *** CALCULATE STRAINS.CC » > CALCULATE THE LOCAL DEFLECTIONS FROM THE GLOBAL O N E S . ..
SPL1=0.SPL2=0.SPL3=0.DO 20 ID=1,3S P L 1=SP L 1+SP K I D ) C F (ID)3rL2=SPL2+SP2IID)*CF(ID)Sr L3=3PL3+SP3 (ID) vijr (ID)
144
C » > NOW THE TOTAL LOCAL STRAINS.X1-0.25*FL X2=0.75*FL CO NST-1./(FL*FL)S T R N 1-CONST*( (X 1-3.* F L )* S PL1+4.*(F L -2. 1 )*SFL2
1 +(4.* X 1- F L )* S F L 3 )STRN2-CONST*((X2-3.xFL)* S P L 1+4.*<FL-2. * X 2 )*SPL2
1 +«4.*X2-FL)*SFL3)20 CONTINUE CC ririr IF TEMPERATURES ARE P R ESEN T, CALCULATE C THERMAL STRAINS AND SUBTRACT THEM.
IF (ITFLG .EG. 0) GO TO 22 CALL INTNP (NEL>LCASE)
CC » > AVERAGE TEMPERATURES.
TAVGl-(Tl+T2)/2.TAVG2-(T2+T3)/2.
CC » > CALCULATE THE THERMAL STRAINS AND SUBTRACT THEM.
T S T R N 1-TA VG1*THE XCO TSTRN2=TAVG2*THEXC0 ISO TO 23
22 TSTRN14).TSTRN2-0.
23 ST 11 - (STRN 1 -T STRN 1) *•£S T 2 1 = (STRN2-TSTRN2)*E F C M 0 0 . * A B S ( S T 1 1 ) / S Y F C 2 - 1G O .*AB S(S T 2 1)/SY
25 CALL ELSTR RETURN END
APPENDIX G
DRIVER PROGRAMS AND OUTPUT
C PROGRAM DRITM6C TEST DRIVER FOR TM6 ELEMENT STIFFNESS AND STRESS ROUTINES..CC LOAD WITH K T N 6 jCROSS?DOT?SA3* jSTM6?ELSTR'K'j C ELTDNS?EIGEN,INVNM,SECOND,GRUNT.
DIMENSION TC(3,27)COMMON /D AT/TO(3? 12),TX(3? 12)?TV(3? 12),SK( 12,12),F L O )COMMON /5KF/MAF(3,27),TCI(3),TC2(3),TC3(3),TC4(3),TC50),TC6(3) 1 ,TC7(3),TCX(3,20),SM(6,6)COMMON /MAT/MATPTR, MT, YM, F R O , STY COMMON /THS/ITYPE,IPTRCS.TH
COMMON /RES/ E S K (27,27),V E C (27,27},V A L (27)COMMON /PAR/LHV(11),L G L (5),I P X (14),NDSTCOMMON / D T S /T CSO ,27) ,SP ( 18),SFX(6,13),ST(7,10),LCASE,NSTPTEQUIVALENCE (TC1(1 ),T C(1, D)TU-O.OPEN 12," t l F T ",ATT- H S O P "
C PREPARE ELEMENT GEOMETRY ETC.DO 1 1=1,3 DO 1 J - 1 , t>
1 MAP(I,J)=I+3*J MATPTR-1 MT=1 YM=1.E7 FR=0.30=0. j * Y M / (1.+ P R )ITYPE = 3 IPTRCS=0 TH=0.05 TC1(1)=10.TC1(2)=0.TC1(3)=0.TC3(1)=20.TC3(2)=0.TC3(3)=0.TC6(1)=20.TC6(2)=10.TC6<3)-0.
CA TC1(1)=0.
145
146
Cfi T C K 2 H .CA TCl(3)-0.CA TCo(l)-24.CA TC3(2)-0.CA TC3(3)-0.CA T C 3 U ) - 2 4 .CA TC6(2)-18.CA TC6(3)-0.
