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Lidq- l I L“ 4 R‘ . f: .1 #13225} .5 I‘ “1‘: { , . ‘13!th ~ ‘1 15 ‘L L WV‘CFI,‘ \ .5 *1- " 9:121. '7““.Exi.‘% " L‘ 1.1.»: _ ”I“ 17,-} 4'“- ‘1 {L25 ._ ‘. 711-. ‘ ,\5 h 1 ‘1'" i" R? I 5 53¢": ‘L l .5132 "3L? m} Lvfi“‘;‘{ . .191 " ‘ 1- . 95w? . L “it“ L1“: LL33; 1_- «4‘ H1 i :4 . ‘u‘ i x I 3%?qu i VI}; 1 fikfii “ ; K4. . l 1“ 4 m.) , LIBRARY ; MichiganState , University MSU LIBRARIES 3“ .‘W‘ “WJ I i I W” “J —— RETURNENG MATERIALS: P1ace in boo§“&?35 to remove this checkout {ram your record. fifiNE§_wi13 be charged 1? book 13 returned a ter the dafe stamped be ._. ~v--....._ ’. W--- M ___#<_ -_.. “H-_—'Hw . .._.- -- ...a . THESIS This is to certify that the dissertation entitled FINITE DISPLACEMENT BEHAVIOR OF PRESTRESSED AND UNPRESTRESSED ARCH-FRAMES presented by Mostafa Tavakoli has been accepted towards fulfillment of the requirements for Ph.D. Civil Engineering degree in 7a): ' Major professor William A. Bradley Date August 27, 1984 MSU is an Affirmative Action/Equal Opportunity Institution 0-12771 FINITE DISPLACEMENT BEHAVIOR OF PRESTRESSED AND UNPRESTRESSED ARCH—FRAMES BY Mostafa Tavakoli A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Civil and Sanitary Engineering 198“ 3.5835298” © 1985 MOSTAFA TAVAKOLI All Rights Reserved it ABSTRACT FINITE DISPLACEMENT BEHAVIOR OF PRESTRESSED AND UNPRESTRESSED ARCH—FRAMES BY Mostafa Tavakoli The behavior of prestressed and unprestressed arches is studied. The possible means of achieving better design through cables or end couples is also examined. The arch being studied is approximated by a series of bars connected by rotational springs. The axial deformations occur in the_ bars which are incapable of bending. The banding is taken by the springs at the nodes. The reactions and the rotation or the moment at the left support are guessed as initial values. The equilibrium equations are written at each node and the forces and moments at each element and node are found. The method proceeds to the right end where errors in horizontal and vertical displacements and either moment or rotation may occur. Having the errors and the initial values, the latter are improved by Newtonian iteration and the calculations are iterated until convergence is achieved. Mostafa Tavakoli The method is capable of taking into account both symmetrical and unsymmetrical force application, physical properties, geometry, and deformations. The method is applied to several problems with known solutions and the results are compared in Chapter 3; these include the buckling of a half circle arch with a concentrated load at the crown, a straight beam with a distributed load over half of the span, a quarter circle arch with the distributed vertical and radial load over the entire span, and the non-prismatic arch under uniform load over the entire span. The results obtained compare favorably with the known solutions presented in the literature. In Chapter 4 the two principal problems of this thesis are investigated: the effect of wind load and the combination of arch and cables at different positions; and prestressed arches with concentrated load at the crown and uniform load over the entire and half of the span. It is concluded that the cables do not reduce the crown ‘ displacement except under the wind load. Prestressing the arch is also not effective in achieving a better design.' The method of extrapolation to reduce the need for a large number of elements was used and proved to be effective. To the members of my immediate family whom I love very much, ACKNOWLEDGEMENT I would like to express my most sincere gratitude to my major professor, Dr. W. A. Bradley, professor of civil engineering for his encouragement, constant help, and guidance throughout the study which made this thesis possible. Thanks are also expressed to the other members of my guidance committee; Dr. R. K. Wen, Dr. L. Segerlind, and Dr. N. Hills. Additional mention is also due Dr. J. Lubkin whose knowledge in computers has been a great help to me. I also owe my appreciation to all my friends for their moral support throughout my college years,including; Ms. Luz E. Quinones and her sister Simply, Dr. A. Emami, his wife Cherie, and Mr. Reza Emami, Mr. M. Kh. Amoli and his wife Patricia, Mr. A. Kh. Amoli and his wife Angie, Mr. A. Ghods, Mr. M. Ghobadi, Mr. A. Golian, and.Mr. A. Sadeghi. And finally, the patience and understanding of my parents and other members of my family must be included in any acknowledge- ment of assistance. Their steady confidence and support have certainly contributed to the successful completion of this project. *3 I ' . TABLE OF CONTENTS CHAPTER PAGE I. INTRODUCTION 1 1.1 General 1 1.2 Related Past Work 3 1.3 Notations 5 II. THEORY AND FORMULATION 8 2.1 Physical Model and.Assumptions 8 Made 2.2 Procedure . 12 III. CASES WITH KNOWN SOLUTIONS 16 3.1 Arch with.Concentrated Load at 16 the Crown . 3.2 Half Span Loaded Beam :7 25 3.3 Distributed Load 26 3.b Nonprismatic Parabolic Arch 3h IV. EFFECTS OF CABLES AND PRESTRESSING 39 “.1 Cable—Supported Arches 39 “.2 Prestressed.Arch 57 “.3 Prestressing by Axial Force 65 V. CONCLUSIONS AND FUTURE STUDY 85 5.1 The Problem.Summary and Conclusions 85 5.2 Future Study 86 APPENDICES 88 A . Newton's Algorithm 88 B. Computer Program Listing to Find 90 the Peak Value of Load-Displacement Curve C. Listing of the Main Computer Program 93 D. Alternative Method for 2-D Problems 100 E. Space Arch 101 BIBLIOGRAPHY 109 10. 11. II LIST OF TABLES Comparison of Results for Cantilever Loaded at the End Comparison of Results for Case 3.1 (Circular Arch with Concentrated Load at the Crown) ~Results for Case 3.2 (Half Span Loaded Beam) Comparison of the Results for Quarter Circle with Vertical Distributed Load Comparison of the Results for Quarter Circle with The Radial Distributed Load Comparison of the results for Non- prismatic Arch fl Maximum Moment for Straight Bar Bent into an Arch ’ Optimumdfor the Prestressed Arch of Table 7 Optimum Span and Maximum Moment for the Prestressed Arch of Fig. ho Buckling Loads (for the Arch Prestressed by Axial Force Maximum Moments and Axial Forces (for the Arch Prestressed by Axial Force) PAGE 12 18 26 28 31+ 36 6h 62; 65 67 68 FIGURE 2. 3. h. 5. 6; 7. 8. 9. 10. 11. 12. 13. la. 15. 16. 17. 18. 19. 20. 21. 22. 23. 2h. 25. 26. III LIST OF FIGURES Modeling the Arch A Typical Bar and a Typical Node Sign Convention for Axial Deformation and Rotation Cantilever with End Couple Arch Under General Load Flow Chart of the Computer Program The Arch of Case 3.1 Load-Deflection Curve for Case 3.1 Arch.Deflected Shapes, Case 3.1, h Elements Arch Deflected Shapes, Case 3.1. 6 Elements Arch Deflected Shapes, Case 3.1, 8 Elements Arch Deflected Shapes, Case 3.1, 20 Elements Quarter Circle with Vertical Distributed Load Quarter Circle with Radial Distributed Load Load-Displacement Curve for Vertical Distributed Load Deflected Shapes of Arch with Vertical Distributed Load Load-Deflection Curve for Radial Distributed Load Deflected Shapes of Arch with Radial Distributed Load Load—Deflection Curve r . . Parabolic Arch Under Uniform Load Deflected Shapes of NonéPrismatic Arch Load Displacement Curves for NonéPrismatic Arch Projection of Cables Load-Deflection Curves for an Arch with Cables having Different Cross-Sections Deflected Half Circle.Arch with Cables Load-Deflection Curve for Half Circle Arch with Cables, 8 and 12 Elements PAGE 10 10 10 11 12 15 16 20 21 22 23 2h 28 29 30 31 32 33 34 37 38 39 1+5 #6 1+7 FIGURE 27. 28. 29. 30. 31. 32- 33- 3#. 35- 36. 37. 38. 39. #0. #1. #2. #3. 1m. #5. #6. #7. __ .1 (CONT.) LIST OF FIGURES Deflected Shapes of’Arch with Cables. 8 and 12 Elements Different Cable Base Position (a) Arch Deflected Shape: (b).Effect of Base Position Half Circle Arch with # Pairs of Cables Deflected Shapes of’Arch with # Pairs of Cables Arch Deflected Shapes - Horizontal Loads at Different Nodes LP - Diaplacement Curves LP - Deflection Curve. Wind Loads. with and without Cables Deflected Arch Under*Wind. with and without Cables ' Prestressing the Straight Bar into an Arch Prestressing the Arch. Load-DiSplacement Curve Prestressed Deflected Arch Moment Diagrams for Prestressed Arches Maximum Moment - Span Curve for a Half Circle Prestressed‘Arch Prestressed Arch. Axial Force Load-Displacement Curve for Concentrated Load. H/L = 0.5. Prestressing by Axial Force Load-Displacement Curve for Uniform Load over Entire Span,H/L = 0.5. Prestressed by Axial Force Load Displacement Curve for Uniform Load Over Half of the Span. H/L = 0.5. Prestressing by Axial Porce Load-Displacement Curve for Concentrated Load. H/L a 0.375. Prestressing by Axial Force Load-Displacement Curve for Uniform Load over Entire Span. H/L = 0.375. Prestressed by Axial Force Load-Displacement Curve for Uniform Load over Half of the Span. H/L = 0.375. Prestressed by Axial Force Egg; #8 #9 5o 51 52 53 5# 55 56 57 59 6o 61 63 66 69 70 71 72 73 7a FIGURE #8. #9. 50- 51. 52. 53. 5#. 55- 56. (CONT.) LIST OF FIGURES Load-Displacement Curve for Concentrated Load. H/L = 0.125. Prestressing by Axial Force Load-Displacement Curve for Uniform Load. over Entire Span. H/L = 0.125. Prestressed by Axial Force Load—Displacement Curve for Uniform Load over Half of the Span. H/L = 0.125. Prestressing by Axial Force Moment Diagrams. Prestressing by.Axia1 Force Axial Force Diagrams. Prestressing by Axial Force Arch Shapes. Prestressing bywaial Force Arch Deflected Shapes. Prestressing by Axial Force,H/L = 0.5 ° Typical Mass Point. Forces and Moments Local and Global Axes (Space Arch) 75 76 77 78 80 82 83 101 102 .CHAPTER I INTRODUCTION 1.1 General Arches are among the oldest ferms of structures and devices that have been used for many centuries. Bows as a hunting device and a weapon have been used since the ancient times. The sport of modern archery started even before the 13th century (12). Arches have had other important applications. In many cases such as bridges, buildings, decorative structures, mechanical bodies, and the approximation of shells by a series of arches, they have fermed a major structural component. In lightweight structures, arches are being used, sometimes in connection with cable nets or fabric covering. For ease in erection, these arches are sometimes made from initially straight bars, bent into an arch shape, secured at the ends by clamping or pinning, and then loaded. It is this extensive application of the arch-frames that motivated this investigation. This thesis is concerned with the behavior of an arch under large enough loads to produce large deflections. The study is done using a finite number of discrete elements to approximate the arch. This method makes it possible to consider problems such as arches with variable section where the application of exact methods would be -2- difficult, if possible. Even if it was possible, to solve the problem using the methods available in statics and mathematics would be often not easy. In this investigation, a rigid bar spring node model is used. Both axial deformations, taken by the bars, and rotations, taken by the springs at the nodes_are allowed. Using the equilibrium equations and Newton's algorithm, an iteration procedure is developed to find the forces and moments in all elements and nodes. The final configuration is also found. The model has been used to solve the following problems for which known solutions are available: 1. The symmetric and unsymmetric behavior of a half circle arch, pinned at the ends and loaded at the crown with a concentrated load applied vertically. 