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Computer analysis of planar and spatial grid frameworks. Kinra, Ravindar Kumar 1964

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COMPUTER ANALYSIS OF PLANAR AND SPATIAL GRID FRAMEWORKS by RAVINDAR KUMAR KINRA B. Tech. (Hons.), Indian Institute of Technology, Kharagpur, India. 1962 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF M. A .Sc . in the Department of CIVIL ENGINEERING We accept this thesis as conforming to. the required standard THE UNIVERSITY OF BRITISH COLUMBIA April, 1964 In p r e s e n t i n g t h i s t h e s i s i n p a r t i a l f u l f i l m e n t of the requirements f o r an advanced degree at the U n i v e r s i t y of B r i t i s h Columbia, I agree that the L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r reference and study.'. I" f u r t h e r agree that per-m i s s i o n f o r extensive copying of t h i s t h e s i s f o r s c h o l a r l y purposes may be granted by the Head of my Department or by, h i s r e p r e s e n t a t i v e s . I t i s understood t h a t - c o p y i n g or p u b l i -c a t i o n of t h i s t h e s i s f o r f i n a n c i a l gain s h a l l not be allowed without my w r i t t e n p e r mission. Department of C i v i l Engineering The U n i v e r s i t y of B r i t i s h Columbia, Vancouver 8, Canada i v ABSTRACT The application of the stiffness approach to the exact analysis of both planar and spatial grid frameworks of any complexity and high degree of sta-t i c a l indeterminacy i s presented. Ordinarily, even the simplest planar grid is such a highly redundant structure that i t cannot be analyzed rigorously by manual methods without recourse to some simplifying assumptions at the expense of accuracy. In general, most authors neglect the effect of member torsional r i g i d i t i e s i n order to reduce the size of the problem and make use of the plate theory for the purpose of evaluating deflections. Matrix methods of analysis, however, remove the necessity for resorting to any such approxi-mations and prove extremely convenient for computer application. The fundamentals of the stiffness approach are explained in complete detail and applied to the analysis of rectangular planar grids. For the pur-pose of comparison, example grids given by Ewell, Okubo e Abrams^. and 2 Woinowsky-Krieger have been analyzed by the stiffness method and the compara-tive results are tabulated. The principle of orthogonal transformation, which i s an essential part of the analysis of diagrids and spatial grids is f u l l y described and i t s application demonstrated by various numerical examples including a skew bridge, a cantilever diagrid and a hyperbolic paraboloid space grid. The application of stiffness analysis has been further extended to problems invol-ving temperature changes and support settlements and, also, the procedure to reduce the size of symmetrical structures is described. A special successive elimination and matrix partitioning technique has also been introduced i n order to enable the solution of extremely large numbers of simultaneous equa-tions within the limited core memory capacity of d i g i t a l computers, by taking V advantage of the band form of the stiffness matrices of structures. A com-plete Fortran II computer program for the IBM 1620 and a 1405 disk f i l e i s given, as well as, sample inputs and outputs of the IBM 1620 and 7090. After the f i r s t attempts by Engessers i n 1889 and Zschetzsche i n 1893, a great variety of hand calculation methods have been developed for the ana-• 3 lys i s of planar grid frameworks. Among these, Hendry e Jaeger's harmonic analysis and C. Massonet's^ anisotropic plate theory methods are the most convenient and easily applicable, i n the opinion of the author. The basic assumptions and underlying principles of both these methods are outlined and the procedure of analysis i s il l u s t r a t e d by means of a numerical example i n each case. . Furthermore, i n order to obtain an idea of their accuracy several planar grids with 2 to 6 longitudinals have been analyzed by the stiffness method, harmonic analysis and anisotropic plate theory. In every case, two solutions have been performed, assuming the constituent members of the grid to possess f i r s t zero and then maximum values of torsional r i g i -dity. The comparative values of the load distribution factors for the longi-tudinal and transversal bending moments have been tabulated. i i i ACKNOWLEDGEMENT The author i s deeply indebted to DR. SEMIH S. TEZCAN for his constant encouragement, inspiring guidance and supervision through-out the preparation of this thesis. Dedicated.to my father Karam Narain Kinra v i TABLE OF CONTENTS Page INTRODUCTION 1 CHAPTER 1. PLANAR RECTANGULAR GRIDS 1. Basic assumptions and definitions 5 2. Stiffness equation of a straight structural member 6 3. Stiffness matrix of a straight member with variable moment of inertia 11 4. Stiffness matrix of a straight member including ' torsional rotations 13 5. Stiffness matrix of a straight member considering length changes 15 6. Stiffness matrix of a member considering the effect of shear deflections 16 7. Stiffness equation of a,structural system 17 8. . Evaluation of the stiffness matrix of a system 18 9. Concept of joint loads 22 10. Procedure of analysis for planar rectangular grids 23 11. Numerical example 24 .12. Comparative results with other methods 28 CHAPTER 2. ORTHOGONAL TRANSFORMATION OF AXES 1. Axis transformation of forces and deformations at a point 32 2. Transformation of member stiffness matrices 33 CHAPTER 3. SPATIAL GRIDS 1. Stiffness matrix of a spatial grid member 35 2. Axis transformation of a spatial member 36 3. Direction cosines of the member axes with respect to common axes 38 4. Procedure of analysis for spatial grids 41 5. Numerical example (Hyperbolic paraboloid space grid) .43 CHAPTER 4. PLANAR DIAGRIDS 1. Stiffness matrix of a diagrid member 50 2. Axis transformation of a diagrid member "'.51 3. Direction cosines of a diagrid member, 52 4. Transformed stiffness matrix of a diagrid member ' 53 5. Procedure of analysis for diagrids 54 6. Numerical examples 54 • i i Page CHAPTER 5. SPECIAL TOPICS 1. Reduction due to symmetry 68 2. Thermal e f f e c t s 70 3. Support settlements 72 h. Solution of extremely large structures by band matrices 73 CHAPTER 6. HARMONIC ANALYSIS 1. B r i e f outline 77 2. Numerical example- Two girder bridge 80 . 3* Comparison of Harmonic and S t i f f n e s s analysis d i s t r i b u t i o n factors 86 CHAPTER ?. ANISOTROPIC PLATE THEORY 1. B r i e f ' o u t l i n e 10k 2. Five girder bridge example 106 CONCLUSIONS - 115 NOTATIONS 117 BIBLIOGRAPHY 119 APPENDIX APPENDIX A Computer programme f o r s t i f f n e s s analysis of grids APPENDIX B Torsional constants 120 129 - 1 -INTRODUCTION A grid framework, which is basically defined as a system of beams inter-connected at the nodal points either i n a plane or i n space is an extremely . redundant structure. The degree of redundancy may be expressed as N - 3 J - R for planar grids, and, as N = 6 J - R for spatial grids, where J represents the number of joints and R the number of restraints at the supports. This follows from the fact that, in general, each joint of a planar grid undergoes two rotations and a transla-tion, while that of a spatial grid i s subject to three rotations and three translations, upon the application of external loads. Prior to the advent of di g i t a l computers, an exact analysis of even a simple grid was, understandably, considered.impractical. Several approximate methods, however, based on one or more simplifying assumptions, have been developed over the past 60 years. A brief survey of these methods is des-cribed below. M. Hetenyi^ outlined a method for determining the deflections i n a grid-work, assuming that the individual beams deflect without rotation at their intersections. The simultaneous differential equations written for the de-flections, on this assumption, are solved to yield the joint displacements. Neglecting the torsional rotations of the members, Melan e. Schindler^ used a procedure of equating the beam deflections at the joints to arrive at a system of homogeneous linear equations which involve complicated eigen value problems. Based on an analogous assumption that no bending or twisting 7 moment i s transmitted at beam intersections, S. Timoshenko employed trigo-nometric series to define the elastic curves developed by the individual grid beams. The plate theory has been widely used i n connection with grid framework analysis. Neglecting torsional stiffnesses, the theory was successfully 8 9 L. applied f i r s t by Guyon and then by Greenberg . Subsequently, C. Massonet and Taraporewalla"''^ used anisotropic plate theory to include the effects of torsional r i g i d i t i e s . Massonet has compiled a f a i r l y detailed set of co-efficients for the evaluation of longitudinal and transversal bending moments for torsional and non torsional cases. Finite difference equations have been applied to rectangular and skew slabs divided into a network of points by Henri Marcus"^ and Vernon P. 12 Jensen . By dividing the platejtheory fourth-order differential equation into two parts, two sets of linear equations, containing moments and deflec-tions as unknowns respectively, are obtained and solved simultaneously to yiel d the required moments and deflections. If the f i n i t e difference equi-valents are determined directly from the fourth-order equation, a solution i s required for only one set of linear equations. The la t t e r procedure, however, i s not recommended by Marcus. 13 Semih S. Tezcan approached the problem in a different manner. F i r s t , neglecting torsional stiffnesses, the reactions at each joint are determined by equating the deflections at beam intersections. The resulting s t a t i c a l l y determinate system i s solved to obtain the bending moments and shears. Next, anisotropic plate theory i s applied to determine the effect of torsional r i g i d i t i e s alone and these values, added to the results of the f i r s t stage to yi e l d the required exact stress resultants. 2 Ew'ell, Okubo and Abrams employed an auxiliary force system for con-t r o l l i n g the vertical displacements of the joints and a moment and torque distribution process for transmission of the displacement effects. After the bending and torsional moments have been distributed over the grid, only one series of linear equations need be solved to define the deflection pattern as the equations are written in terms of unknown deflections produced by auxiliary forces at the grid joints. Once the deflected surface i s found, i t i s a simple matter to evaluate the bending and torsional moments. The above moment distribution or relaxation method i s applicable to any kind of framework, but, i t ' s general usage is limited as very lengthy computational work i s entailed. The unique method of harmonic analysis developed by Hendry e Jaeger^ has proved to be very efficient for practical design purposes. In their approach, Hendry e Jaeger replace the transverse grid members by a uniform spread medium, which may or may not cover the f u l l span. Differential equations are written for the loading on each longitudinal, including terms due to rota-tion and twist, and solved by harmonic analysis to obtain the amplitude of the harmonics of the deflection or bending moment curve for each girder. Plastic theory has been recently applied to planar grid analysis for the f i r s t time by Shaw^^. It is independent of the support conditions and configuration of the structure, but, requires that the torsional stiffnesses of i t ' s constituent members be negligible. Without exception, a l l the hand calculation methods of gridwork analysis outlined above, involve a great deal of tedious arithmetical calculations. Moreover, i n view of the simplifying assumptions indicated, the results obtained can at best be approximate, except when using the relaxation proce-dure. The common assumption made by most authors i s to neglect the torsional stiffness of members. For steel grids this i s not a serious approximation; however, when working with reinforced concrete, considerable errors ranging from 10 to 40$ may occur. Further, most of the methods are directly applicable - 4 -only to grids with a regular beam layout and particular support conditions. When matrix analysis i s used, however, a complete and exact solution i s possible for grids of any complexity and support conditions. The effect of torsional r i g i d i t i e s , length changes, shear deformations, temperature changes and support settlements may be easily taken into account. Moreover, while the analysis of spatial grids i s impossible by any one of the hand c a l -culation methods mentioned above, the mathematical formulation of the s t i f f -ness analysis of spatial grids remains the same as that of planar grids except that an orthogonal transformation i s required and the size of the individual member matrices i s increased from 6 by 6 to 12 by 12. Above a l l , the stiffness approach allows complete automation, so that the analysis of planar or spatial grids of any degree of redundancy is reduced merely to the cl e r i c a l job of preparing the input data for the computer and absolutely no calculations are required from the engineer. CHAPTER 1 . PLANAR RECTANGULAR GRIDS 1 . Basic Assumptions and Definitions The following assumptions are made i n the stiffness matrix analysis method presented: (a) the structure i s stable, (b) the structural material has a linear stress strain diagram, (c) the displacements of the structure vary linearly with the applied forces, ' (d) the effect of the displacements on the forces and moments i n the struc-ture are negligible, and (e) static forces are applied. A structural member undergoes an elastic deformation 6, when subjected to an axial force p. The assumed linear.force deformation relation can be expressed as p = k 6 where k i s the axial force required to produce a unit axial deformation. By the same principle, i f a member has .end deformations i n more than one direction, the total force p^ along direction i due to deformations 6j i n each direction i s given by i n which k. . i s defined as the force required along direction i to main-tain a unit deformation along direction j only, while deformations along a l l the other specified directions are prevented. The factor k ^ i s termed a "stiffness influence, coefficient" and Eq. • 1 is the basic "stiffness equa-tion" i n the stiffness method of linear structural analysis. - 6 -Sign Convention Clockwise rotations and downward translations acting on a member are assumed to be positive. Similarly, clockwise moments and downward shears are positive. 2. Stiffness Equation of a Straight Structural Member The possible end rotations and displacements of a straight member i j are indicated by arrows numbered 1, 2, 3 and k along the assumed positive directions i n Fig. 1. The stiffness influence coefficients of a member are evaluated by con-sidering a unit deformation along each one of the specified directions and calculating the. forces required along a l l the specified directions. From the stiffness coefficients, the stiffness equation of the individual member may be derived i n 5 stages as f u l l y i llustrated i n Fig. 2. Stage 1 The member i s considered to be clamped at both ends and the forces along the assumed directions i.e. the fixed end moments and reactions are computed. s, Fig. 1 End deformations of'a straight member - 7 -Fig. 2 Stiffness coefficients of a straight member - 8 -Stage 2 The forces required along the specified directions, to maintain a unit deformation along direction 1 only, are Ic^, kg^, k^^ and as shown i n Fig. 3. 2 L / 3 Fig. 3 Unit deformation along direction 1 By the First Moment Area Theorem, the change i n slope between the tan-gents at the ends of the elastic curve i s equal to times the area of the bending moment diagram. Therefore, for • 1 <21 *irL k 2 i , L 2EI " 2EI (2) By the Second Moment Area Theorem, the tangential deviation between the ends i j j i s equal to ^ times the moment of the bending moment diagram about j . . Consequently, - 9 -k l l L . 2L k21' L . L L 1 ( 3 ) 2 3 2 3 Solving Eqs. :2 and 3, 4 EI , . 2 EI 1 " — a n d k21 " — For equilibrium, the required shears at the ends of the member are obtained by dividing the sum of the end moments by the length of the member. Therefore, . k l l * k21 6 EI If C3 '"• ' • S 3 ••' i i 31 L L2 ' ^1 * k21 - 6 EI k U " 1 " • - — Stage 3 For a unit deformation along direction 2, from symmetry we obtain , 2 EI , 4 EI , . ' 6 EI ' k 1 2 " — '> 22 " L a n d k 3 2 " - k 4 2 " — Stage 4 A unit deformation along direction 3 may be considered as a rotation of jr units along directions 1 and 2. Therefore, using the results of the f i r s t two stages, Similarly, . 4 EI 1 a 2 EI 1 6EI ^3 " L • L + "T" • L " ~ 1, L, 6 E I k23 " k13 " — k „ ^ 3 * k23 12EI *33 L L3 k 0 k13 * k23 . 1 2 E I k43 L L3 - 10 -Stage 5 For a unit deformation along direction 4 , from symmetry we obtain 6 EI , 14 < 2 4 - - 7 ? a n d k 3 4 - - k t t " 3 f 3 r Fig. 4 Final end reactions and deformations Now, superimposing the results of each stage, the f i n a l end moments and shears at the ends of the member, shown i n Fig. 4 , are obtained as 6 EI . FEM, FER^ + FER. -4 4 EI 2 EI f. 6 EI L 6 1 + L 62 T L 2 2 EI K * 4 EI JL 6 EI L &1 + L 62 L 2 6 EI 6 EI 12 EI L 2 61 + L 2 62 + I? 6 EI t 6 E I 12 EI L 2 61 I 2 62 I? 6o -6 „ + L 6 EI 2 L 12 EI L 3 12 EI 3 \ i n which FEMn, FEM„, FER0 and FER. are the fixed end moments and reactions 1* £ 1 k due to the external loading. The matrix form of the above equations, called the "stiffness equation", i s given by In short matrix notation {p} - W W + M (5) where, p - column vector of f i n a l stress resultants, k = stiffness matrix of the member, 6 •• column vector of deformations , F = column vector of fixed end moments and reactions. The stiffness matrix [k] i s symmetrical, as is the case with a l l structural members, due to "Maxwell's Theorem of Reciprocity". 3. Stiffness Matrix of a Straight Member with Variable Moment of Inertia The elastic behaviour of a member with variable moment of inertia can be conveniently characterized by the rotational stiffness factors of i t s ends. Essentially, a member i j has three basic stiffness factors, namely a i j * a j i a n <* b i j * w ^ c * 1 a r e <* e ; fi n ed as below. EI a^j -rj- " the moment required at end i , to develop a unit rotation at i only. ( I Q i s the minimum moment of inertia of the member). EI • the moment required at end j , to develop a unit rotation at j only. EI b. . - the moment required at end j , when a unit rotation is developed, at i only, ( b ^ = b ^ by Maxwell's Reciprocity Theorem). - 12 -k 2 | = b. EIo k 3 , sfcj j j+bjj)-^ k 4 , =-(ai;+bii) EIo VI2 y c k 3 2 = (a j j -^bj j )^ k 4 2 a-Ca-j +bij)- j j EIo "43" w i j , v J i ' " T W -(c1:-»c,.) E l k33s ( V c J i ) _i3 S 4 - i k„„ =(c,.+c..) EI< 44 % w i j r w j i ' ,3 Fig. 5 Stiffness coefficients of a variable moment of inertia member For the customary forms of variation, the basic stiffness factors of a member are available i n standard t a b l e s . ^ ' ^ Now, proceeding exactly i n the same manner as described above i.e. by giving unit deformations separately along each of the 4 specified directions and calculating the forces required to maintain them, the stiffness matrix of a member with variable moment of i n e r t i a , as il l u s t r a t e d i n Fig. 5, becomes . \ b u S J i symmetrical M • - r * e,ii. L L L " Ci.1 + C 1 l Ci.1 + °1i L L L 2 L 2 (6) i n which ' i j j i a. . + b. . i j i j a j i + b i J For a member with constant moment of inertia a i j " a j i " U'°-> b i j = 2 ' ° a n d C i j " C j i = 6 ' ° 4. Stiffness Matrix of a Straight Member including Torsional Rotations The possible end rotations and displacements of a typical member are indicated by numbered arrows as shown i n Fig. 6. Evidently the f i r s t 4 x 4 portion of the required stiffness matrix i s identical with the stiffness matrix of the straight member given by Eq. .2,;, (a) R. Guldan, "Rahmentragwerke und Durchlauftrager". Spring e Verlag, Wien 1943, pp. 278-351. (b) "Handbook of Frame Constants", Publication of Portland Cement Associa-tion, 1958. - 14 -because the specified deformations 1 to 4 are the same. Further, when a unit torsional rotation i s given along direction 5, no forces are required i n directions 1 to 4. Therefore, ^5 = k25 = k35 3 k45 " ° The force required along direction 5 i s k . GJ ' 55 L (a) where GJ i s the torsional r i g i d i t y of the member. ' 8, Fig. 6 Flexural member with torsion For equilibrium, the force required along direction 6 is k - - k -65 55 L Similarly, for a unit torsional rotation along direction 6 h6 " k26 - k 36 " k46 " 0 1, G J A v G J k56 a n d k66 T Assembling the values obtained for the stiffness influence coefficients, the stiffness matrix of the member when torsional rotations are considered,is (a) Torsional r i g i d i t y constants for various cross sections are given i n Appendix B. - 15 -5. Stiffness Matrix of a Straight Member Considering Length Changes The possible rotations, vertical deflections and axial deformations of a typical member are shown by numbered arrows i n the positive direction i n Fig. 7. Fig. 7 Flexural member subject to length changes Again, the f i r s t four specified deformations are the same as those of the straight member i n section 2 and, therefore, the f i r s t 4 x 4 portion of the required stiffness matrix i s as i n Eq. 4. For a unit axial deformation along direction 5 the forces required i n each of the specified directions are, ^5 " k25 " k35 " k45 " ° 4 2 6 L 6 L 0 0 6 L - 6 L 0 0 12 L 2 - 12 12 L 0 Symmetrical L 0 GJ EI 0 " ^ - GJ GJ EI EI (7) - 16 -k = - 45 ' and k -k55 L a n d k65 T where A is the cross sectional area of the member. Similarly k l 6 " k26 " k 36 = k46 " 0 . AE . , AE k56 " " T a n d k66 = T Arranging these coefficients i n matrix form, the stiffness matrix of a mem-ber, taking i t ' s length changes into account,becomes M EI L 4 2 6 L 6 L Symmetrical 6 L 6 L 0 0 12 L 2 12 L 2 0 12 I 2 0 A T A I (8) 6. Stiffness Matrix of a Member Considering the Effect of Shear Deflections Though the overall effect of shearing strains i s generally small, i t is sometimes desirable.to take them into account. The typical member for this case remains the straight member of section 2, shown i n Fig. 1. (a) I f , instead of the Moment Area Theorems, the unit load theorem i s used to evaluate the member stiffness influence coefficients and the strain energy due to shear forces i s considered, the required stiffness matrix i s (a) Hall, S.A. and Woodhead, R.W., "Frame Analysis", John Wiley e Sons, Inc. 1961, pp 31 - 33 and pp,150 -152. obtained as EI L 3e + 1 3E - 1 3e + 1 Symmetrical 6e 6e 12e L L I 2 6e - 6c - 12e 12E L L L 2 L 2 (9) i n which the shearing strain parameter 1 e i s e -i • 12 A - S -L AG (10) At any member cross section, the numerical factor A i s used to multiply the average shearing stress to obtain the shearing stress at the centroid. For rectangular sections /\ = 1,5, while for circular sections A = 4/3. When shear deflections are neglected i t i s assumed that AG « oo , so that e equals 1. On substituting unity for e i n Eq. 9 , the resulting matrix is identical with that of Eq. 4. obtained for a similar member with-out taking into account the effect of shear deformations. 