DO 20 1-1,3TC 2(I)-0.3»(TCl(I)+TC3(I)}TC4(I)-0.5*(TCl(I)+TC6(I))
20 TC5(I)4).5*(TC3(I)+TC6(I))Du 30 J— 1,6 DO 30 1— 1,3
30 TC3(I,J)-TC(I,J)CALL SECOND ( I D CALL KTM6 CALL SECOND (T2)TU=TU+T2-T1
100 FORMAT v TM6 TIME - d F 1 0 . 5 )WRITE (12,100) TU
X WRITE (12,101) ( E S K d , I ) , 1-1,13)101 FORMAT (' SK',/,(1X,3E15.7))
CALL EIGEN (ESK,VEC,VAL,27,27)WRITE (12,910) ( V A L d ) , 1=1,27)
910 FORMAT (' EIGENVALUES ',/,(1X ,3E15.7)) NDST=3 STY=10000.0 CALL STM6WRITE (12,900) (N,(ST(I,N),I=1,7),N=1,3)
900 FORMAT (/ STRESSES ',/,(1X,I2,7E10.2)) CLOSE 12 STOP END
14?
C PROGRAM DPTM6C TEST DRIVER FOR TM6 ELEMENT THERMAL FORCE ROUTINE.CC LOAD WITH PTM6,CROSS,DOT,ELTEM,SECOND,GRUNT,INVNM.
DIMENSION TC(3,27),IC(3)COMMON /DAT/T0(3,12),TX (3,12),TY (3,12) ,SK( 12,12),F L O )COMMON /BKF/H AP(3,27),T C I (3),T C 2 (3),T C 3 {3),T C 4 (3),TC5(3),TC6(3) 1 ,TC7(3),T C X (3,20),SM(6,6)COMMON /MAT/MATPTR,MT,YM,PR,G,STY,DEN,THEXOCOMMON /TH3/ITYPE,IPTRC3,THCOMMON /RES/ ESK(27,27),VEC(27,27),VAL(27)COMMON /P AR/L HV(1i ),L G L (5),L P X <14),NDST COMMON /DMTo/TCM(3,27),E S N (6)COMMON /DFT6/LCASE,TCP(3,27),TEMP-27),PG(13)DATA IC/1,3,6/EQUIVALENCE (TC1(1 ),T C(1, D)TU-O.OPEN 12,HSLPT",ATT="SOPH
C PREPARE ELEMENT GEOMETRY ETC.DO 1 1=1,3 DO 1 J=l,6
1 MAP(I,J)=I+3*J MATPTR-1 MT=1 YM=1.E7 PR=0.3G=0.5*YM/(1.+PR)DE M=1.E-5THEX0=1.E-3ITYPE = 3IPTRCS=0TH-0.03TC1(1)=0.6TC1(2)=0.9TC1(3)=2.875TC3(i)-2.2TC3(2)=0.3T C 3 (3)=0.95833333TC6(1)=0.4TC6(2)=2.2
148
TCch 3)-1.53333333 CA TC1(1)=0.CA TCl(2)-0.CA TCl(3)-0.CA TC3(1)=24.CA TC3(2)-0.CA TC3(3)-0.CA T C 6 U ) - 2 4 .CA TC6(2}-18.CA 706(3)4).