2. A straight beam, pinned at the supports and subjected to a uniform load over half of the span. 3. A parabolic arch with variable moment of inertia under uniform load over the entire span. u. A quarter circle arch pinned at the ends and loaded radially or vertically over the entire span. The problems investigated in Chapter U are as follows: 1. A circular arch pinned at the ends, supported by cables at different positions and loaded horizontally or vertically. 2. Circular arcs with various initial spans prestressed into an arch with a fixed span, 3, then pinned at the ends and loaded. A straight bar was also prestressed into arches with different spans, fixed at the ends and loaded. _3- 3. A straight bar prestressed into an arch by an axial force, fixed at the ends, and then loaded with 3 different loadings; concentrated at the crown, uniform over the entire span, and uniform over half of the span. The results are compared with the unprestressed parabolic fixed arch with the same span and the same loadings. The program written is in BASIC and was run with the TBS-80 micro-computer. Included as the appendices of the thesis are; the main computer program, the method used to improve the initial values, a computer program used to find the peak values of the load-displacement curves, and an alternative method for 2-dimensional problems and methods to solve 3-dimensional problems which did not lead to satisfactory results. 1.2 Related Past Works The buckling of curved structures has been investigated by many researchers. Austin (1) has summarized the state of the knowledge of the in-plane bending and buckling of arches. Austin and Ross (3) have compared numerical procedures for elastic analysis of arches by large deflection, 2nd order and classical theories using repeated numerical integrations similar to the Newmark procedure for beams and beam- columns. Following Watwood and Harts (#0), Gallert and Laursen (1#) have used equilibrium models to study an elastic arch of arbitrary geometry and loading by a finite element method based on a mixed variational principle. They removed the restriction of prior satisfaction of equilibrium on the trial set of unknowns, i.e., the -u- stress field was directly obtained by analysis. Using the variational approach, Schreyer and Masur (33) wrote the equilibrium equations in radial and tangential directions to find the equations of buckling load and loadodisplacement curves for a clamped arch loaded by a concentrated force at the apex. They concluded that the symmetric snap. through buckling always governs. Kerr and Soifer (20) gave an analysis of the effect of linearizing the prebuckling state for clamped shallow arches, overestimating the snap-through load. Using Koiter's initial post buckling theory, Dym (11) considered a symmetric buckling from a linear prebuckling stage and its postbuckling aftermath. Oran and Bayazid (29) analysed the stability of uniformly loaded circular arches without the assumption of shallow and inextensional arch and showed that the critical load (both limit and bifurcation) can be expressed in terms of a combined problem parameter in the form of asymptotic formulas. Sheinman (35) has developed a numerical procedure modifying Newton's method and by finite difference based on large deflection, small strain, and moderately small rotations. The equilibrium equations admit shear deformation and geometric imperfection. Using the Newton-Raphson method to solve non-linear equilibrium equations, Wood and Zienkiewicz (#5) have worked on geometrically non-linear analysis of elastic in—plane oriented bodies in a total Lagrangian coordinate system, developing a paralinear isoparametric element. A non-linear elastic finite element for a beam initially curved in one plane but deformable in the three dimensional space has been presented by Wen and Lange (#2). In this work the quadratic and -5.— linear eigenproblems were formulated to calculate the in-plane and out of plane buckling loads of arches. Wen and Rahimzadeh (#3) investigated non—linear elastic frames including arches, approximated by a series of finite elements and using different coordinate systems such as Euler and Lagrange (small rotations and updated). Other studies of interest include, OJalvo and Newman (28), Ojalvo and Demuts (27), Bathe and Bolourchi (#), Chajes (5), DaDeppo and Schmidt (9), Dawe (8), Nempner and Patrick (#1), Yamada and Ezawa (#7), Harrison (16), and Sabir and Look (32). The survey article by Schmidt and DaDeppo (3#) provides additional historical comments and a more complete bibliography than that undertaken here. 1.3 Notations The following symbols have been used in this investigation: A:: Cross Sectional Area of Each Element; A = Cross Sectional Area of Cables; AD: Axial Deformation of Each Element; AF: Axial Force in Each Element; CB: Combined Components of the Axial Forces of the Pair of Cables in the Plane of the Arch; CD: Cable Axial Deformation (Positive if Compression); CF: Cable Axial Force; CE: Horizontal Component of CB; CL: Cable Original Length; CN: Cable Final Length=CL-CD; -6- CV: Vertical Component of CB; E = Modulus of Elasticity of Each Member; Modulus of Elasticity of Cables; GR: Guessed Value for Rotation of the First Element; HL: Applied Horizontal Load at Each Node; H H .Moment of Inertia of Each Member; Element Length; I" ll LF: Load Factors of the Distributed Horizontal Load; M = Moment of Each Node; htmaf Maximum moment; M): Prestressing Moment at Each Node (When Bar is PreStressed By Couples Applied at the Ends); N = Number of Elements; Concentrated Applied Load; '0 II Bifurcation Load (Where Unsymmetric Buckling May Occur); ’6 ll Critical Load (Maximum P on the Symmetric Part of Load-Vertical ‘0 ll Displacement of the Crown Curve); 'V II Limit Applied Load After Which the Deflection of Crown Gets Larger By Adding Cables; R=Radius of {he-Arch; SC= The Angle Between the Two Neighboring'Elements,-Unless:0therWi8€ "Specified; SF= The Shear Force’in Each Element; ‘5: Vertical Displacement of the Crown Corresponding to Bifurcation Load; 2:: Vertical Displacement of the Crown Corresponding to the Critical -7... Load; VL: Applied Vertical Load at Each Node; Vertical Displacement of the Crown Corresponding to the Critical Bll Distiuted Load; Distributed Load Per Unit Length; 1: n W = Bifurcation W (Where Unsymmetric Buckling May Occur); Critical W ( Max. w on The Sym. Part of Load—Displ. Curve); 1': ll W - Radial Distributed Load Per Unit Length That Changes Direction to 2' Stay Normal to the Arch During and After Deformation; WNN=Radial Distributed Load Per Unit Length That Does Not Change Direction and Stays Radial; x = x- Coordinate of Each Node; Y = Y-Coordinate of Each Node; AG: Rotation of Each Element; 6 = Angle between the Element and 3 Vertical Line 0 Measured Clockwise From The Vertical; 6 = -8- CHAPTER II THEORY AND FORMULATION 2.1 Physical Model and Assumptions Made The arch is modeled as a combination of rigid bars and rotational springs. The bars are considered to have axial elastic properties, but are incapable of bending. The rotations are taken by the springs at the nodes (Fig. 1). It is also assumed that the material is isotropic and linearly elastic. No shear deformations are allowed. Fig. 2 illustrates the forces and moments on a typical bar and a typical node. On the first node and element there are also the horizontal and vertical reactions in the positive x and y directions, respectively. The applied horizontal load is considered to be positive if in the positive x direction. The positive applied vertical load is in the negative y direction. For a fixed support a moment also exists at the support. If the forces and moments on the element (or node) i-1 are known, the forces and moments on element (or node) i can be found using the following equilibrium equations at node i. SF(i):-AF(i—1))Sin(SC)+SF(i-1)Cos(SC)+HL(i)Cos(6)+VL(i)3in(6) (a) AF(i)=AF(i-1)Cos(SC)+SF(i-1)Sin(SC)+HL(i)Sin(6)-VL(i)Cos(8) (b) (2.1) AD(1)=AF(1)L(1)/E(1)I(i) (e) -9- M(i)=M(i-1)+SF(i-1)L(i-1) (d) 06(1)=M(i) { L(i)/2E(i)I(i)+L(i-1)/2 E(i—1)I(i—1)} (e) Axial deformation and rotation of a node are shown in Fig. 3 in the positive sense. As with most finite element solutions, taking more elements leads to a better accuracy. The results obtained using this model agree well with the ones previously obtained by others. Table 1 shows the results for a cantilever beam loaded at the free end with a vertical downward load. As can be seen, with extrapolation between 6, 8 and 10 elements exact results (23) are obtained. The #, 6 extrapolation differs from the exact value by only .061. The #, 6, 8 and 10, 20 extrapolations are more accurate than the solution with 100 elements. Fig. # shows the results for a cantilever beam with end couple. The results are very good again compared with the exact ones. The effectiveness of this model and the procedure used will be compared with some other previously worked out models and procedures later in the next chapter. -10.. Fig.1. Modeling The Arch SF(i) M(i+1) [{FUL) \ 6(1) Ages SF(i-1) Fin-1) Fig.2. Typical Bar And Node \ép( L(i Node 1 ' i \/ AD(i)=AL(i) Fig.3. Sign Convention For Axial Deformation And Rotation massed was sues uw>mHHucmo s\x N. H. w n . a Hm DWIIIIIIIIIIAU mummEme OH oauommu .. H. nvunluuulluuxu wucwemflw ON Hm . nVIIIIIIIIIAu mucoswsm on muswfl IIIIIIIIIIII mucwEme m . . 1 l . m w z m . Newman. mmaaas.- sumxm sstsH. HQNH¢H...ON.omwco m ca mmmmsa. «mason...os.m an H etuxm . scones. anNaH. - mucoEmHm om fiancee. Noumea. - mucuswam on seemed. maooom. - wuawsmsm m . A\vcm» a\ecmx muHm\sz toe A\> .s .me Table 1 - Comparison of Results For Cantilever Loaded At The End -12- 2.2 Procedure x y 'l of elements free end free and slope (free end) 8 60.3019 73.3951 1.20618 6 60.8807 72.2761. 1.21116 8 61.018 71.8838 1.21298 10 61.098 71.7018 1.21383 20 61.2028 71.8597 1.21898 1oo 61.2358 71.3823 1.21535 1 8,6 61.27178 71.3809 1.21518 3 8,8 61.2567 71.3735 1.21525 2 6,8 51.28596 71.37 5 1.21532 2 , 8,10 61.28022 71.37896 1.21538 PL /EI 5 10,20 61.23773 71.37900 1.21536 #,6,8 61.23736 71.3777 1.21538 6,8,10 61.23700 71.37921 1.21535 xact (23) 61.237 71.379 1.21537 cf. (13) #el. 61.29 71.91 1.223 ef. (13)10e1. 