W515 I IV RSITY LIBRARIES Ililllfllllllllllllflllllll m l m 1293 00567 4357 llllGQbaaa LIBRARY mchigan State University This is to certify that the dissertation entitled A NONT.'|'_NEAR FINITE ELEMENT FOR CURVED BEAMS presented by Bambang Suhendro has been accepted towards fulfillment of the requirements for Ph.D. degree in Civil Engineering (Structures) / E Dr. Robert K. Wen Major professor Date February 20 , 1989 MSUl'ennAfflr-nnt' A ‘ F. '1‘” ' v ' ‘ 0-12771 MSU LIBRARIES ~§ RETURNING MATERIALS: Place in book drop to remove this checkout from your record. FINES will be charged if book is returned after the date stamped below. _ w . n u m u m M Dep; .A NONLINEAR FINITE ELEMENT FOR CURVED BEAMS By Bambang Suhendro A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Civil and Environmental Engineering 1989 A procedur curved beam st beam finite el dimensional 3; second order p nonlinerities the longitudina displacement f the increment; assuming that t p0lymmials Whi unchanged. Th: over the elemer expression yi. second Order 11 element ASSuming equilibrium e potential ener usmg load inc- 567575¢ ABSTRACT A NONLINEAR FINITE ELEMENT FOR CURVED BEAMS W Bambang Suhendro A procedure for the computation of nonlinear elastic response of curved beam structures is presented. The structure is represented by beam finite elements curved in one plane but deformable in three dimensional space. The curved axis of the element is represented by a second order polynomial in the curvilinear coordinates. Geometric nonlinerities are considered by including the effect of rotations on the longitudinal strains. In deriving the linear stiffness matrdjc, the displacement functions are approximated by cubic polynomials. However, the incremental (or nonlinear) stiffness matrices are derived by assuming that the longitudinal displacements are interpolated by linear polynomials while the interpolations for the other displacements remain unchanged. The nonlinear terms in the strain expression are averaged over the element length. Differentiation twice othflue strain energy expression yields the linear stiffness matrix ,Hd, and the first and second order incremental stiffness matrices ,[nl] and [n2], of the element. Assuming that the system is elastic and conservative, the equilibrium equation is obtained from the first variation of the potential energy. The problem is solved by the Newton-Raphson method. using load increments. A compute nonlinear equi involving arche of geometry, 10 Numerical based on a fixe “small displa< as for "interme For all t deflection cun represent the the arch (four tw0 elements, Compariso' iIldicate that the others. The SOlt System is als displacements In additi Was also im displaCEment a displacement amplifiCation Bambang Suhendro A computer program was prepared for the implementathniof the runninear equilibrium solution. Numerical results were obtained involving arches with in-plane and out-of—plane behavior. Various types of geometry, loading, and support condition were considered. Numerical results indicated that the proposed method, which is basedn> iv Page iii vii viii t—‘CO\I\J\Jl\)t—‘ 15 15 15 17 18 19 19 22 24 25 27 28 3O 31 31 32 CHAPTER III. IV. NONLINEAR NUMERICAL 4.1 GENEI 4.2 LINE 4.2.: 4.2.1 4.3 NONL SW 4.3. 4.3. 4.4 NONI INTE 1.4, 4.4. 4.4. 4.1., 4.4 4.4 4.4 4‘5 STRI CHAPTER III. IV. NONLINEAR EQUILIBRIUM ANALYSIS OF CURVED BEAM STRUCTURES 3.1 GENERAL 3.2 NEWTON~RAPHSON METHOD 3.2.1 CONCEPT NEWTON-RAPHSON METHOD FOR FIXED COORDINATES CONVERGENCE CRITERION STRESS COMPUTATION ER PROGRAM 0 I o E§$~unv 3 2. 3 2. 3 2. 3.3 COMP NUMERICAL RESULTS 4.1 GENERAL 4.2 LINEAR EQUILIBRIUM PROBLEMS 4.2.1 CONCENTRATED INPLANE LOAD AT CROWN 4.2.2 CONCENTRATED OUT-OF-PLANE (TRANSVERSE) LOAD AT CROWN 4.3 NONLINEAR LOAD-DISPLACEMENT BEHAVIOR FOR SMALL DISPLACEMENT PROBLEMS 4.3.1 INPLANE PROBLEMS 4.3.1.1 A 9O°-HINCBD CIRCULAR ARCH SUBJECTED TO UNIFORM RADIAL LOAD 4.3.1.2 A HINGED PARABOLIC ARCH SUBJECTED TO UNIFORM LOAD ON HORIZONTAL PROJECTION 4.3.2 OUT-OF-PLANE PROBLEMS 4.3.2.1 A 9O°-HINCBD CIRCULAR ARCH SUBJECTED TO UNIFORM RADIAL LOAD 4.3.2.2 A HINGED PARABOLIC ARCH SUBJECTED TO UNIFORM LOAD ON HORIZONTAL PROJECTION 4.4 NONLINEAR LOAD-DISPLACEMENT BEHAVIOR FOR INTERMEDIATE DISPLACEMENT PROBLEMS 4.4.1 A 28°-CLAMPBD CIRCULAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN 4.4.2 A 60o-CLAMPED CIRCULAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN 4.4.3 A 60o-CLAMPED CIRCULAR ARCH SUBJECTED TO A SKEW CONCENTRATED LOAD AT CROWN 4.4.4 A CLAMPED MULTIPLE RADII CIRCULAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN 4.4.5 A HINGED SEMI-CIRCULAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN 4.4.6 A CLAMPED SEMI-CIRCULAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN 4.4.7 ARCHES WITH DIFFERENT PROFILES 4.4.7.1 SEMI-ELLIPTIC ARCH 4.4.7.2 CIRCULAR ARCH 4.4.7.3 PARABOLIC ARCH 4.4.7.4 SINUSOIDAL ARCH 4.5 STRESSES AND AMPLIFICATION FACTORS V Page 34 34 34 34 36 38 38 39 42 42 43 43 44 44 45 45 46 46 46 47 48 48 49 50 50 51 51 52 53 53 53 54 54 CHAPTER V. DISCUSSIO 5.1 DISC 5.1. 5.1. 5.1. 5.2 SUMM TABLES FIGURES LIST OF REFEREN( APPENDICES A. NEWTON-R A.1 GENi A.2 HIT A-3 UPD. B. INCREMEN THE AVER CHAPTER V. DISCUSSION AND CONCLUSION 5.1 DISCUSSION 5.1.1 COMPARISON WITH PREVIOUS WORKS 5.1.2 APPROACHES OF NONLINEAR ELASTIC ANALYSIS 5.1.3 NATURE OF [n1] AND [n2] MATRICES 5.2 SUMMARY AND CONCLUSION TABLES FIGURES LIST OF REFERENCES APPENDICES A. NEWTON-RAPHSON METHOD FOR UPDATED COORDINATES A.1 GENERAL A.2 INITIAL STRAIN STIFFNESS MATRIX, [ k6 ] A.3 UPDATED COORDINATES PROCEDURE ° B. INCREMENTAL STIFFNESS MATRICES, [n1] AND [n2], BASED ON THE AVERAGE AXIAL STRAIN MODEL B.1 THE FIRST ORDER INCREMENTAL STIFFNESS MATRIX, [n1] B.2 THE SECOND ORDER INCREMENTAL STIFFNESS MATRIX, [n2] C. COMPUTER PROGRAM GENERAL DESCRIPTION OF SUBROUTINES VARIABLES USED IN THE COMPUTER PROGRAM INPUT DATA ARRANGEMENT COMPUTER PROGRAM "NANCURVE" OOOOO UivL-‘UONH D. INCREMENTAL STIFFNESS MATRICES, [n1] AND [n2], BASED ON THE QUARTIC AXIAL STRAIN MODEL vi Page 57 57 57 59 60 66 68 73 97 101 101 102 107 109 109 117 128 128 129 134 138 198 i'u NBLE 2-1 4-1 4-2 4-3 ACCURAC FOR PAR LINEAR TO A CC LINEAR A CONCI LINEAR TO A Cl TABLE 2-1 4-1 4-2 4-3 LIST OF TABLES ACCURACY OF THE GEOMETRIC REPRESENTATION FOR PARABOLIC ARCH (RISE-9.6", SPAN-48") LINEAR EQUILIBRIUM OF A SEMI-CIRCULAR ARCH SUBJECTED TO A CONCENTRATED IN-PLANE LOAD AT CROWN LINEAR EQUILIBRIUM OF A PARABOLIC ARCH SUBJECTED TO A CONCENTRATED IN-PLANE LOAD AT CROWN LINEAR EQUILIBRIUM OF A SEMI-CIRCULAR ARCH SUBJECTED TO A CONCENTRATED LATERAL LOAD AT CROWN vii Page 68 7O 71 72 FIGURE 1-1 2-1 2-2 2-3 2-4 2-5 3-1 4-1 4-2 4.3 4-4 4.5 4-6 4-7 4.3 LOAD-DE BEAM E1 CROSS-E COORDII TYPICA] TYPICA] COORDI] NEWTON LINEAR TO A C LINEAR A CONC LINEAR WAC A 90°. RADIAL A time 0N HOE A 90°. RADIAU A HIN< 0N H01 A 28°. A VERT FIGURE 1-1 2-1 2-2 2-3 2-4 2-5 3-1 4-1 4-2 4-3 4-4 4-5 4-6 4-7 4-8 LIST OF FIGURES LOAD-DEFLECTION RELATION BEAM ELEMENT (Curved In The x-z Plane) CROSS-SECTION OF PRISMATIC MEMBER COORDINATE SYSTEMS TYPICAL ELEMENT TYPICAL ELEMENT AFTER TRANSFORMATION T0 ELEMENT COORDINATE SYSTEM NEWTON-RAPHSON ITERATION LINEAR EQUILIBRIUM OF A SEMI-CIRCULAR ARCH SUBJECTED TO A CONCENTRATED IN-PLANE LOAD AT CROWN LINEAR EQUILIBRIUM OF A PARABOLIC ARCH SUBJECTED TO A CONCENTRATED IN-PLANE LOAD AT CROWN LINEAR EQUILIBRIUM OF A SEMI-CIRCULAR ARCH SUBJECTED TO A CONCENTRATED LATERAL LOAD AT CROWN A 900-HINGED CIRCULAR ARCH SUBJECTED TO UNIFORM RADIAL LOAD (IN-PLANE BEHAVIOR) A HINGED PARABOLIC ARCH SUBJECTED TO UNIFORM LOAD ON HORIZONTAL PROJECTION (IN-PLANE BEHAVIOR) A 900-HINGED CIRCULAR ARCH SUBJECTED TO UNIFORM RADIAL LOAD (OUT-OF-PLANE BEHAVIOR) A HINGED PARABOLIC ARCH SUBJECTED TO UNIFORM LOAD ON HORIZONTAL PROJECTION (OUT-OF-PLANE BEHAVIOR) A 28°~CLAMPED CIRCULAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN viii Page 73 74 74 75 75 76 77 78 79 80 81 82 83 84 85 FIGURE 4-9 4-10 4-11 4-12 h-13 4-14 4-15 4-16 4-17 “~18 4-19 A~1 A 600-01 CONCENTI A 60°-c1 CONCENTI A CLAMP? TO A VEI A HINGE A VERTI A CLAMP A VERTI ARCH AN SINUSOI TRIANGL A CONCE STRESSI POINT I AXIAL I STRESS! STRESS} CONFIG ELEMEN' IACRAN FIGURE 4-9 4-10 4-11 4-12 4-13 4-14 4-15 4-16 4-17 4-18 4-19 A-l A 60°-CLAMPED CIRCULAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN A 60°-CLAMPED CIRCULAR ARCH SUBJECTED TO A SKEW CONCENTRATED LOAD AT CROWN A CLAMPED MULTIPLE RADII CIRCULAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN A HINGED SEMI-CIRCULAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN A CLAMPED SEMI-CIRCULAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN ARCH AND FRAME PROFILES SINUSOIDAL, PARABOLIC, CIRCULAR, SEMI-ELLIPTIC ARCHES, TRIANGULAR FRAME, AND RECTANGULAR FRAME SUBJECTED TO A CONCENTRATED LOAD AT THEIR CROWNS STRESSES AND AMPLIFICATION FACTORS AT THE QUARTER POINT A ‘ AXIAL FORCE AT CROWN STRESSES AND AMPLIFICATION FACTOR AT CROWN STRESSES AND AMPLIFICATION FACTORS AT THE CROWN CONFIGURATION OF A TWO DIMENSIONAL CURVED BEAM ELEMENT AT SUCCESSIVE LOAD INCREMENTS IN UPDATED LAGRANGE FORMULATION ix Page 86 87 88 89 90 91 92 93 94 95 96 101a I; 1.1 GENERAL The concep gained increasi computation of t displacement re nonlinear anal: Emal)’$is was 8111; amount of compt HOWever, current increasingly afi Nonlinear t which represex response, or to relatIOn. In the pre: elasm1 respon nonlinearity is an efficient In CHAPTER I INTRODUCTION 1.1 GENERAL The concept of basing structural design on ultimate strength has gained increasing acceptance in recent years. In general, the computation of the ultimate strength of a structure would involve load- displacement relationships that are nonlinear. In other words, nonlinear analysis of structure becomes necessary. In the past, such analysis was shunned by engineers because it usually implies a large amount of computations (in addition to theoretical complexities). However, current developments in computers are making such analysis increasingly affordable for engineering practice. Nonlinear behavior of structures may be due to geometric changes, which represent the effect of distortion of the structure on its response, or to material properties such as a nonlinear stress-strain relation. In the present study a procedure for the computation of nonlinear elastic response-of curved beam members is presented. Only geometric nonlinearity is considered. This study was originated from a search of an efficient method of nonlinear elastic analysis of arches or curved structures in two and three dimensional space. "H. I. This chapt work, a literatul the subsequent a1 1.2 OBJECTIVE A] Many engine as curved beams. frames, horizont frames, ship fra Figure 1-1 structure (her: interchangeably) equilibrium equg "fundamental pa load is a relat: Properties of before the limi POInt A'). 1mm path, the Struc Summary patt limit Point, ch If the bifurc; would have bt (48,22)*. WW This chapter describes the objective and scope of the present work, a literature review of related studies, and the notation used in the subsequent analysis. 1.2 OBJECTIVE AND SCOPE Many engineering structures have components that may be considered as curved beams. Several examples are the ribs of'aunfln'bridges, arch frames, horizontally curved highway bridges, the components of aircraft frames, ship frames, and vessel frames. Figure 1-1 illustrates a load—displacement curve of a general arch structure (herein the terms "archfi euui "curved beam" are used interchangeably) which can be obtained by the solution of the nonlinear capfilibrium equations of the system. The curve "0CD” is called the “fundamental path". The point (C) on the fundamental path at which the load is a relative maximum is called a "limit point". Depending on the properties of the arch and loading, a point of "bifurcation" may occur before the limit point (i.e., point A) or after the limit point (i.e., point A'). Immediately beyond the bifurcation point on the fundamental path, the structure is unstable, so that the response could follow the secondary path AB or A/B'. If the bifurcation point occurs before the limit point, the buckling shape would be “antisymmetrical” (sidesway). If the bifurcation point occurs after the limit point,CL the arCh would have buckled at C in a "symmetrical" mode (snap through) (48,22)*. * Numbers in parantheses refer to entries in the list of references. N It should assume that, u; structure would the prebuckling adjacent equilib in magnitude. Consideral on the developme of curved beam linear or stab: studies that h; buckling analy rePresents a s 0n the aSSumpti displacement an those situatiOn small. For mor not Small, it b Problem and Obt the ultimate 1c Nonlinea formulate and studies that ‘ limited to beh; The Obj dlmensional n01 It should be noted that the "classical bucklirgfl theory would assume that, up to the point when buckling takes place , A", the structure would maintain its original undeformed shape. In other words, the prebuckling deformation.is neglected. At buckling, it goes into an adjacent equilibrium configuration, B", which would then be unspecified in magnitude. Considerable amount of work has been done (see literature review) on the development of suitable finite element models for the analysis of curved beams. Most of the previous works have dealt with their linear or stability analyses in the plane of the structure . Past studies that had considered out-of—plane behavior have been limited to buckling analysis (as an eigenvalue problem). Such an analysis represents a short cut procedure to obtaining the ultimate load based on the assumption of no displacement, or a linear relation between displacement and load, prior to buckling. Its application is limited to those situations where the displacement at the incipient buckling,is small. For more general cases, i.e., when the latter displacement is rumzsmall, it becomes necessary to solve the nonlinear equilibrium problem and obtain the corresponding load-displacement curve from which the ultimate load could be determined. Nonlinear equilibrium analysis is, in general, difficult to formulate and expensive to carry out the numerical solution. Past studies that dealt with such analysis of curved structures have been limited to behavior in the plane of the arch. The objective of the present study is to develop a three dimensional nonlinear curved beam finite element which is applicable to both linear am three dimensions The curved of the previous analysis. The have been simpl: the use of the substantially m1 curved axis of in the curvili matrix, the ( Polynomials. Th derived by int respectively by 0f the arch dis The Preser outlined by M; terms of the dj are °°nSidere( displacements FurthermOre’ (nonlinear) te: functims th derived. This ‘ Cllbic ' and q Fields the 1111 both linear and nonlinear analyses of arbitrary geometry in two and three dimensional space. The curved beam element developed herein represents an improvement of the previous model presented by Wen and Lange (45) for buckling analysis. Tflue geometric representation and the displacement functions have been simplified for more convenient application. However, through the use of the "average axial strain", the newrmxkfl.is found to be substantially more effective and accurate than the previous (nus. The curved axis of the element is represented by a second order polynomial in the curvilinear coordinates. In deriving the linear stiffness matrix, the displacement functions are approximated by cubic polynomials. The incremental (or nonlinear) stiffness matrices are derived by interpolating the transverse and longitudinal displacements respectively by cubic and linear polynomials. A Lagrangian description of the arch displacements is used. The present study uses the "incremental stiffness matrices" method outlined by Mallett and Marcal (24). The strain energy is written in terms of the displacement variables. Geometrically nonlinear effects are considered by including both the linear and quadratic terms of the displacements in the expression for the generalized strains. Furthermore, following Wen and Rahimzadeh (47),tflmaquadratic (nonlinear) terms are averaged over the element length . By using these functions the expression.fOr the strain energy of an element is solve the resulting nonlinear equations. Circular, parabolic, and semi-elliptic arches were analysed to obtain their inplane nonlinear responses. They concluded that their mixed model performance was accurate and less sensitive to variations in the arch geometry than that of the displacement model. Belytschko and Glaum (5) presented a higher ...<2-3> ds R 2 ds R 2 ds in which the quantity in the first parenthesis is the usual linear hoop strain in a curved element, and the next two terms (which are nonlinear), represent the contributions to the strain by the rotations of the centroidal axis about the y- and x—axes, respectively. Substituting Equations (2—2) and (2—3) into Equation (2-l), the expression for the longitudinal strain is obtained : dw u l du w dv l 2 2 6 = 5 = ( __ - ‘ ) + ‘ ( —— + - ) + — ( -— ) z s'g’" . ds R 2 ds R 2 ds fl dzv + n (— - ‘—2) ....(2-4) R ds d2u 1 dw d 1 -§ [-—2+——+w-—<—>1 ds R ds ds R 2.3 STRAIN ENERGY EXPRESSION The expression for the strain energy of the element, UE , may be written as (2-5) where U6 is the strain energy due to the longitudinal strain and Ut 15 that due to the shear strain resulting from St. Venant torsion (Note that warping i to be either so following expre vol in which 6 is t and G are the constant of the Of 118ng a su 9- . a g ’ fls d6; tOtal strain e: 2.4 FINITE m In this geometric rep: ene rgy express 18 that warping is not considered herein as the cross-section is assumed to be either solid or a closed tubular one). They are given by the following expressions E 62 E 62 U6 = f dV = f f dAds (2-6a) vol 2 s A 2 GR 1 c _ 2 b Ut = f “35+ vs) ds ....(2-6) s 2 R in which 6 is the longitudinal strain, A is the cross-sectional area, E and G are the Young’s modulus and shear modulus, and RC is the torsion constant of the cross section. In the preceding equation, the notation of using a subscript to represent a differentiation has been used, e.g., fls = dfl/ds . This notation will also be used subsequently. The total strain energy of the element becomes dAds+f—(fls+—V) ds R (2-7) 2.4 FINITE ELEMENT FORMULATION In this section, the definition of coordinate systems, the geometric representation, the displacement functions, and the strain energy expression of a typical curved beam element are presented. ' . ~.,.: ;"‘--u.‘.u_,;r’ 2.4.1 DEFINITI‘ The global consists of a s the crown of t the Y-axis ver curvature. The means of this illustrated ii its origin loce the y-axis nor the curved cem Node B denotes 2.h.2 ELEMENT Referri coordinates o COOIdinate 5y relative Posit The C°<>rdinate are giVen by : Letting ¢ de end Me A) an angle ¢ is a 19 2.4.1 DEFINITION OF COORDINATE SYSTEMS The global coordinate system used in the analysis (see Figure 2-3) consists of a single set of cartesian axes with the origin located at the crown of the arch. The system is oriented with X-axis horizontal, the'Y-axis vertical, and the Z-axis perpendicular to the plane of curvature. The positions of the nodes of the structure are expressed by means of this system. The local or element coordinate system , illustrated in Figure 2-4, consists of one set of cartesian axes with its origin located at node A, with the x—axis in the radial direction” the yuancis normal to the plane of curvature, and the z-axis tangent to the curved centroidal axis forming an angle ¢A with the global X-axis. Node B denotes the end node of the element. 2.4.2 ELEMENT GEOMETRY Referring to the curved element shown hingure 24g the coordinates of the nodes.A and B , with respect to the global coordinate system are, ( XA’YA ) and ( XB’YB ) respectively. Their relative position is defined by XL = XB - XA and YL = YB - YA . The coordinates ( xB, zB ) of node B in the element coordinate system are given by cos ¢A .... (2-8a) x H - X sin ¢A + Y B L L b = ‘ (2-8 ) ZB XL cos ¢A + YL Sin ¢A Letting ¢ denote the angle between the element z-axis (tangent at the end node A) and the tangent at a given point, Figure 2-5 shows that the angle ¢ is a function of s. It varies from zero at node A to 6 at node B. The radii of by R1 and R2. At any poi dz=- ds dx=ds The curve is ap The boundary ( are the follow 20 Bu The radii of curvature R at nodes A and B are respectively denoted by R1 and R2. At any point along the curve, the following relations hold : dz ds cos d .... (2-9a) dx ds sin ¢ .... (2-9b) The curve is approximated by a second order polynomial in ¢ 2 s = bl ¢ + b2 ¢ .... (2-10) The radius of curvature is obtained by defferentiating Equation (2-10) ds R = —— = b + 2 b o ---- (2'11) dd 1 2 The element length is given by L = b a + b 92 .... (2-12) The boundary conditions used to solve for the coefficients b1 and b2 are the following : xB 6 a (l) x- = f dx = f R sin a d¢ ... (2-13 ) B O O ' I 2B 9 b (2) 2B = f dz = f R cos ¢ do ... (2-13 ) The first bounds From the second The rESUItlng ( (2'15) are b = x 2 [(1 x bls _ Thlls, the gEOI] completely de geometric rep] numbers of ele 21 The first boundary condition gives 6 B f ( b1 + 2 b2 a ) sin a as O N II = ( 1 - cos 6 ) b1 + 2 ( sin 6 - 6 cos 6 ) b2 (2-14) From the second boundary condition it is found that 6 2B = f0 ( bl + 2 b2 a ) cos a as = I ( sin 6 ) b1 + 2 ( 6 sin 6 + cos 6 - l ) b2 ... (2-15) The resulting coefficients b1 and b2 obtained from Equations (2-14) and (2-15) are zB ( l - cos 6 ) - xB Sin 6 2 [(l - cos 6)(6 sin 6 + cos 6 - l) - sin 6 (sin 6 - 6 cos 6)] (2-15a) xB - 2 b2 ( sin 6 - 6 cos 6 ) b b = ... (2-15 )- ( l - cos 6 ) Thus, the geometry of the finite element as given by Equation (2—10) is completely defined by XB , 2B and the angle 6.7Hm2accuracy of the geometric representation presented in this section , for different numbers of element, is given in Table 2-1. h 2.4.3 DISPIACEJ As indica the curved ele displacement f the x, y, z axe For the polynomials in the displacemer AS shown in F Eq, (2‘10). Fo functims beco 22 2.4.3 DISPLACEMENT FUNCTIONS As indicated in Equations (2—4) and (2-7),tflmastrain energy of the curved element considered here depends on four independent displacement functions, i.e., u, v, w, and fl , the displacements along the x, y, z axes and the rotation about the z-axis , respectively. For the finite element, these functions are approximated by polynomials in the variable ¢~ In deriving the linear stiffness matrix, the displacement functions are approximated by cubic polynomials 2 3 u = a1 + a2 ¢ + a3 ¢ + 04 ¢ V = a5 + a6 ¢ + a7 ¢2 + a8 ¢3 (2-16) w = a + a ¢ + a $2 + a ¢3 9 10 ll 12 2 3 5 = “13 + “14 ¢ + “15 ¢ + ”16 2 As shown in Figure 2-5 , d is related to the arch length , s , by Eq.(2-lO). Fbr simplicity, the independent variable ¢ in the preceding ¢ equations may be normalized by defining 7 = — , and the displacement 6 functions become 3 A4 7 A8 73 (2-17) 3 A12 7 A 3 r11 In deriving the while the trar polynomials, th polynomials as l v = 35 w = 59 fl = 511 In the norma written as u s ‘ A1 v s " i ‘5 w s “ A9 1 fl = — 11. The follI AS will - him the give 23 In deriving the incremental stiffness matrices [ml] and [n2] , however, while the transverse displacements are still approximated by cubic polynomials, the longitudinal displacements are approximated by linear polynomials as follows _ 2 -— 3 u = l + a2 ¢ + a3 ¢ + 4 v=as+a6¢+a7¢2+38¢3 (2-18) w = a9 + alO ¢ *9=°‘11+‘"12"S In the normalized variable 7 = ¢/6 , the above equations can be written as - - — 2 — 3 u=Al+A27+A3y+A47 V=XS+X67+X772+3873 (2-19) w = A9 + A10 7 fi=211+127 The following discussion explains why different approximations were used for w and 6 in Equations (2-17) and (2-19). As will be shown in the following sections, the linear stiffness matrix [R] is independent of displacements. It can be obtained directly from the given structural properties and geometry. The shape functions 1" defined in Tim: are essential ( nonessential de the solution 1 related to thc (into global or other finite e2 0n the otl stiffness ma quadratic fun Equation (2-1 be done each increase the web condensat derive [k] a] from US ing qu- 24 defined.iJIIEquatirn1 (2-17) requires 16 degrees of freedom, 12 of them are essential (for equilibrium) and 4 of them are nonessential. The nonessential degrees of freedom can be condensed out, once for all, for the solution process. The resulting 12 by 12 stiffness matrix is related to the essential degrees of freedom. It canlxatransflnmwd (into global coordinates) and combined with the stiffness matrices Cd? other finite elements in the usual fashion. On the other hand, as discussed in later sections, the incremental stiffness matrices, [ml] and hfiH , are respectively linear and quadratic functions of the displacements. If the shape functions Equation (2—17) are used, condensation of the stiffness matrices need be done each there is a displacement change. This nnnild greatly increase the cost of computation. Using Equathm1(2-l9)rwnfld.avoid such condensations. Of course one could justziuue.Equation (2-19) to derive [k] ale“). However, the result would be less accurate than that from using Equation (2-17). 2.4.4 ELEMENT STRAIN ENERGY AND STIFFNESS MATRICES In this section, two strain energy expressions are derived. One of them is based on the quartic axial strain function, Equation (2-4), and another aqi 6q. 0 aqi aq. J J The expressions for the integrands in terms of the q's in Equation (2-28) are too lengthy to be presented here. They are given in the subroutine NUMINT of the computer program in Appendix C. It should be noted that the integrals themselves need be evaluated.rnnmarically by Gauss quadrature. The linear stiffness matrix developed in this section has a size of 16 by 16. As mentioned earlier, the nonessential degrees of freedom would then be condensed out. The resulting 12 by 12 lgumaar stiffness matrix is then compatible with the incremental stiffness matrices, [n1] and [n2], developed subsequently. 