DO 20 1=1,3TC2 (I ) =0.5» (TCI (I ) +TC3 (I )) TC4(I)=0.5»(TC1(I)+7C6(I))
20 TC5(I)=0.5*(T C3( I)+ TC6( D)DO 30 J=l,6 DO 30 1=1,3 TCM(I,J)=7C(I,J)TCP(I,J)=TC(I, J)
30 CONTINUEX WRITE (10,110)X DO 40 1=1,3
II=IC(I)X WRITE (11,120)X READ FREE (11) TCP(1,II),7CP(2,II),TCP(3,II)X40 CONTINUEX110 FORMAT (/," INPUT COORDINATES - X , Y , Z " )X120 FORMAT ( " >")
NDST=3 CALL PTM6
C WRITE (12,930) (PG(I),1=1,13)C930 FORMAT (' GLOBAL FORCES ',/,(1X,3E15.7))
CLOSE 12 STOP END
149
C PROGRAM DMTM6C TEST DRIVER FOR TM6 ELEMENT MASS ROUTINE.CC LOAD WITH KTM6,CROSS,DOT,SECOND,GRUNT,INVNM.C
DIMENSION TC(3,27),IC(3)COMMON /DAT/T O(3,12),TXt'3,12),TY(3,12),SK( 12,12),FL(3)COMMON /B KF/M AP(3,27),T C I (3),T C 2 (3),T C 3 (3),T C 4 (3),T C 5 (3),T C 6 (3)
1 ,TC7(3),TCX(3,20),SM(6,6)COMMON /MAT/MATPTR,NT,YM,PR,G,STY,DEN,THEXOCO MMON /TH3/ITYPE,IPTRCS,THCOMMON /RES/ E S K (27,27),V E C (27,27),V A L (27)COMMON /P AR/L HV(11),L G L (5),L P X (14),NDST COMMON /D MT6/TCM(3,27),E S M <6)COMMON /DPT6/LCASE,TCP(3,27),TEMP(27),PG(13)DATA IC/1,3,6/EQUIVALENCE ( T C1(1 ),T C(1, D)TU-O.OPEN 12,"@LFT",ATT="S0P"
C PREPARE ELEMENT GEOMETRY ETC.DO 1 1-1,3 D) 1 ,>1,6
1 M A P d , J)-I+3#J MATPTR-1 MT=1 YM-1.E7 FR-0.3G-0.5*YM/(l,TpR)DEN-l.E-3THEX0-1.E-3ITYPE ^ 3IPTRCS-0TH-0.05TC1(1)=4.6TCl(2)-0.?TCl(3)-2.375TC3(l)-2.2TC3(2)-0.3T C 3 (3)-0.95833333TC6(l)-0.4
150
TG6(2)-2.2 TC6 (3) -1.58333333
CA TC1(1)=0.CA TC1(2)=0.CA TC1(3)=0.CA TC3(l)-24.CA TC3(2)-0.CA TC3(3)=0.CA T C 6 U ) = 2 4 .CA TC6(2)=18.CA TC6(3)-0.
DO 20 1=1,3T C2 (I ) =0.5 * (T C 1 (I ) + T C3 (I )) TC4(I)=0.5*(TC1(I)+TC6(I))
20 TC5(I)=0.5v(T C3( I)+ TCA( D)DO 30 J=l,6 DO 30 1=1,3 T C M U , J ) = T C ( I , J )TCF(I,J)=TC(I,J)
30 CONTINUEX WRITE (10,110)X DO 40 1=1,3
II=IC(I)X WRITE (11,120)X READ FREE (11) TCM(1,II),TCM(2,II),TCM(3,II)X40 CONTINUEXI10 FORMAT (/," INPUT COORDINATES - X , Y , Z " )X120 FORMAT ( " >")
CALL MTM6WRITE (12,920) (ESM(I),1=1,6)
920 FORMAT (' ELEMENT MASS M A T R I X ',/,(IX,3E15.7)) CLOSE 12 STOP END
151
LOCAL COORDINATES1 -.6666666E 01 -.3333333E 012 -.1666666E 01 -.33333335 013 .3333334E 01 -.33333335 014 .-.1666666E 01 .1666667E 015 .3333334E 01 .1666667E 016 .3333334E 01 .66666678 01
-.11111 H E 00 .4444444E 00-.11U107E 00 .4444440E 00 .4444441E 00-.1111U2E 00-.3333335E-01 .OOOOOOOE 00 .3333335E-01-.1333333E 00 .1333333E 00 .OOOOOOOE 00.OOOOOOOE 00-.1333333E 00-.3333330E-01 .1333333E 00 .OOOOOOOE 00 .333333GE-01.2000001E-01-.4000002E-01 .2000001E-01 .OOOOOOOE GO .OOOOOOOE 00 .OOOOOOOE 00.OOOOOOOE 00 .4000002E-01-.4000002E-01-.4000002E-01 .4000002E-01 .OOOOOOOE 00 .OOOOOOOE 00 .OOOOOOOE 00 .2000001E-01 .OOOOOOOE GG-.4000002E-01 .2000001E-01