61.25 71.86 1.221 J The procedure used in satisfying the boundary conditions for a frame is the so-called "shooting method". It starts out with assuming values for the unknowns at the left support of the arch under any loading condition (Fig. 5). The final configuration of the arch after AU’ / load Arch Under General Load V x Fig. 5. -13- loading is to be determined. The assumed unknowns would include reactions and rotation (for pinned end) or moment (for fixed end). Having this information and using formulas 2.1, we can find the forces and deformation of the next element and moment and rotation of the next node and proceed to the right support. We will end up with certain values for the x and y displacements and moment (for pinned end) or rotation (for fixed end) which are the errors (if not equal to zero). Having the initial guessed values and the final errors and using Newton's algorithm (appendix A), we improve the initial values and iterate until convergence is achieved, i.e., the errors are within acceptable range. In Newton's algorithm, to find the derivative of f(x,y,z) representing each one of the 3 errors at the right support with respect to say, x representing one of the 3 initial values, we find f(x,y,z) and then f(x+Ax,y,z), Alt being a small increment in x. Then LD§(X.y.2) = f(X+AX.y.2)-f(X.y,2) (2'2) ‘bx Ax Ax:.01 has given the best results in all cases. A listing of the computer program is given in appendix C. Fig. 6 shows the flow chart of the program. In the process of inputting data, variable thickness, properties, shape, element length, and loading can be allowed for. The uniform loading is assumed to be lumped at the nodes. Components of the loads at the nodes in x and y directions are input. A double precision program has also been used and the results have not changed significantly from the ones using single precision. -l#- It should also be noted that the choice of the initial values may change the speed of convergence significantly. Another point to consider is that a set of initial values intended for one case of equilibrium, such as symmetric, may lead to another equilibrium configuration, such as unsymmetric. With a little experience, one will have a feeling of what the initial values for a better and a faster convergence in the right direction should be. -15- amLmonm Lousnnoo och uo aL-LU scan .0 stands so: vouoaaso nu aeuanomau on» Huge: no>« 11>«Lov vcwu 0» < on some on non» .o:Hd> Andean“ code 0» .COBohoca HHdEm o>du.s£a«LomH n.cou:oz ”can: nosau Huauacu o>onns seemedau can» egos coda-Loud no Lonny: on n a Locus oanuuaooou easy“: access» ucwun on access Lo\:o«oauon can mucosooaannav oL< l 5 nova: on» he nomacvaOOo vcd novoc a”. was napusoao . < “an no neoduauoz can .nco«uuanouou Hague .nucoeoa .noonou vcdm Illlllllllllllllllll o o . do Lopes: unsoaaa on» and .uaaaa Locno .Auconasn ago” an ceauuaoc vcd acoaaouan Louv nonnosm Huauacu azncH ~16- CHAPTER III CASES WITH KNOWN SOLUTIONS The following cases have been studied. 3.1 Arch With Concentrated Load at the Crown A simply supported half circle arch with a concentrated vertical downward load at the crown was considered (Fig. 7). Fig. 7. The Arch of Case 3.1 The program was run for different initial guesses, all leading to the same results. Convergence was fast, except at the loads very close to the critical load. To overcome this problem a third degree curve fitting was taken passing through # neighboring points on both sides of the peak value. The # points had already been found using the program. So the equation of the curve was written. If the coordinates of the # points are (xlyl), (x2, y2), (X3, y3), and (x4 , Y4) then the coefficients of the third degree polynomias can be found as follows: -17- 3"1 H Xi X1 1 3 2 yz Kg Kg X2 1 (3.1) y3 = x3 x3 x3 1 3 2 j y f4 x4 x4 1 d -3 2 - -1 x1 x1 x1 1 Y1 3 x2 x 1 y = “g g 2 2 (3.2) C X3 X3 X3 1 Y3 2 d _§2 x x4 9 Y4 Knowing the coefficients, the peak value can be determined by solving the following equation for x and finding the corresponding y. The program to do so is listed in appendix B. dy/dx = 3ax2 + 2bx+c The bifurcation load was determined by approximating the symmetric and the unsymmetric branches of the load-displacement curve by two polynomials passed through determined points on the branches and then solving for the intersection of these two curves. Fig. 8 shows the load vs. crown vertical displacement curve. The reference curve (15) is also plotted. Figs. 9—11 show the deflected shapes of the arch under different loads for #, 6, and 8 elements, respectively. Fig. 12 shows the deflected shapes of a 20 element arch with different properties. Table 2 compares the results obtained using this method and other references. The S differences in results are also shown. In this table: Pc : critical load (peak value on the symmetric part of load- displacement curve) -18.. Pb = bifurcation load (load at which unsymmetric buckling may occur) Vp = displacement of the crown corresponding to Pc FIahle 2 u comparison of Results For Case 3.1 ' 2 2 A Z Ditrer nce v - ~ — No. of PbR PCR 9 1n p Rl/EI In P RS/EI In v /R Elements E1 151 R b _ t p 4 6 8 6 8 4.6.8 6,8 4,6,8 6.8 4 ~ 4.66 6.960 .82 6 5.335 8.075 .78 8 5.600 8.915 .765 ,8 6 5.940 10.00 .75 4,6,8 5.960 10.32 .75 (15)50E1. 5.875 10.15 .75 1.45% 1.112 1.672 1.482 02 OZ (42)16E1. 6.766 11.92% 12.212 - (21) 6.540 . 8.872 9.172 (43) 5.700 4.56% 4.21% Fig. 8 is for the arch with 5:200, 1:108, A=105, and R:5000. Table 2 corresponds to the same arch. For an arch with different properties (E:1.O#x107, I:6.21x10_6, A:.0775,R:10), the values for #, 6, and 8 elements shown in Table 2 did not change significantly. It should be noted that the results are for a much fewer number of elements than many others. The larger difference with two of the references in Table 2 is mainly due to the difference in assumptions. In reference (#2) the weakening effect of bending deformations on the axial stiffness is -19- neglected. The displacement of the structure is assumed to increase linearly with the applied load until buckling occurs. The assumptions in reference (21) were based on the cross sections being inextensional. Although the term involving the cross sectional area does not appear in the dimensionless term PRZ/EI, the effect of the area is significant in many cases (#2). The present study is not based on any of the above assumptions. The 3 point Richardson's extrapolation has been done using the following formula: 4 n? A ni A n: A= Af -—-——-———-*-4- - i--*'-—-———- + c'-—jr-——————-——- 2 2 - 2 2_ 2 2_ 2 _ 2 2_ 2 (“f"nc)(n¥—ni) (nf ni)(ni nc) (nf nc)(ni no) (3.3) where Af, Ai, and Ac are the values corresponding to nf ,ni, and nC number of elements, respectively. For a 2 point extrapolation the following formula has been used: 2 n2 n A= Af —2—f—. — Ac—;—~2—— (3.11) (nf—ng) (nf—nc) The energy can also be checked. For a body in equilibrium the energy due to the external forces should be equal to the internal energy. The external energy due to load p is equal to the area under the load-displacement curve from load=0 to the point corresponding to p. This area can easily be found by approximating the curve by some straight lines. The internal energy would be equal to: Internal Energy = 1/2211(1)Ae(1)+AF(1)AD(1)) (3'5) Using Fig. 8, the external energy for a load of 7000 (before the peak value on the curve) would be equal to 1.507376x107 and the PRZ/EI H O -20.. 49y———————7£5 Ref. (15) E}——-—-———E3 8 Elements GP——--———<3 6 Elements e———e 4 Elements Points on the curves do not R = = E = 200 A = 105 represent data points. they are . there to distinguish between the 1 1 times! 1 1 o .2 .4 6 .8 1.0 vc/R Fig. 8, Load — Deflection curve for case 3.1 -21- I nvnunuuunuuanniu nvuluunuluullxu MOHflfi wOHnH m mummEmHm q .H.m ammo .mmdmsm pmuomammo sou< .m .wam oeuuoEEAms: o nudes» o a m on Hmaumaho com um ooom um 1T muamsmam o .H.m sumo .wmammw wouowawoo cuw< .OH .wfim o o D D UHMLmEmCD T’s uawuosamm Hmcwwwno -22- + moeue com-m woenH ooomne masseuse w .H.m ammo .mhamsm emuomsuma noh< .HH .wam -23- T’s 833 0'0 Hafiwfio mosue oomum woeuH ooomum assesses ON .H.m ammo .mwamew ewuuosuun hope .NH .mfim .- Pa. 0 o D B :95 630636. m/M'oflllsfl .. n o .sam .emuumfiuoa old —28- Hmewwaho memo.n< o-osxam.ouH coexq.oanm cane -25- internal energy is equal to 1.5152700x107. There is a 0.52: difference. For a load of 6000, (on the curve passed the peak value) the external and internal energy and the percentage difference are equal to 2.9865619x107, 2.9928650x107 and 0.2:, respectively. These were all for the 8 element case. Therefore the external energy and internal energy are equal, which confirms again that the configurations obtained are the equilibrium configurations. We see that the model and procedure work very well for this case. 3.2 Half Span Loaded Beam A simply supported beam with a horizontal restraint at the end was loaded uniformly over half of the span. Table 3 shows the results for different w using different numbers of elements. The values for 6 divisions seem to be away from the normal accuracy. This is possibly because of the length of elements not being exactly entered as 100/6 in the data for computer program. In the data, instead of 100/6 the length was taken as 16.6667 which caused the difference in the length of 16.6667x6:100.0002=§ difference = .0002. The .0002 extension results in the axial force of (.0002) (EA/L)=.2 which may have caused the difference. As can been seen in Table 3, using double precision did not change the results significantly. With no horizontal restraint at the end (referred to as the exact results in Table 3). -2 v =.651042x10 wL4/EI (a) center (b) M -.0625 wL2 center (3.6) -26- For a small load (Wzl) the results from Table 3 are about 11 Table 3 - Results for Case 3.2 of V31 ":10 Hgso a "CgtJr H VC "C H vC M(3 H I .675887 617.67 -1o.788 8.5075: 8052.86 -88#.128 9.18338 8325.3 -231o.28 6 .658508 618.17 -1o.3183 8.83921 8095.93 -881.573 9.63835 8816.02 -2322.75 8 .652289 618.17 -1o.8073 8.8189; 8111.76 -880.591 9.60818 8853.8 -2326.2 10 .689821 618.28 .10.353 8.8038. 8119.82 -880.08 9.58889 8872.88 -2327.6 20 .685630 610.32 -10.2903 8.38901 8129.79 -879.271 9.56896 8898.81 -2329.29 8,6 .688630 618.57 8,8 .688383 618.38 -10.2808 8.3880: 8131.53 -879.812 9.55779 8896.6 -2331.51 6,8 .688206 618.18 8,10 .688393 618.35 -1o.2565 10,20 .688366 618.35 -1026.98 8.38801 ”879.01:1 -#79.015 9.56232 3501,13 -2329.“ ,6,8 .688066 ,8,10 .688898 .10.20 .688361 618.38 -1o.2719 8.8.10 .688395 618.36 -10.2519 Double Precision 8 .6758516810 617.677859 10.7873779 8.5075603 #052.