7. Stiffness Equation of a Structural System As for a single member, the loads acting on a structure are related to i t s resultant deformed shape by the basic stiffness equation {?} - [K] {D} . (11) * i n which, {p} • the column vector of loads acting on the structure, [K] » the stiffness matrix of the system, {D] • the column vector of the deformations of the system. - 18 -8. Evaluation of the Stiffness Matrix of a System Though the main stiffness matrix of a structure K may be generated following the fundamental definition as i l l s t r a t e d for a single member, this becomes a cumbersome procedure. Several simplified methods have been devel-oped using reduced stiffness matrices^, external strain energy^ and the 17 code number approach . The last mentioned, which is a special application of the strain energy method, i s extremely suitable for efficient computer application and has been used for all'the problems presented i n this thesis. External Strain Energy Method a. Transformation Matrices When a structure deforms under the influence of external loads, the deformations of i t s constituent members, [b^ , must conform with those of the system, {D} . This relation i s expressed by the following compatibi-l i t y equations. . W - [T] {D} (12) wherein the coefficient matrix [TJ i s called a transformation matrix. Each member has i t s own transformation matrix, which depends on i t s location and interconnection i n the structure and can be easily expressed by visual inspection. The compatibility equations for members 1, 2 and 3 of the portal frame i n Fig. 8, with respect to the assumed deformation directions of a typical member are: For member 1, from an examination of the figure, 61 " °* °1 + °* °2 + °* °3 62 " l a D l + °* °2 + °* D3 63 - 0. DX + 0. D2 • 0. bU " °* D l + °* °2 + 1' °3 - 19 -1200 O CM O W7t Fig. 8 Example portal frame In matrix form i n which Wx • .Hx W. Wx 0. 0. 1. 0. 0. 0. 0. 0. o. 0. 1. i s the transformation matrix of member 1. Similarly, for members.2 and 3 [t] 1. 0. 0. 0. 1. . o. 0. 0. 0. 0. 0. 0. and 0. 0. 0. °. 1. 0. 3 0. 0. 0. 0. 0. 1. The generation of the stiffness matrix of a structure from the trans-formation and stiffness matrices of i t s individual members is based on the principle of conservation of strain energy, which states that - 20 -"The total strain energy of a system is equal to the sum of the strain energies stored i n each constituent member". In other words the sum of the work done by the loads, { p| , of a system along i t s possible joint deformations, {D} , must equal, the work done by the stress resultants ( p j of the individual members along their respective end deformations {&} . Therefore, M i ( p 1 D X * p 2 D 2 • — p. D . ) - } V ( P l m 6^ • P 2 r a 6^ • — ) m • J. where M is the number of members. In matrix notation,after cancelling £ from both sides, m M { D J | P J - ^ ( 6 j m (pj (m - 1 M , for a l l members) m • 1 Substituting for {P} , {&} and {p}. from Eqs. 11 , 12. and 1. respectively, ~{Df [K][D} . ± (Df [,£ 1—.—' m = 1 v v -J i 1 must equal It follows that M " * f \ M l Wm Mm ( m = 1 — M ' f 0 r a 1 1 m e m b e r s ) ( l3> • m » 1 wherein i s the required stiffness matrix of the system, b. The Code Number Approach When the directions of the deformations of a system coincide with those of the members, the transformation matrices consist only of zeros and ones. In such a case, the main stiffness matrix can be generated directly without carrying out the t r i p l e matrix products indicated above. Each member i s given a code number which characterizes i t ' s location i n the structure and i s composed of those deformations of the system which coincide with the specified deformations of the member in question. These numbers correspond to the non-zero terms of the transformation matrices and are written down by visual inspection. In writing a code number, i t i s necessary to follow the same sequence as assumed for numbering the end defor-mations of the typical member. One end deformation is taken at a time and the number of the deformation of the system coinciding with that particular member end deformation i s entered into the code number. I f there i s no coinciding system deformation, a zero i s inserted. As an example,consider member (l) of the frame i n Fig. 8. The l e f t and right end rotations of the member coincide with deformations 0 and 1 of the system, whereas the l e f t and right end translations coincide with deformations 0 and 3 of the system. Therefore, i t s code number i s 0103. To generate the main stiffness matrix using code numbers, each number i s taken at a time and coupled f i r s t with i t s e l f and then with the remaining numbers i n the code number. The sequential order i n the code number of the number and of that with which i t is coupled constitute the row and column numbers of the elements to be taken from the individual member stiffness matrix, whereas the corresponding numbers themselves indicate the row and column numbers of the general stiffness matrix into which these elements must be placed. This operation must be repeated for a l l the constituent members of a structure to obtain i t s general matrix. Note that zero code numbers do not contribute to the general stiffness matrix. As'an example, consider member (l) of Fig. 8 again. Its code number 0103 indicates the following information: - 22 -Sequence Number Code Number Element 22 of [k]. » 24 " 44 " •i w » Besides being used to generate the main stiffness matrix of a structure, code numbers also indicate the appropriate deformation of the system to be used during the back substitution process to determine the f i n a l end moments and shears of the individual members as subsequently indicated i n the pro-cedure of analysis. 9. Concept of Joint Loads To f a c i l i t a t e the application of matrix methods, a l l the external loads applied on a structure are replaced by equivalent joint loads acting along the specified directions of deformations. Joint loads may be regarded as the resultant forces transferred to the joints from the ends of a member and, therefore, are obtained for each joint i n the structure by the sum of the fixed end moments and reactions of a l l the members meeting at each joint, with their directions reversed. The fixed end reactions are calcula-ted by assuming the members to be clamped at both ends. Note that no special consideration or modification factor is necessary i n the case of members with hinged ends. The condition of zero moment at a hinged support i s auto-matically satisfied i n the stiffness method by specifying a rotation at the hinge. When determining the f i n a l stress resultants of the individual members i n accordance with Eq. 5, the fixed end reactions are substituted with their proper signs. S LU EJ i s placed into 11 of II 1 3 33 " 31 system - 23 -1 0 . Procedure of Analysis for Planar Rectangular Grids• Step 1 The members and possible joint deformations are numbered on a sketch of the structure to be analyzed. Step 2 The code numbers of each member are written alongside the members. Step 3 The individual stiffness matrices {jcj of each member of the struc-ture are evaluated. Step 4 The main stiffness matrix of the system [KJ i s generated by means of the member code numbers, from the individual stiffness matrices. Step 5 The column vector of joint loads {p} i s determined by taking the sum of the fixed end reactions at each joint, with their signs reversed. Step 6 The basic stiffness equation of the structure, given by Eq. 11 (pj = £ K ] J {D}, is now used to yield the required deformations { D } of the system either, from {»} - M " 1 W or, by applying Gaussean elimination directly to the stiffness equation. In most cases the elimination process requires considerably less computer time than inversion and i s , consequently, preferred. Step 7 Finally, the stress resultants {p} of each individual member are computed by means of back substitution i n the member stiffness equation (Eq. 5), {P } . • [ k ] { . } . . • { , } The column vector of deformations {&} is composed of the appropriate deformations of the system and i s easily obtained using the member code num-bers, by spelling out the information i n i t i a l l y stored i n them. The column vector {F} represents the fixed end reactions of the member. - 24 -The following example planar grid with only three possible joint deformations, i . e . 3 unknowns, i s manually solved to demonstrate the pro-cedure outlined above. 11, Numerical Example 100.0 units 30.0 units Fig. 9 Example planar rectangular grid Step 1 The members and joint deformations of. the grid are numbered as shown i n Fig. 9. Step 2 The code numbers are written alongside each member. Step 3 Individual member stiffness matrices are evaluated. The typical member shown i n Fig. 6, considering torsional rotations, is adopted. Therefore, substituting numerical values into Eq. 7, - 25 -DO 40 20 40 Symmetrical 6 6 1.2 6 - 6 - 1.2 1.2 0 0 0 0 3.0 0 0 0 0 - 3.0 3.0 which i s the same for a l l the four members. Step 4^  Using the member code numbers, the stiffness matrix of the struc-ture i s obtained as [ K ] system ( 3 + 3\ \40 + 40j - 6 +-6N 1.2 + 1.2' 1.2 + 1.2 - 6 + 6 40 3 86 0 0 0 4.8 0 0 0 86 Step 5 Joint loads are obtained from the fixed end reactions. The fixed end reactions of member (1) are shown i n Fig. 10. P L 8 = - . 2 . 5 * ^ 10 ite— u r i3) 2 ° P L 8 = 12.5 k f t r p _ k T - - 5 Fig. 10 Fixed end reactions of member (l) - 26 -Therefore, the joint loads are ? 1 - 0 P 2 - + 5.0 k p3 " - 12.5 k f t . Note that the signs of the joint loads are with respect to the assumed posi-tive directions of the typical member. Step 6 Deformations of the system from Eq. 14 ' \ V 8 6 o ' 0 r ~\ 0 0 - V *• a o V 4 . 8 0 - 5.0 >- - 1.042> D 3 0 0 % 6 - 12.5 s. - .145 Step 7 For each member, the f i n a l stress resultants are evaluated by back-substituting the appropriate joint deformations of the system into the mem-ber stiffness equation i n accordance with Eq. 5. For member ( l ) , Pi 40 20 6 -6 0 0 0 -12.5 -21.65 P 2 20 40. 6 -6 0 0 -0.145 +12.5 + 0.45 - P 3 p 4 6 -6 6 -6 1.2 -1.2 -1.2 1.2 0 0 . 0 .0 -0 1.041 <• + -< - 5.0 - 5.0 - 7.12 • - 2.88 -P5 0 0 0 0 3 -3 o 0 0 • H ^ 0 0 0 0 -3 3 0 0 0 The f i n a l stress resultants of the remaining members are computed i n a similar manner and shown i n Fig. 11. The computer output of the above example,'obtained using the program presented i n Appendix A, i s given on the following page. ID NUMBER - 0 6 6 3 P R I N T E D FOR S . T E Z C A N ON M A R . 31 AT 3 H R . 2 1 . 5 M I N . E X E C U T E F O R T R A N P R O G R A M . 4 3 N O . L E N G T H 1 10.00 2A 10 . 00 E I 1 0 0 . 0 0 100.00 G J 3 0 . 0 0 3 0 . 0 0 UDL 0.00 0.00 L O A D 10.00 0.00 U 5.00 C . C C 0 3 CODE 3 0 C 2 NUMBER 2 0 0 1 1 0 3 1 0 . O C . 4 .10.00 1 0 0 . C C • 1 0 0 . 0 0 3 0 . 0 0 3 0 . CO 0.00 0.00 O . O C 0.00 O . O C 0.00 0 1 1 0 0 2 2 0 0 3 3 C D E F O R M A T I O N S 1 0 . OF THE S Y S T E M COOC 2 1 . 0 4 1 6 A 3 - . 1453 .F I N A L END. MOMENTS AND M l M2 R E A C T I O N S R l - •R2 ( K i p s ) — ^ 2 . 8 7 7 - . 3 7 7 T l T2 (K.ft) 1A -2 ! . 6 5 6 2 .4 36 (K.ft) . 4 3 6 3 . 3 4 3 -(Kips) " - 7 . 122 .377 (K.ft) C . O O O 0.000 ""(K'.-ft")' 0.000* 0.000 i ro -a i 3 - 6 . 2 4 9 4 6 . 2 4 9 - 6 . 2 4 9 . 6 . 2 4 9 - 1 . 2 ^ 9 1 . 2 4 V 1 . 2 4 9 1 . 2 4 9 .4 36 - . 4 3 6 . - . 4 3 6 . 4 3 6 P R O G R A M C A M E TO NORMAL END A L L DATA C A R D S WERE R E A D BY T H E PROGRAM . ^ • END OF T H I S RUN AT 3 H R . 2 5 . 7 M I N . m • • • :  , 1 . 1 12 " • . - ; •_ LL 2 v . 10 9 . ^ : . _ . 3 8 ' ' ,. 7 •. 4 6 . __: __ 5 5 v 4 3 • ' •  6- 28 -Fig. 11 Final stress resultants 12. Comparative Results with Other Methods (a) Woinowsky - Krieger's^" method The four girder grid shown i n Fig. 12, originally presented by Woinowsky-Krieger, was analyzed by the stiffness method s t r i c t l y for the purpose of comparison. In his method,Woinowsky-Krieger neglects the torsion-a l r i g i d i t i e s of the grid members and, further, replaces the bending s t i f f -ness of. the cross girders by that of a continuous torsionless plate of the same length as the main girders. Based ,on these assumptions, the longi-tudinal bending moments are expressed using Clapeyron's equations i n terms - 29 -of unknown joint deflections. The joint deflections i n turn, as well as, the external loads are expressed i n Fourier series and back substituted into Clapeyron's equations to form a system of homogeneous linear equations. The eigen value solutions of these equations yi e l d the required f i n a l moments i n the form of a sine series. Fig. 12 Woinowsky-Krieger's grid For the grid shown in Fig. 12, Woinowsky-Krieger gives the longitu-dinal bending moments at the mid spans of the main girders due to loads and P^. These values are tabulated i n Table 1 along with the corres-(a) ponding stiffness analysis and Stahlbau-Kalenderv ' results. (a) Vgl. Stahlbau-Kalendar 1934, S . 342. Berlin 1943. - 30 -GIRDER p l " 1 k at midspan P2 = 1 k at midspan NO. Girder 1 Girder 2 Woinowsky Woinowsky Krieger Stahlbau Stiffness Krieger Stahlbau Stiffness 1 6.68 6.70 6.694 2.94 2.92 2.928 2 1.76 1.74 1.757 2.82 2.84 2.830 • 3 0.43 . 0.42 0.411 1.54 1.56 1.563 4 - 0.87 - 0.86 - 0.861 0.70 0.69 0.684 Table 1. Comparative Midspan Moments (b) Method of moment distribution by Ewell, Okubo e Abrams Ewell, Okubo and Abrams eliminate two of the three unknown quantities at each joint by a moment and torque distribution process. The unknown deflections are then expressed i n terms of auxiliary forces at the grid joints and, therefore, only one series of linear equations has to be solved i n order to define the deflection pattern. Torsional r i g i d i t i e s are duly considered and the procedure is applicable to grids of any configuration and support conditions. However, for large unsymmetrical grids the arithmetical calculations become tedious. A separate solution must be made for each joint displacement, followed by. the corresponding reactions at each joint*, and, f i n a l l y , to obtain the deflections a set of simultaneous equations which satisfy the shear relations must be solved. In some cases, the problem may be reduced by adopting the. method of successive shear corrections, analogous to that used for planar frames with (a) sidesway, as suggested by Scordelis . The symmetrical grid shown i n Fig. 13, presented by Ewell, Okubo e Abrams,•consisting of members of equal length and cross section, has been (a) Scordelis, A.C, Discussion of "Deflections i n Gridworks and Slabs" ASCE Transactions, Vol. 117, 1952, p. 869. - 31 -analyzed for comparison by the stiffness method. The results are compared with those evaluated by Scordelis i n Fig. 13, the latte r values being indi-cated by parantheses. B C G F Fig. 13 Ewell, Okubo e Abrams grid The only value computed by Ewell, Okubo e Abrams is M.. = 181.27 FL - 32 -CHAPTER 2 ORTHOGONAL TRANSFORMATION OF AXES 1. Axis Transformation of Forces and Deformations at a Point A system of forces acting at a point 0, represented by components along coordinate axes x, y, z, may be transformed into an equivalent system along another set of axes x', y', z' with the same origin 0, as shown i n Fig. 14. z S y s t e m " a " S y s t e m " b " Fig. 14 Transformation from x, y, z to x', y 1, z" To write the relations between the original and transformed sets of forces, referred to as "a" and "b" respectively, consider for example the - 33 -transformed force which, i n terms of the original forces may be expressed as Pi - 1 Pi + m p. + n p„ *_ • x r l x 2^ v ^ 'x *1 ~x 42 x r3.where l x , m^  and n^ are evidently the direction cosines- of axis x' with respect to the original axes x, y, z. Writing similar equations for the remaining forces the desired matrix relation between the two systems i s obtained as Pi i P2 t P3 p4 i p5 p6 i.e. 1 m n X X X 1 m n y y y l m n z z z 0 1 m n X X X 1 m • n y 7 7 l m n _z z z P l P2 - >< Po r P4 p5 a A . (15) (16) b •«- -"a v ^ a Note that the transformation matrix £lQ consists entirely of the direc-tion cosines of x', y', z' with respect to x,y, z ; ' As forces and deformations are identically related to the coordinate system and a deformation can be expressed as a vector, a system of deforma-tions "a" can be transformed to an equivalent system "b" by a similar equation. M b - J>1 Ma <-7' 2. Transformation of Member Stiffness Matrices To enable the analysis of diagrids and spatial grids i t is necessary to obtain the stiffness matrix of the structure with respect to the chosen . - 34 -common axes. Consequently, the stiffness matrices of each individual member have to be transformed from their member axes to the common axes before generating the main stiffness matrix of the structure. •Let us assume that x, y, z represent the common axes of the system while x', y', z' are the member axes. The work done by the forces of the common axes along the corresponding deformations must be equal to the work done by the st a t i c a l l y equivalent forces of the member axes, along the cor-responding deformations. This may be written as xyz v 'xyz ^ J x'y'z' L 'x'y'z' From.the basic stiffness equation of a member (Eq. 5)j (pY - [kl {5] -t^cyz -L -bcyz •<•  J xyz ^P'^ x'y'z' x'y'z' x'y'z' From Eq. 16, after applying the rule of matrix transposition {&«} T = [b] ' [ T ] T (19) x'y'z' x'y'z' Therefore, substituting the above values of {pj- , jp'J Y |&'J. i n Eq. 18 and cancelling 7; from both sides, J.6i T [k] {6} - h]T Trf [ki] T T ] {6} 1 Jxyz J_j|xyz  K" Jxyz  L Jxyz t L J L ^x'y'z' L - 1/ Jxyz (20) Comparing the l e f t and right hand sides of the above equation, the transformed stiffness matrix k of the common axes is obtained from the xyz stiffness matrix k , , , of the member axes i n the following form x'y'z' 0 [k] - [ T ] T [k'J [ T ] ' (21) L—'xyz L " i 1 'x'y'z' 1 1 - 3 5 -CHAPTER 3 SPATIAL GRIDS 1. Stiffness Matrix of a Spatial Grid Member The number of possible end deformations of a spatial grid member is twelve as depicted i n Fig. 15. Fig. 15 End deformations of a spatial grid member Sign convention: Translations are positive i n the positive sense of the 'axes and rotations are positive i n accordance with the  right hand screw rule. Deformations 6 .^ 6 ^ correspond to flexural rotations and deflec-i « ' tions i n relation to the major axis x ; and 6^  represent torsional i t rotations; 6^  6^ are the.directions of flexural rotation and deflec-tion with respect to the minor axis z 1, while 6^ and 6^ a r e a x i a l deformations. In other words, the stiffness matrix of a space member con-sists of the stiffness matrices of flexural members acting about the major and minor axes, a pure torque member and that of a member subject to axial deformations only. A somewhat irregular pattern of numbering was adopted for the defor-mation directions i n order to obtain the corresponding stiffness matrix - 36 -i n a very regular and convenient form as shown below. Mx'y 'z ' : I 2 3 4 "4EI ~"LX sym 2EI '4EI 6 E I ^ 6Elj 1210^  -I2E^ -6EI-6SI.-I2ELI2E1 0 0 0 • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 6 7 8 I. 0 0 I 0 0 I ' 0 0 I 0 0 • 0 0 1 0 0 I 0 0 10 0 TE"-A_!~ ~" 7" — T 1 0 0 •AE AE 1 N N I 4 £ r z 0 0 . I T 9 0 0 0 0 0 0 0 0 0 0 , 2 . 0 0 0 0 I 0 0 0 0 I 0 0 I sym i 0 0 "XL '6EI7 6EI 0 0 i L2 a . o o l2EI rl2EIz' 6Er2-6£I7-l2£I212EI7l I G I P G I P * I L L 0 . 0 i 0 0 0 0 0 0 . 0 0 0 (22) 2. Axis Transformation of a S p a t i a l Member A t y p i c a l space member with i t s possible deformations marked along the member and common axes i s shown i n F i g . 16, with the same order of numbering as previously adopted f o r i t ' s s t i f f n e s s matrix. The end forces and deformations given with respect to the member and common axes may be r e l a t e d to each other by means of a transformation matrix. T i n accordance with Eqs. 15.and 16. To correspond with the new order of numbering f o r the member end deformations as shown i n F i g . 16, the transformation matrix i s rearranged and given by Fig. 16 A space member with respect to member and common axes - 38 -1 p l 1 X 0 0 0 m X 0 0 n X 0 0 0 0 r n  p i P2 0 \ 0 0 0 m X n X 0 0 . 0 0 0 P2 P3 0 0 n z 0 0 0 0 0 0 l z m z 0 P3 p; 0 .0 0 n z o. 0 0 • 0 1 z 0 0 m z p4 P5 1 y 0 0 0 m 7 0 0 n y 0 0 . 0 0 p5 . P 6 p7 0 0 1 7 1 z 0 0 0 0 0 0 m y m • z n y n z 0 . 