2.4.4.1.2 INCREMENTAL STIFFNESS MATRICES The incremental stiffness matrices, [n1] and [n2] , are derived by assuming tflurt the transverse and longitudinal displacements are respectively interpolated by cubic and linear polynomials given.in Equations (2—19). As before, by substituting Equations (2-19) into Equations (2-25) and (2-26), U and U become fhuuztions of the 3 4 coefficients ‘:i of the polynomials. The degrees of freedom chosen to replace the coefficients Xi are : u A, LIB, V A7 VB) “7A! WB’ flA) flB! 6 . YA , 6yB , 6XA , and 6XB . These degrees of freedom W111 be denoted collectively by vector { E ). The relation between {.3 l and {-3 } can be obtained by use of Equations (2—19) and the above definitions. Thus, ' 4 1 U3 and U4 car freedom { ii i U - l 3 0 U - f 4 o in which E3 of the q’s, The inc: calculated frc [n1]. [n2]. The exPressio (2‘31) and integrals nee give“ in AP matrices pull of the di SDI; 29 U3 and U4 can be expressed in terms of the element nodal degrees of freedom { a } U3 - f £3 ({ q }) d7 .... (2-29) 0 U4 - i 1?4 (r a }) d7 .... (2-30) in which f3 and f4 are, respectively, cubic and quartic functions of the E's. ] The incremental stiffness matrices , [n1] and [n2] , may be calculated from : ‘ 32113 1 32 £3 [nil-inlijl-[T—Z'l-[f -_——_-d71..<2-31) aqi aqj o aqi aqj 62114 1 32 f4 [ n2 1 - [ n2ij 1 - [ -:f-:f 1 - i f '-:7——:f d7 ] .- (2-32) aqi aqj o aq1 aqj The expressions for the integrands in terms of the q's in Equations (2-31) and (2-32) are also too lengthy to be presented here . The integrals need be evaluated numerically by' Gauss quadrature. They are given in.Appendix D. It should be noted that the elements of the matrices [n1] and [n2] are respectively linear and quadratic functions of the displacements. 2.4.4.2 AVERAl An alter section is to ‘ their average axial strain 1 As before, it Can be obtai‘ Equation (2-1 By me of Eq‘ element can I 30 2.4.4.2 AVERAGE AXIAL STRAIN MODEL An alternative to the strain energy derived in the preceding section is to replace the nonlinear terms in the strain expression by their averages over the element length. Thus, the expression for the axial strain is rewritten from Equation (2-3) as dw u l L du w 2 62 IO = ( —— - — ) + ——— f ( —- + — ) ds ds R 2 L 0 ds R l L dv 2 + — f (—) as ....(2-33) 2 L 0 ds As before» ina‘terms of 7 , the expression for the longitudinal strain can be obtained by substituting Equations (2-2) and (2-33) into Equation (2-1) w u l u w 7 l 7 2 e = ( -—— - — ) + ——— f ( ——— + — ) R 9 d7 R 6 R 2 L o R a R 1 fl v7 2 fl v77 +— (—)R0d7+n(‘-—“—-V7) 2 L o R a R R262 7 55 u W W 77 7 ) g ( + u 7 + + ~ 7 .... (2-34) R262 7 55 R2 6 R 57 By use of Equations (2-5), (2—22), and (2—34), the strain energy of the element can be written as (2-35) l in which U2 U4 are given bj =2 U3 E U=— 1* 2 where H:- 2 It represents 2.4.4.2.1 LI Since th average axia axial Strain the same as integrating r be C°ndensed 2.4.4.2.2 1] By use 4 and using ex the exPIP—SS lntegrals Ca‘ 31 in which U2 remains the same as given by Equation (2-24), and U and U4 are given by the following expressions 1 U3 =EA £(w7-6u) Md7 ....(2-36) EA 1 _ 2 U4= fR6 (M) d7 ....(2-37) 2 o where l u w v 1 ’Y 2 72 M =— f[(———+—) +(—)]R6d7...(2-38) 2L 0 R6 R R6 It represents the average of the nonlinear part of the axial strain. 2.4-4.2.1 LINEAR STIFFNESS MATRIX Since the quadratic term of the strain energy U2 , based on.the average axial strain model, remains the same as thatcflftbe quartic axial strain model, the resulting linear stiffness matrix is exactly' the same as that discussed in Section 2.4.4.1.1. As before, after integrating numerically, the nonessential degrees of freedom can then be condensed out to get a 12 by 12 stiffness matrix. 2.4.4.2.2 INCREMENTAL STIFFNESS MATRICES By use of Equations (2—36) and (2-37) for U3 and U4, respectively, and using exactly the same procedure as described in Section 2.4.4.1.2, the expressions for [nl] and [n2] {Q} - Lo] m (2-40) 2 6 12 The first var equation : ([Kl+ This represent first parenths The equa‘ from the seco: ([Kl+ in which (6} equilibrium displacement Parenthesis j denotes the at {Q} a {6 Equatii t° dWelop a beam Structu 33 The first variation of the potential energy produces the equilibrium equation : ( [K] + - [N1] + - [N2] ) { Q } = { P } ... (2-41) 2 3 This represents a set of nonlinear algebraic equations. The term in the first parenthesis is called the SECANT STIFFNESS MATRIX. The equations governing the linear incremental.lmflurvior follow from the second variation of the potential energy and are given by : ( [K] + [N1] + [N2] { no } = 1 AP 1 ... (2-42) >16} in whixfln (O) denotes the displacements at a reference (or the current) equilibrium position , and { AQ ) and { AP } are the incremental displacement and load vectors, respectively. The temnirrthe first parenthesis is called the TANCENT STIFFNESS MATRIX. The subscript {6} denotes tine fact that the tangent stiffness matrix is to be evaluated at {Q} {6}. Equations (2-41) and (2—42) will be used in the following chapter to develop a procedure for nonlinear equilibrium analysis; i) would get sufficiently close to Q. The load displacement relation can be written as { P } = { f (Q) } Using a first order Taylor series expansion about Qi we have a f { P } = { f (Q.) } + {- } { A Q. } 1 8 Q Q 1 j i in which, [ f (Qi) ] may be interpreted as representing the ' 6 f resistance of the structure corresponding to Qi , and { 6 Q. J the tangent stiffness at Qi . Then the modification to Qi is at -1 { A Q. } = { } { P - f (Qo) } 1 a Qj Q1 1 6f -1 ={ } {ARi} 3 Qj Qi in which { A Ri } is the unbalanced force vector at stage Qi' The modified displacement is (3-1) (3-2) elastic ) as Q. l <3-3) <3-4) The proce sufficiently illustrated g] The prec1 increment. Gr‘ the load in each incremen At the b may not be up nonlinear ar large displac large disp] involved. 811 arches that problem type However, it "initial st, Procedure _ 3'“ NEW In thj Steps 0f the 1) Giv¢ dis] res loa 36 The process may be repeated until either A Qi+k or A Ri+k is sufficiently small. For one degree of freedom system, the process is illustrated graphically in Figure 3-1. The preceding discussion was for the load applied as a single load increment. Greater accuracy in the solution may be obtained by applying the load in increments (e.g., A P, 2 A P, 3 A P, ...... , etc ). For each increment the concept described previously applies. At the beginning of the increment the geometry of structure may or may not be updated. As will be discussed in chapter IV, the problems of nonlinear analysis of arches are divided into small, intermediate, and large displacement categories. The updated procedure is necessary if’ large displacements (more tharu say, 25% of the arch span) are involved. Since practical designs in civil engineering would resuflJ: in arches that fall either in the small or intermediate displacement problem type, the updated procedure is not presented in this chapter. However, it is given in Appendix A together with the derivation of the "initial strain stiffness matrix" , [k6 ] , which is needed by the o procedure. 3.2.2 NEWTON-RAPHSON METHOD FOR FIXED COORDINATES In this case the geometry of the structure is not updated. The steps of the calculation are as follows 1) Given the current state displacement {Q} = {6} resistance {R} = {R} = [ KS ] {6} load {P} = {f} 2) Checl stop. 3) Form ret and 37 2) Check if the intended total load has been applied. If it has, 3) 4) 5) 6) 7) 8) 9) stop. Otherwise, increase load to {P}. Form the structural tangent stiffness matrix , [KT] , as [KT]=[K]+[N1<{6})]+[N2<{Q}>] Solve for { A Q } from the linear equations, [KT]{AQ}={AR}={P}-{E} Add {A Q} to the latest {Q} to obtain anew {Q} = {Q} + {A Q}. Based on the new {Q} from step 5, evaluate Nl({Q}) and N2({Q}). Form the tangent stiffness matrix [KT], secant stiffness matrix [KS] , and resistant force vector as [ KT ] = [ K i + [ Nl({Q}) ] + [ N2({Q}) ] l l [ K ] = [ K ] + “‘ [ Nl({Q}) ] + “' [ N2({Q}) ] S 2 3 Resistant force vector { R } = [ KS ] { Q } Evaluate the unbalanced force vector { A R } as { A R } = { P } - { R } If the unbalanced force vector, { A R }, is sufficiently small, return to 2 . Otherwise , set { Q } = { Q }, and { R } = { R }, and return to 3. 38 3.2.3 CONVERGENCE CRITERION In.imp1ementing the above Newton-Raphson method, a convergence criterion based on unbalanced force vector has been used. A tolerance 5f ,vflfich.has the unit of force or moment, is prescribed for each group of components (i.e., force or moment) of the unbalanced force vector. After the evaluation of the unbalanced force vector in each iteration the absolute value of each component of the vector is independently compared with the prescribed tolerance. Convergence is considered achieved if, for each component, this absolute value is less than or equal to the tolerance. 3.2.4 STRESS COMPUTATION Referring to the calculation steps presented in Section 3.2.2 , element end forces and stresses at every load increment can be obtained at the beginning of step 9 as follows a) When the unbalanced force vector , {AR}, is sufficiently small, form the current element displacement vector, {q}, and the current element secant stiffness matrix, [ks], as l . l [ ks l = [ k ] + ‘ [ nl({q}) ] + — [ n2({q}) ] 2 ‘ 3 b) Element end forces, { r } = [ ks ] { q } c) Letting P , M , M , c , and c denote , respectively, the xx yy x y axial force, the moment about x-axis, the moment about y-axis, the distance from the y-axis to the extreme fiber , and the 39 distance from the x-axis to the extreme fiber , the stress at any element end , a , can be computed as P M cX xx c 0 g _ + _XI___ + ___—.1 A I I YY XX 3.3 COMPUTER PROGRAM In this section, a general description of the computer program developed for this study is presented. The program was named NANCURVE , whiCh stands for Nonlinear Analysis of Curved Beam Structures. As discussed previously, the program has a capability of solving nonlinear equilibrium problem of arbitrary curved beam structures either in two or three dimensional space. The program can also be used for solving linear equilibrium problems when the [N1] and [N2] nuurrices are set equal to zero. The program itself and the corresporfling input data example are given in Appendix C. The major steps 111 due program are described in the same order in which they are executed l) The basic information concerning the physical description of the arch is input . This information includes the number of elements, the number of nodal points, and the type of arch. 2) Parameters which specify whether both [N1] and [N2] are to be used , or [Nl] only , or neither of them in the solution of linear equilibrium problems, are then input. I 3) The global coordinates of the nodes are input with the parameters defining the boundary conditions of the arch. 4O 4) Next , the maximum number of iterations , the unbalanced force 5) 6) 7) 8) tolerance , the applied loads ( initial , increment , and total loads ) and their orientations are specified. The element data is input . This includes the element number , the node numbers at element ends , the modulus of elasticity , the shear modulus , the cross sectional area , the moments of inertia about the two principal axes , and the torsion constant of the cross section. From the information input in l and 3 , the slopes of the tangent at the end nodes of each element are calculated . For arbitrary arch type, however, these slopes are to be input (for convenience, these are input in step 3 together with the global coordinates of the nodes) . Next , the coefficients b1 and b2 for defining the geometry of each curved element are computed. The radius of curvature at each node and the element lengths are then evaluated. The number of Gauss points , which is needed in the numerical integration for evaluating the linear stiffness matrix, [k], is specified. The numbers of Gauss points available in the program are 2, 3, 4, 5, 6, 10, and 15. From the information input in l and 3 also , the semi bandwidth of the structural stiffness matrix is computed. The element linear stiffness matrices are then evaluated and assembled into the linear stiffness matrix of the structure. This matrix is assembled in banded format and due to symmetry only the upper semibandwidth is constructed. 4l 9) Based on the initial applied load input in 4, a linear analysis 10) ll) 12) of the arch is performed to obtain the displacements of the nodal points . The displacements so determined are used to compute, for each element, the matrices [hi] and [n2] which are then assembled ( also in banded format ) into the structure incremental stiffness matrices [N1] and [N2]. The rest of the steps , which are given previously in section 3.2.2 , can then be performed to obtain the nonlinear response of the arch . In performing those steps , possible instability along the solution path is tested by checking the determinant of the tangent stiffness matrix, [KT], at every load increment. Load-displacement relations can then be computed and the critical or limit load of the structure can also be determined. By using the end displacements and the secant stiffness matrix of each element obtained in 10, the element end forces at every load increment can be computed. Finally , the stresses' ( axial and total ) can be evaluated. 4.1 GENERAL In this of-plane behe the finite with analyti reliable for When so PIOblems '1 Categories. displacemer about 2% of the deflec1 the Problem Numer out'0f‘plan arch tyPe, into the ef response, having the the “Umbe investigate CHAPTERIV NUMERICALIRESUETS 4.1 GENERAL In this chapter a number of numerical examples of inplane and out- of-plane behavior of arches are considered. Firstly, a comparison of the finite element solutions of linear equilibrium problems was made with analytical solutions to show that the method presented is also reliable for linear case. When solving nonlinear load-displacement problems, we divide the problems into "small", "intermediate", and "large" displacement categories. This is a relative classification. What we mean by a "small displacement" problem is the case in which the deflection is less than about 2% of the arch span. "Intermediate displacement" problem means the deflection is of the order of 2 - 25% of the arch span. Beyond 25% the problem is called "large. displacement" problem. Numerical results were obtained involving arches with inplane and out-of—plane behavior. Various types of loading, support condition, arch type, and arch geometry are considered. To provide some insight into the effects of variations in the arch profile on its nonlinear response, semi-elliptic, circular, parabolic, and sinusoidal arches having the same rise to span ratio were considered. The influence of the number of elements on the accuracy of the results was also investigated. In addit: in the struc1 factors for ratios of non For the have been giv consistent; be inches an concentrated in}, 1b., a: 4.2 LINEAR Two typ problems f0 Plane load a 4.2.1 CONCl The so] Parabolid were used t. The i CirCular at 4‘1 Shows the crown a of elements The c' solutiOn c HOWeVer, ti Very Smal: 43 In addition to the displacement response, the response of stresses in.the structure is also investigated. Furthermore, amplification factors for displacement and stresses, which are defined1x>be the ratios of nonlinear response to linear response, are studied. For the:rnnnerical results presented herein, no units of the data have been given. The dimensions of the various (punitities are self- consistentn :i.e., if the basic units of length and force are taken to be inches and pounds, then the values of area, moment of inertia, concentrated load, and distributed load given would have units of in.2 in.4, 1b., and lb/in., respectively. 2 4. 2 LINEAR EQUILIBRIUM PROBLEMS Two types of problems were solved. They are linear equilibrium ‘problems fOr arches subject to a concentrated inplane load and out-of- plane load at the crown. 4.2.1 CONCENTRATED INPLANE LOAD AT CROWN The solution was obtained for Um) types of arches, circular and parabolic. In.both cases the symmetry of the load and of the structure were used to reduce the number of equations. The first problem investigated was linear analysis of semi- circular arch subjected to a concentrated inplane load at crown. Figure 4-1 shows tiue(){l% of P (P = 56.25 q) applied at the crown in +Z direction was introduced. Because of the symmetry of the geometmfif, loading, and ‘buCkling mode, only one half of the arch was considered. For different number of elements, the resulting load-displacement curves are shown in Figure 4-7. The results were obtained with a load increment of 20 (equivalent to q a 0.3556) and an unbalanced force tolerance e = 1.0% f of the lxuui increment. As in the previous case, the load-displacement curves indicated ultimate loads very close to the buckling loads obtained by (from eigenva 4.4 NONLINEI DISPIAC] As def: class of pro] the arch spa‘ of arches fa the small di The nu: because the different me made to ex Problems ch4 What may } influence 0 also invest 4.4.1 A 23 LOAE The p1 tree of s incremental} be used to The 1. with a lo. 1°ading Wa approxima 48 obtained by Wen and Lange (45) as well as by TokarzemuiSandhu (40) (from eigenvalue problem solutions). 4.4 NONLINEAR LOAD-DISPLACEMENT BEHAVIOR FOR INTERMEDIATE DISPLACEMENT PROBLEMS .As defined previously, the intermediate displacement problem is a class of problems where the deflection is of the order of 2 - 25% of the arch span. It should be noted that the common practical proportions of arches fall either in this category or in the previous one, i.e.: the small displacement problem. The numerical examples presented in this section were chosen because they had been solved by other investigators using various different methods of nonlinear analysis. Thus, a comparison can then be made to examine the accuracy of the proposed method. The example problems chosen also include a range of rise to span ratios covering ‘what may'tne regarded as "shallow" as well as "deep" arches. The influence of the number of elements on the accuracy of the results was also investigated. 4.4.1 A.28o-CLAMPED CIRCULAR.ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN The problem, which is illustrated in Figure 4-8, falls into the type of shaldxno arch (5). Because this arch remains stable, an incremental load procedure described in the proposed method could still be used to determine the entire response. The loadedisplacement curves shown in Figure 4—8 were obtained with a lxuui increment of 2000 and ef= 0.5% of the load increment. The loading was continued until the apex had displaced an amount equal to approximately 1.5 times the initial rise or 9.5% of the arch span. The configuratior Different m half of the 4 It can 1 used to rep displacemen Glaum (5) wi solution w elements we] the figure. Figure obtained b; results, ho method with 4.4.2 A 6c COM The p] load incr Critical 1. increment displaceme Diff the one he Figure 1,. curves f0] elements Those Cur 49 configuration remains symmetric about the apex throughout deformation” Different ruaners of element, i.e.: 2, 4, and 8, to represent the one half of the arch were considered. It can be seen in Figure 4-8 that even when only two elements were used to represent the one half of the arch, the resultjxug load- {displacement curve was close enough to that obtained by Belytschko and Glaum (5) with 10 elements, which is extremely close tx>ema analytical solutirniiflnich may be considered to be "exact" (5). When 4 or 8 elements were used, the results were of course better, as also shown in the figure. Figure 4~8 also shows the resulting load-displacement curves obtained by Belytschko and Glaum (5) with 2 and 5 elements. These results, however, are less accurate than that obtained by the proposed. method with 2 elements. 4.4.2 A 6OO-CLAMPED CIRCULAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN The problem is illustrated in Figure 4~9. It was solved with.a load increment of 50 and ef= 1% of the load increment. Near the critical load level, where the displacement increases rapidly, the load increment was halved to get more data points for drawing the load displacement curve. Different number of elements, i.e.: 2, 4, 8, and 16, to represent the one half of the arch were considered. The results are given in Figure 4—9. As shown in the figure, the resulting load-displacement curves for 4, 8, and 16 elements have no significant differences. For 2 elements the curve is somewhat stiffer but is stihlcndte accurate. Those curves agree very well with that obtained by Calhoun and,DaDeppo (8). The c] displacement DaDeppo and for their eig deflection 1: 4.4.3 A 60° LOAD The pro the same 10 Because of t considered t The re: element, 1 curves agrew The result DaDePPO and 4.25% of th 4.4.4 A c] CON! This 1 methOd am As discuss input are The a radii hav concentrat 50 (8). The critical load obtained and the corresponding crown displacement also agree very well with those obtained analytically'by' DaDeppo and Schmidt (11) and Austin and Ross (2) who used 24 elements for flufix eigensolution. In this problem, the order of the maximum deflection is approximately 5% of the arch span. 4.4.3 .A GOO-CLAMPED CIRCULAR ARCH SUBJECTED TO A SKEW CONCENTRATED LOAD AT CROWN The problem, which is illustrated in Figure 4-10, was solved with the same load increment and tolerance 6f as in the previous example. Because of the present of the horizontal load, the entire arch.was considered to obtain the response. The resulting loadedisplacement curves for different numbers of element, i.e.: 4, 8, and 16, are shown in Figure 4-10. All of the curves agree very well with that obtained by Calhoun and DaDeppo (8). The resulting critical loads also agree well with that obtainedlxy DaDeppo and Schmidt (11). The maximum deflectjxna is of the order of 4.25% of the arch span. The buckling mode is antisymmetry. 4.4.4 A CLAMPED MULTIPLE RADII CIRCULAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN This problem demonstrates the extended capability of the proposed method and.the computer program for solving an arbitrary arch profile. As discussed in Section 3.3, the only geometric data whicirrurve to be input are the nodal coordinates and the end slopes of the elements. The arch has two different radii, R1= 200 and R2= 100. The two radii have a.common tangent point at the crown of the arch. A vertical concentrated load is applied at the crown, as shown in Figure 4-11. Since t considered. 1 to solve thl the load incr are shown i; Calhoun and maximum defl 4.4.5 A BIT TOAD The prc symmetrica Proposed m1 increment, representin be seen in by using 15 order of th 4.4.5 A or com The p] the arch i was S°lved The reSul- (equivalen repreSent 51 Since the geometry is not symmetry, the entire arch should be considered. Two different numbers of element, i.e.: 4 and 8, were used to solve the problem. A load increment of 50 and a tolerance of 2% of the load increment were used. The resulting load-displacement curves are shown in Figure 4-11. Both curves agree well with that obtained by Calhoun and DaDeppo (8) . The resulting buckling mode is symmetric. The maximum deflection is of the order of 6.75% of the arch span. 4.4.5 A HING- SEMI-CIRCUIAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN. The problem is illustrated in Figure 4-12. In this example, its symmetrical response was analysed. The problem was solved by the proposed method with a load increment of 1 and - 1% of the load 6:? increment. The resulting load-displacement curves for 4 and 8 elements representing the one half of the arch are shown in Figure 4-12. As can be seen in the figure, the results agree very well with that obtained by using 16 straight beam elements of Wen and Rahimzadeh (47) . The order of the displacement is approximately 25% of the arch span. 4.4.6 A CLAMP- SEMI-CIIRCULAR ARCH SUBJECTED TO A VERTICAL CONCENTRATED LOAD AT CROWN The problem is illustrated in Figure 4-13. Two thickness ratios of the arch were considered, namely h/R - 0.05 and 0.005 . The problem was solved by the proposed method with a load increment of 0.384 EI/Rz. The resulting load-displacement curves for 3 , 4 and 8 elements (equivalent to 7 , 10 , and 22 degrees of freedom, respectively) representing the one half of the arch are shown in Figure 4-13. The results agrt (equivalent mixed formull approximatel‘ 4.4.7 ARCHE To pr solutions, sinusoidal 1 sectional p1 at their bo crown. The different . increment o increment. where the d The p] be regardec Symmetric; Straight or structures Figur. it is see: that of th higher th that of t} 52 results agree very well with that obtained by using 6 elements (equivalent to 3.7 degrees of freedom) of Noor et a1 (27) based on a mixed formulation of finite element. The order of the displacement is approximately 20% of the arch span. 4.4.7 ARCHES WITH DIFFERENT PROFILES To provide further comparisons of results with existing solutions, arches with semi-elliptic, circular, parabolic, and sinusoidal profiles having the same rise to span ratio, span, and cross sectional properties were considered. All arches are hinged supported at their both ends and are subjected to a concentrated vertical load at crown. The symmetrical buckling of those arches were analyzed by using different numbers 'of element. All problems were solved with a load increment of 100 and an unbalanced force tolerance ef- 1% of the load increment. The load increment was halved near the critical load level, where the displacement increases rapidly. The profiles of a "rectangular frame" and a "triangular frame" may be regarded as the limiting cases of the arches mentioned above. The symmetrical responses of these frames were also investigated by using straight beam elemnts of Wen and Rahimzadeh (47). The profiles of these structures are shown in Figure 4-14. Figure 4-15 shows the load-displacement curves. From the figure it is seen that the stiffness of the sinusoidal arch is higher than that of the parabolic arch; the stiffness of the parabolic arch is higher than that of the Circular arch, which is in turn higher than that of the semi-elliptic arch. Furthermore, the stiffness of the "triangular frame" is higher than that of the sinusoidal arch, and the stiffness of elliptic art shorter the stiffness. T proportion o and thus gre 4.4.7.1 SE! The p1 freedom) of element. T1 (23 degrees half of the As sho Obtained] methods. Th freedom), Span. 4.4.7.2 c] Using Huddlestox 20 element For Was Solved curves ag 4'15- More load Obta order of 1 53 stiffness of the "rectangular frame" is lower than that of the semi- elliptic arch. Thus it appears that for this type of loading, the shorter the total curved length of the structure, the greater the stiffness. The reason seems to be that shorter length implies a greater proportion of the load being carried by axial force than by bending, and thus greater stiffness. 