#50 -888.1298 9.78332 8325.3089 -231o.279 8 .65228988 618.179029 10.8063318 8.81890 8111.750? -880.5910 9.6081376 8853.76837 -2326.211 59888 5177 91658 220 9 68805 8,8 .68838207 L618.386gg 10.279316 8.3880216 8131.51762 -879.8116S 9.5577809 8896.5895 .2331.5228 Exact .651082 | 625 one md on roller T dif 1.022 1.063 8,8,10 | Le1og , 5:10 1:100 1:10 different from the exact ones obtained using formulas 3 J- 6- Of course the non~linearity causes a small horizontal reaction whicfli makes the two cases (Table 3 and formulas 3.6) slightly different- 3.3 Distributed Load load over the entire span was considered (Fig. 13). First a quarter circle with pinned ends and a vertixzal distributed -27- 6 E=4.l76x10 I=2.03125 A=2.7 Fig. 13,Quarter Circle with Vertical Distributed Load The loads on the two half lengths of the two elements adjacent to each node were taken as the concentrated load at that node. Fig. 15 shows W-Vc curve where Vc=vertical displacement of the crown. Fig. 16 shows some deflected shapes of the arch. Let Wb be the point where the symmetric and unsymmetric curves meet (buckling load), Wc be the critical load which is the max. value of w corresponding to the symmetric curve, and Vb and Vm be the corresponding vertical displacements of the crown. Table # compares the results obtained using this and other methods. As shown with extrapolation between 6 and 8 elements, the maximum difference between this and other procedures is about 21. -28- Table 4- Comparison of The Results for Quarter Circle With. Vertical Distributed Load J 7 Di fpronr U11 Mn 3 o. of El. wbR /EI 631, RP] 6.8 ch /EI Vb/R Vm/R 6 15.27 25.55 .0125 .08#5 8 15.5 26.98 .0127 .0875 6,8 15.78 28.83 .0130 .091” (#2)8El. 15.#3 1’ 11 2% (2) 15.62 2’ . 11 1’ Next a quarter circle with pinned ends and radial distributed load was considered (Fig. 1#). E=107,I=.8789x10—2 A=.l875 Fig. 1#. Quarter Circle With Radial Distributed Load The load may or may not stay normal to the arch. It was again distributed among nodes as concentrated loads. Fig. 17 shows the load displacement curves for 8 and 10 element arch. Table 5 compares the results for the case when the load stays normal to the arch. With extrapolation between 8 and 10 elements the difference is 61 from the one obtained by Wen and Lange(42) and .42 from that obtained by Timoshenko (37). -29_ mHuuHu umuumao wmoq vousnfiuumfio HmUHuuw> now w>u=u ucmeowHawHanmog .Ezwc .9» .wac .8» D U) D U) .mucoEmHm w .mucmEmHm w .wucmEmHm o .mucvamam o a.Nn< nNHmo.NuH oosxoafi.qnm N.meum m\0 > N. _ _ .na .wfim o rmH Is/gan -30_ umoq wouspauuwan Hmuauuw> so“: suy< mo moamsm wmuuoawoo .ea .wam Afiflgv UHHUOEMCD To Amnzv owuuoeszm nYllllllllllAu HNCHMHHO I waoufio youumso N .N"< nNHmo.~uH m OOHXQRH. .VHN N.®Hmflm vmoq Haavmm vmusnauumao you o>u=o coauumawwolvmoa .NH .wam . Eu . . . 0H > «a NH 0H mo. 00. so. No. 0 .II+\ “ i u w u n u o v No a 23E. - s ownwoo.uH my Realm wauuao kuumso . omum a A ll 0- x o 3 AqulllllLAW momma o>oam wzu «0 Han you .Emw olllllllllllo coauumuaw wwcmno uoc moon vmoq ..Hmw AWw.w. lelllllllmw HmEuo: mzmum wzm3am vmoa ..Hmm .Emmca mVIlllllllmw aoauowufiv owcmco uo: mwov vmoq ..Hm0H olllllllllo awake: wzmum mamaam vmog .Hmoa mot/52m smog ewusaauumaa Hmaemm cuss cot< mo magmam uwuumauoo .wH .wam mama. u< “OH um oaouau “obtuse mmawoo. "H on um a 35838 I chfimfiuo GIG Ama .wamv 0 .ug .soauemos .m mans Avv “ma .wfimv m a .ua .mmuvmoq -32- N woo: Any -33.. 0>HDU GOHUUUHuun I vflOA omH .w._n..H mks NH. ca. mo. so. so. No. Q _ . _ p . p O _ u u d — . 3608on m N. 62803.2. «wanna go: moo—u v3..— . 11 ad ovozv n Au «waxy .. e. an «nozv namH.-< o uuao “outage «wamoo.uH a II V noalm 3 a. . .i c . onlm é IQ/cXM -3u- iable 5- Comparison of Results for Quarter Circle,Radial Dist. Load J ‘ _ No. of "N R3 . 1 Difference With “NN Elements 3031 8 £1. 10 El. 8,10 3051 8 .N70 - - - .503 10 .NBO - - - ~51 N=Normal to Arch 8,10 .u98 — - - .522 Lu2)16£1. .5291 13$ 10: 61 NN=Not Normal L37) .50 6: M .131 to Arch The deflected shapes are shown in Fig. 18 (a and b). As can be seen, this case corresponds to one of the many modes of vibration. Fig. 18 (c and d) also shows the deflected shapes for 2 other modes of vibration. The corresponding load deflection curves are shown in Fig. 19. 3.” Nonprismatic Parabolic Arch The buckling of a nonprismatic parabolic arch, pinned at the ends and subjected to a uniform loading on a horizontal projection over the entire span was investigated for two different variations of the cross sectional area. In both cases the moment of inertia at a cross section is I=Ic sec 0 (3.7) where Ic=the moment of inertia at the crown; é: the angle between the tangent to the arch axis and the horizontal (Fig. 20). {1“ J. J, .1, ifiq ch 10 Ic=.l 5:107 keee 20 ~' Fig. 20, Parabolic arch Under Uniform Load -35... The two variations of the cross sectional area are: A :10 I A =10,000I The rise-span ratio is taken as 0.5, that of a half circular arch. The nodes were taken in such a way that they were on the arch axis and the lengths of the elements were equal. Fig. 21 illustrates the deflected shapes of the arch. The load- crown vertical displacement curves are shown in Fig. 22. For the parabolic arch with uniform load, the axial forces are the main cause of displacement. As can be seen, the curves for the symmetric case do not have a decreasing segment. For the case of A=10,000 I, the symmetric curve is very close to a vertical line. The large cross sectional area causes small axial deformations and small displacement. The critical values of the axial compressive force at the quarter points of the span can be expressed by the following equation, as is done by Austin (1). P = m—Ele— (3.8) e 2 where S = one half the length of the arch axis; x : a coefficient. The horizontal component of the thrust at buckling can be expressed as EI He= P where C (3.9) L2 -36- L = span length; and P = a coefficient. The values of x and P obtained using this method are compared with the reference (1) values in Table 6. The 3 point extrapolation gives a very good set of results, specially for the case of A=1OI. ,Part of the difference in the results can be due to the possible difference in the form of the variation of the cross—sectional area which is not specified in the reference literature. Table 6 — Comparison of the Results for Non-Prismatic Arch No. of Aelo I A=1o4 I A=10 I A=1r4 T Elements u ZDiff;. ' Diff. Diff. F I Diff u . / Ref. / Ref~ P F/ Ref. W/ Ref. 6 11.323 1.70 11.078 3.66 13.522 3.19 13.3252 3.50 8 11.521 .83 11.277 2.95 13.788 1.31 13.5822 2.79 10 11.552 .59 11.3252 2.53 13.8665 .89 13.8301 1.07 6,8,10 11.5725 .31 11.3323 2.33 13.93 .30 15.3721 3.15 ref.(1) 11.62 - 11.62 - 15.0‘ - 15.0 -37.. sow< ofiumswfiumncoz mo moampm wouomauoo .HN .wam QlllllLlllo 0’0. H3... I 833.5 75.3 New .11 owaonmwmm TIIImMHMIIIL m geomUHlH 2H [3» 2. .. HUD .. NOHHN sow< afiumamfiwmncoz wow mo>w=o ucoswomfiawfio 1 umoq .NN .mam A\u> mo. «0. mwoos em. 1.) 4. com F2 QIIIIIlIlIldoauquE>wca mYIIIIIIIlmw ofiuuwsszm HOHu< m . Awowuu0555mca 1.00m HQOHI< Tl... m owuuoeaam aucoaoam 0H oaaonmuwm Tllllkmnnrlx euomoHIH nu OHH A/Di OH‘IUH ZHHH Boa-m 1. 8e ween ’13/51b —39_. CHAPTER IV EFFECTS OF CABLES AND PRESTRESSING 3.1 Cable Supported Arches A slight modification in the program used for case 3.1 makes it possible to apply it to the cable supported arches. To keep it an in- plane problem, two cables, identical in property, length, and strength but on two opposite sides of the arch and symmetric with respect to the plane of the arch were attached to any desired node (Fig. 23). Fig. 23. a) Projection on a b) Projection of Cables in Horizontal Plane The Plans of Arch -30- The symmetry would allow us to find the in-plane component of the axial force caused by extension in each cable and double it to find the total force due to the pair of cables on the node to which the cables are attached. The components of the cables axial forces in the direction perpendicular to the plane of the arch would be equal in magnitude but opposite in direction, therefore they cancel. In the modified program, if there are any cables at a node, the coordinates of the node are found as if the cables did not exist. _Having the new coordinates and VD (Fig. 23), new CC can be found. cc=[yD2+ y(i§Jl/2 (3.1) The new length of each cable is then equal to CN=(CC2+ BD 2)“2 (3.2) So if CL = the original length of the cable, the change of length for each cable is (compression is positive) CD= CL-CN (”'3’ Now, if E(:and A(:are the cable modulus of elasticity and cross- sectional area,respectively, the axial force in each cable would be CF=CD-E -A /CN ("'") C C If CP = the angle between the cable and the arch plane, the components ofCF of the two symmetric cables in the plane of the arch would be CB=2(CF)Cos(CP) ("'5) Then CV and CH, the components of CB in vertical and horizontal directions are (3.6) CV=-(CB) Sin(BK) ~31- CH=(CB) Cos(BK) (“-7) These are added to the applied vertical and horizontal loads. Having new forces, moment and axial deformation of the corresponding node and element can be found leading to the new coordinates of the node. Then, we proceed to the next node. Note that cables are useless when compressed so they should be allowed to work in tension only. To examine the effect of cables, a pinned-ends half circle arch with vertical and horizontal (wind) loads was studied. with a vertical load at the crown and 16 elements, the program was tested for a set of weak cables. The critical load of 10.0 and the displacements agree well with the previously obtained results (in Table 2). Then the same arch with 3 elements and a set of strong cables at nodes 2 and 3 was studied.Fig. 23 shows the load-vertical deflection (of the crown) curves for cables with different cross sectional area or modulus of elasticity. As can be seen when the cables are strong, the pattern,in which after the critical load (right half of the curve) decreasing the load would increase the deflection, may not exist. Displacements are smaller and the critical load is bigger than without cables. When the cables are too strong, the corresponding node tends to stay where it was before loading. The neighboring nodes will be displaced large amounts.causing discontinuity,and therefore making convergence very hard or impossible. Fig. 25 shows the deflected arch. The 3 element arch can not be a good representation of the problem because the arch together with the cables acts as a truss type -32- structure. For this reason an eight bar arch was considered. The cables helped the structure by decreasing the deflection up to a limit load, 2‘, smaller than, but close to the critical load, as can be seen from Fig. 26. At P: g ,the cables cause an increase in the crown deflection up to a certain load after which the cables go into compression. Changing the position of the base of the cables did not improve it very much. It was thought that this increased deflection was caused by the fact that for a large cross sectional area and after a certain deformation occurs, not much axial deformation is possible (compared with rotations) so the crown starts deflecting more as we add the cables. But with different cross sectional areas,the basic pattern still was unchanged (Figs. 26 and 27 ) . However, it was noticed that the change in geometry due to the cables, for loads greater than Bi? increased the-horizontal reaction