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Stresses in pitch-cambered glulam beams Thut, Walter K. 1970

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STRESSES IN PITCH-CAMBERED GLULAM BEAMS  by WALTER K. THUT Diploma Swiss  Degree  ( C i v i l Engineering)  I n s t i t u t e o f Technology, 1 9 6 5  A THESIS SUBMITTED IN PARTIAL FULFILMENT OF " .' THE REQUIREMENTS FOR THE DEGREE OF MASTER OF APPLIED SCIENCE in  t h e Department of  C I V I L ENGINEERING  We a c c e p t t h i s t h e s i s to  the r e q u i r e d  The' U n i v e r s i t y  as c o n f o r m i n g standard  of British  Columbia.  In.presenting this thesis i n p a r t i a l f u l f i l m e n t of the requirements f o r an advanced degree at the U n i v e r s i t y o f B r i t i s h Columbia, I agree t h a t  the L i b r a r y s h a l l make  it  and study.  f r e e l y a v a i l a b l e f o r reference  I further  agree that p e r m i s s i o n f o r e x t e n s i v e copying o f t h i s t h e s i s f o r s c h o l a r l y purposes may be granted by the Head o f my Department o r by h i s r e p r e s e n t a t i v e s . that  I t i s understood  copying or p u b l i c a t i o n o f t h i s t h e s i s f o r f i n a n c i a l  g a i n s h a l l not be allowed without my w r i t t e n  W. K. Thut  Department o f C i v i l The. U n i v e r s i t y  of B r i t i s h  Vancouver 8 , Canada April 1970 -  Engineering Columbia  permission.  ABSTRACT  The  s t r e s s d i s t r i b u t i o n near the c e n t e r l i n e apex  o f pitch-cambered glulam beams i s analysed. i s done w i t h an o r t h o t r b p i c  t r a p e z o i d a l f i n i t e element u t i  l i z i n g the s t i f f n e s s approach. various  g r i d s i z e s against  The element i s t e s t e d w i t h  known s o l u t i o n s and shown to  give an o v e r a l l accuracy to w i t h i n :  Radial due  The a n a l y s i s  five  percent.  tension stresses perpendicular  t o moment, were c a l c u l a t e d f o r v a r i o u s  p l o t t e d as a d e s i g n a i d .  These s t r e s s e s  to grain  geometries and  can be s e v e r a l  times those found i n a uniform curved beam. The  e f f e c t o f changing the e l a s t i c moduli o f  the whole beam or i n d i v i d u a l l a m i n a t i o n s and  was i n v e s t i g a t e d  found t o be unimportant. R a d i a l s t r e s s due to shear f o r c e s was found t o  be  unimportant. Radial  s t r e s s f o r one t y p i c a l geometry due to  moisture change was i n v e s t i g a t e d i n d e t a i l .  This  showed  that t e n pounds p e r square i n c h t e n s i l e s t r e s s , was gener a t e d f o r a one p e r c e n t change i n moisture content.  •  Twenty two.tests were performed on t e n s i o n pendicular  t o g r a i n to i n d i c a t e that the a l l o w a b l e  would be .near t h i r t y  per-  stress  f i v e pounds p e r square i n c h .  A n u m e r i c a l example and d e s i g n recommendations are  included.  ii.  TABLE OF CONTENTS i  ABSTRACT TABLE OF CONTENTS  i i  LIST OF FIGURES AND TABLES  iv  LIST OF MAJOR SYMBOLS  vi  •  ACKNOWLEDGEMENT  x  1.  INTRODUCTION  1  2.  DERIVATION OF A TRAPEZOIDAL FINITE ELEMENT I N PLANE STRESS  5  2. 1.  GENERAL  5  DERIVATION OF THE FINITE ELEMENT  8  .2. 2. 2. 33.  4.  TRANSFORMATION  INTO POLAR COORDINATES  COMPUTER SOLUTION PROCEDURE  17 ; 21  3- 1.  