0 0 0 0 0 0 0 .0 0 •< p6 P7 -p8 l z 0 0 0 m z 0 0 n z 0 0 0 0 p8 P 9 0 0 0 n X 0 0 0 0 1 X 0 0 m X P9 *io 0 0 n X 0 0 0 0 0 0 Xx m X 0 p10 0 0 n -7 0 0 0 0 0 0 1 y m y 0 p l l .V. J x'y' z» 0 0 0 n y-0 0 0 0 1 y c 0 ni y pl 2 J 2 3. Direction cosines of the member axes with respect to the common axes In general, the common axes are normally chosen for a structure to be along the transverse, longitudinal and gravity directions. These are shown i n Fig. 17 by x, y and z respectively. . The direction cosines o f the member axes x', y 1, z' with respect to the common axes x, y, z, may be obtained i n terms of the coordinates of the ends of the member as follows. F i r s t , the direction cosines of the member centre line axis y 1, by definition, are - 39 -i n which the member length L, is Fig. 17 Spatial grid member Next, i t is assumed that the x' axis of the member i s horizontal i.e. parallel to the x, y plane, as i s normally the case for most structures. Therefore, "x = 0 ( 2 6 ) The remaining direction cosines of the x' and z' axes are evaluated i n terms of l y , my and ny from the conditions of normality and orthogonality. Conditions of normality: 2 . _ 2 2 x ,2 2 ^ 2 1 + m + n y y y n 2 2 2 1 + m + n z z z Conditions of orthogonality: 1 1 + mm + n n x y x y x y 1 1 +mm + n n x z x z x z From Eq. 30, 1 1 +mm + n n y z y z y z 1 1 x y ni. ~ x m y Solving Eqs. 27 and 33, m "Sc Q and 1 x Q 2 2 where Q = /1 + m •7 J Substituting the above values int 1 1 x z m = z . m x or, using Eq. 33, ' 1 z m = m •=— z y 1 ' y - hi -From Eq. -32,after replacing for m from Eq. 37, m 1 1 + 1 + n n = 0 7 z ,2 z y z whereby, n 1 y y Z Q2 (38) Substituting for m and n from Eqs. 37 and 38 into Eq. 29 z z ' 1 2 . . 1 2 ,2 2 z A 1 + nv — r + z y T 2 _ 2, 2 z Q2 n 1 y y Solving this for 1 , 1 n 1 = z Q (39) The negative value of the solution for 1 i s used i n order to conform with z . the directions of the assumed right-hand reference system. Finally, from Eqs. 37 and 39, . m n m - y v m = — 'L " z Q Similarly, from Eqs. 38 and 39-, n = Q z (AO) (U) A l l the required direction cosines have now been determined and are shown in Table 2. Major Axis x' Centre line axis y» Minor Axis z« i x - V Q • m x = - W n x - 0 V = ( X2 " *1>/L ray= ( y 2 - y i ) A j n y = ( Z2 Z l ) / L 1 = - 1 n/Q z y / m = - m n /Q z y y n = Q z Table 2. Direction cosines of member axes x', y 1, z' with respect to common axes x, y, z. - 42 -4. Procedure of Analysis for Spatial Grids Except for a fev; important differences, the procedure remains the same as described for planar grids. However, for the purpose of cl a r i t y , the steps in the analysis are f u l l y enumerated below. Step 1 The members, joints and possible joint deformations are numbered on a sketch of the structure. The joint deformations are indicated by arrows along the positive sense i.e., translations i n the direction of the axes and rotations according to the right hand screw rule. Step 2 The following input data i s prepared: a. Number of members and deformations. b. Coordinates of the joints vdth reference to a chosen common axes. c. Rigidities of each member, EI , EI , GJ : and AE. X It d. Code numbers of each member as illustrated i n the following, numerical example. e. External loads acting on the structure. Step 3 The individual member stiffness matrices C^X-i i z t °^ ^2, with respect to the member axes are evaluated numerically for each member. Step 4 The transformation matrices of each member of Eq. 23, consis-ting of the direction cosines of i t s member axes are computed. The required direction cosines are defined entirely by the coordinates of the ends of the member with reference to the common axes, as shown in Table 2. Step 5 The transformed stiffness matrix fkjxyz °^ e a c ^ member is obtained by carrying out the t r i p l e matrix product of Eq. 21 Step 6 The main stiffness matrix of the system \_K], is generated by means of code numbers from the transformed individual stiffness matrices. - 43 -Step 7 The column vector of joint loads £P}, acting along the assumed directions of deformation of the joints, is determined by taking the sum of the fixed end reactions at each joint, v;ith their signs reversed. Step 8 The basic stiffness equations of the structure, M - I*] {•>} :are now used to yield the required deformations £D} of the system along  the common axes either, by inversion from .{D. ' W " 1 {P} or by Gaussean elimination. Step 9 The f i n a l stress resultants ' {p}— '^ of each member along the common axes are evaluated by means of back substitution into the member stiffness equation, Eq. 5, R c y z " Hxyz Wxyz + IF) The column vector of deformations |&}, is composed of the appropriate deformations of the system and is obtained by means of the member code num-bers„ The column vector £ F J represents the fixed end reactions of the member. Step 10 In case the f i n a l stress resultants along the member axes { p ' } x i y T 2 i i are required, the following transformation of Eq. 16 i s applied to {p}XyZ« M - M . W ^ x'y'z' xyz v  Jxyz 5. Numerical Example (Hyperbolic Paraboloid Space Grid) A iine diagram of the grid i s shown i n Fig. 18. The structure is symmetrical and has 40 members and 126 joint deformations. The beams are assumed to be •.• ' 18 WF 114. The torsional r i g i d i t y of this section was - 44 -. calculated from the corresponding formula given i n Table B , Appendix B. As an example the code number of member 3, i n accordance with the order of numbering of the deformations shown i n Fig. 15, is 1 13 . 2 14 3 15 16 4 17 5 6 18 Fig. 18 Line diagram of hyperbolic paraboloid space grid - i*5 -The external loads and the f i n a l bending moments along the member axes of the members for a quarter of the structure are shown i n Fig. 19. The par-t i a l computer output corresponding to the f i n a l stress resultants of members 1 to 21 i s also given in the following pages. The stress resultants of the remaining members may be easily obtained from symmetry. • z Fig. 19 Hyperbolic paraboloid space grid (Moments i n k.ft.) X 28= - , 0 0 0 0 0 1 0 9 X 29= 5 . 0 9 5 4 0 4 5 0 x _ 3 Q _ 0 0 0 0 0 1 u - 0 •• UNI T NO.- 1 - LOAD NO. 1 X 1 = - . 5 9 4 7 5 4 1 f X 2 = - 7 . 2 5 3 6 5 8 4 0 X 3= - . 2 8 6 0 7 3 0 2 X 4 = . 3 5 4 0 9 6 5 2 X 5 = . 4 5 6 1 4 7 0 5 X 6 = - 5 . 0 5 7 6 9 6 2 0 X 7 = - . 2 8 6 0 / 3 6 0 X 8 = - 7 . 2 5 3 6 7 6 7 0 X 9 = - . 5 9 4 7 5 4 1 2 X 10 = - . 3 5 4 0 9 4 9 6 X 11 = 5 . 0 5 7 6 6 0 9 0 X 12 = - . 4 5 6 1 5 5 3 0 F I N A L S T R E S S RESULTANTS (IN KIPS AND KIP-FT) £; F I R S T L I N E IS WRT COMMON,SECOND L I N E IS WRT MEMBER AXES ' MXT~ MX2 I) ~. 12 MY'1 MY2 MZ 2 MZ 1 X2 XI YI Y2 MEMBER NUMBERS - '— : • : -UNIT NO. 1 LOAD NO. 1 - 1 . - 1 4 . 9 4 9 -17 .7 -61 9 . 9 9 9 - 9 . 9 9 9 8 6 . 1 3 7 2 9 . 6 8 9 - 1 8 . 5 8 6 - 1 6 . 1 14 2 . 4 1 6 - 2 . 4 16 2 . 3 1 3 - 2 . 3 1 3 1 8 6 . 1 3 7 2 9 . 6 8 9 5 . 6 1 b - 5 . 6 1 8 - . 1 7 5 . 175 - 2 5 . 7 0 7 - 2 1 . 9 8 0 - 2 . 3 1 3 2 . 3 1 3 8 . 6 1 7 - 8 . 6 1 7 2 8 6 . 1 3 7 2 9 . 6 8 9 9 . 9 9 9 - 9 . 9 9 9 - 1 4 . 9 4 9 - 1 7 . 7 6 1 1 8 . 5 8 6 1 6 . 1 1 4 2 . 3 1 3 - 2 . 3 1 3 2 . 4 1 6 - 2 . 4 16 2 8 6 . 1 3 7 2 9 . 6 8 9 5 . 6 1 8 - 5 . 6 1 8 .175 - . 1 7 5 2 5 . 7 0 7 2 1 , 0 8 0 2 . 3 1 3 - 2 . 3 1 3 8 . 6 1 7 - 8 , 6 1 7 U N I T N O , 2 LOAD NO. 1 3 - ..3 1 8 - 4 . 4 3 5 3 . 3 0 6 - 3 . 3 0 6 -1 1 . 5 5 7 6 0 . 7 0 0 - T T . 401 - . 7 6 5 . 0 7 8 - . 0 7 8 .8811 • **-3 - 1 1 . 5 5 7 6 0 . 7 0 0 3 . 0 5 1 - 3 . 0 5 1 . 0 1 7 - . 0 1 7 - 1 2 . 2 3 4 - . 8 2 9 - . 8 1 1 .811 1 .276 - 1 . 2 7 6 4 1 8 . 0 7 9 -6 1.071 6 . 6 9 2 - 6 . 6 9 2 - 1 8 . 1 3 2 - 1 4 * 9 2 5 - 15 .716 l9.~351 " 2 . 3 3 7 - 2 . 3 3 7 1.502 - 1 . 5 0 2 4 1 8 . 0 7 9 6 1.07 1 3 . 8 3 9 - 3 . 8 3 9 . 0 8 0 - . 0 8 0 2 1 . 6 7 4 2 6 . 5 1 9 2 . 3 3 7 - 2 . 3 3 7 5 . 6 8 3 - 5 . 6 8 3 '"-5 - 1 8 . 1 3 2 - 1 4 . 9 2 5 6 . 6 9 2 - 6 . 6 9 2 1 8 . 0 7 9 6 1 . 0 7 1 - 1 5 . 7 1 6 - 1 9 . 3 5 2 1 .502 - 1 . 5 0 2 2 . 3 3 7 - 2 . 3 3 7 5 1 8 . 0 7 9 6 1 . 0 7 1 3 . 8 3 9 - 3 . 8 3 9 - . 0 8 0 . 0 8 0 - 2 1 . 6 7 4 - 2 6 . 5 1 9 - 2 . 3 3 7 2 . 3 3 7 5 . 6 8 3 - 5 . 6 8 3 i .t-6 -1 1 .557 6 0 . 7 0 0 3 . 3 0 6 - 3 . 3 0 6 - . 3 1 8 - 4 . 4 3 5 -o i T l . 401 . 7 6 5 .811 „ - . R 1 1 . 0 7 8 - . 0 7 8 6 - 1 1 . 5 5 7 6 0 . 7 0 0 3 . 0 5 1 - 3 . 0 5 1 - . 0 1 7 ' . 0 1 7 1 2 . 2 3 4 . 8 2 9 .811 - . 8 1 1 1.276 -1 . 2 7 6 U N I T , N O . 3 LDAO NO. 1 - • 7 - 4 . U 3 5 - . 3 1 8 - 3 . 3 0 6 3 . 3 0 6 - 6 0 . 6 9 9 1 V . 5 5 7 1 . 7 6 5 11.401 . 0 7 8 - . 0 7 8 - . 8 1 I . 8 11 1 v 12 7 - 6 0 . 6 9 9 1 1 . 5 5 7 - 3 . 0 5 1 3 .051 - . 0 1 7 . 0 1 7 XL . 8 2 9 1 2 . 2 3 4 .811 - • 8 1 1 1.276 - 1 ^ 276 ?. . 10 9 8 8 . 8 7 0 6 7 . 3 9 6 6 . 6 1 3 - 6 . 6 1 3 0 .000 0 . 0 0 0 3 3 0 .000 0 . 0 0 0 0 .000 0 . 0 0 0 1 .622 - 1 . 6 2 2 4 7 6 8 8 . 8 7 0 6 7 . 3 9 6 3 . 6 9 9 - 3 . 6 9 9 0 . 0 0 0 0 . 0 0 0 .5 5 0 .000 0 . 0 0 0 0 . 0 0 0 0 .000 5 . 7 1 7 - 5 . 7 1 7 4 3 9 - 3 . 6 8 2 - 3 . 6 9 3 1 . 6 9 2 - 1 . 6 9 2 - 4 2 . 4 6 3 5 2 . 7 2 9 6 -9.451 -9.U27 -10.147 -10.12 I _ _ V Q z ^ - ^ 2 527729 9.4.51 9.428 " 10 -1T27462 5277 2~9 . 10. 147 10.12 1 r r _ OTO'CO 07000 0.000 0.000 • _ _ n 87870 677396 0.000 0.000 12~ ^607700 7 1T7557 -.765 -1 1.401 . . ___ — .^-^ 077o 0- ~rr."55 7 -.829 -12.234 WlT~N07~4 "~- ~ LOAD NO. 1 ___ _______ _______ - 16.114 18.586 | VlT. ^29.689 -86. 137 ... . -2 1.980 25.707 1 12 • i i ~14 ~T8• 079 5"l 707 T z" ,o --15.716 -19.352 9 . '_ • 1 :, 8 14 18.079 61 .071 7 -21.674 -26.519 4 6 ; : 5 15 -3.693 -3,682 ,• , 9.428 9.451 6 2,581 -2.581 1.258 -1.258 "76 3 7 -763 7 0700 0 0700 0 1.258 1.258 3.020 -3.020 17692 =T769_ -37682 ~"=37-»~9~3~ 1.258 -1.258 2.581 -2.581 "76 37 7^537 ~~ 0700 0 07000 1.258 -1.258 3.020 -3.020 67613 -676T3 ~ 87870 ~ 677396 1.622 -1.622 0.000 0.000 3T?f9"9 -376"9'9 07000 07000 0.000 0.000 5.717 -5.717 3730 6 "37306 -4743 5 7^31 8 -.81 1 .811 .078 -.078 £ -3". 05 1 370 51 ~7D17 =7017 -.811 .fill 1.276 -1.276 •"97999 97"9~99 =2"976"8'9~ -8"67~137 2.416 -2.416 -2.313 2.313 R5 .618 5T6T8 771757 ^7175 2.313 -2.313 8.617 -8.617 "67692 -6TW2 rTB7T3T T47925 -2.337 2.337 1 .502 -1 .502 3.839 -3.8 39 -.080 •.080 2.337 2.337 5.683 -5.683 -1.692 1.692" -52.729 42.463 2.581 -2.581 -1.258 1.258 15 - 5 2 . 7 2 V 4 2 . 4 6 3 1 0 . 1 2 1 1 0 , 1 4 7 "16 ^60". 0 08 3 577 5"0 0 . 0 0 0 0 . 0 0 0 16 -60.008 35.750 0.000 0.000 ~T7 OTOOO OTOdO 0.000 0.000 ~1"7 ^607009~ 357750 0.000 0.000 ~18 -5"277 29 42 74~6T -9.428 -9.451 - 18 - 5 2 . 7 2 9 4 2 . 4 6 3 - 1 0 . 1 2 1 - 1 0 . 1 4 7 19 - I 8 . V 3 2 147925 - 1 5 . 7 1 6 1 9 . 3 5 2 19 1 8 . 0 7 9 6 1 . 0 7 1 - 2 1 . 6 7 4 2 6 . 5 19 20 - 2 9 . 6 8 9 - 8 6 . 1 3 8 - 16 . -114 - 1 8 . 5 8 6 20 - 2 9 . 6 8 9 - 8 6 . 138 v - 2 1 . 9 8 0 - 2 5 . 7 0 7 -1 ' UNIT NO. 5 ' • LOAD NO. 1 21 = /3 86 . 1 3 7 2 9 7 6 8 9 - 1 8 . 5 8 6 ' - 1 6 . 1 1 4 — 2 1 ^ / 3 8"6T1 37 297689 - 2 5 . 7 0 7 - 2 1 . 0 8 0 n 10 9 -.637 .637 0.000 • 0.000 1.258 -1.258 3.020 -3.020 67000 07W0 b'7000 oTodo" 0.000 0.000 4.139 - 4 . 1 3 9 T7506 7J7506 OTWO "OTOOO" 0.000 0.000 3.R55 * -3.855 OTCrOD : OTWO =6"0T0O"9 3~57T5"0~ 4.139 - 4 . 1 3 9 0.000 0.000 T.506 1 .506 OTOOCT ~~0.000" 0.000 0.000 3.855 -3.855 17692 T7692 -37693 ~37682~ 1.258 - 1.258 2.58 1 -2.58 1 =7637 7 637 0.000 1.258 1.258 3.020 "678"92 —=6TG92 18707 9 1.502 -1.502 -2.337 3783^T~ -37839 ~ TWO T~08~0~ 2.337 -2.337 5.683 -5;683 "97999 9T"99~9 1^77761 -1"4T9"W -2.313 2.313 2.416 -2.416 -57618 ST6T8 — - 7 V 7 5 " ~TT75 -2.313 2.313 8 . 6 1 7 -8.617 0.000 • -3.020 ^ • _ g _ 1 _ _ 0 . 7 1 2^ 337 9 ~ 9 9 9 — ^ 9 T o 99 1479 49 1777 6 T 2 , 3 1 3 2 . 3 1 3 2 . 4 1 6 - 2 . 4 1 6 57618 "^"576T8 =7175 7T75 2 . 3 1 3 2 . 3 1 3 8 . 6 1 7 - 8 . 6 1 7 - 5 0 CHAPTER 4 PLANAR DIAGRIDS 1. Stiffness Matrix of a Diagrid Member A diagrid may be regarded as a special case of a space grid. Some of the possible beam outlines of planar diagrids are illustrated i n Fig. 20 L a t t i c e M o d e l D i a g r i d S k e w b r i d g e Fig. 20 Common types of planar diagrids As for rectangular planar grids, the ends of the constituent members of a diagrid are subject to bending rotations 6^, &2, vertical transla-tions 6 0, 6, and torsional rotations 6C, b, as shown by numbered arrows j k ? o ' i n Fig. 2 1 . The directions of the end deformations are not the same as those shown i n Fig. 6 for rectangular grid members, but, instead, they coincide with the directions of the corresponding deformations adopted for space members i n Fig. 15. Accordingly, translations are specified along the positive sense of the coordinate axes and rotations obey the right hand screw rule. Nevertheless, following the basic definition, the stiffness matrix [ k ] x , y l z I of a diagrid member is obtained i n exactly the same form as Eq. 7 for rectangular planar grids. - 51 -AZ EI, GJ, L Fig. 21 End deformations of a diagrid member 2. Axis Transformation of a Diagrid Member > The possible end deformations of a typical diagrid member are shown i n Fig. 22 along the member and common axes. Since the end deformations of a .diagrid are identical with the f i r s t 6 assumed end deformations of a' typical spatial grid member, the matrix £T] required for the purpose of transforming forces from the common axes to the member axes of a diagrid is exactly the same as the f i r s t 6 by 6 submatrix of Eq. 23. That i s , i '3 i 4 5 to °Jx'y'z' _ -1 X 0 0 0 m x 0 p l 0 l X 0 0 0 m x p2 0 0 n z 0 0 0 -P3 0 0 0 n z 0 0 P4 1 y 0 0 0 m y 0 P5 0 \ 0 0 0 m y p6 (42) Jxyz - 52 -Memb8r a x e s x', y) z ' C o m m o n a x e s x , y , z Fig. 22 Member and conimon axes of a diagrid 3. Direction Cosines of a Diagrid Member The direction cosines of a diagrid member, as required by Eq. 42, may be obtained from those of a space member using Table 2. However, i t i s possible to further simplify them as follows. The direction cosines of the member centre line axis, which always l i e s i n the xy plane, are - 53 -m y2 " 71 n = 0 y i n which Since n_ i s zero, i n order to satisfy the condition of normality of Eq. 28, 1 2 + m 2 \ - 1 y y Using this value of Q, the remaining direction cosines reduce to the values shown i n Table 3. Major Axis x' Centre line Axis y' Minor Axis' z» 1 • m • i x2 " x l , 1 * z =• 1 1 « 0 X y L z m = - 1 7 2 " y l m - 0 X y L z n_-0 n = 0 y ». - 1 Table 3. Direction cosines of member axes x'y'z' of a diagrid 4. Transformed Stiffness Matrix of a Diagrid Member The transformed stiffness matrix [ k ^ _ i s obtained by carrying out the t r i p l e matrix product of Eq. 21. Gj Representing the ratio of — by q , the required transformed matrix |k| i s obtained as l Jxyz 54 -W !xyz EI L * 7 6ra 6m L L 6m - 6m L L 1+ ami/ f- 2mi\ \- qmiy 2ml r ^ /- 4m_\ 1+ qml J 12 L 2 12 L 2 61 L §1 L Symmetrical. 12 L 2 61 L 61 L U T qm J qm/ 41< qm'1 (43) In order to save computer time, the transformed stiffness matrix v/as fed into the diagrid programme directly i n the above form, thus eliminating the time otherwise required to carry out a t r i p l e matrix product for each member. 5. Procedure of Analysis for Diagrids The procedure of analysis, which, remains exactly the same as for spatial grids, has been ill u s t r a t e d below by several numerical examples. 6. Numerical Examples a) Two beam grid Step 1 The members, joints and possible joint deformations of the structure are numbered as shown i n Fig. 23. Step 2 The code numbers of each member are written alongside the members. Step 3 Individual member stiffness matrices are obtained by substituting the numerical values of EI, GJ and L into the member stiffness matrix of Eq. 7. These matrices for a l l the' members are - 55 -WXlylZl 40 20 40 Symmetrical 6 6 1.2 - 6 - 6 - 1.2 1.2 0 0 0 0 3.0 0 0 . 0 0 - 3.0 3.0 EI = 100.0 units G J = 30.0 units L = 10.0 ft. © (10,17.30) Fig. 23 Two beam diagrid Step 4 The transformation matrix [ T J of Eq. 42 has to be evaluated for each member. The required direction cosines of members 1 and 2 as taken from Table 3, are 1 - X ? = a _ 0.5 A L 10 • ? m 7 3 " 72 m g.65 - 0 L 10 0.865 - 56 -Therefore, 0.865 0 0 0.. 0.5 0 0 0.865 0 0 0 0.5 0 0 . * 0 0 1 0 0 1 0 0 0 0 - 0.5 0 0 0 0.865 0 0 - 0.5 0 0 0 0.865 for members 3 and 4 • 1 - x3 - x l L 10 3E 0.5 m » v} - y x L „ 8.6? " 0 0 - 0.865 0.865 0 0 0 - 0.5 0 0 0.865 0 0 .0 - 0.5 Prl 0 0 1 0 0 0 L T J 4 = 0 0 0 1 0 0 0.5 0 0 0 0.865 0 0 0.5 0 0 0 0.865 Step 5 The original member stiffness matrices [ k ] x i y t z l o f s t e P 3 are transformed to the common axes by carrying out the t r i p l e matrix product of Eq. 21. Alternatively, the transformed stiffness matrix of Eq. 43 yields the same results as follows, - 57 -M - H and,similarly, 30.76 U.25 30.76 symmetrical 5.20 5.20 1.20 - 5.20 - 5.20 - 1.20 1.20 16.05 9.97 3.01 - 3.01 12.28 9.97 16.05 3.01 - 3.01 2.76 12.28 ~30.76 —1 U.25 30.76 symmetrical 5.20 5.20 1.20 - 5.20 - 5.20 - 1.20 1.20 - 16.05 - 9.97 - 3.01 3.01 12.28 - 9.97 - 16.05 - 3.01 3.01 2.76 12.28 Step 6 U3ing the code numbers shown i n Fig. 23, the stiffness matrix of the structure i s obtained from the transformed member matrices as 1 2 3 symmetrical - 5.2 + 5.2' r 5.2 + 5.2 A . 2 + 1.2\ ^1.2 + 1.2J 9.97 • 16.05 \ /- 3.01 + 3.0l\ / 12.28 • 12. - 16.05 - 9-97/ V- 3- 0 1 + 3.01/ \ 12.28 • 12. 28 28 123.04 0 0 0 4.8 0 0 0 49.12 Step 7 A single concentrated load i s applied in a direction opposite to that of deformation 2. Therefore the joint loads are P = P e 0 1 • 3 P- = - 10.0 - 58 -Step 8 Deformations of the system: D, 1 . D„ 2 D3 - J 1/123.04 0 0 0 1/4.8 0 0 0 1/49.12 0 10 0 0 - 2.08 0 Step 9 The fi n a l stress resultants{p}^- with respect to the common axes, using the member stiffness equation,are,given for member 1 by 30.76 0 "b" 10.81" 14.25 30.76 symmetrical 0 0 10.81 5.20 5.20. 1.20 0 0 2.50 -5.20 -5.20 -1.20 1.20 -2.08 0 -2.50 ' 16.05 * 9.97 3.01 -3.01 12.28 0 0 6.25 9.97 16.05 3.01 -3.01 2.76 12.28 0 0 . J 6.25 -* P5 ^x.y_ •-The required f i n a l stress resultants {p'} xt~t_i w i t h respect to the member axes are given by Eq, 16. 1 p l .865 0 0 0. .5 0 10.81 12.50 P2 0 .865 0 0 0 .5 10.81 12.50 P3 -< " i > 0 .0 0 0 1 0 0 1 0 0 0 0 -2.50 -2.50 -2.50 -2.50 p; - .5 0 •0 0 .865 0 6.25 0.0 ?&. x'y'z» 0 - .5 0 0 0 .865 6.25 xyz 0.0 1 x" y' z' The stress resultants of the remaining members are similar to those of the f i r s t member because of symmetry. - 59 -b) Cantilever Diagrid The cantilever diagrid shown i n Fig. 2U was analyzed by the stiffness method, both for zero and for an assumed torsional r i g i d i t y , GJ • 0.2 EI. The f i n a l bending moment values along the member axes obtained i n the two cases as well as the error percentages caused by neglecting torsion are shown i n Table 4. In addition, the complete computer output for the torsional analysis i s also presented i n the following pages. Fig. 2L Cantilever diagrid - 60 -Member No. Bending Moment (Kip ft.) Error % Considering Torsion GJ - 0.2 EI Neglecting Torsion GJ = 0 Left Right Left Right Left Right 1 103.27 - 42.18 98.66 - 62.52 - 4.5 A8.2 2 87.55 - 35.70 98.66 - 35.84 12.7 0.4 3 - 72.18 66.44 - 83.00 67.20 15.0 1.1 4 - 67.46 34.36 - 67.20 13.31 - 0.4 - 61.3 5 61.31 16.06 86.32 27.29 40.8 69.9 6 .10.64 13.36 - 13.31 23.88 - 225.1 78.7 7 35.03 - 10.53 35.84 ' - 11.08 2.3 5.2 8 68.00 - 18.18 76.05 - 11.46 11.8 - 37.0 9 - 43.90 66.38 - 45.94 71.67 4.6 8.0 10 61.96 52.20 - 71.67 59.73. 15.7 14.4 11 - 52.08 40.30 - 59.73 50.19 14.7 24.5 12 - 45.42. 13.09 - 50.19 30.52 10.5 133.2 13 - 16.18 - 16.22 - 13.65 -.11.94 - 15.6 - 26.h 14 - 14.38 17.84 - 23.88 24.33 66.1 36.4 15 12.21 10.28 11.08 16.08 - 9.3 56.4 16 21.42 - 3.85 11.46 15.47 - A6.4 - 501.8 17 43.51 - 56.54 29.70 - 76.07 - 31.7 . 34.5 18 - 29.68 36.75 - 15.47 40^44 - 47.9 10.0 19 • - 36.76 18.52 - 40.44. • 24.33 10.0 31.4 20 - 5.27 - 7.72 - 9.82 - 0.96 86.3 - 87.6. 21 - 6.95 U.13 - 16.08 2.15 . 131.4 - 47.9 22 37.01 - 5.97 48.40 - 1.92 30.8 - 67.8 Table 4. Bending Moments of Cantilever Diagrid I 0 NUMBER 0 6 6 3 P R I N T E D FOR S. T E Z C A N * ON MAR. 32 AT 0 HR. 38.9 MIN. / C A N T I L E V E R D I A G R I D FOR T H E S I S (TORSIONAL CASE) E X E C U T E FORTRAN PROGRAM. 2D 3 1 1 3 UNIT S I Z E MEMBERS UNCOMMON MEMBERS !A 2 3A 6 8 U 12 13 .8 12 10 10 X A A N D Y C O O R D I N A T E S OF T H E J O I N T S 1 C C C C 2 C O 20.0 3 5 ' 5.CT " T O .'0 ""6 '0. C" — C VC 7 9 15.C 1C.0 HI 15.G 30 ,0A i l 13 20.C UC.O 0.0 UC.O h 5.0 -itJTC—20T0 fi 10-.-0-2C.0 0.0 '2 20.0 10.0 -UC-.C-20.0 U N I T 1, J0 = 2 1 U N I T 2. J0 = Y !2 NO . U 6 ••— 1 — 2 3 —4 5 7 8 —9-10 1 1 -12 E I 2 ^ ~—•• looo.oo 1000.00 1000.00 1000 v o o — 6 9 9 i i 7 1000.00 »OO-OTOCT— 1000.00 1000.00 1 C O O T o o — 100C.CC 1000.00 U N I T 3. J0 = 7 10 13 - I ' U — 15 16 "17 18 19 10 CO"." C O — 12 11 12 10 1000.00 I'OOCTOO-— 1000.00 1000.00 12 8 "TOCO" 1000 1000 CO -GJ .. — 2 CO VOO"' 200.00 200.00 2CCV00-7 3 5 200.00 —2-C-CrvGC-200.00 .200.00 200.00 200.00 200.00 2OCT0 f l -iC 8 13 200.00 200; 00" 200.00. 2C0.0G -—200 VOO -200.00 200.00 13 12 13 CODE N U M B E R S 1 C 4 7 4 0 7 4 7 1 G T 3 ~ I 4 A ~1T 10 10 "~G~ 7 1 6 T 6 ~ 1 3 0 7 W 1 0 2 5 J O I N T L O A D S —0" 19 2 5 " 2 2 " 2 8 O F 0 ~2~ 5 2 _ 0_ i 1 1 1 17 "210 2 0 U N I T 2 ~5~ 0 b T 7 " 1 4 0 2 0 2 6 2 3 " 2 9 T 6 3 -~cr 1 2 1 2 no"" 9 18 3 ~ 6 " 0 V T B ~ 15 0 - 0 " 2 1 2 7 • 2 T ~ 2 " 4 ~ 2 7 3 0 l 20 21 "22" 2 ~ S " 8 2 - 5 -8 1 1 2 5 1 0 0 0 . 0 0 1 0 0 0 . 0 0 " 1 0 0 0 T C 0 " 2 0 0 . 0 0 2 0 0 . 0 0 " 2 0 C T C C T 0 — r o i -1~6-7 13 4 Q -0 10 0 0 " 2 " 0 2 '"r 8 1 4 5 -o-0 i 1 ~o-i i 0 0 - 3 -0 3 1 - 8 " o 15 u 0 12. - Q -1 2 3 _ 6 _ 3 ~ 6 ~ 9 1 2 4 4 -t-3-7 1 6 0 —0-13 0 n 5 _ 2 _ 5 8 17 0 0 1 Ji 17 0 0 1 0 1 9 0 2 5 T 9 " " 2 8 " 1 1 0 "2'0" 2 0 2 6 " 2 9 ' 12 0 2 1 21 2 7 " 3 0 " 3 6 -0-13 2 2 2 2 ~ 2 2 2 5 " 2 8 -14 2 3 "2 3 -2 3 2 6 2 V L O A D I N G N O . 0 . 0 0 0 f: P n r : 0 . 0 0 0 0.000 0 . 0 0 0 0 . 0 0 0 6 . _ 3 _ 6 •15---9 18 1 5 2 4 - 2 ' J I ~ 15 " T 8 -0 0 2 4 2 7 • ' 3 0 -ON ro J O I N T L O A D S O F U N I T 2 L O A D I N G N O . 1 c.crocr 0.000 TOTO'0'0" 0.000 J O I N T L O A D S O F U N I T 3 0 . C O C T O . O C O "OTCuO-C . 0 0 0 :0T0Tr0- trrocro- 0. oco "0T00T Y 12 J I 1.0 9 8 7 6 . 5 4 3 L O A D I N G N O . 1 0 . 0 C C G . 0 0 0 0 7 0 0 0 — - — O T O C C T J 01 N T D E F O R M A T I O N S 0 . 0 0 0 C . G O C 0 . 0 0 0 -OTCOO-0 . 0 0 0 0 . 0 0 0 O . O O C G . 0 0 0 U N I T N O . 3 T 0 X 0 ~ N 0 7 _ T X 19= - 1 . 6 2 6 6 1 4 3 0 X 2 0 = - 4 3 . 7 3 3 2 1 2 0 0 - K 2 1 ~ - — R T 7 2 6 Y 5 3 " X 2 2 = - 1 . 3 9 3 4 14 1 0 X 2 3 = - 2 4 . 6 1 8 3 6 7 0 0 • X 2 - = - . 6 5 2 6 _ _ 7 6 X 2 5 = - 1 . 0 C C 5 2 3 2 0 ~X2(r= - • 9 T 5 T C " 6 ' 7 - 0 T 0 " • X 2 7 - . • - . 6 0 8 0 1 . 5 5 6 X 2 8 - - 1 . 4 3 0 3 0 7 7 0 " ^ " 3 4 ~ r 9 - T 3 i r 5 8 C . C ~ -X2 9"F" X 3 C = - . 8 6 5 0 2 2 ' 5 U NI T~N0. L O A D 'NO. X 7 = - 1 . 9 0 7 0 5 9 8 0 _x__g _ _ —5-2-7 7"6"9i*5"0"0" X 9 = - . 8 7 7 4 7 7 8 4 X 1 0 = ' - 1 . 6 8 . 7 3 8 0 0 ~XTT_ = T 0 T 8 T Q 5'5'2"C 0~ X I 2 = - . 