causing larger moments therefore larger deflection as we add the cables. V As the number of elements increases, PL gets closer to the critical load. As seen in Fig. 26, for 8 elements PL/Pc= 6.1/8.915=.6842 and for 12 elements PL/Pc= 8.7/9.71=.896 Fig. 27 shows the deflected shapes of the arch with 8 and 12 elements. Figs. 28 and 29 show that changing the position of the bases of the cables or properties of the arch does not change the general behavior of the arch much. Figs. 30 and 31 show that with more cables the same pattern still exists (deflection of the crown increases with cables after a certain load 2‘). —33- One of the main applications of cables is when wind forms an important part of the loads. Therefore, we now examine the behavior of the arch-cable combination under wind load. The horizontal component of the wind on a circular arch is more important. For this reason, the horizontal load was considered even though it is just a matter of inputting the magnitudes of the vertical components as the vertical loads in the computer program to take into account the total effect of the wind loads. A pair of concentrated loads were applied first. The two loads were equal in magnitude and opposite in direction, applied on two sides of the arch at the same position relative to the two supports to make I it a symmetric problem. As expected,the lower part of the arch moves in and the upper part (including the crown) moves outward (Fig. 32). The horizontal uniform load was distributed as concentrated loads among the nodes according to half of the total vertical length of the two elements adjacent to each node. These ratios are called wind load factor (LF). To increase the horizontal load, LP of each node is multiplied by a constant number. For 12 element arch the load factors for nodes 1-7 are 12.93, 25, 22.315, 18.7, 13.385, 6.825, and 1.785 (Radius of the half circle arch = 100). Fig. 33 shows the LF - crown displacement (vertical and horizontal) curves without cables. The curve for LP - horizontal displacement of node 3 in a 12 element arch is also shown in Fig. 33. Node 3 follows the same pattern in terms of horizontal displacements except that they are bigger than the horizontal displacements of the crown fbr a certain load. ~33- In Fig. 33 the LF - horizontal displacement curves with and without cables are shown. The load is scaled on the horizontal axis for a better illustration. The curves tend to approach a horizontal limit. Fig. 35 shows the deflected shapes of the arch for different loads, with and without cables. As can be seen from Figs. 33 and 33, attaching cables would help the arch have less displacements and a greater horizontal limit of LF- displacement curve. Therefore, when horizontal (or wind) loads are applied, cables are effective and useful. mcofiuoomlmmomu mammaum> wcfi>mm moanmo Lugs cuu< am you ww>wso coauomamma 1 one; .qN .wfim o m\ > q.H N.H o.H w. o. c. N. o _ _ _ _ _ _ . . _ . n u o . > .Hm vIIIIIIIIIIJ I I nucoEon Na ecu m .:u~< uaucau “an: new no>usu ecuuuuauoa no.0; ~.~ u .2 .nudneu Lea: 0ou>> .Hm NA .aUHpnu Lu“: .Hm o scan-u ex) .Hm Na .aoHn-u o\: .um m .noanau :u«: .au .Hu ma am NH m\u> o.a .2 A: o Load=0 Load=2000 ad: 300 rL_¥ Cgidf20001 Load=2800 -u3_ 0 m__0 Original o——————o ‘w/o cables 0——0 with cables /2\ \/ 90 igo 90 QU'QHLv:9H °______° w/o cables 0 ___0 with‘ cables P Eg=1000 A =0.5 c For (a) and (b) R=100, E=Z9000,I=102, A=10.8 Fig. 27. Deflected Shapes of Arch, Different Number of Elements . uu ooo can u < —39- 0 Q coon-5 oooel; 0’0 3.33 ox: COHuwwOm Dawn QHDQU uCOHUuuwn uQN 11d 111/z -50- eoaeaeoe eeem eaaeo no eeeccm any eeeam evanescea see< any .aN .uee 5.5 cl n. o 4 q a q o :owmmouano. Apv any uoc aoamcma cw Hawum one eoapeo news: you eeoa.xez naee . . oaue< N m mw O l 83.. m m 11¢ _ :o _ 1Il\\\\\\\\\\\ ooomlvuoq ..m a QIIIIIIIIG oomlo> , + .183 do“ once 2 com A m mam. ..umnw noal< .QOHIH Anv vac Amy you cow-m .ooonlm m eeaeeu mo eeaee_e sees aee< eHeeao can: .on .wne > ~.H o.a m. e. b q a v. N. o b q a ab 0 n.o. «.oooauem m.oa.<.~oH.H oooa~um.ooa.a _+ -51- 01 o 328 fit. nvu, 11nu nuance o\3 < m No.mug v? in 1 OH on ow om , swung-«J... ma «H e m mooz um madame mo wwwnm c .muaoamam 0H IS/zad -52- non .wae so a .eev oomwueeoa o O a U m on < oooau m w.oau¢ .NoamH oooamum .ooaum A meaeeo no enema e seas eee< co meemem eeeeoaceo .Hm .wae .Arl + u on .wae so u .emv. commuumoa m 0, o moHnmo sues mYI1 nu moanmu o\3 mW1 flu Heeawaeo _ \r 1 yr .2. mooauemoq cm 11 oe1: om ouvmoqm OH. -53... moanmo 02 .mmqu uaouowmua um cocoa Hmucounom .moamnm monomauoa £uw< .Nm .wwm oonumnmoq .ahm .uawo: mo comm m.oau¢ NOHIH ooommlm cor-m _ Aw Q S .N .39. a. 38+ . AU w.o memos Humvee mWr 1am on .q moves Hemvm._ 0 ‘0 m.n mmvoa um 23. _ Hecamauo 0 -5»- mo>uso ucosmuoaamwa c h «\ua I kg N. mm .wam u amumaamwo aeowuuo>n> unmamumaamao .uauomn: w.oal< .NOHIH ooomNIm .ooalm _ - IB/ZH (31) 09 -55- Ska 2.: o... :28 26.3.. v... 5:. 2.3 v3: .296 3:888 1 5 .3 J: [m e.“ N." S e e e N o u < a a O 1‘ N 0 e .1 F > c :1 .f. n 3 I a nausea-Hanan .uwuomun 7.2 087.» n .O~I< NOHIH OOOowlm Godly— +. r O .m. add 1. cm , mYIIIIIIIIIo an ova: u. uqu-u s elIIIIIIIIIJ “a .o come: no asap-u o o H a: .1 a w cm , «Ma 0 n mYIIIIIIIImw soap-o ox) weapon usomuas van Luna .umog mafia nova: suu< vouuoawoo .mm .me wouumw vmoqnhq mo mu=He> mundane“ no>u=u ecu co muonasz AVIIIIIIIIIIIAU canoe o\3 oIlIIllIlIIIé Ha um moanmu Ollo 3 .m on magma mu . . mu Heanwaeo -56- o.Huu< .oooauem w.oH-< .NOHIH o ooooNIm .ooaum -57- “.2 Prestressed Arch A straight bar may be bent into an arch, then fixed and loaded. To take the effect of this prestressing into account, we have to find the prestressing moment, Mo, and add it to the moments at the nodes. Fig. 36, Prestressing The Straight Bar Into An Arch If n=number of elements, then the angle between element 1+1 and i would be “/n (Fig. 36). To form this angle ,the prestressing moment M0 is applied. M “BEE—I (”08) o n L Then if GR=rotation of the first element due to the applied load, the moment at node 1 is (u.9) M(l)=Mo+(GR)-%l§ For a typical node i, if SFzshear force M(i)=M(i-1)+SF(i-1) L(i-1) ‘ (n.10) Because M C)is included in M(i), 50(1) in Fig. 3 would now include SC also. Note that for an unprestessed arch with fixed ends, at node 1 we will still have a rotation,GR, in the first half of element one which is lumped at node one. But this GR is related to the moment at the first node through the following formula: -53- E1 M(1)=(GR)-f7§ (H.11) The rest of the program is the same as for the pinned end arch except the error term in the right support corresponding to rotation is EM= M(n+l)- Mo—[qq(n)-.—0(n)][%] (”'12) where 9N(n)=0(n)+AO(n) (Fig. 3) An 8 element arch was studied. As shown in Figs. 37 and 38, fixing the supports will not help increasing the critical load compared . with pinned supports but will reduce the displacements significantly. . Prestressing the arch will not change the forces and displacements but it will change the moments at the nodes. With different modulus of elasticity of the arch ,the same behavior was observed. Next,a fixed length bar was bent into arches with different central angles (or) and span 3. It was then fixed at the two ends and loaded vertically at the crown with a load of 500. Table 7 shows the resulting maximum moment for 14, 6, and extrapolation between 18 and 6 elements. The length of the straight unloaded bar is 500017 : 15707.9633. The optimum case is when prestressing starts decreasing the maximum moment in absolute value sense. The optimum value for o: is shown in Table 8 which is obtained using Table 7. As can be seen at «4114.1, the max. prestressed moment is smaller in absolute value than the max.non-prestressed moment for the load = 500. umwmmhwmmnmtcoz new ummwmuvwowm x\u> c. ~.H o.H m. _ .wucm Umxwm can um::fim .wm>nzo uzmsmomamwwoacmog .un .mwk [I . 0. ~ — u c 1 u n n #0 m. o .o>Hfio sown nmxflm nonwonpmmnmuao: on» ow Hmoapcoufi wwoean wH noun mcco cmxfim umwmwnpmoun m2» op mcflvnonwouhoo w>H=o one -59.. noal< QOHIH cowl» ooonlm IUGOEUHN Q m 0 even voxau Q IQ even woes: IT IZ/zud nou< vommouumoum vuuuoauua .wn mum .Hmofipcoufi «woan_owm.oomwowpmonmuco: can commohvwoum.momd£m cownoa muse onfim nOHI< moHIH oonlm coonlx . /.J\ 11¢ 360 vuccu mm .wII‘ 11$ «use voxw 0. .10. even coca.“ 0T 10 menu vex.“ 003.33 D! in Hqcfimnuo -61— 6 Element 0 o o H b N O N w \0 \J‘ . N D—l \O L.) )—‘ L.) O \l O‘ ,_. H H b N ex 5 H m H H VI \1 w b N )_4 P=1000 22001. 19384 36 5 4461 61 833 E 062 11 45 m 855 1 29 E 66 885 8 2 1142 1 97 3 ll 5 468 10 21 m w 49 2710 o n o m 8 m M Q q v—« v-1 H H H II H II II II U) m ”1 m u; & varying chord lengths,s, Circular arcs with fixed length,150, 100,pinned & loaded prestressed into arches with span 104, I=1oo, A=10 Moment diagrams for Prestressed Arch,E Fig. 39. -62— r .4716 715 452 2216 018 .1773 1234 513 3851 263 1: D. j.8098_ 71423 904 ’45:} 101. u r y g 688 6373 0/77 7805 8565 g m 3 ES 668 2947 275 5663 2129 AL ‘0 _ N — 13329—— O O O N 8551 6'. -16 L :1 712 2328 160 4310 I 140 5661 0* 13141—— 854 357 /6958 O .1 514 541 8862 5%: C II' 33’ 85 79 4952 an .9: 3 75238 m. _ m . I o- O O N ll 9-. 1123113 s=100 s=105 s=110 s=115 s=120 Fig. 39. Continued nou< commouumoum odoufio wamx w you mw>uso swam .m> ucmsoz.xmz .oe .wnm .mpcfiom name on» mmpdowvcfi E d p > C d «F In .« on.< .oodnH. ooo.o~.m .omnm m oma owa oma oma oHH ooH cm _ . . n n w 0000 swam + ooo~nvwog ooo~uvmoq .fioooofi coca-smog coca-vac; nxooosfi an E X mu 0.. S -booomm. m. 9 A p. T. a: 1.000% : mu -6L1 fable 1- Mal. Noncnc (or Straight lar lent Into Ace» With £-100.l-l0a A-Igf, Load-500’ Max. Non~fresnresstng Nonenl Max. Prestressing Mouth! 1- cl. u 6 £1. at 4.0 4 c1. 84 6 cl. at (x node 3 nod! 4’ nodes 2.‘ nodes I.‘ ". deg. 180 -)6l)67 -1§1801 ‘ -198148.2 4.1568286 4.2267166 4.1227156 120 -169780 -185876 -198752.8 2.944886 1.86022E6 2.792556z6 90 -182609 -191219 -401707 2.2641156‘ 1.!!85E6 2.l27996£6 30 = -581679 -492641 -419810.6 ' 1.00817E6 901174 815777.: 25 - -721215 -540195 -194119 952191 ' ll6562 7o7898.6 21 -857666 - - 955124 - - - 22.1 -972455 - - 972404 - - 22.19 -976455 - - 911111 - 20 - -651811 - - 169156 - 17.71 - -817271 - - 817869 - ’17.7_ - -817271 - - 818620 - 17.5 - -857512 - - 817945 1 - Table 8 - Optimum x {or the Prestressed Arch of Table 7. No. of Optimum flax. Non- flax. Pre. Elements in Degrees Prestressed Prestressed Moment Hement 9 22.17 -972935 972W! 6 17.7 -818899 818620 41,6 111.7211 -69607o.2 695592.8 Finally.circular arcs of fixed length of 150 and varying chord lengths. s.was prestressed into arches with a span of 100.Then the ends were pinned and the arch was loaded at the crown with a vertical load. The program used is the same as the one used for problem I, except in the data Processing the coordinates of the nodes of the arches with different sPans were inputted and the span was set equal to 100 for all arches in the calculations. F18- 39 shows the max moments for 6 and 8 element arches. Fig. 90 -55- is obtained using Fig. 39 to plot the absolute values of the max. moments vs. the original. span. From this, Table 9 was obtained. This shows that the values of s for which the maximum moment is the lowest,are 111.1“ and 122.79 for the loadtof 1000 and 2000,respectively. The corresponding values of moment are 5975 and -13854,respectively, Table 9 - Optimum Span and Hax.Moment for the Prestressed Arch of Fig. N0 ~ Optimum Span Optimum Max. Homent No. of Elements Load:1000 Load=2000 Load:1000 Load=2000 6 110 120.5 7077 -17100 I , 8 110.5 121.5 6595 -15680 6,8 111.11 122.79 5975 -13854 _. _ ._ “.3 Prestressing by Axial Force In section 9.2 an arch prestressed by moments applied at the ends was studied. In this section, a study is made of a straight bar compressed axially to fern an archJThen the arch is fixed at the ends and loaded (Fig.