LATERAL S T A B I L I T Y OP RECTANGULAR BEAMS BY Leon A l e x a n d e r B e l l B.A.Sc. U n i v e r s i t y o f B r i t i s h C o l u m b i a , I960 A THESIS SUBMITTED I N P A R T I A L FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF A P P L I E D SCIENCE I N THE DEPARTMENT OF CIVIL ENGINEERING We a c c e p t t h i s to the r e q u i r e d t h e s i s as conforming standard THE U N I V E R S I T Y OF B R I T I S H COLUMBIA 1966 In presenting this the requirements British for Columbia, I available for mission for agree reference extensive representatives. cation without of this that copying It of agree that of freely per- scholarly my D e p a r t m e n t financial gain Columbia s h a l l make i t thesis for understood that of the U n i v e r s i t y I further the Head of Is at Library this permission. The U n i v e r s i t y o f B r i t i s h V a n c o u v e r 8, C a n a d a Date the and s t u d y . thesis for my w r i t t e n in partial fulfilment an advanced degree p u r p o s e s may be g r a n t e d b y his thesis or copying or s h a l l not be by publi- allowed ABSTRACT This buckling thesis presents a method f o r f i n d i n g t h e l a t e r a l l o a d o f a s t r a i g h t beam l o a d e d longitudinal axis with loads perpendicular to i t s a p p l i e d a t a n y e c c e n t r i c i t y . The beam c a n be r e s t r a i n e d a t a n y number o f p o i n t s a l o n g i t s length a g a i n s t d e f l e c t i o n s and r o t a t i o n s . The beam i s d i v i d e d i n t o a number o f s e g m e n t s a n d j o i n t s f o r which equations i n v o l v i n g s t a t i c s , continuity are written. e l a s t i c i t y , and The r e s u l t i n g g r o u p o f n o n l i n e a r simultaneous equations i s solved f o r s e v e r a l magnitudes o f the l o a d i n g p a t t e r n f o r which a b u c k l i n g load i s desired. of t h e determinant of the s t r u c t u r e m a t r i x y i e l d s the c r i t i c a l Included A graph versus the load level load. i n t h e t h e s i s I s a complete l i s t i n g o f the computer program used i n t h e s o l u t i o n technique. I t was w r i t t e n f o r u s e o n t h e IBM 7OI4.O i n s t a l l a t i o n a t t h e U n i v e r s i t y o f British Columbia. Included a l s o are curves g i v i n g the l a t e r a l b u c k l i n g s t r e s s o f a beam s i m p l y c a n t i l e v e r e d over a f l e x i b l e top f l a n g e c o l u m n a t t h e o t h e r e n d , a n d whose i s laterally restrained. load together lever. s u p p o r t e d a t one e n d , a n d with a concentrated The beam c a r r i e s a uniform l o a d a t t h e end o f t h e c a n t i - I n a d d i t i o n , some d a t a i s i n c l u d e d on t h e e f f e c t o f tension flange bracing on t h e b u c k l i n g load. TABLE OF CONTENTS Page No. LIST OF TABLES i . LIST OF FIGURES i i . NOTATION i i i . ACKNOWLEDGEMENTS v i . 1 INTRODUCTION DERIVATION OF EQUATIONS 2.1 Assumptions 3 2.2 D i f f e r e n t i a l Equations 3 2.3 Development 6 2.4 C o n s t r a i n t s - General 2.5 Summary o f C o n s t r a i n t Types of S o l u t i o n Procedure 18 19 SOLUTION OF SIMULTANEOUS EQUATIONS 3.1 T r i p l e Block Matrices 25 3.2 Size of Blocks 30 DETERMINATION OF CRITICAL LOAD 31 k.l A p p l i c a t i o n o f Loading k.2 Determinant Method f o r F i n d i n g C r i t i c a l Load 32 k.3 E f f e c t of I n i t i a l k.k Computation Time Guess on R e s u l t s 39 39 EXAMPLES CONCLUSIONS 35 - k& APPENDIX I - DATA INPUT-OUTPUT 51 APPENDIX I I - ADJUSTMENT OF PROGRAMME FOR SPECIAL CASES 57 APPENDIX I I I - COMPUTER PROGRAMME LISTING 60 i. L I S T OF TABLES Page Table I D a t a I n p u t Sequence $2 i i . L I S T OF FIGURES Page Fig. 1 Fig. 2a Co-ordinate 2 Axis 5 F r e e - b o d y D i a g r a m o f Beam Segment F i g . 2b P o s i t i v e D i r e c t i o n s o f Moments and Fig. 3 F r e e - b o d y D i a g r a m s o f a J o i n t and Shears 5 a 9 Segment 11 Fig. k Matrix Fig. 5 Matrix Fig. 6 Equation Fig. 7 Beams R e s t i n g on V a r i o u s Fig. 8 E x a m p l e o f S t r u c t u r e M a t r i x f o r Beam and Vector and z\ - Z Vector P±zJ + R Z^ 12 + 1 16 C Column S u p p o r t s 22 B r o k e n i n t o F o u r Segments ( F i v e J o i n t s ) 26 Triple Block Matrix Notation 28 F i g . 10a P l o t of Determinant Versus Load 36 F i g . 10b E f f e c t i v e Range f o r I n i t i a l G u e s s Fig. 9 of C r i t i c a l Load 36 F i g . 11 R e s u l t s f o r Case 1. kl Fig. 12 R e s u l t s f o r Case 2. kk Fig. 13 Results f o r Various T y p i c a l Beam-Column Connections F i g . Ik i|7 Example Used f o r D e s c r i b i n g D a t a Input 55 Procedure F i g . 15 F i g . 16 Coding Sheet f o r Example Describing Data Input 56 F l o w D i a g r a m f o r C o m p u t e r Programme 59 ill. NOTATION TEXT PROGRAMME i i n texib = DEFINITION K i n computer programme d D length of segment »1 XDET1 value of determinant after f i r s t cycle D XDET2 value of determinant a f t e r second cycle \ XDETJ4. value of determinant after third cycle E E modulus of e l a s t i c i t y CE(K) modulus of e l a s t i c i t y f o r column at i Ell(K) distance Wl i s applied from xz plane, (seg. i ) E21(K) distance PI " " " " " , (joint i ) E31(K) distance HI " " " *' " , (joint i ) E12(K) distance W2 " " " " " , (seg. i ) E22(K) distance P2 E32(K) distance H2 " G G shear modulus h H beam depth H1(K) concentrated h o r i z o n t a l load applied a t joint i 2 E Ci e li 6 2i e 3i H2(K) *1< '3l 11 " " " " " " t h joint " , (joint i ) " , (joint i ) > concentrated horizontal load created by cons t r a i n t at j o i n t i Hll(K) distance Wl i s applied from xy plane, (seg. i ) H21(K) distance PI " " H31(K) distance HI " " H12(K) distance W2 " " H22(K) distance P 2 H32(K) distance H2 " 11 " " " 11 " " " " " " " , (joint i ) " , (joint i ) " , (seg. i ) " " , (joint i ) 11 " , (joint i ) iv. TEXT PROGRAMME 7 J DEFINITION XI2 moment of i n e r t i a of beam about y axis XII moment of i n e r t i a of beam about z axis CI(K) moment of i n e r t i a about x axis of column at j o i n t i XJ t o r s i o n a l constant J = tb.3(l - # 63^)/3 t o t a l length of beam L CL(K) length of column to underside of beam at joint I ai XMA(K) i n t e r n a l t o r s i o n a l moment about "a" axis (right face of joint i ) bi XMB(K) i n t e r n a l moment about "b" axis, (right face of joint I) XMC(K) i n t e r n a l moment about "c" axis, (right face of joint i ) M, XMOA(K) t o r s i o n a l moment about x axis applied to joint i M obi XMOB(K) moment about y axis applied to joint i o i XMOC(K) moment about z axis applied to joint i XM0A2(K) t o r s i o n a l moment about x axis created by constraint at j o i n t i XM0B2(K) moment about y axis created by constraint at joint i XM0C2(K) moment about z axis created by constraint at joint i N number of joints P1(K) concentrated v e r t i c a l load applied at j o i n t i P2(K) concentrated v e r t i c a l load created by cons t r a i n t at joint i L Ci M ,M a M ,M b M ,M c f C °i cr Ci designation f o r generalized c r i t i c a l load V. TEXT PROGRAMME Q2. DEFINITION d e s i g n a t i o n f o r g e n e r a l i z e d i n i t i a l l y appliefe load Q2>Q3 designation f o r generalized adjusted load S slenderness t T u^ V VB(K) b 1 V. c VC(K) 1 ratio = J Lh/t 2 width of beam h o r i z o n t a l component of displacement of the centre of g r a v i t y of t h e beam shear f o r c e p a r a l l e l t o "b" a x i s ( r i g h t f a c e of j o i n t i ) shear f o r c e p a r a l l e l to "c" a x i s ( r i g h t f a c e of j o i n t i ) v e r t i c a l component of displacement centre of g r a v i t y o f the beam Wj ^ of the W1(K) uniform v e r t i c a l l o a d a p p l i e d a l o n g segment i W2(K) uniform v e r t i c a l l o a d a p p l i e d along segment i BETA(K) angle of t w i s t of beam a t j o i n t i , measured from i n i t i a l p o s i t i o n ^ B 0 generalized horizontal deformation generalized r o t a t i o n slope of beam i n the xz plane at j o i n t i 0 ^, V slope o f beam i n the xy plane at j o i n t i k c r critical U i l a t e r a l buckling stress ACKNOWLEDGEMENTS The Dr. R. P. Hooley to him during also for author expressed for the to wishes the for the use Vancouver, B.C January, 1966 thank h i s encouragement development of this of its and to computer. supervisor, and guidance work. the N a t i o n a l Research some f i n a n c i a l s u p p o r t , Centre to Gratitude Council the U . B . C . given of is Canada Computing 1. INTRODUCTION The solutions available f o r the e l a s t i c l a t e r a l buckling problem are limited to those beams with simple loads and support conditions Tl]« a A search of the l i t e r a t u r e f a i l e d to y i e l d any general d i f f e r e n t i a l equations f o r the l a t e r a l buckling of a beam continuous over supports and a r b i t r a r i l y braced. The lack of such a set of equations l e d the author to look more deeply into the problem, with the r e s u l t that a general solution procedure, involving the e l e c t r o n i c computer, was developed. The method developed by the author i s suitable f o r the analysis of a continuous rectangular beam under almost any combination of constraints, horizontal and v e r t i c a l loads perpendicular to the longitudinal axis, and moments about any p r i n c i p a l axis. At present, the following l i m i t a t i o n s exist as to what type of problem can be analysed: 1. The beam must be of constant, rectangular cross-section. 2. Loads p a r a l l e l to the longitudinal axis cannot be applied. 3. P i n joints cannot exist anywhere except at the extreme ends of the beam. a. Numbers i n square parentheses r e f e r to the bibliography. 2. N Section A-A FIGURE I. COORDINATE AXIS 3. DERIVATION OF EQUATIONS 2.1 ASSUMPTIONS 1. The trigonometric functions used i n the derivation of the equations were approximated i n accordance with the small d e f l e c t i o n theory. 2. The curvatures i n the ac and the ab planes, F i g . (1) were assumed to be equal to the curvatures i n the xz and the xy planes respectively. This was i n ac- cordance with the small d e f l e c t i o n theory. 3. Deformations of the beam due to shear stresses were neglected. k. The material must be e l a s t i c , that i s , I t must obey Hooke's Law. Loads applied v e r t i c a l l y were assumed to remain verti c a l when the beam was i n i t s deflected p o s i t i o n . S i m i l a r l y , h o r i z o n t a l l y applied loads were assumed to remain h o r i z o n t a l . 2.2 DIFFERENTIAL EQUATIONS Two sets of co-ordinate axes, F i g . (1), are used i n developing the required set of d i f f e r e n t i a l equations. The xyz system, with i t s o r i g i n at the centre of gravity of the l e f t end of the beam, refers to the beam i n i t s unstressed, undeformed p o s i t i o n . From this system are measured u, the horizontal displacement of the centre of gravity of the beam, v, the v e r t i c a l displacement of the centre of gravity, and/3, the r o t a t i o n of the cross-section of the beam. The co-ordinate system based on the abc axis has i t s o r i g i n at the centre of gravity of the beam section i n i t s deformed p o s i t i o n at the l o c a t i o n x = x. Because the beam i n i t s deformed posi- t i o n i s curved, the orientation of these axes varies with x. The d i f f e r e n t i a l equations were written f o r a beam with no d i s c o n t i n u i t i e s . This was i n a n t i c i p a t i o n cf the solution procedure which would require that the beam be divided into a large number of equal lengthed segments over which no d i s c o n t i n u i t i e s would e x i s t , and j o i n t s , where a l l discontinui t i e s due to concentrated loads or to constraints would be applied. Prom the free body diagram of the beam segment dx, drawn i n i t s deformed p o s i t i o n , P i g . (2a), f i v e equations of s t a t i c s were obtained. The i n t e r n a l forces were defined to be always acting In the plane of the deformed member. The following equations were obtained: 1. Sum of moments about the "a" axis at x • x, y i e l d s , -dM 4. M d2u + M d v 2 a d 2. c b dx x dx 2 + 0 m Sum of moments about the "b" axis at x » x, y i e l d s , 12, dM + Md>3 + M djv + V b c dx 3. wdi-^ e-^) = 2 a dx dx c + w(h + e^Jdv-js x 0 dx 2 Sum of moments about the "c" axis at x • x, y i e l d s , -dM c + dx Mfcd^ - Madjj-u - V dx dx b - w(h 1 + e-^Jdu = dx 0 FIGURE 2 b . POSITIVE DIRECTIONS FOR MOMENTS AND SHEARS k. Sum of shears p a r a l l e l to the "b" axis at x = x, y i e l d s , dx 5. - V£ + w = o <w dx Sum of shears p a r a l l e l to the "c" axis at x = x, y i e l d s , dx dx The symbols h^ and e-^ designate the horizontal and v e r t i c a l eccentricities of the load w, as measured from the centre of gravity of the cross-section. In deriving the f i v e equations, only terms of the f i r s t order of magnitude (M, V, u, v,$ , dM dv du dv d>3 d ^ d v* „ . , . 2 5' d P dx' d P dP' W 6 r e r e t a l n e d - Since the f i v e equations contained eight unknowns, three further equations were required f o r a unique solution. The three additional Independent equations were obtained from e l a s t i c i t y considerations and are: *hS?Z -- - M b (6) dx^ EI„d v 2" " dx 2 GJd# 4r dx = M c (7) -M a (8) The p o s i t i v e directions f o r moments i s shown In P i g . ( 2 b ) . 