AiIXXjIXYiIYY.7769E 02 .4316E 03 .2158E 03 .4316E 03
ELEMENT MASS MATRIX.9536738E-12 .1294S08E-04 .1473194E-1G.1294S07E-04 . V294307E-04 . 3337360E-11
LOCAL COORDINATES1 -.11067978 02 -.36893235 012 -.31622765 01 -.36S9323E 013 .47434175 01 -.3639324E 014 -.2371707E 01 .1844662E 015 .5533986E 01 .13446625 016 .6324556E 01 .73736485 01
-.1111105E 00 .4444430E 00-.1111106E 00 .4444449E 00 .4444438E 00-.1111109E 00 -.2108136E-01 .54S3245E-07 .21G8184E-01-.843272SE-01 .8432740E-C-1 .342702SE-07 .3011700E-02-.1204676E 00-.3312854E-01 .1325145E 00-.120467SE-01 .3011692E-01 .799999SE-02-.1599997E-01 .7999986E-02 .OOOOOOOE 00 .1261057E-07-.2277779E-08
-.2285722E-02 .2742857E-01-.2514284E-01-.22S5717E-01 .2285715E-01 .1348769E-07 .1332699E-03-.3591820E-02 .1975508E-01 .3265321E-02-.359183SE-01 .16326535-01
A,IXX,IXY,IYY.1102E 03 .7502E 03 .6430E 03 .1699E 04
ELEMENT MASS MATRIX.2995132E-10 .183726IE-04 .20027155-10.18372685-04 .18372625-04 .12374605-10
O O
FUNCTION DOT (V1,V2)*** WRITTEN JUNE 17, 1975 BY MICHAEL W. MCCABE
C THIS FUNCTION COMPUTES THE DOT PRODUCT OF VI AND V2.C
DIMENSION V1(3),V2(3)C
DOM.DO 5 1-1,3
5 DO T=DOT+Vl(I)*V2(I)RETURNEND
SUBROUTINE CROSS (VI,V2,VC,FL)C *** UPDATED 3/20/74 BY MWMC *** WRITTEN JUNE 27, 1974 BY MICHAEL W. MCCABECC SUBROUTINE TO COMPUTE CROSS PRODUCT VC = VI C FL IS LENGTH OF RESULTANT VECTORC VC CONTAINS DIRECTION COSINES OF RESULTC
DIMENSION V1(3),V2(3),VC(3),JK(4)DATA JK /2,3,1,2/
CC *** COMPUTE CROSS PRODUCT
FL=0.DO 5 1-1,3 J - J K (I }K=JK(I+1)VC (I ) - V 1 (J ) *V2 (K ) - V 1 (K ) *V2 (J )
5 FL-FL+VC(I)*VC(I)FL=SQRT(FL)
CC NORMALIZE RESULT
IF (FL .EQ. 0.) RETURN DO 10 1=1,3
10 VC(I)=VC(I)/FLRETURN END
SUBROUTINE ElGEN (A,T,B,AIK,L,N,EF2,MS)C A MATRIX TO FIND EIGENVALUES AND EIGENVECTORS FOR C T MATRIX CONTAINING EIGENVECTOERS COLUMNWISEC B VECTOR CONTAINING EIGENVALUES IN ASCENDING ORDERC AIK VECTOR USED FOR TEMPORARY STORAGEC L VECTOR USED TO REORDER EIGENVALUESC COMPILER DOUBLE PRECISION
INCLUDE "DINCGM.FR"DIMENSION A <MS,MS),T(MS,MS),A I K (M S ),B (M S ),L (M S ) INCLUDE "FILER.PR"DATA EPS3/I.E-6/ITMAX-N DO 2 I-1,N L(I)-I DO 2 J - l ,N
2 T(I,J)-0.NM4HSIGMAl-0.
CX WRITE (NLP,707) ( ( A ( I , J ) , J ^ 1 , N ) , M , N )CX707 FORMAT (' ElGEN 5 A MATRIX ',/,(4E15.7))CX WR I T E (NLP,747) N, MS,E PS!,EFS2,EPS3CX747 FORMAT C N,MS ',215,' EPS (1 3) ',3E15.7)
DO 5 1=1,N3IGMA1-SIGMA1+A(I,I)**2T(I,I)=1.