5 9 0 7 7 3 18 X 1 3 = - 1 . 2 2 8 8 8 C 8 0 - X T 1 T - - - = h - 3 T 7 7 3 5 4 2 0 0 " X 1 5 = - . 3 4 7 6 9 4 0 3 X I 6 = - . 6 1 3 9 2 2 4 9 U J " X T 7 = = _ T 2 ' n " 9 ' 9 97 0~ X 1 8 = - . l b 0 6 1 5 7 4 12 J l 10 9 8 7 6 , 5 4 3 T T N T T -L O A D " N O V N O . X 1 = - X — 2 = -X 3 = X .= - X - 5 - -X 6 = - 1 . 4 5 i i 5 . 9 3 0 - - i - 6 v 5 8 i 7 8 - 3 0 0 -- . 0 9 6 2 9 8 5 8 - . 8 1 8 3 3 5 8 9 — 4 T 3 9-1-6-65 V0-- . 0 9 6 0 4 9 8 6 "FrN7 ._ '~STRFSS~"R :ESUrTAMTS F I R S T . L I N E I S W R T C O M M O N , S E C O N D M T M 2 L I N E I S W R T M E M B E R A X E S ' ;Z 1 Z 2 (K.ft) U N I T NO T W O " . 0 1 1 (K.ft) (Kips) (kips) Y 1 (K.ft) 1 0 3 . 2 7 I 1 0 3 . 2 7 1 - 4 2 . 1 8 3 . - 4 2 . 1 8 3 3 . 0 5 4 3 . 0 5 4 - 3 . 0 5 4 - 3 . 0 5 4 . 9 6 2 . 9 6 2 Y 2 (K.ft) - . 9 6 2 - . 0 6 2 ro o- o ."O cc- o- o r- o p • i • • CO cc 1 <V in ro ro CM bo o o ro I rO rO o lc -d- la-in ;o r> i n u-i u. ..ro CM O CM m CO CM • o o CM O C\i o in in CM I I -3 £ in i n r— f- _ J _r ro ro - f-•c }c in, in i i CNJ CNJ _> CE co —• •o CM i C\i J l •c ro •c i_t I rO ro o _n — cc •c o * L* ;n a. CM i i n in — cr-oc o o m - i i o o •c _ o o CM CM _ jo CM CM I jl - X) f. CM r-o I ro in c ro ro ro I r- f-i ii OC CD •c rc-o o o o ro ro p o —. r~ CM -o o _ _r i — o <-> < _ o _> - J - 64 -O ro . O O c in CM — r^ r— . cc ro CM —• CO CM in ro o o ro in - -o ro lc O CM O it Wi CM J o-— I I -C ro •C ir, o r— i Ii _t o CM r — ro — r-. i r— CM -C CM oo ro ro o- 0 if-1 i f rn ro j i r CM "M I X> X> -c <J a.' cc-ro o CM CM I ro r-CM i m m in in it r- r--t o -r in . I . r- fM CM |l I i O p o b CM CM O «c |o ro CM O CM ro i t CM CM CM I ro ro 00 CO a b-CM - s o fc LT, ro -t ft i i m m m i!n' — i Js-CM CM ^t J CM CM I I G O o b • i -CM CM co i i ro ro in, i n o o r- Is-x co i iO ro m do o o I cc o cc: ro, •c in, m o ro cc ro o • cc o ro i c CM V) •- -C CM O-I — r- ro r ^ - IS-VI • cc CM I o I O CM m o o r-n r— ro r-o o o o X. 00 «o cc 00 o o I o c O ro l>- cc m, in, r-i M n •O CM •o o> ro b-C> CM in p ro I tr — — o o — ro, X CM C-X CC CM CM CM I >o CO' ro oo 00 CM I o o CO X — T CM CM o c~ X CC CM CM I ro h .-> in rO tl-I CM O in b--=»• c ro CM CM cn oo N IO i n v fM - ro 13 -16.177 - 1 6. 2 i 8 -3.; 14 . . . . . . . ( u > , 1 2 -14.381 17.837 15 13.687 6.419 2 15 12.205 10.275 . 2.1 16 • 23.812 -8.096 IT' 16 2V.417 -3.847 !.! 17 4 3.507 -56.5uh - • i 17 4 3.507 -56.544 —. 1 - 18 -33.956 ' 27.635— 18 36.749 -29.682 • ' 19 - 15.795 32. 105 — V • 1 19 18.524 -36.758 -1 20 -3.9_6 3.9 26 • - ! . 2C -5.275 -7.722 • ~ ! • • 21 -4.568 2.04 b-21 -6.954 4.133 • 22 rw.ei'6 -5.972" _ r ; . , 22 37.006 -5.972 , l . 7 3 W 3. 239 "73'0"9~ -.309 -2 V O I Q ~1T57T -1.57 1 T T S S T .651 -~r6"3"2~ -.632 -rv6_o-i . 630 - n - 2 9 o -I . 299 T2-"2r .252 '--r.-5"5-i--1 .55 1 PROGRAM CAME TO NORMAL END ALL DATA CARDS WERE READ BY. THE PROGRAM -5.6C8 4 . 1 1 8 —0-8-2-6.. 195 1 G . it 1 1 . "6T0'8"0" 6.080 ni}T2*o_--2.429 7 F 9 T 8"3tr 1.728 —5-7-27-3--3.9.26 -6T4-1-3-3.693 -2T570-2.570 5.608 ~~1-T767n~ - 4 . H 9 -10. 136--6.195 75-9-r - 1 0 . 4 1 1 ~6T080~ -6 .080 •117-1 cn-2.429 -17.985--1 .728 -7T-722" 3.926 ---5-7-15-2--3.69 3 —2V5T0— -2.570 0 END OF THIS RUN" AT 1 HR . -14.7 MIN. - 66 -c) F o u r G i r d e r Skew B r i d g e 3 The g r i d shown i n F i g . 25 has b e e n t a k e n f r o m H e n d r y e J a e g e r , page 43, a n d a n a l y z e d b y t h e s t i f f n e s s a p p r o a c h . T h e f i n a l b e n d i n g moments o f a l l t h e members a r e shown i n F i g . 25. I n a d d i t i o n , t h e b e n d i n g moment d i a g r a m s o f g i r d e r s 1 and 2 a n d c r o s s g i r d e r GG, f u r n i s h e d b y H e n d r y e J a e g e r a r e compared w i t h t h o s e o f t h e s t i f f n e s s a n a l y s i s i n F i g . 26. F i g . 25 Skew B r i d g e f r o m H e n d r y e J a e g e r - 67 -c tt> e o c 300 200 272.6 Harmonic — Analysis Stiffness Analysis ® 100 CD 30 40 50 -> S p a n ( f t ) 266.65 260.0 ^ 297.97 Stiffness Analysis 10 20 30 40 50 -> S p a n ( f t ) c a> E o c •5 c 0) CO 10-h -10-t - f t 16. 18 -St i f fness Analysis -Harmonic Analysis L C R O S S G I R D E R G G 10.04 S p a n ( f t ) Fig. 26 Comparative results of stiffness and harmonic analysis - 68 -CHAPTER 5. SPECIAL TOPICS 1. Reduction due to Symmetry In view of the fact that the computer time required for the solution of linear simultaneous equations,or the inversion of matrices, is approxi-mately proportional to the cube of the number of unknowns, any reduction that is possible in their sizes is naturally desirable. For grids with one way symmetry, the number of equations can be halved as follows. Note that .. the loading need not be symmetrical, because any unsymmetrical irregular loading may be considered as the sum of a symmetrical and an antisymmetrical loading. If the numbering of the joint deformations of a symmetrical structure follows a symmetrical pattern similar to that of the structure, the column vectors of joint loads and deformations can be divided into two equal por-tions. In case of symmetrical loading, the deformations and joint loads of the two portions are identical. Applying matrix partitioning to Eq. 11 this may be expressed as p p It follows that {p} - [ K n + K i 2 ] {D) = M s W . where £K ] 3 is the effective stiffness matrix of a symmetrically loaded grid. On the other hand, for anti-symmetrical loading, although the deforma-tions and the joint loads of the two portions are equal, they are opposite hi | h_2 K21 ! K22 (44) - 69 i n sign; Therefore,. Eq. 11 becomes P - P _K13. I K12_ K21 ! K22 j» - D (46) This gives ••. W " [ K i i - "12] {"'}. • HaW ;•• in which [K] denotes the effective stiffness matrix of an anti-symmetri-cally loaded grid. Axis of ' _ / v io Symmetry 9 (a ) (b ) Fig. 27 Reduction due to symmetry There i s no d i f f i c u l t y encountered i n applying Eqs. 45 e 47 to a structure such as shown i n Fig. 27 (a), i n which the joints are equally divided by the axis of symmetry. But, i f the axis of symmetry l i e s along a line of joints as i n Fig. 27 (b), the deformations and joint loads can-not be divided equally. It i s possible, however, to make the latter case - 70 -similar to the former by considering the members along the axis of symmetry to be dummy members as they remain straight upon the application of symmetri-cal or anti-symmetrical loads and, therefore, do not contribute to the . strength of the structure. 2. Thermal Effects The effect of temperature changes on a structure may be taken into account by the introduction of f i c t i t i o u s forces acting on the individual members. These forces are equal to the forces required to deform the mem-bers by an amount equal to that which would be caused by the known tempera-ture change. For a uniform temperature rise At, the elongation of a member is ' AL = a t At L (48) where is the coefficient of thermal expansion of the material and L i s the.length of the member. The axial force required to produce the same elongation i s P = f AL. (49) Substituting for AL from Eq. 48, the f i c t i t i o u s force to replace the effect of the temperature change becomes P = ( cct At) AE (50) Obviously, the force P i s tensile for a temperature rise and compressive for a temperature drop. In case the top and bottom faces of a member are subject to unequal temperature chsnges, At^ and At^ as shown in Fig. 28, the f i c t i t i o u s loads consist of an axial force and a moment. Assuming a linear variation of temperature between the two faces, the elongation of the centre l i n e , from Eq. 48, is - 71 -At + At AL - -«t 1 2 L Consequently, the fictitious axial force is obtained as At.. + At P - a t —^-2 AS (51) (52) ^ • « f ( A t 2 - At, )L -Fig. 28 Non-uniform temperature change The rotation 6 of the section due to the unequal length changes of the top and bottom fibres is a t L (At 2 - At x) e (53) where d is the distance between the faces. The fictitious end moment required to develop the same rotation is obtained from the fundamental expression for flexural strain given by 1 • M_ R " E I (54) - 72 -in which R = | (55) Substituting Eqs. 53 and 55 into Eq. 54, /At - At , M = \L-i-a U E I (56) To simulate the effect of thermal changes, the structure is considered to be loaded by a set of joint loads, which are calculated from the sum of the fictitious member forces given by either, Eq. 50 in the case of uniform, or, Eqs. 52 and 56 in the case of non-uniform temperature change. The rest of the analysis remains' the same as described in the previous chapters. However, note that, as the assumed loads are fictitious, these must be sub-tracted from the calculated stress resultants of the members concerned in order to obtain the actual thermal stresses. . 3. Support Settlements A statically indeterminate structure subject to support settlements may be analyzed by specifying additional joint deformations along the direc-tions of the settlements and enlarging the stiffness matrix of the system to include the rows and columns corresponding to the newly introduced defor-mations. The enlarged stiffness matrix is then multiplied by the column vector of known displacements i.e. the settlements of the structure, to yield a column vector of external loads. These are, by definition, the forces required at the joints to prevent the rest of the structure from de-forming when the support settlements take place. Therefore, to simulate the actual deformed shape of the structure, that is to obtain the final joint deformations due to support settlements, the calculated external loads are reversed in sign to yield the fictitious joint loads. The analysis then - 73 -proceeds in the usual manner, as for an ordinary structure under external loads. However, i t is to be noted that the support settlements are consi-dered as part of the final end deformations of the members concerned. 4. Solution of Extremely Large Structures by Band Matrices Large numbers of equations can be solved within the limited core memory capacity of digital computers, i f the coefficient matrix is obtained in a band form. This is easily achieved by adopting a regular pattern for num-bering the joint deformations. The width of the band is entirely dependent on the number of joints in the transverse direction of the structure. Therefore, in order to obtain the narrowest possible band width, the num-bering of deformations should start from one end of the frame and proceed consecutively along the longer dimension of the structure, covering the f u l l width at each stage, so that the difference between any two adjacent- defor-mation numbers is kept minimum. • When a structure is divided into substructures by transverse planes across it's width, the corresponding stiffness matrix is also divided into sub-matrices, such that for each substructure considered there is a main block located along the diagonal;and two side blocks on either side consis-ting of terms contributed by members common with the adjacent substructures. Therefore, the .stiffness matrix of any structure divided into substructures is of a triple block band matrix form. The mathematical solution of such a matrix can be achieved by applying Gaussean elimination- and matrix parti-tioning as described below. -• A group of unknowns can be eliminated by means of matrix partitioning from a set of equations, represented by Eq. 57, as follows: (57) - 74 -From Eq; 57, K l l 3 1 + K 1 2 V"" P l K 2 1 D 1 + K 2 2 D 2 " P2 Solving for D.^  from Eq. 58, "D 1 = ^ l 1 P l " ^ l 1 K12 D2 (58) (59) (60) and, substituting into Eq. 59 K22 D2 " P2 where K22 " ( K22 K21 ^ l 1 K12 ) ( P2 " K21 hi Pl> (61) (62) (63) The same process of elimination may be applied successively to large band matrices. This i s exemplified below for a structure divided into four units: (64) K l l l K21 ! K12 0 0 h i K23 0 — \ 0 .[ K32 *33 K34 D3 0 j 0 K43 K44 D4 After eliminating ' using Eq. 61, the above equation becomes _ K 2 2 _ | _ K 2 3 _ ° _ -p.-K32 ! K33 K34 — ~D7 • V \ (65) i n which and are the same as i n Eqs. 62 and 63, respectively. Next , eliminating from Eq. 65 i n a similar manner, - 75 -I -L-34 -c ~ D3 * j L J where * * - l N K33 " ( K33 " K32 K22 K23 } 3 * - (P - K K _ 1 P*) 3 ^3 32 22 V Finally, eliminating from Eq. 66, 44 K i n which K 44 ( K44 " K43 Sf K34 } P * = (P| . K j , 4 ; 1 P*)' 4 43 33 3 Solving the last group of unknowns from Eq. 69, (66) (67) (63) (69) (70) (71) (72) These values are now back substituted step by step i n accordance with Eq. 60 to obtain the remaining deformations of the- structure. This gives (73) N * K l ] " 1 (P3 " K34 \ ] {D2) - [4]" 1 K " D 3) l D l ] ' f n ] " 1 ( p l " K12 D2} (74) (75) The chief advantage of the above technique for the solution of large numbers of unknowns, is that i t i s extremely convenient for computer appli-cation, because the calculations follow a uniform pattern and only the - 76 -(a) standard expressions of Eqs. 60 and 61 are repeatedly employed. (a) A computer programme for the "Solution of large capacity band matrices" using triple block matrices, developed by S.S. Tezcan is available through the IBM 1620 library, Computing Centre, The University of British Columbia; Programme No. (C-FTN-P) M6-3. - 77 -CHAPTER 6. HARMONIC ANALYSIS 1. B r i e f Outline Hendry e Jaeger's method of harmonic analysis i s based on the assump-t i o n of a continuous transverse spread medium, i n which the t o r s i o n a l r i g i -d i t i e s of the transversals are neglected. For any given loading w = w (x) applied on a simple beam, Hendry e Jaeger express the load i n terms of sine functions as follows, . TTX s i n 2rr x s i n n u x w = wn s i n •=— + w,, —= + — — + w — ; — + 1 L 2 L n L i n =oo w = X w n s i n ^ - p (76) n = 1 Then, by successive integration, the shear force F, the bending moment M , the slope 6 and the deflection y, are obtained as n =<x> w T . L v— n n TT x , __s TT 2 _ - T C 0 S ~ L ~ ~ ( ? 7 ) n = 1 T 2 n = oo w L ^ — n . n TT x /r7rt-\ TT n = 1 n T 3 n = co v; Q L "v— n n TT x / e - — j X - C O S L ( 7 9 ) EI TT n = 1 n 4 n = w „ L x— n . n TT x , ar.^ EI TT . n = 1 n The constants of integration are zero at each stage because of the assumed hinged ends. Applying Fourier series to Eq. 76, the coefficients w^ , w^  W r are - 78 -giveri by 2 w = — n L w s i n " ^ x dx (81) J F o r a u n i f o r m l o a d w p e r u n i t l e n g t h y E q . 81 r e d u c e s t o w » 0 i f n i s e v e n n , w = — i f n i s o d d n TT S u b s t i t u t i n g t h e s e v a l u e s i n t o ' E q . 7 6 , i+W TTX 1 . 3 TT X /rt„v w = — s i n — + s i n — = — + — ; — . (82) F o r a p o i n t l o a d W a t a d i s t a n c e b ' f r o m t h e l e f t - h . m d s u p p o r t , t h e c o r r e s p o n d i n g h a r m o n i c s e r i e s f o r t h e l o a d i s 2W TT b . TT X 2 TT b . 2 TT X /rt_\ w = — s m -j— s i n - y - + s i n — ^ — s i n — - — + ( 8 3 ; Once t h e r e l e v a n t l o a d s e r i e s i s d e t e r m i n e d , t h e s h e a r , moment, s l o p e a n d d e f l e c t i o n c u r v e s a r e r e a d i l y computed u s i n g E q s . 7 7 - 8 0 . F o r t h e a n a l y s i s o f s i n g l e s p a n i n t e r c o n n e c t e d b r i d g e g i r d e r s w i t h more t h a n one l o n g i t u d i n a l , H e n d r y e Jaeger, e v a l u a t e d i s t r i b u t i o n c o e f f i c i e n t s o f t h e f i r s t a n d h i g h e r h a r m o n i c s o f t h e f r e e d e f l e c t i o n c u r v e o f one o f t h e l o n g i t u d i n a l s , a s s u m i n g t h a t i t c a r r i e s t h e e n t i r e l o a d i n g . b y i t s e l f . T h e s e d i s t r i b u t i o n c o e f f i c i e n t s a r e o b t a i n e d b y c o n s i d e r i n g t h e r e l a t i v e d e f l e c -t i o n s o f t h e ' l o n g i t u d i n a l s i n t h e t r a n s v e r s e d i r e c t i o n , i n t e r m s o f t h e . d i m e n s i o n l e s s g r i d p a r a m e t e r a , w h i c h r e p r e s e n t s t h e s t i f f n e s s o f t h e f r a m e . 4 V h J E I TT where, ELp flexural r i g i d i t y of a transversal, EI = flexural r i g i d i t y of a longitudinal, n = number of cross girders, h =.. spacing of the longitudinals. The distribution factors derived to distribute the free deflection .curve are assumed to be equally valid for distributing the loads, shears and bending moments. Further, the same distribution factor is used for the whole length of a longitudinal, irrespective of the position of the load. This i s not really the case and the accuracy of these factors i s limited, as i s amply illustrated by the examples i n the following section. For grids with 2 - 6 longitudinals, Hendry e Jaeger provide the values of the distribution factor p , for zero (p = 0 ) and f u l l torsion, ( 0 = oo) The torsional parameter, p characterizes the torsional r i g i d i t y of the grid and i s given by in which GJ^ i s the torsional r i g i d i t y of the longitudinals. For inter-mediate values of (3, Hendry e Jaeger recommend the following interpolation function for the distribution factors, To evaluate the transversal bending moments, the deflected shape of the longitudinals and, hence, the relative deflections of. the ends of the transversals are determined. Slope deflection equations then yield the required moments. The transverse moment expressions, i n terms of the longi-tudinal bending moment distribution coefficients and the free bending (85) moment curve, are tabulated i n Appendix IC. In order to demonstrate the method of harmonic analysis and to compare the results with those of the stiffness approach, a ^ o girder bridge is completely analyzed below. 2. Numerical Example - Two Girder Bridge O © le " « >| Fig. 29 Two girder bridge Step 1 Using an average number of transversals i.e. n = less parameters are: 8, the dimension-- 81 -, EI. • y\ outer 1 ., n I a •-. = csr- • 1.0 . > inner hl^ Step 2 By Eq. 86, the interpolation function for |3 = 1.25 i s ( P „ - P 0 ) A2* • / ^ 7 7 _ ° V 3 .+ 1.25 73715 0(1.25) = p o + — "<>' V 3 * 1.25 /3T154 p(1.25) = po + (p*o - P 0 ) °-651 Step 3 Referring to Appendix I-B Pg. 252, the f i r s t harmonic bending moment distribution factors for a load on girder (l) are obtained as follows: For (3 = o, = 1.0 p21 ' 0 , ct For 0 =00, p n 2 = 0.728 1 1 1 + 2a o a n ° 0.272 '21 1 + 2a o where, CZQ = a ( l - - ^ ) = 3.154(1--^) 0.597 TT TT Interpolating for p = 1.25, the required distribution factors become P 1.0 + (0.728 - 1.0) 0.651 - 0.822 11 P 2 1 = 0. V (0.272 - 0) . 0.651 -.0.178 Step 4 The simple bending moment diagram for the grid treated as a single girder is shown i n Fig. 30. Step 5 The longitudinal bending moments are obtained simply by distributing the simple bending diagram i n proportion with the calculated distribution factors. For example the bending moment at the load point, M-^ Q* i s Fig, 3 0 Simple bending moment diagram Step 6 From Appendix I-C pg. 260, the transversal bending moments are K12 = M21 = 1 , 7 9 "^ 2 ( p l " p2) M = » M ( 8 7 ) per unit length of the transverse medium. The distribution factor u- intro-duced here w i l l be referred to as the transverse bending moment coefficient. Substituting the appropriate numerical values into Eq. 87, the required transversal bending moments are obtained as M 1 9 - M0, - 1.79 x - ^ 5 x 3.154 ( 0.822 - 0.178) M - 0.0091 M ^ 100 Hence, for the loaded transversal K c c - 0.0091 x 234.37 x 12.5 - 26.65 k.ft The remaining transversal bending moments are calculated i n lik e fashion, and the complete results of harmonic analysis, as well as, the corresponding stiffness solution are shown i n Fig. 31. In addition, the complete IBM-7090 computer output of the above example for several loading conditions is also presented i n the following pages. -.83..-TWO BE A M GRIP USE D. JO. .CHECK...HE ND.R Y.S ..Dl ST.RI B UTION. FAC TOR S i 2 7 TJ 23 46 32 22 36 1 COOE_NUUB_ERS, ICD( I ] : 1 i 2 0 33 0 3 6__ 8_35__36_ 7_ 9 12 14 38 39 13 15 20 40 41 19 21 4 7 io_ 13 16 18_ 24 30 26 43 44 25 27 32 46 0 31 0 2 _5 8 U_ 14 17 2 _ 8 14 20 26 3 4 33 34 10 36 37 16 39 0 22_41 42_ 28 44 45 19 33 40 3 9 "l 5 2.1_ 5 11 0 23 27 29 2 18 3 6 9 12_ 15 18 4 10 17 22 28 5 6 12 18 24_ 34 35 37 38 0 40 42 43 5 11 0 23 30 45 46 29 21 34 41 4 7 13 19 2 5_ 31 20 19 22 7 13 23 29 35 42 38 45 6 12 22 28 20 23 LENGTHS OF THE MEMBERS 12.5 12,. 5 12.5 12.5 25.0 12. 25, 5 0_ EI VALUES OF THE 10000.0 10000.0 12.5 25.0 9 .15. 12.5 12.5 25.0 25 36 43 31 39 46 8 24 14 30 21 11 27 37 44 1.0 26 12._5_ 12.5 25.0 12.5 12.5 J_5_-fi_ 12. 5_ 12.5 25.0 12. 5_ 12.5 10000.0 ...500.-0 GJ VALUES 35_50._0__ MEMBERS 10000.0 10000.0 500.0 10000.0 500.0 OF THE MEMBERS 3_5 5_Q .0 3550.0 JL 0000.0 J.0000.0 10000.0 10000.0 500.0 500_.0 10000. 0_1_0.000.0 _ipooo.. 0_ 10000.0 10000.0 10000.0 500.0 _ 500.0.._ _3_5.50_._0 3_5_5.0.._0_ 3550.0 167.5 3550.0 16_7_. 5 3550.0 167-_5_ 3550.0 167,5 3550.0 __16.7_,5_ 3.55_0_._0___3_5_5_0.._0 3.5.5i)_._0_ 3550.0 3550.0 3550.0 __167_.5 16X..5 INVERSE OF 33 SG FROM NS+1. (DEFLECTION INFLUENCE SURFACE) 0.2962 0.1025 0.4732 0.1860 0.5363 .0.2368 0.5159 .0_._24_9_.1_ 0.4350 0_2_2_V2_ 0.3122 _0_._1_6.8Q_ 0.1626 _0_._0.8_9_I_ 34 _0.._4_73_2_ 0.1860 ._0_._8_3.21_ 0.3398 _0_._9.8.8A_ 0.4358 _0_._97_0_7_ 0.v4616 _0_._82X6_. 0.4176 .0.5974. 0.3141 _0._3.122_ 0.1680 35 .0. 0, 5363 2368 0.9884 0.4358 1.2663 0.5648 1.2998 0.6046 1.1329 _0___5_5_1_7_ 0.8276 _Q_.4_176_ 0.4350 _0_._22A2 36 _0_.-5_l.5_9_ 0.2491 _0.._9_7_0_7_ 0.4616 _1...2_9.9.8_ 0.6046 1.4284 0.6550 .1.2998 0.6046 _0._9_7_0.7_ 0.4616 _0..515.9_ 0.2491 _3_7_ 0.4350 _0_.__2it.2_ 0.8276 J1.A1J_6_. 1. 1329 Jl.33J.-7_ 38 ..3.1.2.2_ 1680 __0.._59_74_ 0.3141 -0_._82_76_ 0.4176 1.2998 _0.^ 6_0J4_6_ _Q._9_7.0_7_ 0.4616 1.2663 _0_.3.6it_8_ 0.9884 _0_.A33_8_ 0.5363 _0_._23_6.8_ _0_._98.8-4_ 0.4358 _0._8.32i_ 0.3398 .0.4132. 0. 1860 _3_9_ 0.1626 _a._0_83_Z_ 0.3122 JL._L6.8JL 0.4350 _0.._2.2A2_ 0.5159 _0_.^ 4_9_1_ 0.5363 _0_..23_6_8_ 0.4732 _0_..L8.6.Q_ 0.2962 _0_.J.D.2_5_ 40 0.1025 6.2962 0.J860 0.4732 10.2368 0.5363 _0._2_4.91_ 0.5159 0._2242 0.4350 _0_._1680_ 0.3122 J)___089_7_ 0.1626 _41_ 0.1860 _0_._4_13.2_ 0.3398 J)_._83.2_l_ 0.4358 _Q_...9.8.8_4_ 0.4616 _0.._9_7_0_7_ 0.4176 _0_._8.2_7_6_ 0.3141., .0_..5__9_7j4__ 0.1680 _0_..3_122_ 42 .0.2368.. 0.5363 .0_._4.35.8_ 0.9884 0.5648. 1.2663 _0-..6046_ 1.2998 .0.35.17.. 