2.3 DEVELOPMENT OP A SOLUTION PROCEDURE Due to the nonlinear nature of the equations involved, created by the existence of terms made up of products of forces m u l t i p l i e d by deformations, a numerical solution technique was used. The numerical technique was especially advantageous 7. since i t s use made possible the development of a very e f f i c i e n t system f o r adjusting the equations to accommodate constraints and concentrated loads. The beam was subdivided into a large, even number of segments of equal length, each segment being bounded by joints. The joints are not to be taken as mechanical j o i n t s ; rather, they are mathematical points at which a l l constraints and concentrated forces are to be applied. In t h i s way, the segments were kept completely free of d i s c o n t i n u i t i e s . A set of algebraic equations was written f o r each j o i n t and segment. The f u l l group of equations was then solved on a computer. With no d i s c o n t i n u i t i e s e x i s t i n g on the segment, and assuming that the segment size i s small enough so that only n e g l i g i b l e errors arise when a d i f f e r e n t i a l i s replaced by a difference, as f o r example dV being replaced by V ^ g ^ then the previous to (5) l e set of d i f f e r e n t i a l equations can be con- verted Into an equivalent set of algebraic Eqns. (1) - V £ , and Eqn. (8) equations. contain only f i r s t order d i f f e r e n t i a l s and therefore a l l the d i f f e r e n t i a l s occurring In these equations can be expressed by the end values of the one segment under consideration. Eqns. (6) and (7), however, are second order d i f f e r e n t i a l equations and therefore more than two points are required to express the function algebraically. To overcome t h i s , two additional equations are included, defining two new unknowns. These equations are: t 8. du dx - 0u dv dx = 9v (average) (9) (average) (10) Setting derivatives equal to the average of the end slopes of the segment has often been used i n similar applications, with successful r e s u l t s . To systemize the solution, a l l j o i n t s are numbered sequentially beginning with 1 at the l e f t end, while the segment number i s set equal to the lesser of the j o i n t numbers bordering i t . In the following set of algebraic equations the subscript i refers to the joint number, and the superscripts R and L r e f e r to the r i g h t and l e f t faces of the j o i n t respectively. Averaging of moments and shears occurring i n the nonlinear terms was not done, rather, the value at the l e f t end of the segment (right face of the joint) was always used. Referring to P i g . ( 3 ) , the following set of equations was obtained: M ai - <J>\ ' *c»l ± - w i d e i*i (la) i x i (2a) SEGMENT ( i) (EXTERNALLY APPLIED CONCENTRATED FORCES OMITTED FOR CLARITY, REFER TO SECTION A-A) FIGURE 3. F R E E - B O D Y DIAGRAM OF A (UNIFORMLY DISTRIBUTED LOAD OMITTED FOR CLARITY, REFER TO SECTION B-B) JOINT AND A SEGMENT SECTION B-B i i u b Ma £ u R i i+ 1 i c 1 i \ i b , M e - ^ u i 1 + 1 c V 1 i " u + M i+1 a v d l 2GJ = i (3a) b ( i+ 1 „ L ! c + M a u = 6 a ) i + 1 2EIy D d - u ! f + M?, i 2ET.Z A; 1 L - e£ d i 21ly 1 V d i + 1 2EIz (7a) L d 2GJ i+ 1 (8a) d it i 2 (9a) (10a) d i4- 1 2 This block of equations summarized i n matrix notation gives: -R "1+ i z i + 1 + B i <> n L Z^ and Z£_|. ^ are solution vectors containing the unknowns, Pigs. (Ij.) and (5). In the case of Z^, the unknowns are terms on the l e f t hand side of Eqns. (la) to (10a) which possess the subscript i and the superscript R, while f o r + 1 the unknowns 11. R Nl 1 1 N2 Mb; R -d 1 Ma: R -d R -MCj -Mbj -Mbj N2 Mc R I 1 R -VC| Vbj N3 Vc; d 2 1 d 2 1 -d . 2GJ 1 -d 2EIy d 2EI 1 2 R F ix NI = vi-, i -lie ° R F Z-i R N2 = - Mcij - Wj Chii -*-e|i /3j)d R N3 = Vbj + Wj d FIGURE 4. R I MATRIX Fj AND VECTOR R 0v; 12. R R -MCj -Mbj R R Ma; 1 1 R R -Mb; Ma • , R -Vc, R Vbj 1 I I I d 2 -d 2 L Vc i L i+l i+ v L d 2EIy L 1 eu | i+ -d 2EI 1 Z _ L i +I F FIGURE Mb^ Mq+i 1 26 J i+| L -d I Ma x Z L i-H 5. MATRIX Fj+,AND VECTOR 'zj" i + 0V; i+1 13. are the terms on the r i g h t hand side of the equations that possess the subscript i + 1 and the superscript L. The co- e f f i c i e n t s of the terms that are contained i n the solution R R vector make up the square matrix F^, Pig. (if), and the c o e f f i c i e n t s of the terms that are contained i n the vector z i + 1 ^a^® P U t h ® square matrix ^, P i g . (5). In the case of the nonlinear terms, the deformation portion was considered to be the c o e f f i c i e n t . This arrangement was chosen since the solution technique used (section k,2) necessitated guessing the values of the nonlinear c o e f f i c i e n t s and I t was f e l t that the forces could be guessed more exactly than deformations, especially f o r a beam with a complex bracing arrangement. Since two arrangements of the nonlinear terms were possible, the solutions could conceivably converge to two d i f f e r e n t c r i t i c a l loads. In a l l test cases considered, however, r e s u l t s always converged to the anticipated c r i t i c a l load. w i l b d a n d w i » d o c c u r i *i S i n n Two terms, the f i r s t and fourth elements of the vector make up the load vector B^. A set of algebraic equations can be written f o r each joint. Acting on the j o i n t faces are sets of Internal forces equal In magnitude and opposite i n d i r e c t i o n to the forces acting on the faces of the adjacent elements, Pig. ( 3 ) . Acting d i r e c t l y on the j o i n t are a l l the externally applied forces. The equations have been l i m i t e d to the cases i n which u, v , / 3 , 0 u and 8 are continuous over the j o i n t , that i s , the value of the deformation Is the same on the r i g h t and on the Illl e f t face of the j o i n t . A discussion on the adjustment of the equations to accommodate forces of unknown magnitude that are created by any constraints i s Included i n the next section. The following set of j o i n t equations r e s u l t : - V 3 e f H °i V i ^ c =b +i i Z V* " c i P 3 o, " i Vi i - H - P A °i (12) (e, - h 5 ) 9 3 i \ 1 ( a . - h £?)9 +H b ] R °i V -3^l h ± ©i 1 i R h 1 u v R (13) i R i (16) (17) r R± &\ (18) z &l e e u = e i = e L V (19) u (20) i R (2D V 1 i This block of equations summarized i n matrix form give: L i Z R I R = P i i Z + C I ( 2 2 ) and Z^ are the solution vectors as previously defined, and 1 P^ i s a square matrix of c o e f f i c i e n t s , P i g . ( 6 ) . i s the load vector consisting of a l l l i n e a r terms made up of applied loads. To decrease the number of simultaneous equations that have been created by the segmentation of the beam, each equation of type (11) has been combined with an appropriate equation of type (22) ,R *? i z to give: TTL = p „R i i < i i i p i + z + + + w + B. or expanded, "^i +^+lVl l Z + l The r i g h t hand side of Eqn. - - i l°H-l- l (23) y + B (2 3> a f t e r matrix m u l t i p l i c a t i o n and addition contains only l i n e a r terms made up of the applied loads. Since there i s one more j o i n t than segment, the group Moj 1 -Mo j -Mc-bi N4 : N6 ci -Mo 1 Mc: 5i Ma j Mbj R N5 MCj R 1 Vb; R R Mo 1 Mb c 1 Vc -HOJ Vbj -Poj VC| R u i R 1 V N5= Mo N6 = - M o 6 +P ai a j 0J (e 2 i ^ R +h -POJ h -Ho j EQUATION 2 | )+Ho (e 2 i j e 3 i 3 j l\ = P z f + C j s -h3 ^) i bi ci POj - HOJ 0 0 R 1 N 4 = -POJ egj +HOJ -Mo 0 GU; 1 FIGURE j I 9Ui -Mo R 1 01 N6 R 0 0 17. of matrix equations w i l l be made up of N - 1 equations of type ( 2 3 ) , where N i s the number of j o i n t s , and one equation of type ( 2 2 ) , that being f o r j o i n t number one. With ten algebraic equations making up each matrix equation, 10(N4-1) the r e s u l t i n g set of ION algebraic equations unknowns. contains The a d d i t i o n a l ten unknowns can be obtained from a consideration of the l e f t side of j o i n t number one and the r i g h t side of j o i n t number N. Considering j o i n t number one, and remembering that the external loading i s applied d i r e c t l y on the j o i n t rather than on i t s face, the following f i v e equations w i l l always hold: = M £'1 c l D l c l 0 0 = (21;) (25) = 0 (26) = 0 (27) = 0 (28) A s i m i l a r set holds f o r the r i g h t side of j o i n t number N. The ten equations could have been incorporated into the main body of equations, but f o r convenience i n programming the various types of constraints, this was not done. 1 8 . 2.L CONSTRAINTS - GENERAL A constraint s h a l l be defined to be a r e s t r i c t i o n to the movement of the cross-section of a beam, i n a p a r t i c u l a r d i r e c t i o n , at the j o i n t under consideration. Constraints can exist at any or a l l j o i n t s , and from a combination of them the e f f e c t s of bracing or boundary conditions can be created. Since constraints can be applied only at j o i n t s , only the matrix P j , or i n the case of Eqn. (23), the matrix F^P^ w i l l be affected by t h e i r i n c l u s i o n . Before l i s t i n g the constraints that can be accommodated by the solution procedure, i t w i l l be advantageous to discuss i n general the effects of one p a r t i c u l a r type on the o r i g i n a l set of equations. Consider the case of preventing any v e r t i c a l movement of the centre of gravity of the beam at j o i n t i . In equation form this i s : • 0 Including t h i s equation i n the previous ten written f o r j o i n t i gives eleven equations, but only ten unknowns. However, beeause v e r t i c a l movement i s r e s t r i c t e d , a v e r t i c a l reaction of unknown magnitude must be present eleventh unknown. to do t h i s . This force becomes the To accommodate t h i s force the f i r s t ten equations are rewritten, this time, however, instead of including only the v e r t i c a l l y applied force P ^, two forces P c and 0£(unknown) P the previous a r e appl ied » t ^ n e G i equations are s i m i l a r to set except that wherever a function of P ^ existed G 19. alone In the matrix, there now exist functions of P and Q i P~ , . • In addition, where P^ existed i n the C vector, °i(unknown) °i P would now also e x i s t . However, because i t i s an i (unknown) i 4 unknown, i t was removed from the C^ vector and was placed i n the solution vector to become the required eleventh term. The coefficients of P . , create an eleventh column f o r °i(unknown) ftl x the ?^ matrix, making i t an 11 x 11 square matrix. Similar manipulations are used to accommodate the other types of constraints. 2.5 SUMMARY OP CONSTRAINT TYPES The following constraints can be applied to the beam, either alone or, with the exception of certain groups that are l i s t e d at the end of the section, i n combinations to form boundary conditions. 1. u ± z c X i This i s the case i n which the centre of gravity of the beam, at j o i n t i i s r e s t r i c t e d i n i t s horizontal movement to the s p e c i f i c displacement c^. A concentrated horizontal force at i i s Included as an additional unknown. 2. u, 1 = c?B ± £ A 1 (where c * i P i s a constant) This i s the case In which the horizontal displacement of the centre of gravity of the beam at j o i n t i i s d i r e c t l y proportional to the angle of twist. A 20. concentrated horizontal force acting at a distance Cg above the centre of gravity i s included as an additional unknown. I f , f o r example, the top flange of the beam cannot move h o r i z o n t a l l y , a l l other movements being possible, then f o r small angles of twist, 2 2 i V = % This case i s s i m i l a r to case ( 1 ) except that the v e r t i c a l displacement i s r e s t r i c t e d to a movement c_ and a concentrated v e r t i c a l force becomes the additional unknown. v i ~ ~ ij.^i c ( ^ w e r e * s a constant) This case i s s i m i l a r to case (2) except that the v e r t i c a l displacement i s d i r e c t l y proportional to the angle of twist and a concentrated v e r t i c a l force becomes the additional unknown. * i = % This i s the case i n which the twist at j o i n t i i s s p e c i f i e d to be c ^ radians. A twisting moment at i becomes the additional unknown. s - \ This i s the case i n which the slope of the beam i n the XZ plane, at j o i n t i , i s specified to be c b radians. A i A bending moment about the weak axis at i becomes the additional unknown. 21. This i s the case i n which the slope of the beam i n the XY plane, at joint i , i s specified to be c 7 'i radians. A bending moment about the strong axis at i becomes the additional unknown. The f i n a l cases r e l a t e to a beam supported on a f l e x i b l e column, P i g . (7). The beam must be attached to the column In a manner such that the movement of the centre of gravity of the beam at j o i n t i can be related d i r e c t l y to the movement of the column cap. The s t i f f n e s s approach i s used to derive the necessary equations. A u n i t horizontal displacement i s given to the column cap, a l l other movements being held f i x e d , and the forces developed at the cap are calculated. Next, a u n i t r o t a t i o n i s applied at the cap, a l l other movements being held fixed, and again the forces developed are calculated. Neglect- ing v e r t i c a l deformations, the equations r e l a t i n g forces and deformations at the column cap are: H, , = aS + b9 i column (29) Q M Q = a i bS + c G (30) column where a, b, c are s t i f f n e s s c o e f f i c i e n t s . 