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Finite strip analysis of sandwich panels Das, Dhananjoy Kumar 1989

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FINITE STRIP ANALYSIS OF SANDWICH PANELS by Dhananjoy Kumar Das B . Engg. , University of Gauhati , Gauhat i , India M . Engg.(Hons.), University of Roorkee, Roorkee, India A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF T H E REQUIREMENTS FOR T H E D E G R E E OF M A S T E R OF A P P L I E D SCIENCE in T H E FACULTY OF GRADUATE STUDIES CIVIL ENGINEERING We accept this thesis as conforming to the required standard T H E UNIVERSITY OF BRITISH COLUMBIA October,1989 (c) Dhananjoy K u m a r Das, 1989 In presenting this thesis in partial fulfilment of the requirements for an advanced degree at the University of British Columbia, I agree that the Library shall make it freely available for reference and study. I further agree that permission for extensive copying of this thesis for scholarly purposes may be granted by the head of my department or by his or her representatives. It is understood that copying or publication of this thesis for financial gain shall not be allowed without my written permission. Department of C ^ v n . G+J<S, n v r g & g . ) f s T £ , The University of British Columbia Vancouver, Canada Date 11-fo QdC. \*)X°t DE-6 (2/88) Abstract A finite strip analysis of sandwich wood panels is presented. The panels consist of upper and lower plates stiffened by beams (joists) in one direction only. The analysis considers a Fourier series expansion in the stiffeners' direction and a polynomial finite element approximation i n the direction normal to the stiffeners. The number of degrees of freedom is 34, which includes also the lateral and torsional deformation of the joists. This permits consideration of the effect of joist bridging on the maximum deflection and bending stresses. A maximum of 20 joists per panel can be analysed. The modulus of elasticity(E) of the joists may be selected randomly from a distribution, and controlled to be within a given range. The upper and lower plates may possess orthotropic properties. Nails connect the plates to the joists. Nai l ing may be considered either as a continuous or as discrete connectors. The loading may be in the form of an uniformly distributed load spread over the entire panel or over a maximum of 20 smaller (concentrated) areas of the top plate. Numerical investigations have been carried out to verify the program. Parametric studies have been done to understand the behavior of the model. Lastly, the formula for shear lag given by C S A Standard CAN3-086.1-M84 is checked against the shear lag obtained from the current computer program. Contents Abstract ii Tables iii Figures iv Notat ion v i Acknowledgement ix 1 Introduction and Literature Review 1 1.1 General Remarks 1 1.2 The Model and The Existing Methods . . . 2 1.2.1 The Model 2 1.2.2 The Existing Methods: 3 1.2.2.1 Analysis of Box-Girders as Thin-Walled Beams . . . 3 1.2.2.2 Analysis of Box-girders as Folded Plate Structures . 4 1.2.2.3 Plate Idealization of Box Structure 5 1.2.2.4 Grillage Idealization 5 1.2.2.5 Finite Element Methods 6 1.2.2.6 Finite Strip Method 6 1.2.3 Objectives of the Present Study 9 2 Theoretical Formulation 10 2.1 The Model 10 iii 2.2 Derivation of the Stiffness Matrix 14 2.2.1 Strain Energy in the Upper Plate 15 2.2.2 Strain energy in the lower plate 18 2.2.3 Strain Energy in the Joist 20 2.2.4 Strain Energy in the Upper Plate Connectors 21 2.2.5 Strain Energy in Lower Plate Connectors 23 2.2.6 Stiffness Matrix and the Solution of the System of Equations . 24 3 T h e Computer Program 28 3.1 Program Features 28 3.2 Program Structure 30 4 Verification and Numerical Results 31 4.1 Introduction 31 4.2 Verification 31 4.2.1 Verification for flange stresses in the longitudinal direction: . . 31 4.2.2 Verification by comparison with FAP: 33 4.3 Parametric Study 36 4.3.1 Effect due to change in nail stiffness 37 4.3.2 Effect due to the change in the ratio of plate thickness to the depth of the joist: 38 4.3.3 Effect due to the change in the ratio of spacing between the joists to the panel span : 40 4.4 Study of the Shear Lag Effect . . . 50 4.4.1 Effect of the number of longitudinal ribs 52 5 Conclusions and Scope for further research 61 5.1 Conclusion 61 5.2 Scope for further research 62 iv Bibl iography 63 A 67 A . l Shape Functions: 67 A . 1.1 Vector {M0} 67 A.1 .2 Vector { M 3 } 67 A.1 .3 Vector { M 5 } 68 A . 1.4 Vectors {Mx}, { M 2 } , { M 4 } , and {M6} 68 B 69 B . l Shape Functions: 69 B . l . l Vector {M7} 69 B.1.2 Vector { M 1 0 } 69 B.1.3 Vector { M 1 2 } 70 B.1.4 Vectors { M 8 } , { M 9 } , { M u } , and { M 1 3 } 70 C U S E R ' S M A N U A L - F A P P 71 C . l Input 71 D P R O G R A M L I S T I N G - F A P P 83 v Tables C. l Table of Boundary Condition Codes v i Figures 1.1 Wood Panel Assembly 2 2.1 Nodes for I - beam finite element 11 2.2 Joist degrees of freedom 12 2.3 Flange area of a finite strip element 17 2.4 B(i , i) submatrix in the Global stiffness matrix 25 4.1 Element with fixed connections 33 4.2 Joist deflection vs. Nail stiffness 41 4.3 Joist Bending stress vs. Nail stiffness 42 4.4 Stress at top layer of upper plate in x-direction at varying nail stiffnesses 43 4.5 d/HJT vs. Joist Deflection at varying nail stiffnesses 44 4.6 d/HJT vs. Joist Bending stress at varying nail stiffnesses 45 4.7 d/HJT vs. Stress at top layer of upper plate in x-direction at varying nail stiffnesses 46 4.8 d/HJT vs. Joist Deflection at varying s/Lp 47 4.9 d/HJT vs. Joist Bending stress at varying s/Lp 48 4.10 d/HJT vs. Stress at top layer of upper plate in x-direction at varying s/Lp 49 4.11 Probabilitiy of non-conservative XQ at s/Lp = 0.05 55 4.12 Probabilitiy of non-conservative XQ at s/LP = 0.15 56 4.13 Probabilitiy of non-conservative XQ at s/Lp = 0.25 57 4.14 Probabilty levels of XQ at d/HJT = 0.075 58 4.15 Probabilty levels of XG at d/HJT = 0.05 . . . 59 vii 4.16 Panel geometry reduction factor XQ at varying s/LP 60 C . l Correlation between joist M.O.R. and its Modulus of Elasticity . . . . 79 C.2 Loading Area Coordinates 80 C.3 The Nodes of an element 81 viii Notation u axial displacement of plate v lateral displacement of plate w vertical displacement of plate U axial displacement of joist V lateral displacement of joist W vertical displacement of joist. 6 angle of rotation of the joist uin axial displacement at node 1 of the finite-strip element for the nth term of the Fourier series vin lateral displacement at node 1 of the finite-strip element for the nth term of the Fourier series win vertical displacement at node 1 of the finite-strip element for the nth term of the Fourier series u'ln derivative of uXn with respect to x. Similarly u'3n, u'4n k u6n v'ln derivative of vin with respect to x. Similarly v'3n, v'4n & v'6n w'ln derivative of n>ln with respect to x. Similarly w'2n, w'3n, w'4n & w'5n, w'6n £, rj normalised coordinates Mo shape function. Similarly M 1 ? M 2 , M 1 3 F\n product of shape function and deformation vector. Similarly F 2 n ,F 3 „, F6n ix KX,KX\ flexural stiffness in the x direction for upper &; lower plates Ky,Kyi flexural stiffness in the y direction for upper & lower plates KV,KV\ coefficient associated wi th Poisson's ratio for upper & lower plates Dx,Dxi axial stiffness in x direction for upper &; lower plates Dy,Dyi axial stiffness in y direction for upper &; lower plates DV,DV\ coefficient associateds wi th Poisson's ratio for upper &; lower plates DG, DQI In plane shear stiffness for upper and lower plates Ex, Exi modulus of elasticity of upper & lower plates in x direction Ey, Eyi modulus of elasticity of upper & lower plates i n y direction vxy Poisson's ratio, strain i n x direction when stress is applied in y direction (for upper plate) vvx Poisson's ratio, strain in y direction when stress is applied in y direction (for upper plate) G, Gl shear modulus in x — y plane for upper & lower plates Vxy\ Poisson's ratio, strain in x direction when stress is applied in y direction (for lower plate) VyX\ Poisson's ratio, strain in y direction when stress is applied in y direction (for lower plate) Uu strain energy in the upper plate of one strip element UL strain energy in the lower plate of one strip element Uj strain energy in the joist Use strain energy in the upper plate connectors UNL strain energy in the lower plate connectors U total strain energy in one T-beam strip element «5n deformation vector for one T-beam strip element corresponding to the n-th Fourier term Afc global deformation vector for a l l the elements corresponding to the fc-th Fourier term Rk load vector corresponding to the fc-th Fourier term e tolerance factor NJT number of joists d thickness of the upper plate HJT depth of the joist s spacing between the joists Lp panel span XQ panel geometry reduction factor to account for shear lag. x i Acknowledgement In India there is a saying that even i f you learn a single word from someone you cannot repay the debt of acquiring the knowledge in one lifetime. If that be the case, I believe that I w i l l have to take rebirth a couple of times to pay off my debt to my supervisor Professor Ricardo 0. Foschi. I wish to express my deepest gratitude to Prof. Foschi for his invaluable advice and guidance throughout the research work and in the preparation of this thesis. I sincerely feel myself to be extremely fortunate to be able to work under his supervision. I am thankful to Professor Borg Madsen for checking the manuscript of this thesis. I would also like to thank Rajesh Chandra for his support and valuable suggestions when I needed them most. Final ly , the financial support from the Natural Sciences and Engineering Research Counci l of Canada is gratefully acknowledged. Financia l support in the form of a Research Assistantship from the Natural Sciences and Engineering Research Counci l of Canada is gratefully acknowledged. x i i C H A P T E R 1 Introduction and Literature Review 1.1 General Remarks In the early ages of civil ization the existence of stiffened structural forms was perhaps learnt from the great book of nature. Sea shells, leaves and trees - a l l of these are, in fact stiffened structures. The observation of such structures created by nature indicate that strength and rigidity depend not only on the material but also on the form. This was realised by the Egyptians at least around 3000 B .C. and they developed a craft made of planks fastened around a wooden framework. Also the ancient V i k i n g ships were made of planks which were tied on the inside to ribs. However, due to l imited number of strong materials and l imited knowledge about them, the development of stiffened structural elements was very much restricted. Thei r wide use began in the nineteenth century , mainly wi th the application of the steel plates for hulls of ships and wi th the development of the steel bridges and aircraft structures. In addition to the applications above, stiffened plates are also widely used in the shape of ribbed, waffle and sandwich type slabs for floor and roof construction in buildings. The rigidity of the plates is obtained either by various types of stiffeners reinforcing thin sheets or by the geometry of the form of these sheets, as i n the stressed skin panels for the walls of buildings. Very often such sheets derive their stiffness from a corrugated geometry. 1 Chapter 1: Introduction and Literature Review 2 1.2 The Model and The Existing Methods 1.2.1 The Model The type of wood-panel system considered i n the present study is shown i n F i g . 1.1. T w o covers, also called sheathing, are fastened (nailed) to the wood joist at the top and bo t tom. Th i s assembly behaves as a stiffened plate under the action of load. Figure 1.1: Wood Panel Assembly Chapter 1: Introduction and Literature Review 3 The cover may be considered as an orthotropic plate because it may have different modulus of elasticity i n the directions parallel and perpendicular to the joist. The panels making up the cover may have gaps between them which w i l l influence the behavior of the structure. Each of the joists may have different modul i of elasticity because of wood's natural variability, and the fastener's properties w i l l depend on a large number of parameters, some of which are the type of the fastener, nai l diame-ter, type of the nai l coating, depth of penetration of the nails and direction of nail penetration in relation to the grain direction in the joist. A simple way of analysing this structure is to assume that the applied load is carried by the joists only, thereby not taking into account any composite action in the assembly. This approach may be correct i f the bending stiffness of the covers is very small relative to that of the joists and the nai l interval is large. However in realistic conditions, such analysis cannot be proved justified. The model presented here considers the structure as a single/multi cell box-girder deck. Let us at first look into the analytical finite element and finite-strip methods available for solving the problem. 1.2.2 The Existing Methods: 1.2.2.1 Analysis of Box-Girders as Thin-Walled Beams In this method, the actual thin walled space structure is regarded as a single beam. The application of the method is advantageous i f the girder cross-section is not too complicated (single or double box cell) and i f the cross-sectional dimensions are small i n relation to the span. The girder may have a variable cross-section( such as vari-able thickness of webs, or also the height), the various parts of the structure may be made of materials having different properties, and the static system may be fairly Chapter 1: Introduction and Literature Review 4 complex (continuous beam, framework). The solution is based on a number of sim-plifying assumptions and belongs therefore into the group of methods based on the ordinary theory of beams. In accordance with the accepted assumption concerning the small dimensions of the girder cross section in relation to the length of the span, the transverse bending and the torsional effects of the separate walls of the structure may be neglected (one-way action) and a simple longitudinal bending of cross-section assumed. 1.2.2.2 Analysis of Box-girders as Folded Plate Structures If the box-girder is of constant cross-section and i f its end cross-sections, perpendicular to the girder axis , are transversely stiffened against the deformation of their shape, i t is very advantageous to analyse i t as a folded plate structure. Constant cross-section of the structure is the basic prerequisite for the direct application of this method. In some cases of exceptional arrangements , continuous structures of variable cross-section can be analysed also; for the solution of general folded-plate structures of variable cross-section, the iterative procedure is suitable. In this procedure the complexity of the cross-section as well as the ratio of the cross-sectional dimensions to the span are nearly immaterial , because the solution is not based on ordinary girder principles, but on the elasticity plane stress and the plate bending theory. W i t h i n the scope of assumptions of the theory of elasticity, the folded plate theory represents an exact method, because it considers the structure in its actual form as an assembly of rectangular plate-shaped parts forming together a real spatial system. Hence the solution does not make any difference between open and closed cross-sections on the multi-cell structures, nor does differentiate the various kinds of stresses (bending, shear, torsion) ; the loading of the structure may be quite general (horizontal, vertical, longitudinal, arbritrarily distributed on any surface and/or linear, continuous moment loading, concentrated forces etc. ). Chapter 1: Introduction and Literature Review 5 1.2.2.3 Plate Idealization of Box Structure B o x structures, particularly when they are of a massive character represent a transi-tion to plate structures with longitudinal hollows between them, and for this reason the plate model may be used for an approximate calculation. The simplest way of modelling the actual structure is to substitute it by a solid isotropic plate whose thickness can be determined so that the flexural rigidity of plate corresponds to that of the actual structure. Thus the bending behavior of the structure as a whole is relatively well expressed, but the torsional behavior less precisely, and the transverse distribution of the load is very much over estimated in a more thin-walled structure without diaphragms, because the transverse rigidity of the actual structure is formed almost solely by the frame stiffness of the cross-section. Hence this substituton is suitable particularly for the analysis of the effects of a not pronouncedly concentrated loading. Better results are obtained by using an orthotropic plate for substitution , or, bet-ter s t i l l , a sandwich plate composed of three layers : the marginal layers representing the effect of the lower and the upper flanges , and the inner (core) layer transferring only the shear effects which represent the shear and transverse frame behavior of the structure. 1.2.2.4 Grillage Idealization For structures of a greater width composed of a large number of girders, which by themselves do not have a pronouncedly thin-walled character , which may be con-nected by a slab, and which are moreover often skew, a solution according to the folded plate theory or by other methods may be too complicated. In this case very good results are obtained by regarding the actual structure as a grillage. Good re-sults can also be arrived at by using Hamberg's proceeding (Ref. 13) where the individual box-girders , on the basis of their stiffnesses i n bending and torsion, are Chapter 1: Introduction and Literature Review 6 assigned distributing numbers which are then used in the analysis of the grillage by the harmonic-load method. 