1.1329 .0.4176-0.8276 _0.2242_ 0.4350 _43_ 0.2491 0.5159 0.4616 0.9707 0.6046 1.2998 0.6550 1.4284 0.6046 1.2998. 0.4616 0.9707 0.2491 _Q.,5_1.5_9_ 44 0.2242 0.4176 0.5517 0.6046 0.5648 0.4358 .0.2 368 .-84...-0.4350 0.3276 1.1329 1. 2998 1.2663 0.9884 0.5363 45 0.1680 0.3141 0.4176 0. 4616 0.4358 0.3398 0. 1860 46 0.3122 0.5974 0.8276 0. 9707 0.9884 0.8321 0.4732 0.0897 0.1626 0.1680 0.3122 0.2242 0.4350 0. 0. 2491 5159 0.2368 0.5363 0.1860 0.4732 0.1025 0.2962 FINAL END MOM 1 -0.000 ENTS FOR -96.915 10K (IN k.ft) . -0.00035 2 96. 112 -72.079 -0.00028" 3 5 71.835 38.288 -52.805 -26.396 -0.00015 0.00003 4 6 52.884 26.732 -38.036 -16.708 -0.00010 0.00016 7 9 17.080 -0.000 -8.111 -12.461 0.00018" -0.00006 8 10 8.498 13.264 0.000 -21.673 0.00009 -0.00007 LOAD AT 11 13 21.917 24.214 -25.323 -20.482 0.00002 -0.00000 12 '• 14 25.243 20.146 -24.467 -14.544 -0.00000 0.00004 : IA 15 17 14.171 4.052 -7.515 4.052 0.00004 0.00000 16 18 7.128 5.004 0.000 5.004 0.00003 0.00000 19 21 ,4.182 1.869 4. 182 1.869 -0.00000 0.00000 2 0 22 2.956 1.055 2.956 1.055 0.00000 -0.00000 23 1 0.471 -0.000 0.471 -72.954 -0.00060 2 71.560 -149.529 -0.00060*1 3 5 148.809 79.543 -109.996 -54.697 -0.00039 0.00019 4 6 110.004 55.321 -79.128 -34.474 -0.00020 0.00033 7 9 35.195 -0.000 -16.684 -20.799 0.00038 -0.00011 8 10 17.446 22.192 0.000 -37.975 0.00018 -0.00015 11 13 38.694 45.462 -46,259 -39.057 0.00001 0.00003 12 14 46.251 38.434 -45.877 -28.029 -0.00000 . 0.00008 , LOAD AT . IB 15 17 27.309 5.016 -14.568 5.016 0.00006 - 0 . 16 18 13.805 8.218 0.000 8.218 0.00007 0.00000 19 21 7.938 3.999 7.938 3.999 -0.00000 -0.00000 20 22 6.032 2.336 6.032 2.336 0.00000 0.00000 23 15 17 19 21 23 15 17 1.064 1.064 -0.000 -54.014 -0.00066 52.491 -110.709 -0.00066 3 109.572 -175.739 -0.00039 5 126.957 -87.144 0.00010 4 6 7 5 5.6 30 -26.317 0.00048 8~ 9 -0 . 000 -24. 114_^0_. 000 15 10_ 11 46.682 -58.642 0.00003" 12 13 60.550 -53.487 0.'00002 14 175.356 -126.578 -0.00023 _8 7.J9 42 -54 .619 0.00030 ' 27.425 0.000 0.00021 _25.636 - ± 5 . 54 6^0.00015 " 59.025 -60.929 -0.00005 52.689 -39.135 0.00007 38.124 -20.560 0.00008 16 4.204 4.204 0.00000 18 10.055 10.055 0.00000 20 _6 ._491 6. 49 UzO.OOOOO, __22 1. 887 1 ."887  19.452 _7.950 8. 967 4.010 0.000 7.950 8.967 4.010 0.00009 :0.00000 0.00000 0.00000 -0.000 -39.250 -0.00063 37.865 -80.095 -0.00061"! 3 78.912 -127.191 -0.00057 4 5 183.674 -126.425 0.00017 6 7 80.095 -37.864 0.00057 . 8 9 -0.000 -23.252 -0.00015 10 11 46.092 -60.316 -0.00001 12 13 66.333 -61.081 0 « 0 0 0 0 3 14 126.425 -183.675 -0.00031 12 7. 19 0 -78.912 0.000 34 - 39.2 50 0.000 6.000 31 24.638 -44.910 -0.00021 7 61.081 -66.333 6.00000 60.316 -46.092 0.00009 44.9 10 2.980 -24.637 2.980 0.00009 0.00000 19 8.972 8.972 -0.00000 21 8.972 8.972 -0.00000 16 20 22 23.252 6.049 0.000 6.049 10.503 6.049 10.503 6.049 0.00011 0.00000_ 6.00000 0.00001 23 2.980 • 2.980 UNIFORM LOADS ON THE MEMBERS LOAD AT — 1 C — -LOAD-A-T-1D 1.0 0. 1.0 0. 1.0 0. 1.0 0. 1.0 0. 1.0 0. 1.0 0. 1.0 0. — 8 5 — 0. _Q._ _Q... .0 . . SET OF LOADING NO. 1 D E F O R M A TLPN S OF THE SYSTE,M TIMES 2 . 8 6 2 1 2 2 . 5 9 9 9 _ 1 .0296 7 - 1 . 2 8 8 6 . - 1 . 2 8 8 6 12 - 1 . 9 3 9 4 - 2 . 8 6 2 1 17 1 .3047 1 6 " l i " 16_ 21 - 0 . 9 9 1 7 22 0 . 5 0 0 4 26 - 0 . 5 0 0 4 27 - 1 . . 2 8 8 6 3 8 13" 18_ 2 3 28 31 41 46 - 0 , 88, 29, _15. - 0 . 5401 6115. 4153 9097 Ml 002 -32 - 1 . 3 0 4 7 33 _ 3 7 _ 8 J . _ 1 1 2 8 _ 38 42 3 8 . 4 3 5 7 43 M2 CHECK 3 9 3 . 4 9 5 - 0 . 0 0 2 9 9 0 . 0. LA INCHES) - 0 . 5 4 0 1 4 - 0 . 0 0 0 0 9_ - 0 . 9 9 1 7 14 1 .2089 19 Q._ - 1 . 2 8 8 6 - 0 . 9 2 5 3 24 29 1 .9394 - 1 . 3 9 1 9 - 2 . 5 9 9 9 _ -0 .5401_ - 0 . 0 0 0 0 - 0 . 9 9 1 7 5 10 15 20 25 30 - 0 . 9 9 1 7 - 1 . 0 2 9 6 - 0 . 5 4 0 1 __0.9253_ - 1 . 3 9 1 9 - 1 . 2 0 8 9 3 4 . 6 4 9 4 34 6 3 . 3 6 1 2 _39_ 4 1 . 6 0 1 2 44 6 3 . 3 6 1 2 34 .6493_ 3 8 . 4 3 5 6 35 40 45 8 2 . 1 1 2 8 15.9.098 2 9 . 4 1 5 3 3 6 3 9 . 7 5 8 - 7 8 9 . 8 1 4 - 0 . 0 0 2 4 5 4 5___8 35_.082_-786.26 i r . 0 . 0 0 1 4 6 6 "7 6 4 6 . 5 5 1 - 3 8 4 . 1 7 0 6 . 0 0 3 2 5 8 9 - 0 . 0 0 0 _ - 1 5 3 . 3 9 8 0 .00019- _10_ 11 2 9 7 . 7 7 4 - 3 8 2 . 1 0 2 0 . 0 0 0 5 0 12 J 3 4 1 4 . 9 6 2 - 3 8 5 . 6 4 7 0 . 0 0 0 3 7 14 15 17. 19 _21_ 23 7 , 290 , __25_ 55. _ § 5 _ , 25 . 980 - 1 6 2 . 7 1 7 0 . 0 0 0 6 6 16 136 25_._136 - 0 . 0000.2 1JL 012 5 5 . 0 1 2 0 . 0 0 0 0 3 20 _0_1.2 5_5_..O1.2Lj_0..m0.Ol__2_2_ 136 2 5 . 1 3 6 0 Ml 3 8 4 . 1 7 2 - 6 4 6 7 8 6 . 2 6 6 - 8 3 5 789.8.14 j _ _ 3 9 3 9 3 . 4 9 3 0 162.718___-290. 3 8 5 . 6 4 9 - 4 1 4 3 8 2 . 1 0 2 - 2 9 7 1 5 3 . 3 9 8 0 _ 4_3_._9J33 43 5 8 . 6 2 6 _j43.,_9_32_ 58 _4.3. M2 CHECK . 5 5 3 - 0 . 0 0 1 3 7 . 0 8 5 - 0 . 0 0 3 1 8 . 7 5 3 . . _ 0 . 0 0 3 5 6 . 0 0 2 0 . 0 0 1 7 9 .J38.l___0.00.08 3 . 9 6 2 0 . 0 0 0 0 1 . 7 7 3 0 . 0 0 0 5 2 .001 0 . 0 0 0 7 9 ..933 0 . 0 00.00. . 6 2 6 - 0 . 0 0 0 0 3 .J93.2 0.._0_0_0O3. w • 1 k / f t on 1 - 86 -3. Comparison of Harmonic and Stiffness Analysis Distribution Factors Several example grids with 2 - 6 longitudinals at a moderate spacing were selected i n order to provide a relative idea of the results obtained by harmonic and stiffness analysis. In every case, seven transversals were used to simulate Hendry e Jaeger's assumption of a uniform transverse spread medium. Constant values of EL^ « 10,000. and EI_ = 500. units were adopted for the elastic r i g i d i t i e s of the longitudinals and transversals, respec-t i v e l y . . Each grid was analyzed f i r s t assuming the members to possess zero torsional r i g i d i t y (p = 0), and then assuming the torsional r i g i d i t y of each member to be of such a value that the harmonic analysis torsional parameter |3 equals 1.25, which almost corresponds to i n f i n i t e torsional r i g i d i t y . In order to cover a wide range of loading conditions, a point load and a uni-formly distributed load of unit intensity were non-concurrently applied at the joints and along the longitudinals respectively. In the following tables, the harmonic analysis longitudinal and trans-versal bending moment distribution coefficients p and |_, calculated as described i n the preceding numerical example, are tabulated against the cor-responding values of stiffness analysis. It i s clearly seen, that the harmonic analysis distribution factors are entirely independent of the location of the section and the point of application of the load along the longitudinals. This approximation gives rise to considerable inaccuracies, as the exact analysis provides different values for each different section and point of application of the load. These errors range from 7 10 to + 2QO% depending upon the arrangement of the girders and loads. STIFFNESS ANALYSIS 0 00 M I-I o O M E-i Load on Girder 1 2 00 O JM S i- l DER 8 CO Point ] Load at UDL S < < s PTD X •I25L •2.5L •2-75L • 5 L •I2.5L •88.5 1. <b> • 770 1. •&7& 1. 1. •25L •no i . •798 1. ?7o3 1. •613 1. •S30 I. •375L •hi& i . .703 1. •750 1..: •675 1 . •673 1. 1 •5U •609 1. •£3* 1. '(,75 1. •735 1. 1. 1 •G25L •563 1. •5SS 1. •£20 1. •675-1. •673 1. •822 1. • 7 5 L •535 1 . •551 1. 1. 1. 1. •S75L •512 1. •SbS 1. •56i 1. •6p9 1. •719 1. •I25L •Il5 O •250 0 •524 O •'i9l 0 •28| O •25 L •250 O . •202 O •297 O '167 O •3lO 0 •375 L •324 0 •297 O •250 O '325 0 •3i7 O 9 2 • S L •391 O •o£7 0 •525 O •265 O •331 O hi •17? •G25L 0 •4l5 O •580 O •325 O •317 O O •75L o • in 49 0 'U\5 O •367 O •3lO O •875L •Lf82 0 0 •A37 O '391 O •281 O f i j -2 GIRDERS DATA; EL^ - 10,000 units E I1 GJ, 500.0 units L 0 3,550. units GJ T = 177.5 units Table 5. DISTRIBUTION FACTORS FOR LONGITUDINALS (a) (a) The P^j values, when multiplied by the simple moments, yield the longitudinal moments. (b) The lower values are for zero torsion. ( 8 = 0 ) STIFFNESS ANALYSIS O CO 3, NO. o M EH Load on Girder 1 M H a co 0 >-t § 3 •< 2; 3, NO. Point Load at UDL Q CO •IZ5L •2 Si •375L •6 L EC < s X u x : LO^  •25L 0 >> 0 40-7 o O 3.7-5 O 1 •5L 37-8 O ' 38-6 o 38-3 o o 37-6 0 91.0 0 •75L 27.0 O 0 34.2 O ja-7 . o 37. S o •25L M _•_> 0 0 40-7 0 . 38.7 0 37.5 0 2 •5L 37<8 0 3g.£ o -8-3 o o 37.. O 9 i - o 0 •7SL -7.0 0 o 314-2 O 38-7 O 37-5 O Table 6. DISTRIBUTION FACTORS FOR TRANSVERSALSv; (a) The values, when multiplied by the simple moments, yield the transverse moments per unit length of the transverse medium. (b) The lower values are for zero torsion ({3 =0). The torsional values are given for the l e f t hand sides of interior supports. • o STIFFNESS ANALYSIS 40NIC LYSIS • O STIFFNESS ANALYSIS 0 CO S3 o M Load on Girder 1 40NIC LYSIS S3 £H K SECT 101 Load on Girder 2 1—1 H S CO O JH GIRDE CO Point Load at UDL < § / B K SECT 101 Point Load at UDL 63 a: < GIRDE X -I25L • 25 L • S 7 S L •5 L K SECT 101 •I25L • 2 5 L •375L •5 L •I25L - 95&b-•670 •9o 1. •566 •261 •4C2 •8 30 •612 •8 74 •I25L •134 •o?8 •252 •195 • 316 •277 •361 • 3 3 « * ^-/o •252 •2.5L • 6 7 0 • 904 •72 1 •314 • 5S8 •87^ •499 •81(2 •574 •865 •25 L •25"2 -195 -215 •172 •296 •252 •344 . 3 1 4 •238 •270 •3751 •543 •96 1 •598 •874 •654 •891 •554 •860 •553 •8 59 J? •375L •326 •277 •236 •252 •246 •214 •310 •278 •310 •282 0 J •5L •A 69 •830 •499 •81(2 •554 •860 •635 •88? •546 •S59 5.. •665 1 •5 L •361 •338 •31,4 •314 •310 •278 •255 •227 •314 •286 •287 •625 L •420 •81 0 •4:42 •820 •485 •S16 • 554 •860 •553 •853 •863 '625L •574 •380 •366 •361 •348 •328 •310 •278 •310 •28 2 •274 : 7 5 L •390 •797 •40 s •80S •442 •820 •499 •842 •574 •865 •75 U •7>76 •405 •374 •390 •366 • 3 6 / •344 •314; •2=38 •270 -— •<27£L •376 •790 •390 •767 • 4 2 0 •8 10 •469 •810 •612 •874. •875L •373 •1(20 •376 -405 •374 •380 •361 •338 •276 •252 •125L •134 •09g •252 •195 •316 •277 • 361 •339 •276 •252 •I25L •732 •30g •4% •610 •348 •l|tt6 •27g •324 •1.48 • 4 % •25L •252 •195 •215 •172 •296 •252 •344 •3m •298 •270 •25L •496 •610 •5 70 •6 56 •4 08 •496 •312 •372 •404 •460 •375L • 526 •271 •296 •252 •246 •2/4 •310 •278 •310 •282 •375L •3i(8 •408 •496 •SOg •572 • 380 •444 •380 • 416 '426 2 •5L •361 •3}8 •3 •314 •B.IO •272 •765 •227 •314; •286 •287 2 •5L •27g •324 •312 •372 •380 •444 •490 •546 •372 •428 •625L •374 •380 • 3 6 6 •361 •342 •328 • 310 •278 •3 IO •282 •274 •625L •252 •240 •262 •27? . -304 •344 •380 •444 •380 •4 36 '452 'IS L •376 •40S •374 •390 •366 •36 1 •344 •314 •29g • 170 •75 L •248 •190 •252 • na •268 •278 • 312 •372 •404 •460 •875 L •375 •1)20 • 3 7 6 •4o5 •374 •380 •361 •338 •276 •252 •87SL •254 •160 •248 •252 •21,0 •Z78 •324 • 4 M « •496 - 3 GIRDERS DATA: EI^ » 10,000. units E I T 500.0 units GJ L = U,hJ0. units GJ_, - 221.5 units 00 Table 7. DISTRIBUTION FACTORS FOR LONGITUDINALS (a) (a) The j values, when multiplied by the simple moments, yie l d the longitudinal moments. (b) The lower values are for zero torsion. (0 » 0) GIRDER NO. 60 : -J O l t~ UN I -tf' ro UN r-r- y - J — fo cn ro in r- H SECTION l .to <v — in o — • . . o o vJ to _ T i _ SJ o <*\ ON vi> o UJ to CO O l > . l • o c OJ OJ CO 01 Ul r •v o ci r 1 o a. o p* o 3 O H* •i CD 05 •-3 M ^3 **1 CO CO > > CO M CO jo N> O w to X" - N> _> _ in In ON - J «N <T> t O O CO ON ON _ -O O o> DO N cn r~ CO O O cn i GO U> — ro 1 ON ON I UJ. OJ. OO ON 0 o 01 o 1 . IO — (T> ON OJ -CO CO UJ -J Ul f i VO O 1 _ in "1 ON NJ i _i _ i OO ON to O f oi CO ON in vi ON s j 1 ON >J _> 0 IA r c-i-1 N — ON N l _ |N> UN oo 1 jr o> — ~ J i X X 7 Ul o 1 x- Ol — -4 1 Ol N> Ul C» 1 to — ON W a a f If _• CO ^ HARMONIC ANALYSIS OJ GIRDER NO. 00 - J m r-<J r - r •si Ol r-N ON f~ N UN r-H SECTION x- OJ tO -J O "* x-tn ON OJ OJ O X-OJ OJ V _• 00 — Io 0» -J to *M ON — (O V0 ON ON f_ o -_ , Ol oo J> N Ol r -a o H-3 c+ f O P t-1 O P P-o P. ro CO »-3 M ^ W CO CO > > CO M CO X" OJ ° ^ UN ON Co OJ OD —1 o xr Ow tM CN ON — ON OJ OJ — x-4r -c-to IVJ in us fO ON — ho — tO UN — to w N to ul r CH OJ OO - J O x-(jJ OJ DN ON — UN OJ o* N> X-CO CO f j _> -I — oo O H to — x-jr ON N to in _> IN) ON to o> - j — - J ON ic Nl Ul r <JJ OJ OJ ON CO _ O" — x-X - xr oO O rJ ui -4 oi to Oi oo O o- o* — X" x- x-\>J OJ O" ON oo — r to to C" -o Ox |o lj? -4 " J O oo ro OJ on — ON -F to <jJ So to io O 0 0 »o |0 UN -o IO ON a - J CO N X -^HARMONIC ANALYSIS - 06 -Pjj - 3 GIRDERS GIRDER NO. STIFFNESS ANALYSIS HARMONIC ANALYSIS 1 GIRDER NO. K SECTION STIFFNESS ANALYSIS HARMONIC . ANALYSIS GIRDER NO. Load on Girder 1 HARMONIC ANALYSIS 1 GIRDER NO. K SECTION Load on Girder 2 HARMONIC . ANALYSIS GIRDER NO. K SECTIO] Point Load at UDL HARMONIC ANALYSIS 1 GIRDER NO. K SECTION Point Load at UDL HARMONIC . ANALYSIS GIRDER NO. K SECTIO] •125L • 2 5 L •375 L • 5 L HARMONIC ANALYSIS 1 GIRDER NO. K SECTION •I25L •25 L • 375L •5 L HARMONIC . ANALYSIS GIRDER NO. K SECTIO] \i X ] LO^  GIRDER NO. K SECTION U X 10^ 1 • 2 S I 52-5 0 42-7 0 47.2 0 41-6 0 42-4 0 62-6 O 1 •251 -37.1 0 -2C|.3 O - 2 7.0 O -iG-6 0 - 2 1 . 0 0 - 2 3 - 0 O • 5 t 3 ^ 2 O 41.5 O 43-0 O 38-8 0 41.5 0 •&L - 12-0 0 -16-5 O -21-3 O -21-2 0 - 19-0 0 •7SL 0 27-8 O 34-1 0 0 42.4 0 • 7 5 L - 6 - 8 * 0 - 2-2 0 - 5 -8 0 - 1 6 - 6 0 - 2 I ' 0 0 CM • 2 5 L -52-3 - 2 9 - 0 -42-5 -21-8 -Mfe-S - a i - o -1,0.3 -28-2 -41-3 -26-2 -72.3 - 2 0 . 3 2 •25L 53.7 -58-0 43.0 4 7-6 45.2 S g . O 37.0 57-6 39-1 53*6 4 3 0 40«6 CM •5L -37.7 -28-8 -40«2 - 2 8 - 8 -42-0 -21.6 -3S-0 -25-7 -1)5.5 -27.4 • 5 L 33.5 S7.6 37.0 51-6 4o.o 55-2 3(.-7 47.4 38-0 54<S CM •7SL -2I.O - 2 0 ' 6 - 25 - 4 -23-0 -32-2 -29-0 -40-3 -28«& -4I '3 -26.8 - 7 5 L 13-4 41-2 18.4 46-0 26-7 52-2 37.0 57.6 33-1 53-6 3 • 2 5 L -15-4 0 -13-7 O -2o.l O -32-8 0 -21-3 O - 3 9 - 6 3 • 2 5 L -37.| '.O . -29-3 0 -17.0 0 -16-6 0 -21-0 O - 2 3 - 0 0 • 5 1 -27.2 0 - 2 6 ' 2 O -21.8 0 - i6-4 0 0 • S L -12-0 :.o - IS-5 . O -2.1.3-O - 21.X O - 19-0 0 • 7 5 L - 3 o - 6 0 -2.0.0 0 -28-3 a -32-8 O -21.3 0 •75 U - 6 - 8 •O - 2-2 0 -5-8 O -16-6 O -21-0 O Table 8. DISTRIBUTION FACTORS FOR TRANSVERSALS The values, when multiplied by the simple moments, yield the transverse moments per unit length of the transverse medium. The lower values are for zero torsion (0 » 0). The torsional values are given for the l e f t hand sides of interior supports. GIRDER NO. H SECTION STIFFNESS ANALYSIS HARMONIC "^ANALYSIS GIRDER NO. H SECTION STIFFNESS ANALYSIS ^ HARMONIC ANALYSIS GIRDER NO. H SECTION Load on Girder 1 HARMONIC "^ANALYSIS GIRDER NO. H SECTION Load on Girder 2 ^ HARMONIC ANALYSIS GIRDER NO. H SECTION Point Load at UDL HARMONIC "^ANALYSIS GIRDER NO. H SECTION Point Load at UDL ^ HARMONIC ANALYSIS GIRDER NO. H SECTION M 2 5 L •25L •375L •5 L HARMONIC "^ANALYSIS GIRDER NO. H SECTION •I25L •25L •375L •5L ^ HARMONIC ANALYSIS J •I25L • £ 4 0 •8 52 • 5 iS •736 •M4-0 •749 •8 1 2. L •796 1 •I25L • 1 4 I •112 •258 ••2 18 •321 •501) • 3 4 4 •365 * 2.10 •274 *z •2o2 • 297 •25 L •€(,0 •858 •6 30 •275 •553 •815 - 4 7 0 •765 •545 • 7 g e •25L •2<58 •2 IS •217 • 192 •293 •276 •330 '341 • 2 9 0 •294 •375L •518 •796 •559 •815 •629 • o 4 0 •52 6 •792 •525 . •730 •375L •321 •3o4 • 293 •276 •242 •234 •3oo •3o2 •3oo • 3 0 6 •5L •If 1(0 •749 • 470 •765" •526 •112 •610 • 8 3 0 •S 18 •786 •SL »34<V •36* •330 •341 •3oo •302 ' '248 •2-47 <3o3 •310 •625L •392 •7i «f •413 •728 •456 •756 •52.5 •732 •525 •790 •G25 L •34 7 •405 •3/+3 '38 5 •332 •351 •300 •302 • 3oo •3 06 •75 L •364 •622 • 380. •702 •413 •728 • 4 7 0 •765 •545 •793 •75 L •3W •428 •344 ' 414 •343 •385 •330 • 341 • 2 9 o •294-'27SL •35o •675 •35 4 •688 •392 •714 •<J,40 •74-3 •584 •812 •875L •334 •4W •340 '4 IS •347 •4os '344 • 3 6 5 •'270 •274 2 •I25L • l l t l •1 12 •252 •Hg •321 •304 •34 4 •365 •270 •274 L •222. •297 2 •I25L •697 •765 •445 •552 •2<35 • ^ 0 0 •226 •300 •400 •465 •368 •423 •25 L •258 •2lg •217 •192 •2-93 •276 •JJO •341 •290 •294 •25L '445 •552 •52.9 • 6 1 0 •366 •4SI •262 •341 •354' •k22 •375L •321 •lot* •293 •27« •242 •254 •300 •302 • 300 • 3 0 6 •315L •265 •400 •356 •451 •463 •539 •331. •412 •330 •406 •5 L •344 •364 •330 •341 •300 •302. •248 • 21*7 •3oj> •3 10 •5 L •226 •3oo •262 •34/ • 131 •4H. •4i(7 •519 •323 •£(00 '625 L •347 •405 •343 •385 •332 • 3 5 3 •3oo •3o2 •300 •30f •£25 L • 2oS •246 •219 •Z74 •256 •326 •331 •412 •330 • 4 0 6 •75 L •340 •418 •344 •414 •342) •385 • ybo •341 •29o • 2 9 4 •75L •201* •220 •206 •23g •219 •274 •262 •341 •354 •428 •875L •334 •440 •340 •422 •347 •405 •344 '365 •270 •274 •875L •208 •210 •204 •220 •20 5 •246 •226 • 3 0 0 •400 •465 / i i - 4 G I R D E R S U . is' . is'~»i*- i ? ' - J DATA: EI^ • 10,000. units E I T 500.0 units GJ L = 3,9W>. units GJ_ - 197 units Table 9. DISTRIBUTION FACTORS FOR LONGITUDINALS (a) (a) The PJLJ values, when multiplied by the simple moments, yield the longitudinal moments (b) The lower values are for zero torsion. ( 0 = 0 ) -1> OJ GIRDER NO. -J Ul 1--j r-ON N> tn r -vn i — u> oi ru . cn r f\> cn |— ta -J cn r -. j r-CT» M cn r Ul r~ UJ "Ni U| f-(u \r\ f— IO Ul I— 8 SECTION to 6 o CO cOx-. x- ui ' . — O - IN i 6 x- m O J i U* '. . o o IT- fO cn cn o 6 C J 0 c5 s> cD to UJ i _ 6 to S 5 6 — to cn -o O 6 — S .a 1 * • o o - Nl - -vt 1 6 o — V — -c-N Ul f~ o •f o CO cx o CO I — O cS> 00 -c Ui i — o CO _ J o oo » — o in ON CO —1 1 - O N cn cS5 ^ / o o c£> OJ Ul 00 I O c t o o O t-J ^ o ? ^  to -J o — J? 5 1 , O L> O "vi cn O 1 . o O — vi _ NI cn r M 0 CO > _^ o — w — o cn ON ro _ j I- o <-o oo — O — sr to o~> o o cQ cw — <JJ i O o CD Co <J> CO t . o ? o — cn cO x- x-o — OJ Ul o c> — ru (TN C*J o o O vO —' cn o — X- O o — o ]u Co _ U1 - J Ut r t-4 O So ex 3 a • - 6 IT cn O -1 i - 6 H cn <S> ol 1 — o -- x-fo ON i 6 6 cx> U> o oo 1 — o t  or-IO ON 1 . O to vn v£> UJ O _ j j o I- -•• - J 6 -— M "1 CO 2 o io O J 2 ^ CN «> o — • J JT <o _1 d —fO ON -J o r fu c+ p. CD >• CO M CO 1 — 6 o U J 1 1 - 6 O Jr VX> CN 1 — 6 xr <p> 1 — O oi CO 1 — 6 — j r XT CTN 1 1 -c» 1 O l" — —1 2 5 cO 2 N ON CO » • IO to o —-— to OO cO 2 ^ 6 — — o X" oo G a tr 1 t o 6 £ o O UJ HARMONIC ANALYSIS OJ GIRDER NO. OO —J r~ - J r <Tl N> Ul -I— Ul t-rv> u» r -rv> cn r-00 u> r-- J r ro cn t— Ul r u> 1 ON 1— to cn t - cn r H SECTION O u. — y * o O — 6 — -J O 8 CM — i o o - ~j — — • i 6 o _ V " — jr cf\ U> cn <^ w cr> N» to ON vi W to O CD H tv> CC> ffN ON (JJ to to O to UJ N J? ^ ) PO Ui !~ CO o ^ 6 o -x~ oo x- -J S i ro 6 — i 6 6 cn O ^ H — — j -J'JT CO <T\ to ro CO vJN ON CV jo to cD g\ oo — to ro cS> ON ON |N> CO o to ~ o OO ro _C-to to O !o (O Ul r-o 3-o p a . o M a n CO ?> 6 — co x- J T £ - — ' x- —I O — cu vn o tn O — — N d OO s ° 3 O — o — x- o O |0 CW _ cD cn CT\v& CO — t ' to cO Ul to OO to o p _ B -a IO JO £3N j r -00 O to to CD ON <*i CJJ U1 -4 Ul r -t-1 O a . -J o to x-r° - J 2 M ON co 2 O 2 ^ (Tk OO 3 ? to -J o — - i O IW 1^  ff  O v0 *S ^CP ON Qv ^ ) to to ON JT CO — tJ NJ •o O Oi — TN jr CO — |0 S3 co <r* ON ,c> c»J fO — ff^ ° Co Ul r CU p. CD > CO M CO 2 5 •X" u> — K> <T> to 2 .o CO cD 2 c — to CO ^ B ON y o to ON CN) to JT -J JT O O to —J XT to to ro f>» -j x-o o fc" fo ON ^ to j -to ro <r> to t 3 —1 o UJ N> UJ N> i n " X- [O HARMONIC ANALYSIS - £6 -H - 4 G I R D E R S GIRDER NO. STIFFNESS ANALYSIS HARMONIC ANALYSIS GIRDER NO. H SECTION STIFFNESS ANALYSIS HARMONIC | ANALYSIS GIRDER NO. S3 t—, Load on Girder 1 HARMONIC ANALYSIS GIRDER NO. H SECTION Load on Girder 2 HARMONIC | ANALYSIS GIRDER NO. v—1 M fn o Point Load at UDL HARMONIC ANALYSIS GIRDER NO. H SECTION Point Load at UDL HARMONIC | ANALYSIS GIRDER NO. w 00 rx •1251 •2SL •575L HARMONIC ANALYSIS GIRDER NO. H SECTION •I25U •25 L • 375L • 5 L HARMONIC | ANALYSIS GIRDER NO. kl X ] L0^ GIRDER NO. H SECTION M x 10* 1 •25L 53-3 O U3-4 o 46-2 O O Ul-1 0 85-2. O 1 •25L -34-2 0 -2g.6 • Q -Zl-8 O - -^.2 O -15-5 O -2 .V5 O • 5 L 36-0 o 3^.2 0 0 3&-4 0 O • 5 L - 2-5 0 - 1 • 1 O -15-1 O -17.2 O - 1 2 ^ O - 7 5 L 20-7 O 24 .5 O 3I.0 0 39-2 O 41. 1 0 -75 L 14-7 O 10.4 0 2-5 O - 1.2 O -15-5 O . C\J '25L - 5 V | -36 -i. -4S-0 -33-2 -45-2 - 3 7 - S -37.3 -38-6 -31,6 -35-9 -35-0 -36-4- 2 •25 L 5 4.^ 43.1 '50.1 M4.T 52.0 34'4 S4-4 38-0 52-1 4 8 - 6 53-8 C\J • 5 L -33-1 -37.3 -3S'fe -40-4 -36-8 -37.2 -31-5 -38 .4 -3U • 5 U 5 2-8 3M'U 54-4 3S.8 54-1 36-4 4-8 36-3 52-^ C\J •7SL -17.0 - 3 34 -21.1 -35-2 -28-3 -37 .3 -373 -33-4 -3<K -35-1 •7SL io-2 34-7 14.<} 21-2 46-8 34.14 54.4 38.0 52-7 3 •251 -18-7 - 12-1 -I6-U -10.4 -23-1 -16-1 - 2 L 3 -2J-0 -24-8 - iq-o -17-6 3 •25L -38-4 -'3.1 -3o .2 -27.0 - 4-1 -45-5 7-3 -21.1 2-1 52.1 3-1 •5L -31.6 -2S'4 -21-3 -2V0 -25-5 -iq-s -20.7 -15.5 -26-3 -20.4 . • 5 L -io-3 -15-5 7-3 -21-4 2-1 -22-0 - 0.6 -18.8 4-1 •75L -34-0 -32-4 -33-6 -30.8 -32-4 -27-7 - 2 1 3 -23-0 -2U-8 -19.0 •75 L 6 -3 3I.3 2-4-26-4 -4.8 18-2 ' -15-5 7-3 -21.1 2-1 4 •25L o - 5 . 7 O O -II. 1 O -10.1 a -22-1 O 4 •2SL -13.0 O - I I - l 0 -ts.(, 0 -18-2 O -15.5 O -31-2 O . • 5L o -11.1 0 - I0.4 0 -S-k O -10.7 0 - 5 L O -18.1 0 . -16.0 0 -12-8 0 -16-3 0 •75 L -18-0 o - I S O 0 -14.6 O -11.1 0 - l O - l 0 •75L -17. <t O -18.3 O 0 -I8«2 O -15-5 O Table 10. DISTRIBUTION FACTORS FOR TRANSVERSALS The values, when multiplied by the simple moments, yield the transverse moments per unit length of the transverse medium. The lower values are for zero torsion (p = 0 ) . The torsional values are given for the left hand sides of interior supports. / i j - 5 G I R D E R S © © © © © - j J I S ' U - i S ' - * - IS'- X 1 5 ' K -• S3 S T I F F N E S S A N A L Y S I S O CO • 0 'ION S T T F P N R S S A N A L Y S T S CJ CO M M • 0 S T I F F N E S S A N A T . Y S T 5 O CO I-I I-I o ss o M tH Load on Girder 1 IARM0NI LNALYSI 'ION Load on Girder 2 S3 CO O M g d S3 P5 S3 O M Load on Girder 3 HARMON! ANALYS! - g Q O w CO Point Load at U D L IARM0NI LNALYSI w Q Ctf a Point Load at U D L < S3 x <»: w 0 - • B IH O W Point Load at U D L HARMON! ANALYS! • M O CO CO 0 •JC •I2-5L •Z5L -37SL •5 L I 3C •I25L •25 L •375 L •s (_ / • I Z 5 L • 2.51- •37SL • 5 L i •I25L • 755., ( •552 •817 • 400 • 740 •324 •62t) • 762 •I25L '159 •130 •27 3 •2i*7 •914 •332 •312 •384 •260 •294 •I25L • oso -•003 • 108 •012 •IS8 • 0 41+ •194 • 0^7 • 155 •053 •25L •538 •817 •60S •839 •451 •765 •357. •70S •444-•745 •25 L •2.73 •247 •225 •214 •288 •2.96" •hoe, •362. •276 •3IS •25L • 108 • O l 2 •096 •017 • 143 • 04I •ISO •074 •ISO' •052 •375L •400 •740 •4 5-1 •765 •540 •Soo •42£> •735 • mo •735 i •375L •332 •288 •296 •238 •254 •Z86 •323 •282 •3 26 D •375 L • i5« •04 4 •143 •0 41 • 120 •03g •157 •060 •159 •062 1 •5.L O i l ) •631) •357 -7o5 • 420 •73S •52.1 •787 •413 •730 •482. i •SI •312 •384 •306 •362 •286 •3 23 •240 •265 'Tg«f" •330 •276 1 •S L • I9U •067 •180 •074 •15 7 -QiO •127 •0 41 •161 •064 •131 •C25L •28H •644 •3 04 •660 •344 •693 •4^ 0 •738 .4-20 •735 •7^3 •C25L •294 •413 •2<*q •399 •301 •371 •196 •323 •282 •326 •317 •625L •215 •123 •204 •108 •186 •085 -157 •O60 •158 •062-•060 •75L •263 •276 •632 •3o^ •660 •357 •705 •444. •745 •75L •275 •430 •286 •420 •293 •399 •306 •362. •276 •315" •75L •226 •154 «22<? •136 •204-•108 •ISO •07l( • ISO •05 8 •S75L •254 -'600 •263 •6i5 •2.8 4 •644 •324 •624 •488 •762 •8T5.L •263 •436 •275" •A 50 •294-•413 •312 •384 •26 0 •294 •875L •230 •172 •226 •I54 ~l\S~' •123 •194 •067 •135 •053 Table 11. D I S T R I B U T I O N F A C T O R S F O R L O N G I T U D I N A L S ^ ^ (a) The Pij values, when multiplied by the simple moments, yield the longitudinal moments. 