8 # In i k i s case, the beam s i t s on a column that Is f i x e d at the base, P i g . (7a). c o - e f f i c i e n t s are: The s t i f f n e s s original location F i 9- Fig. 7b 7 a FIGURE 7. Fig. 7c BEAMS RESTING ON VARIOUS Fig. 7d Fig. 7e COLUMN SUPPORTS Fig. 7 f 23a = 12 c,Ic E c Equations (31) i and (32) are obtained by substituting the above c o - e f f i c i e n t s (but with negative signs) into Eqns. (29) and (30) and then r e l a t i n g ^ and 9 to the centre of gravity of the beam, Pig. (7e). Negative signs were required since the forces were derived with reference to the column. H 111 oa nknown) ' ^ i(u l " T 2 •'-'Ci k L * * ^ S " ( 3 + VgVt T <^ ^Cj - 0( 2) 3 h Case (8) must always be used i n conjunction with either Case (3) or (k) since v e r t i c a l movement i s always restricted. 9. This case i s similar to Case ( 8 ) except that the column i s pinned at the base, P i g . (?b). The equations become: H ° 1 (unknown) + ^ f o £ £ < H - . „ (33) 2k. M ° «a i (unknown) • 3 V I e V l U * + - 5 E c 1—^c ^< 1 l I 1 1 + 1 2 h±)k* "li 0 (Ik) 0 W Case (9) must also be used i n conjunction with either Case (3) or (k) since v e r t i c a l movement i s always r e s t r i c t e d . 10. Case (10) i s Case (8) combined with Case ( 2 ) , P i g . (7 Here, the top flange of the beam i s prevented from moving h o r i z o n t a l l y and therefore the c o - e f f i c i e n t C £ equals . 5 h . I t has been included as a special case because the boundary condition technique used i n the computer program precludes any p o s s i b i l i t y of combining these two cases i n the usual manner. The i n c l u s i o n of this special case makes i t very easy to adjust the program to accommodate the cases shown i n Pigs. (7d) and (7©). (Refer to Appendix III). The equations become: % (unknown) " W^ 6 2 + V, „ , ! W unk OTO + -rd L C 1 S^i < 3 + 2 = 0 ^'^ - (35) 0 .36, h Case (10) must always be used i n conjunction with Case (3) or (k). Because of the method by which the i n d i v i d u a l cases have been programmed, the following combinations are not possible at a p a r t i c u l a r j o i n t : 25. 1. Case ( 1 ) cannot be combined with Case (2). 2. Case ( 3 ) cannot be combined with Case (k). 3. Cases ( 8 ) , ( 9 ) , ( 1 0 ) cannot be combined with one another. k. Case ( 1 ) or (2) cannot be combined d i r e c t l y with Case ( 9 ) or ( 1 0 ) . 5. Case (5) cannot be combined with ( 8 ) , ( 9 ) or ( 1 0 ) . Except f o r these, p r a c t i c a l l y a l l boundary conditions can be developed from a combination of not more than f i v e of the fundamental cases. To accommodate these f i v e possible additional equations i n the computer program, a l l the matrix equations were augmented by the introduction of f i v e dummy equations of the form: a t = a t (t - 1 , 2, 3 , k, 5) (37) If a p a r t i c u l a r boundary condition or bracing system exists at joint i , then the equations that define this condition replace an equal number of dummy equations. SOLUTION OF SIMULTANEOUS EQUATIONS 3.1 TRIPLE BLOCK MATRICES [2] The t r i p l e block procedure was used t o solve the large group of simultaneous equations. Its main advantages were the ease with which the process could be programmed and the simplicity with which the programme could be made to accept the necessary modifications to the equations caused by the constraints. To use the t r i p l e block method, the structure 5 5 5 <—ak J K — ? 15 15 L -'h A r- 1- 2 F> 2 F 3 P - F_ 30 3 F4 P4 30 - F. F 5 5 P 5 MATRIX A, - 5x5 Unitary matrix obtained from Eqns.(20) to (28) written for joint I. MATRIX A - 5 x 5 Unitary matrix type ( 3 7 ) MATRIX A All other FIGURE 2 3 written for the 5 Eqns. of — 5 x 5 Unitary matrix obtained from Eqns. (24) to (28) written for joint N. blocks have been described in the 8. text. EXAMPLE OF STRUCTURE MATRIX FOR BEAM BROKEN INTO 4 SEGMENTS (5 JOINTS) 27. matrix was broken up into a number of equal sized square matrices (blocks) as shown i n P i g . ( 8 ) . The p a r t i t i o n i n g had to be such that every row and column of the partitioned matrix contained not more than three non-zero blocks, except f o r the f i r s t and the l a s t which could contain two non-zero blocks. blocks were zero matrices. A l l other The 30 x 30 block size was chosen for reasons that w i l l be discussed l a t e r . Since the procedure used i n obtaining the c r i t i c a l load required the valve of the determinant of the structure matrix, i t was necessary to investigate the standard t r i p l e block approach In somewhat more d e t a i l than could be found i n Referring to P i g . ( 9 ) f o r the notation to be the l i t e r a t u r e . used, a step by step analysis of the t r i p l e block procedure follows. The o r i g i n a l structure matrix T can always be formed from the product of a lower triangular matrix L and an upper triangular matrix U. Furthermore, the diagonal blocks of the lower triangular matrix can be made to consist of unitary matrices. For t h i s arrangement, the elements of the blocks making up each triangular matrix are: u l,l ~ u i,i = u i, a i+ 1 l , l a -1 i , i- i + i i - i , a = a u i-i i, i , 14- 1 i - i ( i /, / -i \ ' 08) 1 } ( 3 9 ) li, i-1 = a . uJ. ^ l f 1 1 a lf (k0) 1 121 u 1 0 0 ll U I2 u 22 1 0 *43 u u I 0 0 23 0 33 u Q a 2l 34 0 44 0 u I2 a 22 a 32 0 a 0 0 23 0 a 33 a 34 a 43 a 44 [ L U - T ] FIGURE 9. TRIPLE BLOCK MATRIX NOTATION ( 4 x 4 matrix used for convenience) ro CD 29. The postscript -1 implies the matrix operation of inversion and the subscript I refers to the block locations i n the partitioned matrix, Pig. (9). A disadvantage of the block procedure i s that an inversion i s required. operation triple Such an on the computer i s very time consuming, e s p e c i a l l y i f a large block must be inverted. Because the matrix i s to be inverted, i t must be non-singular, that i s , a l l rows (columns) must be completely independent of one another. off-diagonal blocks may be All singular. Letting Z and W denote respectively the solution vector and the load vector f o r the entire beam, the matrix equation f o r the beam can be written TZ where T i s a 15(N+ 1) = as: (kl) W x 15>(N-t- l ) structure matrix. Letting •T - (k2) LU and substituting into Eqn. LUZ (kl) y i e l d s : = W (k3) Furthermore, l e t t i n g X = UZ and substituting X into Eqn. LX * (kk) (kl) y i e l d s : W (kla) Since L i s triangular i t i s a very simple process to solve f o r X. obtained: Solving, the following set of equations i s 30. l x - w i -1 ^ These values are substituted Into Eqn. (kk) and again, because U i s triangular, the following r e s u l t s are e a s i l y obtained: Z z H+1 2 i 3.2 = = U .-1 N±1, N+1 N+1 2 " 2 " -2~" X u I,i ( x i - i, i i a + z 1 + i>' (i - sgi. .... (¥>) 1) SIZE OP BLOCKS Since p a r t i t i o n i n g of the structure matrix breaks up the set of equations into a r b i t r a r y parts, i t Is possible that while the structure matrix as a whole i s non-singular, the diagonal blocks themselves are not. Because every diagonal block has to be inverted, i t s size must be such that s i n g u l a r i t y would not occur under any set of constraints. I t was possible to decide upon a suitable size f o r matrix u-^ b 7 inspection since i t s inversion was not preceded by any matrix operations. Whether the same size would be s a t i s f a c t o r y f o r the other diagonal blocks was not c e r t a i n . However, i n a l l the tests that were carried out, not once did the 3 0 x 3 0 block become singular. Of the block sizes that would have been satisfactory, the smallest was chosen to decrease the computation time. 31. DETERMINATION OF THE CRITICAL LOAD k.l APPLICATION OF LOADING When a loading pattern consisting of several loads i s applied to the beam, a l l loads are applied simultaneously. Otherwise, due to the non-linear c h a r a c t e r i s t i c s of the system, a d i f f e r e n t buckling load w i l l be obtained. If the load pattern acting on a beam consists either of a set of concentrically applied v e r t i c a l loads or a set of concentrically applied horizontal loads then a value f o r the buckling load w i l l probably not be attained, since the nonl i n e a r i t y of the system w i l l not be triggered. To i n i t i a t e the non-linearity, i t Is necessary to either r e l y on round-off errors that creep i n during computations, or do one of the following: 1. Apply a l l or a portion of the loading at some small eccentricity. 2. Apply a small a u x i l l i a r y couple. Because a l l calculations were carried out on the computer with double p r e c i s i o n (sixteen s i g n i f i c a n t f i g u r e s ) , round-off errors would be i n s i g n i f i c a n t , and therefore a great many i t e r a t i o n s (see Section k.2) would be required before the beam stopped behaving l i n e a r l y . round-off errors. I t i s therefore impractical to r e l y on The alternate methods have been found s a t i s - factory when loads were applied with e c c e n t r i c i t i e s of up to twenty percent of the beam width, or when a u x i l l i a r y couples of about one percent of the magnitude of the p r i n c i p a l bending moment were applied. were not investigated. Larger values of e c c e n t r i c i t i e s and couples 32. k.2 DETERMINANT METHOD FOR FINDING CRITICAL LOAD The procedure used f o r c a l c u l a t i n g the l a t e r a l buckling load evolves from the f a c t that the value of the determinant of the structure matrix at buckling Is equal to zero. The general procedure involves obtaining the value of the determinant f o r three d i f f e r e n t magnitudes of the load pattern f o r which a solution i s required. From these values the determ- inant versus load curve, F i g . (10a) i s approximated i n the general region of the buckling load by a second degree polynomial which can be solved f o r the c r i t i c a l load. Before the actual procedure used f o r obtaining the buckling load Is discussed, an explanation as to how the value of the determinant can be obtained from the t r i p l e block procedure s h a l l be given. For t h i s , the following results from l i n e a r algebra are u t i l i z e d . F i r s t l y , the value of the determinant of the product of m non-singular matrices i s equal L to the product of the determinants of the m o r i g i n a l matrices [3j • Secondly, the value of the determinant of a triangular matrix, partitioned into square blocks of equal size, i s equal to the product of the determinants of the i n d i v i d u a l diagonal blocks [3] , jV|. In the case of the t r i p l e block procedure, since the L matrix has only unitary blocks on I t s main diagonal, the value of the determinant i s equal to the product of the determinants of the diagonal blocks of the U matrix. Since a determinant was required i n the Inversion procedure used i n the computer programme, i t was a simple matter to gather up 33. the determinants of the i n d i v i d u a l u^ ^ matrices and multiply them together to obtain the determinant of the structure matrix. The actual procedure used f o r c a l c u l a t i n g the c r i t i c a l load w i l l now be explained. Let QJJ be a generalized load that i s applied to the beam, and D n the value of the determinant of the structure matrix corresponding to that applied Q^. An i n i t i a l guess Q-^ * &de as to the value of the c r i t i c a l sm load. In addition, since the structure matrix Is non-linear, the terms M , &± M^, M , V^, V c and/3 ± corresponding to a s well as the forces created by the constraints are guessed. An attempt i s made to guess these forces accurately. Accordingly, when a v e r t i c a l load i s applied to the beam, M ^ and c guessed to be t h e i r l i n e a r values. are Using these guesses, the computer solves f o r the solution vector Z. I f the i n i t i a l guesses happened to be correct, then they would be equal to t h e i r counterparts l n the Z vector. w i l l not be the case. In general, however, t h i s Prom preliminary investigations into the choice of a solution technique, i t was found that even though the non-linear terms were not equal to the corresponding terms i n the solution vector, the value of the determinant was within three per cent of i t s true value i f the values of the l i n e a r moments and shears were used as i n i t i a l guesses. This was discovered during the i n i t i a l stages of the development of the procedure when a f u l l i t e r a t i o n technique was used to f i n d the rates of convergence of the various items. The i t e r a t i o n technique consisted of substituting as new guesses f o r the 3k. non-linear items,the appropriate terms from the solution vector, and then repeating the process u n t i l every guessed value was within one per cent of i t s value i n the solution vector. Because the value of D could be obtained quite accurately without i t e r a t i n g , the major portion of the i t e r a t i o n scheme was removed and the technique was somewhat modified. The value of D-^ and the solution vector Z, both corresponding to Q^, e r e calculated without any i t e r a t i n g . w A second value of loading, Q » 2 applied to the beam. equal to l.ZQ.