4.4.7.1 SEMI-ELLIPTIC ARCH The problem has been solved by using 6 elements (38 degrees of freedom) of A.K. Noor et a1 (27) based on a mixed formulation of finite element. The problem was also solved by using 8 straight beam elements (23 degrees of freedom) of Wen and Rahimzadeh (47) representing the one half of the arch. As shown in Figure 4-15, the resulting load-displacement curve obtained by proposed method agrees very well with those of other methods. The problem was solved by using 4 elements (11 degrees of freedom). The maximum deflection was of the order of 12.5% of the arch span. ‘ 4.4.7.2 CIRCUIAR ARCH Using different approaches, the problem had been solved by Huddleston (20) analytically, and Fuj ii and Gong (17). The latter used 20 elements to represent the one half of the arch. For different numbers of element, i.e.: 2, 4, and 8, the problem was solved by the proposed method. The resulting load-displacement curves agree very well with those of other methods, as shown in Figure 4-15. Moreover, the curves seem to asymptotically approach the buckling Load obtained by Austin and Ross (2). The maximum deflection is of the order of 12.5% of the arch span. 4.4.7.3 PAR The pro‘ 6 elements behavior see are very (:1 obtained by resulting c with that 01 order of 12 4.14.7.1: SI The elementS, '1 the result with 4 eleu span. As < each other 4.5 Stuns In pr °°mm0n tha used to e reSPOnse, Which is linear re: In 1 Structurt 54 4.4.7.3 PARABOLIC ARCH The problem was analysed by the proposed method with 2, 3, 4, and 6 elements representing the one half of the arch. For 2 elements, the behavior seems to be too stiff. For 3, 4, and 6 elements, the curves are very closed to each other, and they agree very well with that obtained by Fujii and Gong (17) using 20 elements. As before, the resulting critical load and the corresponding deflection also agree with that obtained by Austin and Ross (2). The deflection is of the order of 12.5% of the arch span. 4.4.7.4 SINUSOIDAL ARCH The problem has been solved by Fujii and Gong (17) with 20 elements. The result is plotted in Figure 4-15. The figure also shows the resulting load-displacement curve obtained by the proposed method with 4 elements. The order of the deflection is about 12% of the arch span. As can be seen in Figure 4-15, both curves are quite close to each other . 4. 5 STRESSES AND AMPLIFICATION FACTORS In practice, especially in the preliminary design stage, it is common that a simpler method so called "amplification factor method" is used to estimate nonlinear response of structure from its linear response. Therefore, information regarding the amplification factor, which is defined to be the ratio of the nonlinear response to the linear response, is very useful. In this section, the response of internal stresses in the structure is first investigated. 'Two types of stresses, i.e., axial stress and ' considered. factor for s amplificatio As nume factors of 4.4.l , and Figure point of a horizontal presented. ' stress, tot the displac amplificat the diSpla reSpectiv elements re Figul Shallow ci: croon. In load is in 27,000. ( as the in Pmblem wa arch. The 55 stress and total stress (axial stress plus bending stress) are considered. The results are then used to obtain the amplification factor for stresses. Similarly, by using the displacement responses the amplification factor for displacement can also be obtained. As numerical examples, the resulting stresses and amplification factors of the problems discussed previously in Sections 4.3.1.2 , 4.4.1 , and 4.4.2 are presented. Figure 4.16 shows the axial and total stresses at the left quarter point of a hinged parabolic arch subjected to uniform load on horizontal projection. Both linear and nonlinear responses are presented. The figure also shows the amplification factors for axial stress, total stress, and displacement. From the figure it is seen that the displacement amplification factor is larger than the total stress amplification factor. Near the critical load level, the magnitude of the displacement and total stress amplification factors are, respectively, 2.3' and 1.3 . The results were obtained by using 8 elements representing the entire arch. Figure 4-17 shows the axial force at the crown of a 28° clamped shallow circular arch subjected to a vertical concentrated load at crown. In this problem, the axial force initially increases when the load is increased. However, when the intensity of the load is about 27,000. (i.e. , when the crown vertical displacement is about the same as the initial rise of the arch), the axial force decreases. The problem was solved by using 8 elements representing the one half of the arch. The result agrees very well with that obtained by Belytschko and Glaum with 10 elements. The lin are shown 1 factors for amplificatic is increas amplificatf and larger ‘ factors ar figure it 1 factors 1‘ intensity c amplifica decrease. A factor is maximum mal factors ar Figur crown of Concentrat factors f Were obtaf arch_ As is larger magnitud. are) rESp 56 The linear and nonlinear responses of the axial and total stresses are shown in Figure 4-18. The figure also shows the amplification factors for axial stress, total stress, and displacement. The amplification factor for axial stress initially increases when the load is increased. When the load intensity is about 27,000. the amplification factor decreases. When the load intensities are 30,000. and larger than 30,000. the magnitude of the axial force amplification factors are, respectively, equal to one and less than one. From the figure it is seen that the total stress and displacemet amplification factors initially increase when the load is increased. When the intensity of the load is about 30,000. (i.e. , when the axial stress amplification factor is equal to one), the amplification factors decrease. As in the previous problem, the displacement amplification factor is larger than the total stress amplification factor. The maximum magnitude of the displacement and total stress amplification factors are, respectively, 3.9 and 2.2 . Figure 4—19 shows the linear and nonlinear total stress at the crown of a 600 clamped circular arch subjected to a vertical. concentrated load at crown. The figure also shows the amplification factors for axial stress, total stress, and displacements. The results were obtained by using 8 elements representing the one half of the arch. As in the previous cases, the displacement amplification factor is larger than that of total stress. Near the critical load level, the magnitude of the displacement and total stress amplification factors are, respectively, 3.8 and 2.0 . 5.1 DISCUSS? In the nonlinear c nonlinear a dimensional obtained, 1 following s< 5.1.1 coup Previo 12 . 4-13 a model cempe a) In th: CHAPTER V DISCUSSION AND CONCLUSION 5.1 DISCUSSION In the preceding chapters the development of a three dimensional nonlinear curved beam element and its applications to linear and nonlinear analyses of arches with various geometry in two and three dimensional space have been presented. From the numerical results obtained, the features of the proposed method are discussed in due following sections. 5.1.1 COMPARISON WITH PREVIOUS WORKS Previous comparisons, as given in Figures 4-8, 4-9, 4—10, 4-ll, 4- 12 , 4-13 and 4-15 , indicate that the proposed "Averaged Axial Strain" model competes very well with the other models. a) In Figure 4-8 it is shown that by using only 2 elements with the proposed method , the accuracy of the result is comparable to that obtained by Belytschko and Glaum (S) with 10 elements, the result of which is very close to an analytical solution. The results of Belytschko and Glaum using 2 and 5 elements indicated considerable differences from the correct results. b) Figure 4-9 indicates that using only 2 elements with the prOposed method , the result is as accurate as that obtained by Calhoun and DaDeppo (8) using 8 elements for the converged 57 answe anah c) For resu elen the d) For res and e) Fig prc 58 answer . That limit load value also agrees very well with an analytical solution . The result of Calhoun and DaDeppo with 4 els. showed considerable distance from the correct solution. c) For' the skew loading problem , as shown in Figure 4-10 , the result obtained by the proposed method with 4 elements is comparable to that of Calhoun and DaDeppo (8) with 8 elements . The resulting limit load also agrees very well with the analytical solution. d) For the multiple radii problem, as shown in Figure 4-11 , the result obtained by the proposed method. using 4 elements is in very good agreement with the converged result of Calhoun and DaDeppo (8) using 8 elements. e) Figure 4-12 indicates that , for the hinged semi-circular arch problem, the result using 4 elements of the proposed method is very close to that of Reference 47 using 16 straight beam elements. f) Figure 4-13 indicates that , for the clamped semi-circular arch problem , the result using 3 elements (7 degrees of freedom) of the proposed method is in very good agreement with that of Noor et a1 (27) with 6 elements (37 degrees of freedom). g) Figure 4-15 indicates that, for parabolic and circular arches, the result using 2 elements of the proposed method is close to that of Fujii and Gong (17) with 20 elements . For sinusoidal arch , 4 elements of the proposed method gives the result comparable to that for 20 elements of Fujii and Gong . For the semi-elliptic arch, 4 elements ( 11 degrees of freedom ) of the proposed method gives the result comparable to that for 6 eleme 8 str In ten preceding con than the ot' clamped semi closest com proposed me formulatior general strt number of reference i: 5.1.2 APPR It is to be too by Suffici membrane 1, T0 ow a) To Th 59 elements (38 degrees of freedom) of Noor et a1 (27) as well as 8 straight beam elements (23 degrees of freedom) of Ref. 47. In terms of number of elements needed for accurate results, the preceding comparisons show that the proposed method is more effective than the others. Among the compared results, those related to the clamped semi-circular and elliptic arches of Reference 27 showed the closest competition . However, the results are still in favor of the proposed method . Furthermore , as discussed previously , the formulation in Reference 27 is not as conveniently adaptable for a general structures computer program as the proposed method, and the number of degrees of freedom per end node of the element in that reference is actually twice that in the proposed one. 5.1.2 APPROACHES OF NONLINEAR.ELASTIC ANALYSIS It is generally known that curved beam elements has the tendency to be too stiff unless the in-plane displacement field is represented by sufficiently high order polynomials. This phenomenon is called membrane locking (37). To overcome this problem, four approaches have been suggested : a) To use higher order polynomials for the displacement fields. This approach was taken by Dawe (l6) , using quintic functions for linear analysis. b) To use a mixed formulation of the finite element , as in the work of Noor et a1 (27). c) To use "reduced integration" , as suggested by Stolarski and Belytschko (37). d) To use the "average axial strain" model, as described herein. Using h analysis woul involve a la that is the 1 (b), as ment: are cons ldEI involved and structures ' carried out evaluation ( “mathematic: does not apj zero energy The pr above diff the other e and the ac< 11.3 NAT The results th the aVera Stiffness, in the £07 a nonline. Con: strain [2 60 Using higher order (e.g. quintic) polynomials in a nonlinear analysis would be much more unwieldy than lfléi linear one. It would involve a large amount of work with no guarantee of success. (Perhaps, that is the reason why it had not been tried.tfluns far.) In approach (b), as mentioned previously, since both nodal displacements and forces are considered as degrees of freedom, matrices of larger size are involved and the formulation is inconvenient for inclusion in a general structures program."In approach (0), the numerical integration is carried out by using only 1 or 2 Gauss points, rather than an accurate evaluation of the integral as defined by the analysis. Thus there is a "mathematical looseness" or computational artificiality involved, which does not appear desirable. Furthermore, the possibility of existence of zero energy mode should be of concern. The present study indicates that approach (d) overcomes all the above difficulties. The procedure is simpler andrmnxaefficient than the other approaches, the integration is carried out an; it should be, and the accuracy of the results is generally better. 5.1.3 NATURE OF [n1] AND [n2] MATRICES The reason why the averaged strain model produces more accurate results than the unaveraged strain model appears to be:t1ue fact that the averaging process reduces the strain energy and thus decreases the stiffness to the correct order of magnitude. An analysis ieslyresented in the following. A similar analysis was also given in Reference 47 for a nonlinear straight beam element. Consider a two dimensional curved beam element with the nonlinear l u w strain term , — ( -1 + - ) , being unaveraged in one case 2 R 0 R and averaged expressions f the two cases The exp models may l as follows : U 3 (unaverage U 3 (averaged Consid 0f the axia varying. Constant w may be ta written as E A 1 2f0 61 and averaged in another . Corresponding to Eqs. (2-20) and (2-34) , the expressions for U2 are the same. Thus the linear stiffness matrices for the two cases are identical. The expressions of U3 for the unaveraged and averaged strain models may be rewritten respectively from Equations (2-25) and (2-36) as follows E A u u w “l w 2 U3 = -- f0 < -1 - - > [ < -1 + - > 1 R 9 d7 (unaveraged) 2 R 6 R R 6 R ...(5-1) B A l w u l l u w 2 U3 = -- f0 < -1 - - ) [ - f0 < -1 + - > R 6 d7 1 R 6 d7 (averaged) 2 R 6 R L R 6 R ...(5-2) w u Consider the quantity ( -—1 - - ) . It represents the linear part R 6 R of the axial strain in the element. Experience shows that it is slowly varying. (In fact, a linear element based on setting it equal to constant was shown to be very effective (1)). Therefore, the qruuntity may be taken.as a constant. Consequently, U (Eq.S-2) may be (averaged) written as follows (noting f3 R 6 d7 = L) l u w “‘ f1 < -31 - 3 > [ - I; < -1 + - >2 R 6 d7 1 R 6 d7 = L 2 O R 6 R R 6 R E A u U E )2 — (1,<—-1-->[<—-”~+ w ] R 6 d7 2 R 6 R R 6 R which is the The exp: be rewritten (unav (aver constant. C l [‘J L in which n Equation Constant, the root. from the finite e] nonlinear EquatiOn illugtrat C0ns follOWing 62 which is the same as U as given in Equation (S-l). (unaveraged) The expressions of U4 for the unaveraged and averaged models cxui be rewritten from Equations (2-26) and (2-37) as follows E A 1 u w 2 2 U =- — f0[(——‘1+-)]Rad7 ...(5-3) (unaveraged) 8 R 6 R E A 1 l 1 u w 2 2 U4 s~—fo[-fO(—1+-)aad7]aady (averaged) 8 L R 6 R ...(5-4) B A l 2 u w 2 Let ( -—- ) / ( ——1 + - ) = f(7) , and note that f(y) is not 8 R 6 R constant. Consequently, l l 2 [- I}, f(7)R0d7 12 < - f5 [ Em] Roch ... L L in which the left term is U4 for Equation (5-4) and the right tmnnn for Equation (5-3). The preceding follows from the fact that, for f('y) as constant, the mean (square root of the left side) is always less than the root-nmuui-square (square root of the right side) (50). It follows from the relationship between the displacement formulation of the finite element method and the principle of potential energy that the nonlinear stiffness as represented by [n2] is lower’fkn: the model of Equation (5-4) tflunn that of Equation (5-3). The above observation is illustrated by the following numerical example. Consider a 2-dimensional curved beam element (circular) having the following properties ( The abov previously The unaveraged av. [n1] a un. 63 E - 107 ; R1 = R2 = 100. A — 2.0 , L = 3.06960000 I = 0.6667 ; 6 = 0.03069600 It undergoes the following displacements uA = 0.01803063 ; uB = 0.08288997 wA = -0.00979038 , wB = -0.01899512 6yA = 0.01269580 ; 6yB = 0.02964475 ( The above data is taken from the solution of the problem discussed previously in Section 4.4.1 at the applied load level of P = 20,000.) The resulting entries of [n1] matrix for both averaged and unaveraged models are, respectively, given below : ’-22860.2 136863 9 -5419.8 27049.9 -l36097.9 -8430.3‘ -4195.8 27966.6 -136928.2 -6.4 -28492.6 -27881 8 6275.6 -27787.1 6636.3 [n1] = av . symmetry -31239.7 l36033.6 7551.9 4183.0 28737.9 ( -27870.2, r-23147.7 136863.5 -6895.5 27337.4 —136093.9 -6450.7‘ -4195.7 27972.9 -136928.0 -6.4 -28508.0 -26706.6 7751.2 -27775.9 7022.8 [n1] = un. symmetry -31527.2 136029.6 5572.2 4183.0 28730.9 -29339.1J \ The r unaveraged II {“2181}:- [02]un . The are L_ 64 The resulting entries of hflH nmtrix for both averaged and unaveraged models are, respectively, given below : l 4674.60 -71 745 1048.38 -4674.6OO -71.745 -135.16‘ 1.101 -16.09 71.745 1 101 2.07 2019.75 -1048.380 -16.090 -597.57 [n2] - av. symmetry 4674.600 71.745 135.16 1.101 2.07 \ 2026.16, l 5217.50 -80.lOO 2280.30 -5217.500 -80 100 -138.66l 1.229 -35.00 80.077 1 229 2.12 3655.20 -2280.300 -35.000 -l621.70 [n2] = un. symmetry 5217.500 80.100 138.66 1.229 2.12 l 8037.801 The entries of the linear stiffness matrix, [k], of the element I are \ “2767012. 57544. 4245573. -2763943. -142440. 4244004. 6515136. -13998. 142440. -6513833. -116310. [ k 1 8687758. -4244003. —116310. 4343067. symmetry 2767012. -57544. -4245573. 6515135. -13998. ( 8687759., It may stiffness. ] [n2] matrix For fut [n2] matri the same cr given in th [k]. [n1] 4 65 It may be observed also that [n1] tends to decrease the total stiffness. It represents the effect of membrane flexure cxnniling. The [n2] matrix tends to increase the total stiffness. For further comparison, the resulting entries of [k], [n1], and Ln2] matrices for nonlinear straight beam element (Ref. 47), which has the same cross sectional properties, length, and displacements, are given in the followings (2766090. 0. 4245396. -2766090. 0. 4245396.1 6515507. 0. 0. -6515507. 0. [ k 1 = 8687777. -4245396. 0. 4343889. symmetry 2766090. 0. -4245396. 6515507. 0. \ 8687777. '-23445.5 l376l6.8 -5997 4 23445.5 -137616.8 -5997.4‘ 0.0 28166.9 -137616.8 0.0 -28329.7 [HI] = -24546.0 5997.4 -28166.9 6136.5 symmetry -23445 5 137616 8 5997.4 0.0 28329.7 \ -24S46.OJ [n2] ' It is [n1](unave versions f1 the curve. this eleme This of c FurthermO‘ element i subtendiny all entrf the strai 5.2 SUHH Inlt element displaceI in°°rP0r. The model, than an 66 f 4745.59 0.000 1065.32 -4745.590 0.000 -127.96 0.000 0.00 0.000 0.000 0.00 2047.01 -1065 320 0 000 -603.78 [n2] = symmetry 4745.590 0.000 127.96 0.000 0.00 l 2048.42 J It is seen that the differences between [n1] and (averaged) [n1] are not substantial whereas those between the two (unaveraged) versions for [n2] are. A comparison of the stiffness nuuzrices between. the curved (averaged model) and straight elements indicates that, for this element with a small value of 6 , the matrices are quite similar. Tflnis of course can not be expected to be the case when 6 is not small. Furthermore, when the magnitude of the radius of curvature of the element is taken to be sufficiently large (and simultaneously the subtending angle is decreased to result in the same element length), all entries of the [k], [n1], and [n2] matrices converge to those for the straight beam element. 5.2 SUMMARY AND CONCLUSION In this dissertation, a three dimensional nonlinear curved.bean1 element has been developed. It has 12 degrees<flffkeedom in 6 displacements (all "essential") per end node. Thus it can readily be incorporated into a general structural computer program. The element, which is formulated based on the average axial strain model, is shown to be more accurate, for same number of elements, than.all methods compared. Accurate load-displacement curve may be obtained by For symmetri need be con: The me‘ system, we the arch sp of the ar Lagrangian necessary involved. The 4 larger tha effects 0 practice, As me nonlineari becomes 5 inWortant 67 obtained by using at most eight elements to represent the entire arch. For symmetrical problems, only one half of the arch (four elements) need be considered. The method, which is based on the fixed Lagrangian coordinate system, works very well for small displacement problems (2% or less of the arch span) as well as for intermediate displacement problems (2-25% of the arch span). The solution procedure based on an updated Lagrangian coordinate system is also presented. The procedure is necessary if large displacements (25% or more of dmeauxfiispan) are involved. The amplification factor for displacements seems tolxealways larger than the amplification factor for stresses. This fact and.its effects on the amplification factor method, that commonly used in practice, need be studied more thoroughly. As mentioned previously, the present study is limited to geometric nonlinearity. For many practical problems, when geometric nonlinearity' becomes significant, effects of material nonlinearity wmikihmcome important at the same time. Thus, future studies of nonlinear analysis of curved beam structures should include these effects. TABLES NO. 0F ] ELEMENTS TABLE 2-1 .ACCURACY OF THE GEOMETRIC REPRESENTATION FOR PARABOLIC ARCH (RISE=9.6", SPAN=48”) NO. OF ELEMENT APPROX. EXACT REMARKS ELEMENTS # VALUE VALUE (0 L 26.3212 26.3575 1 1 R1 24.1510 30.0000 R2 53.8679 63.0067 L 12.3115 12.3127 C) 1 R1 28.7201 30.0000 R2 35.9910 37.4807 (3 2 L 14.0438 14.0447 2 R1 35.2926 37 4807 R2 60.1671 63.0067 L 8.0937 8.0938 1 R1 29.4501 30.0000 R2 32.6650 33.2562 (3 C) L 8.6351 8.6352 3 2 R1 32.5061 33.2562 <3 R2 42.7926 43.6711 L 9.6284 9.6284 3 R1 42.5459 43.6711 R2 61.6663 63 0067 L 6.0398 6.0398 1 R1 29.6947 30.0000 R2 31.4995 31.8178 L 6.2729 6.2729 (D ® 2 R1 31.4457 31.8178 (3 R2 37.0694 37 4807 4 e L 6.7153 6.7153 3 R1 36.9778 37.4807 R2 47.0094 47.5805 L 7.3294 7.3294 4 R1 46.9014 47 5805 R2 62.2310 63.0067 68 NO. 0F ] ELEMENTS TABLE 2-1 ACCURACY OF THE GEOMETRIC REPRESENTATION FOR PARABOLIC ARCH (CONTINUED) NO. OF ELEMENT APPROX. EXACT REMARKS ELEMENTS # VALUE VALUE L 4.8204 4.8204 1 R1 29.8058 30.0000 R2 30.9599 31.1593 L 4.9410 4.9410 2 R1 30.9364 31.1593 R2 34.4867 34.7240 <9 ® (3 L 5.1738 5.1738 5 3 R1 34.4449 34 7240 R2 40.6379 40.9440 L 5.5048 5.5048 4 R1 40.5844 40.9440 R2 49.8112 50.2070 L 5.9175 5.9175 5 R1 49.7525 50 2070 R2 62.5036 63.0067 L 4.0118 4.0118 1 R1 29.8648 30.0000 R2 30.6675 30.8035 L 4.0820 4.0820 2 R1 30.6546 30.8035 R2 33.1014 33.2562 Q) Q) © L 4.2189 4.2189 @ 3 R1 33.0797 33.2562 R2 37.2916 37 4807 6 L 4.4163 4.4163 4 R1 37.2638 37.4807 R2 43.4358 43.6711 L 4.6666 4.6666 5 R1 43.4046 43.6711 R2 51.7889 52.0801 L 4.9618 4.9618 6 R1 51.7524 52.0801 R2 62.6600 63.0067 69 TABLE 4-2 A NUMBER OF ELEMENTS For ( Anal: 9«Di TABLE 4-2 LINEAR EQUILIBRIUM OF A PARABOLIC ARCH SUBJECTED TO A CONCENTRATED IN-PLANE LOAD AT CROWN C.‘ 5" NUMBER 0F* DISPLACEMENT"“ DIFFERENCE*** ELEMENTS AT CROWN (IN. x 10'“) (a) - 1 0.453155953 2.982 2 0 460521696 1.405 3 0.465530466 0.332 4 0.466613128 0.101 5 0.466937636 0.031 6 0 467273494 -0.040 7 0.467232458 -0.031 8 0.467054924 0.006 9 0.467268255 -0.039 10 0.467500213 -0.088 For one half of the arch -4 , Analytical solution = 0.467085 x 10 1n. Analytical - Numerical % Difference = Analytical 71 TABLE 4-3 I l 9: NUMBER OF ELEMENTS For Ana? we TABLE 4-3 LINEAR EQUILIBRIUM OF A SEMI-CIRCULAR ARCH SUBJECTED TO A CONCENTRATED OUT-OF-PLANE LOAD AT CROWN NUMBER OP* DISPLACEMENT** DIFFERENCE*** ELEMENTS AT CROWN (IN.) (a) 2 2.81090450 0.531 3 2.81890678 0.248 4 2.82110500 0.170 5 2 82034874 0.197 6 2.82473183 0.042 7 2.82618713 -0.009 8 2.82285213 0.108 9 2 81906509 0.242 10 2.82073689 0.183 For one half of the structure Analytical solution - 2.825930 " Analytical - Numerical % Difference - Analytical 72 FIGURES 73 A/ CLASSICAL BUCKLING LOAD m B“ C I A BIFURCATION‘ A“ POINTS J D LOAD \ FUNDAMENTAL PATH (NONLINEAR ELASTIC BEHAVIOR) o V DISPLACEMENT Figure l-l : LOAD-DEFLECTION RELATION 74 Yav e 7 l AI’ / * Figure 2-1 : BEAM ELEMENT (Curved In The x-z Plane) I I I I I I l z,w} _ I I I A’)p ........... / y’v’n L1 7 /’j” / S l U V’XIUIQ Figure 2-2 : CROSS-SECTION OF PRISMATIC MEMBER 75 WY Figure 2—3 : COORDINATE SYSTEMS STRUCTURE SYSTEM x,v,z X,Y.Z ELEMENT SYSTEM Figure 2—4 : TYPICAL ELEMENT I I 76 Figure 2-5 : TYPICAL ELEMENT AFTER TRANSFORMATION TO ELEMENT COORDINATE SYSTEM it 77 Load I I ith Solution Point (i+l)th Solution Point Conver ed Solution P'_ 1 ..._.___._Z / g i- I I AR1+1 I f(QIH) k '— IO IO P. + ’u—l Deformation P D '0 - dr- Figure 3-1 : NEWTON-RAPHSON ITERATION EIL . zzomo ea 1am: oaou mz<=mm mz < OH Qmeumhmom mom< M Z3omo 0.0 0.... 0.0 ON 0. P 0.0 _I p _ _ _ — ... u — h _ I— — _ _ 1p _ _ n _ co um.we 1 T111111111111;111L .1mooorm mod n m r m .H 5666.o 1 H - .N 1 a T:.ooo.o~ .oou 1 m T m . 11.ooo ma Mm .\ r x \\ \ 1.80.8 xx. Anacoaoao 0 no 00 posuofi venomoum .\\\\\. r .11/JV1 \\ \ \\\\\Hw\\\\\ 1..ooo.m~ . \1. A.mHoINV eczema venomoum n \. \\\1. o \I\ .111 .11111111 . \\\\\\\\\ .ooo.om .K 1 1 .A\. -111. onesuom eoaxmxeam \\x A.6H61ouv zaago a osmomesumm 1 \x. .\\\\\\\\\ 11111.1111111111 1..ooo.mm xx. .\. xxx/AAW\. 1 \ \ \\\\ Amucmsmam1mv z= < OH QmHomhmDm mom< m 230mm 06: o.m 86 \n ._ . \ . -11.H11\.1\. 731 f S 8 SE95 a 2:053 .7 oom li’ul‘l'" 1“ > F Tmmm 1 2.3 :3: Sdéum a 032.3 F609 \\ a\\\»;1 Amucosmfim1qv ommma ZBOMO 06 06 o.¢ O.N o.o ml b P — n n L1! — n r p m n p n ” lb — 0° .\/ x x 1 \\af-l.\.\// \ / x / r. TIM \ Odom I: \ / o3 u m [.00N \ / 2mm; 1 H T. _ oé 1 < 3.0; OS. 1 m 1.09., m w.o 1 7.03 m j Foam Thoow \ Amucmamam 3 yo my @932: vmmomoum \\ *1 \\\..\\\ A.mHm13 @932: wmmoaoum I. DON. . 1 \\\\.x\ I“ .. . To Ill1l11d1H111h.\\.\\ A.2". 2 Ho 8 8:95 a 2858 con F TE 1 33 :5: SEES a 8.593 1.00m ([VO'I AQUCQEMHN|QV QO£uUE UQmOQOHm .000“ Am . \\\\\\\. AQUEuU-BMUHU 0H 1H0 va CthHMnQdVQ aw z:o~.~1HAVHV ZZOMO H< Q < OH mmHUmmem mom< m zzomo 0.0. Cd 06 oé ON 06 r F 1— h h _ n — h b h n i— n - I» b L? I— .0 T. V/ V. _oom// _ / fi N ¥ // .OONHHM T .03 1 mzwm / O — / \ _ Teen 2: 1 H _ m 22; 1 H 1 o.¢ u < r m 1 11.009 r \ f \. r Amucmsmamlwv vOSHmE tumoaoum \\\ . I \HHHAHV\. AmucmEmHm «V cosumE wmmoaoum .HHU\. T.n:u®~ .\. . 11.1111111V11HU11111HHA/ll $21531 3 .8 3 GEES a 22233 c "l‘l G V O T 89 z30mo H< $04 QMngszZOU A < OH. QmHUMWmDm :05» fiADUmHUIHme mesz 4‘ ” NTLH mmDOHm Hzmzmo ZBOMU 06 o.