#1).For the comparative unprestressed case, a parabolic arch is approximated by a set of elements whose ends are located on the parabola with the same span as the prestressed arch. Ratios of H/L =0.125, 0.375. and 0.5 were used. The locations of the nodes were found using a trial and error method to solve a set of equations resulting from setting the lengths of the elements equal. For the prestressed case,an initially straight bar was used with a length equal to the number of elements times the length of each element as found in the parabolic unprestressed case. Then the program was run for the -66- straight bar (0:90 for all elements) with a unit load at the crown. If this small load is not applied, the vertical reaction would be equal to zero and in the iteration procedure a division by zero would result. After convergence, the new shape is loaded and the program run to give the final prestressed loaded shape, axial forces, and moments. afll— 1‘ L ;l I‘ Fig. 111 - Prestrbssed Arch, By Axial Force The load vs. vertical crown displacement curves are shown in Figs. fl2-50 fcr 3 different loadings: vertical concentrated load at the crown, uniform load over the entire span, and unifOrm'load over half of the span. Three ratios of H/L were studied: 0.125, 0.375, and 0.5. Figs. 5" and 55 show the deflected shapes. These figures also show the curves and shapes corresponding to the parabolic arch with the same span, non-pretressed and fixed at the ends. As can be seen from these figures, prestressing the arch causes greater displacements and smaller buckling loads (Table 10). .Table 10 and Fig. 52 also show that the maximum axial forces corresponding to the loading conditions stated in Fig. 51 get larger with prestressing. This would be expected since the prestressing process produces an initial force. The moment diagrams for certain loads smaller than the buckling load are shown in Fig. 51. Again, prestressing increases the maximum moments. For these loads the moments for 8 and 10 elements are within less than 7‘ difference. These are shown in Table 11. It can be seen that the non-prestressed arch is more desirable. Thus, although there may be constructional advantages in fbrming -67- lightweight curved frames by using an initially straight bar, the moments and axial fcrces are greater than fOr an initially curved bar with the same loadings and the same span. Note that the loads in Table 10 for concentrated loads correspond to the peak values of the symmetrical load-displacement curve,since the bifurcation to an unsymmetrical mode could not be found.This agrees with Hasur's results (33) that the symmetric case always governs. Therefore, the peak values of the symmetrical part of the curves are entered in the table. However, as shown in Pig. #2, unsymmetric equilibrium modes were found. These do not correspond to bifurcation loads fcund in the pinned end arch and will not lead to finding the buckling loads. It should also be noted that the peak values and the bifUrcaticn loads were feund using the polynomial process described in section 3.1. Fig.5} compares the unloaded shapes of the non-prestressed arch and the prestressed parabolic one. Table 10- Buckling Load§ 37L 'Loading Condition Buckling Load"3 “‘ _L? Concent.&.3£_'Uniform I EI NOD’PTQSCe Prest.f h (a) g 51.57 37,5f 5 (b) 193.15 73,55 ° (c) 128.15 100 00 ,____,_ (Z) - 48.95 “' 38792 “ 375 (b) 222.10 93.50 ' (c) 132.05 107.92 --~d’ (a) F " 22.97 22.50 '”' 125 (by 64.50 60.55 ' (c) 75.30 70,50 (a Crown Concentrated (byfiUnifcrm Over Entire Span (c) Uniform Over Half of The Span -68- P, r000. .000. «0.. 00. 00000. 00000. 000.0 000.0. 000.0. 000.0. 00.~0 0000000 aw”. 0000., 0000., 00.0 0.0 000~0. 0000. 000~0. 0.000. ._000. ._000. 00.0. 0000000 gum a. 0000.0 000., 00., 00.. .0000. 0~00_. .00... 00000. 00000. 00000. 00.0 00000060090 . came e00: «000. 0,00. 00.. .0.~ 0~00c. 0.00,. 0000.. 00000. 0~000. 0.000. 00.00 0000000 mpn. 00.0. .00. 00.0 0.0 00000. 0.000. 000~0. ,0000. .0000. ,00001.:00.00 0000000 .1uaaquqnqq1 000... _000. 90.0 .00., 0~000. .000.. 0~00,. 00000. 00000. 00000. 00.0. 000.000.0000 . 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Sym. .nmoom 000mm enmwm Sym. o.wm~I o.nuHI o.mm o.mma mix? 0.00H .prest. mmNHN Vnwomul mmHHN macaw nwoon Hanan mmqwm ooawa mmum est. «(3:04003f rest . V oomomn NQHH moona quNH QMHHI onwwHI moawal nwm owuom JFig. 51. Continued -30- . IJ . . . I 00000 0 nooo0 m m, m, m. 00000 00000 0 00000 00000 00000 00000 5. 00000 00000 00000 00000 00000 0000 m 00000 00000 00000 00000 00000 00000 m . 00000 00000 000000 00000 000003 00000 I: . an n 000 00000 0000.0 000000 00000 LI. 00000 nu. fiI _ nu *lr , WI mw .L 0 O . 0 0 O 0 2 2 o. __ . o. .0 .0 IQ 1 J 14m. . . . madam 4 unnua s m. m, m, 00000 00000 ommum umoma . 00000 00000 5 00000 00000 00000 00000 00000 00000 mu. 00000 00000 00000 mmwwm 00000 000.00 m «Dunn mmoom «enoumwn .. concwu awnmm 0 o n 000 0 0000 0 00000 00000 0000 m1 L 0 _ mi , 00000 0.1. c nu .u t m n. n m mw a 0 3 .0. a a .0. 0 a ,D 2 a 0.. L “M” 0.. L. 0... 0 00 .0. .r , I-O.10, A-lOO, 8 Elements, Parabolic Arch(ETm'Non-prest. case) 52. Axial Force Diagrams, Prestressing by Axial Force -31- 5 .0 2 l o. = L / "n 00000 00000 00000 UH 00000 w++ 00 nu 2 0 k S e r P Sym. 00000 Honmm mmowm wnawm L, l—Prest. Gamma mnoma ooooqa momqwa Prest. . t S e r 9.. Prest. -Prest. Hunwm woman memoc mwnmm «comm ounce mnmmc omonq 52.Continued -32- canon Hmfix< 0m wcawmmuummum .mmamsm nou< .00 .me 00 .000 so 0 vcw .N .0H.mum ou wc0vconmmuuoo Hm 00 .70 .vwmmouumouaaa .mmawxm vmuomamoo .m.ona\m 0‘. 0.00000000 000.0.000 0.00000000 000.0.000 A.ummuac=v m.ouq\x ooau< .oa.oIH .moalm A . ammunmncozv oaflopmumm . _ omlg _ 0H< D c 0 vmmmmuumoum.d -33- v.00 van-uncoucoo cacao . ol 0 IL”. 000 00: 00 000 000 scam ouuucu as» uo>o v.00 shaman: . o. g . n 000 000 00 mw0_0ov 0:0 .000 000 000. 0000.000.0.0 000 0...N\p0 m .900..— 0302 .3 05:93-90.— .3020 v3.3.3.9 no.2 .3. .3..— an m can of no :8. .026 .013 50:05.95} . 0.0 I dd «3 0 comm ouwucu 0:» uc>o v.00 shaman: H” .n.l.~\=.o.~ I y: A03 0 v.3 0039023208 .596 an o“05\:o N." I III!- A.V m. 0. b a 00.00.353.000 4 OIIIIIllOv-voog. 39095 [331.5 . a .98 1.3.05 . “-9095 08. 0.00. 0 OOO. ONO.I n 000. omo. w 000. 000. .00 000. 000. 0W .50. 000. .0 000. 0N0. 0 002.0050 .00 .000 000. N00. 9 . ONO. mmo. .m 33 50:05 can :1: . .0qu 53.0.5 :30 0305 . manna: .pmmnm . . u 00 . . a? 00.00 000.0 .. w. .00 .. M0000 A3 000 .0 .000 00 .0400 .Anv — . 4 8 . . v.3 00.03005 :30 man: 33 van-3:00:00 050.00 .P a... . _. .. .00. E I vow-00.5 .uuoum OIIIIO v0.0.3.5 .unoumccoz 1 1 vow-00c: . 0.0.0.. 0'10 0.3.00.5 . 03.0.7.5: -85- CHAPTER V CONCLUSIONS AND FUTURE STUDY 5.1 The Problem Summary and Conclusions The elastic non-linear behavior of arches has been investigated using a rigid bar-spring node physical model. The influence of bending and axial deformation has been included. The load displacement curves and the deflected shapes for a variety of loads and fOr fixed and pinned supports have been obtained.’ The effect of cables at different positions has also been studied. Prestressing a straight bar into' circular arches with different spans, and also circular arches with different spans into a half circle arch has been investigated. The results obtained are in good agreement with the known solutions, when such solutions are available. For a pinned end arch with a concentrated load at the crown and a pinned support beam with a distributed load over half of the span, the results are in errors by less than 51 and in most cases about 11 compared with the ones obtained by others. The error for a quarter circle arch loaded radially or vetically is about 1%. The results for a non-prismatic parabolic arch with uniform load over the entire span are with less than ”.51 error with the previously obtained solutions. Prestressing a straight bar by axial force to form an arch fixed at the ends results, in general, in lower buckling loads and greater forces in the arch and therefore does not achieve a better design. -35- This was studied for a concentrated load at the crown and a uniform load over the entire or half of the span. Attaching the cables to the arch loaded vertically does not help the arch at loads near the critical load. However, cables are effective and useful for the horizontal loads.As summarized in Tables 7-9 (Section l+.2),a study was made of circular arcs of afixed length but initially with different radii .and chord lengths and then“ prestressed into arches of the 'same span.It was found that this technique can.reduce the taxinmm moment in the £ina-lzlocated arch. The optimum initial length was found to vary depending on the load,being greater for larger loads. Based upon these results it may be concluded that the model and procedure used are accurate and that it is not necessary to take more than 10 elements, eleven nodes, when two or three point extrapolation is employed to produce good results. Sometimes this extrapolation gives more accurate results than when 100 elements are used. The method is also economical. All the work to solve the two dimensional problems was done using a TBS-80 micro-computer. It should also be noted that using this model, variable geometry and elastic properties can be accounted for. A case in which the moment of inertia was not constant was studied in this thesis. 