-^ was automatically The non-linear items were taken to be equal to the appropriate terms from the f i r s t solution vector increased by twenty per cent. The computer programme again calculated a solution vector and a corresponding determinant As before, i t e r a t i o n was not carried out. 2 Note, however, that because terms from the f i r s t solution vector were used as the value of D. guesses, should be closer to i t s true value than was D^. To guess a load Q closer to the c r i t i c a l load, the determinant-load curve, P i g . (10a), was approximated by a straight l i n e passing through the two points defined by (Q^, D^) (Q » 2 D 2 ) * '^ iie i ^ Q r ^ c t i o n of this l i n e with the horizontal axis yielded Q 3 and which has the value: = KQ 2 = (5 + S Dj 6(D X )Q 2 (W -D ) 2 With t h i s load, and appropriate terms from the solution vector m u l t i p l i e d by K, a t h i r d solution vector and a determinant were calculated. Without changing the load, the terms from 35. this solution vector were used as guesses for an additional t r i a l , and a fourth determinant cycle was was obtained. This extra carried out to obtain a more exact value for the determinant, since i t s magnitude near the c r i t i c a l load i s comparatively small. Using the three points (Q^, D-^), (Q , D ) 2 D j^) a 2 and polynomial of the form: Q = AD 2 4- BD C 4- (k8) can be written to approximate the determinant-load curve. A, B, and C are constants which can be evaluated from the three Setting D = 0 , as i s the sets of available information. condition at Q , the value of the c r i t i c a l load equals the cr constant C, whose value i s : c * ^cr • Q 3< * x ., 1 2 \ , J ( D - D ) ( D - D )(1.2D - D ) D k If the value of 1 D k 2 1 (1*9) 2 was indeed close to Q-^ ^hen the r e s u l t i s quite exact. k.3 THE EFFECT OF THE INITIAL GUESS ON RESULTS The I n i t i a l guess as to the magnitude of the c r i t i c a l load i s a most Important consideration for obtaining an accurate solution. If the i n i t i a l guess i s too low, less than about .1Q._, then the r e s u l t w i l l usually be inaccurate. This i s because at small loads the beam behaves almost l i n e a r l y , and i s somewhat e r r a t i c i n Its attempt to take on i t s buckled shape. 36. FIGURE 10b EFFECTIVE RANGE FOR INITIAL GUESS OF CRITICAL LOAD 37. In this case, D 1 and D 2 are p r a c t i c a l l y equal, and since t h e i r difference i s required i n Eqn. (k7), a small error i n either would tend to make the value of Q too d i f f e r e n t from Q , j> cr 0 thereby making any extrapolation inaccurate, (Section k.2). Furthermore, there was no attempt during the f i r s t two cycles to Iterate onto the exact solution, and therefore, as described e a r l i e r , the value of the determinant could be i n error by about three per cent. Since the value of the difference of two determinants may also be small f o r small loads, the possible error i s obvious. I t must be noted that the use of approximate values f o r 2 low. and D affects results only i f the guess i s too With larger loads the non-linearity i s magnified, causing the values of the determinants to be quite d i f f e r e n t from one another, and since and D 2 are much larger than the value of the determinant near the c r i t i c a l load, an error i n them has l i t t l e e f f e c t on establishing the approximate equation of the determinant-load curve near the c r i t i c a l load. It has been found that i f the values of and D 2 are within about eight per cent of one another, then the value obtained f o r the c r i t i c a l load w i l l be u n r e l i a b l e . for The reason this Is that such values f o r D means that either the guess as to the value of the c r i t i c a l load i s too low, or that the beam i s extremely r e s i s t a n t to buckling, both of which imply that the beam acts almost l i n e a r l y , (a l i n e a r system has only one value f o r i t s determinant). Because Eqn. (k7) requires the difference D-^ - D , which i n the present case i s very small, 2 38. a s l i g h t error i n either determinant w i l l cause a major change i n the l o c a t i o n of Q,^. A further test on the r e s u l t s can be based on the value of D^. I f the value of i s less than about .lD-^, or .5>D, then a good r e s u l t may be expected, since 2 only a small amount of extrapolation i s required to reach the horizontal a x i s . An e s p e c i a l l y good r e s u l t w i l l be attained i f small i n comparison to and D 2 and negative. best results w i l l be obtained i f the value of at not less than .25Q -,. rt Is Normally, the can be guessed I f the value of the i n i t i a l guess i s cr greater than Q , cr but i n the range defined by S i n Pig. (10b), then the same conclusions as to accuracy can be applied. The s t i f f e r that a beam i s against buckling, the greater w i l l be the p o s s i b i l i t y that a second or even a t h i r d t r i a l w i l l be required to determine the buckling load. Is due to the nature of the determinant-load curve. This If the beam i s very s t i f f , then the curvature of the curve near the horizontal axis i s very great, and In f a c t , i f the beam cannot buckle, then the curve would not even cross the axis. On the other hand, i f the beam Is susceptible to buckling, then the change i n slope near the horizontal axis i s almost zero, and i t crosses as a straight l i n e . Because of the great curvature, the i n t e r s e c t i o n with the horizontal axis i s at a very f l a t angle, the curve must be approximated with much greater for s a t i s f a c t o r y r e s u l t s . accuracy To overcome this problem, the com- puter programme can be revised to develop the determinant-load 39. curve exactly. This i s done by inputting an i n i t i a l guess Q]_ and i t e r a t i n g u n t i l the values i n the solution vector equal those i n the previous solution vector. The load and appropriate terms i n the solution vector would then be increased by some percentage and the procedure repeated. This would be continued u n t i l enough points were obtained to establish the c r i t i c a l load. k.k COMPUTATION TIME The t o t a l time required to run one case on the IBM 70k0 I n s t a l l a t i o n at the University of B r i t i s h Columbia can be obtained from the following approximate formula: TIME (minutes) = .57(N+1) + 2.0 This equation i s developed f o r the case i n which an object deck i s used. Without such a deck an extra s i x minutes should be added to the computation time. I t was found that the computation time was n e g l i g i b l y affected by the complexity of the loading and the number of constraints. I t depended almost e n t i r e l y on the number of j o i n t s , the time to read the object deck, and the time i t took to s t a r t the machine. The l a t t e r two items make up the two minutes that were added to the equation. EXAMPLES To emphasize the value of the theory that has been developed, several numerical examples of a more or less general kO. nature have been included. Furthermore, to make the information presented more general, i t has been summarized i n dimensionless form whenever convenient. The c r i t i c a l stress, G . must be a function of the cr following items: G c r = function (50) JE, G-, L, h, tj In addition, I t w i l l depend upon the type of loading and the type of boundary conditions. Since the s i x items i n Eqn. contain only the dimensions (50) of force and length, they are linked by any four dimensionless parameters which contain a l l s i x terms. One set of parameters i s : lc> = function pL,hh t, Gj L cr ~"E~ (CQ) J E The following three cases I l l u s t r a t e how '•gp/E varies with these parameters. CASE 1. This i s the case of a simply supported beam loaded at i t s neutral axis by either a concentrated load applied at the centre l i n e of the span, or a uniformly d i s t r i b u t e d load, F i g . (11). Constraints are applied at the centre of the span which prevent twist of the section, but allow horizontal and v e r t i c a l movement. I t seems reasonable to assume that the behaviour of t h i s beam i s governed by the same parameters as a simply supported beam with no diaphragm. simpler case Equations of this 16] are: concentrated load) L 2 kl. w - P -4 locn. of diaphragm diaphragm _L 2 30 2 curve for unbraced beam carrying cone, load P - curve for unbraced beam carrying distributed load w uniformly 20 IOOO | ^_result for diaph. and cone, load P R 10 G/E =.58 10 15 20 25 slenderness FIGURE II. RESULTS FOR CASE I. ratio 30 35 40 45 50 42. w cr = 2 8 * jEIyGJ 3 (uniformly d i s t r i b u t e d load) L3 These can be transformed to: 1000£ E cr = k.200 = 3514-0 S2 [G (concentrated load) and 1000 £er E where S s S 2 i* E (uniformly d i s t r i b u t e d load) jLh/t2 and i s c a l l e d the slenderness r a t i o . £ C R /E These equations show that depends d i r e c t l y on ]G/E, a r e s u l t that cannot be predicted by dimensionless analysis. Furthermore, E > / c r E depends not on L/h and t/h Independently, but rather on S. F i g . (11) shows the r e s u l t s f o r Case 1. plotted against S (for G/E = . 5 8 ) and compared with G beam with no diaphragm. c r f o r a simply supported As expected, \o i s considerably higher cr when a diaphragm has been added. However, more computer runs are necessary to firmly establish that ^ / E cr proportional to S 2 and i s i n f a c t Inversely JE/G. CASE 2. Because very l i t t l e data exists that quantitavely describes the extent to which continuous bracing of the tension flange against horizontal displacement by a roof deck affects the magnitude of the l a t e r a l buckling load, the b e n e f i c i a l affects of such bracing i s usually disregarded i n design. was decided to investigate t h i s procedure with the Idea of It 43. determining whether a thorough investigation would be i n order. For this case, the constraints making up the bracing are assumed pinned to the tension flange so as to offer no t o r s i o n a l r e s t r a i n t . It was anticipated that a simply supported beam whose ends are prevented from twisting, and whose tension flange i s braced against horizontal displacement would be governed by the same parameters as that of a plate with similar boundary conditions, F i g . ( 1 2 ) . I t was desired to analyse the beam separately f o r equal end moments, a concentrated load, and a uniformly d i s t r i b u t e d load, the l a t t e r two acting on the free edge of the beam. Since no exact solution could be found for a plate loaded i n a l i k e manner, an energy s o l u t i o n was with which r e s u l t s could be compared. ^> c a n developed found by D e cv equating the work done by the external loads during buckling to the increase i n s t r a i n energy V, where V = .5( f ? JQJ-h/2 L h / [ Dx L T W x 2 2D,W + 1 xx yy D WL. +- kD y yy ^xy xy J x x v are s t i f f n e s s c o - e f f i c i e n t s . r' J Here, W i s the displacement i n the z d i r e c t i o n and D , D (5) W ? l dxdy D,, D , The expression assumed f o r the deflected shape was: W . k(h - y) slnTrx TT 2~ where x and y are measured from the centre of g r a v i t y of the l e f t end of the beam, F i g . ( 1 2 ) , and k i s a constant. The values of D_ and D are respectively E t 3 / 1 2 and Gt3/12, x xy the values of and D y are not required since W Y T s o. values obtained f o r the work done by a concentrated load, while The FIGURE 12. RESULTS FOR CASE 2 a uniformly d i s t r i b u t e d load, and equal end couples, are respectively, Pk h/2, wk^a/k, and MkVah/k. 2 The r e s u l t i n g equations are: tor = ^ ) \ —E~" IT h EL ( 6 c p = t 2 t ( 2 G 2 } 1+ JZ (k) lf| 12 V Gj G [14-^ 2 ( h 2 } B] (concentrated or u n i f ormly d i s t r i b u t e d loads) (equal end couples) E These equations show that f o r a plate i n which h/L i s small, ^ C I / E depends d i r e c t l y on G/E and ( t / h ) ^ and i s independent of h/L, r e s u l t s that cannot be predicted by dimensionless analysis. for The reason why the equation i s the same either a uniformly d i s t r i b u t e d load or a concentrated load i s that the energy solution i s only approximate, since the deflected shape W i s only a guess. Referring to Pig. (12), i t Is seen that the computer r e s u l t s f o r a concentrated load and f o r equal end moments compare almost exactly with those of the energy solution, while the r e s u l t for a uniformly d i s t r i b u t e d load does not. The reason f o r t h i s may be that the assumed d e f l e c t i o n function was incorrect f o r the uniformly d i s t r i b u t e d load. A l l cases showed that the tension flange bracing substantially increased the c r i t i c a l buckling stress over that of an unrestrained beam and i t i s f e l t , therefore, that this i s an area i n which further investigation would be valuable. k6. CASE 3. In a great many instances, the roofs of many of the smaller i n d u s t r i a l and storage buildings are constructed of timber decking supported on a series of bents made up of glued laminated beams r e s t i n g on columns located such as to d i v i d e the width of the building into three equal bays. The centre span of such an arrangement often consists of a simply supported beam resting on a cantilevered projection of the sidespan. Such a system, Pig. (13), braced along the top flange to prevent horizontal displacement of that flange, i s investigated t o determine the e f f e c t s of various beam to column connections on the c r i t i c a l stress. Because only a l i m i t e d number of computer runs were made, not enough to consider a l l the variables, no new theory i s presented. The c r i t i c a l stress i s plotted against the slend- erness r a t i o S only to give some basis f o r comparing the various connections. Connections to cause the actions shown i n Pigs. (7d) and (?e) and i n addition the case of a diaphragm located at the.column l i n e and preventing horizontal and v e r t i c a l movement, as well as twist of that section, were studied. P i g . (13) ex- presses