¢ o.n ON 0; 0.0 _111 1 L _ L h H _ . _ .0 Tuop IN #0# n m .OH H m - mmnwood u H .n 326 1 < 1 m 13¢ 1 T 11.0 H. 7.\. r CIVO'I m zzomo H< Q < 09 ml. QmHUMhmDm mom< m::’,H ””,, " Proposed method (4 or 6 elszl__ 5 5 \\Proposed method 8 elements) Proposed meth (Z-EIS.) J) AUSTIN & ROSS (limit load = 1269) ./ I ...—u-- / f..— 47(8- €15./):p-'/ A// //Q /<:——/ Proposed method (4- els. ) \—-N00R (6-els.) / I - Ref. / x / ¢$f 40 160 g - .flfl'~_fi_ - \ ~/””’/’/” '——~ "RECTANGULAR FRAME” (Ref. 47) = 1.0 0.08333 ' A ./ I ////// E = 107 I I l I 10.0 15.0 CROWN VERTICAL DISPLACEMENT QINUSOIDAL PARABOLIC, CIRCULAR, SEMI ELLIPTIC FIGURE 4-15 : . , ARCHES TRIANGULAR FRAME, AND RECTANGULAR FRAME SUBJECTED TO A COVCENTRATED LOAD AT THEIR CROWNS «r\ A 20.0 mmmhum HNHN< HQQfiHHfiOZ .I m 1|..W0UUQN .HQEN WWUHUW WWHNAV U QQOQ EMOHHZD 1QN 93 < HzHom mmam<=o mma H< mm0H0 (0"; 0&0 “£0 ‘0‘ Qxfb' {$0 Yjvléxl. «'06:». f 99 Limit load 953 . Q // e / ® I A [—-(DaDeppo & Schmidt) l 800.— 600.-‘- _ __.Nonlinear Total Stress -—— Linear Total Stress 4OO.—1 P ’I 26.795L 200. pH--- _- 200 % 0' | I I f f —[ r “r l r I —r r i I I 0. 5000. 10000 15000. STRESS ___l I l I 4f 1 T r I I *1 1 r’ l T 1 -1 I x —1 l- 2. 3. 4 5. AMPLIFICATION FACTOR FIGURE 4-119: STRESSES AND AMPLIFICATION FACTORS AT THE CROWN REFERENCES 10. 11. Ash Ring and .1 oh Aus 102 Bat Ba1 Pal C01 10. ll. 12. 13. LIST OF REFERENCES Ashwell , D. G. , " Strain Elements , With Applications to Arches, Rings , and Cylindrical Shells " , Finite Elements for Thin Shells and Curved Members ( Edited by D. G. Ashwell and R. H. Gallagher), John Wiley & Sons, 1976. Austin , W. J. , Ross , T. J. , " Elastic Buckling of Arches Under Symmetrical Loading" , Journal of Structural Division , ASCE, Vol. 102, No.ST5, May, 1976, pp.lO85-1095. Bathe , K.J., Wilson , E.L., " Numerical Methods in Finite Element Analysis", Prentice Hall Inc., Englewood Clift, N.J., 1976. Batoz , J. L. , " Curved Finite Elements and Shell Theories with Particular Reference to the Buckling of A Circular Arch " , Short Communications, 1979, pp.774-779. Belytschko , T. , Glaum , L. W. , " Applications of Higher Order Corotational Stretch Theories to Nonlinear Finite Element Analy- sis", Computers and Structures, Vol.10, pp.l75-182, 1979. Belytschko, T. , Hseih, B.J., " Nonlinear Transient Finite Element Analysis with Convected Coordinates " , International Journal for Numerical Methods in Engineering, Vol.7, pp.255-271, 1973. Billington , D.P. , "Thin Shell Concrete Structures", 2nd Edition, Mc.Graw Hill Book Co., New York, 1982. Calhound , P.R., DaDeppo, D.A., "Nonlinear Finite Element Analysis of Clamped Arches", Journal of Structural Engineering , ASCE, Vol. 109, No.3, March, 1983, pp.599-612. Chajes , A., Churchill, J.E., " Nonlinear Frame Analysis by Finite Element Methods", Journal of Structural Engineering, ASCE,Vol.ll3, No.6, June, 1987, pp.1221-1235. Cook, R.D.,"Concepts and Applications of Finite Element Analysis", 2nd Edition, John Wiley and Sons, New York, 1981. DaDeppo , D.A. , Schmidt, R. , "Large Deflections and Stability of Hingeless Circular Arches Under Interacting Loads", Journal of Applied Mechanics, ASME, December, 1974, pp.989-994. DaDeppo , D.A., Schmidt , R. , "Sidesway Buckling of Deep Circular Arches Under a Concentrated Load " , Journal of Applied Mechanics, ASME, June, 1969, pp.325-327. DaDeppo , D.A. , Schmidt , R. , "Stability of Two Hinged Circular Arches With Independent Loading Parameter ", Technical Notes, AIAA Journal, Vol.12, No.3, March, 1974, pp.385-386. 97 14. 15. 16. 17. 18. 19. 20. 21. 23. 24. 25. 26. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 98 Dawe, D.J., "A Finite Deflection Analysis of Shallow Arches by The Discrete Element Method " , International Journal for Numerical Method in Engineering", Vol.3, pp.529-552, 1971. Dawe , D. J. , "Curved Finite Elements For The Analysis of Shallow and Deep Circular Arches " , Computers and Structures , Vol.4, pp. 559-580, 1974. Dawe,D.J.,"Numerical Studies Using Circular Arch Finite Elements", Computers and Structures, Vol.4, pp.729-740, 1974. Fujii , F. , Gong, 8., "Field Transfer Matrix for Nonlinear Curved Beams " , Journal of Structural Engineering , ASCE, Vol.114, No.3, March, 1988, pp.675-691. Gallert , M., Laursen, M.E., "Formulation and Convergence of Mixed Finite Element Method Applied to Elastic Arches of Arbitrary Geometry and Loading " , Computer Methods in Applied Mechanics and Engineering, Vol.7, 1976, pp.285-302. Gong , S. , Fujii , F. , "Blending Functions Used in Discrete and Nondiscrete Mixed Methods for Plate Bending " , Computers and Structures, Vol.22, No.4, pp.565-572, 1986. Huddleston , J. V. , "Finite Deflections And Snap Through Of High Circular Arches " , Journal of Applied Mechanics , ASME, December, 1968, pp.763-769. ' Kikuchi , F. , "On The Validity Of The Finite Element Analysis Of Circular Arches Presented By An Assemblage Of Beam Elements", Computer Methods In Applied Mechanics And Engineering , Vol.5, pp. 253-276, 1975. Lange,J.G., "Elastic Buckling Of Arches By Finite Element Method", thesis presented to the Department of Civil And Environmental Engineering , Michigan State University , in 1981 , in partial fulfillment of the requirements for the degree of Doctor of Philosophy. Mak, C.K., Kao,D.W., "Finite Element Analysis Of Buckling And Post Buckling Behavior Of Arches With Geometric Imperfections",Computer And Structures,Vol.3,pp.l49-161,1973. Mallett, R.H., Marcal, P.V., "Finite Element Analysis Of Nonlinear Structures " , Journal Of The Structural Division, ASCE, ST.9, pp. 2081-2105, September, 1968. Mebane, P.M.,Strick1in,J.A., "Implicit Rigid Body Moyion In Curved Finite Elements ", American Institute Of Aeronautics _And As- tronautics, Vol.9, No.2, 1971, pp.344-345. Medallah,K.Y., "Stability And Nonlinear Response Of Deck Type Arch Bridges " , thesis presented to the Department of Civil and Environmental Engineering , Michigan State University, in 1984, in 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. part of E N001 Anal Ode Edi 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 99 partial fulfillment of the requirements for the degree of Doctor of Philosophy. Noor, A.K. , Green,W.H., Hartley, S.J. , "Nonlinear Finite Element Analysis Of Curved Beams " , Computer Methods in Applied Mechanics and Engineering, Vol.12, pp.289-307, 1977. Oden, J.T., Ripperger, E.A., "Mechanics of Elastic Structures",2nd Edition, McGraw Hill Book Company, New York, 1981. Ojalvo, I.U., Demuts, E.,and Tokarz, F., "Out of Plane Buckling of Curved Elements", Journal of the Structural Division, ASCE,Vol.95, No.STlO, October, 1969, pp.2305-2316. Ojalvo, I.U.,Newmann, M.,"Buckling of Naturally Curved and Twisted Beams", Journal of the Engineering Mechanics Division,ASCE,Vol.94, No.EM5, October, 1968, pp.lO67-1087. Rahimzadeh , J. , "Nonlinear Elastic Frame Analysis by Finite Elements " , thesis presented to the Department of Civil and Environmental Engineering , Michigan State University, in 1983, in partial fulfillment of the requirements for the degree of Doctor of Philosophy. Reddy,J.N., "An Introduction to the Finite Element Method", McGraw Hill Book Co., New York, 1984. Richardson, G.S. , "Arch Bridges", Structural Steel Designers Hand Book (edited by F.S. Merritt),l976. Sabir, A.B., Lock, A.C., "Large Deflexion, Geometrically Nonlinear Finite Element Analysis of Circular Arches", International Journal of Mechanical Sciences, Vol.15, pp.37-47, 1973. Segerlind, L.J. , "Applied Finite Element Analysis", 2nd edition , John Wiley and Sons, New York, 1984. ' Shames , I.H. , Dym , C.L., " Energy and Finite Element Methods in Structural Mechanics", McGraw Hill Book Company, New York, 1985. Stolarski , H. , Belytschko , T. , " Membrane Locking and Reduced Integration for Curved Elements " , Journal of Applied Mechanics ASME, Vol.49, March, 1982, pp 172-176. ’ Stolarski,H.,Be1ytschko, T., "Shear and Membrane Looking in Curved C Elements" , Computer Methods in Applied Mechanics and Engineer- ing, Vol.41, pp.279-296, 1983. Timoshenko,S.P.,Gere,J.M., "Theory of Elastic Stability", 2nd ed., McGraw Hill Book Co., New York, 1961. Tokarz , F.J., and Sandhu , R.S. , " Lateral Torsional Buckling of Parabolic Arches", Journal of the Structural Division,ASCE,Vol.98, No.STS, May, 1972, pp.ll61~1179. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 100 Walker , A. C. , " A Nonlinear Finite Element Analysis of Shallow Circular Arches ", International Journal of Solids and Structures, Vol.5, pp.97-107, 1969. Weaver, W., Gere., J.M., " Matrix Analysis of Framed Structures ", 2nd edition, D. Van Nostrand Co., New York, 1980. Wen, R.K.,"Incrementa1 Stiffness Matrices for Nonlinear Structural Analysis " , Eleventh Southeastern Conference on Theoretical and Applied Mechanics (SECTAM XI). Wen , R. K. , " Introduction to Nonlinear Structural Analysis " , unpublished notes on CE890 , Department of Civil And Environmental Engineering, Michigan State University, 1988. Wen , R. K. , Lange , J.G., "Curved Beam Element for Arch Buckling Analysis", Journal of Structural Division, ASCE, Vol.107, No.ST11, November, 1981, pp.2053-2069. Wen , R. K. , Medallah , K., " Elastic Stability of Deck Type Arch Bridges ", Journal of Structural Engineering, ASCE, Vol.113, No.4, April, 1987, pp.757-768. Wen , R.K., Rahimzadeh , J., " Nonlinear Elastic Frame Analysis by Finite Element ", Journal of Structural Engineering, ASCE,V01.109, No.8, August, 1983, pp.l952-1971. Yabuki , T. , Vinnakota , S. , " Stability of Steel Arch Bridges - A State of the Art Report", Archives 9, 1984, pp.llS-158. Zienkiewicz , O.C. , " The Finite Element Method", 3rd ed., McGraw Hill Book Co., 1977. .— Bowman, F., and Gerard, E.A., "Higher Calculus", Cambridge Univer- sity Press, 1967, Cambridge, England, p. 265. APPENDICES A.1 GE Tl Lagrang proced (displz treat small Lagrar incren new i coord the ' matri matri Sect poss radi Whic APPENDIXHA NEWTON-RAPHSON METHOD FOR UPDATED COORDINATES A.1 GENERAL The formulation discussed in Chapter II, which is refered to Lagrange small rotation formulation, and the corresponding solution procedure described in Chapter III can not be used for large rotation (displacement) problems. For this class of problem, however, we may treat the problem as consisting of a series of increments involving small rotations (displacements). For each increment the concept of Lagrange small rotation formulation applies. At the beginning of each increment the geometry of structure should be updated. For atypical new increment, although the initial displacements in the new coordinates are zero, the strains are not. The strain, which is called the "initial strain", would lead to an "initial strain stiffness matrix". The procedure how to obtain the initial strain stiffness matrix is presented in the next section. It should be noted that the geometric representation described in Section 2.4.2 makes the updated procedure for curved beam element possible. Without it we would be faced with the problem of defining the radii of curvature at the end nodes of the updated curved beam element, which are required for evaluating the updated stiffness matrices. 101 101a '+1A 1 1+1¢ 1+1C Figure A-1 : CONFIGURATION OF A TWO DIMENSIONAL CURVED BEAM ELEMENT AT SUCCESSIVE LOAD INCREMENTS IN UPDATED LAGRANGE FORMULATION 102 A.2 INITIAL STRAIN STIFFNESS MATRIX, [ k6 ] 0 Consider a two dimensional curved beam element illustrated in Figure A-l. The X and Y axes represent the global coordinate system, the 1x and 1z axes denote the member coordinates, and 1C the member configuration at the beginning of the ith load level. From the stage from configuration 1C to 1+10 , the former configuration should be thought as fixed and the latter variable. The displacement components of i+1C are measured with respect to the member coordinates of 1.-C. They are the generalized coordinates at the stage of the analysis. (See the displacement 1u and 1w in Figure A-l.) A A - . '+l ’ Let l+leo denote the total strain corresponding to 1 C , 160 ‘ 1+1 1 denote the total strain at 1C , and e - co - 60 The strain energy is E 2 E U f 1+1. 2 dV - f ( e + 150 ) — av o 2 2 E . . H f —— ( .2 + 2 1.0 e + 1502 ) dV (A-l) 2 Since is is independent of the generalized coordinates or o displacements, the last term of U may be dropped. The expression for U can then be written as U - U + U ‘ ' .... (A-2) 103 where U - f —-— 52 dV .... (A-3) iU - f E is e dV .... (A-4) lfiua tangent stiffness matrix is obtained following the procedure discussed in Section 2.4 . Its m-n entry is equal to : 62 U 02 U a i i (A-5) 8 qm a qn a q a q 6 q 6 q The first term is exactly as before, with 1qm replacing qm, resulting in 1[k] , j'[nl] , and 1[n2] . For the second part , we encounter something new. At the beginning of the ith increment, the initial strain is : i-l dw u 1 du w 2 is (S.§’.n) - 2 j[<—- -)+— f1;(—+ -> as + ° j-O ds R 2 L ds R 2 1 L dv 2 6 d v —fo(—)dS+n('-—2) - 2 L ds R ds d2u l dw d 1 C 1 “‘5 + ‘ -‘ + W '— ( ‘ ) l ] ...(A-6) ds R ds ds R in which j denotes the stage of the configuration. 104 By substituting Equation (A-6) into Equation (A-4) we have the following : 1-1 u l w L 2 1U - E f [ 2 j{ ( w - — ) + ——- f ( u + - ) ds + 60 vol j-o S R 2 L o s R 1 13 — f:(V)2dS+n(--vss)- 2 L R l d 1 § ( u + - w + w ——— ( - ) ) } ] * SS R S ds R u 1 1 L 2 [ ( w - - ) + -—— f ( u + — ) ds + S R 2L ° 3 R 1 79 ——— f: ( vS )2 ds + n ( ‘ - vSS ) - 2 L R 1 d l g ( u + — w + w “' ( - ) ) ] dV ...(A-7) 55 R 3 ds R For two dimensional problem, the previous expression becomes : i-l . u 1 L w 2 1U — E f [ Z J{ ( wS - - ) + “‘ f0 ( us + - ) ds - 60 vol j-o R 2 L R 1 d 1 {(u +-w+w—('))}]* 35 R 3 ds R u 1 w ' L 2 l — - a w - “ ) + f ( u + ) d5 [(5 R 2 o S R l d l g (u + - w +w -(-))] dV ...(A-8) It should d + w -— ( ds stiffness Therefore : 105 . u . 1 be noted that the terms : l( w - - ) , and l( u + - w 5 ss 3 R R 1 - ) ) have no contribution to the " initial strain R because they are linear in the generalized coordinates. i-l u l w E f [ z j{ ( ws- - ) + -—— f: ( 05+ — )2 ds - vol j-o R 2 L R 1 d 1 g ( u + — w + w “' ( ‘ 7 ) 7 l * SS R 3 ds R l w 1[ -—— fL ( u + - )2 ds ] dV 2 L ° 3 R (A-9) In terms of 7 , and realizing that ds - R d¢ - R 0 d7 , the above equation can be written as : iUe - E f [ I J.{( O i 1 dw u 1 du w vol j-o R 9 d7 R 2 L R 0 d7 R ‘ C ( R202 d12 d7 ds2 R R9 d7 R ds d7 1 du w 2 — ) R 9 d7 ] dV m rd 0 73 Q: 0.. .4 W 106 i—l 1 1 1 1U6 ' E f f 1 E Jl “' ( w - 9 u ) + ——- 1 ——— ( u + 0 w )2 d7 0 s A j-o R 0 7 2 L o R 0 9' 1 1 -— —- +R9 +- ~1- R 0 ( R 9 u17 u 135 R w + 6 w 757 ) }] . l 1 1[ “‘ I: “‘ ( u + a w )2 d1 ] dA ds 2 L R o 7 (A-lO) Since the cross section is presumed to have two axes (xf symmetry , IA f ( .... ) dA - 0. Therefore the third part of the first term in Equation (A-10) may be dropped from the equation. 1 1 1-1. 1 1 1 l 2 U HEA£[zJ{—(w-6u)+—fo—(u7+9W)d7H* Eo j-o R o 7 2 L R 0 . 1 1 1 2 1[ I ___ ( u + a w ) d7 ] R 9 d7 0 *1 2 L R 9 (A-ll) . i . The initial strain stiffness matrix [ ke ] 18 equal to the 0 second derivative of 1U with respect to the generalized coordinates e , o i qm - 107 A.3 UPDATED COORDINATES PROCEDURE The loads are applied in increments. At the beginning of each increment.the geometry of the structure is updated. In addition to the usual stiffness matrices [k], [n1], and [n2], there is the initial strain stiffness matrix as explained previously. The steps of the calculation are as follows : 1) Set load increment ( and check if the intended total load has been applied ). 2) Determine the most up-to-date geometry of the structure by using the latest joint displacements and rotations , and update the linear stiffness matrix. 3) Form the tangent stiffness matrix , [KT] , according to the one of the following cases : a) For the first load increment : [KT] - [K] + [Nl({Q})] + [N2((Q})] b) For other load increments : [KT] - [K] + “(.1 + [Nl({Q})] + 1112001)] 0 in which O[K ] is the structure initial strain stiffness e 0 matrix. 4) Solve for {AQ} from : { A Q } - [ KT ].1 {load increment vector} 108 5) Add {AQ} to the latest {Q} to obtain a new {Q}. 6) Based on the new {Q}, evaluate [Nl({Q})] and [N2({Q})]. 7) Form tangent and secant stiffness matrices and resistance force vector as a) For the first load increment : [KT] - [K] + [Nl({Q})] + [N2({Q})] l 1 [KS] - [K] + — [Nl({Q})] + -- [N2({Q})] 2 3 b) For other load increments : [KT] — [K] + “(.1 + [Nl({Q})] + [N2<1Q1)1 1 l [K ‘ [K] + [K I + “’ [Nl({Q})] + ‘g‘ [N2({Q})] 2 Resistant force vector - [KS] { Q } 8) Evaluate the unbalanced force vector {AR} as : { A R } - load increment vector - resistance force vector 9) If the unbalanced force vector , {AR} , is sufficiently small , return to 1 . 10) Return to 4 but use the unbalanced force vector as the load increment vector. APPENDIX,B INCREMENTAL STIFFNESS MATRICES, [n1] AND [n2], BASED ON THE AVERAGE AXIAL STRAIN MODEL 3.1 THE FIRST ORDER INCREMENTAL STIFFNESS MATRIX, [n1] Only nonzero terms are given. If} p. = [ (18 A -12 A +8 A6) A0 + 6 {(3 A2-2 A4) A2 + 1,1 3 5 m l" (6 A3-2 A5) A3 + (3 A5-4 A6) A4 + 6 (3 A2-3 A3) Ag + o (3 A3-A5) 110} 1 F! 11> Q: 1,2 = a L [ (3 A2-2 A4) A6 + (6 A3-2 A5) A7 + (3 A5-4 A6) 18 ] 02 -4 A6) A0 + (1-;;) {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A5-4 A6) A4 + 0 (3 A2-3 A3) A9 + 02 6 (3 A3-AS) 310} - —;— {(1.5 AZ'A4) A2 + (3 A3-A5) A3 m > H 6 (-6 A -2 A4+6 A 1,3 3 5 N [... + (1.5 A3-A4+1.5 A5-2 A6) 14 + a (1.5 A2-l.5 A3) 19 + 6 (1.5 A -0.5 A5) A10} ] 3 ['11 > = R1 6 [ (-3 A2+12 A3+2 A4-7 A5+4 A6) A0 + m t" 6 --{(-12 A +15 A 12 1 2 (-6 A4+15 A5 0 (-3 A2+15 A3-4 A5) A10} ] E A R1 02 = --::7;—- [ (2 A1-2 A2+A4) A6 + (AZ-4 A3+A5) A7 + -8 A4) 32 + (-6 A2+30 A3-8 A5) A3 + -16 A6) A4 + 6 (-12 Al+15 A2-12 A3) Ag + 1,6 (AA-2 A +2 A6) 18 1 5 109 n1 n1 n1 n1 n1 n1 n1 n1 n1 n1 1,11 1,12 2,2 2,3 2,5 2,6 2,7 2,8 2,9 2,11 E A "‘ ( - 18 A + 12 A - 8 A ) A 2 L 3 5 6 o - n1 1,2 E A 3‘; [ 6 ( - 9 A3 + 6 AS - 4 A6 ) A0 + 02 (1 - I; ) {(2 A4-3 A2) A2 + (2 A5-6 A3) A3 + (4 A6-3 A5) A4 + 6 (3 A3-3 A2) A9 + 6 (AS-3 A3) A10} - 02 —;— { (1.5 A2-A4) A2 + (3 A3-A5) A3 + (1.5 A5-2 A6) A4 + 6 (1.5 A2-1.5 A3) A9 + 6 (1.5 A3-0.5 A5) A10 } ] E A R2 0 9 ---——- [ (6 A -5 A +4 A ) A + ___ { (3 A -4 A ) A + 2 L 3 5 6 o 12 2 4 2 (6 A3-4 A5) A3 + (3 A5-8 A6) A4 + 6 (3 A2-6 A3) A9 + 6 (3 A3—2 A5) A10 } ] E A R2 92 ———— [ (AA-A2) A6 + (AS-2 A3) A7 + (2 A6-AS) A8 ] 4 L E A “" ( 18 A - 12 A + 8 A ) A 2 L 3 5 6 o E A 62 -—-— (1-——) [(3 A2-2 A4) A6+ (6 A3-2 A5) A7+ (3 A5-4 A6) A8] 2 L 12 E A R1 92 [(3 A2-2 A4) A6+ (6 A3-2 A5) A7+ (3 A5-4 A6) A8] 24 L E A R1 6 "_——--' ( 3 A2 - 12 A3 - 2 A4 + 7 A5 - 4 A6 ) A0 2 L n1 1,2 - n1 2’2 - n1 2,3 2 E A R2 6 24 L 110 [(2 A4-3 A2) 16+ (2 A5-6 A3) 17+ (4 A6-3 A5) 18] n1 n1 n1 n1 n1 n1 n1 2,12 3,3 3,5 3,6 3,7 3,8 3,9 111 E A R2 6 "————-' ( - 6 A + 5 A - 4 A ) A 2 L 3 5 6 o E A 2 -—- [ 6 ( 1.5 A3 + 2 A4 - 3 A5 + 2 A6 ) A0 - 2 L 62 ( 1 - I2 ) 6 {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A3-2 A4+3 A5-4 A6) A4+ 6 (3 A2-3 A3) A9+ 6 (3 A3-A5) A10 E A R1 6 . ‘—---- [ 6 ( 1.5 A2 - 4.5 A3 - 2 A4 + 3.5 A5 - 2 A6 ) A0 2 L 62 - - { (3 A2 - 2 A4) A2 + (6 A3 - 2 A5) A3 + 24 (3 A3-2 A4+3 A5-4 A6) A4+ 6 (3 A2-3 A3) A9+6 (3 A3-A5) A10} 92 - ( l - I2 ) { (2 A1-2 A2+A4) A2 + (AZ-4 A3+A5) A3 + (AA-2 A5+2 A6) A4 + 6 (2 A1-2 A2+1.5 A3) A9 + 6 (0.5 A2-2 A3+0.5 A5) A10 } ] EAR16 62 ----- ( 1 - -— ) [ (2 A1-2 A2+A4) A6 + (AZ-4 A3+A5) A7 2 L 12 + ( A4 - 2 A5 + 2 A6 ) A8 ] E -- [ 6 ( 6 A3 + 2 A4 - 6 A5 + 4 A6 ) A0 - 2 L 2 0.25 6 { (3 A2—2 A4) A2 + (6 A3-2 A5) A3 + (3 A3-2 A4+3 AS 02 - (1- 1;) {(3 Az-z A4) A2 + (6 A3-2 A5) A3 + (3 A5-4 A6) A4 -4 A6) A4+ 6(3 A2-3 A3) A9+ 6 (3 A3-A5) A10} + 6 ( 3 A2 - 3 A3 ) A9 + 6 ( 3 A3 - A5 ) A10} ] - n1 2’3 . 2 E A 0 0 l 5 A A A 2 L [ 6 (3 A3+A4-3 A5+2 A6) A0 + (1-12) ( . 3- 4). 4 ] n1 n1 n1 n1 n1 n1 3,11 3,12 5,5 5,6 5,7 5,8 112 E A R2 6 2 L [ 6 (-l.5 A3 - A4 + 2.5 A5 - 2 A6) A0 + 62 -— { (3 A2-2 A4) A2 + (6 A3-2 A5) A3 + 24 (3 A3-2 A4+3 A5-4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 ( 3 A3- A5) A10 } + 02 (l- :3) {(A2-A4) A2+ (2 A3-A5) A3+ (AS-2 A6) A4+ 6 (AZ-1.5 A3) A9 + 6 (A3-O.5 A5) A10} ] 2 E A R2 6 6 __;_;_H_.(l_ 12) [ (A4-A2) 16+ (AS-2 A3) 17+ (2 A6-A5) A8 ] E A ___ 2 ( R1 6 ) [ (2 Al- 4 A2+ 8 A3+ 2 A4- 4 A5+ 2 A6) A0 - 2 L 6 g {(2 Al-2 A2+A4) A2 + (AZ-4 A3+A5) A3 + (A4-2 A5+2 A6) A4 + 6 (2 Al-2 A2+1.5 A3) A9 + 6 (0.5 A2-2 A3+0.5 A5) A10} ] E A R12 63 [ (2 A1 - 2 A2 + A4) A6 + (A2 - 4 A3 + A5) A7 + 24 L ( A4 - 2 A5 + 2 A6 ) A8 ] E A R1 6 [ (3 A2-12 A3-2 A4+7 A5-4 A6) A0 - 2 L 6 I; {(12 A1-9 A2+4 A4) A2 + (6 A2-18 A3+4 A5) A3 + (6 A4-9 A5+8 A6) A4 + 6 (12 A1-9 A2+6 A3) A9 + 6 (3 A2-9 A3+2 A5) A10} ] E A R1 92 24 L [(2 A4-3 A2) A6+ (2 A5-6 A3) A7+ (4 A6-3 A5) A8] n1 n1 n1 n1 n1 n1 n1 n1 n1 5,9 5,11 5,12 6,6 6,8 6,9 6,11 6,12 2 L 113 E A R1 6 2 L [ 6 (1.5 A2-6 A3-A4+3.5 A5-2 A6) A0 + 22 (1- :2) {(2 A1-2 A2+A4) A2 + (A2-4 A3+A5) A3 + (A4'2 A5+2 A6) A4 + 6 (2 A1-2 A2+1.5 A3) A9 + 6 (0.5 A2-2 A3+0.5 A5) A10} - 22 24 {(3 A2-2 A2) A2+ (6 A3-2 A5) A3+(3 A5-4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 (3 A3-A5) A10} ] E A 2 -—— R1 R2 6 [ (-A2+4 A3+A4-3 A5+2 A6) A0 + 2 L 6 I2 {(2 Al-A2) A2 + (A2-2 A3) A3 + (AA-A5) A4 + 6 (2 Al-A2) A2 + 6 (0.5 A2-A3) A10} ] E A 6 2 R1 R2 6 [ (AA-A2) A6 + (AS-2 A3) A7 + (2 A6-A5) A8] 24 L E A 2 2 -- R1 6 ( 2 Al - 4 A2 + 8 A3 + 2 A2 - 4 A5 + 2 A6 ) A0 2 L E A 2 -—- R1 6 [ (2 Al-2 A2+A4) A6 + (A2-4 A3+A5) A7 + 4 L ( A4 - 2 A5 + 2 A6 ) A8 ] - n1 2,6 E A 62 -- R1 6 (1- -) [ (-2 A1+2 A2—A4) A6 + (-A2+4 A3-A5) A7 + 2 L 12 (-A4+2 A5-2 A6) A8 ] E A 6 2 R1 R2 6 [ (-2 A1+2 A2-A4) A6 + (-A2+4 A3-A5) A7 + 24 L (-A4+2 A5-2 A6) A8 ] -— R1 R2 6 ( - A2 + 4 A3 + AA - 3 A5 + A6 ) o n1 n1 n1 n1 n1 n1 n1 n1 n1 n1 7,7 7,8 7,9 7,11 7,12 8,8 8,9 8,11 8,12 9,9 tr} 3° N [_4 6 {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A5-4 A6) A4 114 [ ( 18 A - 12 A + 8 A ) A0 - 3 5 6 + a (3 A2-3 A3) 19 + a (3 A3-A5) A10} 1 -- [ 6 ( 9 A - 6 A + 4 A6 ) A0 + 6 (1- ——) {(3 A2-2 A4) A2+ (6 A3-2 A5) A3+ (3 A5-4 A6) A4 1 02 “‘ {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 AS-A A6) A4 4 E A R2 6 2 L 6 ““ {(9 A2-8 A4) A2 + (18 A3-8 A5) A3 + (9 A5-16 A6) A4 12 E A ___ 2 _ _ R2 9 [ (AA-A2) A6 + (A5 2 A3) A7 + (2 A6 A5) 18 1 4 L n1 2’2 n1 2’3 ' “1 2,11 ' “1 2,12 E A 6 2 L 62 (1- ——) {(3 A2-2 A4) 12+ (6 A3-2 A5) A3+ (3 A5-4 A6) AA 12 2 2 3 5 + 6 (3 A2-3 A3) A9 + 6 (3 A3-A5) A10} - + a (3 A2-3 A3) A9 + 6 (3 A3-A5) A10} ] [ ( - 6 A3 + 5 A5 - 4 A6 ) A0 + + 6 (9 A2-12 A3) 29 + 6 (9 A3-4 A5) A10} ] 6 2 [ - ( 9 A3 - 6 A5 + 4 A6 ) A0 + + a (3 A2-3 A3) A9 + a (3 A3-A5) 110} 1 “1 9,11 a “1 9,12 ° n111,11 ’ nl11,12 ' n112,12 ' in which : 115 E A R2 6 6 [-(-6A+5A-4A 2 L 2 3 5 6 62 g; {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A5-4 A6) A4 + )Ao+ 6 ( 3 A2 - 3 A3 ) A9 + 6 ( 3 A3 - A 02 (l- 1;) {(A4-A2) A2 + (AS-2 A3) A3 + (2 A6-A5) A4 + s 10} 6 ( 1.5 A3 - A2 ) 19 + 6 ( 0.5 A5 - A3 ) 110} 1 E A R2 6 62 --;-;T-' (1- 1;) [ (AZ-A4) A5* (2 A3'A5) 37+ (A5‘2 A6) *8] E A 2 2 -——— R2 6 1 ( 2 A - 2 A + 2 A ) A - 2 L 3 5 6 o 6 -;; {(2 A2-2 A4) 12 + (4 A3-2 A5) 13 + (2 A5-4 A6) 14 + 6 ( 2 A E A 6 2 2 R2 6 1 (AZ-A4) A6 + (2 A3-A5) A7 + (AS-2 A6) 18 1 2 - 3 A3 ) A9 + 6 ( 2 A3 - A5 ) A10} ] 24 L E A 2 2 —R26(2A-2A+2A)A , 2 L 3 5 6 o 1 . 1 ¢~ H o‘ H on! H + h) 0‘ no % Q6 N <1.» 0‘ 116 2 3 2 3 b1 3 b1 3 b1 bl+2 b2 6 A5 ' 2 [1 " + 2 " 3 log (——)] 6 b2 4 6 b2 (2 6 b2) (2 6 b2) b1 4 b 2 b2 . 4 b3 1 1 1 A6 ‘ 2 [ 1 ' + 2 ' 3 + 8 6 b2 6 6 b2 (2 6 b2) (2 6 b2) 4 b4 b + 2 b 6 1 1 1 2 4 08 ( ) 1 (2 6 b2) 61 For circular arch, the expressions of Al’ A2, A3, A4, A5, and A6 become 1 2 A = ““ A — -——— 1 ’ 2 6 b1 6 b1 4 l A = A - .5 A 3 ’ a 3 3 6 b1 3 9 A - — A a 5 ’ 6 6 b1 5 6 b1 The expressions for A0, A1, .......... A12 are given by : 6 62 R1 62 6 62 R2 62 A = - - u - (1———) w - 6 - — u + (1-——) w + 6 ° 2 A 12 A 12 VA 2 B 12 B 12 VB ’\1 7' uA A2 = - 6 WA + R1 6 6yA A3 = - 3 uA + 2 6 WA - 2 R1 6 6yA + 3 uB + 6 WB - R2 6 6yB A4 - 2 uA - 6 WA + R1 6 6yA - 2 uB - 6 WE + R2 6 6yB "5 = vA B.2 and n2 n2 n2 1,1 1,3 117 A7 a - 3 VA + 2 R1 6 6xA + 3 VB + R2 6 6x3 A8 = 2 VA - R1 6 6xA - 2 VB - R2 6 6xB A9 3 WA A10 - - WA + wB A11 flA THE SECOND ORDER INCREMENTAL STIFFNESS MATRIX, [n2] Only nonzero terms are given. The expressions for A1, A2, .....A6 A0, A1, ......... A12 remain the same. E A 6 = 4 L2 (bl+6 b2) [ (31+32+B3+B4) (18 A3-12 A5+8 A6) + {(2 A4-3 A2) A2 + (2 A5-6 A3) A3 + (4 A6-3 A5) A4 + 2 6 (3 A3-3 A2) A9 + 6 (AS-3 A3) A10} ] E A 6 1,2 - 4 L2 (b1+6 b2) [ { (2 A4-3 A2) A6 + (2 A5-6 A3) A7 + (4 A6-3 A5) A8 1 { (2 A4-3 A2) A2 + (2 A5-6 A3) A3 + (A A6'3 A5) 14+ 6 (3 A3-3 A2) 19+ 6 (AS-3 A3) 110} ] E A 6 - 4 L2 (b1+6 b2) 6 [ (31+BZ+B3+B4) (-6 A3-2 A4+6 A5-4 A6) + 0.5 {(3 A2-2 A4) A2+(6 A3-2 A5) A3+(3 A3-2 A4+3 A5-4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 (3 A3-A5) A10} { (2 A4-3 A2) A2 + (2 A5-6 A3) A3 + (4 A6-3 A5) A4 + 6 (3 A3-3 A2) A9 + 6 (AS-3 A3) 110} ] n2 n2 n2 n2 1,6 1,7 1,8 1,9 (b1 +6 b 2) R1 6 [ (B1 +3 -7 A + (A4'2 A 6 (0.5 A 118 +3 3+34) (-3 A +12 A +2 A4 2 2 3 5+4 A6) + {(2 A1-2 A2 +A 4) A2 + (A2 -4 A3+A5 ) A3 5+2 A6) A4 + 6 (2 Al-2 A2+1.5 A3) A9 + -2 A +0.5 A5) A10} {(2 A4-3 A2) A2 + 2 3 (2 A5-6 A3) A3 + (4 A6-3 A5) A4 + 6 (3 A3-3 A2) A9 + 6 (AS-3 A3) A10} ] (b1 +6 b 2) R1 6 {(- 2 A1+2 A2 A) A6 + (- -A2 +4 A3- A5) A7 + (-A4+2 A -2 A6) A8} {(2 A4-3 A2) A2 + 5 (2 A5-6 A3) A3 + (4 A6-3 A5) A4 + 6 (3 A3-3 A2) A9 + 6 (A573 A3) A10} ] (bl+6 b2) [ (B 1+B2+BB+B4) (-18 A3+12 A5-8 A6) + + ((3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A5-4 A6) A4 + 6 (3 A2-3 A3) A9+ 6 (3 A3-A5) A10} {(2 A4-3 A2) A2 + + (2 A5-6 A3) A3 + (4 A6-3 A5) A4 + 6 (3 A3-3 A2) A9 6 (AS-3 A3) A10} ] (b1+6 b2) [ {(3 A2-2 A4) 16 + (6 A3-2 A5) A7 + (3 A5-4 A6) A8} {(2 A4-3 A2) A2 + (2 A5-6 A3) A3 + (4 A6-3 A5) A4+ 6 (3 A3-3 A2) A9+ 6 (AS-3 A3) A10) 1 (b1+6 b2) [ (31+BZ+BB+B4) 6 (- 9 A3+6 AS- -4 A 6) + 6 2 + + + {(3 A2-2 A4) A2+ (6 A3-2 A5) 13+ (3 A5-4 A6) 14 6 (3 A2-3 A3) 19+ 6 (3 A3-A5) A10} ((2 A4-3 A2) 12 (2 A5-6 A3) A3 + (4 A6-3 A5) 14 + 6 (3 A3-3 A2) 19 6 (AS-3 A3) A10} ] 119 E A 6 n2 1,11 = 4 L2 (b1+6 b2) R2 6 [ (31+BZ+B3+B4) (6 A3-5 A5+4 A6) + {(A4'A2) A2 + (AS-2 A3) A3 + (2 A6-A5) 14 + 6 (1.5 A3-A2) A9+ 6 (0.5 A5-A3) 110) {(2 A4—3 A2) 12 + (2 A5-6 A3) A3 + (4 A6-3 A5) A4 + 6 (3 A3-3 A2) A9 + 6 (A5'3 A3) A10) ] E A 6 n2 1,12 = 4 L2 (b1+6 b2) R2 6 {(AZ'A4) A6 + (2 A3-A5) A7 + (AS-2 A6) A8} {(2 A4-3 A2) A2 + (2 A5-6 A3) A3 + (4 A6-3 A5) A4 + 6 (3 A3-3 A2) A9 + 6 (A5'3 A3) A10} E A 6 1 n2 2,2 = 4 L2 (bl+6 b2) [ (31+32+B3+B4) (18 A3-12 A5+8 A6) + 2 {(2 A4—3 A2) A6 + (2 A5-6 A3) A7 + (4 A6-3 A5) A8} ] E A 6 6 n2 2 3 - 2 (bl+6 b2) - {(3 A2-2 A4) 12 + (6 A3-2 A5) A3 + ' 4 L 2 (3 A3-2 A4+3 A5-4 A6) A4 + 6 (3 A2-3 A3) 19 + 6 (3 A3-A5) A10) {(2 A4-3 A2) A6 + (2 A5-6 A3) A7 + (4 A6-3 A5) A8} E A 6 n2 2,5 = 4 L2 (b1+6 b2) [ R1 6 {(2 A1-2 A2+A4) A2+ (AZ-4 A3+A5) A3 - l + (A4'2 A5+2 A6) A4 + 6 (2 A1 2 A2+1.5 A3) A9 + 6 (0.5 A2-2 A3+0.S A5) A10) {(2 A4-3 A2) A6 + (2 A5-6 A3) A7 + (4 A6-3 A5) A8} ] E A 6 2 n2 2,6 = 4 L2 (bl+6 b2) R1 6 [ (B1+B2+BB+BQ) (3 A2-12 A3— A4+7 A5 -4 A6) - {(2 A1—2 A2+A4) A6 + (AZ-4 A3+A5) A7 + (A4'2 A5+2 A6) A8} {(2 A4-3 A2) A6 + (2 A5-6 A3) A7 + (4 A6-3 A5) A8} ] n2 n2 n2 n2 n2 n2 2,7 2,8 2,9 2,11 2,12 3,3 120 (b1+6 b2) {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A5-4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 (3 A3-A5) A10} {(2 A4-3 A2) A6 + (2 A5-6 A3) A7 + (4 A6-3 A5) A8} ] (b1+6 b2) [ (B1+B2+B3+B4) (-18 A3+12 A5—8 A6) + {(3 A2-2 A4) A6 + (6 A3-2 A5) A7 + (3 A5-4 A6) A8} {(2 A4-3 A2) A6 + (2 A5-6 A3) A7 + (4 A6-3 A5) A8} ] 6 (b1+6 b2) [ g [(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A5-4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 (3 A3—A5) A10} {(2 A4-3 A2) A6 + (2 A5-6 A3) A7 + (4 A6-3 A5) A8} ] (bl+6 b2) [ R2 6 {(A4'A2) 12 + (AS-2 A3) 13 + (2 A6-A5) A4 + 6 (1.5 A3-A2) k9+ 6 (0.5 AS-A3) A10} {(2 A4-3 A2) A6 + (2 A5-6 A3) A7 + (4 A6-3 A5) A8} ] (b1+6 b2) R2 6 1 (B1+B2+B3+Ba) (-6 A3+5 A5-4 A6) + {(A2-A4) A6 + (2 A3-A5) A7 + (AS-2 A6) A8} {(2 A4-3A2) A6 + (2 A5-6 A3) A7 + (4 A6-3 A5) A81 1 62 (b1+6 b2) -;— [ (B1+B +3 +34) (3 A3+4 A4-6 A5+4 A6) + 0.5 {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A3-2 A 3 A 5 2 3 -4 A6) 14+ 6 (3 A2-3 A3) A9 + 6 (3 A3-A5).A10} 4 2 + 1 n2 3,5 n2 3,6 “2 3,7 n2 3,8 n2 3’9 121 2 R1 6 (b1+6 b2) 7 A5 (A4-2 A5 6 (0.5 A2 -2 A 3 [ (Bl+B +B +B 2 3 4) (3 A2-9 A3 +2 A6) A4 + 6 (2 A1-2 A2+1.5 A3) A9 + +0.5 A5) A10} {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A3-2 A4+ 3 A5-4 A6) A4 + 6 (3 A2-3 A3) 19 + 6 (3 A3-A5) 1101 1 R1 6 (bl+6 b2) 2 2 {(-2 A 1 -4 A + 4 -4 A6) + {(2 A1-2 A2+A4) A2 + (AZ-4 A3+A5) A3 + +2 Az-Aa) A6 + (—A2+4 A3- A5) A7 + (-A4+2 A5-2 A6) A8} {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A3—2 A4+ 3 A5-4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 (3 A3-AS) 110} ] 6 (b1+6 b2) _2— [(B1 2 +3 +B +B 2 3 4) (12 A 3 +4 A 4 -12 A 5+8 A6) + {(3 A2-2 A4) 12 + (6 A3-2 A5) 13 + (3 A5-4 A6) A4 + 6 (3 A2-3 A3) A9+ 6 (3 A3-A5) A10} {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A3 -2 A4+3 A 5 -4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 (3 A3-A5) 110} ] 6 (b1+6 b2) ; {(3 A2-2 A4) A6 + (6 A3-2 A5) A7 + (3 A5-4 A6) A8} {(3 A2—2 A4) A2 + (6 A3-2 A5) A3 + (3 A3-2 A4+3 A5-4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 (3 A3-A5) A10} ] 6 2 (b1+6 b2) -;— [ (31+BZ+B3+B4) (6 A3+2 A4-6 A5+4 A6) + 0.5 { (3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A5-4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 (3 A3-A5) A10} {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A3-2 A4 +3 A5- 4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 (3 A3-A5) A10} ] 122 EA6 R262 b ( l+6 b2)— 2 [(B1+B 2+3 3+B4) (- 3 A3- 2 A4 +5 A5- —4 A 6) :3 h) II N) 3,11 4 L + {(A4-A2 ) A2 + (A5 -2 A3) A3 + (2 A6-A5) A4 + 6 (1. 