1.2.2.5 Finite Element Methods For the type of problem we are dealing with, either a 2D or a 3D finite element model can be used. A simpler form of two dimensional analysis is the grillage analogy and attempts have been made to employ this method incorporating the effects of the transverse cell distortion in a cellular structure (Ref. 1,14,16,23). A formulation with shear-weak plate elements is also available (Ref. 7). In this case the shear stiffness of the plate is assumed to represent the distortional stiffness of the actual cell. But this approach does seem suitable for single-cell or twin-cell structures (Ref. 8,25). An alternate way in which an attempt has been made to utilise the advantages of the three-dimensional approach whilst at the same time reducing the number of degrees of freedom per element (and thus reducing the overall size of the problem to an equivalent two-dimensional model) is by making suitable simplifying assumptions, for example, omitting the local bending deformations at each node {9x,9y,9z) and retaining only the three translational degrees as freedom (u, v, w) (Ref. 24) . The use of shear web element coupled with the elimination of the rotational degrees of freedom for nodes associated with the flanges can also lead to an equivalent two-dimensional model of a cellular structure (Ref. 12) To formulate a three-dimensional model, quadrilateral or traingular elements are used (Ref. 10, 15 , 26, 29) 1.2.2.6 Finite Strip Method The Finite element method is known as the most powerful and versatile tool of so-lution in structural analysis. However, for many structures having regular geometric Chapter 1: Introduction and Literature Review 7 plans and simple boundary conditions, a full finite element analysis is very often un-necessary. For this reason, the finite strip method was developed. In this method the structure is divided into two dimensional (strips) or three-dimensional (prisms , layers) subdomains in which one opposite pair of sides (2D) or one or more opposite pairs of faces (3D) of such a sub domain are in coincidence wi th the boundaries of the structure. The finite strip method can be considered as a special form of the finite ele-ment procedure using the displacement approach. Unl ike the standard finite element method, which uses polynomial displacement functions i n a l l directions, the finite strip method calls for the use of simple polynomials in some directions and contin-uously differentiable smooth series in the other directions, wi th the stipulation that the series should satisfy a priori the boundary conditions at the end of the strips or prisms. The general form of a displacement function is given as a product of polyno-mials and series. Thus for a strip, in which a two dimensional problem is reduced to a one-dimensional problem, ™= £ fm(y)Xm (1.1) m=l where, fm(y) is the polynomial expression wi th undetermined constants for the mth term of the series and Xm is the series which satisfies the end conditions and deflected shape in the x-direction. Cheung (Ref. 3 , 4) formed a rectangular flat strip by combining a bending strip and a plane stress strip. There are four displacement components (u, v, w, 9) at each end. He analyzed a right box-girder bridge wi th this element. Loo and Cusens (Ref. 18) have demonstrated that improved accuracy can be obtained by using the auxiliary nodal line ( A N L ) procedure in analyzing different internal stresses i n box-girder structures. They also suggested that there is no justifi-cation in using more than one A N L per strip. The A N L technique allows higher order functions to be incorporated in the finite-strip formulation but leaves the bandwidth Chapter 1: Introduction and Literature Review 8 of the overall matrix equation unchanged. The 'plate-strip elements' developed by the authors mentioned above were based on Kirchoff's plate theory. However, when the members of the structure cannot be classified anymore as 'thin' within the context of Kirchoff's hypothesis , the Kirchoff's plate theory may produce inaccurate numerical results. Mindlin's plate theory (Ref. 20) which takes into account the effect of transverse shear deformation (thus making it valid for thick plate cases) has been extended by Benson and Hinton (Ref. 2) in the context of the finite strip method for the static and dynamic analysis of plates using the 3-noded strip element. Onate (Ref. 21) extended the same to deal with straight box-girder bridges. Onate and Suarez (Ref. 22) have shown that plates, bridges and axisymmetric shells of uniform transverse cross-section can be treated in a simple and unified form using two-noded Mindlin linear strip elements with a single quadrature integrating point. Cheung and Fan (Ref. 5) have done static analysis of right box-girder bridges by the spline finite-strip method, in which the function series which satisfies the end conditions a priori in the longitudinal direction has been replaced by a spline function. The converged answers are nearly correct when compared with experimental results and finite element solutions. Thomson, Goodman and Vanderbilt (Ref. 27) presented a paper which considers a model with a top cover and joists underneath. The paper takes into account the composite action between the cover and the joists. The floor is assumed as a system of crossing beams, with a series of T-beams and sheathing strips perpendiucular to these T-beams. Each of the T-beams represent a joist with a portion of 'contributory' cover, which may contain a few layers of sheathing. A finite element analysis is done by further subdividing the T-beams and the sheathing strips into smaller elements and satisfying the compatibility requirement in the form of defiormation at the points of intersection. A computer program FEAFLO based on this model has been widely used Chapter 1: Introduction and Literature Review 9 i n simulating the floor behavior under different levels of variability in joist properties (Ref. 28). The program takes into account the slip between the cover and the joists due to the nai l deformation. In the paper mentioned above, the width of the T-beam flange, which is similar to the idea of effective width, is difficult to estimate beforehand. This is due to the fact that it is dependent upon the ratio of floor span to joist separation. When the ratio is large, the effective width tends to become equal to the joist spacing and this is the value often used in the F E A F L O model. This approach may not be accurate for short spans wi th comparatively wide opening for joists. Moreover an effective width is always defined wi th an objective, for example, to equate the deflection of the T-beam wi th the actual deflection of the stiffened structure. In general, a different effective wid th would be obtained if the objective is changed to equality of maximum bending stresses. In view of these setbacks, Foschi (Ref. 11) has proposed an improvement in modelling by using the finite strip model implemented in the program F A P . A linear behavior for fastener's has to be assumed. In the present study Foschi's model has been taken into consideration except that, as mentioned i n A r t . 1.2.1, a bot tom cover has been fastened to the joists. 1.2.3 Objectives of the Present Study The objective of the study is to look into the behavior of the stiffened structure, with double (top and bottom) covers, and to compare the design provisions for Stressed Skin Panels as given in the C S A Standard CAN3-086 .1 -M84 "Engineering Design in W o o d (L imi t States Design)". In this context the verification of the expression for the shear lag effect, as given by Clause 8.6.3.2 of that Code, is to be done and its dependence on the number of longitudinal ribs studied. C H A P T E R 2 Theoretical Formulation 2.1 The Model A short general description of the finite strip method has been given i n the previous chapter. The panel is modelled as consisting of I - beam finite strips i n the x-direction. The nodes considered for each finite strip element are shown i n the F i g . 2.1. The deformations of the plate in the y - direction are approximated by a one-dimensional finite element using the nodes shown in the F i g . 2.1. In the x - direction (longitudinal), a Fourier series approximation is employed. Using this model, the deflections of the plates and the joist can be expressed as : Upper plate displacements (middle surface) : (2.1) (2.2) (2.3) 10 f Chapter 2: Theoretical Formulation 11 1 o Figure 2.1: Nodes for I - beam finite element where N is the number of terms taken in the Fourier series. Since w = v = 0 at x = L , the assumed functions imply simply supported boundary conditions at the ends. The functions Fin(y) , •f72n(y) and Fzn(y) are determined from a finite element polynomial approximation in y - direction. These functions are expressed in terms of the nodal degrees of freedom at points 1, 2 and 3 in Fig. 2.1. Joist displacements: The displacement functions for joists are expressed as follows N Vertical displacement: W(x) = £ Wnsin^~- (in z - direction) (2.4) n=l N Axial displacement: U(x) = £ t/ncos^ y^ (in x - direction) (2.5) N Lateral displacement: V(x) = £ Vns'm—j- (in y - direction) n=l (2.6) Chapter 2: Theoretical Formulation 12 Figure 2.2: Joist degrees of freedom N Rota t ion: 0(x) = £ 0 n s i n T17TX (2.7) n= l where Wn , Un , Vn and 6n are joist degrees of freedom as shown i n F i g . 2.2. Lower plate displacement ( middle surface ): N Vert ica l displacement: w(x,y) = ^ F 4 n ( y ) s i n nirx (2.8) (z - direction) n= l N A x i a l displacement: u(x,y) = i 7 1 5 n (y)cos IXTTX (2.9) (x - direction) n= l N Lateral displacement: v(x,y) = y] F 6 n ( y ) s i n 717TX (2.10) n= l (y - direction) Chapter 2: Theoretical Formulation 13 Here, again, the functions F4n(y) , FSn(y) and Fen(y) are determined from a finite element polynomial approximation in y - direction. These functions are expressed in terms of the nodal degrees of freedom at points 4 , 5 and 6 i n F i g . 2.1. For each strip, a total of 34 degrees of freedom are identified associated wi th each n (term in Fourier series). The elemental nodal degrees of freedom vector is shown below. For the n-th Fourier term there are six degrees of freedom at each of the {>ln w'lnS Uln UlnS Vln VlnS W4n w's U 4 n UAnS V4n v's U2n V2n W2n... Wn Un Vn 9 u'6ns v6n v'6ns} nodes 1 , 3 , 4 and 6 , three at node 2 and 5 and four node B . For example , w'ln is the component denoting the derivative of w(x,y) wi th respect to y at node 1 , multiplied by the joist spacing V , to make the vector dimensionally consistent. Using the shape functions shown i n the Appendix A , and the non-dimensional variable £ = 2y/s , the functions Fln(y), F2n(y), F3n(y), F4n(y), FSn and F 6 n ( y ) can be writ ten i n terms of {Sn} : Fm(0 = {Mo(0) T{6n} (2.11) F2n(0 = { M 3 ( 0 } T { U (2.12) F3n(0 = {Ms(0} T{Sn} (2.13) F4n(0 = W O f R } (2.14) {*«} = F5n(0 = {M10(0} T{6n} (2.15) Chapter 2: Theoretical Formulation 14 *en(0 = W 2 ( 0 ) T { < U (2.16) Furthermore, these equations permit us to obtain the following derivatives : = {Afi(0>T{M (2-17) d?Fln(() {M2(t)}T{6n} (2.18) d( ^f1 = W ( 0 } T { U (2-19) = {M«(0>T{^} (2-20) dF4n(Q {MsU)}T{Sn} (2.21) {M9(0}T{Sn} (2.22) = Wi(0}T{U (2-23) = {M13(0}T{Sn} (2.24) W i t h the displacement functions and their derivatives thus defined , the strain energy of a strip element can be computed and the stiffness matr ix derived as explained in the next section. 2.2 Derivation of the Stiffness Matrix The stiffness matr ix is obtained from the total strain energy of the system . In order to obtain the total energy , we have to consider contributions from the various components of the floor system i.e. the plates, the joists and the connectors between the joist and the plates. Chapter 2: Theoretical Formulation 15 2.2.1 Strain Energy in the Upper Plate Using the small deflection orthotropic plate theory , the strain energy per unit area of the upper plate can be expressed as: + 2KG(^f + (Dx/2)(^)2 + (DY/2)(^)> + Dv{duldx){dvldy) + DG/2 {(du/dy) + (dv/dx)}2 (2.25) KX = flexural stiffness i n the x - direction per unit length Exd? 12(1 - VXyVyX) Ky = flexural stiffness in the y - direction per unit length KV = coefficient associated wi th Poisson's ratio = vxyKx d 3 KQ = torsional stiffness = G— Dx = axial stiffness i n x direction = Exd (1 - VxyVyx) E Dy = axial stiffness in y -direction = Dx—^-Ex Dv = coefficient associated wi th Poisson's ratio = vxyDx DG = In plane shear stiffness = Gd Ex = Modulus of elasticity in x - direction Ey = Modulus of elasticity in y - direction Vxy vyx = Poisson's ratio, strain i n x direction when stress is applied i n y direction = Poisson's ratio, strain in y direction when stress is applied in x direction Chapter 2: Theoretical Formulation 16 G = Shear modulus i n x-y plane The total strain energy, Uu in the upper plate finite strip can be obtained by inte-grating over the area as shown F i g . 2.3. Thus the upper plate strain energy can be expressed as : Uu = £ £ NJU dxdy (2.26) O n substituting the expressions for displacements given by eqns. (2.1) to (2.3) into eqn. (2.25) and integrating over the span length , we obtain, Um = £ (KJ2) {^fdx = (Km/2) £ F?n(y) ^ (2.27) a result i n which the orthogonality of the trigonometric functions has been used. Similar ly , Uu2 = £ (Ky/2) (§^)2dx = (Ky/2) £ (F"ln (y)) 2 (L/2) (2.28) Um = i ""(j^ O = ~K" £ F l " ( ! , )  F > M I T ( 2-29 ) V m - 2 K ° t & d * ' 2 K° | TT <2-30» Um = I (Dx/2) ( | | ) 2 dx = (Dx/2) £ (F2n(y)Y fj- (2.31) Uu* = fL (Dy/2) ( §^ ) 2 dx = (Dy/2) £ (F3n (y))2 (L/2) (2.32) J 0 "ii n=l UU7 = fL Dv ( d x = -D„ £ F2n (y) F3n (y) ( ^ ) (2.33) Jo ox if y ~L I Vm = / 1 ( C o / 2 ) ( ^ + ^ ) ^ x Jo dy ox = (Do/2) £ [ Fi (y) + (n r/L)F3n (y) ] 2 L/2 (2.34) Chapter 2: Theoretical Formulation 17 L t -mmmrnmmm •t mmmmmmt I wmmmm i 1 t * s . ' * — < i i i i i i l l i l ;x i l l l i l l l l ••••I Y G a p d I h— s —"I y Cover t 1 t „ h — 1 1 ; | S J o l s * ^ Z Figure 2.3: Flange area of a finite strip element Substi tut ing the expressions for the functions F i„ ( y ) , F2n(y) and i ^ ^ y ) and their derivatives, as given by eqns.(2.11) to (2.13) and eqns. (2.17) to (2.20) , integrate over the joist spacing , and we get Uvi = E( K^ 2)jJFs £ {<U T {Mo (£)} {Mo (Of R } di (2.35) Uu2 = E ( ^ / 2 ) ^ f {6n}T{M2(0}{M2(0}T{6n}d( (2.36) n=l 3 Chapter 2: Theoretical Formulation 18 Um = £ (-/C/2) I {Sn} T {Mo ( 0 ) {M2 (Of R } dt (2.37) ^ 4 = E 2 ^ / ' {<« T {Ma ( 0 ) W (Of {Sn} dt (2.38) Efo = £ (Dm/2) T R } T {M3 (0} {M3 (Of {6n} dt (2.39) n=l * ^  J _ 1 Uue = E(D9/2)L/s £ {6nf {M6 (t)} {M6 (Of {6n} dt (2.40) = £ (-A, » */2) lX {6nf{M3(t)}{M6(0f{6n}dt (2.41) n=l - 7 - 1 = J2(DG/2)(sL/4) f {6nf(2/s{M4(t)} + (n ir/L) {M5(t)}) n=l + (2/5 {M4 ( 0 } T + (n *IL) {M5(t)f) {6n} dt (2.42) Finally, the strain energy in the upper plate of one strip element is : Uu = Uui + Uu2 + Uus + Uu4 + UU5 + Uue + UU7 + UU8 (2.43) 2.2.2 Strain energy in the lower plate Using the same small deflection orthotropic plate theory , the strain energy of the lower plate per unit area can be expressed as : / r , !~s,d 2w.~ ,_, In,,d 2w^ .s.d 2w..d 2w. = ^ - / 2 ) ( ^ ) 2 + ( ^ / 2 ) ( ^ ) 2 + (K„)(j^)(w) + 2 KG, ( | ^ ) 2 + (Dxl/2) ( g ) 2 + (DYl/2)(^) 2 + D^du/dx) (dv/dy) + DGI/2 {(Ou/dy) + (dv/dx)} 2 (2.44) where: Kxi = flexural stiffness in the x - direction per unit length Exld* 12(1 - VxylVyzl) Kyi = flexural stiffness in the y - direction per unit length Chapter 2: Theoretical Formulation 19 KV\ = coeffient associated wi th Poisson's ratio (2-45) KGI = torsional stiffness = Gl-jr X & Dxi = axial stiffness in x direction = ~^ x l 1 (1 - vxyXvyxX) Dyi = axial stiffness in y -direction = Dx\ -=^-D„i = coeffiecient associated wi th Poisson's ratio = vxy\Dx\ DQ\ = In plane shear stiffness = G\d\ Exi = Modulus of elasticity in x - direction Eyi = Modulus of elasticity in y - direction Vxyi = Poisson's ratio, strain in x direction when stress is applied in y direction vyx\ = Poisson's ratio, strain in y direction when stress is applied in x direction G\ = Shear modulus i n x-y plane Now the total strain energy in the lower plate finite strip can be obtained by integrating over the same area as shown in F i g . 