the importance of a proper beam t o column connection. When the beam i s attached as i n Pig. (7e), for the c r i t i c a l stress, the range of slenderness r a t i o s investigated, remained a constant. The bounding case, that of the beam with a diaphragm, was found to have a c r i t i c a l stress many times greater than the 47. /3wL i \' w\r \> v if \> i_ Jr_ " t=5 ' =4cfirT . h A v di. bracing (continuous) j Plane of critical stress ( c r ) and location of * diaphragm "X cr diaphragm p i n n e a DIAPHRAGMED CASE 20 1000 'cr CONNN AS IN FIG. ( 7d ) 10 CONN'N AS IN FIG. (7e) Qi 20 For the 25 points plotted ft = .3 Lh , , b/h = I / 8 CL/d =5.1 G/E = .58 , E/CE = 1.0 a =.2 30 CI =.025 FIGURE 13. RESULTS FOR VARIOUS TYPICAL BEAM-COLUMN CONNECTIONS k8. l a t t e r condition. This case i s not affected by the column properties, and any set of constraints that r e s t r i c t s the beam i n the same manner can be substituted. When the beam i s connected as i n Pig. (7d), the r e s u l t s l i e between the two previous sets of values. The condition that the c r i t i c a l stress increases with increased slenderness r a t i o can be explained by the f a c t that as the beam i s Increased i n length while i t s section properties remain constant, the r e l a t i v e s t i f f n e s s of the column i s increased, u n t i l i n the l i m i t , i t becomes equivalent to the diaphragmed case. The graphs adequately express the importance of the need f o r a prpper beam to column connection. CONCLUSIONS The intention of t h i s thesis was to develop some theory by which i t would be possible t o obtain the c r i t i c a l buckling load of a beam loaded and constrained i n a most general way. With the a i d of an e l e c t r o n i c computer, this has been accomplished. The procedure involving an approximation to the load-determinant curve proved to be accurate, as w e l l as r e l a t i v e l y f a s t . In a l l cases tested, whenever a comparison could be made against e x i s t i n g solutions [ l ] , the r e s u l t s were found to be within one per cent of the published values, with only a single t r i a l required. The determination of the buckling load of a r e l a t i v e l y s t i f f beam usually required more than one t r i a l . This i s k9. because the load-determinant curve exhibits very large curvatures i n the v i c i n i t y of the horizontal axis, and the interpol a t i o n scheme i s r e l a t i v e l y poor. i f a good r e s u l t f o r Q c r Great accuracy i s necessary i s to be expected. To a l l e v i a t e t h i s problem, the programme can be e a s i l y modified to develop the curve almost exactly. Such a procedure, i n f a c t , was used during the preliminary stages of development, at which time i t was done to study the rates of convergence of the various non-linear terms. Though the l a t t e r technique i s most accurate, i n most instances i t Is uneconomical to use because of the computation time involved, about eight minutes f o r each point calculated to define the curve. The included examples show just a few of the many facets of the l a t e r a l buckling problem that can be studied by programming the equations developed In the theory. Extension of the equations can lead to the i n c l u s i o n of a x i a l forces, various cross-sections, and possibly decreasing computation time. 50. BIBLIOGRAPHY 1 Flugge, D., "Handbook of Engineering Mechanics," McGrawH i l l , 1st ed., 1962. 2 G a l l e t l y , G. D., discussion of "Computer Design of a Multistory Frame Building," by A. M. Lount, Journal of the Structural D i v i s i o n , A.S.C.E., Vol. 86, No. ST 6, June, I960. 3 Murdoch, D. C., "Linear Algebra f o r Undergraduates," John Wiley & Sons, 1st ed., 1957. k Ayres, F., "Schaum s Outline of Theory and Problems of Matrices," Schaum Publishing Company, 1st ed., 1962. 5 Timoshenko, S. P., and Woinowsky-Kreiger, S., "Theory of Plates and S h e l l s , " McGraw-Hill, 2nd ed., 1959. 6 Hooley, R. F.., and Madsen, B., "Lateral S t a b i l i t y of Glued Laminated Beams," Journal of Structural D i v i s i o n , A.S.C.E., June, 196k. 1 51. APPENDIX I - DATA INPUT-OUTPUT In the data input description of Table I, i t has been assumed that the reader i s f a m i l i a r with computer programming. To further c l a r i f y the procedure, antexample, Pigs. (Ik) and (15) have been included. For convenience of description, the beam has been divided i n t o only four segments. A l l information input Into the programme i s i n units of feet and kips. The information that the programme p r i n t s out includes the following: 1. The value of the c r i t i c a l load acting at each j o i n t . If the c r i t i c a l load i s made up of more than one type of load, as f o r example a concentrated load and a uniformly distributed load, then both values are printed out. 2. The values of the determinants D^, D , T>y D^ - these 2 are printed out to be used as a check on the accuracy of the r e s u l t i n g c r i t i c a l load. 52. TABLE I - DATA INPUT SEQUENCE SET DATA FORMAT DETAILS G.l N,E,G,D,T,H IIO,5PIO.O - 1 card G.2.1. XMA(K), XMB(K), XMC(K), VB(K), VC(K), BETA(K) 6F10.0 - N-l cards, one for each j o i n t from 1 to N-l. - each card contains a guess as t o the magnitude of the item at the j o i n t under consideration caused by the i n i t i a l load Q-j^ G.2.2 BETA(N) F10.0 - 1 card - card contains a guess as to the value of at joint N caused by the i n i t i a l load Q^. One set of G.3 data cards i s required f o r each j o i n t G.3.1. (NBC(5K), 1*1,10), 1115 JLN(K) - 1 card per j o i n t . - the f i r s t ten f i e l d s indicate which constraints, 1 to 10 respectively, as described i n Section 2 . 3 . 3 exist a t the j o i n t . I f , for example, constraint types 3 and 8 exist, then a 1 i s placed In f i e l d s 3 and 8 , while zeros are placed elsewhere (card 1 of example). - the 11th f i e l d indicates whether cards G.3.k, G.3.5* G . 3 . 6 , G.3.7 are to be included. I f the integer 2 (card 7 of example) i s placed i n the f i e l d , then a l l four cards are to be included. I f any other integer i s used, then only G.3.7 i s to be included (card l k of example). 53. TABLE I SET (cont'd) DATA DETAILS FORMAT Such a s i t u a t i o n may occur when a l l data on G . 3 . 4 , G.3.5j G.3.6 has zero values. G.3.2 BC(I,K) F10.0 - 1 card f o r each integer 1 occurring on card G . 3 . 1 . I t gives the value of c^ , (Section 2 . 5 ) . The vtlue 0 . 0 i s to be used i f a 1 occurs i n f i e l d s 8, 9, or 1 0 , (card 9 of example). G-.3.3 CI(K), CL(K), CE(K) 3F10.0 - 1 card i s read when a 1 occurs i n any of f i e l d s 8, 9, 1 0 of G . 3 . 1 . I t gives the physical properties of the column, (card 1 1 of example). G.3.4 P1(K), E21(K), H21(K), P2(K), E 2 2 ( K ) , H22(K) 6F10.0 - 1 card - contains data describing any v e r t i c a l l y applied load as well as guesses as to value of the v e r t i c a l force created by any constraints that may be acting. G.3.5 H1(K), E 2 1 ( K ) , H21(K), P2(K), E 2 2 ( K ) , H22(K) 6F10.0 - 1 card - contains data describing any h o r i z o n t a l l y applied load as well as. guesses as t o the value of the horizontal forces created by any cons t r a i n t s that may be acting. G.3.6 XMOA(K) , XMOB(K), 6 F 1 0 . 0 XMOC(K), XMOA2(Ki XMOB2(K), XM0C2(K) i 1 card - contains data describing applied moments, and guesses as to value of any moments created by any constraints that may be acting. 5k. TABLE I (cont'd) DETAILS DATA FORMAT G.3.7 W1(K), E l l ( K ) , H l l ( K ) , W2(K), E12(K), H12(K) 6F10.0 - 1 card (for j o i n t 1 to N - l only) - data f o r v e r t i c a l l y applied, uniformly d i s t r i b u t e d load along segment K. G.k.l ITT(L), L-1,6 615 - 3 cards - describes which externally applied loads are to be increased during each c y c l e . If a zero i s placed i n the f i e l d , then that p a r t i c u l a r loading i s not increased. If a 1 i s placed i n a f i e l d , then that loading i s increased. ! SET 1 1 1 1 1 1 in field " " " " " " " " " 11 1—increase 2— " 3— " ij.— " 5— " 6— " XMOA(K) XMOB(K) XMOC(K) P1(K) H1(K) W1(K) 55. I0 Pl(3) =10' A- XMOC(l)= 30k-ft XM0C(5) = 30k-ft 'I Vi •i/ i/ )f ]/ >]/ r 1r .25' •!/ <— 25' A J (2) Joint No. (I) Section A-A (4) (3) (5) L = 36.0' COLUMN BEAM - fixed at base - CL(I) = column length = 12.0 ft -CE(I)= " modulus = 260,000 ksf - CI (I) = " inertia = .025 f t 4 N = 5 joints E = 260,000 ksf G = 150,000 ksf D = 9.0ft (segment length) INITIAL GUESS Pl(3) = 10 (acting at top of beam at an eccentricity H2l(3)=.25ft ) XMOC(I) =XM0C(5) = 30 k-ft Wl(l) = WI(2) = WI(3) = WI(4) =.l k/ft (acting at centre of gravity of beam) FIG. 14 EXAMPLE USED FOR DESCRIBING DATA INPUT PROCEDURE 56. 5 COL 5 COL CARD 5 260 000.0 150 000.0 9.0 .417 2.5 0.0 0.0 0.0 5.0 0.0 0.0 0.0 0.0 -15.0 5.0 0.0 0.0 0.0 0.0 -60.0 -5.0 0.0 0.0 0.0 0.0 -15.0 -5.0 0.0 0.0 G.2.1 0.0 0 0 0 0 0 0 .025 12.0 260 0.0 0.0 0.0 -5.0 -1.25 0.0 0.0 0.0 0.0 .1. -1.25 0.0 0.0 0.0 -30.0 1.0 0.0 0.0 0.1 0.0 0.0 0.0 0.0 0.0 0 1 0 0 2 0.0 0 0 0 0 0.1 0 0 0 0.0 0 0 0 0 0.0 0 0 0 0 0.0 0 0 0 3 0.0 0 0 0 0.0 2 10.0 1.25 .25 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.1 0.0 0.0 0.0 0.0 0.0 0 0 0 0.1 1 0 000.0 0 0 0.0 1 0 0 0 0.0 1 0 0 0 0.0 0 0 0 3 QO 0 0 0.0 2 0.0 G.3.2 0.0 0.0 0.0 0.0 0.0 -5.0 0.0 0.0 0.0 0.0 0.0 .1 0.0 0.0 0.0 0.0 30-0 0.0 0.0 0.0 0 0 1 1 0 0 0 0 I 1 0 0 0 0 1 1 0 0 r G.4.1 FIG. 15 CODING SHEET FOR EXAMPLE DESCRIBING DATA INPUT 57. APPENDIX II - ADJUSTMENT OP PROGRAMME FOR SPECIAL CASES The computer programme written to solve the l a t e r a l s t a b i l i t y problem was written i n Fortran IV language f o r the IBM 70k0 computer i n s t a l l a t i o n located at the University of B r i t i s h Columbia. Appendix I I I . A copy of the programme i s included i n A l i s t of the relevant symbols Is included i n the Notation section at the beginning of the paper. To include the column connections shown i n Figs. (7d) and (7©), a series of revisions must be made to the o r i g i n a l programme. In both cases, the new cases replace constraint type 10. To develop the condition shown i n F i g . (7d), the l i n e s l a b e l l e d a., b., c , d. In the programme l i s t i n g must be replaced successively by the following: 930 A3(6,8) - -3.-JrCE(N2)*CI(N2)*H%(l.+ CL(N2)/H)/CL(N2)^3 A3(8,8) = -A3(6,8)*CL 9314- A3(21,23) » -3.^CE(N3)^CI(N3)^H%(l.^-CL(N3)/H)/CL(N2)<Hr3 A3(23,23) = -A3(21,23)*-CL To develop the condition shown i n F i g . (7e), the following revisions must be made. Replace the statements labelled a., b. successively by: 930 A3(6,6) = P2(N2)/CL(N2) A3(6,8) = -.5*P2(N2)*H/CL(N2) and Immediately after b., add: Al(26,13) - 0.0 Al(27,10) = Al(27,10) - XM0A2(N2) Al(28,9) Al(27,10) A3(3,13) 0.0 A3(k,10) A3(4,10) - D2*XM0A2(N2) A3(5,9) A3(5,9) + D1*XM0A2(N2) Replace the statements l a b e l l e d c , d. successively by: 934 A3(21,21) = P2(N3)/CL(N3) A3 (21,23) = -.5-"'P2(N3)-«-H/CL(N3) and immediately after d., add: A3(ll,28) a 0.0 A3(12,2£) = A3(12,25) - XM0A2(N3) A3(13,24) = A3(12,25) A3(13,28) =0.0 A3 (19,25) s A3 (19, 25) - D2»XM0A2(N3) A3(20,2k) » A3(20,2k) -|- D1*XM0A2(N3) Remove statements l a b e l l e d e., f . Replace statements l a b e l l e d g., h. by the single 946 statement: IF(NBC(5,K).EQ.0.AND.NBC(8,K).EQ.0.AND.NBC(9,K).EQ.O)GOT0947 S i m i l a r l y replace the pairs of statements l a b e l l e d i . , j . : k., 1«: m., n.: o., p.: and q., r . successively 951 IP( ... as above by the statements: )G0T0952 1023 IF( ... as above )G0T0102k 1028 IF( ... as above )G0T01029 1034 IF( ... as above )GOT01035 3039 IF( ... as above )G0T010k0 59. FLOW c DIAGRAM Start ( Read Data Card 6.1 ) Adjust matrix a-,\ and ajj for boundary conditions Calculate Constants /Reoad Data Card y s e t G.2 Store ajj on tape Calculate terms for Matrix a,, Fig. 9 " Calculate XJ (Eqn.45) during process obtain determinant of UJJ <z Backsubstitution to calculate Z, Eqn. (46) and determinant of structure matrix Adjust a,, for Boundary Conditions Invert a ,obtain its determinant and store inverted matrix on tape I n •a i o a> Calculate terms for matrix Q;J (j <i) o Adjust WJ for boundary conditions 2 , 3 r d ,4 run Read card G.4, Increase applied loads and appropriate terms of load vector by 2 0 % o Calculate terms of Ist load vector w; n d st t h run ist run Read data card set G.3 for next 2 joints^T" Calculate terms of next load vector wj o a> Read card G.4, Increase applied loads and appropriate terms in load vector Z by % calculated by Eqn. (49) CL a> a: Adjust appropriate terms of load vector Z Print 4 determinants and critical load Adjust WJ for boundary conditions Stop Calculate terms of next matrix a\\ Calculate terms of next matrix ajj (j>i) FIG. 16 FLOW DIAGRAM FOR COMPUTER PROGRAM "O c OJ •a to v 12 il' IC s 3 7 S I B F T C STAB C L A T E R A L ST A3 I L i TY..QF. RECTANGULAR _B.EAM.S- ... .... C N = NUMBER OF UOINTS = 1+ NUMBER OF SEGMENTS DIMENSION NBC(10,23)»BC(10,23),JLN(23)»ITT(6) DOUBLE P R E C I S I O N Wl (71 ) ,F1 1 (71 ) ,H1 1 (71 1 .W7f ?_1 I . J ^ 2 - L 2 J - 4 - ^ L 4 ^ J _ 2 J - J DOUBLE P R E C I S I O N P K 2 1 ) , E 2 1 ( 2 1 ) , H 2 1 ( 2 1 ) , P 2 ( 2 1 ) , E 2 2 ( 2 1 ) , H 2 2 ( 2 1 ) DOUBLE P R E C I S I O N H 1 ( 2 1 ) » E 3 1 ( 2 1 ) » H 3 1 ( 2 1 ) , H 2 ( 2 1 ) , E 3 2 ( 2 1 ) , H 3 2 ( 2 1 ) . DOUBLE._P.RE_C I.SI.OjN XM0A.(.21 ).?^M0B.(,2L). ,XiMQC(-.2.1 ) ,XMCA2.( 21 )., X X 0 3 2 ( 21 ) DOUBLE P R E C I S I O N X M 0 C 2 ( 2 1 ) , C E < 2 1 ) , C L ( 2 1 ) , C I ( 2 1 ) » V B ( 2 1 ) , V C ( 2 1 ) DOUBLE P R E C I S I O N A ( 3 0 , 3 0 ) , A 2 ( 3 0 , 3 0 ) , A . 