5 CONTINUE SI G M A 1 = SQRT(SIGMAl)EPS2 = EP2*SIGMA1 DO 26 ITER=1,ITMAX
X WRITE(NLP,727) SIGMA1X727 FORMAT C SIGMA1 ',E15.7)
DO 20 1=1,NM!IP1=I+1DO 20 .>IP1,NAIJ = A(I,I)-A(.J,.J)SGNA = 1.IF (AIJ .LT. 0.) SGNA = -1.Q = ABS(AIJ)
CX WRITE(NLP,717) Q CX717 FORMAT (' Q =',E15.7)
155
CX WRITE(NLP,737> A(I,J)CX737 FORMAT (' A(I,J) ^ , E 1 5 . 7 )
IF (ABS(A(I,J)).LE.EPS2) GO TO 20P=2.*A(I,J)*SGHASF6=S6RT(P*P+W*Q)CSA-SQRT ((1.0+Q/S'PQ) /2.0)3 N A - P / (2.O ^ CSAv SPQ)
CX m l IE (NLP, 737) OS A, SNA CX757 FORMAT (' CSA,SNA ',2515.7)
DO 11 K--1,N TKI = T(K,I)TKJ -= T(kSJ)T(KiI)-TKI*CSA+TKJ*SNA
11 T (K ,. J)-TKIvSMA-TKJvCSA CX WRITE(NLP,767) ((T(K,LL),LL=1,N),K=1,N)0X767 FORMAT (' T MATRIX (4E15.7))
DO 16 K-1,NIF (K .GT. J) GO TO 15A I K ( K ) = A U , K >A(I.K)-CSAvAIK(K)+SNA*A(K,J)IF (K.NE.J) GO TO 14 A(J,K)=SNA*AIK(K )- C 3 A * A (J, K )
14 GO TO 1615 TIK -- A(I,K)
T-.IK = A(J,K)A(I,K)=CSA*TIK+SNA*TJK A (J , K ) =SNA*T I rH.:3A*T JK
16 CONTINUECX WRITE(NLP,777) ((A(K,LL),LL=1,N),K=1,N)0X777 FORMAT {' A MATRIX AFTER DO 16',/,(4E15.7))
AIK(J)=SNA*AIK(I)-C SA*A IK(J)DO 19 K— 1,JIF (K.LE.I) GO TO 18A(K,J)=SNA*AIK(K)-CSA*A(K,J)GO TO 19
13 TKI ^ A(K,I)TKJ ^ A(K,J)A (K ,I )-CSA*TKI+SNA#TKJ A (K» J )-SNA*TK I-CSA#TKJ
19 CONTINUECX WRITE(NLP,787) ((A(KK,LL),LL-1,N),KK-1,N)0X737 FORMAT C A MATRIX AFTER DO 19',/,(4E15.7))
20 A(I,J)-0.X WRITE(NLP,808)X808 FORMAT (IX, ' I M O X , ' A ( I , I ) y)X WRITE(NLP,818) (I,A(I,I),I-1,M)X313 FORMAT (IX,I5,5X,E15.7)
SIGMA2=0.DO 21 I-1,N B(I)-A(I,I)
156
21 SI GrtA2-S IGHA2+8 (I ) **2 SIGMA2 ^ SQRT(3IGMA2/
X WRITE(NLF»797) SIGMA2 X797 FORMAT (" SIGMA2 =',E15.7)
IF ((1.O- SIGH A1/31G M A 2 ).G E .E P S 3 ) GO TO 25 WRITE (NLP,204) ITER
C PAUSE 1602NV-1
302 DO 303 1=1,NIF (L(I).EQ.O) m TO 303 Vi1=B(I)NR=IGO TO 304
303 CONTINUEWRITE (NLP,100) I
100 F O R M A T (4 0 X ,29H+++ GE + ElGEN + L(I)=0 I ^,16)C PAUSE 1602C CALL ERROR (0)
304 DO 300 1=1,NIF (L(I).EQ.O) GO TO 300 IF (3(1) .LT. VM) GO TO 300 NR=I VH=B(I)
300 CONTINUE AIK(NV)=B(NR) to 301 1=1,N
301 A(I,NV)=T(I,NR)L(NR)=0NV=NV+1IF (NV.LE.N) GO TO 302 DO 305 0=1,N B(J)=AIK(J)DO 305 1=1,N
305 T(I,J)=A(I,J)RETURN