2.3 for the upper plate. Thus the strain energy in the lower plate can be expressed as : UL = T / 2 jL A UL dx dy (2.46) J-s/2 JO O n substituting the expressions for displacements as given by eqns. (2.8) to (2.10) into eqn. (2.44) and integrating over the span length and thenafter substituting the expressions for functions F4n (y) , F5n (y) and Fen (y) and their derivatives as given by eqns. (2.14) to (2.16) and eqns. (2.21) to (2.24) and at last integrating over the Chapter 2: Theoretical Formulation 20 joist spacing , we get , * „4 IT* ri UL1 = £ j (KTL/2) jjT » £ {*n}r {Mr (0) {M? (Of R} d£ (2.47) UL2 = f {^f {M9 (0) {M9 (Of {SN} d^ (2A8) n=l 5 %s = E ( - ^ i / 2 ) ^ ^ / ' R} T W (0 ) {M9 ( 0 } T R} (2.49) 71=1 S -L ./ —1 Uu = £ 2 ^oi ^ T R ) T W (0 ) {Ms (Of R} (2.50) n=l 5 ^ C/L5 = E ( X W 2 ) ^ ^ f R} r Wo (0) {M10 (Of R) if (2.51) n=l * ^ ^L6 = J2(DYL/2)L/s f 1 {6nf {M13 (0) {M13 (Of {6n}d£ (2.S2) n=l % 7 = E (-Ain*/2) /' R} T { M 1 0 ( 0 ) « 3 (Of R} * (2-53) n=l J - 1 UL8 = E ( U 0 1 / 2 ) ( » i / 4 ) / * { « » } r ( 2 / 3 { A f „ ( 0 } + (n7r/L){M 1 2(£)}) n=l + (2/a {Mn (Of + (n */L) {M12(0f) R} (2.54) Thus the strain energy in the lower sheathing of one element of one element is : UL = ULi + UL2 + ULz + ULA + ULs + UL6 + UL7 + UL% (2.55) 2.2.3 Strain Energy in the Joist Strain energy in the joist consists of two flexural components (one being bending in the vertical plane and the other, lateral bending in the horizontal plane) , an axial component and a torsional component . The strain energy is thus expressed as + EA/2 1f (^-fdx + ~ [L IM/dx)' dx (2.56) Jo dx 2 Jo Chapter 2: Theoretical Formulation 21 where , for a rectangular cross-section, _ bjf_ v " 12 _ htf 2 ~ 12 A = bh It = 13 (h/b) h b 3 and B being the joist width , h the joist depth and /? the torsional constant function of the ratio h/b Using the expressions for W ( x ) , U ( x ) , V ( x ) and 0(x) as given by eqns. 2.4 to 2.7 , the strain energy in the joist is written as: N n 4 it 2 N r> 4 v 4 Uj = EIJ2 + EIz/2 Y , V ^ + E A/2 £ £/*  r-7fj- + G It/2 £ e2n^f- (2.57) 2.2.4 Strain Energy in the Upper Plate Connectors It is assumed here that the upper plate is connected to the joist by uniformly placed non-rigid fasteners along the centerline of the top surface of the joist . The model assumes three types of slips between the plate and the joist. These are : 1. Sl ip parallel to the joist = A u = K - rf/2(^)] - + ft/2 ( ^ ) ] (2.58) 2. Slip perpendicular to the joist = A v = [Vo-d/2(^)} - [V + (h/2) 8} (2.59) 3. Rotat ional Slip = <j> = (2-60) Chapter 2: Theoretical Formulation 22 where d is the thickness of the upper plate , and the displacements u0 , v0 and w0 correspond to node 2 in the upper plate's mid-plane directly over the joist , as shown in the Fig. 2.1. In the above eqns. (2.57) to (2.59) it has been assumed that the vertical displacement of the upper plate w0 and the displacement W of the joist are identical . Letting Km , Kny represent single connector stiffnesses corresponding to the x and y directions , NA is the number of nails per joist , (A u); is the slip in the x direction for the i-th connector , (A t>), is the slip in the y direction for the i-th connector and Kn g the single nail stiffness against the rotational slip , then the strain energy accumulated in the connectors can be expressed as NA UNU = £ [(K„/2) (A u)t2 + (Knv/2) (A v)] + (K^/2) (*)?] (2.61) t=i Eqn. (2.60) considers the fasteners in a discrete manner and may be replaced by using an equivalent continuous connector by taking the slips A u , A v and <f> as continuous functions . In this case , the strain energy , is expressed as follows : where e is the connector spacing. On substituting the expressions for plate and joist displacements from eqns. (2.1) to (2.7) one obtains : UNC = E ( f ^ tfMe = 0) - Un - Wn | | (h + d)f + f f f (^n(£ = o) - vn - en(h/2d/2)y + \ {w'^ ~6n)2} ( 2 - 6 3 ) Eqn. (2.62) can be written in terms of the vector {6n} as given in the pp. 13. To this end, we define the following vectors : {ei3}T = (0 0 1 0 0), where the thirteenth element is unity Chapter 2: Theoretical Formulation 23 {e 1 7}T = (0 0 1 0 0), where the seventeenth element is unity and Similarly, vectors {e16},{ei4},{e18} and {eig} are defined. Thus, the strain energy in the upper connectors is given by: " L Knx T ni:(h + d) UNC = 2s \2 ~2tT {°n> ( e i 3 - e i 7 - e i e ^ ) nir(h + d) T ( e i 3 - e i 7 - e 1 6 — ) {<$„} + 2~2e~ {6n] {ei4- ei8- ei92~s~ ei52~s ) (e 1 4 - eis - e i 9 — - e 1 5 — ) r { £ „ } +f 1 ? * ^ <f - T ) ( T - T )T{S" )} < 2'64' 2.2.5 Strain Energy in Lower Plate Connectors Here also we assume that the lower plate is connected to the joist by uniformly spaced fasteners along the center line of the bottom surface of the joist. The model also assumes three types of slips between the plate and the joist. 1. Slip parallel to the joist = A u = [«o + */2(^j)] -[U - h/2 (^)](2.65) 2. Slip perpendicular to the joist = A v = [vo + Wjjf)} ~ [V ~ (h/2) 0](2.66) 3. Rotational Slip = 4> where d\ = thickness of the lower plate ; and the displacements u0 , v0 and wo correspond to node 5 in the lower plate's mid-plane directly under the joist , as shown in the Fig. 2.1. In the above eqns. (2.64) to (2.66) it has been assumed that Chapter 2: Theoretical Formulation 24 the vertical displacement of the lower plate , u; 0 ,and the displacement , W , of the joist are identical . Following a similar development as previously described for the upper connectors, we obtain the strain energy in the lower plate connectors as follows: TJ fL Knxi , l T , nir(h + d). UNL = 22 1 2 ~2t~ ^  n> ^ 20 ~  617 ~  616 2L ^ (e 2o - e 1 7 - e16 — y {6n} . L Knv r r i Tr h d . + 2 ^ 7 M ( e 2 1 - e i 8 - e i 9 2 ^ - e 2 2 2 l ) ( e 2 i - e 1 8 - e 1 9 — - e 2 2 — ) T{6 n } Z It s s s s where ie2o}T = (0 0 1 0 0), where the twenteeth element is unity i e 2 i } T = (0 0 1 0 0), where the twentyfirst element is uni ty Similarly, vectors {e 1 7 } ,{e 1 8 } ,{e 1 9 } and {e 2 2 } are defined. F ina l ly the total strain energy in one T - beam strip element is given by, U{n) = Uu(n) + UL(n) + Uj(n) + UNU(n) + UNL(n) (2.69) 2.2.6 Stiffness Matrix and the Solution of the System of Equations O n taking the first variation of U(n) wi th respect to { £„} we get an expression of the form [i^„e]{<5„} in which [A' n e ] is the element stiffness matr ix of size 34 * 34. There are 34 unknowns per element , but 12 of them are shared wi th the adjoining element (6 at each of the nodes 3 and 6 of Fig.2.1). The global stiffness matrix for Chapter 2: Theoretical Formulation 25 34-Figure 2.4: B(i , i) submatrix in the Global stiffness matrix each Fourier term is obtained by assembling the stiffness matrix of each element as shown in Fig. 2.4 below. By considering the variation of U with respect to different {£„} , the overall global stiffness matrix can be obtained. This global system has a structure as shown in eqn. 2.69 in which {A/c} are the global vectors and the submatrices [B(i,j)] are the global submatrices. Each global vector {AK} has 22 * NJT + 12 components, and each submatrix, [B(i,j)], is a symmetric matrix of size (22 * NJT + 12). [B(i,j)] is a banded matrix, however, with a bandwidth of 34. The vectors {Rk}(k = 1, 2, , iV) correspond to the first variation of the load potential, UL . UL is the load potential per joist corresponding to the applied load function p(x, y) and is given by, Chapter 2: Theoretical Formulation 26 rs/2 rL UL = / p(x,y) w dx dy J-a/2 JO [ £ ( 1 , 1 ) ] [B( l ,2 ) ] ... [B{1,K)] ... [B(l,N)] [ £ ( 2 , 1 ) ] [ £ ( 2 , 2 ) ] ... [5(2,/0] - [B(2,N)] [B(K,1)] [B(K,2)} ... [B(K,K)} ... [B(K,N)] [B(N,1)} [B(N,2)\ ... [B(N,K)} ... [B(N,N)\ {Ai} {A 2 } {AA-} {A*} > = < (2.70) The global system can easily be solved by an iterative procedure as follows : N {AKy = [B(X )if)]- ,{{i?K}-EW»)]{MW} (2-71) n=l (i = 1 , 2 , ),{n^K)k{K = 1 , 2 , N) with starting vectors { A * } 0 = [B(K,K)]-'{RK) (2.72) Chapter 2: Theoretical Formulation The iterations can be stopped when , for example mod({6KY - {SKY- 1), < e (mod ( { A * } ' " - 1 ) ) , for K = 1,2, , i V and e a small number (e.g.,0.001) and C H A P T E R 3 The Computer Program 3.1 Program Features The Computer Program performs a static analysis of the stiffened structure wi th equidistant stiffeners i n one direction. Nonlinearities in the fasteners or inelastic joist or cover behavior have not been included. The program output gives information about • max imum joist deflection for each joist i n the floor • max imum joist deflection i n the floor • max imum bending stress for each joist • max imum joist bending stress in the whole floor • load sharing factor for bending stress for each joist • max imum load sharing factor for bending stress for a l l the joists • deflections of the cover at nodes 1 and 4. • maximum bending stress in the upper and the lower plate for each element • maximum bending stress in the upper plate for the whole floor • max imum bending stress in the lower plate for the whole floor • max imum upper plate deflection between joists for each joist • max imum lower plate deflection between joists for each joist • maximum upper plate deflection between joists for the whole floor • max imum upper plate deflection between joists for the whole floor 28 t Chapter 3: The Computer Program 29 A few important features of the program are described below : • The modulus of elasticity of the joists can be directly put into the datafile or can be selected from a distribution representing the population of joists' E I (joist stiffness). The distr ibution may be 3 - parameter W e i b u l l , 2 - parameter Lognormal Or Normal . Lower and Upper l imits of the interval wi thin which the modulus of elasticity of the joists should lie can also be specified. • A min imum strength (or Proof level) accepted for the joists can be specified. • There is a provision for gaps in the cover. • The nai l ing pattern may be treated as discrete or as continuous throughout the length of the connection between the joist and the cover. • Load on the floor is uniformly distributed, either over the whole area or over smaller , individual areas, simulating concentrated loads. • Boundary conditions may be imposed on the problem. The dimension statements in the program impose the following limitations : 1. The number of joists in the floor is l imited to 20. 2. The number of Fourier series terms that can be considered for the solution are l imited to a max imum of five terms , so that for symmetrical problems i n the direction of the joists the five terms wi l l be of order 1 , 3 , 5 , 7 and 9. For non-symmetrical problems the five terms wi l l be 1 , 2 , 3 , 4 and 5. 3. There can be a maximum of twenty loaded areas on the floor. 4. The number of gaps in the cover is l imited to 5. These limitations can be easily overcome by changing the dimension statements for the matrices and the vectors. Chapter 3: The Computer Program 30 3.2 Program Structure The program consists of a number of subroutines which find the modulus of elastic-ity of the joist , compute the strength of the joist , evaluate the shape functions , construct the stiffness matrix and the load vector and, finally, solve the system of equations. A brief description of the main subroutines and the operations they carry out is presented here. Subroutine DISTR: This subroutine assigns the modulus of elasticity of the joist depending upon whether the distribution representing the population of joists' stiff-ness has been assumed Weibull , Lognormal or Normal . Subroutine STREND: This routine assigns the strength of the joist, based on a cor-relation with its stiffness. Subroutine GENMTX: This subroutine generates the shape functions and then utilise, them to evaluate the required integrals in the development of the stiffness matrix. The routine computes the integrals numerically using a six point Gauss Quadrature rule, which will be exact for the degree of polynomial utilised for shape functions in the y-direction. Subroutine STIF: In this subroutine , the stiffness matrix is developed by considering only the contribution of the upper and the lower plates and the connections. The contribution of the joist to the stiffness matrix is added later on in the main program. The load vector is also computed. Subroutine DECMP and SOLV: The subroutine DECMP decomposes the stiffness matrix by Cholesky's decomposition method and stores it columnwise. Since the stiffness matrix is symmetrical and banded , only the lower triangular portion of the band is taken into consideration. The subroutine SOLV performs the forward and backward substitution to determine the solution vector. C H A P T E R 4 Verification and Numerical Results 4.1 Introduction In this chapter, the finite-strip element computer program developed i n the previous chapters is verified wi th the floor analysis program, F A P (Ref. 11). Next a parametric study is taken up to observe the behavior of the model under various conditions. The program has been implemented and tested on a mainframe I B M 3081K and an A S T 286 P remium ( I B M P C - A T compatible) at the University of Br i t i sh Columbia. Double Precision (Real * 8) arithmetic is used only in the calculation of the stiffness matr ix . This results i n savings i n terms of computer memory utilised and C P U time. A detailed discussion about the features of the program and its limitations is given earlier in Chapter 3. 4.2 Verification T w o verifications of the present program have been carried out as explained below: 4.2.1 Verification for flange stresses in the longitudinal di-rection: The values of the maximum deflection and the longitudinal normal stress at the top layer of the upper plate obtained from the present computer program are sought to be verified against the known solutions for them from the elementary beam theory. 31 Chapter 4: Verification and Numerical Results 32 A finite-strip element wi th dimensions as shown in F i g . 4.1 is considered. Other properties are -panel span(Z- p): 3800 mm. uniformly distributed load on the upper plate: 0.001916 M P a nai l load-slip modulus (parallel to the joist): 17500000 N / m m . na i l load-slip modulus (perpendicular to the joist): 17500000 N / m m . nai l rotation modulus :44500000 N mm/rad ian boundary conditions: no rotation in the z-y plane modulus of elasticity of plate in x - direction = 12000 M P a modulus of elasticity of plate in y - direction = 12000 M P a modulus of elasticity of joist = 12000 M P a Poisson's ratio = vxy = vyx =0.20 bending stiffness of plates i n x and y directions = 3515625 N m m . axial stiffness of plates in x and y directions = 187500 N / m m . Here very high values of nail load-slip modulus and nai l rotation modulus have been considered so that the connections may be rendered almost fixed. The study was done by considering the first three terms (i.e. n = 1, 3 and 5) of the Fourier series. For the section shown in F i g . 4.1 , following the elementary bending theory, EI = {(40 * 190 3)/12} * 12000 + 2((3375000 * 400) + (180000 * 400) * 102.5 2) = 1.7899 * 1 0 1 2 A r - m m 2 Therefore, the average longitudinal normal stress at top layer of upper plate Moment = ((0.001916 * 400) * 3800 2 )/8 = 1383352W - m m . 1383352 * 12000* 110 = 1.02MPa av 1.7899 * 10 1 2 Chapter 4: Verification and Numerical Results 33 400 1 15 CONNECTIONS ALMOST FIXED 190 15 h 406 H y Figure 4.1: Element with fixed connections . . . J Q 4. * A / c / o o ^ (-001916 * 400) * 38004 Maximum deflection, A = (5/384) * '—— v ' ' 1.7899 * 101 2 = 1.162 mm. From the computer program, Longitudinal normal stress at point A — (rmax — 1.06 MPa and Maximum deflection = 1.193 mm It may be observed that the program computed stress is very close to <jav . The case of deflection is also similar. amax > aav is due to shear lag effect . 4.2.2 Verification by comparison with FAP: In the present program , the model consists of an upper and a lower plate with a joist in between as shown in Fig. 2.1. The plates are connected to the joist by nail connectors. On the other hand, in the program FAP, the model is the same as Chapter 4: Verification and Numerical Results 34 described above except that there is no lower plate. In order to compare the results from the two programs, the thickness of the lower plate is reduced to 0.001 m m . and the nail stiffnesses for the connection between the lower plate and the joist made sufficiently small . A s a result, the lower plate, which is extremely th in , is very loosely connected to the joist and has almost no effect on the behavior of the structure. The datafile for the F A P and the present program have the following properties i n common: panel span(X p ) : 3800 m m . spacing between the joists (s): 400 m m . depth of the joist (HJT): 190 m m . thickness of upper plate (d) : 15mm. uniformly distributed load on the upper plate: 0.001916 M P a nail load-slip modulus (parallel to the joist): 1750 N / m m . na i l load-slip modulus (perpendicular to the joist): 1750 N / m m . nai l rotation modulus :4450000 N mm/rad ian boundary conditions: no rotation in the z-y plane modulus of elasticity of top layer of the plate i n x - direction = 12000 M P a modulus of elasticity of joist = 12000 M P a bending stiffness of plates in x direction = 1017270 N m m . bending stiffness of plates in y direction = 3164840 N m m . axial stiffness of plates in x direction = 59567 N / m m . axial stiffness of plates in y direction = 75335 N / m m . The comparison was done by considering the first three terms (i.e. n = 1, 3 and 5) of the Fourier series. The results were obtained for panels wi th 1 or 4 joists, and were as follows: Chapter 4: Verification and Numerical Results C A S E I: 1 Panel wi th 1 Joist Present Work Joist: Deflection 4.5747 m m . Bending stress 4.2488 M P a Upper Cover: Deflection 4.7027 mm. C A S E II: 1 Panel wi th Element # 1 Joist: Deflection Bending stress Upper Cover: Deflection Element # 2 Joist: Deflection Bending stress Upper Cover: Deflection Element # 3 Joist: Deflection Bending stress 4 Joists Present Work 4.5641 m m . 4.2415 M P a 4.6801 m m . 4.5114 m m . 4.1866 M P a 4.5944 m m . 4.5076 m m . 4.1756 M P a F A P 4.5746 m m . 4.2487 M P a 4.8642 m m . F A P 4.5911 m m . 4.2686 M P a 4.7775 m m . 4.5798 m m . 4.2536 M P a 4.6137 m m . 4.5798 m m . 4.2537 M P a Chapter 4: Verification and Numerical Results 36 Upper Cover: Deflection 4.5458 mm. 4.6201 m m . Element # 4 Joist: Deflection 4.5651 mm. 4.6797 m m . Bending stress 4.2415 M P a 4.3598 M P a Upper Cover: Deflection 4.5858 m m . 4.6137 m m . 4.3 Parametric Study A brief parametric study has been undertaken to investigate the behavior of the model due to the changes i n -• the nai l stiffness • the ratio of the plate thickness to the depth of the joist • the ratio of the spacing between the joists to the panel span. For the purpose of investigation a panel wi th 10 joists is considered. The stiffness of the joists are represented by a 3-parameter Weibul l distribution. The lower and the upper l imits of the modulus of elasticity of the joists are set at 8000 M P a and 14000 M P a respectively. A l l the joists are selected at random but wi th in these upper and the lower l imits . The rotation, in the z-y plane, of the nodes 2 , B and 5 i n the 1st and the 10th element (see fig. C.3) are restricted. For the purpose of comparison, the highest values of the stresses and the deflections in the floor are selected. The conditions mentioned above are kept constant during the whole parametric study as described i n A r t . 