1 ( 3 0 , 3 0 ) , A 3 ( 3 0 , 3 0 ) , A X ( 5 , 3 0 ) DOUBLE P R E C I S I O N A X X (5 , 3 0 U j _ J L _ i n j L J L l J - O C E J 3 0,11 )-,-X-e_X-(-3-Q-, 4^)-,-Z-(-3-Q-,4r4r-)DOUBLE P R E C I S I O N Z M ( 3 0 , 1 1 ) , B E T A ( 2 1 ) ,XMA(21) , X M B ( 2 1 ) , X M C ( 2 1 ) DOUBLE P R E C I S I O N X I 1 , X I 2 , X J , D 1 , D 2 , D E T , A D E T , D J -POUBLE^PREC I SI ON. X0ET 1 ,XDET2., X.DET3 , XDET4 , S..1 ,-S2 ,S3 ,ADJ ,.CL3 .REWIND 1 REWIND 2 READ IOO, N, F , ( _D_ T 100 FORMAT ( I 1 0 . 5 F 1 0 . 0 ) X I I =.08333#T*H**3 „XJ 2 = .,,Q8.3.3 3*H*T**3.. .- • X J = 4 . * X I 2 # ( l . - . 6 3*T/H) DI =.5*D/(E*XI1) D2 = . 5*D/( E*X I 2 ) DJ =.5*D/(G*XJ) NRUN = 1 _COUNT.= 0.0.., . . ........ NM = N - l DO 2 0 0 K=l» NM READ 1 0 2 , X M A ( K ) , XMB ( K ) , XMC ( K ) , VR ( K ) , VC ( K .),, B ETA ( K. 102 FORMAT ( 6 F 1 0 . 0 ) 200 CONTINUE J_EAD_.120.» BE TAJ N )„. . _ . ...... "12 0~ FORMAT ( F 1 0 . 0 ) READ 1 0 3 , ( N B C ( I , 1 ) , 1 = 1,10) , J L N ( 1 ) 103 FORMAT ( 1 1 1 5 ) DO 2 0 2 1=1,10 I F ( N B C ( I , 1 ) .EQ.O ) GO TO 202 > a . H H TI W O- S tt) M CQ- i-3 H READ 1 0 1 , B C ( I , 1 ) .10_L,,_-FP,R_MAJ .15F10..0 L . 202 CONTINUE IF (NBC(8,1).EQ.1.OR.NBC(9,1).EQ.l.OR.NBC(10,1).EQ.1) GO TO 2 3 6 GO TO 2 4 5 236 READ 1 0 4 , C I ( 1 ) , C L ( 1 ) , CE ( 1) 104 FORMAT ( 3 F 1 0 . 0 ) _ . „ 2 4 5 _ READ. 10.2,.PI ( 1) ,E2.1 ( 1 ).,_H2l( 1 ). , P 2 ( . l ) ,E22( 1),H.22 ( 1 1 ... READ 1 0 2 , H 1 ( l ) , E 3 1 ( 1 ) ,H31 ( 1 ) , H 2 ( 1 ) , E 3 2 ( 1 ) , H 3 2 ( 1 ) READ 1 0 2 , X M O A ( 1 ) ,XMOB(1) ,XMOC(1) ,XMOA2(1),XMOB2<1) , X M O C 2 ( 1 ) READ 1 0 2 , Wl( 1 ) ,E1.1 ( 1 ) , m i m »WZM ) ,F1 ? M > .Hi? n ) READ 1 0 3 , (NBC( 1,2) ,1 = 1,10) , J L N ( 2 ) DO 3 0 0 1=1,10 J ...J NBC ( I , 2 ) . EQ . 0 ) .CO ..TO 300... . .. . ._ READ 1 0 1 , BC(I»2) 300 CONTINUE — — 237 295 F I F ( N B C ( 8 , 2 ) . E Q . 1 . 0 R . N B C ( 9 , 2 ) . E Q . 1 . 0 R . N B C ( 1 0 , ? 1 . F O - 1 ) r,n m GO TO 2 3 8 READ 1 0 4 , C I ( 2 ) , C L ( 2 ) , C E ( 2 ) .DO 29 5 1 =1 , 10 _ . .__ , I F (NBC ( I , 2 f. EQ • Y) GO TO 2 9 7 CONTINUE I F ( J L N ( 2 ) . E Q . 2 ) GO TO 297 P 1( 2) =0.0 E2K2) =0.0 _H ?1 ( 2J .=0.0 . ,. . "E2 2 '( ?') =0.0 H22(2) =0.0 HI (2) =0.0 E3K2) =0.0 H31 (2 ) =0.0 E32(2) =0.0 "H3"2"(2) " " = 0 . 0 " " XM0A(2) =0.0 XM0B(2) =0.0 XM0C(2) =0.0 P2(2) =0.0 H2(2) =0.0 . - —- = - p^y - --- •-• - ------~ - . . _ o „ H _ s » » » » 12 '11 8 5 _ S 3 XM0A2(2) = 0.0 ,„XJ10B2J_21, .. XM0C2(2) = 0.0 GO TO 2 9 6 297 READ 1 0 2 , P l ( 2 ) , E 2 1 ( 2 ) , H 2 1 ( ? l , P ? ( ? ) , F ? ? ( ? l . H ? ? ( ? i READ 1 0 2 , H K 2 ) , E 3 1 ( 2 ) , H 3 1 ( 2 ) , H 2 ( 2 ) , E 3 2 ( 2 ) , H 3 2 ( 2 ) READ 1 0 2 , X M O A ( 2 ) , X M O B ( 2 ) , X M O C ( 2 ) , X M O A 2 ( 2 ) , X M O B 2 ( 2 ) ,XMOC2(2 ) 296 READ _10 2.»_ „W1 ',2 ).» E 1 1 J 2 )_,H11(_2 ) ,W2.(.2.)>E12(2) ,H12 (2 ). ... ....... . . 2 99 CONTINUE DET =0.0 ADET =0.0 DO 8 8 2 1= 1 ,30 DO 8 8 2 J= 1,30 . 8 8 2 _ _ ...AH.'J ).. =0 . 0 A ( 1»1 ) = 1.0 A ( 2, 2) = 1.0 A ( 3, 3) = 1.0 A(4,4 ) = 1.0 A(5,5 ) = 1.0 A ( 6 , 1,1 ) = 1,.0 .. . . . . A(7,12) = 1.0 ' A(8,13) = 1.0 A(9,14) = 1.0 A(10,15 ) = 1.0 A(11 , 1 ) = -1.0 A ( 1 1 , 1 6 ) „=1..0 A(11,23 ) =-pl(1 )*E21(1)+H1(1)*H31 ( 1 ) A(11,24) =-XMOC ( 1 ) A(11,25 ) = -XMOR(1 ) A(12,2 ) = -1.0 A ( 1 2 » 17 ) = 1.0 ,A( 12 ,.23.), = XMOC(l.). .... .. . A( 12,2~5) = XMOA<1>+P1(1)*(E21(1)*BETA(1 ) +H21(1) ) 1 +H1(1)*(E31(1)-H31(1)*BETA(1) ) A ( 13 » 3 ) = -1.0 A(13,18 ) = 1.0 A(13,23 ) =-XMOB(l) .-.. ...... | j 1 ! ! 1 1 1 1 1 11 ro i < CS 63. ) rH i < I— LU CQ '' ) rH -—- I Q ,i f r * ——- 1 rH rH r 1 rH CM 1— rH rH LU LU LU •\ + co * — i rH - — rH — L rH ( rH s + .—- i rH CM X rH * rH — rH rH —'• LU rH * O r H • CL rH I II II O; . ' O • o I I, II o •. o • o O II O O O; O rH II . II II II II II II II CM rH + rH — — rH rH — rHj 3 u •CO' * O II + -— rH ! X •— -—- r — rH — . — - ~ rH LP, CM rH u < o s: CM co CM X -— rH Q X X 3 r-f + X 1 M + 1 Q 1 I + 1Q X < I rH II LU II II o II 1 II II II II II II II II I o co ~ CM CM vO. CM f*- in iO vo in rH rH < < 1 r- CM CM CO CM o m o: CM rH CM r~- co oo o j o O O CM CM rH, rH rH rH < ' < < < < < <:< < < M C O O v O CO vt" CM CM CM CO r—I CM CM CM CO <t L P \0 CM CM CM CM CM \o CM CM LP, O CO LP, CM' •—I CM CM CM CO O o CO 00 CO CO CM CM CM CM f~- r- r~ CM- CM CM CM < <;'< < < <<<*<<< CM 1 CO <f- CM CM <•< < < Cil O !N W A ( 3 0 , 2 3 ) =-VB( 1 ) - ( W l ( 1 ) + W 2 ( 1 ) ) * D .___„.NjN _,. " J . _ . i\BC2 900 .880 901 _5Q4_.__CjDNT I 903 904 __05 =1 GO TO 9 0 0 N2 =?*( NN-1. ) N3 =N2+1 I F ( N N . G T . 1 ) GO TO 9 0 1 Q.0.880..I =J.,3C . ... DO 8 8 0 J =1,30 A3(I,J) = A(I» J ) GO TO 9 1 3 DO 9 0 2 1=1,10 I F ( N B C ( I.,N2) . E Q . l ) GO fclUE.. TO . 903 . . ... . . „ GO TO 9 1 3 I F ( N B C ( 1 , N 2 ) . E Q . 0 . A M D . N B C ( 2 , N 2 ) . E Q . O ) GO TO 905 A l (2.6,8) =A1(?6,fll+H?(N7)*H37(N71 A l ( 2 6 , l l ) =-E32(N2) A l ( 2 7 , 10 ) =A1 ( 2 7 , 10 )+H2 (N2 )*( E32 (N? ) -H3 2 (N2 ) »- B E T A ( N 2 ) ) .AJ.J 2.8,9.] „=AJ(2.7,1D) ...... . A 1(29,8 ) =A1(29,8)-H2(N2) A l ( 3 0 , 1 1 ) =-1.0 A3(6,6 ) =1.0 A 3 ( 6 ,11 ) =0.0 A3(3,8) . =A3(3,8)+DJ*H2(N2)#H3 2(N2) jA3J-3 ,11.)... _ = rEJ32 ( N.2 ) * P J . , ... . , . A3(4,10J =A3(4,10)+D?*H2(N2)*(E32(N2)-H32(N2)*BETA(N2) ) A3(5,9) =A3(5,9)-D1*H2(N2)*(E32(N2)-H32(N2)*BETA(N2 > ) I F ( N B C ( 2 , N 2 ) .EQ . 1 ) GO TO 904 GO TO 9 0 5 A3(6,8) =+BC(2,N2) _ I F . J NBC (3.' N2J..EQ. C_..AIH0...NBC { 4 1 H2.UEQ, Q)_.GO_XCL.3xi2-. AK26.8) = AK26,8)-P2(N2)*E22(N2) A K 2 6 . 1 2 ) =-H22(N2) A l < 2 7 , 10 ) =A1 ( 2 7 , 1 0 ) + P ? (N2 )-"-( F72 (N7 )-^RFTA ( M7 ) +H2? f M2 ) ) A1(28,9)=A1(27,10) A l ( 2 9 , 12) =1.0 ._ f OS_ E £ A l ( 3 0 ,8 ) — =A1(30*8)-P2(N2) — , g ! 906 907 AJ(.3»3-) =A3(.3.v8.L-DJ*P2 CN2 ) *E22.(.N2J.... A 3 ( 3 • 12 ) = - H 2 2 ( N 2 ) * D J A3(4,10) = A 3 ( 4 , 1 0 )+D2*P2 ( N2 ) * ( E22 ( N2 )*BETA ( N2 )'+H22 ( N2 ) ) A3 ( 5 * 9 ) =A3(5»9)-D1»P2(N2)»( E22±(i2_)JiBELAJ.N.2-)J--td2.2-(Ji2-)-)__ ' A3(7,7) =1.0 A3(7»12) =0.0 J F (NBC (.4»N.2.) • EQ. 1J. .GO . JO 906. ... .., ... ....... GO TO 9 0 7 A3(7»8) =+BC(4>N2) I F ( NBC ( 5 » N2 ) . EQ. 0 ) GO TO 9 0 8 A l ( 2 6 , 1 3 ) =-1.0 Al<27»10) =A1(27,10)+XMOA2(N2) . _ _A 1.( 2 8 »9 =A1( 28_»9 ).+X.M0A.2.(.N.2L ..... .. . . . . .. . A 3 ( 3 ,13) =-DJ A3(4»10) =A3(4,10)+D2*XMOA2(N2) =1.0( 5 ,9 ) -D1*XM0A2 ( N2 ) A3(8 5,8 9 )) =A3 A3(8,13) =0.0 JL9JL_—,- J -F. N.BC (_6 » N 2 ) . E.Q . Q.) . GO . T0 _ 9Q9 ( . . « . .• -.- • L { -i 8 1 6 11 1 | : —i zl . . ' i = 1 . j ! .. ........ . _ . . . A l ( 2 6 , 1 0 ) =A1(26»10)-XMOB2(N2) Al(27»14) =-1.0 AK28»8) = A 1 ( 2 8 , 8 )-XMOB2(N? ) , A3<3,10) =A3(3,10 )-DJ*XM0B2(N2) A3(4,14) =-D2 ..A3 (.5,8.). =A3(5,8)+D1*XM0B2(N2) , ...... . . ..... . . ..... .. - . A3(9,9) =1.0 A3(9,14) =0.0 909 I F ( N B C ( 7 , N 2 ) . E Q . O ) GO TO 9 1 0 AK26,9) =A1 ( 2 6 , 9 )-XMOC2(N2 ) Al(27,8) =A1(27,8)+XMOC2(N2) •12 _ A l t 2 8 , 1 5 ) =-1.0. .. . . . . .. . . ... . . ii." "A'3('3»9) " = A 3 ( 3 , 9 ) - X M 0 C 2 ( N 2 ) * D J 10 A3(4,8) = A 3 ( 4 , 8 ) +XMOC2 (N2 )*D2 9 A3(5,15) =D1 . 0>8 A 3 ( 1 0 , 1 0 ) =1.0 y 7 A3(10,15)=0.0 S_..„?10_.. I.F_ (NBC ( 8, N2 ) * EQ. 0_. AND.. NBC J 9., N2.) . EQ . 0...AMD ..N.B.C ( 10., N2 ,)...£Q... 0 ) GOT-091-3 • • 5 I 1 ! A l ( 2 6 , 8 ) = A1 ( 2 6 , 8 ) + H 2 ( N 2 ) * H 3 2 < N2) ^^hlJJ&?. XX-L.-=-E3ZAM2A ..... .... - v . , . . „ .. A l ( 2 7 , 10 ) = A 1 ( 2 7 , 1 0 ( + X M O A 2 ( N 2 ) +H 2 ( N 2. ) * ( E 3 2 ( N 7 ) -H 3 2 ( N 2 ) *B E T A ( N 2 ) ) Alt.28,9) =A1(27,10) A l ( 29 ,8 ) = A l ( 79 , 8 )-H? ( N? ) . A l ( 3 0 ,11 ) =-1.0 A l ( 2 6 , 1 3 ) =-1.0 ...____._A2y,»_8J_ .... = A3J__.» 8 ) +J_Jl*tf21N2 ) *H-22 (.i_2.). .. _ .... , A3(3,ll) =-E32(N2)*DJ A3(3,13) =-DJ A3 ( 4 , 10 ) =A3 ( 4 , 1 0 )+ D 7 * H 7 ( M ? )#( F7 7 ( M 7 , - H 7 ? I N ? ' l » R F T A ( M g ) ) 1 911 . +D2*XMOA2 ( N2) A3(5,9 ) =A3(5.9)-D1*H2(N2)*(E2 2(N2)-H32(N2)*BETA(N2>) JL , -D.1*.XM0.A2.(.MZ)„.. . ... , .. . ...„_...____.._ I F <NBC(8»N2 l . E Q . l ) GO TO 9 1 1 . I F ( N B C ( 9 , N 2 ) . E Q . l ) GO TO 9 1 2 I F ( N B C ( 1 0 , N ? ) . F Q . 1 ) GO TO 9 3 0 A3(6,6) =+12.*CE(N2)*CI(N2)/CL(N2)**3 A 3 ( 6 ,8 ) =-6.*CE(N2)*CI (N2)#H*(1.+CL(N2)/H)/CL(N2)##3 . ,_jft3. J &». 6_)._ j = - 6 ..*CE .(.N2 l * C U N 2 . - / C U N 2 ) * t * 2 . . ... .... . . A3(8 , 8 ) =CE(N2)*CI(N2)-*H*(3.+4.*CL(N2)/H)/CL(N2)*#2 GO TO 9 1 3 912 A3(6,6) = + ?.»fFIN7)»n I M ? ) / r i (M?)»»3 , A3(6,8 ) =-1.5*CE(N2)*CI(N2) *H*(1.+2.*CL(N2)/H)/CL(N2)**3 A3 ( 8 , 6 ) = - 3 . * C E ( N 2 ) * C I ( N2 ) /CL ( N2 ) **2 ,.__,A3 (.8 » 8 „_„... = 1..5_£j;,(jN2J * C U J M 2 ) * H * (1..+2-. *CUM2±S.H.WCL (N2*.**-2 — GO TO 9 1 3 930 A3(6,8) =-6.*CE(N2)*CI(N2)*H*(2.+CL(N2)/H)/CL(N2)**3 A3(8,8) = 2.*CE(N2)*CI(N7}*H*(3.+7.»ri (N 7 ) /H ) / f l f M 7 ) » * 7 913 DO 9 1 4 1=1,10 I F ( N B C ( I ,N3 ) .EQ.1 ) GO TO 9 1 5 J914 CONTINUE _ GO TO 92 5 915 I F ( N B C ( 1 , N 3 ) . E Q . 0 . A M D . N B C ( 2 , N 3 ) . E Q . O ) GO TO 9 1 7 A3( 11,23) = A 3 ( 1 1 , 2 3 ) + H 7 ( N 3 ) * H 3 ? ( N 3 ) A 3 ( l l , 2 6 ) =-E32(N3) A3 ( 12 ,25 ) =A3( 1 2 , 2 5 )+H2 (N3 )*( E32 ( N3 ) -H32 (N3 ) -""BETA (N3 ) ) .. .... : ••- • a. fc>. A 3 ( 1 3 , 2 4 ) =A3<12,25) 23. ).... =J\3< 1.4,2 3.)-H2LN3..). .. . ..... .„ __ . . ... A 3 ( 1 5 , 2 6 ) =-1.0 A 3 ( 2 1 » 2 1 ) =1.0 A 3 ( 2 1 , 2 6 ) =0.0 I F ( N 3 . E Q . 1 ) GO TO 916 A3( 18,23) =A3(18,23)+DJ*H2(N3)*H32(N3) - A3 < I?., 2 6.) ._=-D.J*E32 ( M3 ) . _ ........... A 3 ( 1 9 , 2 5 ) = A3(19,2 5 >+D2*H2(N3)*(E3 2 ( N 3 ) - H 3 2 ( N 3 ) * B E T A ( N 3 ) ) A 3 ( 2 0 , 2 4 ) =A3(20»24)-D1*H2(N3)*(E32(N3)-H32(N3)*BETA(N3)) I F ( N B C ( 2 , N 3 ) . E Q . l ) GO TO 931 GO TO 9 1 7 A 3 ( 2 1 , 2 3 ) =+BC<2,N3) _ _ I F„.(_NBC_(_3 » N3J,• EQ• 0 . AND • NBC ( 4 .sJN3 ).. EQ..0 J_.GQ . T 0 _ 9 . 1 9 ™ , A 3 ( " l l , 2 3 ) =A3( 1 1 , 2 3 ) - P 2 ( N 3 ) * E 2 2 ( N 3 ) A 3 ( l l , 2 7 ) =-H22(N3) A3(12,25 ) =A3(12,25)+P2(N3)*(E22(N3)-*RFTA(N3)+H77(N3) ) A3(13,24) =A3(12,25) A 3 ( 1 4 , 2 7 ) =1.0 J M J 15 » 23.) = A3(.15..,23)-P2(N3) . - - .- — — . -_- A 3 ( 2 2 » 2 2 ) =1.0 A 3 ( 2 2 , 2 7 ) =0.0 I F ( N 3 . E Q . 1 ) GO TO 9 1 8 A3(18,27) =-H22(N3)*DJ A3 ( 18 , 23 ) = A 3 ( 1 8 , 2 3 ) - D J * P 2 ( N 3 ) * E 2 2 < N3 ) .A3.(J_M>_ 916 931 917 . - .„ 918 932 919 21 A 3 ( 1 9 , 2 5 ) =A3(.1.9,.2.5.) . + D2*.P N.3 ) * (. E2 2 ( N3 )-+H221H3J « L B £ - T A ( J A 3 ).). -.A3 ( 2 0 , 2 4 ) =A3( 20,2 4 ) - D l * P 2 (N3 ) * ( E 2 2 ( N 3 ) + H 2 2 ( N 3 ) * B E T A ( N 3 ) ) I F ( N B C ( 4 , N 3 ) . EQ. 1 ) GO TO 9 3 2 GO TO 9 1 9 A3(22,23) =+BC(4,N3) I F ( N B C ( 5 , N 3 ) .EQ.O) GO TO 9 2 0 A 3 ( 1 1 , 2 8 ) =-1.0 A3(12,25J'=A3l12,25(+XMOA2(N3) A3(13,24) =A3(13,24)+XMOA2(N3 ) A 3 ( 2 3 , 2 3 ) =1.0 A 3 ( 2 3 , 2 8 ) =0.0 I F ( N 3 . E Q . 1 ) GO TO 9 2 0 A 3 ( 1 8 , 2 8 ) =-DJ 920 A 3 ( 1 9 » 25 ) =A3( 1 9 • 2 5 ) + D 2 * X M O A 2 ( N 3 ) __A3__2 0 i 2 4 ___ =__3_L2f>» 24_L-DJ.* XM0A2,( N3. L I F ( N B C ( 6 , N 3 ) . E Q . 0 ) GO TO 9 2 1 A 3 ( l l , 2 5 ) =A3(ll»25)-XMOB2(N3> A3 ( 13 ,23) = A 3 ( 1 3 , 7 3 )-XMOR7 [ ( M ? ) A 3 ( 1 2 , 2 9 ) =-1.0 A 3 ( 2 4 , 2 4 ) =1.0 .MJJ ..» 2?_). =0.0 4 I F " ( N 3 . E Q . 1 ) GO TO 921 A 3 ( 1 8 . 2 5 ) = A 3 ( 1 8 ,25 ) - X M O B 2 ( N 3 ) * D J A 3 ( 1 9 , 2 9 ) =-D2 A 3 ( 2 0 , 2 3) = A 3 ( 2 0 » 2 3 ) + X M O B 2 ( N 3 ) * D 1 I F ( N B C < 7 » N 3 ) . E Q . O ) GO TO 922 A3 ( 1 1 , 2 4 ) = .A3 ( 11., 24 )-XMOC2 (N3 ) A3(12,23) =A3(12,23)+XMOC2(N3) A 3 ( 1 3 , 3 0 ) =-1.0 A 3 ( 2 5 ,75) =1 .0 A 3 ( 2 5 , 3 0 ) =0.0 I F ( N 3 . E Q . 1 ) GO TO 9 2 2 A 3 J . 1 8 , 2 4 ) = A.3( 18 »24)-DJ*XiMOC2 ( N3 ) . . ... ., . ... A 3 ( 1 9 , 2 3 ) =A3(19,23 J+XMOC2(N3)*D2 A 3 ( 2 0 , 3 0 ) =D1 Lf__LJ_BCJ_S_^ A3(11,23) =A3<11,23)+H2(N3)*H32(N3) A 3 ( l l , 2 6 ) =-E32(N3) A 3 (..12 , 2.5J ..,= A3.U2.*2.5.) +H2 (N3 ) * (..E32XN3 J ( N-3J *BETA ( N34 ) +-X-M0A2-(N3 •) A3(13,24) =A3(12,25) A3(14,23) =A3(14,23)-H2(N3) A 3 ( 1 5 , 2 6 ) =-1.0 A 3 ( 1 1 , 2 8 ) =-1.0 I F ( N 3 . E Q . 1 ) GO TO 9 2 3 A3 (J. 8 , _2 3 )_ .= A3 ( 1 8 , 2 3 ) +DJ#H2 ( N3 ) *H32 ( N3J -. .. A3(18,26) =-E32(N3)*DJ A 3 ( 1 8 , 2 8 ) =-DJ 921 „ 922 . _ A3 ( 1 9 , 2 5 ) 1 A3(20,24) =A3 ... -— • - =- — - ; . . . .. . . ... ( 1 9 , 2 5 )+ D 7> " - H 7 ( N ^ I *( F 3 7 t N 3 ) - m ? ( M ? ) » R F T A . fx| 3 ) ) +D2*XMOA2 (N3 ) =A3(20»24)-D1*H2(N3)*(E32(N3)-H32(N3)*BETA(N3) ) 0>0 0 . 1 -D1*'XM0A2 ( N 3 ) 92 3____IF _i^aC L8-_J_3J l.L.J30_..m_924 I F ( N B C < 9 , N 3 ) . E Q . l ) G O TO 9 3 3 IF (NBC(10,N3).EQ.l) GO TO 9 3 4 924 A3(21,21) =+12.*CE(N3)»CI(N3)/CL(N3)**3 A3(21,23) =-6.*CE(N3)*CI(N3)#H*(l.+CL(N3)/H)/CL(N3)*#3 A3C23,21)=-6.*CE(N3)*CI(N3)/CL(N3)**2 .A3X23.,23..)„5QE (t&X*£JSH3X*ti*l3*JA**£.L M3JsUX£OLU&l**2~-~ GO TO 9 2 5 933. 934 925 926 2 7,2 8.81 • . A 3 ( 2 1 , 2 1 ) =+3 . * C E ( N 3 ) * C I (N3) / C L ( N 3 ) * * 3 A 3 ( 2 1 , 2 3 ) = - 1 . 5 * C E ( N 3 ) * C I ( N 3 ) * H * M .+7..*fl '( N 3 ) / H ) / CI ( N 3 ) * * 3 A3(23»21)=-3.*CE(N3)*CI(N3)/CL(N3)**2 A3(23,23) =1.5*CE(N3)*CI(N3)*H*(1.+2.*CL(N3)/H)/CL(N3)**2 GO T O 9 2 5 . ....... _ A 3 ( 2 1 , 2 . 3 ) =-6. * C E ( N3 ) *C I ( N3 ) * H * ( 2 ."+CL( N3 )"/H7/CL'( N3 ) * * 3 A 3 ( 2 3 , 2 3 ) = ,2.*CE(N3)*CI (N3)*H*(3.+2.*CL(N3)/H)/CL(N3)**2 I F - ( N N . G T . l ) GO TO 9 2 6 G O TO ( 2 7 2 , 2 7 3 ) , N B C 2 CONTINUE PP_88 1,_, I._ .1, . 