1 (b) The lower values are for zero torsion, (p » 0) GIRDER NOr H SECTION STIFFNESS ANALYSIS HARMONIC j ANALYSIS GIRDER NO. * SECTION STIFFNESS ANALYSIS 0 to H M a co 0 / GIRDER NO. H SECTION j STIFFNESS ANALYSIS 0 co H M S CO O S H Hi -1 / GIRDER NOr H SECTION Load on Girder 1 HARMONIC j ANALYSIS GIRDER NO. * SECTION Load on Girder 2 GIRDER NO. H SECTION j Load on Girder 3 GIRDER NOr H SECTION P oint Load at UDL HARMONIC j ANALYSIS GIRDER NO. * SECTION Point Load at UDL GIRDER NO. H SECTION j Point Load at UDL GIRDER NOr H SECTION • I 2 . 5 L • 2 6 L • 3 7 5 L •5 L HARMONIC j ANALYSIS GIRDER NO. * SECTION • I 2 5 L •2SL •375L • 5 L GIRDER NO. H SECTION j • I 2 5 L •23L •375L •5U 2 • I 2 S L . • 1 5 9 , •1-5 O • 2 7 3 •2 4 7 • 3 1 4 •332 •312. •384 •260 •294 k •276 •317 2 M25L •638 •720 • 3 6 I •"•86 •2 29 •334 •19,7 •254 •3<t5 •ij 18 J22. •312 •37»t 2 •1251- '• 132 •I 52 , -) •? O • 256 •245 •294 •232-•28S .204 • 2 4 4 •25L • 2 7 3 •Z47 •225 •214 •222 •2% •3o6 •362 •276 •315 •25L .•361 •486 •465 •56o • 230 •392 •212 •191 •3oo •379 •25L •222 •256 •ISI • 2 ) 1 •226 •268 •232 •zsi •2.IM •254 •375L •3l A ' 332 •288 •236 •238 •254 •323 •282. •3i6 •375L •228 •33k •290 •392 •4OS •430 •224 •2>S2 -273 •357 •J75L •243 •234 •226 • 2 & 8 •186 • 2 2 1 •220 •264 •217 •262 L •22k •256 •5L -312 •384 •3o6 -362 •286 •323 • 240 •265 •2?4 •330 •5L •187 ' •254 •212 •291 •224 •362 •3% •W •27 2 •350 • 273 •357 •5 L •232 •288 •232 •282 •220 •264 •187 • 2 2 2 •21? •26k •625L •291+ •1+13 •263 '393 •301 •371 ?293 •359 •226 • 3 2 3 •306 •362 •282 •32<S •626 L • 185 •213 •18? •238 •212 •282 •224 •362 •625L • 212 •26,2 -220 • 2 7 0 •226 •276 •220 • 264 •217 •263 •75L •275 . • 4 3 0 '286 -.•4 2 0 •276 •3i5 • • 7 5 L •191, •209 •130 •217 •183 •23S •212 •2°A •300 •373 •75L •198 •237 • 2 0 7 . - 2 5 2 •220 •270 •2 32 •2S2 •2/4 •256 •SlSL •263 •43S •275 •430 •2% •4.3 '312 • 3 2 k •2<3U •260 •275 L •202 •20$ •19!) •203 •185 •187 ••34.5 •^18 •8 IbL •190 •220 •138 •237 •212 • 26Z' •232 •288 •2k k 3 •12-5 4 •05o - 'o o3 • 1 0 2 •oi2 •158 •143 •01(1 •194 '067 •135 •053 A, •131 •06o 3 •I25L . • 1 3 5 •152 •222 •2.56. •243 •29M •232 •288 •204. •244. •224 • 256 3 •I25L •630 •702 •340 •463 •M • 3 2 3 •151 •256 •320 •k&g J33 • 2<3 1 •368 '25 L •log •012 •096 •017 •160 • 0 7 c . •150 •058 •25 L •222 •256 •191 . •21 1 •226 •268 .232 •2S2 •214 •256 •25L •340 •*i63 •4U8 •544 •264 •380 •173 •23? •272 -370 '3751 •158 •ouM •143 •041 •120 •oJ5 • 1 5 7 •oio •156 •062 •375L •243 •294 •226 •26? •232 •282 •l$6 •221 •220 •264 •217 •262 •375L •m •323 •26k •380 •391 •482 •2!i6 •35k •250 •351 •5L •19/, •067 •180 •074 •157 •060. -12.1 •047 •161 • 0 5 4 •5 L •232 •282 •220 •264 •IS7 •2 2 2 •218 • 2 * 4 •51 •151 •256 •179 •288 •21+6 •35k •373 •4&5 •242 •345 '625L • 2 1 5 •123 .204 •107 •085 -204 • 1 0 7 • 1 5 7 •060 •158 '062 •625L •212 •262-•220 •270 •226 '276 •220 •liU •217 •262 •625L '146 •230 •152 •24M •176 •280 •21(6 •354 •250 •351 •75L •226 •154 . -220 •136 •iSo • 0 7 * , •150 •135 •o5i • 7 5 L •198 •237 •207 •252 •220 •no •232 •ni •2(4 •256 •75L •/52 •218 •149 •226 •152 .244 •179 •288 •272 •370 •875L -230 • 1 7 2 •226 •154 • 2 1 5 •123 •194 • 0 6 7 •875L • l9o •220 •I9S •237 •212 •262 • 2 32 •288 •2ol» •244 •875L •159 •216 •152 •2l8 •146 •230 •1S1 •256 •320 •403 Table 11. (cont'd) DISTRIBUTION FACTORS FOR LONGITUDINALS 1 — • v O 1 NO. STIFFNESS ANALYSIS HARMONIC . ^ANALYSIS £R NO. • 5 STIFFNESS ANALYSIS O CO M M S3 CO O M ^ 3 ac < •P SR NO. H SECTION STIFFNESS ANALYSIS O CO M t—1 S3 CO O <C S3 SE < •p 1 NO. o (—1 Load on Girder 1 HARMONIC . ^ANALYSIS £R NO. O M fH 8 CO X Load on Girder 2 SR NO. H SECTION Load on Girder 3 GIRDS rH 0 0 Point Load at UDL HARMONIC . ^ANALYSIS Point Load at UDL Q s H SECTION Point Load at UDL GIRDS • I 2 . 5 L •25L •375 L HARMONIC . ^ANALYSIS •I25L • 25L •375L • 5 L H SECTION •125L •2SL •375L •5L •I25L' • 0 2 3 -•015 • O S 2 - •03 I •060 -•0 33 . 1 0 7 - • 0 4 4 •073 - • 039 •1-25 L •oit4 •Oil • 0 4 3 • 0 4 3 • I 35 : . •088 • 16 4 •131 •116 ' 0 8 4 • I 2 S L •135 •152 •222 •256 •245 • 2 9 4 •232 •288 • •204 •2M 4 •25 L •05-2 -•031 •047 -•ozfc • 074 -•041 •099 -•osi •082 -•042 •25L • 0 9 3 • 0 4 3 •082 • 0 4 3 •122 •07 8 •152 •MC •128 •093 •25L •222 •256 •181 •211 •226 •268 •232 -292 •214/ •256 •J75L •OSO -•033 ' 0 7 4 -•041 • 0 6 4 - • 0 3 4 • 0 8 6 -•044 •OS7 -•ot*5 n •375L •135 •o&g •I 22 •078 •102 •O67 •133 •095 •134 •099. •375L •243 - 2 3 4 •226 •268 •186 •221 • 2 2 0 •264 •217 •262 • 2 2 U 4 •S L •107 - • 0 4 4 "•o<fl -•051 •08£ - • 0 4 4 •070 " -•035 • 0 8 3 -•045' •061 4 •5L •164 •131 •116 •135 •095 •108 •075 •!31 • lo2 •P42 •127 4 •SL •232 •222 •232 •_282_ •220 •264_ •187 j222 _ •218 • 2 6 4 •625L •129 -•oS8 ' 1 70 -'OSQ •105 - • o 5 | •086 -•044 •o87 -•ou5 - • 0 4 3 -&2SL •181 •163 •172 • I 4 S "•T57 • 1 3 3 •035 •134 • 0 9 9 •095 •&25L •212 • 2 6 2 • 220 .270 • 2 2 6 •276 ~22o" ;_26 4 •262 •256 • 7 5 L •14 4 -•057 •135 -'007 •120 " -•o5'6 •693' -•051 •082. - • 0 4 2 • 7 5 L M89 •181 ~-l8~4~ ' •169 •172 •138 -------•116 •128 •033 •75L •198 •237 •2o7 •252 • 220 • 270 .232 •282 •2/4 •2 56 •875 L • 1 5 4 - • 0 5 4 • m u -•057 •129 - • 0 5 8 •I07 -•044 •073 -•039 •875 L •192 • 1 3 0 •IS<3 •181 •181 •163 •164 • r 3 l •116 •o84. •875L •150 •220 •138 •237 •212 •262' •232 •288 •204 •244 •I25L • 0 13 - • 0 2 C • 0 3 0 - • 0 4 4 • 047 -•070 •06& - • 0 7 8 • 0 4 3 -•Oil •I25L • 0 2 3 -•015 •052 - •031 • 0 8 0 - • 0 3 3 • I 0 7 - • 0 4 4 • 0 7 3 - • 0 3 9 •125 L • 0 5 0 - • 0 o 3 . • 1 0 8 • 01 2 •159 • 0 4 4 •194 •067 •135 • 0 5 3 •25L • 0 3 0 -•oij 4 '021 -•041 •043 -•o€5 •060 - •O3o "•04"3 -•076 •25L •052 - 0 3 1 •OMT -•oil • 0 7 4 - • 0 4 I • 0 9 9 . - • O S | •082 - • 0 4 2 •25L • loS •012 • 0 9 6 •017 •143 •041 • •ISO •074 •150 • 0 5 8 •375L • 047 -•070 • 0 4 3 -'065 •o38 - • 0 5 7 •052 - • 0 7 8 •052 - • 0 8 0 •OSI •375L • 0 8 0 -•033 •074 -•041 •064 - • 0 3 4 • 0 8 6 - • 0 4 4 •087 - • 0 4 5 •375 L •158 • 0 4 4 •143 •0 41 •120 • 0 3 8 .157 • 0 6 0 •158 •062 -F53 •131 5 •5"L •065 - • 0 7 8 •060 - 'O90 •052 - • 0 7 8 • 0 4 2 -•063 •o54' -•0 20 5 •£ I •107 -•044 •099 -•051 • 0 8 6 -•044 •0 70 -035" ~ 0 8 3 ~ -•045 •061 5. •5 L •194 •1 80 • 0 7 4 •157 • oso •127 • 0 4 7 •161 " j 0 6 4 • 6 2 S L • oso -•122 •074 -•112 •065 -•097 •052 - • 0 7 8 •052 - • 0 8 0 - • 0 7 7 •625L •129 - • 0 5 8 •120 - O S 6 -•OSI •086" -'044 "VosT - • 0 4 5 -•043 •625L •215 •J_23 • 2 0 4 • 1 0 8 •186 •085 •157 • 0 6 0 •158 • 0 6 2 '060 •75L •092 -•142 •085 -•130 •07 4 - • 1 1 2 • o&o -•o9o •049 -•076 •75L •144 -•057 •135 -•057 •120 - • 0 5 6 •099 -•051 "•082 -•042 •75 L •226 ; i 54. •270 . ; . i 36. •204 • 1 0 8 •ISO • 0 7 4 •150 • 0 5 8 . • 135 •0S3 • S7bL •099 -•154 • 092 -* 1 4 X •oso - M 2 2 •065 - - O T 8 •043 -071 •875L •154 - 0 5 4 •144 -•057 •129 - • 0 5 8 •107 -•o4U • 0 7 3 -•039 •275L •230 •172 •226" • 1 5 4 •215 •123 •194 •067 I Table 11. (cont'd) DISTRIBUTION FACTORS FOR LONGITUDINALS vo _ _ . . _ j • STIFFNESS ANALYSIS 0 to M M . 21 CO ill NO. STIFFNESS ANALYSIS O CO H 1 i d STIFFNESS ANALYSIS 4-, HARMONIC ANALYSIS K SECTION Load on Girder 1 ill NO. O M a CO X Load on Girder 2 2: CO O >H < S= K < •P H SECTIOI Load on Girder 3 4-, HARMONIC ANALYSIS GIRDE1 K SECTION Point Load at UDL V—' r 11 a: < f a «-s Point Load at UDL 0 s H SECTIOI Point Load at UDL 4-, HARMONIC ANALYSIS GIRDE1 K SECTION •\Z5l -251 • 3 7 5 1 -5"l_ •I2<5l_ . 2 5 L . •375L •5L H SECTIOI •I25L • 2 5 L •375L •5L 4-, HARMONIC ANALYSIS •I2SL • 0 2 3 --OI5T • G 5 Z - •031 •060 -•033 .107 -•044 •073 -•039 •125 L •044 •0 11 • o93 •043 • I 35 : . •OSS • is 4 •131 •116 '084 •I25L •135 •1 52 •222 • 2 5 6 •243 •294 •232 •288 • •204 •24 4 •25L • osz - • 0 3 l _ •047 -•026 •074 -•04I •o99 -•051 •OB 2 -•042 •25L • 013 •043 •082 •043 •122 •07 8 •152 •U4 -128 •033 ' 2 5 L •222 •256 •181 •211 •226 •268 •232 -2S2 •21 M / •256 •375L •080 - • 0 J 3 , • 074 - • O k J • 064 -•034 •0S6 -•044 • O S 7 —c»45 n •375L •135 •£>&S •122 •072 •102 •O67 •133 •095 •134 •099 n •375L •243 -234 •226 •26S •186 •221 •220 •264 •217 • 2 6 2 P43 •224 4 _\5JL_ •107 -•OHM •039 - •051 •08£ -•044' •070 - ' 0 3 5 •083 -•045' •P4. •061 4 •5L •164 •131 • 152 •116 •133 •095 •IOS •075 •137 • 102 ^42 •127 4 •SL •232 •2S8 •232 _;282__ •220 _.l264__. •187 j222___ •218' •264 •129 -•058 • 120 -•OSG • 1 0 5 -'05I •O86 - • 0 4 4 •087 - •OU5 -•043 •G2SL •181 •163 •172 •148 ; I 2 4 _ •133 •035 •134 •039. •095 •625L •212 • 2 6 2 •220 .270 •226" •276 •220 " • 2 6 4 •217 • 262 •256 •75L •I4U - • 0 5 7 .735"" -•057 "•120 - O S 6 T693" -•031 ~~0~&T -•01,2 •75 L •181 ~I§V •163 •172 •138 • IS 2 •116 •128 •033 •75L •198 •237 •207 • 2 5 2 •220 • 270 • 232 •282 •im •2 56 •875 L •15H -•o5U • 144 - • 0 5 7 -•038 •I07 - ' 0 4 4 •073 -•039 •875 L •192 •130 •189 •18I •l&l •163 •164 •1-3,1 •116 »o&4. •875L •ISO •220 •132 • 2 3 7 •212 • 2 6 2 ' • 232 •288 •20 4 •24k •I25L • 0 13 -•020 • O30 - • 0 4 4 • 047 -•070 •065 -•078 •043 -•071 •I25L •023 -•015 •052 -•021 •O80 -033 • 107 -•C44 •073 -•039 •I2SL •050 —0o3 . - i o 8 •01 2 •159 •O44 •194 • 0 6 7 •135 •053 •25L •03O - •OUM •027 -•041 •043 - • o 6 5 •060 -•O30 •04'3 - • 0 7 6 ' 2 5 L •052 -031 •04T -•o28 •074 -•041 •ogq. - O S | •0S2 -•oi|2 •25L • loS • O i l •09& •017 •143 •041 • •180 •074 • 1 5 0 •058 •37SL •047 - • 0 7 0 •043 -'065 •038 -•057 •052 -078 •052 -•080 •fs, •051 •375L •080 -•033 •074 -•0 41 •O64 -•034 •0E6 -•0144 •087 -045 * 5 2 •061 •375 L •158 •143 •041 •120 •038 • 157 •060 •158 j;.062. -?S3 •131 5 •51 •065 -•078 •060 -'090 •052 -•078 •042 - • 0 6 5 •054--•020 5 •5 L •107 -•044 •099 - •051 •086 -•044 • 0 7 0 -035" •083 - • 0 4 5 5 •5 L •194 •067 •1 80 •074 •157 ' 0 6 0 •127 :°J±L •16.7" '62SL •080 -•122 •074 — 1 12 • 0 6 5 -•097 •052 -•078 •052 -•O80 - • 0 7 7 •625L •129 -•oSS •120 -•056 •ToV -•0S\ •Og6 -•o«,t, •0S7 -•045 -•043 •625L •215 •123 •201) • I 08 •186 •085 •157 •060 •158* M562 ' 0 6 0 •751 • 0 9 2 -M42 •085 -•130 •074 -•112 • 0 6 0 -•080 •0 4(3 -•07& •75L •144 - •057 •135 - • O S T •120 -•056 .053 -•05 \ •082 -•042 •75 L •226 -•J.54_ •220 . .'i.36_ •204 •108 • ISO •074 •ISO •058 -87SL •099 -•154 • 092 - • 1 4 2 •080 -'122 . • 065 -•o78 •043 -071 •875L •154' -•054 •144--•057 •129 -•0 58 •10-7 -•0 4U "•"o73 -033 •230 •172 ;2 26 •154 •215 • 123 •194 •067 • 135 •053 I Table 11. (cont ' d ) DISTRIBUTION FACTORS FOR LONGITUDINALS vc I HU - 5 GIRDERS 1 STIFFNESS ANALYSIS ONIC ISIS NO. ION STIFFNESS ANALYSIS ONIC .YSIS • Q Load on Girder 1 ONIC ISIS NO. ION Load on Girder 2 ONIC .YSIS TIO Point Load at UDL E << :DER SECT Point Load at UDL HARM . ANAL GIRDE SEC •I25L •25L •375L •5L s •1251 •251 •3151 • SL HARM . ANAL GIRDE u x 10 4 X. • jx x 10** •2SL -41-0 -33-3 -HO -3 -33.8 •25L 63.3 et.\ S4-7 •45<4 47-6 2 •5L -3<U -33-2 -31.0 -23-8 -38-1 2 •5L 41.2 ^8-3 44.4 46-5 47.0 •75L -31-6 -33-7 -36-8 -33.3 -37.3 •75L 19.1 12-1 34-2 45-4 47.4 •25L -18-4 -lfe-7 -26-0 -34.0 -23-3> •25L - is- J -11-0 -4-5 0-2 3 •SL -37-2 -34,0 -2^-1 -23-2 -3o-3 -23-8 3 •5L 11.5 £.4 o-3 2-ct 1-3 •7SL -4U-4. -42.8 -33.6 -34-0 -28-3 •751 24.2 lo.M (5-7 0-2 :25L - 5-8 -£•4 -3-2 -13.1 -II.1 •251 -7.0 -5-5 -7-7 -6-1 4 •51 . -lM«3 -13.1 -u-4. -3-3 -11-8 -ll.M 4 •5L -8-8 -7.3 -4-q -un -6-7 •75L -2.2-5 -20-3 -17.0 - 13.1 -ll.1 •75 L 0-4 - o . | -5-t -6-1 Table 12. DISTRIBUTION FACTORS FOR TRANSVERSALSv; The values, when multiplied by the simple moments, yield the transverse moments per unit length of the transverse medium. The values given are for zero torsion only ( 3 = 0 ) . DATA: EL^- = 10,000. units E I T = : 500.0 units GJ L = 7,390. units .GJ- « , 370. units / i j - 6 G I R D E R S CD © © S €>I2'- 60 ' © i NO. 'ION S TIFFNESS ANALYSIS LRMONIC IALYSIS I NO. 'ION S TIFFNESS ANALYSIS O CO STIFFNESS ANALYSIS IONIC ,YSIS i NO. 'ION Load on Girder 1 LRMONIC IALYSIS I NO. 'ION Load on Girder 2 [ARMON] LNALTS] O S3. « Q O M Load on Girder 3 IONIC ,YSIS GIRDEE t-i o w CO Point Load at UDL LRMONIC IALYSIS B Q fi t-i o w Point Load at UDL [ARMON] LNALTS] tH O W CO X. Point Load at UDL £ 5 < S3 GIRDEE •I25L •25L •375L •SL • a . -p CO •I25L •25L •37SL •5L f •1251 • 2SL •375L •5L CC <! f •I25L •6R5 0 1 C •4 35 •7 65 •3O0 •675 •238 • 613 •400 •705 •I25L •178 .•I 5C_ •279 •283 •288 • 360 •260 •400 •242 '3)6 •12SL •066 • 007 • 133 •038 •IfcO •cf\o • 202. •144 •148 •092 •25 I • 4^ >5 •765 •524 •7^3 •356 •702 *2-D G •635 •355 •685 • 25L "•27 9 •283 •224 •242 •271 •330 • 266 ' •3S2 •-2S2 •336 •25L • 133 •038 •H4 •040 •161 •08I •190 •127 •162 •102 •375 L •3oo •6 75 •35 6 •702 •460 •748 •32.9 •676 •332 •672 0-^ 05 0-682 •375L "•288 • 360 " • T j r •330 •225 •276 •26o •346 •253 -346 •156 •375 L • 180 •o9o ~ T & 7 ~ •081 • 133 •071 '•i68 •104 •169 •I08 •iso • 104. 1 •5L . •238 •613 •266 •635 •329 •676 •443 •735 •325 •670 1 •SL •4oo •266 •382 •26o" •346 7 2 2 3 •285. • 254 •350 1 •5L •202. •144 • 190 •127 • 1 6 8 •104 •137 •081 •III •625L •2lO •570 •224 •258 ^624 •329 •676 •33X •672 •625L •230 •415 ""•24 2~ •406 •"258 •386 • 260 •346 •253 •346 •358 •62.5L •206 •189 ~202~ • 169 .",91" •IUO '"•168 ' •I04 '7J6~3~ •108 • 204 •220 •205 •2o! •202 •163 "•190 l i l L "7162 '102 •75L •138 •54l__ •206 •558 •224 •588 • 266 •35 5 •685" •75L • 2og •420 •222 •417 • 242 •406 •266 j 3S2_ • 252 •336 •75L •875 u ; ,"g £ •525 " : I98 •54 1 •210 •570 •138 •2,00 •705 •875L •421 ~ao9~~ •420 •230 .415 •26o •400 •242 •316 •875L • 2O0 •238 •204 •220 '"•206 •lot •I4L) "•148" •092 Table 13. DISTRIBUTION FACTORS FOR LONGITUDINALS (a) (a) The p^^ values, when multiplied by the simple moments, yield the longitudinal moments. (b) The lower values are for zero torsion. (0 = 0) vO t NO. SECTION STIFFNESS ANALYSIS O CO '. NO. "ION STIFFNESS ANALYSIS iONIC ,YSIS . STIFFNESS ANALYSIS O CO t NO. SECTION Load on Girder 1 M H S CO O M '. NO. "ION Load on Girder 2 iONIC ,YSIS O S3 S3 0 r-l Load on Girder 3 M H 2 CO O * M m SECTION Point Load at UDL m Q Point Load at UDL @ Q E-< O W Point Load at UDL § 3 U J tc <c CO O CO EC < X •I25L •25L •375L •5L f •125 L •251 •37 5 L •5L f DC •I25L •25 L •J75L •5L f •I2SL •178 •1 56 •275 •283 •288 •360 •260 •460 •142 •3 16 •I25L •567 •665 • 280 • 412 . 180 •278 •I6S •228 •294 •375 •I25L •~I48~ •171 • 266 • 220 •27 6 •196 •247 •231 •251 •279 •283 •224 •242 .271 •330 •Iii •382 • 252 •336 •25 4 . ISO •412 .11 o5 •S05 • 234 •338 •178 •256 •252 •332 •25L • 221 •266 •178 •215 . • 211 •258 •204 '25 2 • 199 '240 •3751. •288 •360 •271 •330 •225 •276 •260 •344 •253 •346 f x . •256 •538 •375L •180 •278 • 234 • 338 • 358 •445 •225 •310 •234. •218 • 263 •335 •375L •220 •276 •211 •258 •178 •214 •2o4 •246 • n s •242 A 2 •5L -260 •400 • 266 •382 •260 •346 •223 • 285 • 254 •350 2 •5L •166 •228 •178 •256 • 225 •326 •347 •434 •230 •313 2 •5L • 196 •247 •2o4 •252 •204 •2_46_ •177 •212 • 158 •242 ^23 •201 •625L •230 •415 •242 •406 •258 •386 • 260 •346' •253 •346 •625L •174 •218 • 171 •227 •179 •254 -215 •320 • 234 .-III •625L •I78~ •212 •187"" • 227 -----• 200 •244 __ •2o4 •246 •198 •242 •257 •75L •209 •JflO •222 •m7 •242 •40& •266 •382 •252 •33o •75 L •|8l • 225 •176 • 22.X •171 -•227 • 173 •256 •252 •3 28 •75L •163 •187 •2o2 •217 •204 • 252 ~isT~ •21*0 •875L •192 •421 •2o9 •420 •230 •4lS • 260 •400 .242 •316 •'%75L • 183 • 232 .181 • 225 •174-•21$ •166 •228 •294 •37 5 •875L • 166 •174 •169 •187 •178 • 212 •196 •241 •193 •231 •125 L •066 •007 •133 •038 •180 •ogo •202 •1 4 U •148 •092 •I25L •142 •171 •266 •220 •2-76 • 106 • 247 •193 •231 •I25L • S54 •iUO • •252 384 •143 •257 •I2S •2oS .264 •347 •25L • 133 •038 • 114 •040 •161 •081 •l9o . -127 •162 • 102 •25L •21| •2a •178 •215 •211 2§L_ •204 •199 •240 •25-L •252 •384... •381 •4 SO •202 •3i4 •140 •232 • 220 •311 •375L • 180 •090 •161 • o&M •133 _j_0_7l_ •168 •104 ~ l £ 3 ' •108 •375 L • 720 •276 • 211 •258 •178 •214 •204 •146 •19? •142 ?32 •20| •3>75L T 4 F ' •257 '"•202 •314 •331 •423 •192 •292 •202"" •292 f33 ..235 3 • 5L • 102 •144 •190 •127 •168 •104 •137 'OSI •171 •III 131 •150 • 104 3 •5L •196 •2M7 • 204 •252 •204 •246 •177 • 212 •198 •242 3 •5L "•'HG • 208 "TT40""" •232 "•192 •292 •406 "796" • 287 •&25L •206 •I28_ •202 •169 •191 • 140 •168 •104 "•169 •109 •625L "•178" •212 "•nff~ • 227 "~ioa~~ •244 •20M •246 '~.~\W •24? •257 •625U "•'lis •190 "132" ' •200 •226 ~T92~ • 232 ".'"202" •292 •309 •75 L •204 •270 •2 05 " •201 •702 •165 • l9o . . ' J l i _ "•162 " •102 •75L •U9 •187 •I7S •202 • 187 •227 •204 •252 •t<39 '240 •75L •144 • 138 •187 •132 •200 •140 •232 "•22CT •3H •875L •200 •238 •204 •220 • 206" •189 •202 •144 •148 •092 •875L •I6G •174 •187 •178 •212 • 136 • 247 •I9i •231 •g75L •150 " •177 •144 •179 •135 •190 •126 •2o8 •264 '347 Table.13. (cont'd) DISTRIBUTION FACTORS FOR LONGITUDINALS o o * STIFFNESS ANALYSIS HARMONIC ANALYSIS I NO. SECTION STIFFNESS ANALYSIS HARMONIC ANALYSIS O 3 STIFFNESS ANALYSIS O CO :RDER NC O M Load on Girder 1 HARMONIC ANALYSIS I NO. SECTION Load on Girder 2 HARMONIC ANALYSIS 0 r-i £-1 Load on Girder 3 r-i M 23 CO O >H :RDER NC tH s CO Point Load at UDL HARMONIC ANALYSIS 0 0 SECTION Point Load at UDL HARMONIC ANALYSIS Q Pi t—i O w CO Point Load at UDL :RDER NC HARMONIC ANALYSIS SECTION HARMONIC ANALYSIS •—i o •125L •25L •375L • S L -P O X •I25L • 25L •375L •5L •P O •x. •1251 •25L •375L •5L - f •I2SL ' •03k - . 0 1 0 ' 0 73 -•013 •ICS -•02| •13s -•013 - ' 0 10 •I25L * 0 ? 5 •022 •III' •068 •1 47 •1 21 1 r ' •I60 •122 • 104 • I25L •143 •I70 •210 • 26k • 2ul\ •214 • 178 '25k •ISO '236 •25L • 0 1 3 - • 0 1 3 • 065 - •015 •038 -•on •nT' -•012 ~7o5~ - o i l •25L • 1 11 •048 •094 •063 •132 •105 • 156 •143 •133 •115 •25L •210 •26!, •163 '214 • IRT •240 •186 •258 •184 •2i)4 • 3 7 5 L • loS -•021 •098 -•on • 0 8 3 -•on .110 -•oil •110 -•oi l s> •375L •147 •121 •132 •105 • 1 0 3 • 084 • 1 3 8 •112 •139 •122 -P42 •134 •llg •375L . 2 0 4 •276 •137 .260 •164 •217 Tf89 • 252 _ _ '21,4 •fk3 •191 4 •5L •I3fe - 0 1 } •126 - • o n •110 -•011 •0S9 - • 0 0 9 •113 -•on • 0 8 9 4 • S L •165 • I60 •156 •143 •138 M I S • 1 1 3 • 092 •141 •125 4 • 5 L •178 •25k •186 •258 •m •252 •165 • 216 •122 •21(2 •625L • I5l» • 0 0 2 •li+S -•003 •131 - •008 • 110 -•oil •110 - o i l -•Oil •425L • I70 •118 •161 •158-•147 •138 •118 •139 •122 • •625L •158 •228 ' -I6T •140 •182 ~ •252 ~.\W •252 •123 •21,6 •242 • 165 •158 •146 • 105 • 75L • 170 •170 •167 • 156 •133 •75L "•14 9 " ";r&5 "|68~ ~~\W :I84 " •151 • o i l •009 — 0 0 3 - • 0 1 2 -•oil •IS2 • 179 . •147 •T43 • 115 •210 '222 -240 •258 •875L • n o •0 29 • I 6 5 •012 • 15k •Oo2 • •136 - • 0 1 3 " • 0 9 5 - 0 1 0 •g75l •168 •181 •170 •182 • 170 -178 • 165 •160 •122 •Ifilf •875L •144 •201 ~* lk9 • 210 ~ I S 8 ~ •228 " l 78 •25k • ISO •254 •I25L •021: - 0 1 5 • 0 4 6 -•032 •on -•049 •o3£ -•065 • 0 6 5 -047 • I25L •030 •000 •064 •003 •096 •013 . • 123 •03| • 0 8 k •021. •I2SL •055 • 0 2 2 •I il •068 '147 •121 •165 - (60 • 122 •loit •251-• 0 4 6 •0 H2 •665 •083 • 072 •25 L •06k •057 • 087 ..114 • 0 9 3 • 2 6 L Til ~T>9k " •132 TsT" •T33" -•031 -•0x9 - • 0 4 5 -•060 -•051 •003 •oo4 •014 •028 •023 • 0 4 8 . ' • 0 4 3 '105 •143 •115 •071 -01,9 •065 —01,5 . 0 5 6 -•039 • 077 -•052 -076 - • 0 & 3 •275L T09cT • 013 •087 •oik • 074 • 019 •099 • 0 2 3 7o38~ • 024 -fsz •088 •375L •147 • U l •132 •105 ' '109 '084 •138 MIS "• 139 • 1 2 2 S> 5 •5 L • 036 - • 0 6 5 •"o83 —060 •077 -•052 • 0 6 3 -•042 •080 - 0 5 k 75 \ •058 5 •5L •123 •031 •lik •02s ~09r* • 0 2 3 •oS l •0 19 •102 •025 5 • S L • 160 ' -156 ' •14-3 •158 •IIS •113 ' 0 9 2 . •141 '125 153 •134 • 118 • 6 2 S L •112 -•077 •104 -•072 •092 - • 0 6 3 •077 -•052 •076 - • 0 5 3 -051 •425L •137 •052 •123" •04k . . . . . . . . •034 •033 •023 ' • 098" •02k • 0 2 3 •625L •170 •118 •167 •147 • 158 •Ikl •138 IL'JL •139 •122 • 7 5 L "•Till - 0 8 5 -•oSi •104 - 0 1 2 •089 - • 0 6 0 ""•07 2 -•051 •75L •146 •070 •140 •ofco •129 •o4k •Il'4 •028 • 0 9 3 • 0 2 3 •75L •170 •182 " n o -•179 •167 • 167 •156 •143 "•"fJT" '115 •8T5L • 131 - • 0 3 0 • 124 - • 0 ? 5 •112 -•07 7 • 0 9 6 - • 0 6 S '065 -•0 47 •875L •152 •082 •146 •070 •137 • 052 •123 •031 • 0 8 4 •02.1 •875L "•I6~8 " •181 T70 •182 - - - - -•178 "•165 • 1 6 0 •722 •104 Table 13. (cont'd) DISTRIBUTION FACTORS FOR LONGITUDINALS • o S3 STIFFNESS ANALYSIS O CO M M • Q S3 STIFFNESS ANALYSIS O CO r-i H • 0 S3 STIFFNESS ANALYSIS O CO S3 O 1—1 Load on Girder 1 S3 co O ^ S3 P3 O t-H E-< Load on Girder 2 HARMON ANALYS S3 O M E-i Load on Girder 3 H M S3 CO O Q °1 O W CO Point Load at UDL 2 <c w O W CO Point Load at UDL HARMON ANALYS w fi 0 O W CO Point Load at UDL < S3 M C3 X -I2.SL •25L •375 L •5L f • g 5C •I2.5L •25L •375 L •5L OC •I25L •25L •375L •5L a: < -P •U5L •OiS — 016 '03O -•035. •054 -057_. •07O -078 •050 - 0 5 7 •125L •021 -•015 •046 -•032 •071 -0U9 • 0<i6 -•065 •065 -047 •I25L •034 -•010 •073 — OI°J •I0& -•021 •1 36 -•oi3 •095 — o±o__ ' 2SL •02.0 -•035 •032 -•0 33 •05O -052 •0£5 - -072 •056 -061 •25L •046 -•032 •042 -•029 •065 -•04 5 •039 -•060 •072' -•051 •25L •073 -•O 1°, •o65 —•A ( s •09S -•017 -126 -•Ol2 • \oS ZJ0\\_ •375L •051+ -•057 •050 -•052 •043 -•046 •056 -•063 •059 -•O'o4. D •375L •071 -•049 •065 -•045 •056 -'039 .077 -•052 •074 -•o53 ^2 •058 •375L '10^ -021 •093 - o n •083 - • o i l -IIO -•Oil . -110 -•on D 6 •SL •070 -•078 •ObS -•072 •056 -' 0 •01(6 -051 •058 _ K •161 •042 6 •5 L •096 -•065 •029 - 060 •077 -052 •063 -•042 •ogo -•054 6 ' S L '134 -•0 13 •126 -•012 • 110 -•oil •089 -•0o9 •113 :joj_i_ 16 3 •0%9> -•Oil •625L •028 -•018 •082 i ;p90_ •072 -•078 •056 -'043 •059 -•o(>4 -•0 62 •625 L •112 -•077 •104 -•072 • 092 •077 •076 -•o_53 -•05 1 •625L •154 _IP02_ •146 — oo3 '131 -•008 • 110 -•oil •110 -•0 ij •75L •/Oi -MI5_ :09L|" •082 -•090 •06 5 r;_o]2_ •o£6 - '061 •75L •124 -•085 •116 -•081 • 104 -•072 •o&9 -•060 •072 " -'051 •75L •165 •012 """•158 ~ •009 •146"" -•003 •126 -•012 •I05 -•OI| •87SL :Toq -125 •IOI -•115 -•098 •070 -•078 '050" -•057 •875L •131 -•090 •124 --08S •112 -•07 7 '096 -•o65 •065 -•047 •875L •170 •029 •K5 • 012 •154 •002 •136 -'013 •09S -•01O Table 13. (cont'd) DISTRIBUTION FACTORS FOR LONGITUDINALS o Mij - 6 G I R D E R S STIFFNESS ANALYSIS. O CO y M STIFFNESS ANALYSIS 0 CO STIFFNESS ANALYSIS O CO M M NO. s o Load on Girder 1 * S3 CO O *H • 0 S3 ss 0 Load on Girder 2 S3 CO O >H • g S3 O Load on Girder 3 S3 CO O >-< ®t <C S3 cc < GIRDER SECTI Point Load at UDL s < W M fH O Point Load at UDL S t-J S <C < S3 M EH O Point Load at UDL GIRDER SECTI •I25L •25L •375L •5L GIRD W CO •I2SL •25L •375L • S L Q w •1251 •25L • 375L •5 L GIRDER X u x 1 0 4 GIRD X n x 1 0 4 X u x 1 0 4 •> •2SL -45-0 -36-2 -1)1-4 -37.