4.3.1 , 4.3.2 and 4.3.3. Chapter 4: Verification and Numerical Results 37 4.3.1 Effect due to change in nail stiffness A structure wi th (i) panel span (Lp) : 3800 m m . (ii) spacing between joists (s) : 400 m m . (iii) thickness of the plates : 15mm. (iv) u .d . l . on the upper plate : 0.001916 M P a and (v) depth of the joists (HJT) : 190 m m . has been considered. The nai l load-slip modul i i in the directions parallel and per-pendicular to the joist, RKPAL and RKPER respectively, are kept equal. A l l the assumptions made in A r t . 4.3 have also been incorporated. RKPAL (and RKPER) is increased from 0.0175 N / m m . to 17500000 N / m m . i n six steps( see Figs. 4.2, 4.3 and 4.4) keeping the nai l rotation modulus ,RKROT, fixed at 44.5 N mrn. /radian. In the second and th i rd steps , RKROT is changed to 44500 N mm. / r ad ian and 44500000 N mm. / rad ian respectively and RKPAL (and RKPER also) is gradually increased as before. The resulting change in the joist deflection, joist bending stress and the stress i n the longitudinal direction in the upper plate is shown i n F i g . 4.2, F i g . 4.3 and F i g . 4.4. We can observe from the figures that : (i) as the nai l stiffness is increased the joist deflection and the joist bending stress decreases, and the stress in the upper plate increases. However this rate of change decreases wi th increase in the stiffness of the connector. (ii) beyond the value of 17500 N / m m . for RKPAL, there is very small change in the joist deflection, joist bending stress and the stress in the upper plate. (iii) in i t ia l ly the joist bending stress is very high and the upper cover stress in the longitudinal direction is very smal l . Thenafter there is a gradual decrease in the joist bending stress and an increase in and the upper plate stress. A t the end, the stress in the upper plate i n the longitudinal direction is slightly higher than the joist bending Chapter 4: Verification and Numerical Results 38 stress. (iv) the effect of increase of RKROT is much less than that of the increase of RKPAL. (v) there is no use to increase RKROT beyond 44500 N mm./radian. It is clear that as the connector stiffness is increased, more and more load gets transferred to the plates. A limiting case is reached when RKPAL (and RKPER) is 17500 N/mm. and RKROT is 44500 N mm./radian. The effect of increase of RKPAL is much more visible than that of RKROT. 4.3.2 Effect due to the change in the ratio of plate thickness to the depth of the joist: The assumptions made in the Art. 4.3 prevail here too. Other properties of the panel considered are as described below: (i) thickness of the plates: 15mm. (ii) spacing between the joists: 380 mm. (iii) panel span: 3800 mm. (iv) uniformly distributed load on the upper plate: 0.0024 MPa. While keeping the thickness of the plate constant at 15mm., the depth of the joist is increased from 150 mm. to 190 mm., 250 mm., 300 mm. and 375 mm. in four steps. This leads to a decrease in the d/HJT ratio. For each ratio, five cases of nail load-slip modulus (RKPAL and RKPER) and nail rotation modulus (RKROT) were considered as described below: CASE 1 : 0.0175 N/mm. , 0.0175 N/mm. , 44500 N mm/radian CASE 2 : 550.0 N/mm. , 550.0 N/mm. , 44500 N mm/radian CASE 3 : 1750.0 N/mm. , 1750.0 N/mm. , 44500 N mm/radian CASE 4 : 17500 N/mm. , 17500 N/mm. , 44500 N mm/radian CASE 5 : 17500000 N/mm. , 17500000 N/mm. , 44500 N mm/radian The change in the joist deflection, joist bending stress and the maximum stress Chapter 4: Verification and Numerical Results 39 i n the topmost layer in the upper plate (above N O D E 2) in the x - direction, due to the variation in the d/HJT ratio and the nail stiffness, is shown in F i g . 4.5, F i g . 4.6 and F i g . 4.7. In these figures, the values obtained at d/HJT — 0.04 i n each of the cases mentioned above are assumed equal to unity and values at higher d/HJT ratios normalised on this basis. A few observations can be made as follows: • when the nail ing is m i n i m a l , the applied load is carried almost entirely by the joists and, as a result, the joist deflection and the joist bending stress are quite high while the stress in the x-direction at the topmost fibre of the upper plate is very small . • If the nai l stiffnesses are assigned even a small value, the composite action of the plates and the joist comes into play. This is reflected by the sudden increase in the stress in the upper plate. • It can be observed from Case 4 and Case 5 that beyond a particular value of nail stiffness, the change in the deflection and the bending stress i n the joist and the stress i n the upper plate is very small , which shows that the connections are almost rigid. • W h e n the d/HJT ratio is increased, the rate of increase of joist deflection is found to be more rapid for Case 1 than for the other cases. This is very obvious because as d/HJT increases, HJT decreases since we have kept d (thickness of the plate) constant. • Unl ike i n the case of joist deflection (Fig.4.5) the rate of increase of joist bending stress for a l l cases except for Case 1 is almost linear. In Case 1, however, a rapid rate of increase is observed. The increase in the upper cover stress follows the same pattern. Chapter 4: Verification and Numerical Results 40 4.3.3 Effect due to the change in the ratio of spacing be-tween the joists to the panel span : A study has been done to investigate the variation in three parameters, namely, joist deflection, joist bending stress and the maximum stress in the top fibre of the upper plate in x-direction , due to a gradual change in d/HJT ratio (d being the plate thickness and HJT the depth of the joist) when the s/Lp ratio is increased in steps from 0.05 to 0.25. The s/Lp ratio is increased by keeping Lp constant and gradually raising the value of s. Figs. 4.8, 4.9 and 4.10 are drawn assuming the nail load-slip modulii in directions both parallel and perpendicular to the joist as 1750 N/mm and the nail rotation modulus as 4450000 N mm/radian. The assumptions made in Art. 4.3 have also been incorporated. Other properties of the structure are as follows: thickness of the plate: 15mm. panel span: 3800 mm. uniformly distributed load on the upper plate: 0.0024 MPa It may be observed that at higher s/Lp ratios, the joist deflection, joist bending stress and the stress in the upper cover in the longitudinal direction is larger. Fig. 4.9 and 4.10 show that increase in the joist bending stress and the upper cover stress is almost linear with increase in d/HJT ratio. Furthermore we can see from Fig. 4.8 that for a particular value of d/HJT and s/Lp ratios if the corresponding joist deflection is found to be excessive, then in order to achieve a lower level of joist deflection either the s/Lp ratio may be decreased (by decreasing 5 ) or the d/HJT ratio may be decreased (preferably by increasing HJT). Similar inferences can be drawn from Figs. 4.9 and 4.10. >1 Chapter 4: Verification and Numerical Results 41 0.0175 175 550 1750 17500 17500000 RKPAL (= RKPER) In N / mm. Figure 4.3: Joist Bending stress vs. Nail stiffness a. O ul cc X z L U C L a. ZD u. o cc L U O cn cn L U cc 0.0175 RKROT = 44500 and 44500000 N mm./radian 175 550 1750 RKPAL (= RKPER) in N/mm. 17500 17500000 Figure 4.4: Stress at top layer of upper plate in x-direction at varying nail stiffnesses 0.04 0.06 0.08 0.1 THICKNESS OF PLATE / DEPTH OF JOIST Figure 4.5: d / H J T vs. Joist Deflection at varying nail stiffnesses Figure 4.6: d / H J T vs. Joist Bending stress at varying nail stiffnesses Figure 4.7: d/HJT vs. Stress at top layer of upper plate in x-direction at varying nail stiffnesses 0 . 0 4 0 . 0 5 0 . 0 6 0 . 0 7 8 0.1 THICKNESS OF PLATE / DEPTH OF JOIST Figure 4.9: d/HJT vs. Joist Bending stress at varying s/Lp IT) CL 4.5 o LU CC Q « X Z a. cc LU a. a. cc LU 0. O t-z V) <n ui cc i-4 -LU 3.5 3 -2.5 -1.5 1 -0.5 RKPAL = RKPAL1 = 1750 N /mm RKPER = RKPER1 = 1750 N/mm RKROT = RKROT1 = 4450000 N mm./ radian 0.04 I I I 0.05 0.06 0.078 DEPTH OF PLATE / HEIGHT OF JOIST 0.1 Figure 4.10: d/HJT vs. Stress at top layer of upper plate in x-direction at varying s/Lp 3 o o" a c 3 Q o —^. re c Chapter 4: Verification and Numerical Results 50 4.4 Study of the Shear Lag Effect According to the Clause 8.6.3.1 of CSA Standard CAN3-086.1-M84 ' Engineering De-sign in Wood (Limit State Design)' , the factored bending moment resistance along the direction of the webs of a stressed skin panel shall be the least of the factored resistances of the tension or the compression flanges or the web determined as follows: (a) tensionflange MR = <j> tv (KD K'S KT) XJ XG ^E!)E (b) compression flange MR = <f>Pp(KDK'sKT)XjXG (EI)e BA K'S (c) web M r = <f>FBKZBKHKLXG-^^ E KSE °w where <f> = 0.8 ip = specified strength in tension, N/mm (Tables 7.3.A and 7.3.C) pp — specified strength in compression, N/mm (tables 7.3.A and 7.3.C) BA = specified axial stiffness, N/mm (Tables 7.3.B and 7.3.D) FB = fb (KD KSB KT) fb = specified strength in bending of webs, MPa (Tables 5.3.1.1, 5.3.1.2, 5.3.1.3, 5.3.1.4 and 5.3.2 for Sawn Lumber and Table 6.3 for Glulam.) Xj = stress joint factor (Clause 8.3) XG — panel geometry reduction factor (Clause 8.6.3.2) Chapter 4: Verification and Numerical Results 51 E = modulus of elasticity of web, MPa (Tables 5.3.1.1, 5.3.1.2, 5.3.1.3, 5.3.1.4 and 5.3.2 for Sawn Lumber and Table 6.3 for Glulam.) Cu, = greatest distance from neutral axis to outer edge of web, mm. In the formulae above, XQ accounts for shear lag . Clause 8.6.3.2 of the Code defines factor XG as : XG = 1 - 4.8 (s/Lp)2 where, s = clear spacing between the joists (in mm.) Lp = panel span (in mm.) This formula for XG is valid for values of s/Lp ranging from 0.05 to 0.25. In order to verify the formula given by the Code, a study was undertaken to find the values of XQ due to a gradual change in the d/HJT ratio when s/Lp is increased in steps from 0.05 to 0.25. The stiffness of the joists is represented by a 3 parameter Weibull distribution. A limiting condition, given by EMIN and EM AX, is also imposed on the lower and the upper limits of the modulus of elasticity of joists. The study considered 50 panels with 4 joists each. Other properties are -panel span: 3800 mm. thickness of the plates: 15mm. uniformly distributed load on the upper plate: 0.0024 MPa boundary conditions: none (Panel simply supported) The spacing between the joists, s, is varied from 190 mm. to 570 mm. to 950 mm.. Similarly, the depth of the joist, HJT, is increased from 100 mm. to 200 mm. and 300 mm.. The first three Fourier terms (n = 1, 3 and 5) have been considered. For each combination in d/HJT and s/Lp ratios , the stress at the topmost fibre of the upper plate in the x - direction, aav, is found following the elementary bending theory as in Art. 4.2.1, by assuming the average modulus of elasticity of joists as (EMIN + EM AX) / 2 . The stress at the same level, o m a x , is also determined from the computer program. Next, gives the value of XQ for each element. The & max Chapter 4: Verification and Numerical Results 52 min imum value of XQ at each of the 50 panels is found out and then arranged in an ascending order. XQ is plotted against the probability of not being conservative for varying s/LP and d/HJT ratios in F i g . 4.11, F i g . 4.12 and F i g . 4.13. A few comments can be made on the basis of the figures mentioned above. (a) A t s/LP = 0.05 , XQ according to the Code is 0.988. It may be observed in F i g . 4.11 that corresponding to XQ = 0.988 , the probability of being not conservative may range from 0.525 to 0.620 depending upon the d/HJT ratio. (b) A t s/LP = 0.15, XQ as per Code is 0.892, which may be termed quite satisfactory as may be observed i n F i g . 4.12 . (c) In F i g . 4.13 for s/LP = 0.25 we can see that the min imum XQ at zero probability is approximately 0.85, whereas the Code value of XG is 0.7 . To facilitate the study of the shear lag effect in greater detail , an investigation was done to observe the effect of the number of longitudinal ribs (joists). 4.4.1 Effect of the number of longitudinal ribs To observe the effect of the number of longitudinal ribs, a structure wi th the following properties has been considered. Number of panels : 50 Panel span (LP) : 3800 m m W i d t h of each panel : 4000 m m Thickness of the plates (d) : 15 m m U . D . L . on the upper plate of each panel : 0.0024 M P a N a i l load - slip modulus (parallel to the joist) : 17500000 N / m m N a i l load - slip modulus (perpendicular to the joist) : 17500000 N / m m N a i l rotation modulus : 44500000 N mm/rad ian Joist stiffness : represented by a 3 parameter Weibul l distribution Joist modulus of elasticity : selected randomly between 8000 and 14000 M P a Chapter 4: Verification and Numerical Results 53 Boundary condition : none. The nai l stiffness considered here are very high and thus render the connections almost fixed. The number of the longitudinal ribs (joists) considered are 4, 5 , 6 , 7 and 8. A t the same time the spacing between the joists gets reduced because the total wid th of the panel remains constant at 4000 m m . T w o cases are considered, in the first case the depth of the joist (HJT) is 200 m m . and in the second 300 mm. . For each variation i n the number of joists, the value of XQ is calculated in the computer program as mentioned earlier in A r t . 4.4. for each element in the structure. The min imum value of XQ for each panel is found. After being arranged in an ascending order, these XQ values are plotted against the probabili ty of being not conservative . F rom the F i g . 4.14 and F i g . 4.15 we may observe that: (a) increasing the number of joists causes an increase i n XG; (b) when d/HJT = 0.075 and the number of joists per panel is 6, XQ at zero prob-abil i ty of being non-conservative is 0.91. O n the other hand i f d/HJT = 0.05 and the number of joists be the same, at 5 percent probability XQ w i l l be 0.895. Though d/HJT has been changed, at zero probability XG is 0.89. Thus there is a 2 percent change i n XG as a result of the change i n the d/HJT ratio. Similar phenomenon is seen i f the number of joists is changed. (c) W h e n the number of joists are 4, 5 and 6, the s/Lp ratios are given by 0.25 , 0.20 and 0.167 respectively. The proximity of the curve corresponding to 5 joists to the one representing 6 joists i n comparison to the curve representing 4 joists may be attributed to the fact that 0.20 is nearer to 0.167 than to 0.25. Thus it may be concluded that : (a) an increase i n the number of joists decreases the shear lag; (b) XQ is not entirely independent of the d/HJT ratio. (c) XG depends on the s/Lp ratio. Chapter 4: Verification and Numerical Results 54 We may observe from A r t . 4.4 and A r t . 4.4.1 that we should recommend the XQ values to be adapted on the basis of 4 joists being used and the d/HJT ratio giving the min imum XQ- This would be on the conservative side because the maximum shear lag occurs when the number of joists is 4. Furthermore, there is small practical chance of using less than 4 joists per panel. The XG values which may be recommended, assuming 5 percent probability of being not conservative to be acceptable, are as described below: A t s/Lp = 0.05, the value of XQ to be adopted is 0.92. Similar ly at s/Lp = 0.15 and 0.25, XQ should be taken as 0.91 and 0.86 respectively. A plot based on this recommendation is shown in F i g . 4.16. T w o formulae can be put forward from this plot. XG = 0.925 - 0.1 — , for 0.05 < s/Lp < 0.15 XG = 0.985 - 0.5 , for 0.15 < s/Lp < 0.25 (4.1) (4.2) Figure 4.11: Probability of non-conservative >^  at s/L = 0.05 Cn LU > < > CC Ul {/> z o o H o z o z LU CO LL O >-CO < CO o cc 0. 0.9 0.8 H NUMBER OF JOISTS = 4 SIMILARLY OTHERS d/HJT= 15/200 = 0.075 0.7 0.6 0.5 H 0.4 0.3 0.2 H 0.1 {a Mi. c C 0.86 0.94 0.96 0.98 MINIMUM X IN A FLOOR G Figure 4.14: Probability levels of X at d/HJT = 0.075 G 00 Chapter 4: Verification and Numerical Results 60 Figure 4.16: Panel geometry reduction factor XQ at varying s/LP ratios) C H A P T E R 5 Conclusions and Scope for further research 5.1 Conclusion Stiffness of the connectors have a great impact on the behavior of the stiffened plates. The composite behavior is highly dependent on the nail stiffness. Nail stiffness for rotational slip has considerably less significance in comparison to the nail stiffness corresponding to the x and y directions. Even small amount of nail stiffness can affect the stress resultants appreciably. At general levels of connector stiffnesses , the relation between the d/HJT ratio (d being the thickness of the plates and HJT the depth of the joist) and the joist bending stress has been found to be almost linear. The relation between the d/HJT ratio and the upper cover stress in the longitudinal direction is also similar. However, the rate of change of joist deflection increases with increase in the d/HJT ratio. The rate of change is higher for higher s/Lp ratios. The shear lag factor XQ may be assumed independent of the d/HJT ratio. The Code (CSA Standard CAN3-086.1-M84) value for XG has been found to be, for NJT = 4, underconservative and overconservative at s/Lp = 0.05 and 0.25 re-spectively. At s/Lp = 0.15 the Code gives an acceptable value of XG-XQ is dependent on the s/Lp ratio. An increase in the number of joists (i.e. the number of stiffeners) decreases the resulting shear lag. 61 Chapter 5: Conclusions and Scope for further research 62 5.2 Scope for further research In the study for 'effect due to change in the d/HJT rat io ' , 15 m m . thick upper and lower plates and their modulus of elasticity have been selected from experimental data. The plates are made of plywood. The nominal plywood thickness, number of plies constituting them, type of timber from which the plies have originated would obviously affect its bending and axial stiffness and other related properties. Hence further investigation is needed by changing the plywood thicknesses, the number of plies, type of timber constituting them, the depth of the joists, panel spans and the different types of connections between the plate and the joist. The A r t . 