3 , 0 . ' . , ., .... _ _____ . DO 8 8 1 J = l , 3 0 " ' " ' ' A(I,J) =A3(I,J) C A L L I N V ( A , 3 0 , 3 0 ,DET,COND) WRIT E ( 1 ) A ADET =DET DO 883 - 8 8 3. JL= 1 J L 3 . Q A62"(8 I ,8J3)* "=j = 0l.,~30 0 D A 2 ( l , 2 1 ) =-1.0 A2( 1,24) =-.5*D A2(2,22) = -1.0 A2(2,25) =-.5*D 'A2~(3~»16)* ' = D J A2(3,23) = -1.0 A 2 ( 4 , 1 7 ) = D2 A2(5,18) = -Dl A2(4,24) = -1.0 A 2 ( 5 , 2 5 ) =-1.0 ... ... . . . . .,.:„_.,„__, ~~ d , A. ._ t> • f>i _0 »^ < > > to N' '""Ok 70. a +1 col X; +, CXIi t—I CM', *!' • CM! ~ CM o o; o o o o o o II II II o o o o; O o • • •! • o i—I i—I i—ij, •—I II II II n H! it < o oCD ou o X I I o II II , II tl o o o o; o o o II O II - O II 0 ; | t—< • — I o o '—I o rH CM + i'rM a — '.I — — I CMj * CM CM CM —• < ^ — CM CL H C"i — C D U O O rH o o « S'CM a rH X'l 3 X X I X II | II II II II I o z o z in — III II It II p i-i CM? cn <T in Q_ Q- Q_ CL CL i—i co: i—it i—i ro • .j~ CL CLICL CL CL (xi co a CL VO r~ CL CL ) co o o" <-H c\i co CM CM CM a n a i—I ai—I CL CM3CL(\!CL C\Ja C\l CL CL CL r- co ON O CM CM CM CO a. Q_a CL "1" t' O a UJ CM CDO CM CO u_ DO 5 0 1 1=1,10 . _ - I F _ J NB.CU^.N2.) E.Q...l.,).. G.0.1 10. 5 0 2 . CONTINUE GO TO 512 I F ( N B C ( 1 , N 2 ) .EQ.O ) GO TO 503 P(6,NN) =BC(1,N2) I F ( N B C ( 2 , N 2 ) . E Q . O ) GO TO 504 P ( 6, NN.)._ .. JLQ..JL ...... I F < N B C ( 3 , N 2 ) . E Q . 0 > GO TO 505 P(7,NN) =BC(3,N2) I F ( N B C ( 4 , N 2 ) .EQ.O ) GO TO 506 P(7,NN) =0.0 I F ( N B C ( 5 » N 2 ) . E Q . O ) GO TO 507 A 501 502 503 504 505 506 ,.-_~.-JL J. ?. N ) ( 5 07 508 509 510 511 5 12 5 13 5 14 5 15 516 5 17 N = BC (5 , N 2.). I F ( N B C ( 6 » N 2 ) • E Q . 0 ) GO TO 508 P(9,NN) =BC(6,N2) I F ( N B C ( 7 , N 2 ) . E Q . O ) GO TO 509 P(10»NN> =BC(7,N2) I F ( N B C ( 8 » N 2 ) . E Q . O ) GO TO 5 1 0 P ( 6 , N N ) , . =0.0 • P(8,NN) =0.0 I F ( N B C ( 9 , N 2 ) . E Q . O ) GO TO 511 P(6,NN) =0.0 P(8,NN) =0.0 I F ( N B C ( 1 0 ,N2).EQ.O) GO TO 512 .P(6,NN) =0.0 •p(8,NN) =0.0 DO 5 1 3 1=1,10 IF ( N B C ( I , N 3 ) . E Q . l ) GO TO 514 CONTINUE GO TO 5 2 4 I F ( NBC ( 1 ,.N3 ) .EQ.O.) . GO . TO 515 P(2i,NN) =BC(1,N3) I F ( N B C ( 2 , N 3 ) . E Q . O ) GO TO 516 P(21,NN) =0.0 I F ( N B C ( 3 , N 3 ) .EQ.O) GO TO 517 P(22,NN) =BC(3,N3) I F ( NBC ( 4 ,N3 ) .EQ.O ) GO.. TO 5 1 8 . 5_1J_. 5 19 520 521 P(22,NN) =0.0 J.E, J'NBjCJ_l.»..iN3_L. EJ3...0.)« G 0. 10 .5.19-, P ( 2 3 , NN) =BC(5 ,N3 ) IF ( N B C ( 6 » N 3 ) . E Q • 0 ) GO TO 5 2 0 P(24»NN) =BC(6,N3) IF ( N B C ( 7 , N 3 ) . E Q . O ) GO TO 5 2 1 P(25,NN) =BC(7,N3) I F (NBC ( 8 , N 3 ) . E Q . O ) 6 0 , T 0 . 522... ... . P(21»NN> 522 =0.0 P(23,NN) =0.0 I F (NBC(9»N3).EQ.O ) P(21*NN) =0.0 , .- ... . _ GO TO 5 2 3 P(23,NN) =0.0 LJ.NBC.J. 10_jN3J... EQ....0 ) .GO. TO .524 .. , ., . P(21,NN) =0.0 P(23,NN) =0.0 524 GO TO ( 2 7 4 , 2 7 5 ) , N B C 1 274 DO 2 1 9 1=1,30 XP( I , 1 ) =P ( 1,1) _ _219_ CONTINUE, _ __ _ _^ ' COUNT' =COUNT +1.0 NMM =N+1 DO 2 0 3 K=3,NMM,2 I F ( C O U N T . G T . l . ) GO TO 2 6 1 READ 1 0 3 , (NBC(I»K) , 1 = 1,10) • J L N ( K ) DO 2 0 4 1 = 1 , 10 . . .... . . . I F ( N B C ( I , K ) . E Q . O ) GO TO 2 0 4 READ 1 0 1 , B C ( I , K ) 204 CONTINUE IF ( N B C ( 8 , K ) . E Q . l . O R . N B C ( 9 , K ) . E Q . l . O R . N B C ( 1 0 , K ) . E Q . l ) GO TO 2 3 9 GO TO 2 4 8 239 READ 1 0 4 , C I £K ) ,CL ( K ) , CE ( K ) . . ..«___ .. . . ... .. """2 4"8 " DO 2~49 1 = 1,10 I F ( N B C ( I , K ) . E Q . l ) GO TO 2 4 6 249 CONTINUE IF ( J L N ( K ) . E Q . 2 ) GO TO 2 4 6 P1(K) =0.0 523 1 - 1 . -P. — -•• = —" • .... . ..... • - •- • • • ••- • ^ --•— -•- * -j ro 246 260 262 263 264 265 E21 ( K ) = 0.0 - H21 ( K.) ..... =0.0 . _ E2 2 ( K ) = 0.0 H2 2 ( K ) = 0.0 HI ( K ) = 0.0 E3 1( K ) = 0.0 H31 (K ) = 0.0 E32 ( K ) = 0.0 ~H32 (K ) " = 0 . 0 " " " '* " XMOA(K) = 0.0 XMOB(K) = 0.0 XMOC(K) = 0.0 H2 ( K ) = 0.0 P2 ( K) = 0.0 . . . . . . \_ . . .... ._ XMOA2 ( K~) ~ = 0.0 X MOB 2 ( K ) = 0.0 XM0C2(K) = 0.0 GO TO 260 READ 1 0 2 , P 1 ( K ) , E 2 1 ( K ) , H 2 1 ( K ) , P 2 ( K ) , E 2 2 ( K ) , H 2 2 ( K ) READ 102» Hl.U.) ,E31 ( K ) ,H31 (5.) ,H2(.K) ,.E3 2( K.)-»H3_JK) ... R E AD 1 0 2 ,XMOA(K) ,XMOB(K) ,XMOC(K) ,XMOA2(K) ,XMOB2(K) , X M O C 2 ( K ) I F ( K . E Q . N) GO TO 261 READ 1 0 2 , W 1 ( K ) , E 1 1 ( K ) , H 1 1 ( K ) , W 2 ( K ) , E 1 2 ( K ) , H 1 ? ( K ) READ 1 0 3 , ( N B C ( I , K + l ) , 1 = 1 , 1 0 ) , J L N ( K + l ) DO 2 6 2 1= 1,10 Jf ( N B C ( I , K + l ) . EQ.O) GO TO 2 6 2 .... ... _ . . - ... ,, 'READ 1 0 1 , BC( I •K + l ) CONTINUE I F ( N B C ( 8 , K + l ) .EQ. 1 .OR. NBC ( 9, K + l ) . FQ . 1 .OR .NRC ( 1 n . K + 1 ) , EO . 1 ) r,n GO TO 2 6 4 READ 1 0 4 , Cl ( K + l ) , C L ( K + l ) , C E ( K + l ) DO 2 6 5 1=,1,10 . . " I F ~(NBC< I', K + l ) . E Q . l ) GO TO 2 6 6 CONTINUE I F ( J L N ( K + l ) . E Q . 2 ) GO TO 2 6 6 ' PI(K+l) = 0.0 E21(K+l) = 0.0 H 2 K K + 1) =0.0 1 1 i ! i 1 i i i —1 E22(K+1) H2 2 LK + IX-HI ( K + l ) E 3 K K + 1) H3KK+1) E32(K+1) H32(K+1) RQi\(J<,+ D XMOB(K+l) XMOC(K+l) x H2 ( K + l ) 2 67 261 EO_._Q_ P2(K+1) =0.0 XMOA2(K+l) =0.0 . J1QB 2 (»K t 1 > _ = 0.. 0_ _ XMOC2(K+l) =0.0 I F ( K . E Q . N ) GO TO 2 6 1 GO TO 2 6 7 READ 102 »P1(K+l) »E21(K + l ) , H 2 1 ( K + 1 ) , P 2 ( K + l ) , E 2 2 ( K + l ) » H 2 2 ( K + l ) READ 102 «H1(K + l ) • E 3 1 ( K + l ) » H 3 1 ( K + l ) » H 2 ( K + l)»E32(K + l ) »H32(K+l) READ .102 •XMOA (K + l j »XMOB (.<+l) ,XMOC ( K + l ) , XMOA7-( K + L) , XMO B 2- (-K+1 )., 1 XMOC2(K+l) READ 102 »Wl(K + l)» E 1 1 ( K + l)» H 1 1 ( K + l ) » W 2 ( K + l ) • E 1 2 ( K + l ) • H 1 2 ( K + l ) CONTINUE NE = ( N + l ) / 2 NN = ( K + l ) / 2 P_(.1.,,N.N.). . 0..0 . . ... „ _. ......... . .... ... „. ... ...... ... P(2,NN) =0.0 P ( 3 »NN) = + ( X M O A ( K - l ) + P 1 ( K - l ) * H 2 1 ( K - l ) + H 1 ( K - l ) * E 3 1 ( K - l ) )#DJ P(4»NN) =XMOB(K-1)*D2 P(5,NN) =-XMOC(K-l)*D1 P(6,NN) =1.0 P.'I.'.NN) „.=1.0 P(8»NN) =1.0 P(9»NN) =1.0 P(10»NN) =1.0 P( 11»NN) =XM0A(K)+P1(K)*H21(K)+H1(K)*E31(K) + (W1(K-1)*H11(K-1) + 1 W2(K-l)*H12(K-l))*D X 2 66 =0.0 _=_L_0. =0.6 =0.0 =0.0 =0.0 =0.0 _0.Q...__ =0.0 =0.0 S S P ( 12NN) =XMOB(K) -Ei 13 Ml~~.zMQQ.il,)~ ... P ( 14NN) =-Pl(K )-D*(Wl(K-l)+W2(K-l ) f =H 1 ( K ) P ( 15NN) =0.0 P t 16 NN) P ( 17NN) = 0 .0 P ( 18NN ) = (XMO A ( K ) + P 1 ( K ) * H 2 1 ( K ) + H 1 ( K ) * E 3 1 ( K ) ) * D J =.X MOB..(.K.1JLD2 £( I? M L . XMO C ( K ) * D 1 P ( 20 NN ) P.( 21 NN ) = 1.0 P C 2 2 NN) = 1.0 TO 2 6 8 K + l ) + P l ( K + l ) * H 2 1 ( K + l ) + H l ( K + l ) *E31(K+1)+ K ) * H l l ( K ) + W 2 ( K ) * m 7 ( K ) )*r, MOB ( K + l ) MOC ( K + l ) PI ( K + l.) -D*.( W.UK ).+.W21K,)J. 1 ( +K1 ) NN ) NN) HH) 269 NN ) 0.0 P ( 27NN ) = 0.0 P ( 28NN ) 0.0 P ( 29NN.),, "P ( 3 0 NM ) ••0.0 NBC1 = 2 GO T 0 5 0 0 CONT INUE DO 8 84 1 =1 ,30 DO 8 84 J = ' 3_0_ "» J'j" = o."o " AY( i A3 ( ,6 1 ) = 1.0 A3 ( »9 1 ) • = -.5^ D A3 ( »7 2 ) = 1.0 A3 ( » 2i o : =-.5*D A3 ( » 3i ) = DJ : : 269 275 52 c 11 10 e 1 8 84 IA 3 S Cl< IT zi I' I "I A3(3,8) I =1.0 +(-Pl(K-1)*E21(K-l)+Hl(K-l)*H31(K-l>)*DJ JX )_.. _, = -XfViaC(.K.-l L*-Q___ „____._____ A3(3,10) =-XMOB(K-l)*DJ A3(4,2) =D2 A3( > 8 )) =XMOC(K-1 )*D? A3 (4 4,9 = 1 .0 = (XMOA ( K - l ) + P 1 ( K - l ) * ( E 2 1 ( K - l > * B E T A ( K - l ) + H 2 1 ( K - l ) ) A3 ( 4 , 1 0 ) ..+ H1J K- D * (,E3 1 ( K,-1.)--H3JL UC-1) ^BEXA.(X~1_). I-). %D2 . ~ A 3"( 5 V Dl A3 ( 5,8) = X MOB ( -1 K )*D1 A3 ( 5,9 ) A3( 4, 1 0 ) * D 1 / D ? A3 ( 5 , 1 0 ) .0 A3 ( 6 , 1 1 ) .0 •-Q. AJtLI U . 2 J _ A3 (* 8 , 1 3 ) .0 A3 ( 9 , 1 4 ) .0 A3 ( JL 0 f 1 5 L .._C_ A3 ( 11 1) =-1.0 A3 (11 8) =-D*(Wl ( K - l ) - E l l ( K - l )+W2 ( K - l ) * E 1 2 ( K - U ) _A3 ( L l .9), =+XMC(K A3 ( 11 10) =+XMB(K - 1 ) A3 (• 11 16) =1.0 A3 ( 11 2 3 ) =-Pl(K)*E21(K)+H](K)*H31(K A3 ( 11 24 ) = -XMC (K - l ) - X M O C ( K ) A3 ( 11 25 ) K-l)-XMOB(K) = -XMB ( A3.( 12 2 )„. • l . o . A 3 ( 12 5 ) =D A3 ( 12 8 ) XMC ( K - l ) A3 ( 12 10 )•• MA ( K- 1JL+D* W1 ( K -1 i±LtilllX -1 U ^ E I J J _ K ^ 1 - ) ^ +D*W2(K-l)*(H12(K-l)+E12(K-l)*BETA(K-l ) ) A3 ( 12 17) =1.0 12 2 3).. =XMC(K-l.). + X M O C ( K ) . . A3 ( 12 2 5 ) = X M A ( K - l ) + X M O A ( K ) + P 1 ( K ) * ( E 2 1 ( K ) * B E T A ( K ) + H 2 1 ( K ) ) + Hl(K)*(E31(K)-H3KK)*BETA(K)) A3 ( 133) =-1.0 A3 ( 13 4) =D A3 ( 13 =XMB(K-l) s 9 \t., i S "I011 s [IT .Z5 U, .:i> (J". Cy) C~, T \ ' a 77. C r H CM C\J co, X I Q < i— < UJ, H - ca sZ * CM —: CM' Q *: I rH' r H Q. I f\J ^ — CY; O o m' • • ~> r H rH I Q I 1 ; : II II , II II II II CO — vO -co m <f --00 'o—>I CM] i n "oo oCM CM CM CM <f co co. m co <r<J- -4- <j-jm in m in co co ro co co cn co co co co co co < < ' < < < < < <*'< < < < II II II II o ON, CM CM rvo; vo CO co < CO ro co < < o Q: * #.' CM II II CM m CM CM' r- r- r- co CO II II U X U o CM I I , o Q II II CM' .— _ CO X, X II II ' II i—! '—I CO CM CM CM, CM O CO CO 00 CO 00 CO O co ro ro CO CO < < CO < V o — o o SI r H r-l • X rH II X Q —• + II r H CM CM CM II 0 o CO CO <J" • i—I ON OON O o o o O CO CO CO CO' CO CO CO CO CO •» CM CM, O Ol CM CM CM CM CM < < < < < <c. -S ON co o co' 1 s: < i it x I t II ii! «—i ~ CO 1 rH 1 1 II — — vO co <± m'— — r- co <j" in* < < < < < in ; — — — O O' -) D_ CO + UJ Q v v: v I N_ CM I CM I Q' 5_ _> o — — • m — o N_ — O *, * O m o s: on • — o >j • co o CD • m o i n « rH < • x < rH u • I i rH > > rH I | rH I I II rH' | || | + I > rH III | I rH X —' __: I O CM) r H U J ' CO NZ •o ; i r-C — o i —< co ' rH — + I * __, N_~ X + CO UJ < < '-I w < ffl CO CO < < ^ n A3 (20 25 ) = 1.0 .A3 (2J 2,6 ) = 1..0 A3 (22 2 7 ) = 1.0 A3 ( 23 28 ) = 1.0 A3 ( 24 7 9 ) A3 ( 25 3 0 ) =1.0 I F ( N .EQ.NE) GO TO 2 7 0 • A 3 J 2.6 1,6X~=-L«.Q,, . . .. . . - _,,.„_„. 26 2 3 ) = - D * ( W l ( K ) * E 2 1 ( K ) + W 2 ( K ) * E 2 2 ( K ) ) A3 r 26 2 4 ) =XMC(K) A3 ( 26 2 5 ) =XMB(K) A3 (( A3 27 17) = -1-0 A3 ( 27 20 ) = D A3J 27 23 )_ _-X KC ( K ) . . . _ "A 3 (2 7 25 ) XMA ( KT+W 1 ( K ) * ( H I 1 ( K ) +E11 ( K ) -::-BETA ( K ) 1 + W2 ( K ) # ( H 1 2 ( K ) + E 1 2 ( K ) * B E T A ( K ) ) A3 ( 2818 ) = -1_0 A3 (28 19 ) = D A3 (28 23 ) = XMB ( K ) .A3 ( 28 2 4 ) JLA..3(2 7,25.).... . 'A3 ( 29 19) = -l .0 A3 (29 23 ) •-vc( K ) A3 ( 30 20_L ••-1 A3 ( 30 23 ) = - V B ( K ) - D * ( W l ( K ) + W 2 ( K ) ) GO TO 271 A_3 (26 = 1.0. "A 3'("27 17) = 1. 0 A3 ( 28 18) = 1.0 A3 (29 19) = 1.0 A3 ( 30 20 ) = 1.0 DO 8 8 5 1 = 1, 30 _DQ . 8.8 5 ,3. _J = 1 P_ . Al ( I , J .0 ) =6 A l ( 26, =1 I ) .0 A l ( 26, = - P l ( K - l ) * F ? 1 ( K - 1 1+H1 )-:!-H3 1 f K - l ) 8) Al(26,9) =-XMC(K-2)-XMOC(K-l) Al(26,10) =-XMB(K-2)-XMOB(K-l) r 270 271 ~88 5~ CO : ^ 1 I c ^ £ Al(27,2) =1.0 _Al, ?7,j3.) =XMC(K-2J+XMOC(.K.-l ) .._ . . . . . ... ... ...._._ ... .. ...... ... _ .... A l ( 2 7 , 1 0 > =XMA(K-2 J + X M O A ( K - l ) + P 1 ( K - l ) * ( E 2 1 ( K - l ) * B E T A ( K - l ) 1 +H2KK-1 ) ) + H l ( K - l ) * ( E 3 1 ( K - l )-H31(K-l )*BETA(K-1) ) Al(28,3) =1.0 Al(28,8) =-XMB(K-2)-XMOB(K-l) A1(28,9)=A1(27,10) -A.iL29.__.!. = 1.0 „ _._ .. .„„ __. ______ .... „_._... ( Al T29"* 8T~ = -VC"( K-2) -Hl*( K-l") _. * m 2 *~ ------ -» - - ... ' L . .... ... ... . ._. 8 1 ., .... . 5 _.__.„___ !__ !! \Vi« . . _._„ , A l ( 3 0 , 5 ) =1.0 Al(30,8) =VB ( K-2 ) - P I ( K - l ) • NBC2 =2 6 0 TO 9 0 0 c o 2-L tlDNuE _ _ _ _ _ _ _ """WRITE ( _ " ) * A l " " ' " " ........... . _.__._„.. .... , DO 2 7 6 J = l ,30 AX(1,J) =-A(21,J)-.5*D*A(24,J) ! AX(2»J) =-A(22,J)-.5*D*A(25,J) AX(3,J) =DJ*A(16,J)-A(23,J) _ A X ( 4 , J ) =D2_AJ 1 7 , J )-A( 2 4 , J ) _ _ . ..__._. ,._ 276" A X ( 5 , J ) =-D] * A ( 1 8 , J ) - A ( 2 5 , J ) DO 8 8 6 1=1,5 DO 8 8 6 J = l , 3 0 ' 886 A X X ( I . J ) =0.0 , DO 2 7 7 1=1,5 _ _ A X X ( 1,1) =-AX ( 1 , 2 6 ) _ _____ _ . . AXX"( l',2 ) =-AX ( I , 27 ) ' ~~ " " " '"" AXX(I,3)=-AX(I,28) A X X ( I , 4 ) = - A X ( 1,29) AXX( I , 5 ) = - A X ( I , 3 0 ) A X X ( I , 8 ) = - A X ( I , 2 6 ) * A 1 ( 2 6 , 8 ) - A X ( I , 2 7 ) * A 1 ( 2 7 , 8 ) - A X ( I , 2 8 ) * A 1 ( 2 8 ,8 ) , 1 AX ( I , 29 ) *A1 ( 29_,8 )~AX ( J ,30 )*A1 ( 30 ,8 ) ..__. . . ... . . „._. „._,.., _.. .... .. AXX(I-»9) =-AX( I ,26 ) *A1 ( 2 6 , 9 )-AX ( I ,28 )-"-Al ( 2 8 , 9 ) j AXX(I,10)=-AX(I,26)*A1(26,10)-AX(I,27)*A1(27,10) A X X ( I , 1 1 ) = - A X ( I , 2 6 ) * A 1 ( 2 6 , 1 1 )-AX( I , 30 ) * A 1 ( 3 0 , 11 ) , AXX( I , 1 2 ) = - A X ( I , 2 6 ) * A 1 ( 2 6 , 1 2 ) - A X ( I , 2 9 ) * A 1 ( 2 9 , 1 2 ) ^> AXX(I,13)=-AX(I,26)*A1(26,13) AXX(I,14)=-AX(I,27)*A1(27,14) : 1 ^ | I 12 j\ 10 9 B •/ 277 AXX(I,15)=-AX(I»28)*A1(28,15) DO 8 9 3 J = l ,30 IF ( I .GE.6) GO TO 8 9 4 A ( I , J ) = A3 ( T ,.M + A X X ( T . 1 ' GO TO 8 9 3 894 A(I,J) = A3(I,J) 8 9 3..., . CONT INUE ._„ . ...... 'CALL. INV ( A , 3 0 , 3 0 ,DET,COND) WRITE ( 1 ) A 1 DO 2 7 8 1= 1 ,30 X P X ( I , N N ) = 0.0 „ P___2JL9___ = 1.,.5.. . DO 2 7 9 j = 1 , 3 0 279 X P X ( I , NN ) = X P X ( I , N N ) + A X ( I , J ) * X P ( J , N N - 1 ) DO 7 8 0 1= 1 ,30 280 X P ( I , N N ) = P ( I , N N ) - X P X ( I ,NN) 203 CONTINUE _GQ..TO. ( 1 0 01,10.02,1003., 100.4) , NRUN . , .. ., 1001 XDET1 = ADE T PRINT 1006,XDET1 1006 FORMAT ( I X , 3 4 H V A L U F OF D F T F R M T N A M T FTR<;T |Zu.^I-S-,-_-l-6-.-8-/-)GO TO 1 0 0 5 1002 XDET2 =ADET _._Pj?I,NT.. 1P.P.7., X.DE.T2,.. ... .... ..... 1007 F O R M A T ( I X , 3 4 H V A L U E OF DETERMINANT SECOND RUN I S , E 1 6 . 8 / ) GO TO 1 0 0 5 1003 XDET3 =ADET PRINT 1008,XDET3 1008 F O R M A T ( I X , 3 4 H V A L U E OF DETERMINANT THIRD RUN I S , E 1 6 . 8 / ) GO TO 10.05 ... .. . - ..... l"004 XDET4 = A D E T PRINT 1009,XDET4 1009 FORMAT ( I X , 3 4 H V A L U E OF DETERMINANT FOURTH RUN T ? . F1 6 . S / ) ••• GO TO 1 0 1 3 1005 CONTINUE 278 - Co_ P v '. 2 Ji 50 9 _ 3 7 5 . 5 BACKS PAC E 1 READ ,.(1.1 -.A„ DO 88 7 I = 1,30 Z ( I ,E) N = 0.0 DO 88 8 I = 1,30 DO 88 8 J = 1,30 888 Z ( I ,E N) = Z( I , N E ) + A ( I , J ) * X P ( J , N E ) . N N E = N. Erl DO 2 8 1 K K = 'l,NNE K= NN E-K K + l I F (K . LT .NNE) GO TO 2 8 7 BACKS P A C E 2 READ ( 2 ) A l i _ 0.. IP- 28 3 282 2 BACKS P ' A C E 2 BACKS P A C E READ ( 2 ) A l 183 DO 8 8 9 I =1,30 189 ZM(I-,K) =0.0 DO 8 90 1 = 1., 3 0 ... DO 890" J = l ,30 890 Z M ( I , K ) = Z M ( I , K ) + A K I , J ) * Z ( J , K + 1) DO 2 8 4 1=1,30 284 ZM(I,K) =XP(I,K)-ZM(I,K) BACKSPACE 1 BACKSPACE 1 " 'READ ( 1 ) A DO 8 9 1 1=1,30 891 Z ( I , K ) =0.0 DO 89 2 = 1,30 D O 89 2 = 1,30 9 =JJ.»K.