5 A3 A) A 9+ 6 (0. 5 A5 -A3) A10} {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A3-2 A4+3 A5-4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 (3 A3-A5) A10} ] E A 6 R2 62 n2 = 2 (b1+6 b2) -—;—- { (AZ'A4) A6 + (2 A3-A5) A7 + 3,12 (As-2 A6) A8} {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A3-2 A4+3 A5—4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 (3 A3-A5) A10} n2 5,5 = 2 (bl+6 b2) R12 2[(B1+B2 +B 3+B4) (2 A14 A2+8 A3+2 A4 -4 A5+2 A6) + {(2 A1-2 A2+A4) A2 + (AZ-4 A3+A5) A3 + (A4-2 A5+2 A6) A4 + 6 (2 A1-2 A2+1.5 A3) A9 + 6 (0.5 A -2 A 2 2 3+0.5 A5) A10} ] 2 2 n2 5,6 = 2 (b1+6 b2) R1 6 {(2 A1-2 A2+A4) A2+ (AZ-4 A3+A5) A3 + (A4'2 A5+2 A6) A4 + 6 (2 A1-2 A2+1.5 A3) A9 + 6 (0.5 A2-2 A3+0.5 A5) A10} {(-2 A1+2 Az'A4) A6 + (-A2+4 A3-A5) A7 + (- A4+2 AS-Z A6) A8} (b1+6 b2) R1 6 [ (B W+BZ+B 4) (3 A2 -12 A3- -2 A4 +7 A5 :3 A) II no 5’7 4 L -4 A6) + ((2 A1-2 A2+A4) A2+ uxf4.A.§A %,x + (A4'2 A5+2 A6) A4 + 6 (2 A1-2 A2+1.5 A3) A9 + 6 (0.5 A2-2 A3+0.5 A5) A10} {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A3-2 A4+3 A5-4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 (3 A3-A5) A10} ] n2 5,8 = 5,9 “2 5,11 ’ “2 5,12 n2 = 6,6 123 (b1 +6 b2) R1 6 {(2 A1-2 A2 +A 4) A 2+ (A2 -4 A3+A5 ) A3 + (AA-2 A +2 A6) A4 + 6 (2 Al-z A2+1.5 A3) A9 + 5 6 (0.5 A2-2 A3+0.5 A5) A10} {(3 A2-2 A4) A6 + (6 A -2 A5) A7 + (3 A5-4 A6) A8} R1 62 3 (b1+6 b2) [ (B1+B +B 3+B4) (3 A2 -12 A3— 2 A4 + 2 7 A5 (Ah-2 A -4 A6) + {(2 A1-2 A2+A4) A2+ (AZ-4 A3+A5) A3 + 5+2 A6) A4 + 6 (2 A1-2 A2+1.5 A3) A9 + 6 (0.5 A 2 A +0.5 A5) A10} {(3 A2-2 A4) A2 + 2' 3 (6 A3-2 A5) A3 + (3 A3-2 A4+3 A5-4 A6) A4 + 6 (3 A2-3 A3) A9 + 6 (3 A3-A5) A10} ] +A -3 A (bl+6 b2) R1 R2 62 [ (31+B2+B3+ (-A2+4 A3 4 5 BA) +2 A6) + {(2 A1-2 A2+A4) A2+ (A2-4 A3+A5) A3 + (A4-2 A +2 A6) A4 + 6 (2 A1-2 A2+1.5 A3) A9 + 5 6 (0.5 A -2 A 2 3 -A5) A4+ 6 (1.5 A3-A2) A9+ 6 (0.5 AS-A3) A10}1 +0.5 A5) A10} {(A4-A2) A2+(A5-2 A3) A3 +(2 A6 2 (b1+6 b2) R1 R2 6 {(2 Al-2 A2+A4) A2 + (A2 4 A3+ A5) A3 + (A4 -2 A5+2 A 6) A4 + 6 (2 A1-2 A2 + 1.5 A3) A9 + 6 (0.5 A2-2 A3+0.5 A5) A10} {(A2 A4) A6 + (2 A3-A5) A7 + (AS-2 A6) A8} 2 2 (b +6 b2) R1 6 [ (B1+B2 +B 3+B4) (2 A1-4 A2+8 A3+ 1 2 A4 -4 A5+2 A 6) + {(- -2 A1+2 A2— A4) A6 + ( -A2 +4 A3- A 5) A7 + (- -A +2 AS— 2 A 6) A8 }2 ] “2 6,7 = “2 6,8 = “2 6,9 ’ “2 6,11 ’ “2 6,12 ’ “2 7,7 = 124 (b1+0 b2) R1 0 {(-2 A1+2 AZ-Ah) A6+ (-A2+4 A3-A5) A7 + (-A4+2 A -2 A6) A8} {(3 A2-2 A4) A2 + 5 (6 A3-2 A5) A3 + (3 A3-2 A4+3 A5-4 A6) A4 + 0 (3 A2'3 A3) A9 + 9 (3 AB’AS) A10} ] +12 A +2 A - (’3 A2 3 4 (bl+9 b2) R1 9 [ (31+BZ+B3+BA) 7 A5+4 A6) + {(-2 A1+2 A2- 4) A6+ (-A2+4 A3- A5) A7 + (-A4+2 A5-2 A6) A8} {(3 A2-2 A4) A6 + (6 A3—2 A5) A7 + (3 A5-4 A6) A8} ] R1 02 (b1+9 b2) {(-2 A1+2 AZ'AA) A6+(-A2+4 A3«A5) A7 +(-A4+2 AS-Z A6) A8} {(3 A2-2 A4) A2+ (6 A3‘2 A5) A3 +(3 A5-4 A6) A4+ 9 (3 A2'3 A3) A9+ 6 (3 A3‘A5) A10)] {(-2 A 2 (bl+0 b2) R1 R2 0 +2 AZ'AA) A6 + (-A2+4 A3- 1 A5) A7+ (~A4+2 AS-Z A6) A8) {(AA'AZ) A2+(A5-2 A3) A3 +(2 A6-AS) A4+ 9 (1.5 A3’A2) A9+ 0 (0.5 AS’A3) A10}] 2 (bl+0 b2) R1 R2 6 [ (81+B 2+B3 +B4) ( -A2 +4 A3+A4- 3 A5 +2 A6) + {(-2 A1+2 AZ'AA) A6 + (-A2+4 A3- A5) A7 + (’A4+2 A '2 A6) A8) {(Az'Aa) A6 + (2 A3-A5) A7 + 5 (A5'2 A6) A8} ] (b1+0 b 2) [ (B1+B2 +B 3+154) (18 A3- -12 A5+8 A6 ) + {(3 A2 -2 A ) A2 + (6 A3 2 A 5) A3 + (3 A5 4 A6 ) A + a (3 A2-3 A3) A9+ a (3 A3- A5) A10}2 ] “2 7,8 “2 7,9 “2 7,11 “2 7,12 n2 8,8 n2 8,9 “2 8,11 125 (b1+0 b2) {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A5—4 A6) Aa+ 0 (3 A2-3 A33 A9+ 6 (3 A3-A5) A10) {(3 A2-2 A4) A6 + (6 A3-2 A5) A7 + (3 A5-4 A6) A8} 6 (b1+0 b2) ; [ (31+BZ+BS+BA) (18 A3-12 A5+8 A6) + {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A5-4 A6) A4+ 2 0 (3 A2-3 A3) A9+ 0 (3 A3-A5) A10} ] (b1+0 b2) R2 0 [ (31+BZ+B3+BA) (-6 A3+5 AS-A A6) + ((3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A5-4 A6) A4+ 6 (3 A2-3 A3) A9+ a (3 A3-A5) A10} {(Ah-Az) A2 + (AS-2 A3) A3 + (2 A6-A5) A4 + 0 (1.5 A3-A2) A9 + a (0.5 AS-A3) A10} 1 (b1+6 b2) R2 0 {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A5-4 A6) A4+ 0 (3 A2-3 A3) A9+ 0 (3 A3-A5) A10} {(A4-A2) A6 + (2 A3-A5) A7 + (AS-2 A6) A8} (b1+6 b2) [ (Bl+Bz+B3+B4) (18 A3-12 A5+8 A6) + 2 ((3 A2-2 A4) A6 + (6 A3-2 A5) A7 + (3 A5-4 A6) A8} 1 a (b1+6 b2) 2 ( (3 A2-2 A4) A6 + (6 A3-2 A5) A7 + (3 A5—4 A6) A8} {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 AS-A A6) A4+ o (3 A2-3 A3) A9+ a (3 A3-A5) A10} (bl+0 b2) R2 9 {(3 A2-2 A4) 36 + (6 A3—2 A5) A7 + (3 A -4 A6) A8} {(A4-A2) A2 + (AS-2 A3) A3 + 5 (2 A6-A5) A4+ 6 (1:5 A3-A2) A9+ 6 (0.5 AS-A3) A10} “2 8,12 n2 9,9 “2 9,11 “2 9,12 n211,11 n211,12 E A 0 4 L 126 (b1+€ b2) R2 0 [ (B1+B +8 +34) (6 A3-5 A 2 3 +4 A6) + 5 {(3 A2-2 A4) A6 + (6 A3-2 A5) A7 + (3 AS-a A6) A8} {(AA-Az) A6 + (2 A3-A5) A7 + (AS-2 A6) A8} 1 62 (b1+9 b2) -2- [ (31+BZ+B3+BA) (18 A3-12 A5+8 A6) + {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 AS-a A6) A4+ 2 9 (3 A2-3 A3) A9+ 0 (3 A3-A5) A10} ] R2 02 (b1+0 b2) -;- [ (B1+BZ+B3+B4) (-6 A3+5 A5-4 A6) + {(Ah'Az) A2 + (A5'2 A3) A3 + (2 A6-A5) A4+ 0 (1.5 A3-A2) A9+ 0 (0.5 AS'AB) A10} {(3 A2-2 A4) A2 + (6 A3-2 A5) A3 + (3 A5—4 A6) A4 + 6 (3 A2-3 A3) A9 + 9 (3 A3-A5) A10} ] R2 02 (b1+9 b2) {(3 A2-2 A4) A2 + (6 A 32 A E5) i 2 S {(AZ‘AA) A6 + (2 A3-A5) A7 + (AS-2 A6) A8} ] 2 2 (b1+0 b2) R2 9 [ (31+32+R3+34) (2 A3-2 A5+2 A6) + {(Aa-AZ) A2 + (A5-2 A3) A3 + (2 A6-A5) A4+ 2 a (1.5 A3-A2) A9+ o (0.5 A5-A3) A10} 1 2 2 (b1+9 b2) R2 0 {(A4-A2) A2 + (AS'2 A3) A3 + (2 A6-A5) A4+ a (1.5 A3-A2) A9+ a (0.5 AS-A3) A10} {(A2-A4) A6 + (2 A3-AS) A7 + (AS-2 A6) A8} 127 E A 0 2 2 n212,12 = _Z—;5 (b1+0 b2) R2 0 [ (B1+32+B3+Bh) (2 A3-2 AS+2 A6) + 2 {(Az‘Aa) A6 + (2 A3-A5) A7 + (AS-2 A6) A8} ] in which : B - A A2 + A A A + A A2 + A A A + A A A + A A2 l l 2 2 2 3 3 3 4 2 4 5 3 4 6 4 82 = 2 9 Al A2 A9 + 6 A2 A3 A9 + 0.5 0 A2 A2 A10 + 1.5 0 A3 A4 A9 + 6 A3 A3 A10 + 0.5 6 A5 A4 Alo ‘ 2 2 2 2 2 B3 = 9 A1 A9 + 0.5 0 A2 A9 A10 + 0.25 9 A3 A10 B = A A2 + A A A + A A2 + A A A + A A A + A A2 4 1 6 2 6 7 3 7 4 6 8 5 7 8 6 8 APPENDIX C COMPUTER PROGRAM 6.1 GENERAL A general description of the computer program is given in Section 3.3 . A listing of the program, which was named NANCURVE (flonlinear Analysis of Curved Beam Structures), is given at the end of this Appendix. A description of the subroutines used in the program and its corresponding input data examples are presented in the following. C. 2 DESCRIPTION OF SUBROUTINES The computer program consists of a main program called NANCURVE, fourteen subroutines and one function subprogram. The main program NANCURVE directs the flow of the computations by calling the appropriate subroutines for each step of the solution procedure. The subroutine NODDATA reads data regarding the overall geometry of the arch and the nodal degrees of freedom. It generates the coordinates of the nodes and the equation numbers. The subroutine BAND computes the semi bandwidth that the stiffness matrix of the structure will have. This is done by obtaining the largest difference between the equation numbers of the nodes of any element. The subroutine ELEMENT calls the appropriate element subroutine. All basic information concerning the curved elements, i.e. , material, cross-section, and element properties are read by the subroutine CURVED. The subroutine also directs the computation of the geometric 128 129 properties of the curved element, which is accomplished by the subroutine GEOMETRY, the computation of the stiffness matrices of each elenunmt, which is performed by the subroutine NUMINT, and the assembly into the structure stiffness matrices, which is carried out by the subroutine ASSEMBLE. The condensation of the element linear stiffness matrix, from 16 by 16 to 12 by 12, is performed by subroutine REOCON. Subroutine STCOND prints out the element and structure stiffness matrices. ' The subroutines LINSOLN and GAUSSOL solve the system of linear equations by Gauss elimination. Identification of the displacements (flatained from LINSOLN is carried out by subroutine IDENT. The function subprogram DETl evaluates the determinant of the structural tangent stiffness matrix. Finally subroutine STRESS evaluates the element end forces and stresses. It should be noted that in addition to the subroutines mentioned previously, there are some more subroutines contained in the progrann Those subroutines are necessary for buckling (eigenvalue) analysis, which is not discussed in the present study. 0.2 VARIABLES USED IN THE COMPUTER PROGRAM \. The variable names used in the program are listed below in alphabetical order : A1 =- Lower limit of the numerical integration (= 0.0); A2 =- Upper limit of the numerical integration (=- 1.0); A(M) =- Area. of the cross-section.of element M; DETOPTN a Variable controlling the determinant of [ST]; If DETOPTN - 0 , no control on the determinant of [ST]. DI - DELTAl = DELTA2 — DM - EIGVALU - E(N) - G(N) = H a IA(N,I) - IARCH = IB(N,I) - 130 If DETOPTN = l , execution will be terminated if the determinant of [S s 0. T] Total number of horizontal intervals in which the span of the parabolic arch is divided into; Allowable tolerance for force components of unbalanced force vector; Allowable tolerance for moments of unbalanced force vector; Density of the material (set - O in the present study); Set equal to 3 in the present study; Modulus of elesticity of element group N; Shear modulus of element group N; Height of parabolic or arbritary arch; Boundary condition code of node N for its Ith degree of freedom. Initially it is defined as follows IA(N,I) - 1 if constrained; IA(N,I) - 0 if free; After processing, IA(N,I) - 0 if initially a l; IA(N,I) - equation number for the d.o.f if initially=0; Variable that identifies the type of arch being studied If IARCH - 0 , parabolic arch. If IARCH - 1 , circular arch. If IARCH - 2 , arbitrary arch; Additional boundary condition code (in the present study, set - 0 ); ICALl ICAL2 ICAL3 ICALA ICALS ICAL6 ICAL7 IDATA IDIRCN IFIX Variable If ICALl If ICALl Variable If ICAL2 If ICAL2 Variable If ICAL3 If ICAL3 Variable If ICALh If ICALA Variable If ICALS If ICALS Variable If ICAL6 If ICAL6 131 controlling print out. - 0 , entries'of [k], [n1], [n2] are printed. - l , skip; controlling print out. - 0, element geometric properties are printed. - 1, skip; controlling print out. - 0, load vector, [K], [N1], [N2] are printed. - 1, skip; controlling print out. - 0 , initial and nodal loads processed into load vector are printed. - 1 , skip; controlling print out. - 0 , print load vector & displacement vector. - l , skip; controlling print out. - 0 , print element nodal displacements. - l , skip. Set ICAL6 - 2 if ISTRESS a l (to get a nice output); Variable (Set - 1 Variable If IDATA - 0 , If IDATA - l , Set equal to 0 Set equal to l controlling print out of eigenvalue analysis. for the present study); for checking input data. execute the program. data check only, skip all computations; for the present study; for the present study; ILOAD - IPART - ISTRESS ITERCHK IXX(M) IYY(M) JUSTK KT(M) L = LOADDIR MAXITER MP = MSUOPTN NE - NODEI(M)- NODEJ(M)- 132 Set equal to 1 for the present study (load is concentrated at the nodes); Variable controlling print out. If IPART-0, intermediate results of the displacement at every iteration process are printed. If IPART-l, intermediate results of the displacement at every iteration process are not printed; Variable controlling the computations of element end forces and stresses. If ISTRESS - O , skip. If ISTRESS - l , compute end nodal forces and stresses. Set equal 1 for the present study; Moment of inertia about x-axis of the cross-section of element M; Moment of inertia about y-axis of the cross-section of element M; Set equal to O for the present study; Torsion constant of element M; Span of the parabolic or arbitrary arch; Set equal to -1 for the present study; Maximum number of iterations; Number of Gauss points used in the numerical integration (2,3,4,5,6,10 or 15); Set equal to l for the present study; Total number of elements in the structure; Number of node I of element M; Number of node J of element M; NTYPE(N)= NUMEG = NUMEL(N)= NUMNP = N1GOPTN NIOPTIN N20PTIN 133 Element group N (set equal to l in the present study); Total number of element groups (set - l in the present study); Total number of elements in element group N (set equal to NE in the present study); Total number of nodal points in the structure; Set equal to 0 in the present study; Variable controlling the use of matrix [Nl]. If NIOPTIN - 0 , [N1] is not used in the analysis. If NIOPTIN - l , [N1] is used in the analysis; Variable controlling the use of matrix [N2]. If N20PTIN - 0 , [N2] is not used in the analysis. If N20PTIN - 1 , [N2] is used in the analysis; PINT(N,DOF) - Initial load applied at node N, in the DOFth direction; PINC(N,DOF) - Load increment applied at node N, in the DOFth direction; PTOT(N,DOF) - Total load applied at node N, in the DOFth PRIOPTN PROTYPE R - T1,..T8 a T(N) direction; Variable controlling the print out. If PRIOPTN - O , skip. If PRIOPTN - l , intermediate results at every iteration are printed); Set equal to 3 in the present study (fixed Lagrangian); Radius of curvature of the circular arch; Title of problem beeing solved; For circular arch : angle between a node and the center 134 line of the circular arch (in degrees); TT(N) = For arbitrary arch : angle between the slope at a node and global X-axis (in radians); TLDOF - Number of total loads applied to the structure; TOLER - Set equal to 0 in the present study; XS - X coordinate of the left most node of parabolic arch; X(N),Y(N),Z(N) - Global X, Y, and Z coordinates of node N. 0.2.3 INPUT DAIA.ARRANGEHENT The input data are arranged in the following order and formats DATA CARD FORMAT T1,T2,T3,T4,T5,T6,T7,T8 8A10 NE,NUMNP,NUMEG,IDATA,ICALl,ICAL2,ICAL3,ICAL4,ICAL5,ICAL6,ICAL7 1115 IARCH,ILOAD,IDIRCN 315 PRIOPTN,N20PTIN,N10PTIN,ITERCHK,MSUOPTN,NIGOPTN,IFIx:3USTK, 815 TOLER,DETOPTN FlO.5,I5 PROTYPE,EIGVALU,ISTRESS,IPART,LOADDIR 515 R * F15.9 N,(IA(N,I),I-l,6),(IB(N,I),I-1,6),T,Z(N) ** I5,1213,F15.10,F10.6 TLDOF,MAXITER,DELTA1,DELTAZ 2I5,2F10.6 N,DOF,PINT(N,DOF),PINC(N,DOF),PTOT(N,DOF) 215,3F10.4 NTYPE(N),NUMEL(N) 215 E(N),G(N),DM 3310.2 M,NODEI(M),NODEJ(M),A(M),IXX(M),IYY(M),KT(M) 3I5,4E15.6 2F5.2,IS A1,A2,MP 135 This is for circular arch. For parabolic arch : H,L,DI,XS 4F10.S For arbitrary arch : H,L 2F10.5 **This is for circular arch. For parabolic arch : N,(IA(N,I)),(IB(N,I)) 15,12I3 For arbitrary arch : N,(IA(N,I),I=1,6),(IB(N,I),I-l,6),TT(N),X(N),Y(N),Z(N) 15,1213, F9.6,3F10.6 -1M._“' " 136 NONLINEAR ANALYSIS OF BELYTSCHKO'S ARCH (SECTION h.4.1) O Odor—‘00 l—‘kor—‘KOOONmU‘IDWNH ...: O OOOOOOOOr-‘O OWHHO \0 OOI-‘O 1.0E+07 OCDVGUl-DUJNH H OGJNONU'IbUJNH P‘H‘P‘P‘P‘P‘P‘P‘P‘ 1 0 1 0 1 1 1 0 1 1 1 0 1 O 0 1 0 0 1 0 0 1 0 0 1 0 O 1 O O 1 0 l 1 1 10.0 1000.0 A.0E+06 2 3 A 5 6 7 8 9 3 NNNNNNNN l l l 0 -1 h‘h‘h‘h‘h‘h‘k‘k‘k‘ C>C>C>C>C>C>CH3c>c>c>c>c>c>c>c> 10.0 1000.0 0.0E+01 .0000E+00 .0000E+00 .0000E+00 .0000E+00 .0000E+00 .OOOOE+00 .0000E+00 .0000E+00 000000000 1 1 000000000 1 l 0 0 ~14 -8 -7 -5 000000000 OOOOOO 20000.0 OOOOOOOO .6667E+OO .6667E+00 .6667E+00 .6667E+00 .6667E+00 .6667E+OO .6667E+00 .6667E+00 0 -12. 0 -10. -3. -1. 1 0.000 .070 311 553 .794 .035 .276 517 759 0.0 00000000 1 0 000000000 000000000 .6667E+00 .6667E+00 .6667E+00 .6667E+00 .6667E+00 .6667E+00 .6667E+00 .6667E+OO NNNNNNNN .25E+00 .25E+OO .25E+00 .2SE+OO .25E+00 .25E+00 .25E+00 .25E+00 -* NONLINEAR ANALYSIS OF 0 8 9 1 l O 1 3 3 200.0 1 1 2 O 3 O 4 O 5 0 6 0 7 0 8 0 9 O l 90 9 1 1 8 1.0E+O6 1 1 2 2 3 3 a 4 5 S 6 6 7 7 8 8 O 1.0 F‘P‘P‘Ffldrdt414t4 O OHOH HOOOOOOOH OH ' U1 OOHOOOOOOOH U1C>F‘P‘H‘P‘F‘P‘F‘PHH o . 4.0E+05 UJKOWVGWDUJN 4-‘4-‘4-‘4-‘4-‘4-‘4—‘9 DADEPPO'S ARCH (SECTION 4.4.2) 1 l l O -1 ldrdrdh‘h‘hH‘tdré C>C>C>C>C>C>C>CHD an>c>c>CHo<3c> U! 0 o I 0 0 0.0E+01 .OOOOE+00 .0000E+OO .0000E+00 .0000E+00 .OOOOE+00 .0000E+00 .0000E+OO .0000E+00 000000000 1 000000000 137 1 1 0 -26 -18 -15 -3 000000000 000000000 500.0 rdrahJHHArdrdha .3333E+00 .3333E+OO .3333E+00 .3333E+00 .3333E+00 .3333E+00 .3333E+OO .3333E+00 ~30. l 0.00 000 .250 -22. 500 .750 .000 -11. .500 .750 250 0.0 P‘PHAE‘PJF‘PJF‘ l 0 000000000 000000000 .333BE+00 .3333E+00 .3333E+00 .3333E+00 .3333E+00 .3333E+00 .3333E+00 .3333E+00 NNNNNNNN .250E+00 .250E+OO .ZSOE+OO .250E+00 .250E+00 .250E+00 .250E+00 .250E+00 000000000 COMPUTER PROGRAM "NANCURVE" PROGRAM NANCURVE ****************************************************************** THIS PROGRAM USES THE FINITE ELEMENT METHOD TO ANALYZE A CURVED ELEMENT IN THREE DIMENSIONS. OTHER ELEMENTS MAY BE ANALYZED BY ADDING A SUBROUTINE FOR EACH NEW TYPE OF ELEMENT BEING USED. NONLINEAR PROPERTIES ARE TAKEN INTO CONSIDERATION IN THE CURVED ELEMENT. ****************************************************************** REAL IXX,IYY,KT,LENGTH,II,JJ,NlSTTOT COMMON/Cl/NE,NUMNP,NUMEG,NTYPE(3),NUMEL(3),IPAR,ICAL1,ICAL2, + ICAL3,ICAL4,ICAL5,ICAL6,ICAL7 COMMON/C2/NSIZE,NEQ,NCOND,MBAND,IEIGEN . COMMON/C3/IA(37,8),IB(37,8),X(37),Y(37),Z(37),RAD,AC COMMON/C4/SE(16,16) COMMON/C5/E(3),G(3),NODEI(36),NODEJ(36),A(36),IXX(36),IYY(36), + KT(36),L(1,36) COMMON/C6/A1,A2,MP,B1(36),B2(36),B3(36) COMMON/C7/RI(36),RJ(36),PHII(36),PHIJ(36),TETA(36),LENGTH(36), + RIA(36),RJA(36) COMMON/C8/PN(37,8),R(296),PINT(37,8) COMMON/C9/S(296,l6),SP(296,16),IDET COMMON/CIO/D(296),DIO(1184),RC(296),sc(296,16) COMMON/CII/DN(16),U(36,12),w<37,8),V(37,8) COMMON/ClZ/ULOC(36,12),RCOL(9),MSUOPTN,NlGOPTIN COMMON/Cl6/PRIOPTN COMMON/Cl7/A7TOT(36),A7OLD(36),BOL(36,5),BTO(36,S),BE(5) COMMON/Cl8/IARCH COMMON/Cl9/TT(36) DIMENSION DTEMP(296),PTEMP(296),PSTART(296),DTOT(296) DIMENSION PACTUAL(296),PSAVE(296),DACTUAL(296),NlSTTOT(296,l6) DIMENSION SOLD(296,16),SRK(296,16),SRNl(296,l6),PTOT(37,8) DIMENSION REFSTRT<37,8),REFPTMP(37,8>,SRN2(37,16),PINC(37,8) INTEGER PROTYPE,EIGVALU,PRIOPTN,DETOPTN,DOF,TLDOF READ(60,1010) T1,T2,T3,T4,T5,T6,T7,T8 WRITE(61,2020)T1,T2,T3,T4,T5,T6,T7,T8 READ(60,1015) NE,NUMNP,NUMEG,IDATA,ICAL1,ICAL2,ICAL3,ICAL4, + ICAL5,ICAL6,ICAL7 WRITE(61,2010)NE,NUMNP,NUMEG,IDATA,ICAL1,ICAL2,ICAL3,ICAL4, + ICAL5,ICAL6,ICAL7 ........ READ NODAL POINT DATA READ(60,1030) IARCH,ILOAD,IDIRCN WRITE(61,2030)IARCH,ILOAD,IDIRCN READ(60,6971) PRIOPTN,N20PTIN,NlOPTIN,ITERCHK, - ‘ + MSUOPTN,N1GOPTN,IFIX,JUSTK,TOLER,DETOPTN 138 .139 6971 FORMAT(815,F10.5,I5) WRITE(61,6972) PRIOPTN,N20PTIN,N10PTIN,ITERCHK, + MSUOPTN,NlGOPTN,IFIX,JUSTK,TOLER,DETOPTN 6972 FORMAT(10X,8HPRIOPTN=,IZ/lOX,8HN2OPTIN=,12/ + 10X,’NlOPTIN-',I2/10X,'ITERCHK=',IZ/lOX,'MSUOPTN=',12/10X, + 'NlGOPTNa',I2,10X,'IFIXa’,12,10X,'JUSTK=',12, + /1OX,'TOLER=',F10.5,10X,'DETOPTN=',12//) READ(60,1) PROTYPE,EIGVALU,ISTRESS,IPART,LOADDIR 1 FORMAT(SIS) WRITE(61,8761) PROTYPE,EIGVALU,ISTRESS,IPART,LOADDIR 8761 FORMAT(10X,'PROTYPE=',12,1OX,'EIGVALU=',12,10X,'ISTRESS=',12/ + lOX,’IPART =',12,10X,'LOADDIR=',12//) c DX=0. CALL NODDATA (IARCH,DX). c IF(PROTYPE.EQ.2) GO TO 510 c IF(ITERCHK.EQ.1) READ(60,1013) TLDOF,MAXITER,DELTA1,DELTA2 IF(ITERCHK.EQ.1) WRITE(61,1012) TLDOF,MAXITER,DELTA1,DELTA2 1013 FORMAT(215,2F10.6) 1012 FORMAT(' ','TLDOF=',IS,SX,8HMAXITER=,15,SX,7HDELTA1=,F10.6,5X, + 7HDELTA2=,F10.6//) C WRITE(61,409) DO 407 N=1,NUMNP DO 407 I-1,6 PINT(N,I)=0.0 PINC(N,I)-0.0 407 PTOT(N,I)-0.0 I=0 406 CONTINUE I=I+1 READ(60,4OS) N,DOF,PINT(N,DOF),PINC(N,DOF),PTOT(N,DOF) WRITE(61,410) N,DOF,PINT(N,DOF),PINC(N,DOF),PTOT(N,DOF) IF(I.LT TLDOF) GO TO 406 405 FORMAT(2I5,3F10.4) 409 FORMAT(' ',10X,' LOADING CONDITIONS : ’//,6X,'NODE',7X,'DOF',16X, + 'PINT',l6X,’PINC',16X,’PTOT’//) 410 FORMAT(' ',3X,IS,4X,IS,11X,F10.4,lOX,F10.4,10X,F10.4) GO TO 513 c 510 CONTINUE C c ........ READ AND STORE INITIAL LOAD DATA C WRITE(61,2015) WW=O. CALL LOAD (IARCH,ILOAD,IDIRCN,DX,WW) c 513 CONTINUE IF(PROTYPE.NE.3) GO TO 3021 5001 1655 1665 1675 1685 000 9152 909 499 140 SCALE=1.0E+OS DO 5001 I=1,NEQ PSAVE(I)=0.0 DACTUAL(I)=0.0 DTOT(I)=0.0 AUTOMATIC GENERATION OF LODPONl AND CORRESPONDING D.O.F IF(LOADDIR) 1655,1665,1675 IHORZ=1 IVERT=O ILAT=O GO TO 1685 IHORZ=O IVERT=1 ILAT=O GO TO 1685 IHORZ=O IVERT=O ILAT=1 CONTINUE IF LOAD WANTED FOR SPECIFIC LOAD LET ITETO=1 CHOSE THE APPROPRIATE VALUES OF LODPON1,LNODE1,AND LDOFl ITETO=0 IF(ITETO.EQ.0) GO TO 9152 LODPON1=20 LNODE1=8 LDOF1=2 IF(ITETO.NE.O) GO TO 700 D0 200 N=1,NUMNP DO 300 1=1,6 IF(IA(N,I).EQ.0) GO TO 300 IF(PINT(N,I).EQ.O) GO TO 300 IF(I.EQ.1.AND.IHORZ.EQ.0) GO TO 300 IF(IHORZ.EQ.0) GO TO 909 LODPON1=IA(N,I) LNODE1=N LDOF1=I GO TO 700 IF(IVERT.EQ.O.AND.I.EQ.2) GO TO 300 IF(IVERT.EQ.0) GO TO 499 LODPON1=1A(N,I) LNODE1=N LDOF1=I GO TO 700 IF(ILAT.EQ.0.AND.I.EQ.3) GO TO 1093 IF(ILAT.EQ.0) GO TO 1093 LODPON1=IA(N,I) LNODE1=N LDOF1=I 141 GO TO 700 1093 PRINT 13 PRINT 14 13 FORMAT(' PROGRAM CAN NOT CALCULATE THE VALUE OF LODPONl ') 14 FORMAT(' HELP WANTED, PROGRAM STOPPED AT APPR. LINE 266 ') GO TO 900 300 CONTINUE 200 CONTINUE 700 CONTINUE WRITE(61,2900) LODPON1,LNODE1,LDOF1 2900 FORMAT(' ',//11X,'THE D.O.F. IN WHICH LOAD HAS BEEN INCREASED=', + I3,//10X,'AT NODE=',13,5X,'WITH D.O.F =',I3//) DO 3010 I=1,NUMNP DO 3010 J=1,6 3010 U(I,J)=0.0 ICMECR=1 1001 DO 3020 I=1,NUMNP IF(IFIX.EQ.0) X(I)=X(I)+U(I,1) IF(IFIX.EQ.O) Y(I)-Y(I)+U(I,2) IF(IFIX.EQ.O) Z(I)=Z(I)+U(I,3) 3020 CONTINUE IF(PRIOPTN.EQ.0) GO TO 4994 WRITE(61,4995) 4995 FORMAT(/,10X,'NODE',lOX,'X(I)',1OX,’Y(I)',10X,’Z(I)',/) DO 4996 I-1,NUMNP WRITE(61,4997) I,X(I),Y(I),Z(I) 4997 FORMAT(/,10X,IS,3F15.8) 4996 CONTINUE 4994 CONTINUE 3021 CONTINUE C READ AND STORE ELEMENT DATA C AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA'AAAAAAAA'"'k' A AAAAAAAAA C IPAR=1 NUMITER=1 C | IF(ICHECK.NE.1) GO TO 5928 DO 100 N=1,NUMEG READ(60,1020) NTYPE(N),NUMEL(N) CALL ELEMENT(N,IDATA,IARCH) 100 CONTINUE 5928 CONTINUE IF(PROTYPE.NE.3) GO TO 3335 IF(ICHECK.NE.1) GO TO 3333 3335 CONTINUE C COMPUTE SEMIBANDWIDTH OF STRUCTURE STIFFNESS MATRIX nnnnnnnnnnnnnn QILIIDII """"""""""""" AAARRARAARARARAAkkkixnxAnmAAnnmnAmxx*xxxxxx* CALL BAND IF(JUSTK.EQ.1) GO TO 2110 IF(PROTYPE.NE.3) GO TO 2110 142 DO 5745 NN=1,NUMEG IF(NUMEL(NN).EQ.O) GO TO 5745 NAME=NUMEL(NN) DO 5341 K=1,NAME M=L(NN,K) IF(MSUOPTN.EQ.1) A7OLD(M)=0.0 DO 5341 I-1,5 IF(MSUOPTN.EQ.2) BOL(M,I)=0.0 5341 CONTINUE 5745 CONTINUE 2110 CONTINUE ASSEMBLE INITIAL LOADS AND NODAL LOADS INTO LOAD VECTOR SET ARRAYS -8- AND -R- EQUAL TO ZERO IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII llllllllllllllllllllllllllllll XxxxxxwxxxxxwxxxxxxxxxxrxxxxxxxhnanxxxwxAmhmxxkan? \\\\\\\ A AXWXAAKWBK 0000 CALL ASEMBLE (M) 3333 CONTINUE IF(PROTYPE NE.3) GO TO 3336 IF(ICHECK.EQ.1) GO TO 2111 GO TO 2112 2111 DO 5003 I=1,NEQ 5003 PSTART(I)=0.0 J=O DO 415 N=1,NUMNP DO 420 1=1,6 IF(IA(N,I).EQ.0) GO TO 420 J=J+1 PSTART(J)=PINT(N,I) 420 CONTINUE 415 CONTINUE IF(PRIOPTN.EQ.O) GO TO 423 WRITE(61,421) 421 FORMAT(' ',10X,'PSTART :') WRITE(61,422) (PSTART(I),I=1,NEQ) 422 FORMAT(' ',8X,F10 5) 423 CONTINUE i 2112 IF(JUSTK.EQ.1) ICHECK=2 IF(JUSTK.EQ.1) GO TO 3336 IF(ICHECK.EQ.1) GO TO 3336 IPAR=2 DO 2113 I=1,NSIZE D0 2113 J=1,MBAND S(I,J)=0.0 DO 111 N=1,NUMEG CALL ELEMENT(N,IDATA,IARCH) CONTINUE IF(ICAL3.EQ.O) CALL STCONDN [\J H ...