4.3.3 discusses the effect of changing the s/Lp ratio. The study has been done based on a particular set of nail stiffnesses and only three different s/Lp ratios have been looked into. Furthermore, the same type of plate as described above has been used here too. Hence, a more comprehensive study on this topic should be done wi th (i) different set of nai l stiffness values (ii) a few extra s/Lp ratios, and (iii) by changing the properties of plate (e.g. plywood thickness, number of plies, type of t imber constituting the plywood etc.) Bibliography [1] Bakht , B . , Jaeger ,L.G. , Cheung,M.S. and M u f t i , A . A . , 1981, The State-of-the-Art in Analysis of Cellular and Voided Slab Bridges, Canadian Journal of C i v i l E n -gineering, September, 1981, pp. 376-391 [2] Benson, P . R . and Hinton, E . , 1976, A thick Finite Strip Solution for Static, Free vibration and Stability Problems , International Journal of Numerical Methods i n Engineering, V o l . 10, 1976, pp. 665-678 [3] Cheung, Y . K . , 1969 , Analysis of Box-Girder bridges by Finite Strip Method, A C I Publications, SP 26, 1969 , pp. 357-378 [4] Cheung, Y . K . , 1976 , The Finite Strip Method in Structural analysis, Pergamon Press, Oxford. [5] Cheung, Y . K . and Fan, S .C. ,1983, Static Analysis of Right Box-Girder Bridges by Spline Finite Strip Method, Proc. Instn. C i v . Engrs., Part 2, 75, June, 1983 , pp. 311-323. [6] C h u , K . and Dudnik , E . , 1967 , Concrete Box-Girder Bridges Analysed as Folded Plates, Firs t International Symposium on Concrete Brdge Design, A C I Publ ica-tions, SP-23 , 1967 , Detriot. [7] Cope, R . J . , Harris, G . and Sawko, F . , 1971 ,A Quasi Slab Approach to the Anal-ysis of Cellular Bridge Structures, Proc . of the Symposium in Computer Aided Engineering held at the Universituy of Waterloo, M a y , 1971. 63 Bibliography 64 [8] Crisfield, M . A . , 1971 , Finite Element Methods for the Analysis of Multi-Cellular Structures, Proc. Instn. Civ. Engrs., 48(3) , 1971. [9] Fam, A . and Turkstra, C , 1975 , A finite Element Scheme for Box-Bridge Anal-ysis, Computers and Structures, Vol. 5, 1975 , pp. 179-186. [10] Fam, A . , 1986 , A Tapered Finite Element Scheme for Bridge Analysis, Proceed-ings of the Third International Conference on Computing in Civil Engineering, 10-12 August , Vancouver. [11] Foschi, R .O. , 1982 , Structural Analysis of Wood Floor Systems, Journal of the Structural Division, American Society of Civil Engineers, Vol. 108, no. ST7, July 1982 , pp. 1557-1574. [12] Fraeijs De Veubeke, B. , 1965 , Displacement and Equilibrium Models in the Finite Element Method Stress Analysis, John Wiley , London , 1965 , Chapter 5. [13] Hamberg, H . and Trenks, K . , 1962 ,Drehsteife Kreuzwerke, Berlin , 1962. [14] Hambly, E . C . and Pennels, E . , 1975 , Grillage Analysis Applied to Cellular Bridge Decks, The Structural Engineer, July , 1975. [15] Hinton, E . , Razzaque A. , Zienkiewicz O.C. and Davies, J .D. , 1975 , A Simple Finite Element Solution of Plates of Homogenous, Sandwich and Cellular Con-struction, Proc. Instn. Civ. Engrs., Part 2 , 59 , March , 1975 , pp. 43-65. [16] Hook, I.M.A. and Richmond, B. , 1970 ,Western Avenue Extension , Precast Concrete Box Beams in Cellular Bridge Decks, The Structural Engineer, 1970. [17] Kristek, Vladimir, 1979 , Theory of Box Girders, John Wiley &; Sons . [18] Loo, Y . C . and Cusens, A.R. , The Finite Strip Method in Bridge Engineering, Viewpoint Publication, 1978. Bibliography 65 [19] Mindlin, R.D., 1951 , Influence of Rotatory Inertia and Shear on Flexural Motion of Isotropic Elastic PLates, Journal of Applied Mechanics, 18 , 1951 , pp. 31-38. [20] Narayanan, R., 1983, Plated Structures - Stability and Strength, Applied Science Publishers, London and New York , 1983 , pp. 165-194. [21] Onate, E . , 1976 ,Comparisons of Finite Element Methods for the Analysis of Box-Girder Bridges, MSc. Thesis, Civil Engineering Department, University of Swansea, 1976. [22] Onate, E . and Suarez, B. , A Unified approach for the Analysis of Bridges, Plates, and Axisymmetric Shells using the Linear Mindlin Strip Element, Computers and Structures, Vol. 17, no. 3, 1983, pp. 407-426. [23] Sawko, F . , 1968 , Recent Developments in the Analysis of Steel Bridges using Electronic Computers, BCSA Conference on Steel Bridges, June , 1968 , London. [24] Sawko, F. and Cope, R.J . , 1969 , Analysis of Multicell Bridges without Transverse Diaphragms - a Finite Element Approach, The Structural Engineer, Vol. 47, No. l l , 1969. [25] Sawko, F . and Cope, R.J . , 1972 , Discussion to M.A. Crisfield's Paper, Proc. Instn. Civ. Engrs., Vol. 2 , January , 1972. [26] Sisodiya, R . G . , Cheung, Y . , K . and Ghali, A. , 1972 ,New finite elements with application to box girder bridges, Proceedings of the Institution of Civil Engineers , Supplement (X) , 1972. [27] Thompson, E . G . , Goodman, J.R., and Vanderbilt, M.D. , 1975 , Finite Element Analysis of Layered Wood Systems, Journal of the Structural Division, American Society of Civil Engineers, Vol. 101, No. ST12, December, 1975 , pp. 2659-2672. Bibliography 66 [28] Thompson, E . G . , Vanderbilt, M.D. , and Goodman, J.R., 1977 ,FEAFLO: A Program for the Analysis of Layered Wood Systems, Computers and Structures, Vol. 7, 1977 , pp. 237-248. [29] Zhang, S.H. and Lyons, L.P.R., 1984 ,The Application of the Thin Walled Box Beam Element to Multibox Bridge Analysis, Computers and Structures , Vol. 18, No. 5, 1984, pp. 795-802. APPENDIX A A . l Shape Functions: A . l . l Vector {M0} M0 (1) = e - 5 /4£ 3 - l / 2 £ 4 + 3 /4£ 5 M0(2) = ( e - e - e + n/s.o M 0 (16) = 1.0 - 2.0 £ 2 + £ 4 M 0(15) = (Z - 2.0Z3 + e 5)/2.0 Mo (23) = £ 2 + 5/4 e - 1/2 £ 4 - 3/4 £ 5 M 0(24) = ( - £ 2 - ( 3 + e4 + £ 5 )/8 . 0 A l l other components ,M0 (k) = 0.0, k = 1,2, ....,34. A.1.2 Vector {M3} M 3 (3) = l / 4 ( - 3 £ + 4 £ 2 + e - 2 £ 4 ) Ms (4) = ( - t + e + e - e 4) / s.o Af3(13) = 1.0 - 2.0 + £ 4 M 3(25) = 1/4(3^ + 4£ 2 - e - ? ) M3(26) = (-e - e + e + n / s . o A l l other components , M 3 (&) = 0.0, k — 1,2,...., 34. 67 Appendix A: 68 A.1.3 Vector {M5} M5 (5) = M3 (3) ; M5 (6) = M 3 (4) ; M5 (14) = M 3 (13); M s (27) = M 3 (25) ; M 5 (28) = M3 (26) A l l other components , M 5 (fc) = 0.0, fc = 1,2, ....,34. A.1.4 Vectors {Mi}, {M2}, {M4}, and {M6} M 1 ( f c ) = ^ ; M 2 ( f c ) = ^ ; M 4 ( f c ) = ; M6(k) = i n which: = 1,2, ,34 APPENDIX B B . l Shape Functions: B . l . l Vector {M7} M7 (7) = e ~ 5/4f - l /2£ 4 + 3/4£ 5 Mr (s) = ( e - e -t4 + a /8 .o M7(16) = 1.0 - 2.0 £ 2 + £ 4 M7(22) = a - 2.oe + e)/2.o M7 (29) = £ 2 + 5/4 f - 1/2 - 3/4 f M7(30) = ( - £ 2 - t3 + £4 + £ 5 ) / 8 . 0 All other components ,M7 (k) = 0.0, k = 1,2, ....,34. B.1.2 Vector {Mi0} M1 0(9) = 1 / 4 ( -3 e + 4£ 2 + e ~ 2£4) MW (io) = ( - t + e + e - Z4) I 8.0 M lo(20) = 1.0 - 2.0 £ 2 + ( 4 Mao(31) = l / 4 ( 3 £ + 4£2 - £ 3 - f ) M10 (32) = (-e - e + e + t 4 ) / s.o All other components , M 1 0 (k) = 0.0, = 1 ,2 ,34. 69 Appendix B: B.1.3 Vector { M i 2 } Ml2 (11) = M 1 0 (9) ; M 1 2 (12) Ml2 (33) = M 1 0 (31); M 1 2 (34) A l l other components , M i 2 (fc) = M a o ( 1 0 ) ; M 1 2 ( 2 1 ) = M 1 0 = M 1 0 (32) = 0.0, k = 1,2, ....,34. B.1.4 Vectors { M 8 } , { M 9 } , {Mn}, and {Mi 3 } Ms(k) = ^ > ; M 9 ( f c ) = M n (fc) = ; M 1 3 (fc) = in which: fc = 1,2, ,34 APPENDIX C U S E R ' S M A N U A L - F A P P C . l Input The input file should be set up as follows: 1. Enter: on free format N M , NJT, ISYM, INPTE, NFLOR, LPRNT, TOL where N M =maximum order of sine/cosine terms in the Fourier series. There can be a maximum of 5 terms, so that for symmetrical problems in the direction of the joists , NM can be upto 9 ( the five terms wi l l be of the order 1, 3, 5, 7, and 9). If N M = 3, there w i l l be two terms, those for order 1 and 3. For non-symmetrical problems, NM is just the maximum order of sine/cosines and agrees wi th the number of terms: thus, N M = 5 means terms wi th order 1, 2, 3, 4, 5. N J T — number of joists, wi th a maximum of 20. I S Y M = 0 i f non-symmetrical problem along the joists; 1 for symmetrical problems. I N P T E = 4 to pick E - values from a distribution; 2 to enter joist's E - values one by one; 3 to enter joist's /^-values and strengths one by one. N F L O R = numbers of floors to be run. I P R N T = output parameter. If 0, output is just a summary. If 1, output is comprehensive, including detailed stresses and deflections joist by joist. T O L = Tolerance to stop iterations in the solution of the system of equations. 71 Appendix C: USER'S MANUAL - FAPP 72 This is needed only i f NM exceeds 1, and the value T O L = 0.001 may be used. 2. If I N P T E ^ 4 skip to 9. Otherwise, when I N P T E = 4 (Case of Simulations) Enter: Z l , C V S where Zl = random number initializer ( any value can be entered) for the random number generator used by the system. C V S = a number used wi th the min imum selected value wi th in which the E for a l l the joists i n the floor w i l l fall . Thus, i f C V S = 0.10 , a l l the joists' E wi l l fall wi th in 1.0 and 1.1 times the minimum selected E. The use of C V S 7^  0.0 allows the selection of floors wi th joists of similar properties, but w i th the min imum E wi th in a floor s t i l l randomly selected from the entire popu-lat ion. Thus , in the same simulation one is mixing floors wi th joists a l l nearly equally stiff and strong wi th floors ut i l izing joists a l l nearly equally flexible and weak. This procedure can be used to simulate the 'lot sampling' technique, wi th CVS related to the within-lot variability. W h e n CVS is a large number, al l floors w i l l have E picked at random from the entire population. This is a way of doing a normal simulation, without introducing the ' lo t ' concept. W h e n C V S = 0.0 , a l l joists wi th in a floor w i l l be picked at random but wi th in a certain interval for E. 3. If C V S ^ 0.0 skip to 4. W h e n C V S = 0.0, Enter: E M I N , E M A X where E M I N , E M A X = are the lower and upper limits of the interval within which a l l the E 7s in the floor must lie. This option is useful to study the effect of controlling the E of the joists. ( M S R application, for example). In order to do a normal simulation , EMIN and EMAX can be choosen as the Appendix C: USER'S MANUAL - FAPP 73 lowest and the highest E in the population. 4. Enter: K D I S T R , E O , E M , E K where K D I S T R = identifier for the type of distribution representing the population of joists ' EI (joist stiffness). K D I S T R = 1 i f the distribution is a three - parameter Weibul l ; K D I S T R = 2 i f the distribution is a two - parameter Lognormal; K D I S T R = 3 i f the distribution is a Normal . E O , E M , E K = parameters for the corresponding distributions. EO is the mini-m u m value of EI for the three parameter Weibul l , and EO = 0.0 otherwise. 5. Enter: Q A , Q B where Q A , Q B = intercept and slope, respectively, of the regression line betrween M . O . R . and E for the population of joists considered. 6. Enter: E O O , E l l , C V 1 , C V 2 , C V 3 where E O O = min imum value of E i n the population; E l i = maximum value of E in the population; C V 1 , C V 2 , C V 3 = coefficients that permit the calculation of the variability in M . O . R . for a given value of E. These coefficients are obtained by expressing the variabil i ty i n F i g . C . l according to the following equation: CV = CVl *{E- EOO) CV2 * ( £ 1 1 - E) CV3 where C V is the coefficient of variation estimated from F i g . C . l for the value E. 7. Enter: S P R O O F where S P R O O F = M i n i m u m strength (or Proof level) accepted for the joists. This strength refers to 'short term' strength. 8. Enter: S , R L , D , D l where S_ = joist spacing (in) R L = joist span (in) Appendix C: USER'S MANUAL - FAPP 74 D = upper cover (sheathing) thickness (in) D l = lower cover (sheathing) thickness (in) 9. Enter . F A C T O R = 1 . 10. Enter: R K X , R K Y , R K V , R K G , D X , D Y , D V , D G , E C O V , N F A C E & T V . where R K X = bending stiffness of the upper cover in the x-direction, parallel to the joists. For plywood, this is usually parallel to the face grain. R K Y = bending stiffness of the upper cover i n the y-direction, perpendic-ular to the joists. For plywood, this is usually parallel to the face grain. R K V , R K G = parameters for plate bending, connected respectively to Poisson's effect and shear by torsion. D X = in-plane (axial) stiffness i f the upper cover in x-direction. D Y = in-plane (axial) stiffness i f the upper cover in y-direction. D V , D G = parameters for in-plane axial stiffness , related respectively to Poisson's effects and shear. E C O V = modulus of elasticity for the outermost veneer (for plywood) wi th grain perpendicular to the joists. For a non-layered panel, ECOV is the E of the panel. N F A C E = 1 i f cover face grain (plywood) is parallel to the joists; = 2 i f cover face grain is perpendicular to the joists. N F A C E = 1 for non-layered panels. T V = thickness of outermost veneer (in) in the plywood cover. In non-layered panels, TV is the thickness of an arbitrarily defined outer layer. 11. Enter: R K X l , R K Y l , R K V 1 , R K G 1 , D X l , D Y I , D V 1 , D G 1 , E C O V 1 , N F A C E 1 & T V l . where R K X l , R K Y l , R K V 1 , R K G l , D X l , D Y I , D V 1 , D G l , E C 0 V 1 , N F A C E 1 , & T V l are the same properties as mentioned above in 10 but for the lower Appendix C: USER'S MANUAL - FAPP 75 cover. 12. If I N P T E ^ 4 skip to 13. W h e n I N P T E = 4 (case of Simulations), Enter: BJT, HJT, BETA, ALPHA, R E G where B J T = width of joists (in) H J T = depth of joists (in) B E T A = torsion constant which is a function of the ratio HJT/BJT , for rectangular cross-sections. It can be found , for example, in pp. 36.7, Handbook of Engineering Mechanics, W . Flugge, ed. A L P H A = shear deflection constant for rectangular cross-sections, (for parabolic distr ibution of shear stresses , this constant is 1.2) R E G = ratio of E of joist to the shear modulus (E/G ratio). Skip to 16. 13. Enter: for each joist , and consecutively, BJ(I) , 1 = 1 , N J T (width of joists, (in)) HJ(I), I = 1, N J T (depth of joists, (in)) STORE (I , 1), I = 1, N J T (E for each of the joists) 14. If I N P T E ^ 3 skip to 15. Otherwise, when I N P T E = 3 Enter: STREN(I) ,1 = 1, N J T (strength of joists) 15. O N L Y for the first floor, Enter: BETA , ALPHA , R E G where these constants are defined in Step 12. D A T A O N N A I L I N G : (Entered only for the first floor) 16. Enter: ENL, XIN, RKPAL, RKPER, RKROT, NDISCR where E N L = nail spacing (in) along the joist span. Appendix C: USER'S MANUAL - FAPP 76 X I N = distance between support and first nai l along the joist span (in). R K P A L , R K P E R = nail load-slip moduli i , respectively, in the directions parallel and perpendicular to the joist. R K R O T = nai l rotation modulus. N D I S C R = 0 if the actual nailing pattern is transformed to an equivalent continuous connector for the purpose of analysis; = 1 if actual (discrete) nailing pattern is treated as actually is (discrete). The program uses the same type of nailing (nail and pattern) for a l l the joists. 16. Enter: E N L 1 , X I N l , R K P A L l , R K P E R l , R K R O T l , N D S C R l where E N L 1 , X I N l , R K P A L l , R K P E R l , R K R O T l , R K R O T l & N D S C R l are the same properties as described above but for the connection between the lower cover and the bot tom of the joist. D A T A O N G A P S I N T H E C O V E R ( S H E A T H I N G ) : (Entered only for the first floor) 17. Enter: N G A P S N G A P S = number of gaps present in the upper cover. These are gaps perpendic-ular to the joist direction. If NGAPS = 0 skip to 23. 18. Enter: N G T where N G T = 0 i f the gaps are as entered in 19. = 1 i f the gaps are adjusted , automatically, to the nail ing pattern. 19. Enter: from 1 = 1 unti l N G A P S , (for each gap), G A P X ( I ) , G A P ( I ) where G A P X ( I ) = x - coordinate of the beginning of the i th -gap (in) G A P ( I ) = gap width of the i ih - gap (in) 20. Enter: N G A P S 1 Appendix C: USER'S MANUAL - FAPP 77 where N G A P S = number of gaps present i n the lower cover. These are gaps perpendicular to the joist direction. If NGAPSl = 0 skip to 23. 21. Enter: N G T 1 where N G T 1 = 0 if the gas are as entered in 22. = 1 if the gaps are adjusted , automatically, to the nail ing pattern. 22. Enter: from I = 1 unti l N G A P S l (for each gap), G A P X l ( I ) , G A P 1 ( I ) where G A P X l ( I ) = x - coordinate of the beginning of the i th -gap (in) G A P 1(1) = gap width of the i th - gap (in) A P P L I E D L O A D S : (Entered only for the first floor) 23. Enter: N L O A D , N L U , P L O A D , S R E F , W R E F where N L O A D = number of loaded areas, entered one by one (max. 12). If N L O A D = 0 then the load is uniformly distributed over the entire floor and equal to PLOAD. N L U = 1 i f a l l the joists have the same load distribution , in which case one can have upto 12 loaded areas per joist; = 0 i f different joists have different loadings. In the case of N L O A D = 0, N L U is used to distinguish the cases of loaded or unloaded outer flanges for the outside joists. If N L U = 1, al l joists are equally loaded, meaning that even the outside joists receive the same load and thus their outer flanges are also loaded. If N L U = 0, the outside joists have their outer fanges unloaded. In the case of N L O A D ^ 0, the outside joists have the outer flanges unloaded auto-matically, only if N L U = 1. P L O A D = magnitude of the uniformly distributed load. S R E F = calculated stress on a joist (acting alone, wi th no sheathing contribution) under the load PLOAD. Appendix C: USER'S MANUAL - FAPP 78 W R . E F = calculated deflection of a joist (acting alone, wi th no sheathing contri-bution) under the load PLOAD and for an assumed modulus of elasticity E. SREF and WREF are used to compute load sharing factors for stresses and deflec-tions , relating these as occuring in the floor system to those occuring in a single joist w i th no sheathing contribution. If NLOAD = 0 (Uniformly distributed load) skip to 25. 