BASIC STEPS IN THE PROCEDURE  21  3. 2.  SIMPLIFIED FLOW DIAGRAM  24  CHECKING OF THE FINITE ELEMENT WITH VARIOUS . PROBLEMS 4. 1. . 4. 2. 4. 3-  ISOTROPIC CIRCULAR BEAMS AND RINGS ORTHOTROPIC CIRCULAR BEAMS PITCH-CAMBERED ORTHOTROPIC BEAM  29 .  29^  . .'. 40 45  5 .  STRESSES IN PITCH-CAMBERED BEAMS  5  0  PREAMBLE  5  0  5 . 1 .  STRESSES FROM MOMENTS  5  3  5 . 2 .  STRESSES FROM SHEAR LOAD  6  0  5. 3-  STRESSES DUE TO CHANGE IN MOISTURE CONTENT . '  5 . 4 .  VARIATION IN ELASTIC PROPERTIES  . . 7 1  SUMMARY .  7  ;  '  62  3  6 . . EXPERIMENTAL TEST ON STRESS PERPENDICULAR TO GRAIN " 7 . 8 .  .  7  4  NUMERICAL EXAMPLE  7  8  CONCLUSIONS  8  3  8  5  LIST OF REFERENCES APPENDIX:  LISTING OF THE COMPUTER PROGRAM  . . 87  iv  LIST OF FIGURES AND TABLES Pig.  1.  Pitch-cambered glulam beam.  4  Pig.  2.  Arrangement o f nodal p o i n t s .  4  Fig:  3.  Trapezoidal gular  f i n i t e elements assumed w i t h  rectan5  orthotropy.  Fig.  4 a . Element parameters.  Pig.  4b. D i r e c t i o n s o f p o s i t i v e nodal displacements and  . 9  nodal f o r c e s .  9  Fig.  5..  S t i f f n e s s matrix [ k ] .  Pig.  6.  Parameters and c o o r d i n a t e s  Pig.  14 d e t e r m i n i n g the t r a -  pezoid.  18  The  18  two s e t s o f nodal f o r c e s and d i s p l a c e m e n t s .  Pig.  78.  Pig.  9 a . I s o t r o p i c c i r c u l a r beam, shear  Fig.  9b. S t r e s s e s  Pig.  9c. Tangential  22  Approximation o f the upper s t r a i g h t edge. loading.  31 32  s t r e s s e s a l o n g r = R + AR/2.  31  at cj) = 0/2.  I s o t r o p i c c i r c u l a r beam, moment  loading.  34  F i g . 10.. P i g . 11.  Thick  i s o t r o p i c c y l i n d e r under e x t e r n a l  F i g . 12.  Thick  i s o t r o p i c c y l i n d e r , uniform temperature  pressure.  36  gradient.  39  F i g . 13-  C i r c u l a r o r t h o t r o p i c beam under pure moment.  42  F i g . 14.  C i r c u l a r o r t h o t r o p i c beam under shear  44  F i g . 15.  Geometric dimension o f the t e s t e d beam.  P i g . 16.  F i n i t e element g r i d , l o a d i n g c o n d i t i o n moment."  F i g . 17-  Stresses  F i g . .18.  Arrangement f o r d/R = .2 and t a n a = .4.  52  F i g . 19.  Three p o s s i b l e cases o f pitch-cambered beams.  52  P i g . 20. F i g . 21.  Beams No. 2, 9, 24 and 27-  54  load.  46  at c e n t e r l i n e o f the t e s t e d beam.  C h a r a c t e r i s t i c s t r e s s d i s t r i b u t i o n f o r pure bending moment.  P i g . 22.  47 . 48  Stress  coefficient  54 Cp^ f o r maximum r a d i a l  at c e n t e r l i n e due to pure bending.  stress 57  V.  Fig.  23.  Stress  c o e f f i c i e n t s Cijjjyj and CQJVJ f o r t a n g e n t i a l  stresses • F i g . 24.  at c e n t e r l i n e due t o pure bending.  Antimetric  25.  Shear s t r e s s e s  .. • 6.1 at c e n t e r l i n e from pure shear  load.  6  Fig.  26.  Two cases o f a group o f s t r a i n e d elements.  Fig.  27.  Parameters f o r s t r e s s i n element n due t o  28.  3  6  3  6  6  6  7  l i n e o f element 9.  6  8  Assumed l i n e a r change o f moisture content.  7  0  Influence .initial  Fig.  29. 30.  l i n e s f o r r a d i a l stresses  due t o  r a d i a l s t r a i n of 1%.  Influence initial  Fig.  l 6  s t r a i n i n group o f elements m. Fig.  8  l o a d case f o r a pure shear l o a d  at c e n t e r l i n e . Fig.  5  l i n e s f o r r a d i a l stresses  due to  tangential, s t r a i n o f 1%.  