>+A.( I ?J.)*ZM( J , K )...., J-J ? Kx ?, ' C O N T I NUE 81 0 1 0 , 1 0 1 1 , 1 0 1 2 , 1013 ),.N RUN G O T O ( 1 010 R E A D 116 ,( I T T ( L ) , L = 1 , 6 ) 116 FORMAT ( I X , 6 1 5 ) DO 3 0 3 K = l , N __ . ._. _ J JJI ( 1J ..jiQ..0_) GO. TO 3.0.4 1 F Z -OD- J|0__L_ 305 306 307 308 303 _?_f5 9 46 947 V 12 ll 10 9 8 7 5 5 948 949 3 10 — XMOA(K) =1.2*XM0A(K) .JT ( IJ.T.L2) .E,6.0) .GO TG 305 ._ XMOB(K) =1.2*XM0B(K) I F ( I T T ( 3 ) .EQ.O) GO TO 306 XMOC(K) =1.?*XM0C(K) IT ( I T T ( 4 ) . E Q . O ) GO TO 307 PKK) =1.2*P1(K) . IF.. ( ITT(.5 ),EQ.O ) GO ID. ,308.-.. ----- •- -- • -• • -• HKK) =1.2*H1(K) I F ( I T T ( 6 ) .EQ.O) GO TO 303 I F (K.EQ.N) GO TO 3 0 3 Wl(K) =1.2*W1(K) CONTINUE NMN =N-1 DO 3 1 0 K=2,NMN,2 NB = ( K + 2 ) / 2 XMA(K) =1_?#Z( 1 » MR) XMB(K) =1.2*Z(2»NB> XMC(K), =1.2*Z(3»NB) _v3 <.KJ ,.,. = 1,2 * Z (.4 , NB ) , VC(K) =1.2*Z(5»NB) BETA(K) =1.2#Z(8,NB) IF ( NBC ( 1 , K ) ..E-_.J1.._A^^ 1 . E Q . O . A N D . N B C ( 1 0 , K ) . E Q . O ) GO TO 9 4 5 H2(K) =1.2*Z(11,NB) I F ( MBC (.3 , K ) . FO. 0 • AND . NBC ( 4 , X ) » EQ. 0 ). GO T.C. S.46 P2(K) =1.2*Z(12,NB) I F ( N B C ( 5 * K ) . EQ . 0 . AND . NBC ( 8 , K ).EQ.O.AND.NBC(9 *K).EQ.O.AND. 1NBC ( 10,K ) .EQ.O ) GO TO 9 4 7 • XM0A2(K) =1,2*Z(13,NB) I F (NBC(6»K).EQ.O) GO TO 948 XM0B.2(KJ,. = J . 2 * Z U 4 , N B , ) , . I F ( N B C ( 7 , K ) . E Q . O ) GO TO 949 XM0C2(K) =1.2*Z(15,NB) CONTINUE CONTINUE DO 3 1 1 K = l , N , 2 — . . . . .. • - - - - - - - - | - • •-• . ...... . _ - • •. ...... - 1 • - - — 1 : i i - , . ..i ! i ! CO 1 ro • - — • - - - • - - - 1 r 1 J Z u J t V s NN = ( K + l ) / 2 • XMA(K_L.._ L =1..2*.Z( 16.,NJM). . _ • __. . ______ XMB(K) =1,2*Z(17,NN) XMC(K) =1.2*Z(18,NN) VB(K) = 1. 2 * Z ( 1 9 , N N ) VC(K) =1.2*Z(20,NN) BETA(K) =1.2*Z(23»NN. I F ( NBC ( 1 » K ) .EQ.O . AND. NBC.L2 ».K.) ..EQ. Q.AMD .NB.CJ 8.,JC) .EQ. 0„. 1 . E Q . O . A N D . N B C ( 1 0 , K ) . E Q . O ) GO TO 9 5 0 H2(K) =1.2*Z(26,NN) I F ( N B C ( 3 , K ) . E Q . 0 . A N D . N B C ( 4 , K ) . E Q . O ) GO TO 951 950 P2(K) 951 . =1.2*Z(27,NN) I F (NBC< 5 , K ) . E Q . O . A N D . N B C ( 8 , K ) . E Q . O . A N D . N B C ( 9 , K ) . E Q . O . AND. 1 N B C ( 1 0 , K ) . E Q . O ) GO TO 9 5 2 XM0A2(K) =1.2*Z ("28,NN) 952 I F ( N B C ( 6 , K ) . E Q . O ) GO TO 9 5 3 ,XM0B2(K) = 1 , 2 * Z ( 2 9 ,NN ) 953 . I F ( N B C ( 7 , K ) . E Q . O ) GO TO 9 5 4 XM0C2(K) =1.2*Z(30,NN) 954 . CONTINUE.. . . . _. _.. .. , " 3 1 1 ' CONTINUE GO TO 1 0 1 4 1011 CL3 =5./6. + X D E T 1 / ( 6 . * ( X D E T 1 - X D E T 2 ) ) READ 1 1 6 , ( I T T ( L ) ,L = 1,6) DO 1 0 1 5 K=1,N I J ( I TT ( I) . EQ,.0 ).._G0 TO. 1 0 1 6 " X M O A ( K ) =CL3*XM0A ( K )'" 1016 I F ( I T T ( 2 ) . E Q . O ) GO TO 1 0 1 7 XMOB(K) =CL3*XM0B(K) 1017 I F ( I T T ( 3 ) .EQ.O) GO TO 1 0 1 8 XMOC(K) =CL3*XM0C(K) I F ( I T T ( 4 ) . E Q . O ) GO TO 1 0 1 9 i._l. . PI ( K) ' =CL3*Pl*(k j ' "' ' " " 1019 I F ( I T T ( 5 ) . E Q . O ) GO TO 1 0 2 0 HKK) =CL3*H1 ( K ) 1020 I F ( I T T ( 6 ) .EQ.O) GO TO 1 0 1 5 I F ( K . E Q . N ) GO TO 1 0 1 5 W1(K) =CL3*H1(K) L . . - , __.. __ . . . 8 CO. w . . . . . . . . 1015 CONTINUE DO 1 0 2 1 K=2»NMN»2 NB =(K+2)/2 XMA ( K ) = C I 3 * 7 ( 1 •» N R ) XMB(K) XMC(K) =CL3*Z(2 » NB) =CL3*Z(3,NB) _yj3_«.l..)___. -^„=£L3*&A*JlQ.L--.-..-.^-'~-.,.„r VC(K) =CL3*Z(5»NB) BETA(K) =CL3*Z(8,NB) IF ( N B C ( 1 , K ) . E Q . O . A N D . N B C ( 2 , K ) . E Q . O . A N D . M R C ( R-!C ) . F ( j . n . A N n . N R r i Q | f , i 1 . E Q . O . A N D . N B C ( 1 0 > K ) . E Q . O ) GO TO 1 0 2 2 H2(K) = C L 3 * Z ( 1 1 »N B ) - _. . t ._LiL_2____,I.^ 1023 1024 , 1025 1 N B C ( 1 0 , K ) . E Q . O ) G O TO XM0A2(K) = C L 3 * Z ( 1 3 ,NB I F ( N B C ( 6 » K ) . E Q . O ) GO XJ_0B2_(K_), = C L 3 * 7 ( 14.,NB I F (NBC(7,K).EQ.O) XM0C2(K) 1021 ..... .... . . ._ P2(K) =CL3*Z("l2»NB) I F (NBC(5»K).EQ.O.AND.NBC(8*K).EQ.O.AND.NBC(9»K).EQ.O.AND. GO 1074 ) TO 1 0 2 5 ) 1027 1028 • JL .. ... .... .„...._,...., . . . =CL3*Z<15,NB) .- =CL3*Z( 16»NN.) ... ... - .... , . TO 1 0 2 1 CONTINUE DO 1 0 2 6 K=1,N,2 NN =(K+l)/2 XMB(K) XMC(K) VB ( K ) VC(K) BETA(K) ..... - , - .. .... - _ - - . , . — .. . ... =CL3#Z(17,NN) =CL3*Z(18,NN) = C I 3 * Z ( 1 9 .NM ) =CL3*Z(20,.NN) = CL3*Z(23,NN) J J L -UN PC (,1.» .. E.QjfP • A(_.Q..JNJB CJ.2 ,..„ ) ..EQ , O...AND 1.EQ.O.AND.NBC(10»K).EQ.O) GO T O 1 0 2 7 H2(K) = C L 3 * Z ( 2 6 ,N,N ) I F ( N B C ( 3 , K ) . E Q . O . A N D . N B C ( 4 , K ) . F Q . O ) GO P2(K) =CL3*Z(27,NN) .NJLC. 8.,XI ..£iQ . .0. .-AMD . NBCJ(S»K )• TO in?.. I F (NBC(5»K).EQ.O.AND.NBC(8»K).EQ.O.AND.NBC(9»K).EQ.O.AND. - .- CO_ if m. .. 1 NBC( 10 J K ) .EQ.O) GO TO 1 0 2 9 . XM0A2 ( KL) _._ = C L 3 * Z ( 28 ,NN ) . _ _ I F ( N B C ( 6 , K ) . E Q . O ) GO TO 1 0 3 0 """ToiV X M 0 B 2 ( K ) = C L 3 * Z ( 2 9 . N N ) 1030 I F (NBC(7»K).EQ.O) GO TO 1 0 2 6 XM0C2(K) =CL3*Z(30,NN ) 1026 CONTINUE GO TO 1 0 1 4 " " ' "" '" ~~ ""To iT NMN =N-1 "" DO 1 0 3 1 K=2,NMN,2 NB=(K+2)/2 XMA(K) =Z(1,NB) XMB(K) =Z(2,NB) M£ < -L ,=Z(3,NB) _ . ________ VB(K) =Z(4»NB) VC(K) =Z(5»NB) BETA(K) =Z(8tNB) I F (NBC(1» K ) . E Q . O . A N D . N B C ( 2 , K ) . E Q . 0 . A N D . N B C ( 8 , K ) . E Q . 0 . A N D . N B C ( 9 , K ) 1 . E Q . O . A N D . N B C ( 1 0 , K ) . E Q . O ) GO TO 1 0 3 3 H2 ( K) =Z ( 11 , NB) . , " 10 3 3~ I F ( NBC ( 3 ,k ) •EQ•0• AND. NBC "( 4 » K )' • EQ • 6 ) GO f "6" 10 34 ' P2(K) =Z(12,NB) IF (NBC(5,K).EQ.O.AND.NBC(8,K).EQ.O.AND.NBC(9,K).EQ.O.AND. 1034 1 N B C ( 1 0 , K ) . E Q . O ) GO TO 1 0 3 5 XM0A2(K) =Z(13,NB) . _____ _______ __ __, . _1.0_3.5_ I F ( MBC ( 6» K) .EQ.O) GO TO 1 0 3 6 _. " X M 0 B 2 ( K ) "=Z(14»NB) ~~"~ " "" " " ~ 1036 I F ( N B C ( 7 , K ) . E Q . O ) GO TO 1 0 3 1 XM0C2(K) =Z(15»NB) 1031 CONTINUE DO 1 0 3 7 K=1»N»2 NN = ( K + l ) / 2 XMA(K) =Z(16,NN) XMB(K) =Z(17,NN) XMC(K) =Z(18,NN) VB(K) =Z(19,NN) VC(K) =Z(20,NN) BETA(K ) =Z(23,NN) n. . . „. K o*. HP- CO I F ( N B C ( 1 , IO .EQ.O.AND.NBC(2 , K ) . E Q . O . A N D . N B C ( 8 , K ) . E Q . O . A N D . M B C ( 9 ,K 1 . EQ.Q. AM p.,NBC U 0 >K) .EQ. 0 ) GO. T.O 1C3 8. . ..... . . , . . H 2.( < ) =Z(26,NN) 1038 I F ( N B C ( 3 , K ) . E Q . O . A N D . N B C ( 4 , K ) . E Q . O ) GO TO 1 0 3 9 P 2(K) =Z(27,NN) 1039 I F (NBC(5»K).EQ.0.AND.NBC(8»K).EQ.0.AND.NBC(9,K).EQ.0.AND. 1 N B C ( 1 0 , K ) . E Q . O ) GO TO 1 0 4 0 _____ „ __.MQA2.( _)__ = Z $ 2.8 ,NN.) .. ............ .... ~ 1040 I F ( N"BC ( 6 , K ) .EQ.O ) GO TO 1 0 4 1 XM0B2(K) =Z(29,NN) 1041 I F ( N B C ( 7 , K ) EQ.0) GO TO 1 0 3 7 = Z ( 3 0 > NN) XM0C2(K) 1037 CONTINUE GO TO 1 0 1 4 . To 14°' NRUM =NRUN +1 GO TO 9 5 5 PRINT 1054 1013 1054 F O R M A T ( 1 X > 7 4 H * * * * * * T H E FOLLOWING ARE V A L U E S OF THE C R I T I C A L L A T E R A 1L B U C K L I N G LOAD*****-,//) .51 = XOET 1./( X D E T 4 - X D E T 1 ) "S2 = X D E T 2 / ( X D E T 4 - X D E T 2 ) S3 = X D E T 4 / ( 1 . 2 * X D E T 1 - X D E T 2 ) A D J =1. + . 2 * S 1 * 5 2 * S 3 READ 1 1 6 , ( I T T ( L ) , L = 1 , 6 ) DO 1 0 4 2 K= 1 ,N I F _( I T T (.1 ) .EQ.O.) GO TO 1 0 4 3 "XMOA ( K ) = A D J * X M O A ( K ) P R I N T 1 0 4 8 ,XMOA(K) ,K I F ( I T T ( 2 ) .EQ.O) GO TO 1 0 4 4 1043 XMOB(K) =ADJ*XMOB(K) PRINT 1049 ,XMOB(K),K J P _4._.IF J.I TT (.3.)•E.Q.O) .GO..TO 1.0.45 XMOC(K) =ADJ*XMOC(K) PRINT 1050 ,XMOC(K),K 1045 I F ( I T T ( 4 ) .EQ.O) GO TO 1 0 4 6 P1(K) =ADJ*P1(K PR INT 1 0 5 1 , P 1 ( K ) ,K 0 IE. _ii" no -CD- C 1046 I F ( I T T ( 5 ) .EQ.O) GO TO 1047 „ . _~ K) =ADJ*H1 ( K) P R I N T 10 5 2 »H 1 ( K ) » K •1047 I F ( I TT (6 ) .EQ.O ) GO TO 1042 Wl(K) =ADJ*H1(K) PRINT 1053,W1(K),K 1048 F 0 R M A T ( 1 X » 1 5 H C R I T I C A L XMOA = , F 1 6 . 8 , 1 0 H AT 1049 F O R M A T ( I X , 1 5 H C R I T I C A L -X.MQB..= » F.L6. 8., LO.hL-AT. ~10 5 0" FORMAT (" 1 X , 1 5HCR I T I CALXMOC = , F 1 6 . 8 , 1 0 H AT 1051 FORMAT(IX,15HCRITICAL P = ,F16.8,10H AT 1052 FORMAT(IX,15HCRITICAL H = , F16.8,10H AT 1053 FORMAT(1X,15HCRITICAL W = ,F16.'8,10H AT 1042 CONTINUE GO TO_ 298 "9 5 5 CONTINUE COUNT =COUNT+1.0 REWIND 1 H1 ( R E W I N D GO 2-9 8. TO i i J O I N T , I 5) J O I N T ,.1.5 ) JOINT,15) JOINT,15) J O I N T , I 5) JOINT,!5) , 2 299 C O N T I N U E E N D 1 ft -1 41 C 'i i ""c ft '13 S 45 3 C » 7 C 5. 5 ft 4 3 I N V S U B R O U T I N E INV (A ,N ,M,DET,COND) DOUBLE P R E C I S I O N A ( 2 ) , I P ( 1 0 0 ) DOUBLE P R E C I S I O N A I K,AMAX,CSUMA,CSUMB,EPS,T,DET,COND MN1 = M * ( N - l ) EPS = l . E - 2 0 DO 41 I = 1 ,N IP(I) = I I P ( K ) WILL KEEP TRACK OF WHICH ROW ENDS UP AS K*TH P I V O T DET = 1. F I R S T PART OF' COND ' ' " " " " KSW = 1 GO TO 260 CSUMA = CSUMB _ I NVE_RSI ON_STARTS _ _ _ Zl A " S T O P S I B F T C 01 TT ROW _ . 6 M5-7 7 M5-2 8 M5-2M5.-2 . ..9 . M5-2 10 M5-2 11 M5-2 12 M5-2 13 M5-2 14 M5-2. 15... CO -o • — . . . . . . . DO 199 ~KK" = c K K + = 1,N MK ^ " FIND MAXIMUM AMAX IMAX = DABS(A(KK)) = K IF(K.GE.N) KP_= K +l D O 6(T i"~= IK = I + GO ELEMENT TO ™" IN K*TH 65 — — 65 16 M5-2 17 M5-2 19 'z: M5-2 22 .-A - .. I c — . COLUMN KP7N" MK AIK =DABS(A(IK) ) IF(A IK.LE•AMAX) G O TO AMAX = AIK IMAX = 1 "CONTINUE T E S T FOR SINGULARITY IF(AMAX.LT.EPS) G O TO ~ " ~" ' ' " • 4 M5-2 ! C ""'™, 6 ] 1 60 4 .M5-2 M5-2 M5-2 .26. . 27 28 300 LAST = MN1+K I N T E R C H A N G E ROWS K A N D I M A X I F ( Kj. EQ. IMAX ) GOT O 1 0 0 DET = "-DET ' I M A X J = IMAX DO 7 5 K J = K» L A S T > M M5-2 T = A(KJ) A(KJ) = A(IMAXJ) . A U M A X . J ). ,= 75 \l Jl C 100 ~ C • iO 140 T IMAXJ = IMAXJ + M J = IP(K) IP(K) = IP( IMAX) I P ( IMAX) = J COMPUTE DETERMINANT DET =.J?ET*A .KKJ D I V I D E K*TH ROW BY A ( K » K ) T = l./A(KK) A(KK) = 1 . 0 DO 1 4 0 K J = K » L A S T » M A(KJ) = A(KJ)*T 445^2 37- M5-2 39 M5-2 . 41 -CDCO _j , c SUBTRACT A ( I , K ) T I M E S THE K*TH ROW FROM THE OTHER ROWS J20_J__9. ,I„_..b.K „. . __, . I F ( I.EQ.K) GO TO 1 9 9 IK = I + MK T ) A ( I=K )A U=K 0.0 RESTORE PROPER COLUMN ORDER IN THE I N V E R S E . DO 2 5 0 K = 1»N MK = M*K - M C COLUMN NOW OCCUPYING K*TH P O S I T I O N I S A C T U A L L Y 210 J = IP(K) C ...OF THE I N V E R S E . HENCE ... I F ( J . E Q . K ) GO TO 2 5 0 C RELOCATE COLUMN K TO I T S F I N A L P O S I T I O N J ~ M J ~ = M*J-M~" DO 2 2 5 I = 1,N I J = I + MJ IK = I + MK T = A UJ) A( I J ) = A U K ) 225" " "A ( I K ) ="l" ' ' " ~~ C ADJUST I P RECORD .IP(K) =I P ( J ) IP(J) = J C ' AND CHECK NEW K#TH COLUMN. GO TO 2 1 0 """"25 0 CONTINUE KSW = 2 ' C C C A L C U L A T E COND 260 CSUMB = 0.0 DO 2 7 0 1 = 1 ,N — — * 46 -47 - i i i IJ=I _jp_O__190 K J _K_,LAST,M A ( I J ) = A( I J ) - T # A ( K J ? I J = IJ+M CONTINUE 190 199 C ' C — _ M5-2 ! i I M5-? M5-2 M5-2 =^ 54 55 1 - J M _ L _ 2 . ^ - - ~ 3 COLUMN ... M5-2 M5-2 M5-2 6 i i 57 58 59 . .,...._..___„„.. -M5__.2.. .61 .. i i i 66 67 68 69 .. 7.0 M5-2 71 M5-2 72 M5-2 73 M5-2 74 M5-2 75 .._ . _M 5.-2. 76 i M5-2 M5-7 M5-2 M5-2 i i •i i CO • . . _ 1 S ! S c LAST ' /. = MN1+I _JP'0_ 2 7 0___I J_= I »L AS T , M 270 275 C C 300 310 T = A( IJ ) CSUMB = CSUMB + T*T GO TO ( 4 5 » 2 7 5 ) > KSW COND =DSQRT ( C S U M A * C S U M B ) / F L O A T ( N ) RETURN u 5 OJ, M5-2 79 -H-5^2- -8-0- I! _2 83 .84 "PROCEDURE' FOR "SINGULAR" OR ME AR'L Y "S I N G U L A R~ M A T RI*x7 — — " - " — • ""- " { J ^ 85 P R I N T 310» K» AMAX M5-2 86 FORMAT ( 6 H 0 S T E P I 3 , 9 H P I V O T = I P E1 5 . 8 , 7 5 H . INVFR.STON D T S C O N T T N H F D / 1) DET = 0. M5-2 88 COND = l . E 3 j _ "RETURN""*'"' M5-2 90 END M 5 _sO_ O
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Lateral stability of rectangular beams Bell, Leon Alexander 1966
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Title | Lateral stability of rectangular beams |
Creator |
Bell, Leon Alexander |
Publisher | University of British Columbia |
Date Issued | 1966 |
Description | This thesis presents a method for finding the lateral buckling load of a straight beam loaded perpendicular to its longitudinal axis with loads applied at any eccentricity. The beam can be restrained at any number of points along its length against deflections and rotations. The beam is divided into a number of segments and joints for which equations involving statics, elasticity, and continuity are written. The resulting group of nonlinear simultaneous equations is solved for several magnitudes of the loading pattern for which a buckling load is desired. A graph of the determinant of the structure matrix versus the load level yields the critical load. Included in the thesis is a complete listing of the computer program used in the solution technique. It was written for use on the IBM 7040 installation at the University of British Columbia. Included also are curves giving the lateral buckling stress of a beam simply supported at one end, and cantilevered over a flexible column at the other end, and whose top flange is laterally restrained. The beam carries a uniform load together with a concentrated load at the end of the cantilever. In addition, some data is included on the effect of tension flange bracing on the buckling load. |
Subject |
Girders -- Testing Buckling (Mechanics) |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2011-09-23 |
Provider | Vancouver : University of British Columbia Library |
Rights | For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https://open.library.ubc.ca/terms_of_use. |
DOI | 10.14288/1.0050606 |
URI | http://hdl.handle.net/2429/37594 |
Degree |
Master of Applied Science - MASc |
Program |
Civil Engineering |
Affiliation |
Applied Science, Faculty of Civil Engineering, Department of |
Degree Grantor | University of British Columbia |
Campus |
UBCV |
Scholarly Level | Graduate |
Aggregated Source Repository | DSpace |
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