24. W h e n N L O A D ^ 0, Enter: for each loaded area, (1 = 1 , N L O A D ) NLJO(I) , STORE(I,7) then STORE(I,3), STORE(I,4), STORE(I,5), STORE(I,5) where N L J O ( I ) = number of joist loaded; S T O R E ( I , 7 ) = load, uniformly distributed, over the loaded area; S T O R E ( I , 3 ) to S T O R E ( I , 6 ) = coordinate of the loaded area, respectively, X1,X2,Y1,Y2 are as shown i n F i g . C.2. XI and X2 are global coordinates, while Yl and Y2 are local to the joist element. B O U N D A R Y C O N D I T I O N S : (Entered only for the first floor) 25. Enter: NBC where N B C = number of boundary conditions imposed on the problem. If N B C = 0 skip to 27. 26. W h e n N B C ^ 0, Enter: for each boundary condition ( 1 = 1 , N B C ) , IBC(I,1) , IBC(I,2) where IBC(I ,1) = joist number where the boundary condition exists; IBC(I ,2) -= code number for the boundary condition imposed. See F i g . C.3 and the Table C . l . A l l boundary conditions , when imposed , restrict the corresponding displacement Appendix C: USER'S MANUAL - FAPP 79 or rotation to zero. 27. If I N P T E = 4 (Case of Simulations) this is the end of the file. For cases I N P T E ^ 0, Data for another floor must be entered unless only one floor is being analyzed ( N F L O R = 1). Data for the subsequent floors must be entered starting again from step 13. Units: The program works with any set of consistent units. That is, if inches are used, and loads are in pounds, stresses are given in psi and deflections in inches. If millimeters are used for dimensions, Newtons for loads, deflections will be in millimeters and stresses in MPa. M.O.R. E EOO MODULUS OF ELASTICITY E11 Figure C . l : Correlation between joist M.O.R. and its Modulus of Elasticity Appendix C: USER'S MANUAL - FAPP 80 Figure C .2: Loading Area Coordinates Appendix C: USER'S MANUAL - FAPP 1 <> O 3 (x)f Figure C.3: The Nodes of an element Appendix C: USER'S MANUAL - FAPP 82 Table C . l : Table of Boundary Condi t ion Codes POINT CONDITION CODE 1 VERTICAL DISPLACEMENT 1 1 ROTATION IN (z,y) PLANE 2 1 DISPLACEMENT IN x-DIRECTION 3 1 DERIVATIVE OF DISPLACEMENT IN x-DIRECTION w.r.t. x 4 1 DISPLACEMENT IN y-DIRECTION 5 1 DERIVATIVE OF DISPLACEMENT IN y-DIRECTION w.r.t. y 6 4 VERTICAL DISPLACEMENT 7 4 ROTATION IN (z,y) PLANE 8 4 DISPLACEMENT IN x-DIRECTION 9 4 DERIVATIVE OF DISPLACEMENT IN x-DIRECTION w.r.t. x 10 4 DISPLACEMENT IN y-DIRECTION 11 4 DERIVATIVE OF DISPLACEMENT IN y-DIRECTION w.r.t. y 12 2 DISPLACEMENT IN x-DIRECTION 13 2 DISPLACEMENT IN y-DIRECTION 14 2 ROTATION IN (z,y) PLANE 15 B VERTICAL DISPLACEMENT 16 B DISPLACEMENT IN x-DIRECTION 17 B DISPLACEMENT IN y-DIRECTION 18 B ROTATION IN z-y PLANE 19 5 DISPLACEMENT IN x-DIRECTION 20 5 DISPLACEMENT IN y-DIRECTION 21 5 ROTATION IN (z,y) PLANE 22 3 VERTICAL DISPLACEMENT 23 3 ROTATION IN (z,y) PLANE 24 3 DISPLACEMENT IN x-DIRECTION 25 3 DERIVATIVE OF DISPLACEMENT IN x-DIRECTION w.r.t. x 26 3 DISPLACEMENT IN y-DIRECTION 27 3 DERIVATIVE OF DISPLACEMENT IN y-DIRECTION w.r.t. y 28 6 VERTICAL DISPLACEMENT 29 6 ROTATION IN (z,y) PLANE 30 6 DISPLACEMENT IN x-DIRECTION 31 6 DERIVATIVE OF DISPLACEMENT IN x-DIRECTION w.r.t. x 32 6 DISPLACEMENT IN y-DIRECTION 33 6 DERIVATIVE OF DISPLACEMENT IN y-DIRECTION w.r.t. y 34 APPENDIX D P R O G R A M L I S T I N G - F A P P A listing of the program F A P P is presented in this appendix. The Computer Language F O R T R A N is used to develop the program. The user's manual describing the data file to run the program is described i n Appendix C . 83 C ••** FLOOR ANALYSIS PROGRAM, F.A.P ••• C •*** WITH GAPS IN ELEMENTS **** C ••••20 JOISTS LIMIT ***• C DOUBLE PRECISION TEMP,TEMP 1,TEMP2,TEMP3,TEMP4.STIFF C0MM0N/B2/RKX,RKY,RKV,RKG,DX,DY,DV,DG,RL, FACTOR, 1 RKX1,RKY1,RKV1,RKG1,DX1,DY1,DV1,DG1 C0MM0N/B3/ETA(6),H(6),S,NM,NSTEP, NMAX C0MM0N/B4/D,D1,EJT,GJT,BETA,ENL,ENL1,RKPAL,RKPAL1,RKPER,RKPER1 , 1 PLD.X1,X2,Y1,Y2.BJ (20),HJ(20), 2 HJTM, ALPHA, RKROT,RKROTl C0MM0N/B5/ST0RE(20,7),ECOV,ECOV1,NFACE,NFACE1,WP,TV,TV 1 COMMON/B6/GAPX0(5),GAP0(5),GAPX(5),GAP(5),XIN,NGAPS,NAI, NDISCR 1 NGAPS1,GAPX1(5),GAP 1(5),GAP01(5),GAPX01(5),XIN1,NDSCR1,NA11 C0MM0N/B8/STIFF(34,34) C0MM0N/B9/SVEC(6,452) DOUBLE PRECISION XXU52) DIMENSION GSTIF(6,15368),ASTIF{15,30284),DK(15368) DIMENSION F0RCE(6,452),X(452) DIMENSION VECTR(34),V(34) DIMENSION I B C U 6 . 2 ) , NLJ0(20) DIMENSION AJT(20),RIY(20),RIZ(20),RIT(20),SF(20),STREN(20) DIMENSION GAMS(20), GAMW(20), WFT(20), WCOV(20),WCOV1(20) DIMENSION TITLE(20) 1000 F0RMAT(13G20.10) C PI = 4.0D0 * DATAN(1.ODO) ETA( 1 ) = 0.932469514203152D0 ETA(2) = 0.661209386466265D0 ETA(3) = 0.238619186083319D0 H(1) = 0.171324492379170D0 H(2) = 0.360761573048139D0 H(3) = 0.467913934572691D0 DO 1 I = 1, 3 ETA(7-I) = -ETA(I) H(7-I) = H(I) 1 CONTINUE CALL GENMTX C C 21 IS A RANDOM NUMBER GENERATION SEED C CVS = VARIABILITY IN JOISTS' E: IN THIS CASE, E WILL BE C RANDOMLY CHOSEN BUT WITHIN + OR - CVS(%) OF THE FIRST C E. THIS FIRST E IS RANDOMLY CHOSEN FROM THE ENTIRE C DISTRIBUTION C CVS = 0.0 FOR RANDOM SELECTION FROM DISTRIBUTION: C IN THIS CASE, E WILL BE RANDOMLY SELECTED BETWEEN C THE INPUT LIMITS EMIN AND EMAX C NM IS THE MAXIMUM ORDER FOR THE SINE & COSINE SERIES C NM CAN BE UP TO 11 FOR SYMMETRIC PROBLEMS (ISYM = 1) C NM CAN ONLY BE UP TO 6 FOR NON-SYMMETRIC PROBLEMS C NJT IS NUMBER OF JOISTS C ISYM IS 1 FOR SYMMETRIC CASE C INPTE IS 2 FOR INPUTING PROPERTIES OF JOISTS ONE BY ONE C INPTE IS 3 FOR INPUTING PROPERTIES OF JOISTS ONE BY ONE C PLUS THEIR CORRESPONDING STRENGTHS C INPTE IS 4 TO SELECT PROPERTIES OF JOISTS FROM DISTRIBUTION C NFLOR IS THE NUMBER OF FLOORS TO BE RUN C IPRNT IS 0 FOR SUMMARY OUTPUT C NGAPS IS NUMBER OF GAPS IN EACH ELEMENT IN UPPER COVER. THESE GAPS C ARE PERPENDICULAR TO JOIST DIRECTION C GAPX(I) IS X-COORD OF BEGINNING OF I-TH GAP IN UPPER COVER C GAP(I) IS WIDTH OF I-TH GAP (NGT=0) IN UPPER COVER C NGAPS 1 IS NUMBER OF GAPS IN EACH ELEMENT IN LOWER COVER. THESE GAPS C ARE PERPENDICULAR TO JOIST DIRECTION C GAPXKI) IS X-COORD OF BEGINNING OF I-TH GAP IN LOWER COVER C GAP 1(1) IS WIDTH OF I-TH GAP (NGT=0) IN LOWER COVER C NGT = 0 IF GAPS ARE ENTERED BY GAPX(I) AND GAP(I) C = 1 IF GAPS ARE ADJUSTED AUTOMATICALLY TO THE NAILING PATTERN C NGT1= 0 IF GAPS ARE ENTERED BY GAPXKI) AND GAP 1(1) C = 1 IF GAPS ARE ADJUSTED AUTOMATICALLY TO THE NAILING PATTERN C EO.EM.EK PARAMETERS FOR JOISTS' EI DISTRIBUTION C EO = MINIMUM EI FOR THE DISTRIBUTION (LOCATION FOR 3-P WEIBULL C OR ASSUMED 0.0 FOR 2-P LOGNORMAL OR NORMAL DISTRIBUTIONS) C QA, QB ARE PARAMETERS FOR STRENGTH / E CORRELATION C QA = INTERCEPT OF REGRESSION LINE M.O.R / M.O.E C QB = SLOPE OF REGRESSION LINE M.O.R / M.O.E C CV1, CV2, CV3 _ARE COEFFICIENTS OF EQUATION TO COMPUTE THE C COEFFICIENT OF VARIATION OF M.O.R AS A FUNCTION OF M.O.E C EOO AND E11 ARE, RESPECTIVELY, THE MINIMUM AND MAXIMUM M.O.E C USED TO DETERMINE THE COEFFICIENTS CV1, CV2 AND CV3 C WARNING : EMIN MUST NOT BE SMALLER THAN EOO C EMAX MUST NOT BE LARGER THAN E11 C KDISTR = 1 IF THE EI DISTRIBUTION IS 3-P WEIBULL C = 2 IF THE EI DISTRIBUTION IS 2-P LOGNORMAL C = 3 IF THE EI DISTRIBUTION IS NORMAL C STORE(1,1) = E FOR JOIST I C ST0RE(I.2) = G FOR JOIST I C STORE(1,3) = XI FOR LOADED AREA I C STORE(I,4) = X2 FOR LOADED AREA I C STORE(1,5) = Y1 FOR LOADED AREA I C ST0RE(I,6) = Y2 FOR LOADED AREA I C ST0REQ.7) = DISTRIBUTED LOAD ON AREA I FROM X1 TO X2 AND Y1 TO Y2 C TOL = TOLERANCE FOR CONVERGENCE IF NM GREATER THAN 1 C ECOV = E FOR THE UPPER COVER, PARALLEL TO GRAIN (VENEER) C EC0V1= E FOR THE LOWER COVER, PARALLEL TO GRAIN (VENEER) C NFACE = 1 IF UPPER COVER GRAIN PARALLEL TO JOISTS C = 2 IF UPPER COVER GRAIN PERPENDICULAR TO JOISTS C NFACE1= 1 IF LOWER COVER GRAIN PARALLEL TO JOISTS C = 2 IF LOWER COVER GRAIN PERPENDICULAR TO JOISTS C TV = THICKNESS OF OUTERMOST UPPER VENNER C TV1= THICKNESS OF OUTERMOST LOWER VENEER C COVER STRESSES ALL PERPENDICULAR TO THE JOISTS C SPROOF = PROOF LEVEL (OR MINIMUM) STRENGTH FOR JOISTS C NAI =N0. OF NAILS ON EACH JOIST NAILED FROM UPPER COVER C NAI1=N0. OF NAILS ON EACH JOIST NAILED FROM LOWER COVER C ENL =NAIL SPACING ALONG THE SPAN ON UPPER COVER C ENL1=NAIL SPACING ALONG THE SPAN ON LOWER COVER C XIN =DIST. BETWEEN SUPPORT AND FIRST NAIL ALONG THE JOIST SPAN C ON THE UPPER COVER SIDE C XIN1=DIST. BETWEEN SUPPORT AND FIRST NAIL ALONG THE JOIST SPAN C ON THE UPPER COVER SIDE C NDISCR=0 IF THE ACTUAL NAILING PATTERN IN UPPER COVER IS C TRANSFORMED TO AN EQUIVALENT CONTINUOUS CONNECTOR C FOR THE PURPOSE OF ANALYSIS C =1 IF ACTUAL (DISCRETE) NAILING PATTERN IS TREATED AS 86 C ACTUALLY IS C NDSCR1=0 IF THE ACTUAL NAILING PATTERN IN LOWER COVER IS C TRANSFORMED TO AN EQUIVALENT CONTINUOUS CONNECTOR C FOR THE PURPOSE OF ANALYSIS C =1 IF ACTUAL (DISCRETE) NAILING PATTERN IS TREATED AS C ACTUALLY IS C C================================================================: c C READ(1,116) (T ITLE( I ) . 1=1,20) READ(1.1000) NM. NJT, ISYM, INPTE, NFLOR, IPRNT, TOL IF (INPTE.NE.4) GO TO 700 READ(1,1000) 21. CVS IF (CVS.NE.0.0) GO TO 706 READ(1.1000) EMIN, EMAX 706 READ( 1,1000) KDISTR, EO, EM, EK READM , 1000) QA. QB READ(1.1000) EOO, E l l . CV1, CV2, CV3 READ(1,1000) SPROOF 700 CONTINUE READ(1,1000)S,RL,D.D1 READ(1,1000) FACTOR READ(1,1000)RKX,RKY,RKV,RKG,DX,DY,DV,DG READ(1.1000) ECOV,NFACE,TV READ(1,1000)RKXl,RKY1,RKV1,RKG1,DX1,DY1,DV1,DG1 READ(1,1000) EC0V1.NFACE1,TV1 ECOV = ECOV * 10.0**6 EC0V1= ECOV1 * 10.0**6 II = 1 711 IF (INPTE.NE.4) GO TO 701 READ(1,1000)BJT,HJT.BETA,ALPHA, REG QI = BJT * (HJT**3) / 12.0 DO 703 I = 1, NJT BJ(I) = BJT 703 HJ(I) = HJT GO TO 702 701 IF ( I I .GT.1) GO TO 889 READM,1000) (BJ( I ) , 1 = 1, NJT) READ(1,1000) (HJ( I ) , 1 = 1 , NJT) 889 READM.1000) (STORE(I. I ) . I = 1, NJT) IF (INPTE.NE.3) GO TO 780 READM, 1000) (STREN(I), 1 = 1, NJT) 780 IF ( I I .GT.1) GO TO 115 READM, 1000) BETA,ALPHA, REG 702 HJTM = 0.0 DO 709 1 = 1 , NJT 709 HJTM = HJTM + HJ(I) HJTM= HJTM / NJT READM, 1000) ENL, XIN, RKPAL, RKPER, RKROT, NDISCR READ(1.1000) ENL1,XIN1,RKPAL1,RKPER1,RKROT1,NDSCR1 NAI = (RL - 2.0*XIN) / ENL NAI = NAI + 1 NAI1=(RL-2.0*XIN1)/ENL1 NAI1=NAI1+1 READ(1.1000)NGAPS IF(NGAPS.EQ.0)G0 TO 121 READM , 1000) NGT DO 123 1=1,NGAPS 87 READ( 1 .lOOO)GAPX(I),GAP(I) GAPXO(I) = GAPX(I) GAPO(I) = GAP(I) IF (NGT.EQ.O) GO TO 123 NG = (GAPX(I) - XIN)/ENL XG1 = XIN + NG * ENL XG2 = XG1 + ENL IF (XG1.EQ.GAPX(I)) GO TO 824 XG3 = GAPX(I) + GAP(I) GAPX(I) = XG1 GAP(I) = 2.0 * ENL IF (XG2.GT.XG3) GAP(I) = ENL GO TO 123 824 GAPX(I) = XG1 - ENL GAP(I) = 2.0 ' ENL 123 CONTINUE 121 CONTINUE READ(1,1000) NGAPS1 IF(NGAPS1.EQ.OJGO TO 421 READ(1,1000)NGT1 DO 423 1=1.NGAPS1 READ(1,1000)GAPX1(I).GAP 1(I) GAPX01(I)=GAPX1(I) GAP01(I)=GAP1(I) IF(NGT1.EQ.0)G0 TO 423 NG1=(GAPX1(I)-XIN11/ENL1 XG4=XIN1+NG1*ENL1 XG5=XG4+ENL1 IF(XG4.EQ.GAPX1(I))G0 TO 825 XG6=GAPX1(I)+GAP1(I) GAPX1(I)=XG4 GAP1(I)=2.0*ENL1 IF(XG5.GT.XG6)GAP1(I)=ENL1 GO TO 423 825 GAPX1(I)=XG4-ENL1 GAP 1(1) =2.0*ENL1 423 CONTINUE 421 CONTINUE READ(1,1000) NLOAD, NLU, PLOAD, SREF, WREF C PAUSE'STOPPED AT 211' C PLOAD = CHARACTERISTIC LOAD ( E .G . , THE MAXIMUM DISTRIBUTED LOAD) C IF NLOAD = 0 , U.D.L OVER ENTIRE FLOOR (PLOAD PSI) C IF NLU = 0 . OUTSIDE FLANGES ARE UNLOADED C IF NLU = 1 , OUTSIDE FLANGES ARE LOADED (ALL JOISTS C EQUALLY LOADED) C IF NLOAD DIFFERENT FROM ZERO, DEFINE C NLOAD = NUMBER OF LOADED AREAS, ENTERED ONE BY ONE (MAX. 20) C NLU = 0 IF DIFFERENT JOISTS HAVE DIFFERENT LOADING AS ENTERED C NLU = 1 IF ALL JOISTS ARE EQUALLY LOADED (IN THIS CASE. THERE C CAN BE UP TO 20 LOADED AREAS ON A JOIST). C SREF = JOIST ALONE BENDING STRESS UNDER PLOAD (TO COMPUTE LOAD SHARING C FACTOR GAMS(I) C WREF = JOIST ALONE DEFLECTION UNDER PLOAD (TO COMPUTE LOAD SHARING C FACTOR GAMW(I) IF (NLOAD.NE.O) GO TO 11 DO 19 I = 1, NJT STORE(I.3) = 0 . 0 STORE(1,4) = RL 88 ST0RE(I,5) = -S/2.0 STORE(1,6) = S/2.0 STORE(I,7) = PLOAD 19 CONTINUE GO TO 22 11 DO 5 I = 1, NLOAD READM.1000) NLJO(I), STORE(I,7) READ( 1,1000) ( STOREd . J ) , J=3,6) 5 CONTINUE 22 CONTINUE C INPUT BOUNDARY CONDITIONS C NBC = NUMBER OF B.C. C I B C U . U = JOIST NUMBER (ELEMENT NUMBER) C IBC(I ,2) = NUMBER OF THE CONSTRAINT (1-34) IN THE UNKNOWNS' VECTOR READ(1, 1000) NBC IF (NBC.EQ.O) GO TO 9 DO 6 I = 1, NBC 6 READ(1,1000) ( IBC( I . J ) , J=1,2) 9 CONTINUE DO 704 I = 1, NJT AJT(I) = BJ(I) * HJ(I) RIY(I) = BJ(I) * HJ(I)**3 /12.0 RIZ(I) = HJ(I) ' B J ( I ) " 3 /12.0 RIT(I) = BETA * HJ(I) * B J ( I ) " 3 704 CONTINUE PIN2 = PI* '2/(2.0*RL) PIN4 = PI*"4/(2.0*RL**3) NEQ = NJT * 22 + 12 LHB = 34 NA = 34 ' NEQ NA1 = 6 7 * NEQ NSTEP = 1 IF (ISYM.EQ.1) NSTEP = 2 NMAX = NM IF (NSTEP.EQ.2) NMAX = (NM + 1)/2 IF(IPRNT.EQ.0)G0 TO 130 . 116 FORMAT(20A4) C WRITE(2,102)(TITLE(I), I=1,20) C102 FORMAT(1H1,' FLOOR ANALYSIS ' ,5X,20A4./,2X,16(1H*),//) WRITE(2,103)NJT,NM 103 FORMAT(' NUMBER OF ELEMENTS = ' , I5 ,5X , 'N =', I5,/) WRITE(2,550)NGAPS,NGAPS1 550 FORMAT(// , 'NO.OF GAPS(UPPER) =' . 15, 5X , ' NO.OF GAPS (LOWER ) = '•, 15 ,//) WRITE(2,551)S,RL 551 FORMAT('JOIST SPACING=',E13.6,5X,' JOIST SPAN=',E13 . 6 ,/) WRITE(2,104) 104 FORMAT(' PROPERTIES AND DIMENSIONS FOR UPPER FLANGE:' ,/) WRITE(2.105)D,ECOV,NFACE 105 FORMATt'UPPER COVER THICKNESS=',E13.6,/, 1 'MOD. OF ELASTICITY OF OUTERMOST VENEER=' ,E13.6 , / , 2 'DIRECTION OF COVER GRAIN=',13,/) WRITE(2, 106)RKX,RKY,RKV,RKG 106 FORMAT(' K(X) = ' ,E13.6,10X, 'K (Y ) = ' ,E13.6 . &'K(V) =' .E13.6.10X, 'K(G) =' ,E13.6) WRITE(2,107)DX,DY,DV,DG 107 FORMAT(' D(X) = ' ,E13.6,10X, 'D(Y) = ' ,E13.6 . &'D(V) =' .E13.6,10X, 'D(G) =' .E13.6) WRITE(2,552) c 115 CONTINUE IF (IPRNT.EQ.O) GO TO 838 WRITE(2,113) II 113 FORMAT(1H1,' FLOOR NO.= ' , I 3 , / . ' JOIST PROPERTIES'./) 838 CONTINUE IF(INPTE.EQ.4)GO TO 622 GO TO 833 622 DO 648 1 8 = 1 , NJT 625 IF(INPTE.EQ.4) THEN 21 = RAND(21) ELSE Z1 = RAND(I21,IZ2) ENDIF C IF (21.LE.O.O.OR.21.GE.1.0) GO TO 625 CALL DISTR(Z1.KDISTR,01,EO,EM,EK.STORE(18,1)) C WRITE(2.4639) 18 , ST0RE(I8,1) C4639 FORMAT('ELAS(',13, ' ) = ' , 4X.F13.7,/) IF (CVS. EC. 0.0) G'O TO 707 IF (18.EQ.1) GO TO 626 IF (STORE(18,1).GE.ESMIN.AND.STORE(18,1).LE.ESMAX) GO TO 624 GO TO 625 626 IF (ST0RE(IB,1).LT.E00.0R.ST0RE(I8,1).GT.E11) GO TO 625 ESMIN = STORE(18,1)*(1.0 - CVS) IF (ESMIN.LT.EOO) ESMIN = EOO ESMAX = STORE(I8,1)*(1.0 + CVS) IF (ESMAX.GT.E11) ESMAX = E11 GO TO 624 707 IF (STORE(18,1).GE.EMIN.AND.STORE(18,1).LE.EMAX) GO TO 624 GO TO 625 624 CONTINUE C C 708 IF(INPTE.EQ.4) THEN 21 = RAND(21) ELSE Z1 = RAND(IZ1,IZ2) ENDIF C IF (Z1.LE.0.0.OR.Z1.GE.1.0) GO TO 708 CALL STREND(STREN(IB) ,STORE(18,1) ,QA,QB,CV1,CV2,CV3,EOO,E11,21. 1 KOUT,SPROOF) C WRITE(2,4621) 18 . STREN(18) C4621 FORMAT( 'STREN(', I 3,') = ' .4X. F13.7,/) C C WRITE(',*) 'KOUT=',KOUT IF (KOUT.EQ.1) GO TO 625 IF (STREN(IS).GT.SPROOF) GO TO 648 GO TO 708 648 CONTINUE 833 CONTINUE DO 834 18=1, NJT ST0RE(I8,1) = ST0RE(I8. 1)*10.0**6 STORE(I8.2) = STORE(18,1)/REG IF (IPRNT.EQ.O) GO TO 834 WRITE(2.114) 18. ( STORE (13 , J ) , J = 1,2), B J i .6 ) , HJU8) C834 CONTINUE 90 114 FORMAT(' JOIST ' ,12 ,10X. ' E =' .E13.6,10X , 'G =' .E13.6, 110X,'B = ' , F7 .3 ,10X, 'H =' ,F7.3) 834 CONTINUE C C IF (II .GT. 1 ) GO TO 770 RKX = RKX * FACTOR RKY = RKY * FACTOR RKV = RKV * FACTOR RKG = RKG * FACTOR DX = DX * FACTOR DY = DY * FACTOR DV = DV * FACTOR DG = DG * FACTOR RKPAL = RKPAL * FACTOR RKPER = RKPER • FACTOR RKROT = RKROT • FACTOR 770 CONTINUE IF( I I .GT.1)G0 TO 563 RKX1=RKX1*FACT0R RKY1=RKY1'FACTOR RKV1=RKV1'FACT0R RKG1=RKG1*FACT0R DX1=DX1'FACT0R DY1=DY1 * FACTOR DV1=DV1*FACT0R DG1=DG1'FACTOR RKPAL1=RKPAL1 * FACTOR RKPER1=RKPER1'FACTOR RKR0T1=RKR0T1'FACTOR 563 CONTINUE DO 410 I = 1. NJT STORE(1,1) = STORE(1,1) ' FACTOR 410 ST0RE(I,2) = ST0RE(I,2) • FACTOR C PAUSE'STOPPED AT 413' C * ASSEMBLY OF GLOBAL MATRICES ' DO 300 IK = 1,NM,NSTEP DO 300 IN=1,IK,NSTEP IKM = IK INM = IN IF (NSTEP.EQ.1) GO TO 3000 IKM = (IK+D/2 INM = (IN+1 )/2 3000 IF (I '\EQ.IK) GO TO 735 IF ( I I .GT.1) GO TO 300 IKO = (INM-1)'NMAX + IKM - INM'(INM+1)/2 DO 3 I = 1, NA1 3 ASTIFdKO.I ) 0.0 GO TO 260 735 IKO = IKM DO 3001 I = 1, NA 3001 DK(I) = 0.0 DO 4 1=1,NEQ 4 FORCE(IKO,I)=0.0 DO 250 IE=1.NJT EJT=STORE(IE, 1) GJT=ST0RE(IE, 2) KLO = 0 91 ILO = 1 IF (NLOAD.NE.O) GO TO 737 X1 = STORE(IE,3) X2 = STORE(IE,4) Y1 = STORE(IE,5) Y2 = STORE(IE,6) IF (NLU.EO.1) GO TO 736 IF (IE.EQ.1) Y1 = 0.0 IF (IE.EQ.NJT) Y2 = 0.0 736 PLD = ST0RE(IE,7) * FACTOR GO TO 740 737 IF (NLJO(ILO).EQ.IE.OR NLU.EQ.1) GO TO 739 ILO = ILO + 1 IF (ILO.GT.NLOAD) GO TO 742 GO TO 737 739 X1 = ST0RE(IL0,3) X2 = ST0RE(IL0,4) Y1 = ST0RE(IL0,5) Y2 = ST0RE(IL0,6) 738 PLD = ST0RE(IL0,7) * FACTOR 740 CALL STIF(VECTR,IE,KLO,IN,IK,PI) GO TO 741 742 IF (KL0.EQ.1) GO TO 745 PLD = 0.0 CALL STIF(VECTR,IE,KLO,IN,IK,PI) 741 IJ = (IE-1)*22 - DO 204 J=1,34 IF (KLO.EQ.1) GO TO 746 DO 203 K=1.J JK=LHB*(IJ+K-1)+J-K+1 IF (K.NE.J) GO TO 408 IF (J.EO.19) GO TO 404 IF (J.EQ.18) GO TO 403 IF (J.EQ.17) GO TO 402 IF (J.EQ.16) GO TO 401 GO TO 408 401 TEMPI = DK(JK) + STIFF(J,K) GO TO 203 402 TEMP2 = DK(JK) • STIFF(J.K) GO TO 203 403 TEMP3 = DK(JK) + STIFF(J.K) GO TO 203 404 TEMP4 = DK(JK) • STIFF(J.K) GO TO 203 408 DK(JK) = DK(JK) + STIFF(J.K) 203 CONTINUE 746 FORCE(IKO,IJ+J) = FORCE(IKO,IJ+J) + VECTR(J) 204 CONTINUE KLO = 1 IF (NLOAD.EQ.O) GO TO 745 ILO = ILO + 1 IF (ILO.GT.NLOAD) GO TO 745 GO TO 737 745 JK = LHB* (IJ -H5) + 1 DK(JK)=TEMP1+EJT*RIY(IE)*PIN4*IK**4 JK=LHB*(IJ+16)+1 DK(JK)=TEMP2 + EJT"AJT(IE)'PIN2*IK'* 2 JK=LHB' ( IJ+17) + 1 DK(JK)=TEMP3 + EJT *RIZ(IE)*PIN4*IK**4 JK=LHB*(IJ+18)+1 DK(JK)=TEMP4+GJT* RIT(IE)*PIN2*IK**2/(S** 250 CONTINUE C PAUSE'STOPPED AT 497.021' C C WRITE(2,929) (STIFF(17,K),K= 1, 34) 929 F0RMAT(8E15.6) C WRITE(2,930) DK(545) 930 F0RMAT(E15.6) C IF(NBC.EO.O)GO TO 206 DO 205 1=1,NBC NE=IBC(I,1) ND0F=IBC(I,2) M=(NE-1)'22+ND0F J1= 1 MM=M-1 IF(M.GE.34)J1=M-33 DO 17 J=J1,MM JK=(LHB-1)'(J-1)+M 17 DK(JK)=0.0 M1=M+1 M8=M+33 IF (M8.GT.NEQ) M8 = NEQ DO 18 J=M1.M8 JK=(LHB-1)*(M-1)+J 18 DK(JK)=0.0 JK=(LHB- 1)*(M-1)+M DK(JK)=1.0 F0RCE(IK0,M)=0.0 205 CONTINUE 206 CONTINUE CALL DECMP(NEQ,LHB,DK) C DO 3002 I = 1, NA 3002 GSTIF(IKO.I) = DK(I) GO TO 300 260 CONTINUE DO 280 IE = 1, .NJT CALL STIF(VECTR,IE,O.IN,IK,PI) IJ=(IE-1)*22 DO 264 J=1,34 DO 264 K=1 .34 JK = (IJ+K-1)*(2*LHB-1)+LHB+J-K ASTIFdKO, JK)=ASTIF(IKO, JK)+STIFF( J.K) 264 CONTINUE 280 CONTINUE IF(NBC.EQ.O)GO TO 300 DO 288 1=1,NBC NE=IBC(I,1) NDOF=IBC(I,2) M=(NE-1)'22+ND0F M1=M-33 M2=M+33 IF(M1.LE.0)M1=1 IF(M2.GT.NEQ)M2=NEQ DO 285 J=M1,M2 285 288 300 JK=(M-1)*(2*LHB-1 )+J-M+LHB ASTIF(IKO,JK)=0.0 JK=(J- 1)*(2'LHB-1)+M-J+LHB ASTIF(IKO,JK)=0.0 CONTINUE CONTINUE •SOLUTION OF SYSTEM" DO 320 IK = 1, NMAX DO 3003 I = 1, NA 3003 DK(I) = GSTIFdK DO 3004 I = 1, NEQ 3004 X(I) = FORCE(IK.I) I) C C958 C C959 C DEFORMATION VECTOR' ,NEO) , /) 3005 320 322 870 C C C CALL SOLV(NEQ,LHB,DK,X) WRITE(2,958) IK FORMATt/, ' IK='. I2.2X, WRITE(2,959) (X£ I),1 = 1 F0RMAT(E15.6) DO 3005 I = 1, NEQ SVECUK. I ) = X U ) CONTINUE PAUSE'AFTER THE BRANCHING CONDITION AT 595 IF (NM.EQ.1) GO TO 400 IF(NGAPS.EQ.O.AND.NGAPS1.EQ.O)GO TO 400 NFLAG = 1 DO 350 IK = 1, NM, NSTEP IKM = IK IF (NSTEP.EQ. 2) IKM = (IK+D/2 DO 870 I = 1. NEQ XX(I) = FORCE(IKM,I) PAUSE'STOPPED AFTER LINE 603' DO 340 IN = 1, NM, NSTEP IF (IN.EQ.IK) GO TO 340 INM = IN IF (N^TEP.EQ.2) INM= (IN+D/2 IF (IN.GT.IK) GO TO 3260 IKO = (INM-1)'NMAX + IKM - INM*(INM+1)/2 GO TO 3263 3260 IKO = (IKM-1)'NMAX + INM - IKM*(IKM+1)/2 3263 DO 3012 I = 1, NEQ J1 = I - 33 J2 = I + 33 IF (J1 .LE.O) J l = 1 IF (J2.GT.NEQ) J2 = TEMP = 0 . 