—.1 UL) 000000000 H H H C1901 IF(ITERCHK.NE.O) CALL INVTRNS 1901 IF(ICHECK.EQ.3) GO TO 4988 1801 4987 7691 4988 00 3081 3071 C8005 8008 8009 143 IF(NlOPTIN.EQ.0) GO TO 4987 DO 1801 I=1,NSIZE DO 1801 J=l,MBAND S(I,J)=0.0 N=l IPAR=3 CALL CURVED(N,IDATA,IARCH) IF(ICAL3.EQ.0) CALL STCONDN CONTINUE IF(ICHECK.EQ.3) GO TO 4988 IF(N20PTIN.EQ.0) GO TO 4988 D0 7691 I=1,NSIZE DO 7691 J=l,MBAND S(I,J)=0.0 IPAR=4 CALL CURVED(N,IDATA,IARCH) IF(ICAL3.EQ.0) CALL STCONDN CONTINUE IF(ICHECK.EQ.2.0R.PSAVE(LODPON1).EQ.O.) GO TO 5010 DO 9436 1-1,NSIZE DO 9436 J=1,MBAND S(I,J)=0.0 IPAR=7 DO 9437 N-l,NUMEG IF(NUMEL(N).EQ.O) GO TO 9437 CALL KEPSIOl(N) CONTINUE IF(ICHECK.EQ.2) CO TO 5010 D0 3071 I=l,NEQ DO 3081 J=1,MBAND READ(4,lO) RK SRK(I,J)=RK READ(l6,10) RNISTAR SRN1(I,J)=RN18TAR NlSTTOT(I,J)=RleTAR IF(IFIX.EQ.1) S(I,J)=RK IF(IFIX.EQ.1) SOLD(I,J)=RK IF(IFIX.EQ.0) S(I,J)=RK+NlSTTOT(I,J) IF(IFIX.EQ.0) SOLD(I,J)=RK+N1$TTOT(I,J) CONTINUE CONTINUE REWIND 4 REWIND 16' IF(PRIOPTN.EQ.O) GO TO 7233 WRITE(61,8005) FORMAT(///,10X,'KEPSIO MATRIX',/) WRITE(61,8002) ((SRN1(I,J),J=1,MBAND),I=1,NEQ) WRITE(61,8008) FORMAT(///,10X,'K LINEAR STIFFNESS MATRIX',/) WRITE(61,8002) ((SRK(I,J),J=1,MBAND),I=1,NEQ) WRITE(61,8009) FORMAT(///,10X,'S(I,J) MATRIX',/) 144 WRITE(61,8002) ((S(I,J),J=1,MBAND),I=1,NEQ) 7233 CONTINUE GO TO 5011 5010 DO 4071 I=1,NEQ DO 4081 J=1,MBAND IF(NlOPTIN.EQ.1)READ(5,10) RN1 IF(PSAVE(LODPON1).EQ.0.) READ(4,10) RK IF(N20PTIN.EQ.1) READ(16,10) RN2 IF(NlOPTIN.EQ.1)SRN1(I,J)=RN1 IF(PSAVE(LODPON1).EQ.0..AND.N2OPTIN.EQ.1)SP(I,J)=RK+.5*RN1+RN2/3. IF(PSAVE(LODPON1).NE.O..AND.N20PTIN.EQ.1) +SP(I,J)=SOLD(I,J)+.5*RN1+RN2/3. IF(PSAVE(LODPON1).EQ.O..AND.N10PTIN.EQ.O)SP(I,J)=RK IF(PSAVE(LODPON1).EQ.O..AND.N10PTIN.EQ.0)S(I,J)=RK IF(PSAVE(LODPON1).EQ.O..AND.N10PTIN.EQ.1.AND.N20PTIN.EQ.O)SP(I,J)= +RK+.5*RN1 IF(PSAVE(LODPON1).NE.O..AND.N10PTIN.EQ.0)SP(I,J)=SOLD(I,J) IF(PSAVE(LODPON1).NE.O..AND.N10PTIN.EQ.O)S(I,J)=SOLD(I,J) IF(PSAVE(LODPON1).NE.0..AND.N10PTIN.EQ.1.AND.N20PTIN.EQ.O)SP(I,J)= +SOLD(I,J)+.5*RN1 IF(PSAVE(LODPON1).EQ. IF(PSAVE(LODPON1).NE. IF(PSAVE(LODPON1).EQ. +S(I,J)=RK+RN1 IF(PSAVE(LODPON1).NE.O..AND.N10PTIN.EQ.1.AND.N2OPTIN.EQ.O) +S(I,J)=SOLD(I,J)+RN1 ' 4081 CONTINUE 4071 CONTINUE IF(NIOPTIN.EQ.1) REWIND 5 IF(N20PTIN.EQ.1) REWIND 16 IF(PSAVE(LODPON1).EQ.0.) REWIND 4 IF(PRIOPTN.EQ.0) GO TO 5011 .IF(N20PTIN.EQ.0) GO TO 4989 . WRITE(61,7693) ] 7693 FORMAT(///,1OX,11HN2 MATRIX,/) DO 7694 I=1,NEQ 4 DO 7695 J=1,MBAND I READ(16,10) RN2 SRN2(I,J)=RN2 7695 CONTINUE 7694 CONTINUE 4989 CONTINUE IF(N20PTIN.EQ.1) REWIND 16 IF(N20PTIN.EQ.1) WRITE(61,8002)((SRN2(I,J),J=1,MBAND),I=1,NEQ) IF(NIOPTIN.EQ.1) WRITE(61,8004) 8004 FORMAT(//,10X,'N1 NONLINEAR STIFFNESS MATRIX',/) IF(N10PTIN.EQ.1) WRITE(61,8002) ((SRN1(I,J),J=1,MBAND),I=1,NEQ) IF(PSAVE(LODPON1).NE.0.) WRITE(61,8010) 8010 FORMAT(///,10X,'SOLD(I,J) MATRIX',/) IF(PSAVE(LODPON1).NE 0.) WRITE(61,8002) +((SOLD(I,J),J-1,MBAND),I=1,NEQ) WRITE(61,8018) .AND.N20PTIN.EQ.1)S(I,J)=RK+RN1+RN2 ..AND.N20PTIN.EQ.1)S(I,J)=SOLD(I,J)+RN1+RN2 ..AND.N10PTIN.EQ.1.AND.N2OPTIN.EQ.O) OOO 8018 5011 7001 336 00000000 2114 110 1081 1071 1809 9431 6001 8002 5005 145 FORMAT(///,10X,'SP(I,J) MATRIX',/) WRITE(61,8002) ((SP(I,J),J=1,MBAND),I=l,NEQ) WRITE(61,8009) WRITE(61,8002) ((S(I,J),J=1,MBAND),I=1,NEQ) IF(ICHECK.NE.3) GO TO 7001 GO TO 6001 ICHECK=3 GO TO 3339 CONTINUE COMPUTE ELEMENT LINEAR STIFFNESS AND ASSEMBLE INTO STRUCTURE LINEAR STIFFNESS *3]:****‘k************>’c**v'cv’n’r‘kv'n'c7'6)???*‘kv'c*kv'c‘kv'n'o'n'nk‘kvh'c‘k‘k‘k‘k*~k‘k5¥‘k*~k********* DO 2114 I=l,NSIZE DO 2114 J=1,MBAND S(I,J)=0.0 IPAR-2 DO 110 N=1,NUMEG CALL ELEMENT (N,IDATA,IARCH) CONTINUE IF(ICAL3.EQ.O) CALL STCONDN IF(PROTYPE.NE.3) GO TO 3337 DO 1071 I=l,NEQ DO 1081 J=1,MBAND READ(4,10) RK S(I,J)=RK SP(I,J)=RK CONTINUE CONTINUE REWIND 4 IF(NCOND.EQ.O) GO TO 1809 CALL STCONDN CONTINUE IF(PRIOPTN.EQ.O) GO TO 9431 IF(ICHECK.EQ.1) WRITE(61,8008) IF(ICHECK.EQ.1) WRITE(61,8002)((S(I,J),J=1,MBAND),I=1,NEQ) CONTINUE IDET-I FORMAT(lX,6(2X,El9.13),/) CALL LINSOLN DETRMNT-DET1(SCALE) DO 5005 I=1,NEQ DTOT(I)=DTOT(I)+D(I) DACTUAL(I)~DACTUAL(I)+D(I) IF(IFIX.EQ.0) D(I)=DTOT(I) IF(IFIX.EQ.1) D(I)=DACTUAL(I) CONTINUE IF(NCOND.NE.O) CALL RECOVER 5002 5763 8537 2121 2120 3339 1804 3901 2001 2108 2201 2301 5006 8011 146 CALL IDENT IF(JUSTK.EQ.1) GO TO 3339 IF(ITERCHK.EQ.0) CALL INVTRNS IF(ITERCHK.NE.0) GO TO 8537 D0 5763 NN-1,NUMEG IF(NUMEL(NN).EQ.O.) GO TO 5763 NAME=NUMEL(NN) DO 5002 K-1,NAME M=L(NN,K) A7TOT(M)=A7OLD(M)+U(M,7)-U(M,1) CONTINUE CONTINUE ICHECK-2 IF(ITERCHK.NE.0) GO TO 2120 DO 2121 I-1,NEQ R(I)=0.0 PACTUAL(I)=PSAVE(I)+PSTART(I) GO TO 3342 CONTINUE GO TO 1901 DO 2001 I=1,NEQ PTEMP(I)-0.0 IM=I+1 IF(IM.GT NEQ) GO TO 2001 DO 3901 J-2,MBAND IF(SP(I,J) EQ.0.) GO TO 1804 PTEMP(I)=PTEMP(I)+SP(I,J)*D(IM) IM=IM+1 IF(IM.GT.NEQ) GO TO 2001 CONTINUE CONTINUE DO 2301 I=1,NEQ IM=I JM=1 IF(SP(IM,JM).EQ.0.) GO TO 2201 PTEMP(I)=PTEMP(I)+SP(IM,JM)*D(IM) IM=IM-1 JM=JM+1 IF(IM.EQ.O) GO TO 2301 IF(JM.GT.MBAND) GO TO 2301 GO TO 2108 CONTINUE DO 5006 I-1,NEQ IF(IFIX.EQ.0) PACTUAL(I)=PTEMP(I)+PSAVE(I) IF(IFIX.EQ.1) PACTUAL(I)=PTEMP(I) CONTINUE IF(ITERCHK.EQ.0) GO TO 6975 IF(PRIOPTN.EQ.1) WRITE(61,8011) FORMAT(ISX,'I',5X,'PACTUAL(I)',10X,’PTEMP(I)',lOX,'PSAVE(I)',//) IF(PRIOPTN.EQ.0) GO TO 4990 D0 8012 I=1,NEQ WRITE(61,8013) I,PACTUAL(I),PTEMP(I),PSAVE(I) 8012 4990 8013 8547 8541 8542 1003 9001 6975 2115 9731 9732 9733 6976 1571 147 CONTINUE CONTINUE FORMAT(lOX,I5,5X,E21.15,10X,E21.15,10X,E21.15,/) WRITE(61,8547) FORMAT(/,10X,'DTOT(I)',/) IF(IPART.EQ.O) GO TO 1003 DO 8541 MM=1,NEQ WRITE(61,8542) DACTUAL(MM) FORMAT(lOX,E21.15) CONTINUE IF(IHORZ.EQ.1) LODPON2=LODPON1+1 IF(IHORZ.EQ.1) LODPON3=LODPON1+2 IF(IVERT.EQ.1) LODPON1=LODPON1-1 IF(IVERT.EQ.1) LODPON2=LODPON1+1 IF(IVERT.EQ.1) LODPON3=LODPON1+2 IF(ILAT.EQ.1) LODPON1=LODPON1-2 IF(ILAT.EQ.1) LODPON2=LODPON1+1 IF(ILAT.EQ.1) LODPON3=LODPON1+1 WRITE(61,9001)PACTUAL(LODPON1),PACTUAL(LODPON2),PACTUAL(LODPON3), + DACTUAL(LODPONl),DACTUAL(LODPON2),DACTUAL(LODPON3) FORMAT(' ',1OX,11HPACTUAL(X)=,E21.15,10X,11HPACTUAL(Y)=,E21.15, + 10X,11HPACTUAL(Z)=,E21.15//16HDISPLACEMENT(X)=,E21.15, + 10x,16HDISPLACEMENT(Y)=,E21.15,10x,16HDISPLACEMENT(2)=,E21.15//)~ IF(IVERT.EQ.1) LODPON1=LODPON1+1 IF(ILAT.EQ.1) LODPON1=LODPON1+2 CONTINUE DO 2115 I-1,NEQ R(I)=PSTART(I)-PTEMP(I) IF(PRIOPTN.EQ.0) GO TO 6976 WRITE(61,9731) FORMAT(//,20X,'R(I)',15X,'PSTART(I)',15X,’PTEMP(I)',/) DO 9732 IMM=1,3 WRITE(61,9733) R(IMM),PSTART(IMM),PTEMP(IMM) FORMAT(//,10X,E21.15,1OX,E21.15,10X,E21.15,/) CONTINUE IF(ITERCHK.EQ.O) GO TO 3342 D0 2451 NN-1,NUMEC IF(NUMEL(NN).EQ.0) GO TO 2451 NAME=NUMEL(NN) DO 2351 K=1,NAME M=L(NN,K) NI=NODEI(M) NJ=NODEJ(M) D0 2351 K1=1,2 IF(K1.EQ.1) NP=NI IF(K1.EQ.2) NP=NJ D0 2251 1=1,6 IF(IA(NP,I)) 1651,1551,1571 NL:IA(NP,I) REFSTRT(NP,I)=PSTART(NL) REFPTMP(NP,I)=PTEMP(NL) GO TO 2251 1551 1651 1751 1851 1951 2051 2151 2251 2351 2451 6949 6950 2116 2117 3342 .148 REFSTRT(NP,I)=0.0 REFPTMP(NP,I)=0.0 GO TO 2251 IF(IB(NP,I).LT.0) GO TO 1751 NM=IB(NP,I) GO TO 1851 NL=-IB(NP,I)+NEQ REFSTRT(NP,I)=PSTART(NL) REFPTMP(NP,I)=PTEMP(NL) GO TO 2251 IF(IA(NM,I)) 1951,2051,2151 NL=-IB(NM,I)+NEQ REFSTRT(NP,I)=PSTART(NL) REFPTMP(NP,I)=PTEMP(NL) GO TO 2251 REFSTRT(NP,I)=0.0 REFPTMP(NP,I)=0.0 GO TO 2251 NL=IA(NM,I) REFSTRT(NP,I)=PSTART(NL) REFPTMP(NP,I)=PTEMP(NL) CONTINUE CONTINUE CONTINUE DO 6949 NP=1,NUMNP DO 6949 J=1,3 KJJ-J+3 PARTI=ABS(REFSTRT(NP,J)-REFPTMP(NP,J)) PART2-ABS(REFSTRT(NP,KJJ)-REFPTMP(NP,KJJ)) IF(PART1.GT.DELTA1 OR.PART2 GT.DELTA2) GO TO 6950 WRITE(61,2116) PART1,PART2 CONTINUE GO TO 3342 WRITE(61,2116) PART1,PART2 FORMAT(IOX,6HPART1=,E21.15,10X,6HPART2=,E21.15) NUMITER=NUMITER+1 IF(NUMITER.LE.MAXITER) GO TO 2117 GO TO 900 GO TO 6001 CONTINUE IF(ITERCHK.NE.1) GO TO 8945 IF(MSUOPTN EQ.2) GO TO 8945 DO 8538 NN—1,NUMEG IF(NUMEL(NN).EQ.0) GO TO 8538 NAME=NUMEL(NN) DO 8539 K=1,NAME M=L(NN,K) T0=(U(M,8)-U(M,2))/LENGTH(M) SIO=(U(M,3)-U(M,9))/LENGTH(M) TA=U(M,6)-T0 TB=U(M,12)-T0 SIA=U(M,5)-SIO 149 SIB=U(M,1l)-SIO 8539 A7TOT(M)=A7OLD(M)+U(M,7)-U(M,l) ++.5*(T0**2+SIO**2)*LENGTH(M) ++LENGTH(M)*(2.*TA**2-TA*TB+2.*TB**2)/30. ++LENGTH(M)*(2.*SIA**2-SIA*SIB+2.*SIB**2)/30. 8538 CONTINUE 8945 IF(ITERCHK.NE.1) GO TO 4993 IF(MSUOPTN.EQ.1) GO TO 4993 DO 4992 NN=1,NUMEG IF(NUMEL(NN).EQ.0) GO TO 4992 NAME=NUMEL(NN) DO 5344 K=1,NAME M=L(NN,K) ALFA1=U(M,6) ALFA2-2.*(-3.*U(M,2)-2.*U(M,6)*LENGTH(M)+3.*U(M,8)- +U(M,12)*LENGTH(M))/LENGTH(M) ALFA3=3.*(2.*U(M,2)+U(M,6)*LENGTH(M)-2.*U(M,8)+U(M,12) +*LENGTH(M))/LENGTH(M) BETAl=-U(M,5) BETA2=2.*(-3.*U(M,3)+2.*U(M,5)*LENGTH(M)+3.*U(M,9) ++U(M,ll)*LENGTH(M))/LENGTH(M) BETA3=3.*(2.*U(M,3)-U(M,5)*LENGTH(M)-2.*U(M,9)- +U(M,11)*LENGTH(M))/LENGTH(M) BE(1)=(-U(M,l)+U(M,7))/LENGTH(M)+(ALFA1**2+BETA1**2)/2. BE(2)-ALFA1*ALFA2+BETA1*BETA2 BE(3)=(ALFA2**2+BETA2**2)/2.+ALFA1*ALFA3+BETA1*BETA3 BE(4)-ALFA2*ALFA3+BETA2*BETA3 BE(5)-(ALFA3**2+BETA3**2)/2. DO 5343 I-1,5 5343 BTO(M,I)=BOL(M,I)+BE(I) 5344 ‘CONTINUE 4992 CONTINUE 4993 CONTINUE WRITE(61,8649) 8649 FORMAT(/,10X,'DACTUAL(I)',/) DO 8653 I=1,NEQ 8653 WRITE(61,8654) DACTUAL(I) | 8654 FORMAT(lOX,E21.15) WRITE(61,399) PACTUAL(LODPON1),DACTUAL(LODPON1), + DETRMNT,NUMITER IF(DETRMNT.LE.O..AND.DETOPTN.EQ.1) GO TO 900 IF(ABS(PACTUAL(LODPON1)).GE.ABS(PTOT(LNODE1,LDOFl))) CO TO 900 D0 5007 I=1,NEQ PSAVE(I)=PACTUAL(I) DTOT(I)=0.0 5007 CONTINUE IF(JUSTK.EQ.1) GO TO 2118 D0 5281 NN-1,NUMEG IF(NUMEL(NN).EQ.O) GO TO 5281 NAME=NUMEL(NN) DO 5342 K-1,NAME M=L(NN,K) 5342 5281 2118 451 450 2119 9735 9736 9737 6977 3337 801 000 299 4888 4666 601 150 IF(MSUOPTN.EQ.1) DO 5342 I-1,5 IF(MSUOPTN.EQ.2) CONTINUE CONTINUE CONTINUE J=0 DO 450 N=1,NUMNP DO 451 I-1,6 IF(IA(N,I).EQ.0) GO TO 451 J=J+1 R(J)=R(J)+PINC(N,I) CONTINUE CONTINUE DO 2119 I-1,NEQ IF(IFIX.EQ.0) IF(IFIX.EQ.1) CONTINUE IF(PRIOPTN.EQ.O) GO TO 6977 WRITE(61,9735) FORMAT(///,10X,'R(I)',/) DO 9736 IMM=1,NEQ WRITE(61,9737) R(IMM) FORMAT(//,10X,E21.15,/) CONTINUE ICHECK=3 GO TO 1001 CONTINUE A7OLD(M)=A7TOT(M) BOL(M,I)=BTO(M,I) PSTART(I)=R(I) PSTART(I)=R(I)+PSAVE(I) IF(NCOND.EQ.O) GO TO 801 CALL STCONDN CONTINUE SOLVE SYSTEM OF LINEAR EQUATIONS S*D=R IIIDIIIIIIIIIIIIIIII 7.67.67.67.67167"££i££i£iiiixxxxwxxxwxxK7CXn7sKAAAAA IF(PROTYPE.NE.1) GO TO 601 SCALE=1.0E+05 J=O DO 4666 N=1,NUMNP D0 4888 1=1,6 IF(IA(N,I).EQ.0) GO TO 4888 J-J+1 R(J)=R(J)+PINC(N,I) CONTINUE CONTINUE IDET=1 CALL LINSOLN IF(PROTYPE.EQ.2) GO TO 1778 IF(R(LODPON1).EQ.PINT(LNODE1,LDOF1)) CALL IDENT I I I I I I I I I I I I I I I I l I .c A 7:7:vcztvcvcxvcvcwvc*vhk‘kxxx'xx 0000000 000\l0 1555 778 715 1778 000 000000 180 802 190 151 IF(R(LODPON1).EQ.PINT(LNODE1,LDOF1)) CALL INVTRNS ****************************************************************** IN CASE WHICH WE WANT THE END FORCES DUE TO THE LINEAR SOLUTION SUBROUTINE ENDFORC MAY BE CALLED AT THIS STAGE(THE FIRST ITERATION OF THE FIRST LOAD INCREMENT) *A********A*A********AAAAAAAA*AAAAAAAAAAAAAAAAARAAAAAAAAAAAAAAAAAA IF(R(LODPON1).EQ.PINT(LNODE1,LDOF1).AND.ISTRESS E0 1) CALL STRESS IF(PROTYPE.NE.1) GO TO 1778 DETER:DET1(SCALE) WRITE(61,399) PINT(LNODE1,LDOF1),D(LODPON1),DETER,NUMITER GO TO 709 NUMITER=O RECOVER INTERNAL D.O.F."S OF STRUCTURE AA*AAAAAAAAAAAAAAARAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA DO 1555 I=1,NEQ DTEMP(I)=D(I) IF(PROTYPE.NE.1) GO TO 715 NUMITER=NUM1TER+1 CONTINUE NN=MAXITER+1 IF(NUMITER.EQ.NN) GO TO 9999 IF(NCOND.NE.0) CALL RECOVER IDENTIFY DISPLACEMENTS FOUND FROM SOLUTION OF S*D=R AND FROM THE RECOVERY PROCESS ................. *iv‘civ‘civ‘w’viiiififviwb‘ciifiiiv’w‘w’w‘w‘cfiiv‘n‘w‘ckv‘n‘ciixaAAan). nnnnn \ nnnnnnn In. xxxxvnnnsxx CALL IDENT TO HAVE NODAL DEGREES OF FREEDOM IN LOCAL COORDINATES IIIIIIIIIIIIIIIIIIIIIIIIIIII CALL INVTRNS DO 180 I=1,NSIZE D0 180 J=1,MBAND S(I,J)-0.0 N=1 IPAR-B CALL CURVED(N,IDATA,IARCH) IF(ICAL3.EQ.0) CALL STCONDN IF(NCOND.EQ.O) GO TO 802 CALL STCONDN CONTINUE IF(N20PTIN.EQ.O) CO TO 4991 D0 190 I=1,NSIZE DO 190 J-1,MBAND S(I,J)=0.0 IPAR=4 CALL CURVED(N,IDATA,IARCH) 152 IF(ICAL3.EQ.0) CALL STCONDN 4991 CONTINUE IF(NCOND.EQ.0) GO TO 899 CALL STCONDN 899 CONTINUE IF(PROTYPE.EQ.1) GO TO 222 IF(PROTYPE.EQ.2.AND.EIGVALU.EQ.l) CALL EIGENVL(EIGEN,IDATA) IF(EIGVALU.EQ.2) GO TO 444 GO TO 900 222 CONTINUE DO 107 I=1,NEQ DO 108 J=1,MBAND READ(4,10) RK READ(5,10) RN1 IF(N20PTIN.EQ.1) READ(16,10) RN2 IF(N20PTIN.EQ.0) S(I,J)=RK+.5*RN1 IF(N20PTIN.EQ.1) S(I,J)=RK+.5*RN1+RN2/3. IF(N20PTIN.EQ.O) SP(I,J)=RK+RN1 IF(N20PTIN.EQ.1) SP(I,J)=RK+RN1+RN2 108 CONTINUE 107 CONTINUE REWIND 4 REWIND 5 IF(N20PTIN.EQ.1) REWIND 16 IF(NUMITER.EQ.1) GO TO 701 GO TO 702 9999 J=O D0 16 N=1,NUMNP DO 17 1=1,6 IF(IA(N,I).EQ.0) GO TO 17 J=J+1 R(J)=R(J)+0.5*PINC(N,I) 17 CONTINUE 16 CONTINUE DO 333 I=1,NEQ 333 D(I)=DTEMP(I) WRITE(61,1899) PINT(LNONE1,LDOF1),PINC(LNODE1,LDOF1) 1899 FORMAT(ISX,23HLOADINCREMENT IS HALVED/ +15X,6HPLOAD=,F10.5/15X,5HPINC=,FIO.5) GO TO 1777 701 UOLD-D(LODPON1) GO To 778 702 IF(ABS((UOLD-D(LODPON1))/D(LODPON1)).LE.TOLER) GO TO 708 UOLD=D(LODPON1) GO TO 778 708 D0 2555 I=1,NEQ 2555 DTEMP(I)=D(I) IDET=3 WRITE(61,1399) PINT(LNODE1,LDOF1),D(LODPON1),NUMITER 1399 FORMAT(//,10X,5HLOAD=,F10.5/10X,7HDEFLEC=,F15.10/ +10X,11HITERATIONS=,15) DETER=DET1 p C)C)C)C)C)C)C)D‘0 900 C 153 WRITE(61,399) PINT(LNODE1,LDOF1),D(LODPON1),DETER,NUMITER IF(DETER.LE.O..AND.DETOPTN.EQ.1) GO TO 900 J=O DO 161 N=1,NUMNP DO 171 1=1,6 IF(IA(N,I).EQ.O) GO To 171 J=J+1 R(J)=R(J)+PINC(N,I) CONTINUE CONTINUE IF(ABS(PINT(LNODE1,LDOFl)).GT.ABS(PTOT(LNODE1,LDOFl))) GO TO 900 GO TO 299 CONTINUE TO HAVE EIGENVALUE SOLUTION USING DETERMINANT SEARCH METHOD INTHE CASE OF IEIGEN=1 (N1) STIFFNESS MATRIX WOULD BE CONSIDERED IN SUBROUTINE NLEIGNP FOR NONLINEAR EIGENVALUE PROBLEM. FOR IEIGEN=2 (N1+K) WOULD BE CONSIDERED IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII XXX“KWKKKWWKKWXWWXXWKKKKXWWKKKKXWKKKAKWWIVAAAAAAA1C AAAAAAAA W7CXWWARKICKK IEIGEN-l SCALE=1.0E+05 CALL NLEIGNP(SCALE) CONTINUE CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC 10 FORMAT(E21.6) 399 FORMAT(///10X,5HLOAD=,F15.9/10X,'D(LODPON1)=',F15.10/ + 10X,12HDETERMINANT=,E25.15/10X,11HITERATIONS=,15/) 699 FORMAT(6F10.6,15) 1010 FORMAT(A10,A10,A10,A10,A10,A10,A10,AlO) 1015 FORMAT(llIS) 1020 FORMAT(215) 1030 FORMAT(315) 2010 FORMAT(////7X,6HNE =,I3//7X,6HNUMNP=,I3//7X, + 6HNUMEG=,13//7X,6HIDATA=,I3//7X,6HICAL1=,I3//7X,6HICAL2=, + I3//7X,6HICAL3=,I3//7X,6HICAL4=,I3//7X,6HICAL5=,I3// + 7X,6HICAL6=,I3//7X,6HICAL7=,I3) 2015 FORMAT('l',15H INITIAL LOADS//7H NODE,27X,14HLOAD DIRECTION// + 7H NUMBER,9X,1HU,9X,1HV,9X,1HW,9X,1HB,8X,2HTY,8X,2HTX//) 2020 FORMAT('l',A10,A10,A10,A10,A10,A10,A10,A10) 2030 FORMAT(////7X,8HIARCH =,I3//7X,8HILOAD =,I3//7X,8HIDIRCN =,I3) C END C SUBROUTINE NODDATA (IARCH,DX) C C ***************************************k*****k******************** C TO READ AND PRINT NODAL POINT DATA C T0 CALCULATE EQUATION NUMBERS AND CONDENSATION NUMBERS AND C STORE THEM IN ARRAYS -IA- AND -IB- RESPECTIVELY 154 C ******7h???**************>‘<***7‘<7’<********‘k*‘k****‘kv'cv':*7P7k‘k7'n'r‘k‘k‘k‘k7'v‘k7'n'c‘k‘k7k‘k7'c C REAL L COMMON/C1/NE,NUMNP,D1(15) COMMON/C2/NSIZE,NEQ,NC0ND,MBAND,IEICEN COMMON/C3/IA(37,8),IB(37,8),X(37),Y(37),Z(37),R,A COMMON/Cl9/TT(36) C C ........ READ NODAL POINT DATA C WRITE(61,2000) WRITE(61,2010) WRITE(61,2015) IF (IARCH.EQ.O) GO TO 101 IF (IARCH.EQ.2) GO TO 104 READ(60,1010) R 100 READ(60,1000) N,(IA(N,I),I=1,6),(IB(N,I),I=1,6),T,Z(N) PI=4.*ATAN(1.) T=T*PI/180. X(N)=SIN(T)*R Y(N)=R*(l.-COS(T)) WRITE(61,2020) N,(IA(N,I),I=1,6),(1B(N,I),I=1,6),X(N),Y(N),Z(N) IF (N.NE.NUMNP) GO TO 100 . GO TO 103 101 CONTINUE READ(60,1020) H,L,DI,XS A=H/(XS*XS) DX=L/DI DL=0. 102 READ(60,1000) N,(IA(N,I),I=1,6),(IB(N,I),I=1,6),T,Z(N) X(N)=XS+DL Y(N)=A*X(N)*X(N) WRITE(61,2020) N,(IA(N,I),I=1,6),(IB(N,I),I=1,6),X(N),Y(N),Z(N) DL=DL+DX IF (N.NE.NUMNP) GO TO 102 GO TO 103 104 CONTINUE READ(60,1021) H,L 106 READ(60,1001) N,(IA(N,I),I=1,6),(IB(N,I),I=1,6),TT(N),X(N),Y(N) + ,Z(N) WRITE(61,2021)N,(IA(N,I),I=1,6),(IB(N,I),I=1,6),TT(N),X(N),Y(N) + ,Z(N) IF(N.NE.NUMNP) GO TO 106 g ........ PROCESS ARRAYS -IA- AND -IB- TO FIND EQUATION NUMBERS AND C CONDENSATION NUMBERS. STORE NEQ"S AND NCOND"S IN ARRAYS IA AND C IB RESPECTIVELX. C 103 NEQ=0 NCOND=0 DO 125 N=1,NUMNP DO 120 1=1,6 155 IF(IA(N,I).NE.1) GO TO 105 IA(N,I)=0 GO TO 120 105 IA(N,I)=-1 IF(IB(N,I)) 110,115,120 110 NCOND=NCOND+1 IB(N,I)=-NCOND GO TO 120 115 NEQ=NEQ+1 IA(N,I)=NEQ 120 CONTINUE 125 CONTINUE NSIZE=NEQ+NCOND C C ........ WRITE GENERATED NODAL POINT DATA C WRITE(61,2030) WRITE(61,2040) WRITE(61,2050) (N,(IA(N,I),I=1,6),(IB(N,I),I=1,6),N=1,NUMNP) WRITE(61,2060) NSIZE,NEQ,NCOND RETURN C 1000 FORMAT(IS,1213,F15.10,F10.6) 1001 FORMAT(IS,12I3,F9.6,3F10.6) 1010 FORMAT(F15.9) 1020 FORMAT(4F10.5) 1021 FORMAT(2F10.5) 2000 FORMAT('I',33H N O D A L P O I N T D A T A //) 2010 FORMAT(IBH INPUT NODAL DATA //) 2015 FORMAT(7H NODE,26X,36HNODAL POINT BOUNDARY CONDITION CODES,33X, + 23HNODAL POINT COORDINATES/7H NUMBER,21X,7HIA(N,I),33X, + 7HIB(N,I)/11X,2(4X,1HU,4X,1HV,4X,1HW,4X,1HB,3X,2HTY,3X, + 2HTX,10X),9X,4HX(N),8X,4HY(N),8X,4HZ(N)) 2020 FORMAT(I5,6X,12I5,4X,3F12.3) 2021 FORMAT(15,6X,1215,4X,F9.6,3F12.3) 2030 FORMAT(///22H GENERATED NODAL DATA //) 2040 FORMAT(7H NODE,16X,16HEQUATION NUMBERS,22X, + ZOHCONDENSATION NUMBERS/7H NUMBER,21X,7HIA(N,I),33X, + 7HIB(N,I)/11X,2(4X,1HU,4X,1HV,4X,1HW,4X,lHB,3X,2HTY,3X, + 2HTX,10X)) 2050 FORMAT(I5,6X,1215) 2060 FORMAT('-',6HNSIZE=,I3,3X,4HNEQ=,I3,3X,6HNCOND=,I3) C END C SUBROUTINE LOAD (IARCH,ILOAD,IDIRCN,DX,WW) C C **%%%*%*%%%%%%%%%%2171713171717??? 7'; :'w'n'w'cw'n'n'w'w'c A A???» A 7's 7'. A A 717's A7':‘kv’:*7’¢k7’<7’€*7’<*7‘€‘k***** c To READ AND STORE INITIAL LOAD DATA """""""""""""""" C *AAiAiAAAAAAAAAAAAAAAAAAAARA£4888kikkkikfii" A. A "~“AAAWWWACAAC C REAL LENGTH 00 000 100 200 300 00 0000 350 400 1023 1020 2020 2030 2040 O C C I O O o . COMMON/Cl/NE,NUMNP,Dl(15) COMMON/C3/IA(37,8),IB(37,8),X(37),Y(37),Z(37),RAD,AC COMMON/C5/E(3),G(3),NODEI(36),NODEJ(36),D5(180) COMMON/C7/RI(36),RJ(36),PHII(36),PHIJ(36),TETA(36),LENGTH(36), + RIA(36),RJA(36) COMMON/C8/PN(37,8),R(296),PINT(37,8) ....CHECK TYPES OF LOADS TO BE READ ILOAD.EQ.0 , LOAD IS UNIFORMLY DISTRIBUTED ILOAD.EQ.1 , LOADS ARE CONCENTRATED IF (ILOAD.EQ.0) GO TO 200 READ(60,1023) MN,(PN(MN,I),I=1,6) WRITE(61,2020) MN,(PN(MN,I),I=1,6) IF(MN NE.NUMNP) GO TO 100 RETURN CONTINUE READ(60,1020) ww IF (IDIRCN.EQ.0) WRITE(61,2030) WW IF (IDIRCN.EQ.1) WRITE(61,2040) WW DO 300 NM=1,NUMNP DO 300 I=1,6 PN(NM,I)=O. DO 400 M=l,NE NI=NODEI(M) NJ=NODEJ(M) ....CHECK IF DISTRIBUTED LOAD IS VERTICAL OR HORIZONTAL THEN CONCENTRATE IT AT THE NODES IN COMPONENTS IDIRCN.EQ.0 , VERTICAL IDIRCN.EQ.1 , HORIZONTAL IF (IDIRCN.EQ.1) GO TO 350 IF (IARCH.EQ.1) DX=ABS(X(NJ)-X(NI)) PN(NI,1)=PN(NI,l)+DX/2.*W*COS(PHII(M)) PN(NI,3)=PN(NI,3)+DX/2.*W*SIN(PHII(M)) PN(NJ,l)=PN(NJ,l)+DX/2.*W*COS(PHIJ(M)) PN(NJ,3)=PN(NJ,3)+DX/2.*W*SIN(PHIJ(M)) GO TO 400 CONTINUE PN(NI,2)=PN(NI,2)+W*LENGTH(M)/2. PN(NJ,2)=PN(NJ,2)+W*LENGTH(M)/2. CONTINUE WRITE(61,2020) (N,(PN(N,I),I=1,6),N=1,NUMNP) RETURN FORMAT(15,6F7.1) FORMAT(F10.5) FORMAT(IS,6X,6F10.3) . FORMAT('-',38HUNIFORMLY DISTRIBUTED VERTICAL LOAD W=,F10.5//) FORMAT('-',40HUNIFORMLY DISTRIBUTED HORIZONTAL LOAD W=,F10.5//) END 00000 200 0 C)C)C)C)C)C)CIO 700 800 900 2000 158 SUBROUTINE ELEMENT (N,IDATA,IARCH) AAAAAAAA*****************A*A*A*A***AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA TO CALL THE APPROPRIATE ELEMENT SUBROUTINE AAAAAAAAAAAAAA*AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA COMMON/Cl/NE,NUMNP,NUMEG,NTYPE(3),D12(11) IF (NTYPE(N).GT.1) GO TO 200 CALL CURVED (N,IDATA,IARCH) RETURN RETURN END SUBROUTINE BAND kk**xxk"*xx**k*x**x*x*x**xxx*xxx***xxxxxxxannkkxAkklAAxAAwwfinxxxnk TO COMPUTE SEMIBANDWIDTH OF STRUCTURE STIFFNESS MATRIX DONE BY FINDING THE MAXIMUMN DIFFERENCE BETWEEN THE EQUATION NUMBERS ASSOTIATED WITH THE NODES OF A PARTICULAR ELEMENT xA*******AAAA**A***A*A**A*A*AAAARAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA COMMON/Cl/NE,NUMNP,Dl(15) COMMON/Cz/NSIZE,NEQ,NC0ND,MBAND,IEICEN COMMON/C3/IA(37,8),D3(409) COMMON/C5/E(3),G(3),NODEI(36),NODEJ(36),D5(180) MBAND=0 DO 900 M=1,NE NI=NODEI(M) NJaNODEJ(M) D0 800 1=1,6 IF (IA(NI,1).LE.0) GO TO 800 N1=IA(NI,I) D0 700 J=l,6 IF (IA(NJ,J).LE.0) CO T0 700 N2-IA(NJ,J) MB=N2—N1 IF (MB.LT.0) MB=-MB+1 IF (MB.GT.0) MB=MB+1 IF (MB . GT . MBAND) MBAND=MB CONTINUE CONTINUE CONTINUE WRITE(61,2000) MBAND RETURN FORMAT(’l’,20HSEMIBANDWIDTH MBAND=,I3) END 0000 C C C.... C 100 C C.... C 105 C C.... C + + + 159 SUBROUTINE CURVED (N,IDATA,IARCH) AAAAAAA*AAAAAAAAAAAAAAAAAA*AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA CURVED ELEMENT SUBROUTINE A********A*AAAAAAAAARAA*AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA REAL IXX,IYY,KT,II,JJ,LENGTH COMMON/Cl/NE,NUMNP,NUMEG,NTYPE(3),NUMEL(3),IPAR,ICAL1,ICAL2, ICAL3,ICAL4,ICAL5,ICAL6,ICAL7 COMMON/Cz/NSIZE,NEQ,NCOND,MBAND,IEICEN COMMON/C4/SE(16,16) COMMON/C5/E(3),G(3),NODEI(36),NODEJ(36),A(36),IXX(36),IYY(36), KT(36),L(1,36) COMMON/C6/A1,A2,MP,B1(36),B2(36),B3(36) COMMON/C7/RI(36),RJ(36),PHII(36),PHIJ(36),TETA(36),LENGTH(36), RIA(36),RJA(36) COMMON/C8/PN(37,8),R(296),PINT(37,8) COMMON/C9/S(296,16),SP(296,16),IDET COMMON/ClO/D(296),DlO(ll84),RC(296),SC(296,16) COMMON/C11/DN(16),U(36,12),W(37,8),V(37,8) COMMON/C12/ULOC(36,12),RCOL(9),MSUOPTN,N1GOPTIN COMMON/Cl9/TT(36) GO TO (100,200,300,400),IPAR ....READ MATERIAL INFORMATION WRITE(61,2000) NTYPE(N) READ(60,1010) E(N),G(N),DM WRITE(61,2020) NUMEL(N),E(N),G(N),DM ....READ ELEMENT AND CROSS SECTION INFORMATION WRITE(61,2021) K=0 READ(60,1020) M,NODEI(M),NODEJ(M),A(M),IXX(M),IYY(M),KT(M) WRITE(61,2022) M,A(M),IXX(M),IYY(M),KT(M) K=K+1 L(N,K)=M IF(K.NE.NUMEL(N)) GO TO 105 ....READ AND CALCULATE ELEMENT GEOMETRIC PROPERTIES IF (ICAL2.EQ.0) WRITE(61,2025) DO 110 KK=1,K M=L(N,KK) CALL GEOMTRY (M,IARCH) CONTINUE RETURN CONTINUE 0 000 C.... 215 220 C C.... C C 230 300 C C.... C C 160 ....CALCULATE LINEAR STIFFNESS MATRIX OF EACH ELEMENT AND STORE IN ARRAY SE(M,I,J). READ LIMITS OF INTEGRATION AND THE MP-POINTS OF INTEGRATION TO BE USED IN THE GAUSS-LEGENDRE QUADRATURE IF (IDATA.EQ.0) READ(60,1030) A1,A2,MP IF (ICAL1.EQ.0) WRITE(61,2030) A1,A2,MP ....OBTAIN LINEAR STIFFNESS FOR EACH ELEMENT. INTEGRATE NUMERICALLY INUMEL=NUMEL(N) D0 220 K-l,INUMEL M=L(N,K) CALL NUMINT (N,M) IF(ICAL1.NE.O) GO TO 215 WRITE(61,2032) M WRITE(61,2034) ((SE(I,J),J=1,6),I=1,6) WRITE(61,2036) WRITE(61,2034) ((SE(I,J),J=7,12),I=1,6) WRITE(61,2038) WRITE(61,2034) ((SE(I,J),J=1,6),I=7,12) WRITE(61,2040) WRITE(61,2034) ((SE(I,J),J=7,12),I=7,12) WRITE(1,10) ((SE(I,J),J=1,12),I=1,12) CONTINUE REWIND 1 ....ASSEMBLE LINEAR STIFFNESS OF EACH ELEMENT INTO LINEAR STIFFNESS OF STRUCTURE DO 230 K=l,INUMEL M=L(N,K) READ(1,10) ((SE(I,J),J=1,12),I=1,12) CALL ASEMBLE (M) CONTINUE REWIND 1 REWIND 9 REWIND 8 WRITE(4,10) ((S(I,J),J=1,MBAND),I=1,NSIZE) REWIND 4 RETURN CONTINUE ....OBTAIN NONLINEAR STIFFNESS -SE1- FOR EACH ELEMENT. INTEGRATE NUMERICALLY. DO 320 K=1,INUMEL M=L(N,K) NI=NODEI(M) NJ=NODEJ(M) DO 305 ID=1,6 DN(ID)=U(NI,ID) 305 315 320 00 000 330 400 C C.... C C 405 415 161 DN(ID+6)=U(NJ,ID) CALL NUMINT (N,M) IF(ICAL1.NE.0) GO TO 315 WRITE(61,2050) M WRITE(61,2034) ((SE(I,J),J=1,6),I=1,6) WRITE(61,2036) WRITE(61,2034) ((SE(I,J),J=7,12),I=1,6) WRITE(61,2038) WRITE(61,2034) ((SE(I,J),J=1,6),I=7,12) WRITE(61,2040) WRITE(61,2034) ((SE(I,J),J=7,12),I=7,12) WRITE(2,10) ((SE(I,J),J=1,12),I=l,12) CONTINUE REWIND 2 ....ASSEMBLE NONLINEAR STIFFNESS SE1 OF EACH ELEMENT INTO NONLINEAR STIFFNESS OF STRUCTURE, Sl. STORE MATRIX 81 DO 330 K=1,INUMEL M=L(N,K) READ(2,10) ((SE(I,J),J=1,12),I=1,12) CALL ASEMBLE (M) CONTINUE REWIND 2 WRITE(5,10) ((S(I,J),J=1,MBAND),I=1,NSIZE) REWIND 5 RETURN CONTINUE ....OBTAIN NONLINEAR STIFFNESS -SE2- FOR EACH ELEMENT. INTEGRATE NUMERICALLY. DO 420 K=l,INUMEL M=L(N,K) NI=NODEI(M) NJ=NODEJ(M) D0 405 ID=1,6 DN(ID)=U(NI,ID) DN(ID+6)=U(NJ,ID) CALL NUMINT (N,M) IF(ICAL1.NE.O) GO TO 415 WRITE(61,2055) M WRITE(61,2034) ((SE(I,J),J=1,6),I=1,6) WRITE(61,2036) WRITE(61,2034) ((SE(I,J),J=7,12),I=1,6) WRITE(61,2038) WRITE(61,2034) ((SE(I,J),J=1,6),I=7,12) WRITE(61,2040) WRITE(61,2034) ((SE(I,J),J=7,12),I=7,12) WRITE(3,10) ((SE(I,J),J=1,12),I=l,12) 420 0 000 430 10 1010 1020 1030 2000 2020 2021 2022 2025 2030 2032 2034 2036 2038 2040 2050 2055 0 00000 + + + + + 162 CONTINUE REWIND 3 ....ASSEMBLE NONLINEAR STIFFNESS SE2 OF EACH ELEMENT INTO NONLINEAR STIFFNESS OF STRUCTURE, $2. STORE MATRIX S2 D0 430 K=1,1NUMEL M=L(N,K) READ(3,10) ((SE(I,J),J=1,12),I=l,12) CALL ASEMBLE (M) CONTINUE REWIND 3 WRITE(16,10) ((S(I,J),J=1,MBAND),I=1,NSIZE) REWIND 16 RETURN FORMAT(E21.6) FORMAT(3E10.2) FORMAT(3IS,4E15.6) FORMAT(2F5.2,IS) FORMAT(’l',23HG R O U P N U M B E R ,12//2X,6HNUMBER,6X,7HMODULUS ,11X,5HSHEAR,8X,7HDENSITY/4X,2HOF,11X,2HOF,12X,7HMODULUS/ 1X,8HELEMENTS,4X,lOHELASTICITY) FORMAT(I6,3E17.6) - FORMAT(//8H ELEMENT,9X,4HA(M),10X,6HIXX(M);9X,6HIYY(M),9X,SHKT(M)) FORMAT(I6,5X,4E15.6) FORMAT(//8H ELEMENT,3X,8HNODEI(M),3X,8HNODEJ(M),9X, 19HRADIUS OF CURVATURE,18X,17HNODAL SLOPE ANGLE/8H NUMBER/ 39X,5HRI(M),10X,5HRJ(M),lSX,7HPHII(M),8X,7HPHIJ(M),7X, 7HTETA(M)) FORMAT('l',21HLIMITS OF INTEGRATION,3X,3HA1=,F3.l,3X,3HA2=,F3.1// 26H QUADRATURE FORMULA POINTS,3X,3HMP=,12) FORMAT('l',44HUNCONDENSED LINEAR STIFFNESS (SE) OF ELEMENT,I3/// 9H BLOCK 11) FORMAT(//1X,6F15.6) FORMAT(///9H BLOCK IJ) FORMAT('l'//9H BLOCK JI) FORMAT(///9H BLOCK JJ) FORMAT('I',48HUNCONDENSED NONLINEAR STIFFNESS (SE1) OF ELEMENT,I3 + ///9H BLOCK 11) FORMAT('l',48HUNCONDENSED NONLINEAR STIFFNESS (SE2) OF ELEMENT,I3 ///9H BLOCK II) END SUBROUTINE GEOMTRY (M,IARCH) OOOOOOOOOOOOOOOOOOOOO 7.67%7'Ck7k72ili7'fik(1’717:7':7'C7.V7'€*************** 00000000000000000000000000000000000000000000 163 REAL IXX,IYY,KT,II,JJ,LENGTH COMMON/C1/NE,NUMNP,NUMEG,NTYPE(3),NUMEL(3),IPAR,ICAL1,ICAL2, + ICAL3,ICAL4,ICAL5,ICAL6,ICAL7 COMMON/C3/IA(37,8),IB(37,8),X(37),Y(37),Z(37),R,AC COMMON/CS/E(3),G(3),NODEI(36),NODEJ(36),A(36),IXX(36),IYY(36), + KT(36),L(1,36) COMMON/C6/A1,A2,MP,B1(36),B2(36),B3(36) COMMON/C7/RI(36),RJ(36),PHII(36),PHIJ(36),TETA(36),LENGTH(36), + RIA(36),RJA(36) COMMON/Cl9/TT(36) XI=X(NODEI(M)) YI-Y(NODEI(M)) XJ=X(NODEJ(M)) YJ-Y(NODEJ(M)) C C ........ READ ELEMENT GEOMETRIC PROPERTIES C IF (IARCH.EQ.O) GO TO 100 IF (IARCH.EQ.2) GO TO 190 RI(M)=R RJ(M)=R DY=XI/SQRT(R*R-XI*XI) PHII(M)-ATAN(DY) DY=XJ/SQRT(R*R-XJ*XJ) PHIJ(M)=ATAN(DY) GO TO 200 100 CONTINUE D2Y=2.*AC DY=2.*AC*XI RI(M)=(1.+DY*DY)**1.5/D2Y PHII(M)=ATAN(DY) DY=2.*AC*XJ RJ(M)=(1.+DY*DY)**1.5/D2Y PHIJ(M)=ATAN(DY) GO TO 200 190 CONTINUE PHII(M)-TT(NODEI(M)) PHIJ(M)-TT(NODEJ(M)) 200 CONTINUE XL-XJ-XI YL=YJ-YI TETA(M)=ABS(PHII(M)-PHIJ(M)) =TETA(M) C ........ CALCULATE NODAL LOCAL COORDINATES AFTER ROTATION ZR=XL*COS(PHII(M))+YL*SIN(PHII(M)) XR=-XL*SIN(PHII(M))+YL*COS(PHII(M)) C ........ CHECK DATA GENERATION 164 IF (ICAL2.EQ.0) WRITE(61,2010) M,NODEI(M),NODEJ(M),RI(M),RJ(M), + PHII(M),PHIJ(M),T C C ........ SOLVE SYSTEM OF EQUATIONS IN CLOSED FORM , C OBTAIN VARIABLES Bl, B2, RIA(M), RJA(M), AND LENCTH(M), C FIRST CALCULATE COEFFICIENTS OF THE VARIABLES C AA11-1.-COS(T) AA12-2.*(SIN(T)-T*COS(T)) AA21=SIN(T) AA22=2.*(T*SIN(T)+COS(T)-1.) C C ........ CALCULATE B1,B2,RIA(M),RJA(M),LENGTH(M) C BZ(M)-(AA11*ZR-AA21*XR)/(AA11*AA22-AA12*AA21) B1(M)-XR/AAll-AA12*BZ(M)/AAll LENGTH(M)=Bl(M)*T+B2(M)*T*T RIA(M)=B1(M) RJA(M)=B1(M)+2.