-) -37,9 •25 L 63^ 51-3 49-G 3 4.9 41.4 >25L -4-5 -i.c\ 7-7 z o . l 10.8 2 •5L -36-2 -37/? -18.^ -34-8 -37-6 -37.7 2 •5L 28-2 34-1 42-3 Li 1.2 38-6 4o-o 2 -SL 26.0 20.1 11-3 5-2 14.1 12- 4 •75L -26-o -28-5 -32-2 -37.1 -37<<? •7SL 5-5 io-8 20-6 34-1 41.4 •75 L n 3 29.3 27.2 2o,| 10,8 •251- -25-8 -22^ -33-2 -41.6 -34'4 •25L -13-1 -10-3 -2.8 7-6 -0- 1 •25L 69. 0 57.0 59.0 49 ' ! 53'9 3 •SL -44-£ -41.0 -25-1, -2S.4 - 3 U -35.4 3 •51 -n-2 7.6 0.2 -3-1 2.7 1.3 3 •5 L 43-1 49.| 55-0 S2'0 51-8 52-8 •15L -44-5 -46-2 -45.0 -41-0 -34-4 • 75L _I5.| 15.3 13-5 -7.6. -0.1 •75 L 35-0 36-6 40.9 49-1 5 3-? •2SL -11.3 -10.s -17.3 -2^1 -20. | •25L -12-2 -II.7 -10-9 -9-2 ^ S I - -6-6 •-4-1 6.0 iq.i 4 •SL -262 -23-<| -20-6 -16.8 -21-5 -20-8 4 •5L -10.^ - f l.o - 9.6 - 7.0 4 'S L 24-3 n.| 10.5 •^4 13.-1 11-4 I •75L -37.5 -344 -30-0 -23..<j -20.I •7SL - 1.0 - z . 4 -1.0 -10-9 -9-2 . •75 L 32»<i 31,4 27- 4 19'! <M •25L -3-7 -3 .5 -5<<3 -8.4 - 7 < 2 - •25L -3-S -5-6 -7.1 - 5 - 8 •25L -7,0 -43 -4'5 -1.5 -1-3 5 •51 -1-1 -8-4 -7.4 -6-0 -7-6 -7.4 5 •5L -8-1 -7 . ' l -6-0 -4.8 -6-3 -(,•1 5 •5L - c o -1-5 -1-5 -0.8 - L 2 -1.3 •75L -14.1 -13-3 -II- 0 -8.4 - 7.2 •75L -7-8 -8-1 -8-0 -7-1 -5-8 •75L 13-7 4.0 -1-5 -1.3 Table U . DISTRIBUTION FACTORS FOR TRANSVERSALS ' (a) The values, when multiplied by the simple moments, yield the transverse moments per unit length of the transverse medium. (b) The values given are for zero torsion only (p = 0 ) . 8 - 102, -CHAPTER 7. ANISOTROPIC PLATE THEORY 1. Brief Outline 8 Guyon's method of planar grid analysis, based on anisotropic plate theory, was generalised by Massonet^" to include the torsional resistance of the constituent members. for the evaluation of longitudinal and transverse moments, M and M due to the application of a concentrated load P at any point on the grid, Massonet derives the following equations i n terms of Fourier series. oo ,, / s P 1 Y~ 1 v I \ m TT c . m TT X , (x, y) = — 2 _ , ~ Km ( e ' y ) S i n ~ T ~ S i n ~~T~ m • TT m = 1 m oo „ / v 2P b V " • m TT c . m TT x /t>_x M y (x, y) = — iim sm sin — ( 8 9 ) where, 1 = length of the grid1,-' b = half width of the grid, K = transverse distribution coefficients of the load, m 3 c = distance of the load from the origin of axes, x = distance of the section at x^hich the moment is calculated, u = characteristic coefficients for transverse bending moments, m The values of K m and n m for a particular grid depend on the grid factor G and i t s torsional parameter . a , which are given by - 105 - -. In the above expressions, Pp = elastic r i g i d i t y per unit length i n the longitudinal direction, = elastic r i g i d i t y per unit length i n the transverse direction, jlfp = torsional r i g i d i t y per unit length i n the longitudinal direction, = torsional r i g i d i t y per unit length i n the transverse direction. Evidently, for zero torsional r i g i d i t y a = 0 , and for f u l l torsional r i g i d i t y a .= . 1 . For these boundary values, Massonet has prepared tables giving K and u. values for 9 ,= 0 to 3 . 1 6 2 . For intermediate values of a, the interpolations recommended are = K Q +. ( K X - K Q ) / O ? ( 9 2 ) = ^ 0 + ( | i l " ^JS*. ( 9 3 ) and, a linear interpolation i s considered adequate for intermediate values of 9 . Note, however, that for the Fourier series terms of Eqs. 38 and 8 9 , = m 9 . Therefore, i t is only possible.to consider the f i r s t 2 or 3 terms without f a l l i n g outside the range of 0 given by Massonet. After the coefficients . K and \x have been determined, the required longitudinal and transverse bending moments may be computed from Eqs. 88 and 8 9 . .However, as the Fourier series tends to converge somewhat slowly for the bending moments of loaded girders, i t is advisable, i n the author's opinion, to r e s t r i c t the use of these equations to unloaded girders only. The bending moments of the loaded girders are obtained readily by subtracting the sum of the bending moments of the unloaded girders from the free bending moment, regarding the entire grid as a simply supported beam. For the purpose of investigating the accuracy of the anisotropic plate theory coefficients given by Massonet, a five longitudinal grid with seven transverse beams was analyzed for several loading conditions, f i r s t neglecting - 106 -and then including the torsional r i g i d i t i e s of the members. The comparative results of stiffness analysis and plate theory are summarized i n Tables 17 and 18. In order to i l l u s t r a t e the procedure of Massonet's method, the typical calculations involved i n the example are given below. 2. Five Girder Bridge Example 4\ x / / Fig. 32 Five girder bridge for the comparison of stiffness and anisotropic plate theory results Step 1 The grid-parameters, for E = 1 and G = 0.5, are E I L 10000 ?9- e 15 =^ 666.67 units - 107 -E I T - 337 12 .5 = 27.0 u n i t s k 666^67 m Q M G J L G J t 0.5 x 7740 0 .5 x 250  15 12 .5 2• ^ 666;67 x 27 ' -= 1.0 S t e p 2 T h e l o n g i t u d i n a l a n d t r a n s v e r s a l b e n d i n g moment d i s t r i b u t i o n f a c t o r s K a n d u-, d e t e r m i n e d f o r v a r i o u s v a l u e s o f t h e g r i d f a c t o r 8 b y i n t e r -p o l a t i o n f r o m M a s s o n e t ' s c h a r t s , a r e l i s t e d i n T a b l e 15. L o a d o n G i r d e r G r i d F a c t o r T o r s i o n a l P a r a m e t e r K f o r G i r d e r s x 10~^ f o r t r a n s -v e r s a l s e OC 1 2 3 4 5 1 2 3 4 5 1 0:669 0 1 5.98/+ 2 .811 2.113 1.511 0.120 0.733 -0 .517 0.364 -0.692 0.202 0 0 -2123 - '591 -1418 - 419 -435 -213 0 0 y " = - b 1.338 0 1 12.257 5.608 0.223 1.387 -0.562 O.295 -0 .093 O.O67 +0.113 0.016 0 0 - 873 - 216 - 256 - 62 - 73 - 15 0 0 2 0.669 0 1 2.113 1.511 1.919 1.477 1.055 0.995 0.195 0.580 -0 .517 O.364 0 0 1306 30 - 103 - 322 -175 -221 0 0 b y 1.338 0 1 0.237 1.387 3 .115 2.297 0.736 0.828 - 0 . 1 0 0 0.190 -0 .093 O.O67 0 0 892 - 60 - 150 - 77 - 6 0 - 23 0 0 O.669 0 1 0.120 0.733 1.055 0.995 1.618 1.248 1.055 0.995 O.120 0.733 0 0 266 - 86 1828 1096 266 - 86 0 0 3 y = 0 1.338 0 1 -0.562 0.295 0.736 0.828 3 .008 2.125 0.736 0.828 -0.562 0.295 0 0 - 62 - 84 962 606 - 62 - 84 0 0 2.C07 0 1 r 0 . 0 f i 6 0.050 +0.120 0.55 4.460 3.125 +0.120 0.55 -0 .086 .0.050 0 0 - so - 30 596 350 - 80 - 30 0 0 T a b l e 15. A n i s o t r o p i c P l a t e T h e o r y D i s t r i b u t i o n C o e f f i c i e n t s S t e p 3 T h e l o n g i t u d i n a l b e n d i n g moments M , p e r u n i t w i d t h o f t h e g r i d , K due t o a s i n g l e c o n c e n t r a t e d l o a d , P = 10 , a p p l i e d on g i r d e r 1 a t c = 1 /8 , - 108 -as shown i n Fig. 32, i n accordance with Eq. 88, are L F / \ 10 X, 100 f v Tf . . TT X K2 . TI' . 2 TT X S . K x ( X> y ) = 2 ( K 1 S l n 8 S i n T" + 4" S I N 4 S l n ~T~ + T S i n J U TT 3 TT . 3 TT X s . . _ _ S L N _ _ + ) Substituting X = ^/8, for the transverse section A, K K ' / T / R T \ ~ O R T / „ TT . TT 2 . TT . TT 3 • 3 TT • 3 TT M x ' ' > y ' = 3 , 3 ^ 1 S I N 8 S I N 8 + T s i n 4 S I N 4 + ~ S I N ~ S I N ~8~ Using the K values given i n Table 15, the bending moments of the unloaded girders at section A, for a = 0, are M2A = 3 , 3 8 X 1 5 ( 2 ' 1 1 3 x °-3S22 + 2*221 X 0.7072) = 17.20 k.ft. M3:A = 3 * 3 8 X 1 5 ( ° * 1 2 0 x °«3822 - x 0.7072) = - 2.68 k.ft. M4A = 3 , 3 8 x 15(^°.517 x 0.3822 - •°-I|23 x 0.707 2) = - 4.44 k.ft. Mr. = 3.38 x 7.5(-0.692 x 0.3822 + X 0.7072) = - 2.21 k.ft. +" 7.87 k.ft. The bending moment of the loaded girder at section A i s now obtained by . subtracting the above sum of the loaded girder moments from the simple moment. Hence, M. = 10 x 12 5 x 87.5 _ 7 > 8 7 = 101.50 k.ft. 1A 100 Step 4 The transverse bending moments M , per unit length of the grid, k T due to the same loading i.e. P = 10 at c = 1-1/8 on girder 1, using Eq. 89, are „ • / \ 2 X 10 X 30 / Tf . TT X . TT . 2 TT X x M y (x, y) 1 0 0 ^ sin -g sin — + ^  sin ^  sin + ) - 109 -For the transverse section A, at which the load i s applied, s u b s t i t u -t i n g x = L/3, M y (L/8, y) =6.0 ( n x s i n | s i n | + ^ s i n ^ s i n ^ + . ) Referring to Table 15 f o r the nontorsional [i values and taking the f i r s t two terms of the Fourier s e r i e s of the above equation, the bending moments of the transversals are obtained as M Al M, _ = 0 A 5 H A 2 = 6.0 x 12.5(- .2123 x 0.3B22 - .0873 x 0.7072) A3 6.0 x 12.5(- .1418 x 0.3822 - .0256 x 0.7072) M., = 6.0 x 12.5(- .0435 x 0.3822 - .0073 x 0.7072) A 4 - 7.56 k . f t . - 2.52 k . f t . - 0.75 k . f t . Step 5 For the purpose of comparison, the transverse d i s t r i b u t i o n f a c t o r s , p at section A, are c a l c u l a t e d f o r both the plate theory and s t i f f n e s s analyses from the respective f i n a l l o n g i t u d i n a l bending moments and shown i n Table 16. Girder Longitudinal moments ( k . f t ) P D i s t r i b u t i o n f a c t o r s K Massonet S t i f f n e s s Massonet S t i f f n e s s 1 K 1 A 101.50 94.90 92.8$ 86.7$ . 2 M2A 17.20 21.69 15.7? 19.9%' 3 M3A - 2 .68 - 2 .00 . - 2.5% - 1.8$ 4 \ A - 4.44 - 3.18 - 4 . 0 $ - 2.9$ 5 , K 5 A - 2.21 - 2.04 - 2.0% - 1.9$ M f r e e 109.37 109.37 100. % 100. % Table 16. Comparison of transverse d i s t r i b u t i o n f a c t o r s f o r s ection A . - 110 -Likewise, the comparative distribution factors of a l l the transverse sections were calculated for 12 different loading conditions, for both tor-sional and non torsional cases, and are.shown i n Table 17. In addition, the bending moments of the loaded transversals, as obtained by the plate theory and stiffness methods, are directly compared i n Table 18. SECTION P-l on Beam 1 at .12% P=l on Beam 1 at .25L P=l on Beam 1 at .375L P=l on Beam 1 at ..5L roRS SECTION DISTRIBUTION FACTORS FOR GIRDERS DISTRIBUTION FACTORS FOR GIRDERS DISTRIBUTION FACTORS FOR GIRDERS DISTRIBUTION FAC • FOR GIRDERS SECTION 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5. .125L • see •922 • 157 - -0 18 -•025 -•oz<=\ -•OM -•oiq •739 •834 •377 •333 -•012 -.036 -.044 -'025 -•041 —045 •6 31 •714 •50S •504 • 024 -•01? -•01<i -•'090 — -II5 -•045 -•075 •552 •5?9 "•587" '614 • 590 •634 •556 -599 '077 '045 -12S -I53 -<o9o -.108 -•OS} -•too .2% •739 •834 •377 •333 -•Oil —• o 36 -•o&4 -'0S5 -•041 -•045 • 7 70 -832 •327 •3o3 -•000 -•022 -•059 -•076 -•03S - • o B •662 •727 •462 •443 •023 -•007 -•04.0 -•070 •064 •037 _.H7 -M46 -.10 I" .375L •631 •714 •508 •504 •024 —018 -•099 -•127 —065 -•075 • 462 -727 • 462 •443 •023 -•oo'r -•090 -•115 -•o4o -'070 -•083 -.100 •7llf •76o '388 •??6 •026 •003 -• 076 -0% -•053 — 062 •630 •659 • 495 •532 •049 •029 -072 -087 .5 L •552 •589 •5^0 •631) •077 •oi)5' -•I2& -•153 - '090 -•log .581 •6 14 •556 •599. •064 '037 -• 117 -•144 •430 •651 •495 • 532 •043, '02° - • IO I I31 -.072 -•087 •693 •721 -4o6 '434 •037 • 022 -•081 -•108 -.101 -'131 -117 -146 -.0S9 -'070 -"•07T -•oil -•o%i -•loo .625L •415 • 428 •63« •776 • 131 •122 -448 -•iss -II4> --I40 • 520 •480 •6/3 •724 •110 •o96 -•13? -•175 -.104 _ .H2 •565 •561 •569 '64o •077 •o43 - I 2 3 -I55 -•o9o -.109 '630 •659 •495 •532 '649 >0l9 .75L •458 • 291 •£,58 •844 •116 'III, -M89 -•201 -M33 --I6S •480 •362 •458 •191 • &45 •814 •652 •864 •149 •163 -•152 -'192 -•122 --I52 •520 1 •4S0 •613 •724 .110 '096 -•I38 -•I75" -.104 -<\2$ •581 •614 •586 •599 -064 •037 .875L • 437 •202 •66S •906 •204 •227 -•165 -•209 -•146 -•187 •176 '114 -•159 -•202 -•133 -•168 •495 •428 •6 36 •776 •131 '121 -•I48 -I88 -.114 -•140 •552 •58°i •590 •434 •077 •045 -•128 --0^0 -• 153 I--I08 .125L •712. •682. •212 •218 •051 •ot i 'C 20 •025 •004 •607 •467 •403 •366 •32G .113 •134 •0U5 •OGO «0o9 •016 •J!6 •255 • 426 •439 •112 •193 -071 •OS9 •0(6 • 024 •233 •2.2.4 •43o •403 •7.20 •21S •097 -11 3 •Oil _-_o3|_ •020 .019 .25L •46"? •4t>3 •344 •3S6 * (13 •136 'OL.S •060 • 00^ • Ol6 • 546 •5o2 •302 •310 •102 •Uo •041 •054 •009 •o 14 •372 •334 •391 •392 • 156 •172 •o46 •081 .014 •022 •269" • 249 •419 •4o9 •203 • 2 IO • o4o •105 .375L •316 •255 •426 •439 •172 •193 •07| •o?,9 •016 •024 •372 •334 •391 •392 •156 •172. •oU •081 .014 •012. •474 '442 •323 •320 •132 •143 •057 •069 •013 •019 .017" •025 .022 •015 •340 _ ' i i iL • 454 •43o •39o •335 •I76 •182 •07 S •063 •074 ~"o~7S •091 •017 •02S •014 ' O i l .5 L •233 •224 •430 •4o3 •220 '128 •097 • 113 •021 •631 •269 •249 •419 '409 •203 •210 .olo •1 05 •020 .07.9 •340 •314 •390 •385 .176 •in -078 •091 •097. •123 •326 •325 •(42 •I50 TiTtT •I82 .625L •192 •234 -4ofe •357 •153 •241 •120 •131 •027 •03S • 214 •213 •414 •3S? • 27.S •231 •III •123 • 02S •029 •039 •ob2 •042 •2.57 •240 •414 •40S •in -340 •3i4 •39o •385 •017 •025 "020 •029 .75L •174 •202 •383 •36M .273 •2M0 •I3S •14 1 • i5o •I7& •032 •042 •185 •218 -39S •369 • 261 '240 •273 •240 -128 •134 • IIS ' •I4l -214 •223 •414 •3$S •238 •231 • III .123 •025 •015 '249 •249 •4H •4o9 • 203 •2\0 •o9o •105 .875L •165 •150 •367 •4C>1 •283 •728 •OiS •046 •174 •20?. •383 •364 •192 •2*4 .406 •357 •253-•241 •170 •131 •027 •038 •233 •214 •430 •4o3 • 220 •22g •047 •113 •Oil •oil Table 17. DISTRIBUTION FACTORS FOR LONGITUDINALS (a) These factors, when multiplied by the simple moments, yie l d the longitudinal moments. (b) The lower values correspond to anisotropic plate theory, and the upper values correspond to s t i f f n e s s analysis. 0 0 - 0 U J vo —J <_n -IT — vJ O O CO - 1 ON: to to U J to x- vn in w )o fj O os i ° — i O ro — Ul Os —J -X> CO to to O >o — O O O -j Cjs — _o - a O ty- U J o o OS os crs 0 [O UI fo to CO -C-to >J OS 00 U J CO O o VoO UJ U* o UJ 4r jr Co XT co X" Oi CT <TS ro -c CP UJ Vo - o O _D O o OJ O IJ r o o o -X> OS — o <s> — ev <rs S • o O l/i i -OJ o CO JO jo to w to T— us 0? u> JO yJ O OS fo CO -O UJ V O Us —O GS to e> U J —! cr- U J 0 Wi ° -O O | 0 O i tr-sCT- os 0!-» xr to IO CO s-» Os 0 0 -4 o o o cs iTi -UJ o oa . to to 0- OO CO O CS cs - J to © o US JT M UN CjJ UJ —J -D O o to f O (JN us c oo UJ to X" us Ui Ul UN — Ul in VP Ul O O O CO co _£> [O O O 3 ^ J3 Co UJ U> J T 6 S CO CO IO (O S I U' CS - J U J ro o t° tO (O JD (/i u i (O to CO CT-us CO to fO —1 03 _C C° U J o — W — _ -O o OO JJ T~ UJ -J o O 0 I O O U J U J o o N O - 4 -J O 00 -J —t o o c o U co Jo -o VJJ u> U O U J cr- cs O -J to VO Ul US OS oo X" -Cr Cs _ O O O J OS _t> OS cs O 1° 3 ^ to JO o — us U J OJ OS o o U> ,o crs ui UJ to O C to to VTS vi O oo U J fO ^ Ul O O UJ — U> to UJ U to IO CO OJ UJ — -JO —J o x-ui ir-^ J ! to to ro 1 to iw to f j ; f l J ,o 5s COio"- CV5 V." oCi CO o 0 0 *j 1 cr-O Ui —J o o : o 0 _1 _0| ( O Ul JD JD \f> Ul OS 0^  IV UJ - 0 Ul to (J OS x- r -ON — o o •c UJ OS _D ^ D u- UJ -O o — U J ui|J-~ o I |o I-' „ , w -0 -—J JO r CO 4T U J u' 1 OJ cs cs y. oto ,o 0 1 'JN ON oS e g M N OS CO UJ CO sr ess: o O ^ U J ON _0 O O 1-° V, Co - J O Ul CO Jrr- UJ U> Ky <JJ sr. OJ UJ OS 00 to to O - J JO 0 0 U* Ul VP _ — o a c» Vol r 4 Os ro r 01 O 1 (fi UJ (O CO -N -o Co co U» U J to jo o — VJJ o — UJ o S - o ' UJ 0J U J o o C o — (?s -0 o ~1 U J M - -0 -0 (tl U J U J [o 0 0 UJ Uj Ss _D OS O 0 VJJ ¥5 M hJ Ul ui (J X r- r -Ui UJ o UJ o o U J JT (O o 6 <^  Ul V o ro U3 0~- Ul O J - J to ~0 to si Ul Co o to VJ UJ U J to co O O —o 00 O O JT ( J to ^1 o o to — SECTION ro a M 00 i-3 5^ W O r-l S3 Cd c : O >-3 S O > o t-3 O CO CO XT U J i_o to (_)J UJ to tO O — to po U J _ o U J o _ o O CO — c U J o to UJ UJ to ^ o _o o o -o —1 crs os U J cy UJ UJ uJ O UJ UJ U> "JN o o 00 (T-- O . _0 XT -c-— O ui ro U J U J 0 ?• o x; o o u i —J -8 Ul -4 UN H (O N> to CT^ —J 6 ^ UJ s r Ul Ul U l -c-CO "1 Ul ffS — jr Cfs ^ ~1 UJ O 6 ^ U J CO O o UJ JO as ui UI CO to to W uJ fO co O o •02 Co X- yJ to jo ro VJO CO 3 S O i-3 M M w o o s w > o >-3 O 3: UJ U J 00 ~ " co CO V-'J to as c> 1 0 OS U J ro JO CO O U ) VO jo x-° ^ 0 0 oO _JO 00 jr to to U J U J to _ K !° Ul Ul N x-U J tv VJ U J — JS-IjJ U J £T Vi CO O fe vV x-OT CJS vA es u> O -J ON C CP Jr X* X" U\ v.i> ro ^ t,' 0 p O *s UJ 10 O c-U J u, to IO tO _ l -O — CO to to — tJ. U l UJ to to crs _j UJ U" a ° O (_o S J J 0 — 0 — - 1 0 -O UO 0 0 (T- 0 0 U l j-r 0 6 ua — 0 6 U J x W O 0 O U J x-UN ui 0 0 U J Jr. JO £-I I 6 0 - 0 -J x- x -1^  t 0 0 OO t>s. -4-0 . < 1 i 0 — 1 t)s Oo 1 1 £ U N 1 1 0 0 J > (JJ Ui Co \ 1 ° 0 1_ 1 6 6 » s ui \_o 0 o i o M c o 1-3 r-l M ?3 O C3 S M f» ^ c o > o t-3 , O VJJ CO UJ NJ UJ N» i to to 1 >o to =" —" ffs x- — C O CO ofj - J (jj u t-' cs r -ffs co UJ N 0 —' 0 0 0 VJJ to — _o -o cn r- 1 U> (jJ —^ cr-to c-J r - c-— 0 Ul to r - x-CO crs — Ol U"i u i •» Ul r x-0 0 "s — UN x- x-_ 0 US lO UJ io -O ffs to co ro. \ J J uJ CP OS U J UJ 0 0 0 fr to to -J —1 — CO IO K> crs OJ to IO -4 —1 — CO UJ UJ 0 0 0 J^ -UJ uJ oo cK OJ 0 0 ss 00 _ i i ITS 0 0 u-» —1 -O Ul 0 0 CO — 0 6 uJ x---4 ~J 0 0 X- cs CO — -D UN, 0 0 os 0 0 -£> (js f> l_ 1 0 0 ~ J 0 -Cjs-x: 1 f 6 c —j Ul 1 1 c> O Crs- _ 1 1 1 1 6 6 1 1 Q 0 M 'JS UJ J 3 1 0 r> 0S_r-ON a -0 M II CO r-» i-3 0 O H 3 WCd G co O H CO M r - I 03 w o a s ro C 0 > 93 0 «-a 0 • CO R II - SIT -SECTION P=l on Beam 3 at ;12% P=l on Beam 3 at .2% P=l on Beam 3 at .37% P=l on Beam 3 at . % SECTION DISTRIBUTION FACTORS FOR GIRDERS DISTRIBUTION FACTORS FOR GIRDERS DISTRIBUTION FACTORS FOR GIRDERS DISTRIBUTION FACTORS FOR GIRDERS SECTION 1 2 3 h 5 1 2 3 u 5 1 2 3 h 5 1 2 3 h 5 .125L _ -O 12 •141 •114 •732 •715 •141 •114 - -oog - • 0 1 2 - -ooC - • 6 18 - < 000 - ' O i l •238 •232 • 5 1 5 •572 •232 •23 2 - • 0 0 6 -•018 • 0 1 2 -009 • 3 o 3 • 3 o o •376 •420 • 3 o i • 3 o o . 0 1 2 - • 0 0 4 • 0 2 2 •316 • 3 1 8 •212 • 3 2 0 • 516 •318 •038 • 022 .2% - o o t - -o 13 •2.33 •232 •51s •572 • 2 5 8 •232 -•00& -•013 • 2 0 8 • 1 9 3 •584 • 6 3S • 2 o 9 • 193 -•000 - • o n • 0 1 2 -•O03 •*77 • 2 6 6 •U15 • 415 •277 - 2 6 6 • o i l - • 0 0 3 •032 •019 • 3 o 4 • 3 0 0 • 3 2 8 • 3 6 4 • 3 0 4 • 2 . 0 0 • 0 3 2 • 0 1 ^ .375L • O i l - • 0 0 3 • 3 o 3 • 3 o o •370 • 420 •303 • 3 0 0 • 012 -•oo9 •OI2 -•00b •277 • 2 6 4 •425 •475 • 2 7 7 • 2 6 4 • 0 1 2 -•O63 •01 J • 0 0 2 • 2 2 3 • 2 1 5 •516 •56o •223 •215 • 0 1 3 • 0 0 2 • 0 2 4 • 0 1 5 •278 • 271 • 3 9 6 •4 29 - 2 7 8 •211 •02-U •Ol 5 .5 L •032 • 0 22 •316 • 3 i 8 •212 • 3 2 0 •316 •318 •038 • 0 2 . 2 •03Z • 3 o 4 • 3 o o • 3 2 8 •364 •3o4 • 3 o o -032 • 019 • 0 2 4 • 0 15 •278 •271 • 3 9 6 •429 -27S •271 • 0 2 4 • 0 1 5 • 0 1 8 •Oil •231 •216 • 501 •546 •211 • 2 / 6 •OI8 • O i l .625L •066 • 0 6 1 • 3 o 6 •300 •257 •281 •306 • 3 0 0 •06 6 •061 •oss •04g •3o5 •3o3 •2 76 •29S •30 5 • 3 0 3 •055 • 0 4 8 •03$ •031 •3o2 •292 - 3 1 3 •354 •302 •192 •038 •031 • 0 2 4 • 0 ( 5 •272 •271 •396 •429 • 2 7 8 •271 •O2.4 • O l 5 .75L • 0 8 8 .101 • 2 4 0 •233 •242 •320, •210 •2 33 •O88 • 1 0 7 • 0 7 5 •o?i •199 • 2 8 0 •253 • 2 7 5 '299 •2So - 0 7 5 •OSI •OS5 •01,8 • 3 0 5 • 3 0 3 -276 • 2 9 8 • 3 o 5 * 3o2 -055" • 0 4 3 - 0 3 2 •01 9 • 3 o 4 | -323 •300 j - 3 6 4 • 3 0 4 • 3 0 0 -032 • O i l .87% •102. • 1 4 4 -•213 • 2 0 2 • 2 3 9 • 1 7 8 i -102 • 2 0 1 | • 1 44 • 0 8 8 • 1 0 7 •2.qo • 2 3 3 •2.41 '350 • 2 4 0 •233 •OS a - 1 0 7 •06fa •061 •3o6 •300 - 2 5 7 • 2 8 1 • 3 0 6 •3oo • 0 6 6 • 0 6 1 •038 j ' 3 i 6 j -212 •022 I -31<S ! - 3 2 0 •3(6 • 3 l S • O J 8 • 0 2 2 .125L ' 02.6 • 0 3L, • 1 3 4 • l 16 •678 •701 •136. •116 •6 2.6 •034 • 056 •068 •237 •226 . 4 1 4 • 4 1 3 • 2 3 7 •n6 •056 • 0 4 8 • 0 8 6 •016 •2"8I •283 •267 • 2 ^ 2 •281 •283 •o16 • 1 1 0 •114 •2&8 • 1 6 3 • 2 0 5 • 2 4 6 • 2 8 S • 1 4 3 • 1 1 0 • 114 .25L •056 • 0 6 8 •2il • 1 2 6 • 4 1 4 . 4 1 3 •237 •224. •056 .068 •051 • 0 6 0 • 111 •191 •505 •499 •197 • n i •051 • O 6 0 •07 8 • o s 6 •158 •260 • 3 3 0 • 3 o 8 • 2 5 8 • 2 6 0 •078 • 0 8 6 • l O l • 1 05 •180 •248 - 2 3 8 • 2 5 4 •280 • 2 6 8 • 1 0 2 • 1 0 5 .375L •086 •"096 • 2 8 1 •283 • 2 6 7 •242 •281 • 2 8 3 • 0 8 6 •096 • 0 7 8 •086 • 2 S & •260 • 3 3 0 •108 •2 58 •260 • 0 7 8 • O S 6 • O 6 6 • 0 7 2 • 2 1 3 • 2 0 6 • 4 4 0 •441 • 2 1 3 •20 6 • 0 6 & •072 • 068 • O i l •25S •256 • 3 0 7 • 3 0 6 •258 •256 • 0 8 8 •oq\ .5 L •no • 114 •288 •263 •205 • 2 4 6 • 2 8 8 • 2 6 3 • \ 10 .114 • 1 0 2 • 105 •230 •268 •238 •254 • 2 8 0 •268 •102 •105 •088 • o i l •25S •256 • 3 0 7 • 3 o 6 • 2 5 8 •256 •01% • o i l -071 •075 •216 •214 •425 • 4 2 3 • 2 1 6 .214 • 0 7 1 •075 .62% • 1 2 7 •121 • 2 7 8 •269 •11.1 •222 •278 -269 1 ••121 • i l l •116 • 2 S 0 •273 • 2 0 2 • 2 2 2 •2SO •273 -114 •H6 • 106 •119 •118 •270 • 2 3 3 • 2 4 8 •278 • 210 •106 - u q •OSS •091 •258 •256" • 3 0 7 •306 • 2 5 8 • 2 5 6 •0%2 ' • o i l .7% •117 •120 •267 •202 •144 •35& •267 • ^ 0 2 •137 ' •I IO •131 •110 •273 •134 •143 •2*7 •273 • 2 3 6 .12.1 • 1 2 0 •M1 • M 6 •230 • 7 7 3 • 2 0 2 • 2 2 2 •-2-SO •273 • 119 •116 •102 . •105' •280 • 2 6 8 •2 3<J •254 • 280 • 2 6 8 • 1 0 2 •I05 .87%. •142 . •114 •258 •140 •2O0 •493 •25g • 1 4 0 •142 • 1 1 4 •137 • 1 2 0 •267 •2.02. • 114 •358 •2C,7 • 2 0 2 • 117 • 120 •117 • 121 •278 • 2 6 9 •111 •22.2 • 2 7 8 •261 •127 • 121 •1 10 .114 •V6% • 2 6 3 • 2 o 5 • 2 4 6 •2S& • 2 6 3 •1 IO •114 cc = 0 a = 1 Table 17. (cont'd) DISTRIBUTION FACTORS FOR LONGITUDINALS LOAD POSITION TRANSVERSE BEAM BENDING MOMENTS OF LOADED TRANSVERSALS Nontorsional Case (a = 0) BENDING MOMENTS OF LOADED TRANSVERSALS Torsional Case (a = l ) TRANSVERSE BEAM 1 2 3 5 1 2 3 k 5 TRANSVERSE BEAM Right Left Right Left Right Left Right Left Right Left Right Left Right Left Right Left .P-10 kips on Beam 1 at section .12% A . o o -5-1,0, -7-56 - S-56 -7-5C - i - i i - 2 - 5 2 -I-II -2-52 - 0 . 3 5 - o-75 -o-35 -0.75" 0. 0 7.6 1 0 -1.01 - 1 -77 -0.24 - 1 . 7 7 - 1-77 - 0-69 - 0.15 - 0-61 - o-sg -0-2°j — 0 • o~t - 0.23 -0.14 0 .P-10 kips on Beam 1 at section . 2 % B o o - II-3L, - 15-10 -11.34 - 15-90 - 4-56 - 7-24 - 4-56 -7-24 - 1.18 - 7. .13 -1-18 - 2 - 3 3 0 0 II .11 0 -15-22 - 4-o2 0 . 3 1 -4-o2 - S-2C - 2-o4 -O .33 - 2.04 - 1-17 - o-95 - 00.3 -0.95 - o-i,8 0 .P-10 kips on Beam 1 at section .37% C o o -14-21 -17-33 - 14.21 -n-38 - 7.7 1 -10-00 - 7-71 - 10.00 - 2-2.3 - 4 - U - 2-23 -0 0 12-59 O -16-75 - 4-65 0 .qo -4-65 -7-62 - 2-92 -0-46 -2-92 - 5-28 - 1-43 - O. I S -1-43 - o-63 O .P-10 kips on Beam 1 at section .51 D o o -14.37 -17.60 - 14 -17 - 17-66 -$.