0 DO 330 J = J l , J2 TEMP 1 = SVEC(INM.J) IF (IN.GT.IK) GO TO IJ = ;j-1)*(2*LHB-1) GO TO 3266 NEQ 3265 LHB 3265 IJ = (I -1)*(2*LHB-1) + LHB + J - I 3266 TEMP = TEMP + ASTIF(IKO.IJ) * TEMPI 330 CONTINUE XX(I) = XX(I) - TEMP 3012 CONTINUE 340 CONTINUE C C PAUSE'STOPPED AFTER LINE 628' C DO 3013 I = 1, NA 3013 DK(I) = GSTIF(IKM.I) DO 3014 I = 1. NEO 3014 X(I) = XX(I) CALL SOLV(NEQ,LHB,DK,X) TEMP = 0 . 0 TEMPI = 0 . 0 DO 3015 I = 1, NEQ TEMP = TEMP + X(I)**2 3015 TEMPI = TEMPI + SVEC(IKM,I)* * 2 TEMP = DSQRT(TEMP) TEMPI = DSQRT(TEMPI) EPS = DABS((TEMP - TEMPI)/TEMP1) NAC = 1 IF (EPS.GT.TOL) NAC = 0 NFLAG = NFLAG'NAC DO 873 1 = 1 , NEQ 873 SVEC(IKM.I) = X(I) 350 CONTINUE IF (NFLAG.EQ.1) GO TO 400 GO TO 322 C C ' OUTPUT ' C NLOAD =N0. OF LOADED AREAS .ENTERED ONE BY ONE ( MAX. 12 ) C IF NLOAD=0 THEN THE LOAD IS UNIFORMLY DISTRIBUTED C OVER THE ENTIRE FLOOR AND EQUAL TO PLOAD C NLU =1 IF ALL JOISTS HAVE THE SAME LOAD DISTRIBUTION, IN C WHICH CASE ONE CAN HAVE UPTO 12 LOADED AREAS PER JOIST C =0 IF DIFFERENT JOISTS HAVE DIFFERENT LOADINGS C PLOAD= MAGNITUDE OF THE UNIFORMLY DISTRIBUTED LOAD C SREF=STRESS ON A JOISKACTING ALONE WITH NO SHEATHING CONTRIB-C -UTION) UNDER THE LOAD PLOAD C =6'w'S*L 'L/8*B'H*H, w - UNIFORMLY DISTRIBUTED LOAD C S - SPACING BETWEEN JOISTS C B*H*H - SECTION MODULUS C =6'P*L/4*B*H*H, P - LOCALISED LOAD C C WREF=DEFLECTION OF A JOISTUCTING ALONEWITH NO SHEATHING CONTR C -IBUTION) UNDfR THE LOAD PLOAD AND FOR AN ASSUMED MOD. OF C ELASTICITY E. C =(5/384)(w*S*L**4/E ,I) C C GAMS(I)=LOAD SHARING FACTOR FOR MAX. BENDING STRESS OF I-TH C JOIST IN THE FLOOR C C GAMSM=MAX. LOAD FACTOR IN THE CASE ABOVE AMONG ALL THE JOISTS C IN THE FLOOR C C GAMW(I)=LOAD SHARING FACTOR FOR MAX. DEFLECTION OF I-TH JOIST C IN THE FLOOR C C GAMWM=MAX. LOAD SHARING FACTOR IN THE CASE ABOVE AMONG ALL THE C JOISTS IN THE FLOOR C C PAUSE'BEFORE CALL OUT ' 400 CALL 0UT(NEQ,NJT,WMAX,SMAX,SF.WFT.WC0V.WC0V1,WCM,WCM1,CMA,CMA1, & IPRNT.PI,WSM) C PAUSE 'AFTER CALL OUT' C WRITE(3.627) II, WMAX, SMAX, WSM ,(SF(I),1=1,NJT) C627 FORMAT ( // , FLOOR NO. . 18 , / / , C 1 'MAX. JOIST DEFLN. IN FLOOR=' ,G20. 10,//. C 2 'MAX. JOIST BENDING STRESS IN FLOOR=',G20.10,//, C 3 'MAX. STRESS=',G20.10,'(NODE 2 - XDIRECTION)',//, C 4 'MAX. BENDING STRESS FOR EACH J0IST=',20G20.10,//) C IF(ABS(WMAX).GE.A8S(WMAFL)) WMAFL = WMAX IF(ABS(SMAX).GE.ABS(SMAFL)) SMAFL = SMAX IF(ABS(WSM).GE.ABS(WSMFL)) WSMFL = WSM IF(II.EQ.NFLOR) THEN WRITE(5.575) WMAFL , SMAFL , WSMFL 575 FORMAT(//,'JOIST DEFLECTION=',G20.10,//, 1 'JOIST BENDING STRESS=' ,G20. 10,// , 2 'UPPER COVER STRESS (X - DIRECTION)',G20.10) ELSE GO TO 580 ENDIF 580 CONTINUE C WSTAT = WMAX * SIN(PI*(0.5 • 24.0/RD) WSTAT = 25.4 * WSTAT 888 FORMAT(F6.3) C WRITE(3,632) (WFT(I). 1 = 1 , NJT) C632 FORMAT(/,"MAX. DEFLN. FOR EACH JOIST IN FLOOR='.20G20.10,//) GAMSM = 0 . 0 GAMWM = 0 . 0 DO 784 1 = 1 . NJT GAMS(I) = SF(I) / SREF GAMW(I) = WFT(I) / WREF IF (ABS (GAMS (I) ) . GT . GAMSM) GAMSM = ABS(GAMSd)) IF (ABS(GAMW(I)).GT.GAMWM) GAMWM = ABS(GAMW(I)) 784' CONTINUE C PAUSE'AFTER DIVISION BY SREF AND WREF' C WRITE(3,639) GAMSM, (GAMS(I),1=1,NJT) C639 FORMAT(//, 'MAX. LOAD SHARING FACTOR FOR BENDING STRESS FORALL C 1 THE JOISTS='.4X.G20.10,//, 'LOAD SHARING FACTOR FOR BENDING C 2 STRESS FOR EACH JOIST=' .4X,20G20.9 ,//) C WRITE(3,640) GAMWM, (GAMW(I), 1 = 1 . NJT) C640 FORMAT(/,'MAX. LOAD SHARING FACTOR FOR JOIST DEFLN=',4X,G20.10, C 1 /.'LOAD SHARING FACTOR FOR BENDING STRESS FOR EACH JOIST=', C 2 4X.20G20. 10,//) C WRITE(3,641) CMA,CMA1, WCM.WCM1, (WCOV(I), 1 = 1 , NJT) C641 FORMAT(/,'MAX. BENDING STRESS IN UPPER SHEATHING=' ,X,G20 . 10,//, C 1 'MAX. BENDING STRESS IN LOWER SHEATHING=',2X,G20.10,//. C 2 'MAX. UPPER SHEATHING DEFLN. BETWN. JOISTS=' ,2X,G20. 10 , // , C 3 'MAX. LOWER SHEATHING DEFLN. BETWN. JOISTS='.2X,G20.10,//, C 4 'MAX. UP. SHEATH. DEFLN.BET. JOISTS FC~ EACH JOIST='. C 5 20G15.6) c IF (ABS(WCM).GE.ABS(WCMFL)) WCMFL = WCM IF (ABS(WCM1).GE.ABS(WCMIFL)) WCM1FL = WCM1 IF (II.EQ.NFLOR) THEN WRITE(5,585) WCMFL , WCM1FL 585 FORMAT(// , 'MAX. UPPER COVER DEFLN. BET. JOISTS=',2X,G20.10,//, 1 'MAX. LOWER COVER DEFLN. BET. JOISTS=',2X,G20.10) ELSE GO TO 590 ENDIF 590 CONTINUE C C WRITE(3.643) (WCOVKI). 1= 1, NJT) C643 FORMAT( C 1 //, 'MAX. LR. SHEATH. DEFLN.BET. JOISTS FOR EACH JOIST=', C 2 20G15.6) IF (INPTE.EQ.2) GO TO 782 SFMAX = 0 . 0 DO 781 I = 1. NJT SF(I) = SF(I) / STREN(I) 781 IF (SF(I).GT.SFMAX) SFMAX = SF(I) C PAUSE'AFTER LOOP ENDING @ 781; SFMAX FOUND' PMAX = PLOAD / SFMAX C WRITE(3,B42) (SF( I ) , I = 1, NJT) C642 FORMAT(/,'STRESS RATIO FOR EACH J0IST='.4X. 20G20.10.//) C WRITE(3,655) PMAX C655 FORMAT (/.'LOAD PRODUCING FIRST JOIST FAILURE='.4X,G20.10,//) 782 CONTINUE II = II + 1 IF (II.GT.NFLOR) GO TO 629 IF (INPTE.NE.4) GO TO 711 GO TO 115 629 CONTINUE IF (IPRNT.EQ.O) GO TO 771 GO TO 772 771 CONTINUE 772 STOP END C============================================================ C SUBROUTINE DISTR(Z1.KDISTR,QI,EO,EM.EK.ELAS) IF (KDISTR.EQ.3) GO TO 20 IF (KDISTR.EQ.2) GO TO 10 ELAS=EO + EM*(-ALOG(1.0-21))* *(1.O/EK) ELAS = E AS / QI RETURN 10 ELAS = EM + EK * RN0RMI21) ELAS = EXP(EL'S) / 01 RETURN 20 ELAS = EM + EK • RNORMt Z1) ELAS = ELAS / QI RETURN END C===================================================================== C SUBROUTINE STRENDfS. E. A. B. CV1,CV2,CV3,EOO.E11,21.KOUT.SPROOF) KOUT = 0 CV = C V1 * ( (E - E00)*"CV2) * ( ( E l l - E ) " C V 3 ) 97 c C WRITE(',* ) ' CV=' ,CV C SMAX = ( A + B*E)'( 1.0 + 2.0*CV) C C WRITEC.*) 'SMAX=',SMAX C IF (SPROOF.GE.SMAX) GO TO 10 S = ( A + B ' E ) * ( 1 . 0 + C V * RN0RM(Z1); RETURN 10 KOUT = 1 RETURN END C======================================================== c FUNCTION RN0RM(Z1) RMEAN = 0 . 0 RDT = 1.0 IP = 0 IF (Z1.GT.0.5) GO TO 310 GO TO 311 310 IP = 1 Z1 = 1.0 - Z1 311 TEM = SORT(-2.0*ALOG(Z1 ) ) A1 = 2.51557 • TEM * (0.802853 + TEM'O.010328) A2 = 1.0+TEM*(1.432788+TEM*(0.189269+TEM'1.308E-3)) VV = TEM - A1/A2 IF (IP.EQ.1) GO TO 312 GO TO 313 312 VV = -VV 313 RNORM = VV * RDT + RMEAN RETURN END C==========================================; c SUBROUTINE DMAT(II) DOUBLE PRECISION A COMMON/B8/A(34,34) COMMON/B3/ETA(6),H(6) C0MM0N/B10/RM(6,34) , RMK6.34) DO 1 1=1,34 DO 1 J= 1.34 A(I , J)=O.DO 1 CONTINUE DO 2 1=1.6 DO 101 J=1.34 F1 = RM(I,J) IF ( I I .EQ.2.0R.I I .EQ.4) F1 = RMKI. J ) IF (F1 .EQ.O.O) GO TO 101 DO 100 K=1,34 F2 = RMKI.K) IF (I I.EO. 1.0R.II.EQ.4) F2 = RM(I.K) IF (F2.EQ.O.O) GO TO 100 A(J,K) = A(J.K) + F1'F2*H(I) 100 CONTINUE 101 CONTINUE 2 CONTINUE RETURN END C=============================== C SUBROUTINE DECMP(N,LHB,A) DOUBLE PRECISION TEMP,SUM DIMENSION A(15368) C "A IS STORED COLUMNWISE * KB=LHB-1 C 'DECOMPOSITION* TEMP=A(1) TEMP=DSQRT(TEMP) A(1)=TEMP DO 1 1=2,LHB 1 A(I)=A(I)/TEMP DO 20 J=2,N J1=J-1 IJD=LHB*J-KB SUM=A(IJD) K0=1 IF(J.GT.LHB) KO=J-KB DO 5 K=K0,J1 JK=KB*K+J-KB TEMP=A(JK) 5 SUM=SUM-TEMP"2 A(IJD)=DSORT(SUM) DO 18 1=1,KB II=J+I K0=1 IF(II.GT.LHB)KO=II-KB SUM=A(IJD+I) IF(I.EQ.KB)GO TO 15 DO 10 K=K0,J1 JK=KB'K+J-KB IK=KB*K+II-KB TEMP=A(JK) 10 SUM=SUM-A(IK)* TEMP 15 A(IJD+I)=SUM/A(IJD) 18 CONTINUE 20 CONTINUE RETURN END C=============================== C SUBROUTINE SOLV(N,LHB,A,B) DOUBLE PRECISION TEMP, SUM DIMENSION A( 15368) ,6(452) C 'FORWARD SUBSTITUTION' KB=LHB-1 TEMP=A(1) B( 1)=B( D/TEMP DO 30 1=2,N 11=1-1 K0=1 IF(I.GT.LHB)KO=I-KB SUM=B(I) II=LHB'I-KB DO 25 K=K0,I1 IK=KB'K+I-KB TEMP=A ( IK ) 25 SUM=SUM-TEMP*B(K) B ( I ) = S U M / A ( I I ) 30 CONT INUE C "BACKWARD S U B S T I T U T I O N * N1=N-1 L B = L H B ' N - K B TEMP=A (LB ) B ( N ) = B ( N ) / T E M P DO 50 1=1,N1 I1=N- I+1 N I=N- I KO=N I F ( I . G T . K B ) K O = N I + K B SUM=B(N I ) I I = L H B * N I - K B DO 40 K = I 1 , K 0 IK=KB*N I+K-KB TEMP=A ( IK ) 40 SUM=SUM-TEMP 'B (K ) ' B ( N I ) = S U M / A ( I I ) 50 CONT INUE RETURN END C==================================================================== c SUBROUT INE S T R E S S ( N 1 , S T R , W J . W J S , W J T , W C , W C B . W C B 1 , S C 1 , S C 2 , S C 3 SC4 & C Q 1 . C Q 2 , S M 1 , S M 2 , I E , P I , W C 1 ) D IMENS ION S T ( 3 4 ) , S T 1 ( 3 4 ) , W ( 3 4 ) , W 1 ( 3 4 ) , W S ( 3 4 ) , W S 1 ( 3 4 ) D IMENS ION F ( 4 5 2 ) . C K 1 ( 1 9 ) , C K 2 ( 1 9 ) , C L 1 ( 19) , C L 2 ( 19) C 0 M M 0 N / B 2 / R K X , R K Y , R K V , R K G , D X , D Y , D V , D G , R L , F A C T O R . & RKX 1 , R K Y 1 , R K V 1 , R K G 1 , D X 1 , D Y 1 , D V 1 , D G 1 C O M M O N / B 3 / E T A ( 6 ) , H ( 6 ) , S , N M , N S T E P , NMAX C 0 M M 0 N / B 4 / D . D 1 . E J T . G J T . B E T A , E N L . E N L 1 , R K P A L . R K P A L 1 . R K P E R , R K P E R 1 & P L D . X 1 . X 2 . Y 1 , Y 2 , B J ( 2 0 ) , H J ( 2 0 ) , H J T M , A L P H A , R K R O T , R K R O T 1 C 0 M M 0 N / B 5 / S T 0 R E ( 2 0 , 7 ) . E C O V , E C O V 1 , N F A C E , N F A C E 1 , W P , T V , T V 1 C0MM0N/B6/ G A P X 0 ( 5 ) , G A P 0 ( 5 ) , G A P X ( 5 ) . G A P ( 5 ) , X IN , NGAPS. NA I , & N D I S C R , N G A P S 1 , G A P X 1 ( 5 ) , G A P 1 ( 5 ) , G A P 0 1 ( 5 ) . G A P X 0 1 ( 5 ) , X I N 1 , & NDSCR1 ,NA11 C O M M O N / B 9 / S V E C ( 6 . 4 5 2 ) C DO 1 I = 1, 19 W ( I ) = 0 . 0 W S ( I ) = 0 . 0 1 S T ( I ) = 0 . 0 C E J T = J T O R E ( I E , 1 ) G J T = S T 0 R E ( I E , 2 ) B J T = B J ( I E ) H J T = H J ( I E C DO 10 N = 1, NM, NSTEP P I N = N * P I P I N L = P IN / RL IKO = N I F ( N S T E P . E Q . 2 ) IKO = ( N + D / 2 DO 60 J = 1. N1 60 F ( J ) = S V E C ( I K O . J ) 100 J = (IE - 1)* 22 + 16 FAC=(-F(J+1)* PINL +HJT"F(J)*(PINL**2)/2.0)* STORE(IE,1) DO 92 I = 1, 19 XL = SIN(0.05*I'PIN) W(I) = W(I) + F(J)*XL ST(I) = ST(I) + FAC'XL 92 CONTINUE 10 CONTINUE C STR = 0 . 0 WJT = 0.0 DO 12 I = 1, 19 IF (ABS(ST(I)).GE.ABS(STR)) STR = ST(I) IF (ABS(W(I)).GE.ABS(WJT)) WJT = W(I) 12 CONTINUE STR = STR / FACTOR DO 16 I = 1,19 W(I) = 0.0 W1(I)= 0.0 WS(I) = 0.0 CK1(I)=0.0 CK2(I)=0.0 CL1(I)=0.0 CL2(I)=0.0 ST(I) = 0.0 WS1(I)= 0.0 16 ST1(I)= 0.0 C ECC = ECOV /(1.0 - 0.02*0.40) ECC1=EC0V1/(1.0-0.02*0.40) EXC=ECC C C DO 30 N = 1. NM, NSTEP PIN = N*PI PINL = PIN /RL IKO = N IF (NSTEP.EQ.2) IKO = (N+D/2 DO 62 J = 1, N1 62 F(J) = SVEC(IKO.J) IF(NFACE.EQ.2.AND.NFACE1.EQ.2)GO TO 13 UYX=0.02 UYX1=0.02 HD=D/2.0 -TV HD1=D1/2.0-TV1 GO TO 15 13 UYX=0.02 UYX1=0.02 HD=D/2.0 HD1=D1/2.0 15 CONTINUE J=(IE-1)'22 FAC 1 = F(J*6)/S - UYX*PINL*F(J + 3) - HD *(- 46.0 * F(J*1) 1 +14.0'F(J + 23) + 32.0*F(J+16)-12.0*F(J+2)-2.0*F(J + 24) 2 - 16.0*F( J+15) )/(S* *2) + UYX"HD*F(J+1)*PINL"2 C FAC2 = ( - 1 .5*F(J + 5) + 1.5*F(J +27)-0.25*F(J + 6)-0.25*F(J + 28))/S 1 - UYX*PINL*F(J+131+ HD*(8.0*F(J+1)+8.0'F(J+23)-16.0*F(J+16) 101 2 + F ( J + 2 ) - F ( J + 2 4 ) ) / ( S * * 2 ) - U Y X * H D * F ( J + 1 6 ) ' P I N L ' ' 2 F A C 7 = - P I N L * F ( J + 1 3 ) - ( ( D / 2 ) - T V ) * ( P I N L * ' 2 ) * F ( J + 1 6 ) 1 + ( 0 . 0 4 / S ) * ( - . 7 5 * F ( J + 5 ) + 0 . 7 5 * F ( J + 2 7 ) - 0 . 1 2 5 * F ( J + 6 ) - . 1 2 5 ' F ( J + 2 8 ) ) 2 + ( ( D / 2 ) - T V ) " ( 0 . 0 8 / S / S ) * ( 2 . 0 * F ( J + 1 ) + 2 . 0 * F ( J + 2 3 ) - 4 , 0 ' F ( J + 1 6 ) 3 + 0 . 2 5 * F ( J + 2 ) - 0 . 2 5 * F ( J + 2 4 ) ) F A C 8 = - P I N L * F ( J + 3 ) - ( ( D / 2 ) - T V ) * ( P I N L * * 2 ) ' F ( J + 1 ) + 0 . 0 2 * F ( J + 6 ) / S 1 + ( ( D / 2 ) - T V ) * ( 0 . 0 8 / S / S ) * ( 3 . 5 * F ( J + 2 3 ) - 1 1 . 5 ' F ( J + 1 ) + 8 . 0 * F ( J + 1 6 ) 2 - 3 . 0 ' F ( J + 2 ) - 0 . 5 * F ( J + 2 4 ) - 4 . 0 * F ( J + 1 5 ) ) F A C 5 = - P I N L * F ( J + 2 0 ) + ( ( D / 2 ) - T V ) ' ( P I N L * * 2 ) * F ( J + 1 6 ) 1 + ( 0 . 0 4 / S ) * ( - . 7 5 * F ( J + 1 1 ) + . 7 5 * F ( J + 3 3 ) - 0 . 1 2 5 * F ( J + 1 2 ) - . 1 2 5 * F ( J + 3 4 ) ) 2 - < ( D / 2 ) - T V ) ' ( 0 . 0 8 / S / S ) ' ( 2 . 0 * F ( J + 7 ) + 2 . 0 * F ( J + 2 9 ) - 4 . 0 * F ( J - r 16) 3 + 0 . 2 5 * F ( J + 8 ) - 0 . 2 5 * F ( J + 3 0 ) ) F A C 6 = - P I N L ' F ( J + 3 1 ) + ( ( D / 2 ) - T V ) * ( P I N L * * 2 ) * F ( J + 2 9 ) + 0 . 0 2 * F ( J + 3 4 J / S 1 - ( ( D / 2 ) - T V ) * ( 0 . 0 8 / S / S ) * ( 3 . 5 * F ( J + 7 ) - 1 1 . 5 * F ( J + 2 9 ) + 8 . 0 * F ( J + 1 6 ) 2 + 0 . 5 * F ( J + 8 ) + 3 . 0 * F ( J + 3 0 ) + 4 . 0 * F ( J + 2 2 ) ) J 1 = ( I E - 1 ) ' 2 2 + 6 F A C 3 = F ( J 1 + 6 ) / S - U Y X 1 * P I N L ' F ( J 1 + 3 ) - H D 1 * ( - 4 6 . 0 * F ( J 1 + 1 ) 1 + 1 4 . 0 * F ( J 1 + 2 3 ) + 3 2 ' F ( J 1 + 1 0 ) - 1 2 . 0 * F ( J 1 + 2 ) - 2 . 0 ' F ( J 1 + 2 4 ) 2 - 1 6 . 0 * F ( J 1 + 1 6 ) ) / ( S * * 2 ) + U Y X 1 * H D 1 * F ( J 1 + 1 ) * P I N L * * 2 F A C 4 = ( - 1 . 5 * F ( J 1 + 5 ) + 1 . 5 * F ( J 1 + 2 7 ) - 0 . 2 5 * F ( J 1 + 6 ) - 0 . 2 5 * F ( J 1 + 2 8 ) ) / S 1 - U Y X 1 * P I N L * F ( J 1 + 1 4 ) + H D 1 * ( B . 0 * F ( J 1 + 1 ) + 8 . 0 * F ( J 1 + 2 3 ) - 1 6 . 0 * F ( J 1 + 1 0 ) 2 + F ( J 1 + 2 ) - F ( J 1 + 2 4 ) ) / ( S * * 2 ) - U Y X 1 * H D 1 * F ( J 1 + 1 0 ) * P I N L * * 2 FAC 1 = FAC 1 • ECC FAC2 = FAC2 * ECC FAC3 FAC3 * ECC 1 FAC4 = FAC4 • ECC 1 FAC5 = FAC5 • EXC FAC6 = FAC6 • EXC FAC7 = FAC7 « EXC FAC8 FAC8 • EXC DO 20 i I = 1 1 9 XL = I * 0 . 0 5 " RL I F (NGAPS.EQ.O) GO TO 19 DO 1 8 K = 1 , NGAPS ZO = GAPX(K) Z1 = ZO + GAP(K ) I F ( X L . G T . Z O . A N D . X L . L T . Z D GO TO 20 1 8 CONTINUE 19 I F ( N G A P S 1 . E Q . 0 ) G 0 TO 5 6 4 DO 5 6 5 K=1 ,NGAPS1 Z 2 = G A P X 1 ( K ) Z 3 = Z 2 + G A P 1 ( K ) I F ( X L . G T . Z 2 . A N D . X L . L T . Z 3 ) G 0 TO 2 0 5 6 5 CONTINUE 564 XL = S I N ( 0 . 0 5 ' I * P I N ) W ( I ) = W ( I ) + F ( J + 1 ) * X L W 1 ( I ) = W K I ) + F ( J 1 + 1 ) * X L w s m = w s m + F A C I * XL S T ( I ) = S T ( I ) + FAC2 * XL WS 1 (I) = WSKI) + FAC3 ' XL ST1(I)= STK I ) + FAC4 ' XL CK1(I)= CKK I ) + FAC5 * XL CK2(I)= CK2(I) + FAC6 • XL CL1(I)= CLK I ) + FAC7 * XL CL2(I)= CL2(I) + FAC8 * XL CONTINUE CONTINUE RAM=FAC5*SIN(NM'PI*RL/4) RAM1=FAC6 ,SIN(NM*PI'RL/4) WRITE(*,*) 'RAM=',RAM,'RAM1 ='.RAMI WC = 0.0 WC1= 0.0 SC1 = 0.0 SC2 = 0.0 SC3 = 0.0 SC4 = 0.0 CQ1 = 0.0 CQ2 = 0.0 SM1 = 0.0 SM2 = 0.0 DO 40 I = 1, 19 IF (ABS(W(I)).GT.ABS(WC)) WC = W(I) IF (ABS( W1 (I) ) . GT . ABS( WC 1 ) > WC 1 = WKI) IF (ABS(WS(D).GT.ABS(SCD) SC1 = ABS(WS(I)) IF (ABS(ST(I)).GT.ABS(SC2)) SC2 = ABS(ST(I)) IF (ABS(WSKI) ) .GT ABS(SC3) ) SC3 = ABS(WS K I IF (ABS(STKI) ) .GT ABS(SC4)) SC4 = ABS(ST 1 (I IF (ABS(CKKI) ) .GT ABS(CQ1)) CQ1 = ABS(CKK I IF (ABS(CK2(I)) GT ABS(C02) ) C02 = ABS(CK2(I IF (ABS (CLKD) GT ABS1SM1)) SM1 = ABStCLKI IF (ABS(CL2(I)) GT ABSISM2)) SM2 = ABS(CL2(I CONTINUE WRITEC, ' ) "WC^'.WCI WCB = 0 . 0 WCB1= 0.0 IF (IE.EQ.1) GO TO 45 WCB = WC - (WJT + WP)/2.0 WC81= WC1- (WJT + WP)/2.0 WP = WJT RETURN END SUBROUTINE 0UT(N1,N2,WMA,SMA.SF.WFT.WCOV,WCOV1,WCM, & WCM1,CMA,CMA1,IPRNT,PI,WSM) COMMON/B3/ETA(6),H(6).S.NM,NSTEP.NMAX C0MM0N/B9/SVEC(6,452) DIMENSION SF(20), WFT(20), WC0V120), WCOV K 20) DIMENSION F(452) WMA IS THE MAX JOIST DEFLECTION SMA IS THE JOIST BENDING STRESS IN THE FLOOR CMA IS THE MAX BENDING STRESS IN THE UPPER SHEATHING 103 L -THE DIRECTION PERPENDICULAR TO THE JOIST C CMA1 IS THE MAX BENDING STRESS IN THE LOWER SHEATHING -C -THE DIRECTION PERPENDICULAR TO THE JOIST C WCM IS THE MAX SHEATHING DEFLECTION BETWEEN JOISTS -C -IN THE ENTIRE UPPER FLOOR C WCM1 IS THE MAX SHEATHING DEFLECTION BETWEEN JOISTS -C -IN THE ENTIRE LOWER FLOOR C WCOV IS THE MAX SHEATHING DEFLECTION BETWEEN JOISTS FOR C -EACH JOIST IN THE UPPER FLOOR C WCOV1 IS THE MAX SHEATHING DEFLECTION BETWEEN JOISTS FOR C - EACH JOIST I.; THE LOWER FLOOR WMA = 0 . 0 WCM = 0 . 0 WCM1= 0.0 SMA = 0.0 CMA = 0 . 0 CMA1=0.0 DO 2 IE=1,N2 CALL STRESS(N1,STR,WJ,WJS,WJT,WC,WCB,WCB1,SC1,SC2,SC3,SC4, 1 CQ1,CQ2,SM1.SM2,IE,PI.WC1) IF (IPRNT.EQ.O) GO TO 202 WRITE(2,3) IE 3 FORMAT(/,' * ELEMENT ' . 1 2 , ' * ' , / . &9X , 'U (F ) ' , 11X , 'V ( F> ' ,11X , 'W (F ) ' , 11X , 'W ( J ) ' , 11X , , U( J ) ' , 11X , 'V ( J ) ' , &1IX, '0( J ) ' , 1 IX , 'O(F) ' ) WRITE(2,8) 8 FORMAT!1H+.98X,'-',14X, '-') DO 20 I = 1, NM, NSTEP IKO=I IF(NSTEP.EQ.2)IK0=(1+1)/2 DO 40 J = 1, N1 40 F(J) = SVEC(IKO.J) J = (IE - 1)*22 WRITE!2, 18)1 18 FORMAT(' N =•,13) RADN=F(J+19)/S RADF = F(J+15)/S RADF1=F(J+22)/S WRITE(2,4)F(J+3),F(J+5),F(J+1) 4 FORMAT(' 1'.3E15.6) WRITE(2.5)F(J+13),F(J+14),F(J+16),F(J+16),F(J+17),F(J+18), 1 RADN, RADF 5 FORMAT(' 2' .8E15.6) WRITE(2,6)F(J+25),F(J+27),F(J+23) 6 FORMAT(' 3',3E15.6) WRITE(2,566) F(J+9),F(J+11),F(J+7) 566 FORMAT(' 4 ' .3E15.6) WRITE(2.56 7) F(J + 20),F(J + 2 1 ) . F(J+16) . F(J+16 ) . F(J+1 7),F(J+18), 1 RADN,RADF1 567 FORMAT(' 5".8E15.6) WRITE(2,568)F(J+31),F(J+33),F(J+29) 568 FORMAT(1 6 ' ,3E15.6) 20 CONTINUE WRITE(2,7) WJT,STR,WC,SC1,SC2,SM1,SM2,WC1,SC3,SC4.CQ2,CQ1 7 FORMAT(/,' JOIST:',/, 'DEFLECTION= ' . E 1 3 . 6 , ' (BENDING)',/, 1 'BENDING STRESS=',E13.6,//, & 'UPPER COVER: ' ,/ , 2 'DEFLECTION= ' ,E13.6, ' (NODE 1 ) ' , / , 104 3 'MAX. STRESS=',E13.6,'(NODE 1 - Y DIRECTION)',/. 4 'MAX. STRESS=',E13.6,'(NODE 2 - Y DIRECTION)',/, t\ 'MAX. STRESS=' ,E13.6, ' (NODE 2 - X DIRECTION)',/, 6 'MAX. STRESS=',E13.6,'(NODE 1 - X DIRECTION)'.//. 7 'LOWER COVER: ' ,/ , 8 'DEFLECTI0N=',E13.6,'(NODE 4 ) ' , / , $ 'MAX. STRESS=',E13.6,'(NODE 4 - Y DIRECTION)',/, 9 'MAX. STRESS='.E13.6,'(NODE 5 - Y DIRECTION)',/, 5 'MAX. STRESS=', E 1 3 . 