*BZ(M)*T 1F (IARCH.EQ.2) RI(M)-RIA(M) IF (IARCH.EQ.2) RJ(M)=RJA(M) C C ........ CHECK DATA GENERATION C IF(ICAL2.EQ.0) WRITE(61,2020) XR,2R,B1(M),82(M),LENCTH(M), + RIA(M),RJA(M) RETURN C 2010 FORMAT(//I6,5X,IS,6X,IS,6X,2F15.6,6X,3F15.6) 2020 FORMAT(/10X,3HXR-,F15.10,3X,3HZR=,F15.lO//10X,6HBl(M)=,E15.9,3X, + 6HB2(M)-,E15.9,3X,10HLENGTH(M)=,F13.6//1OX,11HRI(APPROX)= + ,F15.9,10X,11HRJ(APPROX)=,F15.9//) END SUBROUTINE NUMINT (N,M) DOUBLE PRECISION A01,A02,A3,A4,A5,A6,BTG,BLG *****************************A*************A***A****************** TO INTEGRATE NUMERICALLY THE TERMS OF THE CURVED ELEMENT STIFFNESS MATRICES SE, SE1, SE2, IT USES THE GAUSS-LEGENDRE QUADRATURE FORMULA. THE ROUTINE NUMINT USES THE MP-POINT GAUSS-LEGENDRE QUADRATURE FORMULA TO COMPUTE THE INTEGRAL OF FUNCTN(GM)*DGM BETWEEN INTEGRATION LIMITS A1 AND A2. THE ROOTS OF SEVEN LEGENDRE POLYNOMIALS AND THE WEIGHT FACTORS FOR CORRESPONDING QUADRATURES ARE STORED IN THE Z AND WEIGHT ARRAYS RESPECTIVELY. MP MAY ASSUME VALUES 2, 3, 4, 5, 6, 10, AND 15 ONLY. THE APPROPRIATE VALUES FOR THE MP-POINT FORMULA ARE LOCATED IN ELEMENTS Z(KEY(I))...Z(KEY(I+1)-1) AND WEIGHT(KEY(I))... WEIGHT(KEY(I+1)-1) WHERE THE PROPER VALUE FOR I IS DETERMINED BY FINDING THE SUBSCRIPT OF THE ELEMENT OF THE ARRAY NPOINT WHICH HAS THE VALUE MP. IF AN INVALID VALUE OF MP IS USED, A TRUE ZERO IS RETURNED AS THE VALUE OF GAUSS. 0000000000000000 165 O x. >9 X- >{' ’4’ >1- ,4. )6. X5 X- x. x. X‘ x. 3(- ,4. x. )1. )4. X- X- x. x. >3 )5. )S. 'r Yr 3(- 5‘ 1' >\ r ,. >1 )‘ vvvvvvvvvvvvvvvvvvvvvvvvvv ................ AA A AAAAAAAAA AAA‘A‘A‘W‘IC‘IC‘I AAAAAAAAAAAAAAAAAA REAL IXX,IYY,KT,II,JJ,LENGTH,L1,L2,K,KK,LL,MM,NN,MS DIMENSION NPOINT(7),KEY(8),Z(24),WEIGHT(24),K(16,16) COMMON/Cl/NE,NUMNP,NUMEG,NTYPE(3),NUMEL(3),IPAR,ICAL1,ICAL2,ICAL3, + ICAL4,ICAL5,ICAL6,ICAL7 COMMON/C4/SE(16,16) COMMON/C5/E(3),G(3),NODEI(36),NODEJ(36),A(36),IXX(36),IYY(36), + KT(36),L(1,36) COMMON/C6/A1,A2,MP,B1(36),B2(36),B3(36) COMMON/C7/RI(36),RJ(36),PHII(36),PHIJ(36),TETA(36),LENGTH(36), + RIA(36),RJA(36) COMMON/C11/DN(16),U(36,12),W(37,8),V(37,8) COMMON/C18/IARCH DATA NPOINT/ 2, 3, 4, 5, 6, 10, 15/. DATA KEY/ 1, 2, 4, 6, 9, 12, 17, 25/ DATA 2 / 0.577350269,0.0 ,0 774596669, 1 0.339981044,0.861136312,0 O ,0.538469310, 2 0.906179846,0.238619l86,0.661209387,0.932469514, 3 0.148874339,0.433395394,0 679409568,0.865063367, 4 0.973906529,0.0 ,0.201194094,0.394151347, 5 0.570972173,0.724417731,0.848206583,0.937273392, 6 0 987992518 / DATA WEIGHT / 1.0 ,0.888888889,0.555555556, 1 0.652145155,0.347854845,0.568888889,0.478628671, 2 0.236926885,0.467913935,0.360761573,0.171324493, 3 0.295524225,0.269266719,0.219086363,0.149451349, 4 0.066671344,0.202578242,0.198431485,0.186161000, 5 0.166269206,0.139570678,0.107159221,0.070366047, 6 0.030753242 / C T=TETA(M) R1=RI(M) R2=RJ(M) L1=R1*T L2=R2*T c C ........ FIND SUBSCRIPT OF FIRST 2 AND WEIGHT VALUE C DO 100 I=1,7 IF(MP.EQ.NPOINT(I)) GO TO 200 100 CONTINUE C C ........ INVALID MP USED C GAUSS=0.0 WRITE(61,2000) GAUSS RETURN C . C ........ SET UP INITIAL PARAMETERS C 200 JFIRST=KEY(I) 166 JLAST—KEY(I+l)-l C-(A2-A1)/2. D=(A2+Al)/2. C C ........ ACCUMULATE THE SUM IN THE MP-POINT FORMULA C CCCC IF (IPAR.GE.3) GO TO 543 D0 249 I-1,16 DO 249 J-l,16 249 K(I,J)-0.0 GO TO 248 543 CONTINUE CCCC DO 250 I-1,12 DO 250 J-1,12 250 K(I,J)-0.0 248 CONTINUE IF(IPAR.CE.3) GO TO 390 D0 500 J-JFIRST,JLAST 1=0 IF (Z(J).EQ.0.) GO TO 350 300 1=1+1 IF (I.EQ.1) GM-Z(J)*C+D IF (I.EQ.2) GM--Z(J)*C+D GO TO 360 350 GM=D 360 AA-6.*GM**2-6.*GM BB-3.*GM**2-4.*GM+1. CC-3.*GM**2-2.*GM DD-12.*GM-6. EE=6.*GM—4. FF=6.*GM-2. GG=2.*GM**3-3.*GM**2+1. HH=GM**3-2.*GM**2+GM II=-2.*GM**3+3.*GM**2 JJ=GM**3-GM**2 KK=l.-GM R-B1(M)+2.*BZ(M)*T*GM GMSS-(-1./(R**3*T))*(2.*B2(M)) GMSG=R*T*GMSS 0 ........ CHECK WHICH PART OF THE STIFFNESS MATRIX IS BEING COMPUTED IPAR-2, COMPUTE ARRAY SE IPAR-3, COMPUTE ARRAY SE1 IPAR-4, COMPUTE ARRAY SE2 0 000 (ACACDrICICACI 167 ........ INTEGRANDS OF CURVED ELEMENT LINEAR STIFFNESS (SYMMETRIC) Cl--E(N)*A(M)*GG/R C2-(E(N)*IYY(M)/(R**3*T**3))*(DD+AA*GMSS*R**2*T**2) C3'(E(N)*IXX(M)*T/R**3)*(-DD/T**2-GMSS*R**2*AA) C4-G(N)*KT(M)*AA/(R**3*T) CS-(E(N)*A(M)/(R*T))*(AA+T**2*HH) C6-(E(N)*IYY(M)/(R**3*T**3))*(-T*EE-GMSS*R**2*T**3*BB+T*AA+ + GMSG*R*T**2*GG) ' C7-E(N)*IXX(M)*T*GG/R**2 C8-G(N)*KT(M)*AA/(R**2*T) C9--E(N)*A(M)*Ll*HH/R C10-(E(N)*IYY(M)/(R**3*T**3))*(L1*EE+GMSS*R**2*T**2*L1*BB) Cll-(E(N)*IXX(M)*T/R**3)*(Ll*EE/T**2+GMSS*R**2*L1*BB) C12--G(N)*KT(M)*L1*BB/(R**3*T) C13—E(N)*A(M)*R1*BB/(R*T) C14-(E(N)*IYY(M)/(R**3*T**3))*(T*R1*BB+GMSG*R*T**2*R1*HH) C15-E(N)*IXX(M)*T*L1*HH/R**2 Cl6-G(N)*KT(M)*L1*BB/(R**2*T) Cl7=-E(N)*A(M)*II/R C18=(E(N)*IYY(M)/(R**3*T**3))*(-DD-GMSS*R**2*T**2*AA) C19=-C3 C20=-C4 C21=(E(N)*A(M)/(R*T))*(-AA+T**2*JJ) C22=(E(N)*IYY(M)/(R**3*T**3))*(-T*FF-GMSS*R**2*T**3*CC-T*AA+ + GMSG*R*T**2*II) C23-E(N)*IXX(M)*T*II/R**2 CZ4=-C8 C25=(-E(N)*A(M)*L2*JJ)/R C26-(E(N)*IYY(M)/(R**3*T**3))*(L2*FF+GMSS*R**2*T**2*L2*CC) 027=(E(N)*IXX(M)*T/R**3)*(L2*FF/T**2+GMSS*R**2*L2*CC) C28--G(N)*KT(M)*L2*CC/(R**3*T) C29-E(N)*A(M)*R2*CC/(R*T) C30=(E(N)*IYY(M)/(R**3*T**3))*(T*R2*CC+GMSG*R*T**2*R2*JJ) C31=E(N)*IXX(M)*T*L2*JJ/R**2 C32=G(N)*KT(M)*L2*CC/(R**2*T) SE(l,1)-C1*T*(-CG)+C2*(DD+AA*GMSS*R**2*T**2) SE(l,2)-O.0 SE(1,4)-0.0 SE(l,6)-0.0 SE(1,8)=0.0 SE(1,lO)-0.0 SE(1,12)-0.0 SE(1,14)-0.0 SE(1,16)=0.0 ' " . .' . . SE(1,3)=CI*(AA+T**2*HH)+C2*(-T*EE-CMSSxRxwszww3wBB+TxAA+ 168 GMSG*R*T**2*GG) SE(l,5)-C1*T*(-L1)*HH+CZ*L1*(EE+BB*GMSS*R**2*T**2) SE(l,7)=C1*R1*BB+CZ*(T*R1*BB+GMSG*R*T**2*R1*HH) SE(1,9)=C1*(-T*II)-C2*(DD+AA*GMSS*R**2*T**2) SE(l,11)=C1*(-AA+T**2*JJ)+C2*(-T*FF-GMSS*R**2*T**3*CC-T*AA+ GMSG*R*T**2*II) SE(l.13)=Cl*(-T*L2*JJ)+C2*(L2*FF+GMSS*R**2*T**2*L2Acc) SE(l,15)-Cl*(R2*CC)+C2*(T*R2*CC+GMSG*R*T**2*JJ) SE(2,2)=C3*(—DD/T**2-GMSS*R**2*AA)+C4*AA SE(2,3)-0.0 SE(2,5)=0.0 SE(2,7)-0.0 SE(2,9)=0.0 SE(2,11)=0.0 SE(2,13)-0.0 SE(2,15)=0.0 SE(2,4)=C3*R*GG+C4*R*AA SE(2,6)=C3*(L1*EE/T**2+GMSS*R**2*L1*BB)-C4*L1*BB SE(2,8)=C3*R*L1*HH+C4*R*L1*BB SE(2,10)=C3*(DD/T**2+GMSS*R**2*AA)-C4*AA SE(2,12)-C3*R*II-C4*R*AA SE(2,14)-C3*(L2*FF/T**2+GMSS*R**2*L2*CC)-C4*L2*CC SE(2,l6)=C3*R*L2*JJ+C4*R*L2*CC SE(3’3>=C5*(AA+T**2*HH)+C6*(-T*EE-GMSS*R**2*T**3ABB+T*AA+ GMSG*R*T**2*GG) SE(3,4)=O.O SE(3,6)=0.0 SE(3,8)=0.0 SE(3,10)=0.0 SE(3,12)-0.0 SE(3,14)=0.0 SE(3,16)=0.0 SE(3:5)=C5*(-T*LI*HH)+C6*(L1*EE+GMS3*R**2*T**2*L1*BB) SE(3,7)=CS*R1*BB+C6*(TARIRBB+CMSCARATAA2AR1AHH) SE(3.9)=C5*(-T*II)+C6*(-DD-CMSSARAA2ATAA2AAA) SE(3’11)'CS*('AA+T**2*JJ)+C6*(-T*FF-GMSS*R**2*T**3*CC_T*AA+ GMSG*R*T**2*II) SE(3’13)=CS*(’T*L2*JJ)+C6*(L2*FF+CMSS*R**2*T**2*L2*CC) SE(3,15)=C5*R2*CC+C6*(T*R2*CC+GMSG*R*T*A2AJJ) SE(4.4)=C7*R*GG+C8*R*AA SE(4,5)=0.0 SE(4,7)-0.0 SE(4,9)-0.0 SE(4,11)=0.0 SE(4,13)-0.0 SE(4,15)=0.0 35(4.5)=C7*(L1*EE/T**2+GMSS*R**2*LlABB)+C3A(_LIABB) SE(4.8)=C7*R*L1*HH+C8*R*L1*BB SE(4,10)=C7*(DD/T**2+GMSS*R**2*AA)+C8A(-AA) 169 SE(4,12)=C7*R*II+C8*(-R*AA) SE(4,14)-C7*(L2*FF/T**2+GMSS*R**2*L2*CC)+C8*(-L2*CC) SE(4,16)=C7*R*L2*JJ+C8*R*L2*CC SE(5.5)=C9*(-T*L1*HH)+C10*(L1*EE+GMSS*R**2*T**2*LlABB) SE(5,6)-0.0 SE(5,8)—0.0 SE(5,10)=0.0 SE(5,12)-0.0 SE(5,14)=0.0 SE(5,16)-0.0 SE(S,7)=C9*R1*BB+C10*(T*R1*BB+GMSG*R*T**2*R1*HH) SE(S,9)=C9*(-T*II)+C10*(-DD-GMSS*R**2*T**2*AA) SE(S,11)-C9*(-AA+T**2*JJ)+ClO*(-T*FF-GMSS*R**2*T**3*CC-T*AA+ GMSG*R*T**2*II) SE(S,13)=C9*(-T*L2*JJ)+ClO*(L2*FF+GMSS*R**2*T**2*L2*CC) SE(S,15)-C9*R2*CC+C10*(T*R2*CC+GMSG*R*T**2*R2*JJ) SE(6,6)=C11*(L1*EE/TAA2+GMSSRRAAQALlABB)+012A<,LlABB) SE(6,7)=0.0 SE(6,9)=0.0 SE(6,11)=0.0 SE(6,13)=0.0 SE(6,15)=0.0 SE(6,8)=C11*R*L1*HH+C12*R*L1*BB SE(6,10)=Cll*(DD/T**2+GMSS*R**2*AA)+C12*(—AA) SE(6,12)=C11*R*II+012*(-R*AA) SE(6,14)=Cll*(L2*FF/T**2+GMSS*R**2*L2*CC)+C12*(~L2*CC) SE(6,16)=C11*R*L2*JJ+C12*R*L2*CC SE(7,7)=Cl3*R1*BB+Cla*(T*R1*BB+GMSG*R*T**2*R1AHH) SE(7,8)=O.O SE(7,10)=O.0 SE(7,12)-0.0 SE(7,14)=0.0 SE(7,16)=0.0 SE(7’9)=C13*('T*II)+C14*('DD-GMSS*R**2*T**2*AA) SE(7,11)-Cl3*(-AA+T**2*JJ)+C14*(-T*FF-GMSS*R**2*T**3ACC-T*AA+ GMSG*R*T**2*II) SE(7,l3)=C13*(-T*L2*JJ)+C14*(L2*FF+GMSS*R**2*T*A2*L2*CC) SE(7,15)=C13*R2*CC+C14*(T*R2*CC+GMSG*R*T**2*R2*JJ) SE(8,8)-C15*R*L1*HH+C16*R*L1*BB SE(8,9)=0.0 SE(8,11)=0.0 SE(8,13)-0.0 SE(8,15)=0.0 SE(8,10)=C15*(DD/T**2+GMSS*R**2*AA)+C16*(-AA) SE(8,12)-C15*R*II+C16*(-R*AA) SE(8,l4)=ClS*(L2*FF/T**2+GMSS*R**2*L2*CC)+Cl6*(-L2*CC) SE(8,16)=C15*R*L2*JJ+C16*R*L2*CC 380 170 SE(9,9)=C17*T*(-II)+C18*(-DD-GMSS*R**2*T**2*AA) SE(9,10)-0.0 SE(9,12)=0.0 SE(9,14)-0.0 SE(9,16)=0.0 SE(9,11)=C17*(-AA+T**2*JJ)+C18*(-T*FF-GMSS*R**2*T**3*CC-T*AA+ GMSG*R*T**2*II) SE(9,13)=C17*(-T*L2*JJ)+C18*(L2*FF+GMSS*R**2*T**2*L2*CC) SE(9,15)-C17*R2*CC+C18*(T*R2*CC+GMSG*R*T**2*JJ) SE(lO,10)-C19*(DD/T**2+GMSS*R**2*AA)+C20*(-AA) SE(10,11)=0.0 SE(10,13)-0.0 SE(10,15)-0.0 SE(10,12)-C19*R*II+C20*(-R*AA) SE(lO,14)-C19*(L2*FF/T**2+GMSS*R**2*L2*CC)+C20*(—L2*CC) SE(lO,16)=C19*R*L2*JJ+C20*R*L2*CC SE(ll:11)=C21*('AA+T**2*JJ)+C22*(-T*FF-GMSSRRRAQATAA3Acc-T*AA+ GMSG*R*T**2*II) SE(11,12)=0.0 SE(11,14)-0.0 SE(11,16)-0.0 SE(11,13)-C21*(-T*L2*JJ)+C22*(L2*FF+GMSS*R**2*TA*2*L2*CC) SE(11,15)-C21*R2*CC+C22*(T*R2*CC+GMSG*R*T**2*R2AJJ) SE(12,12)-C23*R*II+C24*(-R*AA) SE(12,13)-0.0 SE(12,15)=0.0 SE(12,l4)-C23*(L2*FF/T**2+GMSS*R**2*L2*CC)+C24*(-L2*CC) SE(12,l6)=C23*R*L2*JJ+C24*R*L2*CC SE(13,13)=C25*(-T*L2*JJ)+C26*(L2*FF+GMSS*R**2*T**2*L2*CC) SE(13,14)=0.0 SE(13,16)=0.0 SE(13,15)=C25*R2*CC+C26*(T*R2*CC+GMSG*R*T**2*R2*JJ) SE(14,14)=C27*(L2*FF/T**2+GMSS*R**2*L2*CC)+CZ8*(-L2*CC) SE(14,15)-0.0 SE(14,16)=C27*R*L2*JJ+C28*R*L2*CC SE(15,15)=C29*R2*CC+C30*(T*R2*CC+GMSG*R*T**2*R2*JJ) SE(15,16)=0.0 SE(16,l6)-C31*R*L2*JJ+C32*R*L2*CC D0 380 IE=1,16 DO 380 JE=IE,16 K(IE,JE)=K(IE,JE)+WEIGHT(J)*SE(IE,JE) IF (I.EQ.1) GO TO 300 GO TO 500 171 390 CONTINUE C C ........ ENTRIES OF CURVED ELEMENT NONLINEAR STIFFNESS SE1 C C A01,A02,A3,A4,A5,A6, ARE IN CLOSED FORM SOLUTIONS C C IF (1ARCH.EQ.1) CO T0 105 BTG = 2.*T*BZ(M) BLG = LOG((B1(M)+BTG)/B1(M)) A01 = BLG/(T*BTG) A02 = 4.*(l./BTG - B1(M)*BLG/(BTG*BTG))/T A3 = (l.-2.*Bl(M)/BTG+2.*Bl(M)*Bl(M)*BLG/(BTG*BTG))/(T*T*B2(M)) A4 = 1.5*A3 A5 = 2.*(1.-1.5*Bl(M)/BTG+3.*Bl(M)*Bl(M)/(BTG*BTG) + -3.*Bl(M)**3*BLG/(BTG**3))/(T*T*B2(M)) A6 = l.125*(1.-4.*Bl(M)/(3.*BTG)+2.*Bl(M)*Bl(M)/(BTG*BTG) + -4.*B1(M)**3/(BTG**3)+4.*B1(M)**4*BLG/(BTG**4)) + /(T*T*BZ(M)) ' GO TO 106 105 CONTINUE AOl=1./(T*(Bl(M))) AO2=2./(T*(B1(M))) A3=4./(3.*T*(B1(M))) A4=1.5*A3 A5=3./(T*(B1(M))) A6=9./(5.*T*(B1(M))) 106 CONTINUE C C C ........ LAMDAO TO LAMDA12 ARE WRITTEN AS XLDO TO XLD12 C XLDO=-0.5*T*DN(l)-(1:=T*T/12.)*DN(3)-R1*T*T*DN(5)/12.-0.5*T*DN(7) + +(l.-T*T/12.)*DN(9)+R2*T*T*DN(1l)/12. XLD1=DN(1) XLD2=-T*DN(3)+R1*T*DN(5) XLD3=-3.*DN(1)+2.*T*DN(3)-2.*R1*T*DN(5)+3.FDN(7)+T*DN(9)- + R2*T*DN(11) XLD4=2.*DN(1)-T*DN(3)+R1*T*DN(5)-2.*DN(7)-T*DN(9)+R2*T*DN(11) XLD5=DN(2) XLD6=-R1*T*DN(6) XLD7=~3.*DN(2)+2.*R1*T*DN(6)+3.*DN(8)+R2*T*DN(12) XLD8=2.*DN(2)-R1*T*DN(6)-2.*DN(8)-R2*T*DN(12) XLD9=DN(3) XLDlO=DN(9)-DN(3) XLD13=DN(4) XLDl4=DN(10)-DN(4) C C1=E(N)*A(M)/(2.*LENGTH(M)) C2=3.*A02-2.*A4 C3=6.*A3-2.*A5 C443.*A5-4.*A6 I.- nd ...: 3. a W u :- um: I l + + + + + + + + + + + + + + + + + + 172 C5=3.*(A02-A3) C6-3.*A3-A5 C7=3.*A3-2.*A4+3.*A5-4.*A6 IF(IPAR.EQ.4) GO TO 410 SE(l,1)=C1*((18.*A3-12.*A5+8.*A6)*XLDO+T*(C2*XLD2+C3*XLD3+C4*XLD4 )+T**2*(C5*XLD9+C6*XLD10)) SE(l,2)=C1*0.5*T*(C2*XLD6+C3*XLD7+C4*XLD8) SE(l,3)=C1*(T*(-6.*A3—2.*A4+6.*A5-4.*A6)*XLDO+ (1-T**2/12.)*(C2*XLD2+C3*XLD3+C4*XLD4+T*(C5*XLD9+ C6*XLD10))- 0.25*T**2*(C2*XLD2+C3*XLD3+(3.*A3-2.*A4+3.*A5-4.*A6)* XLD4+T*(C5*XLD9+C6*XLD10))) SE(1,4)=0.0 SE(l,5)=C1*R1*T*((-3.*A02+12.*A3+2.*A4-7.*A5+4.*A6)*XLDO+ T/12.*((-12.*A01+15.*A02-8.*A4)*XLD2+(-6.*AO2+ 30.*A3-8.*A5)*XLD3+(-6.*A4+15.*A5‘16.*A6)*XLD4+ T*((-12.*A01+15.*A02-12.*A3)*XLD9+ (-3.*A02+15.*A3-4.*A5)*XLD10))) SE(1,6)=C1*0.5*R1*T**2*((2.*A01-2.*A02+A4)*XLD6+(A02-4.*A3+A5)* XLD7+(A4-2.*A5+2.*A6)*XLD8) SE(1,7)=C1*(-18.*A3+12.*A5-8.*A6)*XLDO SE(1,8)=-SE(1,2) SE(1,9)-C1*(T*(-9.*A3+6.*A5-4.*A6)*XLDO - (1.-T**2/12.)* (C2*XLD2+C3*XLD3+C4*XLD4+T*(C5*XLD9+C6*XLD10)) -O.2S*T**2*(C2*XLD2+C3*XLD3+C4*XLD4+ T*(C5*XLD9+0.5*C3*XLD10))) SE(1,10)=0.0 SE(1,11)=C1*R2*T*((6.*A3-5.*A5+4.*A6)*XLDO+T/12.*((3.*A02-4.*A4) *XLD2+(6.*A3-4.*A5)*XLD3+(3.*A5-8.*A6)*XLD4 + T*((3.*A02-6.*A3)*XLD9+(3.*A3-2.*A5)*XLD10))) SE(l,12)=C1*O.5*R2*T**2*((A4-A02)*XLD6+(A5-2.*A3)*XLD7+ (2.*A6-A5)*XLD8) SE(2,2)=C1*(18.*A3-12.*A5+8.*A6)*XLDO SE(2,3)=C1*(1.-T**2/12.)*(C2*XLD6+C3*XLD7+C4*XLD8) SE(2,4)=0.0 SE(2,5)—C1*R1*T**2/12.*(C2*XLD6+C3*XLD7+C4*XLD8) SE(2,6)=C1*R1*T*(3.*A02-12.*A3-2.*A4+7.*A5-4.*A6)*XLDO SE(2,7)=Cl*T*0.5*(C2*XLD6+C3*XLD7+C4*XLD8) SE(2,8)=-SE(2,2) SE(2,9)=-SE(2,3) SE(2,10)=0.0 SE(2,11)=C1*R2*T**2/12.*(-C2*XLD6-C3*XLD7-C4*XLD8) SE(2,12)=C1*R2*T*(-6.*A3+5.*A5-4.*A6)*XLDO SE(3,3)=C1*(T**2*(1.5*A3+2.*A4-3.*A5+2*A6)*XLDO -(1.-T**2/12.)*T*(CZ*XLD2+C3*XLD3+C7*XLD4+ T*(CS*XLD9+C6*XLD10))) SE(3,4)=0.0 ' ' SE(3,5)=C1*R1*T*(T*(1.5*A02-4.5*A3-2.*A4+3.5*A5—2.WA6)FXLDO + + + + + + + + + + + + + + '+ + + + 173 -T**2/24.*(C2*XLD2+C3*XLD3+C7*XLD4+T*(CSFXLD9+C6*XLD10)) -(1.-T**2/12.)*((2.*A01-2.*A02+A4)*XLD2+(AO2-4.*A3+A5)*XLD3 +(A4-2.*A5+2.*A6)*XLD4 +T*((2.*A01-2.*A02+1.5*A3)*XLD9+(0.5*A02-2.*A3+O.5*AS)*XLD10))) SE(3,6)=C1*R1*T*(1.-T**2/12.)*((2.*A01-2.*A02+A4)*XLD6+ (A02-4.*A3+A5)*XLD7+(A4-2.*A5+2.*A6)*XLD8) SE(3,7)=C1*(T*(6.*A3+2.*A4-6.*A5+4.*A6)*XLDO - T**2/4.* (C2*XLD2+C3*XLD3+C7*XLD4+T*(CS*XLD9+C6*XLD10)) -(1.-T**2/12.)*(C2*XLD2+C3*XLD3+C4*XLD4+ T*(C5*XLD9+C6*XLD10) )) SE(3,8)=Cl*(l.-T**2/12.)*(-CZ*XLD6-C3*XLD7-C4*XLD8) SE(3,9)=C1*T*(T*(3.*A3+A4-3.*A5+2.*A6)*XLDO + (1.-T**2/12.)*0.5*(3.*A3-2.*A4)*XLD4) SE(3,10)=0.0 SE(3,11)=C1*R2*T*(T*(¢l.5*A3-A4+2.5*A5-2.*A6)*XLDO + T**2/24.*(02*XLD2+C3*XLD3+C7*XLD4+T*(C5*XLD9+C6*XLD10)) +(1.-T**2/12.)*((A02-A4)*XLD2+(2.*A3-A5)*XLD3+ (AS-2.*A6)*XLD4+T*((A02-1.5*A3)*XLD9+(A3-O.5*A5)*XLD10))) SE(3,12)=Cl*R2*T*(l.-T**2/12.)*((A4-A02)*XLD6+(A5-2.*A3)*XLD7+ (2 .*A6 -A5)*XLD8) SE(4,4)=O. SE(4,5)=0. SE(4,6)=0. SE(4,7)=0. SE(4,8)-0. SE(4,9)-0. SE(4,10)=0.0 SE(4,11)-0.0 SE(4,12)=0.0 OOOOOO C16=2.*A01-2.*A02+A4 C17=A02-4.*A3+A5 C18=A4-2.*A5+2.*A6 Cl9=2.*A01-2.*A02+1.5*A3 C20=0.5*A02-2.*A3+0.5*A5 C21=0.5*A3-2./3.*A5+2./3.*A6 C22=A5/6.-8./9.*A6+3.*A7 SE(S,5)=Cl*((R1*T)**2)*((2.*A01-4.*A02+8.*A3+2.*A4-4.*A5+2.*A6)* XLDO-T/6.*(C16*XLD2+C17*XLD3+C18*XLD4+T*(C19*XLD9+ C20*XLD10))) SE(S,6)=C1*R1**2*T**3/12.*(C16*XLD6+C17*XLD7+C18*XLD8) SE(5,7)=C1*R1*T*((3*A02-12.*A3-2.*A4+7*A5-4.*A6)*XLDO-T/l2.* ((12.*A01-9.*A02+4.*A4)*XLD2+(6.*A02-l8.*A3+4.*A5)*XLD3+ (6.*A4-9.*A5+8.*A6)*XLD4+T*((12.*A01-9.*A02+6.*A3)*XLD9+ (3.*A02-9.*A3+2.*A5)*XLD10))) SE(5,8)=-SE(2,5) SE(5,9)=C1*R1*T*(T*(1.5*A02-6.*A3-A4+3.5*A5-2.*A6)*XLDO +(1.-T**2/12.)*(C16*XLD2+C17*XLD3+C18*XLD4+T*(Cl9*XLD9+ C20*XLD10))-T**2/24.*(C2*XLD2+C3*XLD3+C4*XLD4+ T*(C5*XLD9+C6*XLD10))) + + + + + + + + + + + ++++ + + 174 SE(5,10)=0.0 SE(5,11)=C1*R1*R2*T**2*((-A02+4.*A3+A4-3.*A5+2.*A6)*XLDO+T/12.* ((2.*A01-A02)*XLD2+(A02-2.*A3)*XLD3+(A4-A5)*XLD4 + T*((2.*AOl-A02)*XLD9+(O.5*A02-A3)*XLD10))) SE(S,12)=C1*R1*R2*T**3/12.*((A4—A02)*XLD6+(A5-2.*A3)*XLD7+ (2.*A6—A5)*XLD8) SE(6,6)=C1*R1**2*T**2*(2.*A01-4.*A02+8.*A3+2.*A4-4.*A5+2.*A6)*XLDO SE(6,7)=C1*R1*T**2/2.*(C16*XLD6+C17*XLD7+C18*XLD8) SE(6,8)=-SE(2,6) SE(6,9)=-SE(1,6)*(1.-T**2/12.)*2./T SE(6,10)=0.0 SE(6,11)--SE(1,6)*R2*T/6. SE(6,12)=C1*R1*R2*T**2*(-A02+4.*A3+A4-3.*A5+2.*A6)*XLDO SE(7,7)-C1*((18.*A3-12.*A5+8.*A6)*XLDO-T*(CZ*XLD2+C3*XLD3+ C4*XLD4+T*(C5*XLD9+C6*XLD10+C7/3.*XLDll+C8/3.*XLD12))) SE(7,8)=-SE(2,7) SE(7,9)=Cl*(T*(9.*A3-6.*A5+4.*A6)*XLDO+(1.-T**2/12.) *(C2*XLDZ+C3*XLD3+C4*XLD4+T*(CS*XLD9+C6*XLD10)) -T**2/4.*(C2*XLD2+C3*XLD3+C4*XLD4+T*(CS*XLD9+C6*XLD10))) SE(7,10)=0.0 SE(7,11)=C1*R2*T*((-6.*A3+5.*A5-4.*A6)*XLDO+T/12.*((9.*A02-8.*A4) *XLD2+(18.*A3—8.*A5)*XLD3+(9.*A5-l6.*A6)*XLD4+ T*((9.*A02-l2.*A3)*XLD9+(9.*A3-4.*A5)*XLD10))) SE(7,12)=C1*R2*T**2/2.*((A4-A02)*XLD6+(A5-2.*A3)*XLD7+ (2.*A6-A5)*XLD8) SE(8,8)-SE(2,2) SE(8,9)=-SE(2,9) SE(8,10)=0.0 SE(8,11)--SE(2,11) SE(8,12)=-SE(2,12) SE(9,9)=C1*T*(T*(4.5*A3-3.*A5+2.*A6)*XLDO +(l.-T**2/12.)*(C2*XLD2+C3*XLD3+C4*XLD4+ T*(C5*XLD9+C6*XLD10))) SE(9,10)-0.0 SE(9,ll)=Cl*R2*T*(T*(-3.*A3+2.5*A5—2.*A6)*XLDO + T**2/24.* (C2*XLD2+C3*XLD3+C4*XLD4+T*(CS*XLD9+C6*XLD10)) + (l.-T**2/12.)*((A4-A02)*XLD2+ (AS-2.*A3)*XLD3+(2.*A6-A5)*XLD4+T*((1.5*A3-A02)*XLD9 +(0.5*A5-A3)*XLD10))) SE(9,12)=C1*R2*T*(l.-T**2/12.)*((A02-A4)*XLD6+(2.*A3-A5)*XLD7+ (A5-2.*A6)*XLD8) SE(10,10)-0.0 SE(10,11)=0.0 SE(10,12)=0.0 SE(ll,11)-Cl*R2**2*T**2*((2.*A3-2.*A5+2.*A6)*XLDO+T/6.* ((A4-A02)*XLD2+(A5-2.*A3)*XLD3+(2.*A6—A5)*XLD4+ C.... C + 175‘ + T*((1.5*A3-A02)*XLD9+(0.5*A5-A3)*XLDIO))) SE(ll,12)--SE(1,12)*R2*T/6. SE(12,12)-C1*R2**2*T**2*(2.*A3-2.*A5+2.*A6)*XLDO DO 400 IE-1,12 DO 400 JE-IE,12 K(IE.JE)-SE(IE,JE) GO TO 500 CONTINUE ....INTEGRANDS ON CURVED ELEMENT NONLINEAR STIFFNESS SE2 AS-AO1*XLD2**2+A02*XLD2*XLD3+A3*XLD3**2+A4*XLD2*XLD4 +A5*XLD3*XLD4+A6*XLD4**2 BS-T*(2.*AO1*XLD2*XLD9+A02*XLD3*XLD9+0.S*A02*XLD2*XLD10+ + l.5*A3*XLD4*XLD9+A3*XLD3*XLD10+0.5*A5*XLD4*XLD10) CS-T**2*(A01*XLD9**2+O.5*A02*XLD9*XLD10+0.25*A3*XLDlO**2) DS-AO1*XLD6**2+A02*XLD6*XLD7+A3*XLD7**2+A4*XLD6*XLD8 +A5*XLD7*XLD8+A6*XLD8**2 MS=AS+BS+CS+DS ' DQl-(-C2*XLD2-C3*XLD3-C4*XLD4-T*(CS*XLD9+C6*XLD10)) DQ2-(-C2*XLD6-C3*XLD7-C4*XLD8) DQ3-0.5*T*(C2*XLD2+CB*XLD3+C7*XLD4+T*(C5*XLD9+C6*XLDlO)) DQ4-0.0 DQ5-R1*T*((2.*AOl-2.*A02+A4)*XLD2+(A02-4.*A3+A5)*XL03+ (A4-2.*A5+2.*A6)*XLD4+T*((2.*A01-2.*A02+1.5*A3)*XLD9+ (0.5*A02-2.*A3+0.5*A5)*XLD10)) DQ6-R1*T*((-2.*A01+2.*A02-A4)*XLD6+(-A02+4.*A3-A5)*XLD7+ (-A4+2.*A5-2.*A6)*XLD8) DQ7--DQl DQ8--DQ2 DQ9-0.5*T*(C2*XLD2+C3*XLD3+C4*XLD4+T*(CS*XLD9+C6*XLDlO)) DQlO-0.0 DQll-R2*T*((A4-AO2)*XLD2+(A5-2.*A3)*XLD3+(2.*A6-A5)*XLD4+ T*((l.5*A3-AO2)*XLD9+(0.5*A5-A3)*XLDlO)) DQ12-R2*T*((A02-A4)*XLD6+(2.*A3-A5)*XLD7+(A5—2.*A6)*XLD8) C01-Cl*T*(B1(M)+BZ(M)*T)/(2.*LENGTH(M)) SE(1,l)-C01*(MS*(18.*A3-12.*A5+8.*A6)+DQ1**2) SE(l,2)-C01*DQl*DQ2 SE(l,3)-COl*(T*MS*(-6.*A3-2.*A4+6.*A5-4.*A6)+DQ1*DQ3) SE(1,4)-0.0 . ‘ SE(I,5)-C01*(R1*T*MS*(-3.*A02+12.*A3+2.*A4-7.*A5+4.wA6)+DQIxD05) SE(1,6)4C01*D01*DQ6 SE(1,7)--SE(1,1) SE(1,8)--SE(1,2) . SE(l,9)-C01*(MS*T*(-9.*A3+6.*A5-4.*A6)+DQI*DQ9) + 176 SE(1,10)=0.0 SE(l,11)=C01*(R2*T*MS*(6.*A3-5.*A5+4.*A6)+DQ1*DQ11) SE(1,12)=C01*DQ1*DQ12 SE(2,2)=C01*(MS*(18.*A3-12.*A5+8.*A6)+DQ2**2) SE(2,3)=C01*DQZ*DQ3 SE(2,4)-0.0 SE(2,5)=C01*DQ2*DQ5 SE(2,6)=COl*(R1*T*MS*(3.*A02-12.*A3-2.*A4+7.*A5-4.*A6)+DQ2*DQ6) SE(2,7)=-SE(1,2) SE(2,8)--SE(2,2) SE(2,9)=C01*DQZ*DQ9 SE(2,10)=0.0 SE(2,11)=C01*DQZ*DQ11 SE(2,12)=C01*(R2*T*MS*(-6.*A3+5.*A5-4.*A6)+DQ2*D012) SE(3,3)=C01*(T*T*MS*(1.5*A3+2.*A4-3.*A5+2.*A6)+DQ3**2) SE(3,4)-0.0 SE(3,5)=COl*(R1*T*T*O.5*MS*(3.*A02-9.*A3-4.*A4+7.*A5-4.*A6)+ DQ3*DQ5) SE(3,6)=C01*DQ3*DQ6 SE(3,7)=:SE(1,3) SE(3,8)=-SE(2,3) SE(3,9)=C01*(T*T*MS*(3.*A3+A4-3.*A5+2.*A6)+DQ3*DQ9) SE(3,10)-0.0 SE(3,ll)=COl*(R2*T*T*MS*(-1.5*A3-A4+2.5*A5-2.*A6)+DQ3*DQ11) SE(3,12)=C01*DQ3*DQ12 SE(4,4)=O. SE(4,5)=0. SE(4,6)=0. SE(4,7)=O. SE(4,8)=O. SE(4,9)=O. SE(4,10)=0.0 SE(4,11)=0.0 SE(4,12)=0.0 OOOOOO SE(5,5)-COl*((Rl*T)**2*MS*(2.*A01-4.*A02+8.*A3+2.*A4-4.*A5+2.*A6) +DQS**2) SE(5,6)=C01*DQS*DQ6 SE(5,7)=C01*(R1*T*MS*(3.*A02-12.*A3-2.*A4+7.*A5-4.*A6)+DQ5*DQ7) SE(5,8)=C01*DQ5*DQ8 SE(5,9)=C01*(R1*T*T*MS*(1.5*AO2-6.*A3-A4+3.5*A5-2.*A6)+DQ5*DQ9) SE(5,10)-0.0 SE(S,ll)=C01*(R1*R2*T*T*MS*(-A02+4.*A3+A4-3.*A5+2.*A6)+DQS*DQll) SE(5,12)=C01*DQ5*DQ12 SE(6,6)=C01*((R1*T)**2*MS*(2.*AOl—4.*A02+8.*A3+2.*A4-4.*A5+2.*A6) +DQ6**2) SE(6,7)=C01*DQ6*DQ7 SE(6,8)=-SE(2,6) 420 C 500 C C.... CCCC CCCC 550 CCCC 177 SE(6,9)-C01*DQ6*DQ9 SE(6,lO)-0.0 SE(6,ll)—C01*DQ6*DQ11 SE(6,12)-C01*(R1*R2*T*T*MS*(-A02+4.*A3+A4-3.*A5+2.*A6)+DQ6*DQ12) SE(7,7)-SE(1,1) SE(7,8)-SE(1,2) SE(7,9)--SE(1,9) SE(7,10)-0.0 SE(7,11)--SE(1,11) SE(7,12)--SE(1,12) SE(8,8)-SE(2,2) SE(8,9)--SE(2,9) SE(8,10)-0.0 SE(8,11)--SE(2,11)' SE(8,12)--SE(2,12) SE(9,9)=C01*(T*T*MS*(4.5*A3-3.*A5+2.*A6)+DQ9**2) SE(9,lO)-0.0 SE(9,11)=COl*(R2*T*T*MS*(-3.*A3+2.5*A5-2.*A6)+DQ9*DQll) SE(9,12)-C01*DQ9*DQl2 SE(10,10)-0.0 SE(10,11)-0.0 SE(10,12)=0.0 . 1 SE(11,11)-C01*((R2*T)**2*MS*(2.*A3-2.*A5+2.*A6)+DQ11**2) SE(11,12)-C01*DQ11*DQ12 SE(12,12)-C01*((R2*T)**2*MS*(2.*A3~2.*A5+2.*A6)+DQlZ**2) DO 420 IE-1,12 DO 420 JE-IE,12 K(IE,JE)-SE(IE,JE) CONTINUE ....MAKE INTERVAL CORRECTION AND’RETURN IF (IPAR.GE.3) GO TO 1 IQ=16 GO TO 2 IQ=12 C=l.0 CONTINUE DO 550 I-1,IQ DO 550 J-I,IQ SE(I,J)-C*K(I,J) SE(J,I)-SE(I,J) 178 C IF (IPAR.EQ.1) CALL REOCON IF (IPAR.EQ.2) CALL REOCON C CCCC RETURN C 2000 FORMAT('l',15HINVALID MP USED///7H GAUSS=,F4.1) END 00000 10 20 3O 35 40 50 60 70 80 C C.... C 100 105 + 179 SUBROUTINE ASEMBLE (M) ****A****************A*AAAAAAAAAAAAAAAAA*AAAAAAAAAAAAAAAAAAAAAAAAA TO PROCESS AND ASSEMBLE ELEMENT STIFFNESS MATRICES AND NODAL LOAD VECTORS INTO THEIR CORRESPONDING STRUCTURE ARRAYS. *AAAAAAAAAAAAAAA*AAAAAAAARRAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA COMMON/C1/NE,NUMNP,NUMEG,NTYPE(3),NUMEL(3),IPAR,ICAL1,ICAL2,ICAL3, ICAL4,ICAL5,ICAL6,ICAL7 COMMON/C2/NSIZE,NEQ,NCOND,MBAND,IEIGEN COMMON/C3/IA(37,8),IB(37,8),D31(1l3) COMMON/C4/SE(16,16) COMMON/C5/E(3),G(3),NODEI(36),NODEJ(36),D5(180) COMMON/C8/PN(37,8),R(296),PINT(37,8) COMMON/C9/S(296,l6),SP(296,16),IDET IF (IPAR.NE.1) GO TO 90 PROCESSING OF INITIAL LOADS AND NODAL LOADS INTO LOAD VECTOR IF(ICAL4.EQ.O) WRITE(61,2000) DO 80 N=1,NUMNP D0 70 1=1,6 IF(IA(N,I)) 20,70,10 II=IA(N,I) GO TO 60 1F (IB(N,I) LT 0) GO TO 30 NN=IB(N,I) GO TO 35 II=~IB(N,I)+NEQ GO TO 60 1F (IA(NN,I)) 40,70,50 II=-IB(NN,I)+NEO GO TO 60 II=IA(NN, I) R(II)— —PN(N, I) IF(ICAL4. EQ. 0) WRITE(61, 2010) 11, N, 1 ,R(II) CONTINUE CONTINUE RETURN ....ASSEMBLE ELEMENT STIFFNESS INTO STRUCTURE STIFFNESS NI=NODEI(M) NJ=NODEJ(M) DO 165 K1=1,2 IF(K1.EQ.1) NP=NI IF(K1.EQ.2) NP-NJ DO 160 1=1,6 IF(IA(NP,I)) 105,160,100 II-IA(NP,1) GO TO 115 IF (IB(NP,I).LT.0) GO TO 110 110 111 112 113 115 120 125 130 132 I35 140 145 C C.... C C C.... 0 150 155 160 165 C 2000 2010 C 180 NN=IB(NP,I) GO TO 111 II--IB(NF,I)+NEQ GO TO 115 IF (IA(NN,I)) 112,160,113 II=-IB(NN,I)+NEQ GO TO 115 II=IA(NN,I) CONTINUE DO 155 K2=l,2 IF(K2.EQ.1) ND-NI IF(K2.EQ.2) ND=NJ DO 150 J-I,6 IF(IA(ND,J)) 125,150,120 JJ=IA(ND,J) GO TO 145 IF (IB(ND,J).LT.O) GO TO 130 NN=IB(ND,J) GO TO 132 JJ=~IB(ND,J)+NEQ GO TO 145 IF (IA(NN,J)) 135,150,140 JJ=-IB(NN,J)+NEQ GO TO 145 JJ=IA(NN,I) CONTINUE ....FILL-IN STRUCTURE STIFFNESS MATRIX IN BANDED FORMAT ONLY UPPER SEMIBANDWIDTH INCLUDING DIAGONAL IF (JJ.LT.II) GO TO 150 IF(K1.EQ.1) IE=I IF(K1.EQ.2) IE=I+6“ IF(K2.EQ.1) JE=J IF(K2.EQ.2) JE=J+6 ....CHANGE -JJ- SUBSCRIPT OF FULL MATRIX TO ~JJ- SUBSCRIPT OF BANDED FORMAT. LOOP OVER TERMS OUTSIDE OF BAND JJ=JJ-II+1 S(II,JJ)=S(II,JJ)+SE(IE,JE) CONTINUE CONTINUE CONTINUE CONTINUE RETURN FORMAT('l',43HINITIAL AND NODAL LOADS PROCESSED INTO LOAD, 12H VECTOR R(I)//) FORMAT('O',2HR(,I3,4H)-P(,I2,1H,,12,2H)=,F16.6) END 00000 C.... 50 6O 90 2015 2020 2021 181 SUBROUTINE STCONDN ****************************************************************** TO WRITE THE UNCONDENSED STRUCTURE STIFFNESS ACCORDING TO THE VALUES OF IPAR=2,3,4 FOR S, SI, 82 RESPECTIVELY ***************************************************************k** COMMON/C1/NE,NUMNP,NUMEG,NTYPE(3),NUMEL(3),IPAR,ICAL1,ICAL2,ICAL3, + ICAL4,ICAL5,1CAL6,ICAL7 COMMON/C2/NSIZE,NEQ,NCOND,MBAND,IEIGEN COMMON/C8/PN(37,8),R(296),PINT(37,8) COMMON/C9/S(296,16),SP(296,16),IDET IF (IPAR.NE.2) GO TO 40 IF (ICAL3.EQ.O) WRITE(61,2050) IF (ICAL3.EQ.O) WRITE(61,2060) (I,R(I),I=1,NSIZE) ....WRITE UNCONDENSED STRUCTURE LINEAR STIFFNESS -S- OR UNCONDENSED NONLINEAR STIFFNESS 81 OR 82 DEPENDING ON VALUE OF IPAR IF (ICAL3.NE.O) GO TO 90 IF(IPAR.EQ.2) WRITE(61,2030) IF (IPAR.EQ 3) WRITE(61,2040) IF (IPAR.EQ.A) WRITE(61,2045) Kl=1 K2-8 K3=MBAND~K1 . IF (K3.LE.7) GO TO 60 WRITE(61,2015) K1,K2 WRITE(61,2020) ((S(I,J),J=K1,K2),I=I,NSIZE) K1=K1+8 K2-K2+8 K3=MBAND-K1 IF(K3.LE.7) GO TO 60 GO TO 50 WRITE(61,2015) K1,MBAND IF (K3.EQ.0) WRITE(61,2027) ((S(I,J),J=K1,MBAND),I=I,NSIZE) IF (K3.EQ.1) WRITE(61,2021) ((S(I,J),J=K1,MBAND),I=l,NSIZE) IF (K3.EQ.2) WRITE(61,2022) ((S(I,J),J=K1,MBAND),I=I,NSIZE) IF (K3.EQ.3) WRITE(61,2023) ((S(I,J),J=K1,MBAND),I=I,NSIZE) IF (K3.EQ.4) WRITE(61,2024) ((S(I,J),J=K1,MBAND),I=1,NSIZE) IF (K3.EQ.5) WRITE(61,2025) ((S(I,J),J=K1,MBAND),I=1,NSIZE) IF (K3.EQ.6) WRITE(61,2026) ((S(I,J),J=KI,MBAND),I=I,NSIZE) IF (K3.EQ 7) WRITE(61,2020) ((S(I,J),J=KI,MBAND),I=1,NSIZE) CONTINUE REWIND 4 REWIND S REWIND 16 RETURN FORMAT('-',7HCOLUMNS,IA,7HTHROUGH,IA) FORMAT('O',8E16.5) FORMAT('O',2E16.5) an 3 1, ... 3“ 2022 2023 2024 2025 2026 2027 2030 2040 2045 2050 2060 0 00000000 C C.... C C 110 C C.... C C C.... C C C.... C 140 182 FORMAT('O',3E16.S) FORMAT('O',4E16.5) FORMAT('O',5E16.5) FORMAT('O',6E16.5) FORMAT('O',7E16.5) FORMAT('O',E16.5) FORMAT('l',45HUNCONDENSED LINEAR STIFFNESS OF STRUCTURE (8)) FORMAT('l',49HUNCONDENSED NONLINEAR STIFFNESS OF STRUCTURE (81)) FORMAT('l',49HUNCONDENSED NONLINEAR STIFFNESS OF STRUCTURE (82)) FORMAT('l’,28HUNCONDENSED LOAD VECTOR R(I)//) FORMAT(' ',2HR(,13,2H)=,F16.6) END SUBROUTINE LINSOLN *************************************************AAAAAA TO SOLVE SYSTEM OF LINEAR EQUATIONS S*D=R BY CALLING THE APPROPRIATE SUBROUTINE S= STUCTURE"S LINEAR STIFFNESS D- VECTOR OF D.O.F."S Ra LOAD VECTOR *********************************************k*********k IIOIIOIIIOI AAKK‘A AAAAA W ****>‘:~k*7‘:7’<7‘< COMMON/C1/NE,NUMNP,NUMEG,NTYPE(3),NUMEL(3),IPAR,ICAL1,ICAL2,ICAL3, ICAL4,ICAL5,ICAL6,ICAL7 COMMON/C2/NSIZE,NEQ,NCOND,MBAND,IEIGEN COMMON/C8/PN(37,8),R(296),PINT(37,8) COMMON/C9/S(296,16),SP(296,16),IDET COMMON/ClO/D(296),D10(1184),RC(296),SC(296,I6) ....FILL-IN ARRAY D(I) WITH VALUES OF LOAD VECTOR R(I) AFTER SOLUTION D(I) WILL CONTAIN THE DISPLACEMENT VALUES 00 110 I=1,NEQ D(I)=R(I) ....CHECK DATA GENERATION FOR SOLUTION OF EQUATIONS IF(ICAL5.EQ.O) WRITE(61,2020) IF (ICAL5.EQ.O) WRITE(61,2010) (I,D(I),I=1,NEQ) ....SOLVE SYSTEM OF -NEQ- LINEAR EQUATIONS CALL GAUSSOL ....CHECK DATA GENERATION IF (ICALS.NE.O) GO TO 140 WRITE(61,2000) WRITE(61,2010) (I,D(I),I=1,NEQ) RETURN , , ‘ ' ta 1* ~10, 4:” ‘ . PM: Niki") ~x ..- .: .x. .1, ,, g r ., 1. 1 1 Cu, a. .. . ‘ 4:": wt . . .1 \ I ; J :! , . ' CM “'9 ‘ .- .