•.39 -10-53 .-8-3 9 -10-53 -1 -88 - 3 - 5 7 -l-SS-- 3 - 5 7 0 0. .1 2 • (6 1 O -16-99-. - 4-81 1 . l S -4-%l - 7-7o - 3.IU - 0-3>l _ 2,. 1 4 -2 .61 - 1 'CO - p . 21' - I .60 - o-6cl O U N O OH - H • H & - p O CQ to II c 0 - O - P a .12% A 0 0 4-80 1 4.00 4-80 1 4-00 --2- 2 o - i - 4 o - 1- 2 0 - l-4o - O-EI -0 -47 - 0 . 2 1 - o-42 O O -2-CJ3" 0 "4-6 6 -0-5-| ' 5 - 7 7 - 0-51 - 4 - 7 1 - 0 -65 0-og -0-65" - I • 06 - 0 - 3 3 -o-33 - 0 - 2 4 O .2% B 0 0 4-06 1 5 -00 I . 0 6 l5-oo - 3 . 1 0 - 1-&5 - 3 - 1 0 - 1-85 - 1 • 1 0 -1-10 - 1 ' 1 0 - 1 , 1 0 0 O -3.41 0 7-43 -o-52 IO-I 2 - 0 - S 2 - 7 . 4 5 - 1 . 8 0 c s o -|.80 - 3. OS - I -00 - 0 . 1 5 - i.oo -0-7S O .375L C 0 0 10-36 16 - 2 o ' 10-36 16-5.0 -.2.65 - L4o - 2-65 - 1 . 4 0 - 1 -63 - 1-34 - 1-63 -1-34 O 0 -2-66 0 7.%8 -o.oq 1 1-47 - 0 . 0 1 -7.71 - 2 - 3 4 I • 4 0 -2-34 -4-37 - I-50 - OM5 - I.SO - 1 - 2 7 0 .5 L D 0 0 IO-53 l6-6o IO -S3 16-60 - 1 . 5 5 -1. 4 4 - 1-55 - 1 . 4 4 - 1 . 4 2 -1-31 - 1 .42 - 1-31 O 0 -2-34 0 7-89 - o - i 5 1 1-75" - 0 - I 5 -7-76 -2 -41 I-5S -2-44 -4-73 - I-65 _o-U| - I-65 - H U 5 O Table 18. COMPARATIVE RESULTS OF TRANSVERSAL BENDING MOMENTS (k.ft.) (a) (a) The upper values correspond to St i f f n e s s analysis, and the lower values correspond to Anisotropic plate theory. 1 I - 115 -CONCLUSIONS The principles of the stiffness method have proved to be extremely-efficient for the general analysis of grid frameworks. The technique employed i s readily applicable to the analysis of both planar and spatial grids of any shape and complexity and, as no approximations are resorted to, the results obtained are exact. Furthermore, without any considerable change i n the formulation of the problem, the effects of torsional, axial and shear deformations, variable moments of inertia and support settlements can be easily taken into account. It is also a point of interest that the same for-mulation remains valid for the analysis of l a t t i c e modelwbrks into which a plate or she l l may be idealized. In contrast, the numerous hand calculation methods developed so far are based on highly theoretical and complicated backgrounds, such as anisotropic plate theory and harmonic analysis. However, realizing the complexity of 3 li 6 13 their own methods, several authors ' ' ' provide numerical tables of. moment distribution factors covering a limited range of problems. For a grid f a l l i n g outside the range of these tables, extremely tedious calculations are involved and the pos s i b i l i t y of a general solution becomes impractical. Hendry e Jaeger's and Massonet's moment distribution factors are found to be reasonably accurate only i n the immediate v i c i n i t y of the applied loads. • In unloaded areas, the errors are substantial despite the fact that the example grids analyzed for comparison were especially arranged to have closely spaced transversals, i n order to meet the authors' assump-tion of a uniform transverse medium. In harmonic analysis the errors arise partly because the moment dis-tribution factors are obtained without regard to the position of the load - 116 - , along the girders and, further, they are assumed to remain constant at every transverse section. This assumption i s completely unjustified by-stiffness analysis, because the exact distribution factors are different at every transverse section for every position of the load, as clearly indicated i n Tables 5 to lk. For instance, i n the torsional case of the five beam grid of Table 11, 35 different stiffness distribution factors ranging from .254 to .755 are tabulated against Hendry e Jaeger's single value of Pn = 0.UZ2. ' . . Although Massonet's anisotropic plate theory approach provides d i f f e r -ent distribution factors at every section for each load position, the degree of accuracy i s s t i l l insufficient. Table 17 shows that, i n loaded areas the errors normally l i e within 10 - 20,?, but, away from the loads discre-pancies from 50 - 100$ are not uncommon. • < Besides being approximate, most of the hand calculation methods are limited to ' specific types of problems. Consequently, the same method may not be applicable to any two different grids. The stiffness method, however, is completely general and is not restricted to any particular type of grid. Moreover, the stiffness approach is so versatile that, once a general computer programme has been developed, the analysis of any type of grid i s reduced merely to the c l e r i c a l job of f i l l i n g out the basic geometric and elastic data on a standard form. In view of the facts stated above about the relative merits of s t i f f -ness analysis, the author is convinced that-, upon the a v a i l a b i l i t y of a d i g i t a l computer, hand calculation methods of grid analysis are rendered obsolete. - 117 -NOTATIONS STIFFNESS ANALYSIS: A = Cross sectional area; a,b,c = End s t i f f n e s s c o e f f i c i e n t s of a member with variable moment of i n e r t i a ; = Column vector of joi n t deformations of the sytem; d = Depth of a prismatic member; -EE = E l a s t i c r i g i d i t y of a member; {F} = Column vector of f i x e d end reactions of a member; FEM,FER = Fixed end moments and shears, respectively; GJ = Torsional r i g i d i t y of a member; J = Number of j o i n t s i n a structure; [K] = S t i f f n e s s matrix of the sytem"; [k] = S t i f f n e s s matrix of an i n d i v i d u a l member; L = Length of a member; l,m,n = Direction cosines of the member axes with respect to the common axes; - Number of joint deformations; = Column vector of joint loads of the structure; = Column vector of f i n a l stress resultants of a member; = Number of support r e s t r a i n t s ; or, radius of curvature; = Transformation matrix; = Common and member axes; = Co e f f i c i e n t of thermal, expension; -= Column vector of f i n a l end deformations of a member; = • Change i n temperature; = Shearing s t r a i n parameter; = Shearing stress numerical fact o r ; HARMONIC ANALYSIS; EI, Elfj, = E l a s t i c r i g i d i t i e s of the longitudinals and transversal respectively; F = Shear force; GJ T = Torsional r i g i d i t y of a l o n g i t u d i n a l ; - 118 -h = Spacing of the long i t u d i n a l s ; L v = Length of the g r i d ; M = Bending moment; n = Number of cross girders; w = Load; y = Deflection; OC = Grid s t i f f n e s s parameter;' = Torsional parameter; = Ratio of EI , / EI, outer inner P V. 9 - Slope ; = Transverse d i s t r i b u t i o n f a c t o r s ; ANISOTROPIC PLATE THEORY; b = Half width of the g r i d ; c = Distance of load from the o r i g i n of coordinate axes; d,e = Spacing of transversals and lon g i t u d i n a l s , respectively; K , = Longitudinal and transversal bending moment c o e f f i c i e n t s , m * m a. • T respectively; 1 = Length of the g r i d ; x = Distance of the section at which the moment i s calculated; OC = Torsional parameter; , X_ = Torsional r i g i d i t y per unit length i n the lo n g i t u d i n a l and r Ci j _ , . , . 9 transverse d i r e c t i o n ; = Grid factor; = E l a s t i c r i g i d i t y per unit length i n the lo n g i t u d i n a l and . transversal d i r e c t i o n ; APPENDIX B ; J = Torsional constant; - 119 -BIBLIOGRAPHY . 1. EWELL, W.W., OKUBO, S. and ABRAMS, J.I., "Deflections i n Gridworks and Slabs", Trans. A.S.C.E.,.Vol. 117, 1952, page 869. 2. WOINOV/SKY-KRIEGER, S., "Zur Statik und Kinetik der Tragerroste", Ingenieur Archiv, XVII. Band, 5. Heft, 1949, Springer-Verlag, Berlin und Heidelberg. 3. HENDRY, A.W. and JAEGER, L.G., "The Analysis of Grid Frameworks and Related Structures, Prentice-Hall, 1959 4. MASSONET, C , "Methode de Calcul des Ponts a Poutres Multiples Tenant Compte de leurs Resistance a l a Torsion", Publications of the International Assn. for Bridge and Structural P^ ng., Vol. X, 1950, pages 147 - 182. ; • 5. HETENYI, M., "Beams on an Ela s t i c Foundation", The Univ. of Michigan Press, Ann Arbor, Mich., 1946, pages 185 - 192. 6. MELAN, E. und SCHINDLER, R., "Die genaue Berechnung von Tragerrosten, Wien 1942, Springer-Verlag. 7. TBiOSHENKO, S.S., "Uber die Biegung von Tragerrosten", Zeitschrift fur angewandte Mathematik und Mechanik, Band 13, 1933, page 153. 8. GUYON, Y., "Calcul des Ponts Larges a Poutres Solidarisees par des Entretoises", Ann. Pet Ch., Sept. - Oct., 1946, pages 553 - 612. ' 9. GREENBERG, H.J., "Moments and Deflections i n Steel-Grid Bridge Floors", Trans. A.S.C.E., Vol. 121, 1956, page 410. -10. TARAPOREWALLA, K.J., "Design of Grid and Diagrid Systems on the Analogy of Design of Plates", The Structural Engineer, April 1958, pages 121-128. 11. MARCUS, H., "Die Theorie Elasticher Gewebe und ihre Anwendung auf die Berechnung biegsamer Platten", Julius Springer, Berlin, 1924. 12., JENSEN, Vernon P., "Analysis of Skew Slabs", Bulletin Series No. 332, Univ. of I l l i n o i s , Urbana, 111. 1941. 13. TEZCAN, Semih S., "Iskara K i r i s l i Sistemlerin Genel Cozumuy Technical University of Istanbul, Kutulmus Matbaasi, Istanbul, i960. A concise English reprint is published in the Bulletin of the Technical University of Istanbul, Vol. 14, 1961. 14. SHAW, F.S., "Limit Analysis of Grid Frameworks" Journal of the Struc-tural Division, A.S.C.E., Oct. I963. 15. V/RIGHT, P.M., "Simplified Formulation of Stiffness Matrices", Journal of the Structural Division, A.S.C.E., April 1963. 16. ARGYRIS, J.H. and KEL3EY, S., "Energy Theorems and Structural Analysis", Butterworths, i960. 17. TEZCAN, Semih S., Discussion of "Simplified Formulation of Stiffness Matrices", Journal of the Structural Division, A.S.C.E., Dec. I963. - 120 -APPENDIX A. Computer Programme for Stiffness Analysis of Grids B. Torsional Constants - 121 - • APPENDIX A • COMPUTER PROGRAMME FOR STIFFNESS ANALYSIS OF GRIDS 1. General Remarks The maximum size of structures that can be handled i s limited by the core memory capacity of the computer used. I f no special technique is employed, the limiting number of unknowns, for example, is about 50 for the IBM 1620 and 160. for the IBM 7090. When the method of t r i p l e block band matrices described i n Chapter 5 i s used, the capacity can be increased to as high as 1200 unknowns with the IBM 1620 and well over 10,000 with the IBM 7090. No accurate relation has been arrived at to estimate the execution time required for a particular grid problem. The time taken for the solu-tion of linear simultaneous equations varies approximately with the cube of the number of unknowns, while for most of the remaining parts of the analysis the time required varies proportionately with the number of members. A relative idea can be formed from the fact that the execution time was 20 mins. for 22 unknowns and 60 mins. for 44 unknowns using an IBM 1620 com-puter with a li+05 Disk F i l e and 1940 Printer. On the other hand, an execu-tion time of only 68 seconds was required to analyze a grid with 68 unknowns using an IBM 7090. The FORTRAN II programme presented i n the following pages i s capable of analyzing planar rectangular grids with a maximum of 47 unknowns and 43 members. For the general analysis of larger grids, diagrids and spatial grids, special large capacity programmes using t r i p l e block band matrices were developed. However, a discussion of these is beyond the scope of this thesis. 2. Input Data Cards Card Variables Definition Example Data for. Problem of Fig. 9 FORMAT 1 ME, N Number of members and deformations 4 3 2413 2 (S(I), I = 1, ME) Lengths of the members 10. 10. 10. 10. 13F6.0 3 (EI(I), I = 1, ME) Elastic Rigidities of the members 100. 100. 100. 100. 13F6.0 4 (GJ(I), I = 1, ME) Torsional r i g i d i t i e s of the members 30. 30. 30. 30. 13F6.0 5 (W(I), 1=1, ME) Distributed loads on the members 0. 0. 0. 0. 13F6.0 6 (C(I), 1=1, ME) Point loads on the members 10. 0. 0. 0. 13F6.0 7 (U(I), I =1, ME) Distances of point loads to the l e f t hand support 5. o. 0. 0. 13F6.0 8 ((ICD(I,J), J=l, 6), 1=1, ME) Identification code numbers of each member (see Ch. l ) 0.3 0 2 0 1 ^  3 0 2 0 1 0 0 1 0 2 0 3 1 0 2 0 3 0 ^  • 2W3 3. Other Symbols Used in the Programme Symbol Definition FLK(I,J) Jth fixed end reaction of member I S l ( l , j ) Individual stiffness matrix of a member (6 by 6) P(D Joint loads SG(I,J) Stiffness matrix of the system (N by N) D(I) Deformations of the system DEL(I) Deformations of a member F(I) Final stress resultants of a member Ju FORTRAN_.IJ?^ I D NUMBER - 1 6 4 3 P R I N T E D FOR R . K I N R A ON M A R . 3 1 AT 1 H R . 4 6 . 7 M I N . Y 12 J l 10 9 F O R T R A N 2 C O M P I L F . D I M E N S I O N S ( 4 3 ) , E I ( 4 3 ) , I C D ( 4 3 , 6 ) , W ( 4 3 ) , C ( 4 3 ) , U ( 4 3 ) , F I X ( 6 , 4 3 ) , IP ( 4 8 ) , S I ( 6 , 6 ) , G J ( 4 3 ) : - COMMON ME , N , W , C , U , S , I C D , E I , G J R E A D 2 t ME t N lBB.IM__2.»iSE. _ . R E A D 2 0 2 , ( S ( I ) , I = 1 •, ME ) R E A D 2 0 2 , ( E I ( I ) , I = I t M E ) _R-F_AD_2.02_,JAJJ_LL,J^_l . t_M_EJ_ R E A D 2 0 2 , ( W ( I ) , I = 1 , M E ) R E A D 2 0 2 , ( C ( I ) , 1 = 1 , M E ) -R.E AD . ..2 0 2 , X J J J J J _ , J_=l.,.f iE_L R E A D 2 , ( ( I C D ( I , J ) t J = 1 , 6 ) , 1= 1 ,ME> W R I T E D I S K I 0 0 0 , 2 , ( ( I C D ( I , J ) , J = 1 , 6 ) t I = 1 t M E ) P R I N T 2 1 5 :  DO 8 0 8 I = i , ME - H 8 0 8 P R I N T 2 1 4 , I , S ( I ) , E I ( I ) , G J ( I ) ,W( I » , C ( I ) , U ( I ) , ( ICO ( I , J ) , J= 1 , 6 ) <J3 _ 2 A FJie^/_j:ijr^jij_3j. •_ - „ _ _ . _ _ . L_ 2 02 FORMAT ( 1 3 F 6 . G ) . . . . . . 2 14 F O R M A T ( I 2 , F 8 . 2 , 4 F l 0 . 2 t F 9 . 2 » 3 X » 6 I 3 ) _2_15_ EORMAT ( / 3 H N O . , I X , 6 H L E N G T H , 6X , 2HE 1 ,8X , 2 H G J , 8X , 3HUDL t 6X , 4 H L O A D , 7X , I 1 H U , 9 X , 1 1 H C O D E N U M B E R ) • . C A L L F I X M O M C M I. S I T O R S ' C A L L L I N K ( 1) END 3 » T A B L E OF MEMORY A L L O C A T I O N S . 7 < 6 P.R0GP.AM-. .STARTS. . AT 15.5Jj.6_ 5 P R O G R A M E N D S AT 1 7 7 4 6 5'.- 4 LOWER END OF COMMON AT 36 119 . 3. _ 6 FORTRAN 2 COMPILE. SUBROUTINE FIXMOM __ oiMENS-I-ON-S (-U3-)-,-E-I-U 3 -IXD.( U.3.,.6 L, W . . L.U3 .U3 .bJXLMLt .FJ jMAfJ t lL i VP ( 4 8 ) , S I ( 6 , 6 ) , G J ( U 3 ) COMMON ME,N,W,C,U,S,ICD,EI ,GJ DO_209 I = 1 , ME , ' _ ^  . • B = S ( I ) - U ( I ) FIX ( 1 , I)=-(W(I )*S( I.) »S( I ) )/»2.- C( I ) » U ( I ) * B * B / { S ( I.)*S ( I ) ) •'• FIX ( 2_JJ___ + W( I )«SCI )» S ( n / J 2 . +_C { I ) « UJ_Ij « U U ) *B /_(_S ( I ) «S ( I )_) FI x"( 3 » i ) =-c. 5*w (T) *s (TT-c ( IT*B/STTT + '" (Ti xTTTT) +FTXT2 ,'Tf) 7s (TT F I X ( 4 , I )=-W( I )*S( I )-C( I ) - F I X ( 3 , I ) F I X ( 5 , I)=C. : F I X ( 6 , I ) = 0 . 2 09 CONTINUE LJ_X_=200jO •_ _ ' _ :  WRITE DISK LFX , 7 5 , ( ( F I X ( j , I ) , J = 1 , 6 ) , 1 = 1,ME j D O . 4 5 1 = 1 ,N 4 5 A P ( I ) = 0 . DO 3 2 5 1 = 1 , M E DO 3 2 5 L= 1 , N M M DO 3 2 5 J= 1 , 6 1 I F ( I C D ( I , J ) - L ) 3 2 5 , 3 2 2 , 3 2 5 • . . . . • ' , 3 22 P ( L ) = P ( L ) - F I X ( J , I ) • 3 25 C O N T I N U E L P = 2 0 C C + M E W R I T E D I S K L P , 7 5 , ( P ( I ) , I = 1 , N ) 7 5 F O R M A T ( 6 E 1 4 . 8 ) R E T U R N END T A B L E OF MEMORY A L L O C A T I O N S . P_RO.GR A M__S JAB_LS_..AX„ • 15602 P R O G R A M ENDS AT 1 9 6 8 8 LOWER END OF COMMON AT 36 119 F O R T R A N 2 C O M P I L E . S U B R O U T I N E S I T O R S D I M E M S - I O N - S ( 4 3 ) ,-E-I ( -43)- , I C D X 4 3 . , 6J_i> f (J l3 . ) . J £ . (A3J . „ U J J l ^ ^ l P ( 4 8 ) , S I ( 6 , 6 ) , G J ( 4 3 ) COMMON M E , N , W , C , U , S , I C D , E I , G J L S I = 3 0 0 0 . DO 2 6 1 = 1 , M E P S I = 2 . * E I ( I ) / S ( I ) S I ( 1 , 1 ) = 2 . * - P S I . S I ( 2 , 1 ) » P S I S I ( 3 , 1) = ( 3 . / S ( I ) ) * P S I S I (4 , 1 ) = - S I ( 3 , 1) S I ( 2 , 2)-= S I ( i , l ) S I ( 3 , , 2 ) = = S I ( 3 , 1 ) S I ( 4 , , 2 ) = = S I (4 , 1 ) S I ( 3 , , 3 ) = = 6 . * P S I / ( S ( I ) * S ( I ) ) S I ( 4 , i 3 ) = = - S I ( 3 , 3 ) S I ( 4 , ,4 ) = S I ( 3 , 3 ) ' DO 2 2 K = 5 , 6 DO 2 2 J = 1 , 4 2 2 . S I (J i t . J ) = C ' • S I ( 5 , 5 ) = G J ( I ) /sT( i " ) ~ S I ( 6 , 5 ) = - S I ( 5 , 5 ) S I ( 6 , 6 ) = S I ( 5 , 5 ) DO 2 3 L = l , 5 L P = L + 1 DO 2 3 J = L P , 6 2 3 S I ( L , J ) = S I ( J , L ) L S I = L S I + 6 W R I T E D I S K L S I , 5 3 1 , ( ( S I ( K , J ) , J = 1 , 6 ) , K = 1 , 6 ) 5 3 1 F 0 R M A T ( 6 E 1 4 . 8 ) . 2 6 C O N T I N U E R E T U R N END __ ... T A B L E OF MEMORY A L L O C A T I O N S . PROGRAM STARTS AT PROGRAM ENDS AT -LOV-.'ER END -OF-COMMON AT 15612 19264 3 6 1 1 9 - - - - - - - - -FORTRAN 2 COMPILE. L _NK( 1 ) B-I - M F N S I ON —i-G-D-(-U3-,-6-)-',-S 1(6., 6-)_,.SG.L4X,_4_8.) COMMON ME,N DO 2 1=1,N lD.CL_2_i._J.T-'. : 2 S G ( I , J ) = 0 . READ DISK 1000, 2 0 7 , ( ( ICD( I , J ) , J = 1,6) , 1=1,ME) LJ_JL__LWLO. . _ : DO 8 41 1=1,ME L S I = L S I + 6 READ D T SK I S I , 234 , ( ( S I ( K , J ) , J= 1 , 6 ) , K= 1 , 6.) DO 841 K=1,6 IF( I C O ( I , K ) ) 842,841 ,842 .8.42 W.N=J.C.D.(J..,J<J , DO 8 4 U J = 1 ,6 I F ( I C O ( I , J ) ) 8 4 3 , 8 4 1 , 8 4 3 I - I n - i . u i 1 | J I SG(NN,MM)= SG(NN,MM ) +SI ( K , J ) r-NJ O 841 CONTINUE WRITE DISK 6000,234, ( ( S G( I , J ) » J = ' , N ) ,I = 1,N) 1 2 07 FORMAT(24 13) 2 34 FORMAT(6E 14 .8 ) CALL L I N K ( 2 ) END 1 12 TABLE OF MEMORY ALLOCATIONS. 11 2 10 . PROGRAM STARTS AT 36366 9 '. P_ROGRA.M ENDS AT 38326. 3 8 LOWER END- OF COMMON AT 39989 7 4 6 Lsu.B!2.R0.G.R.A.M.S._.CALL..E.D L J M ' 5 5 4 LIBRARY FUNCTIONS CALLED. 3 6 F O R T R A N 2 C O M P I L E . L I N K . 2 ) G S 0 L-U-T-I 0 N - 0 F—L-I N E-A R _ E 0 U A T.I 0 M S D I M E N S I O N S G ( 4 7 , 4 8 ) , P ( 4 8 ) , D ( 4 8 ) COMMON M E , N _ _ _ _ P.?. 20.00 + ME R E A D D I S K L P » 2 4 5 , ( P ( I ) » I = 1 , N ) R E A D D I S K 6 0 0 0 , 2 4 5 , ( ( S G ( I » J ) , J = 1 »N) , 1 = 1 , N ) N P = N + 1 N M = N - 1 DO 2 1 = 1 , N ... out 1 i " r \ l 1 DO 3 1=1,MM I P - I + 1 DO 3 J = I P , N R = S G ( J , I ) / S G ( 1 , 1 ) DO 3 K = I P , N P 3 S G ( J , K ) = S G ( J , K ) - S G ( I , K ) * R i D ( N ) = S G ( N , N P ) / S G ( N , M ) DO 4 1= 2 , N ro J = N - 1 +1 • • 1 J P = J + 1 D ( J ) = S G ( J t N P ) DO 5 K = J P , N 5 D ( J ) = D ( J ) - D ( K ) * S G ( J , K ) 4 D ( J ) = D ( J ) / S G ( J , J ) W R I T E D I S K 8 5 0 0 , 2 4 5 , (D( I ) ,I = 1 , N ) P R I N T 8 3 2 P R I N T 8 3 3 , ( I,D( I ) , 1 = 1 , N ) 8 33 F O R M A T ( 5 ( 1 6 , F 1 0 . 4 ) ) 2 4 5 F O R M A T ( 6 E 1 4 . 8 ) 8 32 F O R M A T ( / 2 6 H D E F 0 R M A T I O N S OF THE S Y S T E M ) C A L L L I N K ( 3 ) END PROGRAM STARTS AT 35696 PROGRAM ENDS AT 38724 JlAB.LE_.O.E_.M.E>iOJiy__A.L_ LOWER END-OF- COMMON AT—399*9 F O R T R A N 2 C O M P I L E . • L I N K . 3 ) •G- F-I -N A L--E-N-Q—M OMEN T-S—A N. D _R.E A .CXLO N S „ AF__E AJ l ! l _M £ J * B J _ „ . - . ; D I M E N S I O N I C D ( 4 3 , 6 ) , F I X ( 6 , 4 3 ) , S I ( 6 , 6 ) , D ( 4 8 ) , D E L ( 6 ) , F ( 6 ) COMMON ME , N RE_AD_p„I_SK H)C_C , 2 2 4 , ( ( ICJ ) L L i . - i_ L ' L ' l r 1 _ M JL_ R E A D D I S K 2 0 0 C , 4 4 7 , ( (F I X ( J , I ) , J = 1 » 6 ) , I = I , M E ) . R E A D D I S K 8 5 0 0 , 4 4 7 , (D( I ) , 1 = 1 , N ) L _ L _ 3.0.0 0_ P R I N T 9 6 -DO 6 1 = 1 , M E DO 3 L= 1 . 6 ; : ; '  I F ( IC D ( I » L ) ) 3 1 , 3 1 , 3 2 D E L ( L ) = 0 . GJQ_J_L_3. ; . . M = I C D ( I , L ) D E L ( L ) = D ( M ) C O N T I N U E - _. i _ L S I = L S I + 6 H 1 R E A D D I S K L S I , 4 4 7 , ( ( S I ( K , J ) , J= 1 , 6 ) , K = ! , 6 ) » DO 4 J - 1 , 6 u •_ _ : i F ( J ) = 0 . ' " T " DO 5 K = 1 , 6 5 F ( J ) = F ( J ) + S I ( J , K ) * D E L ( K ) 4 ' F ( J ) = F ( J ) +F I X ( J , I ) P R I N T 2 , I , ( F ( J ) , J = 1 , 6 ) 6 A C O N T I N U E 2A 9 6 F O R M A T ( I 2 , 6 F 9 . 3 ) FORMAT(//31HFINAL END MOMENTS AND R E A C T I O N S / 8 X , 2 H M 1 , 6 X , 2 H M 2 , 7 X , 1 2 H R 1 , 7 X , 2 H R 2 , 8 X , 2 H T 1 , 7 X , 2 H T 2 / ) 1 2 24 F O R M A T . 2 4 1 3 ) •_ y 4 47 F O R M A T ( 6 E 1 4 . 8 ) r 12 S T O P i i END 2 _ 10 9 3 3 7 4 6 5 PROGRAM STARTS AT 17016 5 4 PROGRAM ENDS AT 19712 3 . T_AB.L E_O.F__M.EEQR_Yl_A.L.L OCA T J _ 0 N _ LOWER. END -OF COMMON - AT--39989 6 3 1 3 2 3 - 129 -APPEMDIX B  TORSIONAL CONSTANTS' The torsional moment of inertia of rectangular sections i s , i n general, given by J = T| b 3 h in which h i s the depth of the section, b is the width and is the (a) torsional factor given i n Table A for values of h/ b between 1 and CO . n - Vb 1 1.2 1.5 2 2.5 3 4 5 10 oo 1 .141 .166 .196 .229 .249 .263 .281 .291 .312 .333 Table A. Torsional Constants for Rectangular Sections Values of the torsional moment of inertia for various other types of sections are given i n Tables B^b) a n c j c ^ . Steel . | Sections i •—Ir— • f i t , 7'.iL_ . C. Webere~~~ j F-:rsh.-Arb.Heft249 o d'J £• Schn)i?dene._,_ ;1 Z.anqo'-y. Math Mech Bd.JD!j9?0) — Table B. Torsional Constants for Steel Sections (a) Timoshenko, S., "Theory of E l a s t i c i t y " , McGraw H i l l Inc., 1939, page 2/,8. (b) Poschl, Theodor,"Elemanter Mukavemet", Technical University of Istanbul, 1952. (c) "The Strength of Aluminium^ Aluminium Company of Canada, Ltd., page 76. Section 130 -Torsion Constant, J Factor for Maximum Stress, C , Factor for Ultimotc Torque, C„ 1. Thin Walled Open Section J L t — — 2. Thin Walled Closed Section 3. Thin Walled Closed Section, Uniform Thickness 3 4 A 1 4A't U t J t = maximum thickness 1 2At 2" 2At f = minimum thickness. 1 2At 4. I .Beam b ' T — i 4 k _ 1—Jj r i 5. Channel or Z Section 6. Rectangular • Hollow 7. Square Hollow 8. Thin Wolled Tube 9. Solid Rectangle 10. Solid Square 11. Solid Round •Ul/* .t,n " 7 t,— T 0 . _ t , » + + 2 [ n t l . 0 J 5 ( t 1 - t 1 ) ] ' Sec Fig. C for value of n. 2 M_» + _b£ s + 2^nt 1 + 0.45ft,-t.)J See Fig. C for value of n. 2 o V b't 2 IT R*t ¥ [ l - 0,3 i 4-0.052 ( - I - ) ' ] ' a < b 0.141 b* TTR-2 ['•H(-Hc-)]T l < f r < 2 \<i<2 i 2abt 2Ar 2ott t = minimum thickness. 1 2b't 2"wR't 3(b+0._a) o'b' 1 0.208b' 2 2b"t 2wR't •j-(3l)-a) 2 7 T R ' Nomenclature: A=areo enclosed by the median line of the wall of a closed section, U = the perimeter of the median line of a closed section. The highest shear stress in a shape is obtained by multiplying the applied torque by C t . The ultimate torque for a shape is given by the product of the ultimate shear stress ond C u . TABLE C — TORSION CONSTANTS FOR STRUCTURAL SHAPES 

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