6 , '(NODE 6 - X DIRECTION)'./, 6 'MAX. STRESS=',E13.6, 1(NODE 5 - X DIRECTION)') C 202 IF (ABS(WJT).GE.ABS(WMA)) WMA = WJT IF (ABS(STR).GE.ABS(SMA)) SMA = STR IF (ABS(SC1).GE.ABS(CMA)) CMA = SCI IF (ABS(SC2).GE.ABS(CMA)) CMA = SC2 IF (ABS(SC3).GE.ABS(CMA1)) CMA1=SC3 IF (ABS(SC4).GE.ABS(CMA1)) CMA1=SC4 IF (A8S(WCB).GE.ABS(WCM)) WCM = WCB IF (ABS(WCB1).GE.ABS(WCM1)) WCM1 = WCB1 IF (ABS(SM1).GE.ABS(WSM)) WSM = SMI SF(IE)=STR WFT(IE) = WJT WCOV(IE) = WCB WCOV1(IE)= WCB1 2 CONTINUE IF(IPRNT.EQ.O)GO TO 204 WRITE(2,10)SMA,WMA,CMA,CMA1,WCM,WCM1 10 FORMAT(/,' MAX. JOIST STRESS' ,E15.6,/, 'MAX. JOIST DEFLECTION', 1 E 1 5 . 6 , / . ' MAX. BENDING STRESS IN THE UPPER SHEATHING ' . E 1 5 . 6 . / . 2 'MAX. BENDING STRESS IN THE LOWER SHEATHING',E15.6,/, 3 'MAX. COVER DEFLN BETWN JOISTS IN ENTIRE UPPER FLOOR' ,E 15.6 ,/ , 4 'MAX COVER DEFLN BETWN JOISTS IN ENTIRE LOWER FLOOR',E15.6./) 204 CONTINUE RETURN END C=======================================™ C SUBROUTINE ZERO COMM0N/B10/RM(6 , 34) , RMK6.34) DO 1 I = 1. 6 DO 1 J = 1, 34 RM(I,J) = 0.0 1 RM1(I,J) = 0.0 RETURN END 0========================================================================= C SUBROUTINE GENMTX DOUBLE PRECISION AA COMMON/B1/RM00(34,34),RM22(34,34),RM02(34,34),RM20(34,34) , 1 RM33(34,34),RM66(34,34),RM36(34,34),RM63(34,34),RM44(34.34), 2 RM45( 34,34),RM54(34,34).RM55(34,34).RM1 1(34,34),RM7 7 ( 34,34) , 3 RM99(34,34),RM79( 34,34).RM97( 34,34),RM1010( 34 , 34) . 4 RM1313(34,34),RM1013(34,34),RM1310(34,34),RM1111(34,34), 5 RM1112(34,34).RM1211(34,34),RM1212(34,34),RM88(34,34) C0MM0N/B8/AA(34.34) C0MM0N/B3/ETA(6),H(6) COMMON/B10/RM(6,34) . RMK6.34) CALL ZERO C MO AND M2 MATRICES DO 1 I = 1 , 6 ETA2 = ETA(I)**2 ETA3 = ETA(I)**3 ETA4= ETA(I)•*4 ETA5 = ETA(I)**5 RM(I,1 )=ETA2-5.0*ETA3/4.0-ETA4/2.0+3.0*ETA5/4.0 RM(I,23)=ETA2+5.0*ETA3/4.0-ETA4/2.0-3.0*ETA5/4.0 RM(I,16)=1.0-2.0*ETA2+ETA4 RM(I,2)=(ETA2-ETA3-ETA4+ETA5)/8.0 RM(I,24)=(-ETA2-ETA3+ETA4+ETA5)/B.0 RM(I,15)=(ETA(I)-2.0'ETA3+ETA5)/2.0 RM1(I,1)=2.0-15.0'ETA(I)/2.0-6.0*ETA2+15.0*ETA3 RMKI,23)=2.0+15.0*ETA(I)/2.0-6.0 ,ETA2-15.0-ETA3 RM1(I,16)=-4.0+12.0'ETA2 RM1(I,2)=(2.0-6.0*ETA(I)- 12.0*ETA2 + 20.0*ETA3)/8.0 RM1(I,24) = (- 2.0-6.0*ETA(I) + 12.0* ETA2+20.0*ETA3)/8.0 RM1(I,15)=(-12.0*ETA(I)+20.0*ETA3)/2.0 1 CONTINUE CALL DMAT(1) DO 4500 I = 1,34 DO 4500 J = 1,34 4500 RMOO(I.J) = AA(I, J ) CALL DMAT(2) DO 4501 I = 1,34 DO 4501 J = 1.34 4501 RM22(I,J) = AA(I, J ) CALL DMAT(3) DO 4502 I = 1,34 DO 4502 J = 1,34 4502 RM02(I,J) = AA(I, J ) CALL DMAT(4) DO 4503 I = 1.34 DO 4503 J = 1.34 4503 RM20(I.J) = AA(I,J) CALL ZERO C M3 AND M6 MATRICES DO 2 I = 1, 6 ETA2 = ETA(I) '*2 ETA3 = ETA(I)*"3 ETA4 = ETA ( I ) " 4 RM(I.3)=-3.0*ETA(I)/4.0+ETA2+ETA3/4.0-ETA4/2.0 Rf. !I.25)=3.0*ETA(I)/4.0+ETA2-ETA3/4.O-ETA4/2.0 RM(I,13)=1.0-2.0-ETA2+ETA4 RM(I,4)=(-ETA(I)+ETA2+ETA3-ETA4)/8.0 RM(I,2P>=(-ETA(I)-ETA2+ETA3+ETA4)/8.0 RM1(I,fc;=-3.0/4.0+2.0*ETA(I)+3.0"ETA2/4.0-2.0'ETA3 RM1(I.27) = 3.0/4.O+2.0'ETA(I )-3.0'ETA2/4.0-2.0'ETA 3 RM1(I,14)=-4.0*ETA(I)+4.0'ETA3 RM1(I,6) = (- 1.0+2.0'ETA(I)+3.0"ETA2-4.0*ETA3)/8.0 RM1(I,28)=(-1.0-2.0*ETA(I)+3.0*ETA2+4.0*ETA3)/8.0 2 CONTINUE CALL DMAT(1) DO 4504 I = 1,34 DO 4504 J = 1,34 4504 RM33(I,J) = AA(I,J) CALL DMAT(2) DO 4505 I = 1,34 DO 4505 J = 1,34 4505 RM66(I,J) = AA(I.J) CALL DMAT(3) DO 4506 I = 1,34 DO 4506 J = 1,34 4506 RM36(I,J) = AA(I.J) CALL DMAT(4) DO 4507 I = 1,34 DO 4507 J = 1,34 4507 RM63(I,J) = AA( I , . ' » C M5 AND M4 MATRICES DO 3 I = 1, 6 RM(I,5) = RM(I,3) RM(I,3) = 0.0 RM(I,27)=RM(I,25) RM(I,25) = 0.0 RM(I,14)=RM(I,13) RM(I,13) = 0.0 RM(I,6)=RM(I,4) RM(I,4) = 0.0 RM(I,28)=RM(I,26) RM(I,26) = 0.0 RMKI.3)=RMKI,5) RMKI.5) = 0.0 RM1(I,25)=RM1(I,27) RMKI.27) = .0.0 RMKI , 13)=RM1(I, 14) RM1(I,14) = 0.0 RM1(I,4)=RM1(I,6) RMKI ,6) = 0.0 RM1(I,26)=RMKI,28) RMKI,28) = 0 . 0 3 CONTINUE CALL DMAT(2) DO 4510 I = 1.34 DO 4510 J = 1 , 34 4510 RM44(I.J) = AA(I.J) CALL DMAT(4) DO 4511 I = 1 , 34 DO 4511 J = 1 . 34 4511 RM45(I,J) = AA(I.J) CALL DMAT(3) DO 4512 I = 1,34 DO 4512 J = 1 ,34 4512 RM54(I,J) = AA(I , J) CALL DMAT(1) DO 4513 I = 1 ,34 DO 4513 J = 1 , 34 4513 RM551I.J) = AA(I.J) CALL ZERO C M1 MATRIX DO 5 I = 1. 6 ETA2 = ETA(I)**2 ETA3 = ETA(I)**3 ETA4 = ETA(I) * *4 RM(I,1)=2.0*ETA(I)- 15.0*ETA2/4.0-2.0-ETA3+15.0*ETA4/4.0 RM(I,23)=2.0*ETA(I)+15.0*ETA2/4.0-2.0*ETA3-15.0'ETA4/4.0 RM(I,16)=-4.O'ETAtI)+4.0*ETA3 RM(I,2)=(2.0*ETA(I)- 3.0'ETA2-4.O'ETA3+5.0*ETA4)/8.0 RM(I,24)=(-2.0*ETA(I)- 3.0'ETA2+4.0•ETA3+5.0*ETA4)/8.0 RM<I,15)=(1.0-6.0"ETA2+5.0'ETA4)/2.0 5 CONTINUE CALL DMAT(1) DO 4515 I = 1,34 DO 4515 J = 1,34 4515 RM11(I,J) = AA(I,J) CALL ZERO C M7 AND M9 MATRICES DO 58 1=1,6 ETA2=ETA(I)'*2 ETA3=ETA(I)**3 ETA4=ETA(I)**4 ETA5=ETA(I)*'5 RM(I,7)=ETA2-5.0*ETA3/4.0-ETA4/2.0+3.0*ETA5/4.0 RM(I,29)=ETA2+5.0*ETA3/4.0-ETA4/2.0-3.0'ETA5/4.0 RM(I.16)=1.0-2.O'ETA2+ETA4 RM(I,8)=(ETA2-ETA3-ETA4+ETA5)/8.0 RM(I.30)=(-ETA2-ETA3+ETA4+ETA5)/8.0 RM(1,22) = (ETA(I)- 2.0 *ETA3+ETA5)12.0 RM1(I,7)=2.0-15.0*ETA(I)/2.0-6.0*ETA2+15.0*ETA3 RMKI,29)=2.0+15.0*ETA(I)/2.0 - 6.0*ETA2-15.0*ETA3 RMKI,16) = -4.0+12.0*ETA2 RMKI ,8) = (2.0-6.0*ETA(I) - 12 .0'ETA2 + 20. 0*ETA3)/8 .0 RMKI . 30) = ( -2.0-6 .O'ETAt I) + 12 . 0 ' ETA2 + 20 . 0* ETA3)/8 . 0 RMKI,22) = (-12.0*ETA(I)+20.0*ETA3)/2.0 58 CONTINUE CALL DMAT(1) DO 4600 1=1,34 DO 4600 J=1,34 4600 RM77(I.J)=AA(I,J) CALL DMAT(2) DO 4601 1=1,34 DO 4601 J=1,34 4601 RM99(I,J)=AA(I,J) CALL DMAT(3) DO 4602 1=1,34 DO 4602 J=1,34 . 4602 RM79(I,J)=AA(I,J) CALL DMAT(4) DO 4603 1=1,34 DO 4603 J=1.34 4603 RM97(I,J)=AA(I,J) CALL ZERO C M10 AND M13 MATRICES DO 59 1 = 1 ,6 ETA2=ETA(I ) "2 ETA3=ETA(I)'*3 ETA4=ETA(I)**4 RM(I,9)=-3.0'ETA(I)/4.0+ETA2+ETA3/4.0-ETA4/2.0 RM(I,31)=3.0'ETA(I)/4.0+ETA2-ETA3/4.0-ETA4/2.0 RM(I,20)=1.0-2.0*ETA2+ETA4 RM(I,10) = (- ETA(I)+ETA2+ETA3-ETA4)/8 .0 RM( 1,32) = ( -ETA(I) - ETA2+ETA3 + ETA4V8.0 RMKI,11)=-3.0/4.0+2.0*ETA(I )+3.u"ETA2/4.0-2.0'ETA3 RMKI,33)=3.0/4.0+2.0'ETA(I)-3.0*ETA2/4.0-2.0*ETA3 RM1(I,21)=-4.0*ETA(I)+4.0*ETA3 RM1(1,12) = (- 1.0+2.0'ETA(I)+3.0*ETA2-4.0*ETA3)/8.0 RM1(I,34) = (- 1.0-2.0'ETA(I)+3.0*ETA2+4.0*ETA3)/8.0 59 CONTINUE C c CALL DMAT(1) DO 4604 1=1.34 DO 4604 J=1.34 4604 RM1010(I.J)=AA(I.J) CALL DMATt 2) DO 4605 1=1.34 DO 4605 J=1,34 4605 RM1313U. J)=AA(I, J) CALL DMAT(3) DO 4606 1=1.34 DO 4606 J=1.34 4606 RM1013(I,J)=AA(I,J) CALL DMAT(4) DO 4607 1=1.34 DO 4607 J=1,34 4607 RM1310(I,J)=AA(I,J) C M12 AND M11 MATRICES DO 60 1=1,6 RM(I,11)=RM(I,9) RM(I,9)=0.0 RM(I,33)=RM(I,31) RM(I.31)=0.0 RM(I.21)=RM(I,20) RM(I,20)=0.0 RM(I.12)=RM(I,10) RM(I,10)=0.0 RM(I,34)=RM(I,32) RM(I,32)=0.0 RM1(I,9)=RM1(I,11) RM1(I,11)=0.0 RM1(I,31)=RM1(I,33) RM1(I,33)=0.0 RM1(I,20)=RM1(1,21) RM1(I,21)=0.0 RM1(I,10)=RM1(1,12) RM1(I.'12)=0.0 RM1(I,32)=RM1(I,34) RM1(I.34)=0.0 60 CONTINUE C C CALL DMAT(2) DO 4610 1=1,34 DO 4610 J=1,34 4610 RM1111(I,J)=AA(I,J) CALL DMAT(4) DO 4611 1=1.34 DO 4611 J=1,34 4611 RM1112(I,J)=AA(I,J) CALL DMAT(3) DO 4612 1=1,34 DO 4612 J=1,34 4612 RM1211(I.J)=AA(I.J) CALL DMAT(1) DO 4613 1=1,34 DO 4613 J=1,34 4613 RM1212(I,J)=AA(I,J) CALL ZERO C M8 MATRIX DO 61 1=1,6 ETA2=ETA(I)**2 ETA3=ETA 11)* * 3 ETA4=ETA(I ) "4 RM( 1,7) =2.0' ETA(D- 15.0' ETA2/4. 0-2.0 * ETA3+15 . 0* ETA4/4 . 0 RM(I,29)=2.0*ETA(I)+15.0'ETA2/4.0-2.0*ETA3-15.0*ETA4/4.0 RM(I,16)=-4.0'ETA(I)+4.0'ETA3 RM(I,8)=(2.0'ETA(I)-3.0*ETA2-4.0*ETA3+5.0 ,ETA4)/8.0 RM(I,30) = (- 2.0* ETA(I)- 3.0'ETA2 + 4.0*ETA3+5.0'ETA4)/8.0 RM(I.22)=(1.0-6.0'ETA2+5.0'ETA4)/2.0 61 CONTINUE CALL DMAT(1) DO 4615' 1 = 1,34' DO 4615 J=1,34 4615 RM88(I.J)=AA(I , J) RETURN END FUNCTION Z(N,M,Z1,Z0,RL,NX,PI) IF(N.EQ.M)GO TO 10 S1=SIN((N-M)*PI*Z1/RL) S2=SIN((N-M)*PI*Z0/RL) S3=SIN((N+M)*PI*Z1/RL) S4=SIN((N+M)'PI*Z0/RL) D1=(S1-S2)/(2.0*(N-M)) D2=(S3-S4)/(2.0*(N+M)) IFfNX.EQ. DGO TO 5 Z=RL«(D1-D2J/PI RETURN 5 Z=RL*(D1+D2)/PI RETURN 10 S1=SIN(2.0'N*PI*Z1/RL) S2=SIN(2.0'N ,PI*Z0/RL) D1=PI*(Z1-Z0)/(2.0'RL) D2=(S1-S2)/(4.0*N) IF(NX.EQ.1)G0 TO 12 Z=RL*(D1-D2)/PI RETURN 12 Z=RL*<D1+D2)/PI RETURN END C SUBROUTINE STIF(V,IE,KLO,IN,IK,PI) DOUBLE PRECISION STIFF DIMENSION V(34). AD(34,34) ,AD1(34,34) COMMON/B1/RMOO(34,34),RM22(34,34),RM02(34,34),RM20(34,34), 1 RM33(34.34),RM66(34,34),RM3F'34,34),RM6 3(34.34),RM44(34.34) 2 RM45(34,34),RM54( 34,34).RM55134.34).RM1 1 ( 34.34),RM77 ( 34 . 34) 3 RM99(34,34),RM79(34,34),RM97(34,34),RM1010(34,34), 110 4 R M 1 3 1 3 ( 3 4 , 3 4 ) , R M 1 0 1 3 ( 3 4 , 3 4 ) , R M 1 3 1 0 ( 3 4 , 3 4 ) , R M 1 1 1 1 ( 3 4 , 3 4 ) , 5 R M 1 1 1 2 ( 3 4 , 3 4 ) , R M 1 2 1 1 ( 3 4 , 3 4 ) , R M 1 2 1 2 ( 3 4 , 3 4 ) , R M 8 8 ( 3 4 , 3 4 ) C 0 M M 0 N / B 8 / S T I F F ( 3 4 . 3 4 ) C 0 M M 0 N / B 2 / R K X , R K Y , R K V , R K G , D X , D Y , D V , D G , R L , F A C T O R , R K X 1 , RKY 1 , 1 R K V 1 , R K G 1 , D X 1 , D Y 1 , D V 1 , D G 1 C 0 M M 0 N / B 3 / E T A ( 6 ) , H ( 6 ) , S . N M , N S T E P . NMAX C O M M O N / B 6 / G A P X O ( 5 ) , G A F D ( 5 ) , G A P X ( 5 ) , G A P ( 5 ) , X I N , N G A P S , N A I , N D I S C R , 1 N G A P S 1 . G A P X 1 ( 5 ) , G A P 1 ( 5 ) , G A P 0 1 ( 5 ) , G A P X 0 1 ( 5 ) . X I N 1 . N D S C R 1 , N A I 1 C 0 M M 0 N / B 4 / D . D 1 . E J T . G J T , R E T A , E N L , E N L 1 , R K P A L , R K P A L 1 , R K P E R , R K P E R 1 , 1 P L D . X 1 , X 2 , Y 1 , Y 2 , B J ( 2 0 ) , H J ( 2 0 ) , H J T M , ALPHA, RKROT,RKR0T1 I F ( I E . G T . 1 ) GC TO 30 I F ( K L O . E Q . 1 ) GO TO 30 P 2 = P I * * 2 P 4 = P I * ' 4 C * UPPER AND LOWER COVER * DO 1 1 =1 ,34 DO 1 J = 1 . 3 4 1 S T I F F ( I . J ) = 0 . 0 F A C = 0 . 0 I F ( I N . E Q . I K ) F A C = R K X * ( I K * * 4 ) * P 4 * S / ( 4 . 0 * R L * * 3 ) I F ( I N . E Q . I K ) F A C 1 = R K X 1 * ( I K * * 4 ) * P 4 * S / ( 4 . 0 * R L * * 3 ) I F ( N G A P S . E Q . 0 ) G 0 TO 3 I F ( N G A P S 1 . E Q . O J G O TO 3 DO 2 1=1.NGAPS 2 0 = G A P X ( I ) Z 1 = 2 0 + G A P ( I ) R = 2 ( I N , I K , 2 1 , 2 0 , R L , 0 . P I ) 2 F A C = F A C - R K X ' P 4 * S * R * ( I N * I K ) * • 2 / ( 2 . 0 * R L * • 4 ) DO 2 0 2 1=1,NGAPS1 2 2 = G A P X 1 ( I ) 2 3 = 2 2 + G A P 1 ( I ) R = Z ( I N , I K , 2 3 . Z 2 . R L , 0 , P I ) 2 0 2 FAC1=FAC1-RKX1 * P 4 * S * R * ( I N * I K ) * * 2 / ( 2 . 0 * R L * * 4 ) 3 DO 3 5 0 0 I = 1 . 3 4 DO 3 5 0 0 J = 1 . 3 4 3 5 0 0 A D ( I , J ) = R M O O ( I . J ) CALL A D D ( A D . F A C ) DO 3 5 2 0 1=1,34 DO 3 5 2 0 J = 1 . 3 4 3 5 2 0 A D ( I . J ) = R M 7 7 ( I . J ) CALL A D D ( A D , F A C 1 ) F A C = 0 . 0 I F ( I N . E Q . I K ) F A C = R K Y * 4 . 0 * R L / ( S * * 3 ) I F ( I N . E Q . I K ) F A C 1 = R K Y 1 * 4 . 0 * R L / ( S * * 3 ) I F ( N G A P S . E Q . O ) G O TO 5 I F ( N G A P S 1 . E Q . O ) G O TO 5 DO 4 1=1.NGAPS 2 0 = G A P X 0 ( I ) Z 1 = 2 0 + G A P 0 ( I ) R = 2 ( I N . I K , 2 1 . Z 0 . R L , 0 . P I ) 4 F A C = F A C - R K Y ' 8 . 0 * R / ( S * * 3 ) DO 2 0 4 1=1 ,NGAPS 1 Z 2 = G A P X 0 1 ( I ) Z 3 = 2 2 + G A P 0 1 ( I ) R = Z ( I N , I K , Z 3 , Z 2 , R L . 0 , P I ) 2 0 4 F A C 1 = F A C 1 - R K Y 1 ' 8 . 0 * R / ( S * * 3 ) 5 DO 3 5 0 1 I = 1 , 3 4 DO 3 5 0 1 J = 1 , 3 4 3501 AD(I.J) = RM22(I,J) CALL ADD(AD.FAC) DO 3521 1=1,34 DO 3521 J=1 ,34 3521 AD(I,J)=RM99(I,J) CALL ADD(AD.FACI) FAC=0.0 IF(IN.EO.IK)FAC=-RKV*P2/(S*RL) IF(IN.EQ.IK)FAC2=-RKV1*P2/(S*RL> IF(NGAPS.EQ.O)GO TO 7 IF(NGAPS1.EQ.OJGO TO 7 DO 6 1=1,NGAPS ZO=GAPX(I) Z1=Z0+GAP(I) R=Z(IN,IK,Z1,ZO,RL,0,PI) 6 FAC=FAC+RKV*2.0*P2*R/(S*RL**2) DO 206 1=1.NGAPS1 22=GAPX1(I) Z3=Z2+GAP1(I) R=Z(IN,IK,Z3,Z2,RL.0.PI) 206 FAC^=FAC2+RKV1•2.0•P2'R/(S•RL•*2) 7 CONTINUE FAC1=FAC*(IK'*2) FAC3=FAC2'(IK**2) DO 3502 I = 1.34 DO 3502 J = 1,34 3502 AD(I.J) = RM02(I.J) CALL ADD(AD,FAC1) DO 3522 1=1.34 DO 3522 J=1,34 3522 AD(I,J)=RM79(I,J) CALL ADD(AD,FAC3) FAC1=FAC"(IN"*2) FAC3=FAC2*(IN*'2) DO 3503 I = 1,34 DO 3503 J = 1.34 3503 AD(I.J) = RM20(I,J) CALL ADD(AD,FAC1) DO 3523 1=1,34 DO 3523 J=1,34 3523 AD(I,J)=RM97(I,J) CALL ADD(AD.FAC3) FAC=0.0 IF(IN.EQ.IK)FAC=RKG*4.0*P2'IK"2/(S*RL) • IF(IN.EQ.IK)FAC1=RKG1*4.0*P2*IK**2/(S*RL) IF(NGAPS.EQ.0)G0 TO 9 IF(NGAPS1.EQ.O)GO TO 9 DO 8 1=1.NGAPS ZO=GAPX(I) Z1=Z0+GAP(I) R=Z(IN,IK,Z1.20.RL,1,PI) 8 FAC=FAC-RKG'B.0 ,P2*R"IK , IN/(S ,RL**2) DO 208 1=1,NGAPS1 Z2=GAPX1(I) Z3=Z2+GAP1(I) R=Z(IN, IK,Z3,Z2.RL.1,Pn 208 FAC1-FAC1-RKG1*8.0*P2 ,h iK*IN/(S'RL' 9 CONTINUE DO 3505 I = 1,34 DO 3505 J = 1,34 3505 AD(I.J) = RM1KI. J ) CALL ADD(AD.FAC) DO 3524 1=1.34 DO 3524 J=1.34 3524 AD(I,J)=RM88(I,J) CALL ADD(AD.FAC1) FAC=0.0 IF(IN.EQ.IK)FAC=DX*S ,P2*IK ,*2/(4.C*RL) IF(IN.EQ.IK)FAC1=DX1*S*P2*IK ,*2/(4.0*RL) IF(NGAPS.EQ.0)GO TO 11 IF(NGAPS1.EQ.O)GO TO 11 DO 10 1=1,NGAPS 20= GAPX(I) Z1 = ZO • GAP(I) R=Z(IN,IK,Z1.Z0,RL,0,PI) 10 FAC=FAC-DX , S*P2 , R , IK*IN/(2.0'RL* , 2) DO 210 1=1,NGAPS1 Z2=GAPX1(I) Z3=Z2+GAP1(I) R=Z(IN,IK,Z3,Z2,RL,0,PI) 210 FAC1=FAC1-DX1 ,S ,P2 ,R*IK*IN/(2.0*RL'* 11 CONTINUE DO 3506 I = 1.34 DO 3506 J = 1.34 3506 AD(I.J) = RM33(I,J) CALL ADD(AD.FAC) DO 3525 1=1,34 DO 3525 J=1,34 3525 AD(I,J)=RM1010(I.J) CALL ADD(AD,FAC1) FAC=0.0 IF(IN.EQ.IK)FAC=DY'RL/S IF(IN.EQ.IK)FAC1=DY1*RL/S IF(NGAPS.EQ.O)GO TO 13 IF(NGAPS1.EQ.O)GO TO 13 DO 12 1=1.NGAPS Z0= GAPX(I) Z1= ZO + GAP(I> R=Z(IN.IK,Z1,Z0,RL,0,PI) 12 FAC=FAC-DY'2.0*R/S DO 212 1=1,NGAPS1 Z2=GAPX1(I) Z3=Z2+GAP1(1) R=Z(IN,IK,Z3,Z2,RL,0.PI) 212 FAC1=FAC1-DY1*2.0'R/S 13 .JNTINUE DO 3508 I = 1.34 DO 3508 J = 1,34 3508 AD(I,J) = RM66(I,J) CALL ADD(AD.FAC) DO 3526 1=1,34 DO 3526 J=1,34 3526 AD(I,J)=RM1313(I.J) CALL ADD(AD,FAC1) FAC=0.0 IF(IN.EQ.IK)FAC=-DVPI/2.0 IF(IN.EQ.IK)FAC2=-DV1'PI/2.0 IF(NGAPS.EQ.O)GO TO 15 IF(NGAPS1.EQ.0)G0 TO 15 DO 14 1=1,NGAPS ZO = GAPX(I) Z1 = ZO + GAP(I) R=Z(IN,IK,Z1,ZO,RL,0,PI) 14 FAC=FAC+DVPI'R/RL DO 214 1=1.NGAPS1 Z2=GAPX1 (I) Z3=Z2+GAP1(I) R=Z(IN,IK,Z.,Z2,RL,0,PI) 214 FAC2=FAC2+DV1'PI'R/RL 15 CONTINUE FAC1=FAC'IK FAC3=FAC2*IK DO 3509 I = 1,34 DO 3509 J = 1,34 3509 AD(I,J) = RM36(I,J) CALL ADD(AD,FAC 1) DO 3527 1=1,34 DO 3527 J=1,34 3527 AD(I,J)=RM1013(I,J) CALL ADD(AD,FAC3) FAC1=FAC'IN FAC3=FAC2*IN DO 3511 I = 1,34 DO 3511 J = 1,34 3511 AD(I.J) = RM63(I.J) CALL ADD(AD,FAC1) DO 3528 1=1,34 DO 3528 J=1,34 3528 AD(I,J)=RM1310(I,J) CALL ADD(AD,FAC3) FAC=0.0 IF(IN.EQ.IK)FAC=DG*RL*S/4.0 IF(IN.EQ.IK)FAC2=DG1*RL*S/4.0 IF(NGAPS.EQ.0)G0 TO 17 IF(NGAPS1 .EQ.OGO TO 17 DO 16 1=1.NGAPS ZO = GAPX(I) Z1 = ZO + GAP(I) R=Z(IN,IK,Z1.Z0,RL,1.PI) 16 FAC=FAC-DG*S*R/2.0 DO 216 1=1.NGAPS1 Z2=GAPX1(I) Z3=Z2+GAP1(I) R=Z(IN.IK.Z3,Z2.RL,1.PI) 216 . FAC2=FAC2-DG1*S*R/2.0 17 CONTINUE FAC1=FAC*4.0/(S'*2) FAC3=FAC2'4.0/(S*"2) DO 3515 I = 1,34 DO 3515 J = 1.34 3515 AD(I.J) = RM44(I.J) CALL ADD(AD.FACI) DO 3529 1=1,34 DO 3529 J=1,34 3529 AD(I,J)=RM1111(I,J) CALL ADD(AD.FAC3) FAC^FAC^ .O ' IN ' P I / f S ' RL ) FAC3=FAC2'2.0*IN*PI/(S*RL) DO 3516 I = 1,34 DO 3516 J = 1,34 3516 AD(I,J) = RM45(I,J) CALL ADD(AD,FAC1) DO 3530 1=1,34 DO 3530 J=1,34 3530 AD(I.J)=RM1112(I,J) CALL ADDUD.FAC3) FAC1=FAC*2.0*IK'PI/(S*RL) FAC3=FAC2*2.0*IK*PI/(S ,RL) DO 3518 I = 1,34 DO 3518 J = 1,34 3518 AD(I.J) = RM54(I,J) CALL ADD(AD.FAC1) DO 3531 1=1,34 DO 3531 J=1,34 3531 AD(I.J)=RM1211(I,J) CALL ADD(AD,FAC3) FAC1=FAC*IK*IN*P2/(RL**2) FAC3=FAC2*IK*IN*P2/(RL**2) DO 3519 I = 1,34 DO 3519 J = 1,34 3519 AD(I.J) = RM55(I,J) CALL ADD(AD,FAC1) DO 3532 1=1.34 DO 3532 J=1,34 3532 AD(I,J)=RM1212(I,J) CALL ADD(AD.FAC3) C * NAILING ON UPPER AND LOWER COVER " DO 18 1=1,34 AD(1,I)=0.0 AD(2,I)=0.0 AD(3,I)=0.0 AD(4.I)=0.0 AD(5,I) = 0.0 18 AD(6.1) = 0.0 AD(1.13)=1.0 AD(1,17)=-1.0 AD(1,16) = -IK*PI*(HJTM + D)/( 2.0 * RL) AD(2.13) = 1 .0 AD(2,17)=-1.0 AD(2.16)=-IN*PI*(HJTM + D)/( 2.0 * RL) AD(3,14)=1.0 AD(3,18) = - 1 .0 AD(3.15) = -D/(2.0'S) A D O . 19) = -HJTM/ (2.0 * S) AD(4,14)=1.0 AD(4.18)=-1.0 AD(4,15) = -D/(2.0*S) AD(4.19) = -HJTM/ (2.0 * S) AD(5,15) = 1.0/S AD(5, 19) = -1 .0/S AD(6.15) = 1.0/S AD(6,19) = - 1.0/S 115 IF (NDISCR.EQ.O) GO TO 423 SPAR=0.0 SPER=0.0 DO 20 1=1,NAI X = XIN + ( I-D 'ENL IF (NGAPS.EQ.O) GO TO 420 DO 415 J = 1, NGAPS X11 = GAPX(J) X12 = GAPX(J) + GAP(J) IF (X.GE.X11.AND.X.LE.X12) GO TO 418 415 CONTINUE GO TO 420 418 GO TO 20 420 X = X*PI/RL SPAR=SPAR+COS(IN*X)*COS(IK*X) SPER=SPER+SIN(IN"X)*SIN(IK*X) 20 CONTINUE GO TO 426 423 IF (IN.NE.IK) GO TO 428 SPAR = RL /(2.0*ENL) SPER = RL /(2.0*ENL) GO TO 426 428 SPAR = 0 . 0 SPER = 0 . 0 426 SPAR = SPAR * RKPAL SPER=SPER*RKPER SROT = SPER * RKROT C DO 518 J=1,6 DO 518 1=1,34 518 AD1(J,I)=0.0 AD1(1.20)=1.0 ADK 1 , 17 ) = -1 .0 AD1{1,16)=IK*PI ,(HJTM+D1)/(2.0*RL) AD1(2.20)=1.0 AD1(2,17)=-1.0 AD1(2,16)=IN*PI*(HJTM+D1)/(2.0*RL) AD1(3,18)=-1.0 ADK3 , 19)=HJTM/(2.0'S) AD1(3,21)=1.0 AD1(3.22)=D1/(2.0*S) AD 1(4,18) = -1.0 ADK4 , 19)=HJTM/(2.0*S) AD1(4.21)=1.0 AD1(4,22)=D1/(2.0*S) AD1(5,19)=-1.0/S \D1(5,22)=1.0/S AD1(6,19)=-1.0/S AD1(6,22)=1.0/S IF(NDSCR1.EQ.O) GO TO 923 SPAR1=0.0 SPER1=0.0 DO 29 1=1.NAI1 X= XIN1 + (1-1) * ENL 1 IF (NGAPS1.EQ.O)GO TO 920 DO 915 J=1,NGAPS1 XI1 = GAPX1(J) X12 = GAPXKJ) • GAP 1 ( J) 116 IF(X.GE.X11.AND.X.LE.X12)G0 TO 91B 915 CONTINUE GO TO 920 918 GO TO 29 920 X=X'PI/RL SPAR1=SPAR1+C0S(IN'X)*COS(IK'X) SPER1=SPER1+SIN(IN*X)*SIN(IK'X) 29 CONTINUE GO TO 926 923 IF(IN.NE.IK)GO TO 928 SPAR1= RL/(2.0*ENL1) SPER1= RL/(. .O'ENL1) GO TO 926 928 SPAR1=0.0 SPER1=0.0 926 SPAR1=SPAR1'RKPAL1 SPER1=SPER1*RKPER1 SROT1=SPER1 * RKROT1 C DO 25 1=1,34 DO 25 J=1.34 STIFF(I,J)=STIFF(I,J)+SPAR*AD(1,I)*AD(2,J)+SPER*AD(3,I) 'AD(4,J) 1 + SROT*AD{5,I)*AD(6,J) 2 +SPER1'AD1(3,I) *AD1(4.J) +SR0T1 *AD1(5,I) *AD1(6,J) 3 +SPAR1 •ADKl . I ) *AD1(2.J) 25 CONTINUE C C * LOAD VECTOR * 30 IFtlN.NE.IK)RETURN DO 35 1=1,34 35 V(I)=0.0 IF(PLD.EQ.O.O)RETURN PINL=PI*IK/RL FACTR=PLD*(Y2-Y1)*(C0S(PINL*X1)- COS(PINL*X2))/(2.0*PINL) XIX=2.0*Y1/S XIY=2.0'Y2/S DO 37 1=1,6 XI=XIX+(XIY-XIX)*(1.0+ETA(I))/2.0 X I2=XI"2 XI3=XI* *3 XI4=XI"4 XI5=XI**5 V(1)=V(1)+(XI2-5.0*X13/4.0-X14/2.0+3.0*X15/4.0)*H(I) V(23)=V(23)+(XI2+5*XI3/4.0-XI4/2.0-3.0*XI5/4.0)*H(I) V(16)=V(16)+(1.0-2.0'XI2+XI4)'H(I) V(2)=V(2)+(XI2-XI3-XI4+XI5)*H(I)/8.0 V(24)=V(24)+(-XI2-XI3+XI4+XI5)*H(I)/8.0 37 V(15)=V(15)+(XI-2.0*XI3+XI5)*H(I)/2.0 V(1)=V(1)*FACTR V(23)=V(23)*FACTR V(16)=V(16)*FACTR V(2)=V(2)*FACTR V(24)=V(24)*FACTR V(15)=V(15)*FACTR RETURN END 117 SUBROUT INE A D D ( A D . F A C ) DOUBLE P R E C I S I O N S T I F F D IMENS ION A D ( 3 4 , 3 4 ) C O M M O N / B 8 / S T I F F ( 3 4 , 3 4 ) I F ( F A C . E Q . O . O ) R E T U R N DO 2 0 1=1 ,34 DO 20 J = 1 . 3 4 20 S T I F F ( I . J ) = S T I F F ( I . J ) + F A C * A D ( I , J ) RETURN END 

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