25 WRITE (NLP,202) ITER C25 PAUSE 1603
26 SIGMA1--SIGMA2 WRITE (NLP,203) ITER
C PAUSE 1604202 FORMAT (Sti ITER' = ,15)203 FORMAT (1H0,4X,21H NO CONVERGENCE, WITH//7X,3H ITER =,15)204 FORMAT (1H0,20H CONVERGENCE AFTER ,I4,6H ITRNS)
RETURNEND
SUBROUTINE INVNM (AA ?M,A C,M S)C *** INVERT WON-SYMMETRIC MATRIX AA IN AC.C AA DESTROYED
DIMENSION A A (M S ,M S ),A C (M S ,M S )NB-MIF(M .IT. 0) GOTO 4012 IF(M .EG. 1) GOTO 120
C *** CREATE UNIT MATRIX IN AC.DO 10 I-1,M DO 9 J=1,M
9 A C ( I jU)-0.10 AC(I,I)=1.X WRITE (12,900) M,MS X900 FORMAT (' INVNM M,MS',216)X WRITE (12,910) ((AA(I,J),J=1,M),I-1,M)X910 FORMAT (' MATRIX ',/,(1X,6E12.5))X WRITE (12,930) ((AC(I, J),.>1,M), I^1,M)X930 FORMAT (' UNIT MATRIX ',/,(1X,6E12.5))
MO-M-1DO 4006 1=1,MO K=I+1DO 4001 J=K,MIF (ABS(AA(J,I))-ABS(AA(1,1))) 4001,4001,4002
C INTERCHANGE ROWS4002 DO 4003 N-I,M
V=AA(I,N)AA(I,N)=AA(J,N)
4003 AA(J,N)-vDO 4010 N=1,NB V=AC(I,N)AC(I,N) = AC(J,N)
4010 AC(J,N)=V 4001 CONTINUE
IF ( A A ( I , D ) 4008,4012,4003 4012 STOP
C NULLIFY REST OF COLUMN 4008 DO 4011 J=K,M
F=AA(J,I)/AA(I,I)DO 4004 N=K,M
4004 AA(J,N)=AA(U,N)-F*AA(I,N)
DO 4011 N - I jNB 4011 A C (J ,N )- A C (J ,N)-F*AC(I,N)4006 CONTINUE
C BACK SUBSTITUTION JD -= M
C ELEMINATE ELEMENTS ABOVE A(JD,JD)117 I-JD-1115 F - A A (I> J D )/A A (J D j J D )
C MODIFY ROW IDO 124 J-1iNB
124 AC(ItJ) - AC(I,J)-F*ACwD,J)1=1-1IF (I .GT. 0) GOTO 115 JD = JD-1IF (JD .GT. 1) GOTO 117
C +++ DIVIDE BY DIAGONAL ELEMENTS DO 119 1=1,M DO 119 J=1,NB
119 AC(I,J)=AC(I,J)/AA(I,I)X WRITE (12,920) ((AC(I,J),J=1,M),I=1,M)X920 FORMAT (' INVERSE ',/,(1X,6E12.5>)
RETURN120 IF (AA(1,1) .EG. 0.0) GOTO 4012
ACM 1,1) = l . O / A A d , 1)RETURNEND
SUBROUTINE ELSTRDUMMYRETURNEND
SUBROUTINE ELTDNS C DUMMY
COMMON /FAR/ L H V ( 1 1 ) , L G U 5 ) , L P X U 4 > , N B S T COMMON /MAT/MTP,MT,E,PR,G COMMON /LM6/DC(3,5),XL(3,6),X0(3)COMMON /DTS/ T C (3,27),S P (13),S P X (6,13),S T (7,10),L C A S E ,NSTPT WRITE (12,901) LCASE
901 FORMAT <///,' LOAD CASE ',5%,13)10=0DO 10 N=l,6 XP=XL(1,N)YP=XL(2,N)GOTO (101,102,103),LCASE
101 UP=-1.E-3*(YP-2.11)VP=1.E-3*(XP+1.24)EX=0.EY=0.EXY=0.GOTO 200