5.2 Future Studies Several additional problems are suggested by the work done here. These include the determination of shear effects. If by) is the shear deformation in element 1 perpendiCular to the element axis ,lumped. at node 1.then '06é(i-1)-% where A68(i-1)=the~rotation of node i-l due to the sheaf deformation and II( i)=the length of element‘ ‘i‘.Then 668(1-1) would be added to the bending rotation” at node i-i to obtain the total -87- rotation at node i-1. The effect of variation in geometry and.elastic properties;can also be studied. A slight change in the program will make it possible to be used to investigate the-behavior.of arches with a combination of different support and loading conditions. The space arch under in-plane and out-of-plane loading may be studied using the same model extended to 3 dimensions. Torsion could be taken by torsional springs at the nodes. Several approaches have been tried as explained in Appendix E, but none of them has led to satisfactory results. The non-elastic behavior or arches can also be studied by modifying the program used fer this thesis. -88- APPENDICES A. Newton's Algorithm The Newton's algorithm can be used to improve the initial values in an iterative manner. For three simultaneous non-linear equations f(a,b,c)=0 (a) g(a,b,c)=0 . . (b) (A.l) h(a,b,c)=0 ' (C) if a0,bo, and 00 are approximate solutions, then better so1utions may be a , b , and c , where l l l f \ -O-l fig 'bfo ‘ofo a1 Lo 08 ‘obo Eco f0 g‘ as 2. , (blrsdbo -%58— .838— Egg-8— 2’0 (A.2) c c 2mg 3h bhg . lj L 0 330 ‘053‘ ECO; ho where fo=f(ao.bo,c0) h0=h(a0,b0,c0) The equivalent terms may be evaluated numerically by three successive repetitions of the chain analysis. That is, the functions f0, 3 , and h corresponding to the three initial values a,b, and c 0 0 can be calculated,f,g, and h represent the three error terms corresponding to a,b, andc, respectively. It should be noted that f,g, and h would include other parameters such as the ones corresponding to -39- som'etry and elastic properties. In the next chain of analysis, a small increment Aa is given to he value of a,and f1,g1.and h1 , the corresponding values of f, ;, and h are found. Similarly, (r2. 5, tn?) and (g, g. kg) corresponding to the increments Ab and Ac can be computed. Then using the approximation of formula (2) in section (2.2) of the thesis regarding the derivatives of the functions, we find the improved values of a,b, and c, 1., a1, b1, and cl, respectively. _ _ :1 a -. 7f1‘f0 £2.32. _f_3_f_9_ f0 fl 0 A a A b A c A a db A c ' c h —h h —h h -h h 61‘ 0 1 o 2 o 3 o o [_Aa Ab AC 0 The procedure may be used to improve a set of more than three initial value s using a larger matrix. -90- Egmputer Program Listing to Find the Peak Value of Load- [isplacement Curve The fellowing symbols have been used: :Displacement of the a points determined from the load- displacement curve (data), 1:1, 2, 3, fl. .splacement value of the peak point with positive determinant in the solution of the quadratic equation resulting from setting dy/ dx=0 .splacement value of the peak point with negative determinant in the solution of the above equation; :terminant of the equation dy/dxzo; :ak load value corresponding to DH; :ak load value corresponding to DN; Coefficients of the polynomial, 1:1, 2, 3, H. .-91- l mmumenmnm TO FIND M. [MD 0F [MD-DISH. CURVE USING 3RD DEGREE MWIAL' 'NIH!!!"IIIIIIIIIH!"lIll"lulu"!!!fllllllfllllllllllfllfllllllllflllllfll' 3 1mm man's- 1211 11:4 ' 1215 D111 «11.111.313.111,o1111,r<111,s<111 1211 PRINT'INPUI 1111); 1:1 10 4- 1221 1119111 o¢11.o<21.o<31,o<41 1231 PRINI'JN’UT LMDS mum T0 4' 1211 11111111 P(1),P(2),P(3),P(4) 1251 FOR 1:1 '10 11 1211 1111.1):1 1271 1111.2):11111 1281 1111.3):011112 1291 A(I,4)=D(l)[3 1295 Lm1m-A(-;1;-,.11=-;A<1,1);A(1,2>;A11,31;A11,4> 131111911 1 . 1311 LPRINT'D(I),(I=I to 11 =-;o111;11121;013);om 1321 LPRINI'P(I),(I=1 TO 41=-;m1;r(2>;1><31;r<41 1331 FOR 121 to 11 . 1341 a11,11=1 1351 11m 1 1531 r1111 J=1 To 11 1532 FOR 1:1 10 a 534 1r «1.11111111911512 5311 11m 1 , 538 LPRINT'SJNSUIAR mmx' 541 61110 1831 542 1011 11:1 111 11 511 T=A(J,K) 541 A(J,K)=A(I,K) 543 1111.1):1 551 14111.10 :52 9(J,x)=a(1,x1 554 au,10=1 551 11911 11 553 124/1111,» 561 m 11:1 10 11 S62 “Jungian.” m s11.x>=£sau,x1 £66 11911 11 Lee r011 1:1 to 11 71 1F 1:1 11191 1512 72 52-1111,.» 74 £011 11:1 10 11 u A(L,X)=A(L,K)1EIA(J,K) re B(L,K)=B(L,K)1EIB(J,K) 11 “11911 x :2 um I. :4 11m .1 a mam-1mm 1111:1159 a 1011 1:1 111 a 1 Lraxm'1=';1;-:-;a(1.11;au,21;a¢1,31;a<1,41 -92- 1512 11911 1 15111 1111 1:1 111 1 1595 mm 15911111 11:1 10 1 15111 5111:51111111.11-11111 1111 11911 11 1112 11911 1 . 1111111111111-11111. 11111111' 1111 1111111 11111 1111 1:1 111 1 1111 u111111~1=°;1;- , ans-35111 1111 11m 1 1151 111111-11 111311. 01 1111111111 LMD, 11111111 ram 01. 1111 11 01111 11191 1711 11171 11111111111111 1115111.?- 1111 11111111 or 1191 19:5111151211111513111011211s1111101131 11115 11111111 1711 11111111°o11=-;o11;-111-° 1111 1:11 1111111'11 111111111511 11511., 1111111 2'111111111 m 1721 11111=2 11191 1171 1731 11:512112-3-s111ns13) 1732 11 11111 11191 1755 1731 1#1-51211511110111/13131111 1731 w1s3121-s1111111111113-51111 1711 115111115131101151211101211311111111131 1712 M5111151311m15121111111211513111111131 1751 11111111117 1113111112 311111011: 1111=';111;° 111=-;m 1752 011111-117 1111111va 511111111): cream-11mm I754 6010 176. I755 [PRINT'DETEMIW TO CALCIIJWE M [S NEMTWE' I76. mnfl'munuunuuuuununnuuuuunuuunun' |830 END C. 493.. Listing of the Main Computer Program A simple modification will make it possible to use this program for a fixed end arch prestressed by axial force. The fo1lowing symbols have been used (other symbols have been defined in Section 1.3): BD CE CM EA EI EH EL EM EP EV GH GV GR HN IC ID IE LN NE NN R RI SC SM TC TH TN VD VN VV XL XM XN YN The Horizontal Distance Between the Cable Base and the Plane of the Arch; Cable Modulus of Elasticity; Cable Cross Sectional Area; Element Cross Sectional Area; Element Moment of Inertia; Error in Horizontal Displacement of the Right Support; Element Length; Error in Moment or Rotation of the Right Support; Error Limit Allowed; Error In Vertical Displacement of the Right Support; Guessed Value of the Horizontal Reaction of the Left Support; Guessed Value of the Vertical Reaction of the Right Support; Guessed Value of the Rotation of the Right Support; Final Horizontal Load at a Node Including the Effect of Cables; Allowed Number of Iterations; 1 If Cable Base to the Left of the Node, and Any Other Number Otherwise; 1 If No Cable For the Node, and Any Other Number Otherwise; Final Element Length After Axial Deformation; Number of Elements; Number of Nodes; Radius of the Half Circle Arch; Rise of the Right Support With Respect to the Left Support; Slope Change of Each Element Final Slope Change of Each Element The Change In Slope Between Elements (I—1) and I; Original Angle Between the Element and the Vertical; Final Angle Between the Element and the Vertical (After Rotation); Horizontal Distance Between the Cable Base and the Node Where the Cable is Attached to; Final Vertical Load at a Node Including the Effect of Cables; Horizontal Distance Between the Left Support and the Node Where the Cables are Attached to; Arch Span; Moment at each Node; Final X-Coordinate of the Node; Final Y-Coordinate of the Node; -94- 1 PRINT'PROG.(HTTHESIS);ARCH W/ CHOICE OF PINNED OR FIXED ENDS;CABLES; & PRESTRESSING' 5 PRINTfNON-LINEAR BEHAVIOR: ARCH-FRAME: RIGID BAR MODEL- VERSION’U’-SP' 6 PRINT'IF FIXED ENDS, TYPE I'IINPUT F 7 PRINT'IF PRESTRESSED, TYPE l‘IINPUT P 10 12 15 16 17 28 21 22 25 38 35 48 41 42 43 44 45 50 55 60 65 7a 75 77 78 79 80' 81 82 83 84 85 86 90 95 180 INPUT'NUMBER 0F ELEMENTS';NE NN=NE+1 DIM EL(NE),TH(NE),E(NE),EI(NE),A(NE>,UL(NN>,HL(NN), XN,AF(NE),SF(NE>,AO, XM,TC(NN),HE(NN+1),VE(NN+1),ME,vD(NN),CL(NN),CO(NN),CE(NN),CM(NN),CF(NN), IE(NN>,IO(NN),UN(NN),HN(NN) DIN EA(NE) INPUT'ERROR LIMIT-EPS';EP PRINT'IF SEMI-CIRCULAR ARCH,TYPE 1':INPUT AC IF AC=1 GOTO 41 PRINT'INPUT ELEMENT LENGTH, ANGLE(THETA) ,MODULUS,I,A' FOR I=I 10 NE PRINT 1: INPUT EL(I) ,TH(I), EU), EI, Emu NEXT I: 0010 45 - INPUT'RADIUS'; RIPRINT'FOR EACH ELEMENT, INPUT E. I a A” FOR I=1 T0 NE PRINT 1: INPUT E1I>,E1.EA:TI=3.I415927/NE: EL(I)=2*R*SIN(TI/2) TH(I)-TI*(I-.5):NEXT I PRINT'INPUT N00E LOADS- VERTICAL, HORIZONTAL' FOR I=I T0 NN PRINT 1:1NPUT UL.HL NEXT 1 FOR U=1 T0 3 PRINT-INPUT SPAN,RISE,N0.0F ITERATIONS,GUESS UERT.R, HOR.R,ROTATION INPUT XL,RI.1c,GU,0H,GR 113=011= U8=GVI RB=GR FOR I=1 T0 NN:VN(I)=VL(I):HN(I)=HL(I> NEXT I II=1:XNcI)=e:YN<1>=e:Tc<1>=a:N=e:Ix=e IF Pa: 0010 82 ELSE 86 IP P=1 0010 83 ELSE 84 HO=(E(1)*EI(1)*(TH(2)-TH(1)))/EL(1):GOTO 85 .MO=0 XM(1)=MO*GR*(E(1)*EI(I)*2/EL(1)):GOTO 98 XM(1)30 TN(1)=TH(1)+GR SN=SIN8 GOTO 155 PRINT'PRINT I IF NO CABLES FOR NODE I'IINPUT IE(1) IF IE(1)=I GOTO 162 PRINT'INPUT 80(1),VDB(I),CL(I),CA(I),CME(1)'IINPUT BD(I),VD(1),CL(I),CM(1),CE(I) IF IE(I)=1 GOTO I62 UU=VD(1)-XN(1):BK=ATN(ABS(YN(I))/VV):CC=SOR((UV)* (VV)+(YN(I))*(YN(I))):CN=SOR((CC)*(CC)+(BD(I))* (BD(I))) CD(I)=CL(I)-CN:CF(1)=((CD(1))*(CE(1))*(CM(1)))/(CN): BC=(BD(1))/(CN):CP=ATN(BC/SOR(-BC*BC+1)):CB=2*(CF(I)) *(COS(CP)):IF CD(1)>8 GOTO I62 CH=-(CB)*(COS(BK)):CV=-(CB)*(SIN(BK)):VN(I)=VL(1)+CU: HN(I)=HL(I)*CH AF(I)=((GV-UN(I))*CS)+((GH+HN(I))*SN):SF(I)=((VN(1)- GV)*SN)+((GH+HN(I))*CS) AD(I)= ((AF(I)*EL(I)))/(E(l)*EA(I)): LN(1)= EL(I)-AD(I) XN(2)=LN(I)*SN: YN(2)= LN(1)*CS FOR I= 2 TO NE: J=I- I: K?I+1 PRINT'I=';I;'SF(J)=';SF(J);'LN(J)=';LN(J);'XM(J)='; XH(J) TC(I)=TH(I)-TH(J) SC=((SF(J)*LN(J)/2.8 +XM(J))*LN(J))/(E(J)*EI(J)) IF F=1 GOTO I72 ELSE 173 IF P=I GOTO I75 TN(I)=TN(J)+TC(I)+SC:GOTO 186 TN(I)=TN(J)+SC CA=COS(TN(I)):SA=SIN(TN(I)) IF F=1 GOTO 182 ELSE 185 IF P<>I GOTO 185 CS=COS+SM CA=COS(TN(I)):SA=SIN(TN(I)) IF F=1 GOTO 232 ELSE 235 IF P<>I GOTO 235 CS=COS(SM):SN=SIN(SM):GOTO 248 CS=COS(SM+TC(I)):SN=SIN(SM+TC(I)) AF(I)=AF(J)*CS+SF(J)*SN+HN(I)*SA-UN(I)*CA -96- 245 SF(1)=-AF(J)*SN+SF(J>*CS+HN(I)*CA+VN(I)*SA 258 XM(I)=XM(J)+SF(J) w 255 AD<1)=(AF(I)*EL(I))/(E(I)*EA(1)) 268 LN(1)=EL(I)-AD(I) 261 PR1NT'LN(';I;')=';LN(I) 262 XN=XN<1>+LN(1)*SA 263 YN(K)=YN(1)+LN(1)*CA 264 IF 1x<>8 GOTO 278 265 PRINT'PRINT 1 IF No CABLES FOR MODE -;1 266 INPUT 15(1) 267 IF IE<1>=1 GOTO 288 268 PRINT'PRINT BD(I),VDB(I),CL(I),CA(I),CME(I)':INPUT 80(1),vo<1>,CL,CM,CE(1) 269 PRINT'PRINT 1 IF CABLE BASE TO THE LEFT OF THIS NODE': INPUT 10(1) 278 IF IE=1 GOTO 288 271 IF ID-vo(1> 274 BK=ATN*(CC)+)*(BD>8 GOTO 288 276 IF ID(I)=1 GOTO 278 277 CH=-(CB)*(COS(BK)):GOTO 279 278 CH=)/(E(J)*EI(J)+E(l)*EI(1)): TN(I)=TN> 284 AF(1)=AF(J)*CS+SF(J)*SN+HN(1>*SA-VN(I)*CA:SF(I)=— AF§SN+SF(J)*CS+HN(I)*CA+VN(I)*SA 285 XM(I)=XM(J)+SF(J)*LN(J) 286 AD(I)=(AF(I)*EL(I))/(E(I>*EA(1)): LN(I)=EL(1)-AD(I) 287 PRINT' LN(';I;')=';LN(I): XN(K)=XN(I)+LN(1)*SA. YN(K)= YN(I)+LN(I)*CA 288 NEXT I 289 Nw=NN-1:XM(NN>=XM(Nw)+SF1 GOTO 293 292 EM=XM(NN)-(H0-(TN(NE)-TH(NE))*(E)) 293 Ix=1 295 1F8 GOTO 318 ELSE 388 388 IF(ABS(EV)-EP))8 GOTO 318 ELSE 385 385 318 315 316 317 318 319 328 322 325 338 335 348 345 358 355 368 365 378 375 388 385 398 395 488 485 418 415 428 425 438 435 448 445 458 455 468 462 465 478 475 488 485 498 495 498 588 -97.. IF(ABS(EM)-EP))8 GOTO 318 ELSE 498 IF (N)=8 GOTO 315 ELSE 348 IF(II-IC)(8 SOTO 322 ELSE 316 PRINT-CYCLE NO.-;II;~ERROR H=-;EH;-ERROR v=-;Ev; 'ERROR H=';EM PR1NT'GH=';GH;' GU=';GU;' GR=';GR PRINT'DOES NOT CONVERGE AT CYCLE';11 PRINT'* ’5‘) 1* 1* «I Q 1!" SOTO 498 PRINT'CYCLE N0.';11;' ERRORH=';EH;' ERRORv=-;Ev; ' ERRORM=';EM PRINT'GH=';GH;' GU=';GU;' SR=-;SR PRINT'DOES NOT CONVERSE AT CYCLE';II PRINT"? 