- 183 2000 FORMAT(’I’,34HDISPLACEMENTS FROM LINEAR SOLUTION//) 2010 FORMAT(' ',2HD(,I3,2H)=,E25.15) 2020 FORMAT('l',31HLOAD VECTOR FOR LINEAR SOLUTION//) C 000000 C.... 750 780 790 C C.... C 820 830 C C.... C 850 860 END SUBROUTINE GAUSSOL **************************************************9*************** GAUSS ELIMINATION EQUATION SOLVER, BANDED FORMAT FROM BOOK BY ROBERT D. COOK, FIG. 2.8.1., PAGE 45 CONCEPTS AND APPLICATIONS OF FINITE ELEMENT ANALYSIS ********************************************************k********* COMMON/C2/NSIZE,NEQ,NCOND,MBAND,IEIGEN COMMON/C9/S(296,l6),SP(296,16),IDET COMMON/ClO/D(296),D10(1184),RC(296),SC(296,16) ....FORWARD REDUCTION OF MATRIX (GAUSS ELIMINATION) DO 790 N=1,NEQ DO 780 L:2,MBAND IF (S(N,L).EQ.O.) GO TO 780 I-N+L-1 C=S(N,L)/S(N,1) J=O DO 750 K-L,MBAND J=J+1 S(I,J)=S(I,J)-C*S(N,K) S(N,L)=C CONTINUE CONTINUE ....FORWARD REDUCTION OF CONSTANTS (GAUSS ELIMINATION) DO 830 N=1,NEQ DO 820 L=2,MBAND IF (S(N,L).EQ.O.) GO TO 820 I=N+L—l D(I)=D(I)-S(N,L)*D(N) CONTINUE D(N)=D(N)/S(N,1) ....SOLVE FOR UNKNOWNS BY BACK SUBSTITUTION DO 860 M=2,NEQ N=NEQ+l-M DO 850 L=2,MBAND IF (S(N,L).EQ.O.)GO TO 850 K=N+L~1 D(N)=D(N)-S(N,L)*D(K) CONTINUE CONTINUE RETURN 0 C)C)C)C)C) C.... 150 155 160 170 180 190 184 END SUBROUTINE IDENT *************************************************A**************** TO IDENTFY THE DISPLACEMENTS FOUND IN THE SOLUTION OF EQUATIONS S*D=R AND THE ONES FOUND IN THE RECOVERY PROCESS ********************************************** 1000000 xxkawww ************ COMMON/C1/NL,NUMNP,NUMEG,NTYPE(3),NUMEL(3),IPAR,ICAL1,ICAL2,ICAL3, + ICAL4,ICAL5,1CAL6,ICAL7 COMMON/CZ/NSIZE,NEQ,NCOND,MBAND,IEIGEN COMMON/C3/IA(37,8),IB(37,8),D31(113) COMMON/C5/E(3),G(3),NODEI(36),NODEJ(36),A(36),IXX(36),IYY(36), + KT(36),L(1,36) COMMON/C10/D(296),010(1184),RC(296),SC(296,16) COMMON/Cll/DN(16),U(36,12),W(37,8),V(37,8) ....IDENTIFICATION OF DISPLACEMENTS IF (ICAL6.EQ.0) WRITE(61,2000) 00 230 NN-1,NUMEG INUMEL:NUMEL(NN) 00 230 K=l,INUMEL M=L(NN,K) IF (ICAL6.EQ.O) WRITE(61,2010) M NI=NODEI(M) NJ=NODEJ(M) 00 230 K1-1,2 IF(K1.EQ.1) NP=NI IF(K1.EQ.2) NP-NJ 00 220 I=1,6 IF(IA(NP,I)) 160,155,150 NE=IA(NP,I) U(NP,I)=D(NE) IF (ICAL6.EQ 0) WRITE(61,2020) NE,NP,I,U(NP,I) GO TO 220 U(NP,I)=O. IF (ICAL6.EQ O) WRITE(61,2020) NE,NP,I,U(NP,I) GO TO 220 IF (IB(NP,I) LT 0) GO TO 170 NM=IB(NP,I) GO TO 180 NE=-IB(NP,I)+NEQ U(NP,I)=D(NE) IF (ICAL6.EQ.0) WRITE(61,2020) NE,NP,I,U(NP,I) GO TO 220 IF (IA(NM,I)) 190,200,210 NE--IE(NM,I)+NEQ U(NP,I)=D(NE) IF (ICAL6.EQ.O) WRITE(61,2020) NE,NP,I,U(NP,I) GO TO 220 - j; A r NYSE-1&2; ' HI‘HI‘IVH‘“ .31. ' -':-'-1'-'m gre E‘L'JUUU 200 210 220 230 2000 2010 2020 0 0000 100 C C.... C C C.... C 110 120 185 U(NP,I)=0. IF (ICAL6.EQ.0) WRITE(61,2020) NE,NP,I,U(NP,I) GO TO 220 NE=IA(NM,I) U(NP,I)=D(NE) IF (ICAL6.EQ.0) WRITE(61,2020) NE,NP,I,U(NP,I) CONTINUE CONTINUE RETURN FORMAT('I',35HNODAL DISPLACEMENTS ON EACH ELEMENT) FORMAT('-',7HELEMENT,I3//) FORMAT(’ ',2HD(,I3,1H),5X,2HU(,IZ,1H,,Il,2H)=,E25.1S) END SUBROUTINE STRESS ***************************k***********R1k;IAAkngRRkFx********* TO COMPUTE NODAL FORCES AND STRESSES IN THE STRUCTURE *********************************Rk**xxRAAAARk********R********* DIMENSION D(l6) COMMON/C1/NE,NUMNP,NUMEG,NTYPE(3),NUMEL(3),D11(8) C0MM0N/C4/SE(16,16) COMMON/C5/E(3),G(3),NODEI(36),NODEJ(36),A(36),IXX(36),IYY(36), + KT(36),L(1,36) COMMON/C8/PN(37,8),R(296),PINT(37,8) COMMON/Cll/DN(16),U(36,12),W(37,8),V(37,8) WRITE(61,2000) 00 100 N=1,NUMNP DO 100 I=1,6 PN(N,I)-0.0 ....PROCESS EVERY ELEMENT OF EACH ELEMENT GROUP DO 200 K=1,NUMEG INUMEL=NUMEL(K) DO 190 KK=1,INUMEL M=L(K,KK) NI-NODEI(M) NJ=NODEJ(M) READ(1,10) ((SE(I,J),J=1,12),I=1,12) ....IDENTIFY NODAL DISPLACEMENTS ON EACH ELEMENT N-NI DO 110 1=1,6 D(I)=U(N,I) N-NJ DO 120 I=7,12 D(I)=U(NI,I«6) C C.... C 140 145 155 160 C C.... C 190 200 C C.... C C 10 2000 2010 2020 2030 C 0000000 + + 186 ....OBTAIN RESULTANT LOADS N=NI DO 145 I=1,6 DN(I)=O. DO 140 J=1,12 DN(I)=DN(I)+SE(I,J)*D(J) PN(N,I)=PN(N,I)+DN(I) N=NJ DO 160 I=7,12 DN(I)=O. DO 155 J=1,12 DN(I)=DN(I)+SE(I,J)*D(J) PN(N,I-6)=PN(N,I-6)+DN(I) ....WRITE RESULTANT LOADS OF THE NODES OF EACH ELEMENT WRITE(61,2010) M,(DN(I),I=1,6),(DN(I),I=7,12) CONTINUE CONTINUE ....WRITE NODAL RESULTANTS OF STRUCTURE WRITE(61,2020) WRITE(61,2030) (N,(PN(N,I),I=1,6),N=1,NUMNP) REWIND 8 RETURN FORMAT(E21.6) FORMAT('I',31HRESULTANT LOADS ON EACH ELEMENT///) FORMAT('-',BH ELEMENT,I3//4X,6HNODE-I,6E15.9// 4X,6HNODE-J,6E15.9) FORMAT('l',48HSTRUCTURE RESULTANTS DUE TO LINEAR DISPLACEMENTS/// 1x,4HNODE,15x,3HFN1,15x,3HFN2,15x,3HPN3,15x,3HFN4, 15X,3HPN5,ISX,3HPN6//) FORMAT('O',IS,4x,6E15.9) END SUBROUTINE EICENVL (EIGEN,IDATA) ,,,,,,, I IIII . ,,,,,,,,,,,, L! ' ' 3"k**%********7z***i****k7'VI'SA*7:kAVEkAgciI'CAAAAAAAAAAAAA A A A A A A AI\AAA7\AK7\ TO SOLVE EIGENVALUE PROBLEM S*X=-(LAMBDA)*SI*X WILL OBTAIN ONLY THE LOWEST EIGENVALUE AND CORRESPONDING EIGENVECTOR. USES INVERSE VECTOR ITERATION WITH THE RAYLEIGH QUOTIENT. 0000000000000000000000 O D I, L """""""""""""" A A A A ‘A' A A A W7\AA"A‘7\“A‘**7\3\* COMMON/CI/DUMMY(16),ICAL7 COMMON/C2/NSIZE,NEQ,NCOND,MBAND,IEIGEN COMMON/C9/S(296,16),SP(296,16),IDET COMMON/C10/XB(296),YB(296),X(296),Y(296),EIGNVTR(296) C C... C C 100 C C.... C C 105 107 C C.... C 110 120 130 C C.... C 140 150 160 C C.... 187 ..... ASSUME STARTING SHIFT, STARTING VECTOR, AND MAXIMUN NUMBER OF ITERATIONS ALLOWED. WRITE(61,2010) READ(60,1000) MAX,EPSI,RHO WRITE(61,2000)MAX,EPSI,RHO DO 100 I=1,NEQ X(I)=l. ....OBTAIN VECTOR Y(I) FROM Y(I)=Sl(I,J)*X(I) FIRST CHANGE SIGN OF MATRIX Sl READ(5,10) ((S(I,J),J=1,MBAND),I=1,NEQ) REWIND 5 00 107 I=1,NEQ 00 105 J=1,MBAND S(I,J)=-S(I,J) CONTINUE CONTINUE WRITE(5,10) ((S(I,J),J=1,MBAND),I=1,NEQ) REWIND 5 ....HORIZONTAL SWEEP OF Sl(I,J)*X(I), DIAGONAL NOT INCLUDED DO 130 I=1,NEQ Y(I)=O. II=I+1 IF (II.GT.NEQ) GO TO 130 D0 120 J=2,MBAND IF (S(I,J).EQ.O.) GO TO 110 Y(I)=Y(I)+S(I,J)*X(II) II=II+1 IF (II.GT.NEQ) GO TO 130 CONTINUE' CONTINUE ....DIAGONAL SWEEP OF SI(I,J)*X(I) DO 160 I=1,NEQ II=I JJ=1 IF (S(II,JJ).EQ.O.) GO TO 150 Y(I)=Y(I)+S(II,JJ)*X(II) II=II-l JJ=JJ+1 IF (II.EQ.0) GO TO 160 IF (JJ.GT.MBAND) GO TO 160 GO TO 140 _ CONTINUE ....START ITERATION PROCEDURE BY STABLISHING THE SYSTEM OF Ll? 188 C EQUATIONS S(I,J)*XB(I)=Y(I) AND SOLVING FOR XB(I). C STORE VALUES OF Y(I) INTO XB(I) FOR GAUSS SOLUTION. C DO 300 K = l,MAX DO 165 I=1,NEQ 165 XB(I)—Y(I) IF (IDATA.EQ.0) READ(4,10) ((S(I,J),J=1,MBAND),I=1,NEQ) IF (IDATA.EQ.1) READ(7,10) ((S(I,J),J=1,MBAND),I=1,NEQ) REWIND 4 REWIND 7 IF (ICAL7.NE.O) GO TO 176 C C ........ PRINT DATA SENT TO SUBROUTINE GAUSSOL C WRITE(61,2100) K K1=1 K2=6 K3=MBAND-K1 IF (K3.LE.7) GO TO 174 172 WRITE(61,2110) K1,K2 WRITE(61,2115) ((S(I,J),J=K1,K2),I=1,NEQ) K1=KI+6 K2=K2+6 K3=MBAND-K1 IF (K3.LE.7) GO TO 174 GO TO 172 174 WRITE(61,2110) K1,MBAND IF (K3.EQ.O) WRITE(61,2120) ((S(I,J),J=K1,MBAND),I=I,NEQ) IF (K3.EQ.1) WRITE(61,2121) ((S(I,J),J=K1,MBAND),I=I,NEQ) IF (K3.EQ.2) WRITE(61,2122) ((S(I,J),J=KI,MBAND),I=1,NEQ) IF (K3.EQ.3) WRITE(61,2123) ((S(I,J),J=K1,MBAND),I=I,NEQ) IF (K3.EQ.4) WRITE(61,2124) ((S(I,J),J=K1,MBAND),I=1,NEQ) IF (K3.EQ.5) WRITE(61,2125) ((S(I,J),J=K1,MBAND),I=1,NEQ) C IF (K3.EQ.6) WRITE(61,2126) ((S(I,J),J=K1,MBAND),I=1,NEQ) C IF (K3.EQ.7) WRITE(61, 2115) ((S(I,J),J=KI,MBAND),I=l,NEQ) WRITE(61, 2130) WRITE(6I, 2135) (XB(I), I=—1,NEQ) 176 CONTINUE C C ........ SOLVE SYSTEM OF EQUATIONS S(I,J)*XB(I)=Y(I) C CALL GAUSSOL IF (ICAL7.EQ.0) WRITE(61,2030) C C ........ OBTAIN VECTOR YB(I) FROM YB(I)=Sl(I,J)*XB(I) C READ(5,10) ((S(I,J),J=1,MBAND),I=1,NEQ) REWIND 5 C C ........ HORIZONTAL SWEEP OF Sl(I,J)*XB(I), DIAGONAL NOT INCLUDED C DO 200 I=1,NEQ 180 190 200 C C.... C 210 220 230 C C.... C 240 250 C C.... C 260 300 C C 189 YB(I)=O. II=I+1 IF (II.GT.NEQ) GO TO 200 D0 190 J-2,MBAND IF (S(I,J).EQ.O.) GO TO 180 YB(I)=YB(I)+S(I,J)*XB(II) II-II+1 IF (II.GT.NEQ) GO TO 200 CONTINUE CONTINUE ....DIAGONAL SWEEP OF Sl(I,J)*XB(I) 00 230 I=1,NEQ II=I JJ=1 IF (S(II,JJ).EQ.O.) CO TO 220 YB(I)=YB(I)+S(II,JJ)*XB(II) II=II-l JJ=JJ+1 IF (II.EQ.O) GO TO 230 IF (JJ.GT.MBAND) GO TO 230 GO TO 210 CONTINUE ....COMPUTE RAYLEIGH QUOTIENT RQ=RHO Q1=0.0 02=0.0 00 240 I=1,NEQ Q1=Q1+XB(I)*Y(I) Q2=Q2+XB(I)*YB(I)‘ RHO=Q1/Q2 00 250 I=1,NEQ Y(I)=YB(I)/(Q2**.5) ....CHECK CONVERGENCE TO DESIRED EIGENVALUE CHECK=ABS(RHO-RQ)/RHO IF (CHECK.LE.EPSI) GO TO 310 EIGEN=RHO DO 260 I=1,NEQ EIGNVTR(I)=XB(I)/(Q2**.5) IF (ICAL7.NE.O) WRITE(61,203S) K,EIGEN IF (ICAL7.NE.0) GO TO 300 WRITE(61,2040) K,RHO,CHECK,EIGEN WRITE(61,2050) (XB(I),YB(I),Y(I),EIGNVTR(I),I=1,NEQ) CONTINUE C ........ OBTAIN EIGENVALUE AND CORRESPONDING EIGENVECTOR 310 320 C 10 1000 2000 2010 2030 2035 2040 2050 2100 2110 2115 2120 2121 2122 2123 2124 2125 + + 12X,7HEIGNVTR///) 190 EIGEN=RHO 00 320 I=1,NEQ EIGNVTR(I)-XB(I)/(Q2**.5) ILAST=K WRITE(61,2070) ILAST WRITE(61,2080) EICEN WRITE(61,2090) (EIGNVTR(I),I=1,NEQ) RETURN FORMAT(E21.6) FORMAT(15,2F10.6) FORMAT('-',4HMAX-,I3///6H EPSI=,FIO.6///6H RHO=,F10.6) FORMAT('I',45HLINEAR EIGENVALUE PROBLEM (INVERSE ITERATION)//) FORMAT('l',38HINVERSE VECTOR ITERATION WITH SHIFTING///2X,1HK,9X, 2HXB,16X,2HYB,16X,3HRHO,14X,5HCHECK,15X,1HY,15X,5HEIGEN, FORMAT('-',2HK—,I3,5X,6HEIGEN=,E15.9) FORMAT('-',I3,39X,E15.9,3X,E15.9,21X,E15.9) FORMAT(' ',6X,El5.9,3X,E15.9,39X,E15.9,21X,E15 9) FORMAT('l',34HDATA FOR GAUSSOL S(I,J) AND XB(I)//1X,2HK=,I3//) FORMAT('-',7HCOLUMNS,I4,10H THROUGH,IA) FORMAT('O',6E16.8) FORMAT('O',E16.8) FORMAT('O',2E16.8) FORMAT('O',3E16.8) FORMAT('O',4E1628) FORMAT('O',5E16.8) FORMAT('O',6E16.8) C 2126 FORMAT('O’,7E16.8) 2130 2135 2070 2080 2090 C C 000000 100 FORMAT('-',37HVECTOR Y(I), SENT TO GAUSSOL AS XB(I)//) FORMAT('O',10X,E15.9) FORMAT('l',5X,10HEIGENVALUE,9X,11HEIGENVECTOR,5X,6HILAST=,I3) FORMAT(' ',E15.9) FORMAT(' ',20X,E15.9) END SUBROUTINE NLEIGNP (SCALE) *********************************************************U THIS ROUTINE WILL COMPUTE THE EIGENVALUE OF THE QUADRATIC EIGENVALUE PROBLEM (K+L*N1+L*L*N2)*X=O IT USES THE MODIFIED REGULA FALSI METHOD ************************************************************* > - $ x. x. EXTERNAL DET REAL L READ(60,1000) XTOL,FTOL,NTOL,DINCR WRITE(61,2010) XTOL,FTOL,NTOL,DINCR WRITE(61,2030) A=0. FA=0ET(A,SCALE) 191 WRITE(61,2020) A,FA IF (FA.LT.O.) GO To 110 A=A+DINCR GO TO 100 110 CONTINUE B-A A-A-DINCR CALL MRCFLS (DET,A,B,XTOL,FTOL,NTOL,IFLAG,SCALE) IF (IFLAG.GT.2) GO TO 500 LP ' -‘ L‘- 3..me- -~__n.1w;x) 2'23: :"ui-I'J. . . .- 3 0 3 C 10 1010 2000 2010 2020 2025 2030 2040 C C CCCC 0 220 230 240 250 0 195 CALL EIGENVL (EIGEN,IDATA) RETURN FORMAT(E21.6) FORMAT(IS,F10.5) FORMAT(’l',l8H TILTED LOAD CASE //) FORMAT(////13H DECK BRIDGE///9H IDECK =,12///6H HD =,F10.5) FORMAT(////16H THROUGH BRIDGE///9H IDECK =,I2///6H HD =,F10.5) FORMAT(////21H HALF-THROUGH BRIDGE///9H IDECK =,12/// + 6H HD -,F10.5) FORMAT(////25H LOAD ON EACH COLUMN IS ,FlO 5//// + 16H COLUMN LENGTHS//) FORMAT('O’,7H H(,I2,3H) =,F7.3) END SUBROUTINE REOCON COMMON/C4/SE(16,16) DIMENSION ID(4),XA(14) CONDENSE FROM 16 BY 16 TO 14 BY 14 N=16 NC=2 DO 230 K=1,NC LL=N-K KK=LL+1 D0 230 LPl,LL DUM=SE(KK,L)/SE(KK,KK) D0 220 M=1,L SE(L,M)=SE(L,M)-SE(KK,M)*DUM CONTINUE LL=N-NC DO 250 I=1,LL Do 240 J=I,LL SE(I,J)=SE(J,I) CONTINUE CONTINUE ID(1)=7 ID(2)=8 REORDER COLUMNS AND ROWS N=14 NR=2 NC=2 DO 200 K=1,NR DO 100 J=1,N XA(J)=SE(J,ID(K)) 100 110 120 130 140 150 160 170 180 200 221 231 241 251 CC CCCC C)C) 196 CONTINUE DO 120 L:ID(K),N-1 DO 110 J=1,N SE(J,L)=SE(J,L+1) CONTINUE CONTINUE DO 130 J=1,N SE(J,N)=XA(J) CONTINUE DO 140 J-I,N XA(J)-SE(ID(K),J) CONTINUE DO 160 LaID(K),N-1 DO 150 J-I,N SE(L,J)=SE(L+1,J) CONTINUE CONTINUE DO 170 J-1,N SE(N,J)=XA(J) CONTINUE KK-K+1 DO 180 M=KK,NR ID(M)=ID(M)-1 CONTINUE CONTINUE DO CONDENSATION DO 231 K=1,NC LL=N-K KK—LL+1 DO 231 L-1,LL DUM-SE(KK,L)/SE(KK,KK) DO 221 M-1,L SE(L,M)éSE(L,M)-SE(KK,M)*DUM CONTINUE LL=N-NC DO 251 I=1,LL DO 241 J=I,LL SE(I,J)—SE(J,I) CONTINUE CONTINUE RETURN END FUNCTION DET1(SCALE) *******************************************A v v 7C7: a I 7€3€ ****************** THIS FUNCTION COMPUTES THE VALUE OF THE DETERMINANT OF C)C) 490 450 350 380 390 250 C)C)C) 400 197 THE MATRIX S=K+N1+N2 *************************k*k******************k******************* COMMON/C2/ NSIZE,NEQ,NCOND,MBAND,IEIGEN COMMON/C9/ S(296,16),SP(296,16),IDET IF(IDET.EQ.1) GO TO 250 IF(IDET.EQ 2) GO TO 450 D0 490 I=1,NEQ DO 490 J-1,MBAND S(I,J)-SP(I,J) FORWARD REDUCTION OF MATRIX(CAUSS ELEMINATION) ******************AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA DO 390 LN=1,NEQ DO 380 LL=2,MBAND IF(S(LN,LL).EQ.0.) GO TO 380 I=LN+LL-1 C-S(LN,LL)/S(LN,1) J=O DO 350 KK-LL,MBAND J=J+l S(I,J)-S(I,J)-C*S(LN,KK) S(LN,LL)=C CONTINUE CONTINUE CONTINUE COMPUTE DETERMINANT OF MATRIX s SCALE DOWN"DET1" BY A "SCALE" VALUE AFTER EACH STEP *************************************************************A**** DT-l. D0 400 I=1,NEQ DT=DT*S(I,1)/SCALE CONTINUE DET1=DT RETURN END "l .1}. ' 1‘. .-.n" 3511' I “a. Waugh.“ .-1 CHM-’3 . EWFIG .... .._' 'I “‘1‘ In?“ 5 I if ."-’ --" C14! - .\ u:- "'0 '._nl l . . ' u . '- l..- l .— 1 n i f.’ l'-- v ‘ - I ML, '5 11':- ._. .. .‘u I.“ F "i"; n f '3 Pu n. 535% 'J U '1 l-w --_- na- L' 0000000000000000000 APPENDIX D INCREMENTAL STIFFNESS MATRICES, [n1], AND [n2] BASED ON THE QUARTIC AXIAL STRAIN MODEL The following sUbroutine contains the entries of the [n1] and [n2] matrices based on the quartic axial strain model. SUBROUTINE NUMINT (N,M) IMPLICIT DOUBLE PRECISION (A-H,O-Z) ****************************************************************** TO INTEGRATE NUMERICALLY THE TERMS OF THE CURVED ELEMENT STIFFNESS MATRICES SE, SE1, SE2, IT USES THE GAUSS-LEGENDRE QUADRATURE FORMULA. THE ROUTINE NUMINT USES THE MP-POINT GAUSS-LEGENDRE QUADRATURE FORMULA TO COMPUTE THE INTEGRAL OF FUNCTN(GM)*DGM BETWEEN INTEGRATION LIMITS A1 AND A2. THE ROOTS OF SEVEN LEGENDRE POLYNOMIALS AND THE WEIGHT FACTORS FOR CORRESPONDING QUADRATURES ARE STORED IN THE 2 AND WEIGHT ARRAYS RESPECTIVELY. MP MAY ASSUME VALUES 2, 3, 4, 5, 6, 10, AND 15 ONLY. THE APPROPRIATE VALUES FOR THE MP-POINT FORMULA ARE LOCATED IN ELEMENTS Z(KEY(I))...Z(KEY(I+1)-1) AND WEIGHT(KEY(I))... WEIGHT(KEY(I+l)-l) WHERE THE PROPER VALUE FOR I IS DETERMINED BY FINDING THE SUBSCRIPT OF THE ELEMENT OF THE ARRAY NPOINT WHICH HAS THE VALUE MP. IF AN INVALID VALUE OF MP IS USED, A TRUE ZERO IS RETURNED AS THE VALUE OF GAUSS. ****************************************************************** REAL IXX,IYY,KT,II,JJ,LENGTH,L1,L2,K,KK + + + DIMENSION NPOINT(7),KEY(8),Z(24),WEIGHT(24),K(16,16) COMMON/Cl/NE,NUMNP,NUMEG,NTYPE(3),NUMEL(3),IPAR,ICAL1,ICAL2,ICAL3, ICAL4,ICAL5,1CAL6,ICAL7 COMMON/C4/SE(16,16) COMMON/C5/E(3),G(3),NODEI(36),NODEJ(36),A(36),IXX(36),IYY(36), KT(36),L(1,36) COMMON/C6/A1,A2,MP,Bl(36),B2(36),B3(36) COMMON/C7/RI(36),RJ(36),PHII(36),PHIJ(36),TETA(36),LENGTH(36), RIA(36),RJA(36) COMMON/Cll/DN(16),U(36,12),W(37,8),V(37,8) DATA NPOINT/ 2, 3, 4. 5. 6. 10. 15/ DATA KEY/ 1, 2, 4, 6, 9, 12, 17, 25/ 198 C C.... C 100 C C.... C C C.... C 200 C C.... C CCCC 249 mmwaH 199 ....FIND SUBSCRIPT OF FIRST Z AND WEIGHT VALUE DO 100 I-1,7 IF(MP.EQ.NPOINT(I)) GO TO 200 CONTINUE ....INVALID MP USED GAUSS-0.0 WRITE(61,2000) GAUSS RETURN ....SET UP INITIAL PARAMETERS JFIRST—KEY(I) JLAST-KEY(I+1)-1 C-(A2-A1)/2. D-(A2+A1)/2. ....ACCUMULATE THE SUM IN THE MP-POINT FORMULA IF (IPAR.GE.3) GO TO 543 D0 249 I-1,16 DO 249 J-l,16 K(I,J)-0.0 GO TO 248 DATA 2 / 0.577350269,0.0 ,0 1 0.339981044,0.861136312,0.0 ,0 2 O.906179846,0.238619186,0.661209387,0. 3 0.148874339,0.43339S394,0.679409568,0. 4 0.973906529,0.0 ,0.201194094,0. 5 0.570972173,0.724417731,0.848206583,0. 6 0.987992518 / DATA WEIGHT / 1.0 ,0.888888889,0. 0.652145155,0.347854845,0.568888889,0. 0.236926885,0.467913935,0.360761573,0. 0.295524225,0.269266719,0.219086363,0. 0.066671344,0.202578242,0.198431485,0. 0.166269206,0.139570678,0.107159221,0. 0.030753242 / T-TETA(M) R1=RI(M) R2=RJ(M) L1=R1*T L2=R2*T 774596669, 538469310, 932469514, 865063367, 394151347, 937273392, 555555556, 478628671, 171324493, 149451349, 186161000, 070366047, 543 CCCC 250 248 300 350 360 000000 C 370 C 200 CONTINUE D0 250 I-1,12 DO 250 J-1,12 K(I,J)-0.0 CONTINUE DO 500 J-JFIRST,JLAST I-0 IF (Z(J).EQ.0.) GO TO 350 I-I+1 IF (I.EQ.1) GM-Z(J)*C+D IF (I.EQ.2) GM--Z(J)*C+D GO TO 360 GM-D AA-6.*GM**2-6.*GM BB-3.*GM**2-4.*GM+1. CC-3.*GM**2-2.*GM DD-12.*GM-6. EE-6.*GM-4. FF—6.*GM-2. GG-2.*GM**3-3.*GM**2+1. HH=GM**3-2.*CM**2+GM II--2.*GM**3+3.*GM**2 JJ-GM**3-GM**2 KK-1.-GM R-B1(M)+2.*BZ(M)*T*GM GMSS-(-1./(R**3*T))*(2.*BZ(M)) GMSG-R*T*GMSS ........ CHECK WHICH PART-OF THE STIFFNESS MATRIX IS BEING COMPUTED IPAR-2, COMPUTE ARRAY SE IPAR-3, COMPUTE ARRAY SE1 IPAR-4, COMPUTE ARRAY SE2 GO TO (370,370,390,410),IPAR CONTINUE C ........ INTEGRANDS OF CURVED ELEMENT LINEAR STIFFNESS (SYMMETRIC) C + C1=-E(N)*A(M)*GG/R CZ’(E(N)*IYY(M)/(R**3*T**3))*(DD+AA*GMSS*R**2*T**2) C3-(E(N)*IXX(M)*T/R**3)*(~DD/T**2-GMSS*R**2*AA) C4-G(N)*KT(M)*AA/(R**3*T) C5-(E(N)*A(M)/(R*T))*(AA+T**2*HH) C5‘(E(N)*IYY(M)/(R**3*T**3))*(-T*EE-GMSS*R**2*T**3*BB+T*AA+ GMSG*R*T**2*GG) C7-E(N)*IXX(M)*T*GG/R**2 5"33 3&1 . L" . 51.5}: (‘5': 201 C8=G(N)*KT(M)*AA/(R**2*T) C9--E(N)*A(M)*L1*HH/R ClO-(E(N)*IYY(M)/(R**3*T**3))*(L1*EE+GMSS*R**2*T**2*L1*BB) Cll-(E(N)*IXX(M)*T/R**3)*(L1*EE/T**2+GMSS*R**2*L1*BB) C12--G(N)*KT(M)*Ll*BB/(R**3*T) C13-E(N)*A(M)*RI*BB/(R*T) Cl4-(E(N)*IYY(M)/(R**3*T**3))*(T*R1*BB+GMSG*R*T**2*R1*HH) C15-E(N)*IXX(M)*T*L1*HH/R**2 C16-G(N)*KT(M)*L1*BB/(R**2*T) Cl7--E(N)*A(M)*II/R C18-(E(N)*IYY(M)/(R**3*T**3))*(-DD-GMSS*R**2*T**2*AA) C19--C3 CZO--C4 C21-(E(N)*A(M)/(R*T))*(-AA+T**2*JJ) C22-(E(N)*IYY(M)/(R**3*T**3))*(-T*FF-GMSS*R**2*T**3*CC-T*AA+ GMSG*R*T**2*II) C23-E(N)*IXX(M)*T*II/R**2 CZ4--C8 CZS-(-E(N)*A(M)*L2*JJ)/R C26-(E(N)*IYY(M)/(R**3*T**3))*(L2*FF+GMSS*R**2*T**2*L2*CC) C27-(E(N)*IXX(M)*T/R**3)*(L2*FF/T**2+GMSS*R**2*L2*CC) C28--G(N)*KT(M)*L2*CC/(R**3*T) C29-E(N)*A(M)*R2*CC/(R*T) C30-(E(N)*IYY(M)/(R**3*T**3))*(T*R2*CC+GMSG*R*T**2*R2*JJ) CBl-E(N)*IXX(M)*T*L2*JJ/R**2 C32-G(N)*KT(M)*L2*CC/(R**2*T) SE(l,1)=C1*T*(-GG)+CZ*(DD+AA*GMSS*R**2*T**2) SE(1,2)-0.0 SE(1,4)-0.0 SE(1,6)-0.0 SE(l,8)—0.0 SE(1,10)-0.0 SE(1,12)-0.0 SE(1,14)-0.0 SE(l,l6)—0.0 SE(l,3)-Cl*(AA+T**2*HH)+C2*(-T*EE-GMSS*R**2*T**3*BB+T*AA+ GMSG*R*T**2*GG) SE(I,5)-C1*T*(-L1)*HH+C2*L1*(EE+BB*GMSS*R**2*T**2) SE(l,7)-Cl*R1*BB+C2*(T*Rl*BB+GMSG*R*T**2*R1*HH) SE(1,9)-Cl*(-T*II)-C2*(DD+AA*GMSS*R**2*T**2) SE(l,11)-C1*(-AA+T**2*JJ)+CZ*(-T*FF-GMSS*R**2*T**3*CC-T*AA+ GMSG*R*T**2*II) SE(l,13)-Cl*(-T*L2*JJ)+C2*(L2*FF+GMSS*R**2*T**2*L2*CC) SE(l,15)-Cl*(R2*CC)+C2*(T*R2*CC+GMSG*R*T**2*JJ) SE(2,2)-C3*(-DD/T**2-GMSS*R**2*AA)+C4*AA SE(2,3)-0.0 202 SE(2,5)—0.0 SE(2,7)-0.0 SE(2,9)-0.0 SE(2,11)-0.0 SE(2,13)-0.0 SE(2,15)-0.0 SE(2,4)-C3*R*GG+C4*R*AA SE(2,6)-C3*(L1*EE/T**2+GMSS*R**2*L1*BB)-C4*L1*BB SE(2,8)-C3*R*L1*HH+C4*R*L1*BB SE(2,10)-CB*(DD/T**2+GMSS*R**2*AA)-C4*AA SE(2,12)-C3*R*II-C4*R*AA SE(2,14)-C3*(L2*FF/T**2+GMSS*R**2*L2*CC)-C4*L2*CC SE(2,16)-C3*R*L2*JJ+C4*R*L2*CC SE(3,3)-CS*(AA+T**2*HH)+C6*(-T*EE-GMSS*R**2*T**3*BB+T*AA+ GMSG*R*T**2*GG) SE(3,4)-0.0 SE(3,6)-0.0 SE(3,8)-0.0 SE(3,lO)-0.0 SE(3,12)-0.0 SE(3,14)-0.0 SE(3,16)-0.0 - SE(3,5)-C5*(-T*L1*HH)+C6*(L1*EE+GMSS*R**2*T**2*L1*BB) SE(3,7)-C5*R1*BB+C6*(T*R1*BB+GMSG*R*T**2*R1*HH) SE(3,9)-C5*(-T*II)+C6*(-DD-GMSS*R**2*T**2*AA) SE(3,11)-CS*(-AA+T**2*JJ)+C6*(-T*FF—GMSS*R**2*T**3*CC-T*AA+ GMSG*R*T**2*II) SE(3,13)-C5*(-T*L2*JJ)+C6*(L2*FF+GMSS*R**2*T**2*L2*CC) SE(3,15)-C5*R2*CC+C6*(T*R2*CC+GMSG*R*T**2*JJ) SE(4,4)-C7*R*GG+C8*R*AA SE(4,5)-0.0 SE(4,7)-0.0 SE(4,9)-0.0 SE(4,11)-0.0 SE(4,13)-0.0 SE(4,15)-0.0 SE(4,6)-C7*(Ll*EE/T**2+GMSS*R**2*L1*BB)+C8*(-L1*BB) SE(4,8)-C7*R*L1*HH+C8*R*L1*BB SE(4,10)-C7*(DD/T**2+GMSS*R**2*AA)+C8*(-AA) SE(4,12)-C7*R*II+C8*(-R*AA) SE(4,14)-C7*(L2*FF/T**2+GMSS*R**2*L2*CC)+C8*(-L2*CC) SE(4,16)-C7*R*L2*JJ+C8*R*L2*CC SE(5,5)-C9*(-T*L1*HH)+ClO*(L1*EE+GMSS*R**2*T**2*L1*BB) SE(5,6)-0.0 SE(5,8)-0.0 a: 34 ..h 1.... F... ...I. .1— ....n ..rl. 203 SE(5,10)-0.0 SE(5,12)-0.0 SE(5,14)-0.0 SE(5,16)-0.0 SE(S,7)-C9*R1*BB+C10*(T*R1*BB+GMSG*R*T**2*R1*HH) SE(5,9)-C9*(-T*II)+CIO*(-DD-GMSS*R**2*T**2*AA) SE(S,ll)-C9*(-AA+T**2*JJ)+C10*(-T*FF-GMSS*R**2*T**3*CC-T*AA+ GMSG*R*T**2*II) SE(S,13)-C9*(-T*L2*JJ)+C10*(L2*FF+GMSS*R**2*T**2*L2*CC) SE(5,15)-C9*R2*CC+C10*(T*R2*CC+GMSG*R*T**2*R2*JJ) SE(6,6)-C11*(L1*EE/T**2+GMSS*R**2*L1*BB)+C12*(-L1*BB) SE(6,7)-0.0 SE(6,9)-0.0 SE(6,ll)-0.0 SE(6,13)-0.0 SE(6,15)-0.0 SE(6,8)-C11*R*L1*HH+C12*R*L1*BB SE(6,10)-C11*(DD/T**2+GMSS*R**2*AA)+C12*(-AA) SE(6,12)-C11*R*II+C12*(-R*AA) SE(6,14)-C11*(L2*FF/T**2+GMSS*R**2*L2*CC)+C12*(-L2*CC) SE(6,l6)-C11*R*L2*JJ+CIZ*R*L2*CC SE(7,7)-Cl3*R1*BB+C14*(T*R1*BB+GMSG*R*T**2*R1*HH) SE(7,8)-0.0 SE(7,10)-0.0 SE(7,12)-0.0 SE(7,14)-0.0 SE(7,16)-0.0 SE(7,9)-C13*(-T*II)+C14*(-DD-GMSS*R**2*T**2*AA) SE(7,11)-C13*(-AA+T**2*JJ)+C14*(-T*FF-GMSS*R**2*T**3*CC-T*AA+ GMSG*R*T**2*II) SE(7,l3)-Cl3*(-T*L2*JJ)+Cl4*(L2*FF+GMSS*R**2*T**2*L2*CC) SE(7,15)-Cl3*R2*CC+Cl4*(T*R2*CC+GMSG*R*T**2*R2*JJ) SE(8,8)-C15*R*L1*HH+C16*R*L1*BB SE(8,9)-0.0 SE(8,11)-0.0 SE(8,13)-0.0 SE(8,15)-0.0 SE(8,10)-C15*(DD/T**2+GMSS*R**2*AA)+Cl6*(-AA) SE(8,12)-C15*R*II+C16*(-R*AA) . SE(8,14)-CIS*(L2*FF/T**2+GMSS*R**2*L2*CC)+C16*(-L2*CC) SE(8,l6)-ClS*R*L2*JJ+C16*R*L2*CC SE(9,9)-C17*T*(-II)+018*(~DD-GMSS*R**2*T**2*AA) SE(9,10)—0.0 SE(9,12)-0.0 380 204 SE(9,14)-0.0 SE(9,16)-0.0 SE(9,11)-C17*(-AA+T**2*JJ)+CI8*(-T*FF-GMSS*R**2*T**3*CC-T*AA+ GMSG*R*T**2*II) SE(9,13)-C17*(-T*L2*JJ)+Cl8*(L2*FF+GMSS*R**2*T**2*L2*CC) SE(9,15)-Cl7*R2*CC+C18*(T*R2*CC+GMSG*R*T**2*JJ) SE(lO,10)-C19*(DD/T**2+GMSS*R**2*AA)+C20*(-AA) SE(10,11)-0.0 SE(10,13)-0.0 SE(10,15)-0.0 SE(IO,12)-C19*R*II+C20*(-R*AA) SE(lO,l4)-Cl9*(L2*FF/T**2+GMSS*R**2*L2*CC)+C20*(-L2*CC) SE(IO,16)-Cl9*R*L2*JJ+C20*R*L2*CC SE(ll,11)-C21*(-AA+T**2*JJ)+C22*(-T*FF-GMSS*R**2*T**3*CC-T*AA+ GMSG*R*T**2*II) SE(11,12)-0.0 SE(11,14)-0.0 SE(11,16)-0.0 SE(lI,13)-C21*(-T*L2*JJ)+C22*(L2*FF+GMSS*R**2*T**2*L2*CC) SE(II,15)-CZI*R2*CC+C22*(T*R2*CC+GMSG*R*T**2*R2*JJ) SE(12,12)-023*R*II+C24*(-R*AA) SE(12,13)-0.0 SE(12,15)-0.0 SE(12,14)-C23*(L2*FF/T**2+GMSS*R**2*L2*CC)+C24*(-L2*CC) SE(12,16)-C23*R*L2*JJ+C24*R*L2*CC SE(13,13)=C25*(-T*L2*JJ)+C26*(L2*FF+GMSS*R**2*T**2*L2*CC) SE(13,14)-0.0 SE(13,16)-0.0 SE(13,15)-C25*R2*CC+026*(T*R2*CC+GMSG*R*T**2*R2*JJ) SE(14,14)-C27*(L2*FF/T**2+GMSS*R**2*L2*CC)+CZ8*(-L2*CC) SE(l4,15)-0.0 SE(14,l6)-C27*R*L2*JJ+028*R*L2*CC SE(15,15)-CZ9*R2*CC+C30*(T*R2*CC+GMSG*R*T**2*R2*JJ) SE(15,16)-0.0 SE(16,16)-C31*R*L2*JJ+C32*R*L2*CC DO 380 IE-1,16 D0 380 JE-IE,16 R(IE,JE)-K(IE,JE)+WEIGHT(J)*SE(IE,JE) IF (I.EQ.1) GO To 300 GO TO 500 + 205 CONTINUE ....INTEGRANDS OF CURVED ELEMENT NONLINEAR STIFFNESS SE1 UD=GG*DN(1)+HH*(L1*DN(5)-T*DN(3))+II*DN(7)+JJ*(L2*DN(11) -T*DN(9)) VD-GG*DN(2)-HH*L1*DN(6)+II*DN(8)-JJ*L2*DN(12) WD-KK*DN(3)+GM*DN(9) BD-KK*DN(4)+GM*DN(10) _ UG-AA*(DN(1)-DN(7))+BB*(L1*DN(5)-T*DN(3))+CC*(L2*DN(11) + -T*DN(9)) VG-AA*(DN(2)-DN(8))-BB*L1*DN(6)-CC*L2*DN(12) WG--DN(3)+DN(9) BG--DN(4)+DN(lO) Cl-UG+T*WD C2-WC-T*UD CB-E(N)*A(M)/(R**2*T) SE(l,1)-C3*(-2.*AA*GG*C1+(AA**2*C2)/T) SE(l,2)-C3*(-GG)*VG*AA SE(1,3)-C3*(C1*(-GG*T*(-BB+KK)+(AA*('1-+HH*T**2))/T) +AA*CZ*(-BB+KK)) SE(l,5)-C3*(Cl*L1*(~GG*BB-AA*HH)+(AA*C2*BB*L1)/T) SE(1,6)-C3*GG*VG*BB*L1 SE(l,7)-C3*(C1*AA*(GG-II)-(AA**2*C2)/T) SE(1,8)-CB*GG*AA*VG SE(l,9)-C3*(C1*(-GG*T*(-CC+GM)+(AA*(1.+JJ*T**2))/T) +AA*C2*(-CC+GM)) SE(l,11)-C3*(C1*L2*(-GG*CC-AA*JJ)+(AA*C2*CC*L2)/T) SE(1,12)-C3*CC*GG*L2*VG C4=03/T SE(2,2)—C4*AA**2*CZ SE(2,3)-C4*VG*AA*(-1.+HH*T**2) SE(2,5)-C4*VG*AA*(-HH*LI*T) SE(2,6)-C4*C2*AA*(-BB*L1) SE(2,7)-C4*VG*AA*(-II*T) SE(2,8)--C4*AA**2*CZ SE(2,9)-C4*AA*VG*(1.+JJ*T**2) SE(2,11)-C4*VG*AA*(-JJ*L2*T) SE(2,12)-C4*AA*C2*(-CC*L2) 206 C5-C4*(-1.+HH*T**2) C6-C4*C2*T*(-BB+KK) SE(3,3)—CS*2.*Cl*T*(-BB+KK)+Cé*T*(-BB+KK) SE(3,5)-CS*Cl*BB*L1+C4*Cl*T**2*(-HH*L1)*(-BB+KK)+C6*BB*L1 SE(3,6)-C5*VG*(-BB*L1) SE(3,7)-CS*C1*(-AA)+C4*C1*T**2*II*(BB-KR)-C6*AA SE(3,8)—C5*VG*(-AA) SE(3,9)-CS*C1*T*(-CC+GM)+C4*C1*T*(~BB+KK)*(1.+JJ*T**2) + +06*T*(-CC+GM) SE(3,ll)-CS*Cl*CC*L2+C4*Cl*T**2*(-BB+KK)*(-JJ*L2) + +06*CC*L2 SE(3,12)-CS*VG*(-CC*L2) C7-C4*(-HH*L1*T) C8-C4*BB*L1*CZ SE(S,5)-C7*Cl*2.*BB*L1+C8*BB*L1 SE(5,6)-C7*VG*(—BB*L1) SE(S,7)-C7*C1*(-AA)+C4*C1*BB*L1*(-II*T)+C8*(~AA) SE(5,8)-C7*VG*(-AA) SE(S,9)-C7*C1*T*(-CC+GM)+C4*C1*BB*L1*(1.+JJ*T**2) + +C8*T*(-CC+GM) SE(S,ll)-C7*C1*CC*L2+C4*BB*L1*(-JJ*L2*T)*C1+C8*CC*L2 SE(5,12)-C7*VG*(-CC*L2) C9-C4*(-BB*L1) SE(6,6)—C9*(-BB*L1)*C2 SE(6,7)-C9*VG*(-II*T) SE(6,8)-C9*C2*(-AA) SE(6,9)-C9*VG*(1.+JJ*T**2) SE(6,11)-C9*VG*(-JJ*L2*T) SE(6,12)-C9*CZ*(-CC*L2) SE(7,7)-2.*C4*AA*II*T*C1+C4*AA**2*C2 SE(7,8)-C4*II*T*VG*AA SE(7,9)-C4*(-II*T)*Cl*T*(-CC+GM)-C4*AA*C1*(1.+JJ*T**2) + -C4*C2*T*(-CC+GM)*AA SE(7,11)-C4*(-II*T)*C1*CC*L2+C4*AA*JJ*L2*T*C1-C4*CZ*CC*L2*AA SE(7,12)-C4*II*T*VG*CC*L2 SE(8,8)-C4*AA**2*C2 SE(8,9)-C4*(-AA*VG)*(1.+JJ*T**2) SE(8,11)-C4*AA*VG*JJ*L2*T SE(8,12)=C4*CZ*AA*CC*L2 ClO-C4*(l.+JJ*T**2) 400 C C A C C C C 10 D I O O 207 Cll-C4*T*(-CC+GM)*C2 SE(9,9)-C10*2.*C1*T*(-CC+GM)+C11*T*(-CC+GM) SE(9,11)-ClO*C1*CC*L2+C4*C1*T*(-CC+GM)*(«JJ*L2*T)+C11*CC*L2 SE(9,12)-ClO*VG*(-CC*L2) SE(ll,11)-C4*(-2.*JJ*L2*L2*T*CC*C1+CC*CC*L2*L2*C2) SE(ll,12)-C4*JJ*L2*T*VG*CC*L2 SE(12,12)-C4*02*CC*L2*CC*L2 DO 400 IE-1,12 D0 400 JE-IE,12 K(IE,JE)-K(IE,JE)+WEIGHT(J)*SE(IE,JE) IF (I.EQ.1) GO TO 300 GO TO 500 CONTINUE ....INTEGRANDS ON CURVED ELEMENT NONLINEAR STIFFNESS SE2 WD-KK*DN(3)+GM*DN(9) UG-AA*(DN(1)-DN(7))+BB*(L1*DN(5)-T*DN(3))+CC*(L2*DN(11)-T*DN(9)) VG-AA*(DN(2)-DN(8))-L1*BB*DN(6)-L2*CC*DN(12) C1-E(N)*A(M)*AA/(2.*R**3*T**3) C2-3.*(UG+T*WD)**2+VG**2 CB=2.*VG*(UG+T*WD) SE(1,1)-C1*C2*AA SE(1,2)-C1*CB*AA SE(l,3)-C1*CZ*(-T*BB+T*KK) SE(1,4)-0.0 SE(1,5)-C1*C2*L1*BB SE(1,6)-C1*C3*(-L1*BB) SE(1,7)--SE(1,1) SE(1,8)--SE(1,2) SE(I,9)-Cl*C2*(-T*CC+T*GM) SE(1,10)-0.0 SE(1,11)—C1*C2*L2*CC SE(1,12)-C1*C3*(-L2*CC) C4-3.*VG**2+(UG+T*WD)**2 SE(2,2)-C1*C4*AA SE(2,3)-C1*CB*(-T*BB+T*KK) 'F' 208 SE(2,4)-0.0 SE(2,5)--SE(1,6) SE(Z,6)-C1*C4*(-L1*BB) SE(2,7)—-SE(1,2) SE(2,8)--SE(2,2) SE(2,9)-Cl*c3*(-T*CC+T*GM) SE(2,10)-0.0 SE(2,11)--SE(1,12) SE(2,12)-C1*C4*(-L2*cc) C5-E(N)*A(M)*T*(-BB+KK)/(2.*R**3*T**3) SE(3,3)-CS*CZ*T*(-BB+KK) SE(3,4)—0.0 SE(3,5)-Cs*02*L1*BB SE(3,6)-05*C3*(-L1*BB) SE(3,7)--SE(1,3) SE(3,8)--SE(2,3) SE(3,9)-C5*C2*T*(-CC+GM) SE(3,10)-0.0 SE(3,11)-CS*C2*L2*CC SE(3,12)-C5*C3*(-L2*CC) SE(4,4)-0. SE(4,5)-0. SE(4,6)-O. SE(4,7)-0. SE(4,8)-0. SE(4,9)-0. SE(4,10)-0.0 SE(4,11)-0.0 SE(4,12)-0.0 OOOOOO C6-E(N)*A(M)*L1*BB/(2.*R**3*T**3) SE(5,5)-C6*C2*L1*BB SE(5,6)-C6*CB*(-L1*BB) SE(5,7)-C6*C2*(-AA) SE(5,8)-C6*C3*(-AA) SE(S,9)-C6*CZ*(-T*CC+T*GM) SE(5,10)-0.0 . SE(5,11)-C6*C2*L2*CC SE(S,12)—C6*C3*(-L2*CC) SE(6,6)-C6*C4*L1*BB SE(6,7)--SE(5,8) SE(6,8)--SE(2,6) SE(6,9)--C6*C3*(-T*CC+T*GM) ' .- .' '..' III;- i. . -r n I." ' . ,. '7‘) I I I‘. | . 420 500 C C.... CCCC NH 209 SE(6,10)-0.0 SE(6,11)-SE(5,12) SE(6,12)-C6*C4*L2*CC SE(7,7)-SE(1,1) SE(7,8)-SE(1,2) SE(7,9)--SE(1,9) SE(7,10)-0.0 SE(7,11)--SE(1,11) SE(7,12)--SE(1,12) SE(8,8)-SE(2,2) SE(8,9)--SE(2,9) SE(8,10)-0.0 SE(8,11)--SE(2,11) SE(8,12)--SE(2,12) C8-E(N)*A(M)*(-T*CC+T*GM)/(2.*R**3*T**3) SE(9,9)-C8*C2*(-T*CC+T*GM) SE(9,10)-0.0 SE(9,11)=C8*C2*L2*CC SE(9,12)-CB*C3*(-L2*CC) SE(10,10)-0.0 SE(10,11)-0.0 SE(10,12)-0.0 C9-E(N)*A(M)*L2*CC/(2.*R**3*T**3) SE(ll,ll)-C9*C2*L2*CC SE(11,12)-C9*C3*(-L2*CC) SE(12,12)-C9*C4*L2*CC DO 420 IE-1,12 DO 420 JE-IE,12 R(IE,JE)-K(IE,JE)+WEIGHT(J)*SE(IE,JE) IF (I.EQ.1) GO TO 300 CONTINUE ....MAKE INTERVAL CORRECTION AND RETURN IF (IPAR.GE.3) GO TO 1 IQ=16 GO TO 2 IQ-12 CONTINUE CCCC 550 CCCC CCCC 2000 210 D0 550 I-1,IQ D0 550 J-I,IQ SE(I,J)-C*K(I,J) SE(J,I)-SE(I,J) IF (IPAR.EQ.1) CALL REOCON IF (IPAR.EQ.2) CALL REOCON RETURN FORMAT('I',15HINVALID MP USED///7H GAUSS=,F4.1) END #9 - 5' ‘ i :3” ICHIGRN STQ II III IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII ' . III