102 UP=1.E-3*(XP+YP)VP=2.E-3*(YP-XP)EX=l.E-3 EY=2.E-3 EXY--1.E-3 GOTO 200
103 IP=1.E - 3 * (XP**2+XP*YP)VP=1.E-3*(YP**2-XP*YP)EX=i.E -3*(2.*X F + Y P )EY=1.E-3*(2.*YP-XP)EX Y-1.E - 3 * (X P - Y P )
200 WRITE (12,900) N,XP,YP,UP,VP900 FORMAT (/,' N,XP,YP,UP,VP ,/,I3,2E12.5,3X,2Ei3.6)
S P (I0+1)=U P * D C (1,3)+V P * D C (1,4)
160
SP( 10+2) -UF'sBC (2 > 3) + VP#£iC (214)SP (10+3) -IJr*DC (3,3) +VP*DC (3,4)EF-E/(l.-PRx*2)VEP-PRxEPSX-EP*EX+VEP+EYSY-VEPxEX+EFxEYSXY-GxEXYWRITE (12,910) N ,E X ,E Y ,E X Y ,S X ,S Y ,SXY
910 FORMAT (' STRESSES AND STRAINS M 4 , / , (1X,3E14.6)) 10 I0-10+3
RETURN END
SUBROUTINE ELTEM C DUMMY
COMMON /MAT/MTP,MT,E,PR,G,STY,DEN,THEX COMMON /DPTo/LCASE,TC(3,27),T E H (27),PG(IS)COMMON /IMS/ DC(3,5),XL(3,6/,X0(3)WRITE (12,901) LCASE
901 FORMAT (///,' LOAD CASE ,5X,I3)10=0DO 10 N=l,6 XP=XL(1,N)YP=XL(2,N)GOTO (101,102,103),LCASE
101 TE=100.GOTO 200
102 TE=5.+3.*XP+YP GOTO 200
103 TE=XP**2+XP*YP-YP**2+3.*XP-4.*YP+100.200 CONTINUE
TEM(N)=TE9 WRITE (12,900) N,XP,YP,TE900 FORMAT (/,' N,XP,YP,TEMPRATURE /,/,I3,2E12.5,3X,E13.6)10 10=10+3
RETURN END
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Doherty, W. P., Taylor, E. L., and Wilson, E. L. STRESS ANALYSIS OF AXISYMMETRIC SOLIDS UTILIZING HIGHER ORDER QUADRILATERAL ELEMENTS, Struct. Eng. Lab., Univ. of Calif., Berkley (1969).
Ergatoudis, I., Irons, B . M ., and Zienkiewicz, 0. C. "Curved,Isoparametric, 'Quadrilateral' Elements for Finite Element Analysis," Int. J. Solids Structures, Vol. 4 (1968), pp. 31- 42. Pergammon Press
Fwa, T. F ., Jofriet, J . C., and LeLievre, B . "Comments on theAccuracy of Nine-Noded Isoparametric Finite Elements," Computers and Structures, Vol. 8 (1978), pp. 67-74.
Goodier, J . N. and Timoshenko, S. THEORY OF ELASTICITY, McGraw-Hill(1951).
Hammer, P. C ., Marlow, 0. P., and Stroud, A. H . "Numerical Integration Over Simplexes and Cones," Math Tables Aides Comp 10(1956), pp. 130-137.
Hollister, G. S. and Zienkiewicz, 0. C. STRESS ANALYSIS, John Wiley & Sons Ltd. (1965).
Kamel, H . A. GIFTS MODELLING GUIDE, Univ. of Arizona, Col. of Eng.,AME Dept. (1977).
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