4» + + + + + + +" N=N+1 HE(N)=EH:UE(N)=EV:ME(N)=EH IF8 SOTO 365 ELSE 355 GH=1.81*GH SOTO 81 . IF>8 SOTO 385 ELSE 378 GU=1.01*GV SH=SH/1.81 SOTO 81 IF>8 SOTO 485 ELSE 398 GR=1.01*GR 6v=Sv/1.81 SOTO 81 SR=SR/1.81 REM-JACOSIAN CORRECTION Al=(HE(2)-HE(1))/(.81*GH): A2=(UE(2)-VE(1))/(.81*6H) A3=(ME(2)-HE(1))/(.81*GH): 81=(HE(3)-HE(1))/(.81*GV) 82=(VE(3)-UE(1))/(.81*GU): 83=XM3 TAB(48)AD(I);TAB(52)TN(I) NEXT I FOR I=1 TO NE IF IE(I)=1 GOTO 525 PRINT'I=';I;TAB(18)'CA8LE AX. DEFORM.=';CD(I);TAB(18) 'CABLE AX.FORCE=';CF(I)IGOTO 526 PRINT'I=';I;TAB(15)'NO CABLES FOR THIS NODE' NEXT I PRINT'NODE NO.‘;TAB(12)'X-COORD';TA8(24)'Y-COORD'; TAB<36)'END M' FOR I=1 TO NN PRINT I;TA8(12)XN(I);TAB(24)YN(I);TA8(36)XM(I) NEXT I PRINT'TO OUTPUT RESULTS TO PRINTER, TYPE 1'| INPUT 2 _ IF Z=1 GOTO 558 ELSE 545 PRINT'IF DIFFERENT LOAD OR ERROR LIMIT,TYPE 2': INPUT DL IF DL<>2 GOTO 548 V INPUT'NEw ERROR LIMIT=';EP: GOTO 45 NEXT U END LPRINT'ELEHENT LENGTH THETA E I A“ FOR I=1 TO NE LPRINT I;TAB(10)EL(I)gTAB<28)TH(1);TAB(39)E(1); TAB<40>EI(I):TAB(50)EA(I) NEXT 1 LPRINT'INITIAL GUESSES- GH=';HB;'GU=';UB;'GR=';RB LPRINT' fl * *- LPRINT 'EL.N0.';TAB(10)'AXIAL F';TAB(28)'SHEAR F'; TAB(38)'END M';TAB(40)'AXIAL DEF.';TAB(52)'THETA FINAL' FOR I=1 T0 NE LPRINT I;TAB(18> AF(I);TAB(28)SF(I);TAB(38)XM(1); TAB<4O>AD3TAB<52>TN(I) NEXT 1 FOR I=1 T0 NE 1F IE(I)=1 GOTO 619 LPRIN'I=';I;TAB(10)'CABLE AX.DEFORM.=';CD(I);TAB(18) 'CABLE AX.FORCE=';CF(I):GOT0 628 LPRINT'I=';I;TAB(15)'NO CABLES FOR THIS NODE' NEXT 1 LPRINT'NODE NO.’;TAB(IO)'X-COORO';TAB(20)'Y-COORO'; TAB<30)'END M'3TAB<48)'UL';TAB(50)'HL' FOR 181 T0 NN LPRINT 1;TAB(18)XN(I)3TAB<28>YN:TAB(38)XM(1); TAB<48)UL(1)3TAB<56>HL -99.. 635 NEXT I . 637 LPRINT'EPSB'3EP3'GHF3'yGflg'GUFB'IGUI'GRFB'yGR;'CYCLES=' :11 . ~ 638 LPRINT'SINGLE PRECISICN-VERSIW’U’HIF F<>1 SOTO 641 639 IF P<>1 SOTO 641 648 LPRINT'PRE-STRESSED ARCH-MOMENTB'3HO 641 GOTO 545 ~100- D. Alternative Method For 2-D Problems Ann alternative method for a 2 dimensional arch and rigid bar-spring node model was also developed (31). In this method the displacements u,v in x and y directions , respectively;are originally assumed. Then if(1)] — [u(1)~U(1-1)] s1n[4>(1)] rv(i)=Arctan 1[L(i)+ u(i)-u(i—l)] Cos[¢(i)] +[v(i)—v(i—l)] Sin[dp(i) Assuming u and v and having the above formulas, the fggkég, moments, rotations, and axial deformations can be found. Then we see 'ifsz=oandfiyy=o (the equilibrium equations at each node) within an acceptable range of error :1 x=—N(1)cos(9(1)+4(1)]+N(1+1)c°sEp(1+1)+w(1+1)]- -s(1)s1nE4(1)+ D.5 4(1)]+s(1+1)51n[¢(1+1)+T(1+1)] +P(1)s1ng(1)J 2F =-—N(i)Sin[47(i)+1’(i)]+N(i+l)Sin[9(i+l)+‘Y(i+l) +S(i)Cos[qa(i)+ 11(1) —S(i+1)CosEP(i+l)+’Y(i+l):| —P(1)Cos[E(1)] (D.6) where P=load applied at the node and. é=the angle between P and the vertical axis. -101- (1) 3(1) >X Fig. 55. Typical Mass Point, Forces, and Moments If convergence is not satisfied,the initial guesses for u and v are improved using Newtod?algorithm,and the process is repeated until convergence is achieved; that is,the equilibrium equations are satisfied. This method has not led to a satisfactory result for the model used. E. Space Arch In space arches the same rigid bar-spring node model was used. A basic difficulty exists now because of the fact that the rotational displacements about the three dimensional axes are not commutative. The final configuration of a body depends on the sequence of rotations. That is, when rotated about say,x , then y and finally 2, the body will not be in the same place necessarily as when rotated about say, y, then x and finally z axes. The addition of twist makes space arches more complicated than the plane arches.However, if rotations are small, they are commulative and can be added as vectors. Therefore, the arch can be studied under a small load causing small displacements. The -102- coordinate system will be updated and the new configuration will be the reference one for the next load increment. The procedure continues until the last load increment is applied and the corresponding configuration is then the one corresponding to the final cumulative applied load which may not be small. This method of updated coordinatesa is used for 3-D arches. To study this problem, other than the assumptions made ha the 2-D case, it is also aSsumed that no shear deformations are allowed and that the first and last elements are in one plane and stay in that plane. The latter becomes more of a reality if we‘make the two elements small enough. . The iteration procedure is as follows (7 and 39); (3) Guess the initial values for the reactions in x, y’and 2 directions, Rx, R and R2, moment vectors in local 3 and t directions, 1% and MT, and finally DY, the change in on}, which is the angle between the first element and y axis. 8 is the longitudinal axis (Fig. 56) +1 X Fig. 56. Local and Global axes (Space Arch) (b)If T=direction cosines of the local axes r,s, ad t with respect to x,y, and z,and P and Qare applied forces in (x,y,z) and (r,s,t) coordinate systems, respectively and R is thereaction vector, then .103- {Q(1)}- [T]{P+R} (8.1) AD(1)=QS(1)L(1)/E(1)A(l) For a pinned ends arch M&(1)=0 so; “1 710 M¥}=[T] M (8.2) M2 MT (c) Having the displacements that we find from the forces and moments in step b, final coordinates of node 2 can be found. From then on with an iterative manner we find the rotations about the three local axes having the moments and forces in the provious element (or node). Then the new transformation matrix is found. To do so if we let E be the consines of the angles between element 1 before and after rotations about the three local axes, then ”3:533 [T1211 (1.3) We can also find F, the transformation matrix from the local axes of element 1-1 to the local axes of element 1 [T]=[T(i))[T-l(i-l)] (3.4) (d) Find the new moments and forces due to the applied loads and the forces and moments in element or node i-l. {Q}=[PJ{Q(1-1)}+[T(1)](P(1)} (2.5) Mr(i-l)-Qt(i-1)L(ivl) (M}=[8] Ms(i—l) (8.6) t(1—1)+Qr(1-1)L91-1) -1ou- (e) Find the axial deformation AD and also 8(1), 1(1) and 3(1), the twist and rotations about the two local axes t and r, respectively. p(1)=MS(1)L(1)/2Gj(1)+Ms(1—1)L(1-.-1)/2c;j(1—1) where 63 = torsional property of the element or(i)=Mt(i)[L(i)+L‘(i-l)] /[E(i-l)It(i-1)+E(i)I£i)J (8.7) 3(1)=11r(1)[L(1)+L(1—1)]/[8(12-1)Ir(1—1)+8(1)Ir(1)] (8.8) where I rand a: are the moments of inertia with respect to the local r and t axes, respectively. The new transformation matrices, moments, and forces can be found doing one more iteration. - (f) Having the final deformation, the final coordiantes of node i+lcan be calculated. (8) Proceed to the next element. At the end, the 6 errors at the right support are the three displacements in x, y, and 2 directions, change in the angle with x axis, K , moment about x (coincident with the local r axis) and P(n-l) - q; (n+1) L(n)/ZGj(n) where n=number of elements. fi(n+l) is the total twist. To find p(n+1) we find the components of P(i) on the 3 axis of the last node and add them. Here, due to the small size of the displacements, p was treated as a vector perpendicular to the plane of the node. (h) Having the errors and the guessed initial values,we improve the latter using Newton's algorithm with a 6x6 matrix. The iteration continues until convergence is satisfied. Then the displaced arch is set as the reference configuration and after updating the coordinate system,the procedure will be repeated for the next increment of load -105- until the last load increment has been taken care of. It should be noted that the small rotations have been taken by others (”3) as less than 15? The method was tried for 2 dimensional loads and arch but the values obtained were not satisfactory. Other methods have also been used to solve the 3-D problem including a method using the principle of minimum potential energy. The principle states (7) that among all displacement configurations that satisfy internal compatibility and kinematic boundary conditions, those that satisfy the equations of equilibrium make the potential energy a stationary value. If the stationary value is a minimum, the equilibrium is stable. To apply the principle to the arch problem, after having the coordinates of all nodes, we assume the displacements of each node. From the original and final coordinates of the nodes, we find the bending and torsional rotations,m(i),g(i),and §(i) and aISOIthe axial deformation of each element, AD(i). Then if Rx=Efix/l (a) K6=E18/L ‘ (b) K =Gj/L (c) (E°9) KA=EA/L (d) the total potential energy is equal to n _ 2 2 n U=1/2¥§:%M(1)u (i)+K (i)p(i)+xb(1)§ (i)]+zi§f(i)AD(i)} (E.10) and the element (1, J) of the stifness matrix would be K(i,j)=;°—2—1'- 'bU(i)bU(J) -106- where u(i) and u(J) are the displacements of the i th and J th degrees of freedom. To compute the derivatives of u we can use the method developed in case I of this chapter. From the above formulation,two procedure 8 can be developed. The steps for the first one are as follows: (8) Assume initial values for all displacements of all nodes (b) Find rotations and axial deformations (c) Find the total potential energy, 0 (d) Test for convergence. That is, see if U is a minimum. If PSI-i=0 within acceptable error range for all u(i) then convergence is satisfied. We have the final configuration. If not, go to step (e) (8) Having the initial values and errors and using the Newton's algorithm,we improve the initial values and go back to step (b). The second procedure includes the following steps: (8) Assume initial values for u(i), forming displacement vector D which includes three displacemetns and three rotations about the three global axes for each node. (b) Find the rotations and axial deformations. (c) Find the stiffness matrix,K, which is the second derivative of the total potential energy. (d) Find the force vector due to the displacements F = -Kp (E.ll) (e) Find the unbalanced force F = F+P (E.l3) 1.1 Where P is the vector of the applied forces. (f) Find the displacements due to the unbalanced force D -107- -1 Du = K Fu (E.14) (3) Test for convergence. If not satisfied go to step (h). (h) Find the new displacement vector D =D01d+DU (E.15) new and go back to step (b). To test convergence, one way is first to compute E = tu/uIn (E.16) Where tu is the value of the specified type of deformation of each node due to unbalanced forces and “m is the maximum initial value for that type of deformation. If tu is,say, the value of the displacement in x directiondue to the unbalanced forces,then um will be the maximum value of the displacement in x direction of all nodes in the original assumption (step a). We find the 6 maximum values for E corresponding to the 6 types of deformations. If these 6 values are within acceptable range of error than convergence is satisfied. The above two methods using potential energy have not led to a convergence with a set of satisfactory results. Projecting the arch onto the three perpendicular planes was also considered to deal with the problem. The space arch was projected onto xoy, xoz, and yoz planes. 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