Influence  l i n e s f o r r a d i a l s t r e s s at c e n t e r -  Fig.  3 1 .  Fig.  3 2 . Example o f Testmember.  7  5  Fig.  3 3 . Roof beam.  7  8  1  6  5  6  7  1  7  2  7  6  Table 1 .  Terms o f the s t i f f n e s s m a t r i x .  Table 2 .  Parameters and s t r e s s c o e f f i c i e n t s o f the c o n s i d e r e d beams.  Table 3-  Effect of variation i n e l a s t i c properties ' o f the whole beam.  Table 4.  E f f e c t of v a r i a t i o n i n e l a s t i c in  Table 5 .  properties  one lam.  Results  o f t e s t s on t e n s i o n  to g r a i n .  perpendicular  LIST OP MAJOR SYMBOLS  x , y  coordinates within the f i n i t e element.  u , v  displacement  trapezoidal  i n x and y d i r e c t i o n .  T {a}  = {a ,  ay, T}  stress  vector.  {e}  = (e ,  £y  strain  vector.  a  x  x  5  stress i n x d i r e c t i o n .  x  Gy  stress i n y d i r e c t i o n .  T e ,  shear s t r e s s . s i m i l a r for s t r a i n s .  £y, Y  x  {a}  , {e} o  i n i t i a l stress vector and i n i t i a l o  s t r a i n vector. °"Y » e , X o  NOTE:  i ey , o  i n i t i a l stresses  T  Y  O  initial  strains.  x and y d i r e c t i o n , i n the f i n i t e element, correspond to the tangential and r a d i a l d i r e c t i o n i n the beam.  {6}  displacement  v e c t o r c o n t a i n i n g the  displacements '{f}  element.  f o r c e v e c t o r c o n t a i n i n g the nodal o f the f i n i t e  {f}  o f the f i n i t e  forces  element.  f o r c e v e c t o r due  Q  nodal  to i n i t i a l  strain.  [S]  stress  [k]  s t i f f n e s s matrix  [X]  t r a n s f o r m a t i o n - m a t r i x r e l a t i n g the p o l a r c o o r d i n a t e s to the xy c o o r d i n a t e s .  [k] ,{f}',{f}^,{6} 1  matrix  1  [K],{F},{F} ,{A} Q  matrix.  and  of the f i n i t e  element.  v e c t o r s of the f i n i t e  element  r e f e r r i n g to p o l a r  coordinates.  s i m i l a r matrix  v e c t o r s f o r the  and  structure. {B}  structure load vector.  [D]  elasticity  Ei = E / ( 1 - v y V y ) x  X  matrix.  i n the t a n g e n t i a l d i r e c t i o n of the  X  structure. E E  2  v  = E /(l-Vy v y) v  = xy'E  X  v  x  =  i n the r a d i a l d i r e c t i o n o f the s t r u c t u r e  X  v  y x  -E  y  modulus o f e l a s t i c i t y  i n x direction  (tangential). modulus o f e l a s t i c i t y  i n y direction  (radial). shear modulus. Poisson's r a t i o for stress  (strain i n y direction  i n x direction).  s i m i l a r to v  x  y  .  bending moment. shear  force.  width o f the beam. depth o f the beam. r a d i u s o f c u r v a t u r e o f lower edge o f the beam. d i s t a n c e i n r a d i a l d i r e c t i o n between two adjacent nodes o f the f i n i t e element g r i d . angle between two adjacent r a d i i  descri-  b i n g the f i n i t e element g r i d . r e f e r e n c e angle o f the r a d i i . s l o p e angle o f upper s t r a i g t h edge o f pitch-cambered line.  beam w.r.t. a h o r i z o n t a l  a  .a  tangential  t  '.  r  stress.  radial stress,  x  shear - s t r e s s .  C"rM> C ^ M  stress c o e f f i c i e n t f o r stresses i n r a d i a l and i n t a n g e n t i a l d i r e c t i o n due t o moment  Cy r  s i m i l a r due t o shear  C^Y  3  loading.  C"rpj Ctp  loading.  s i m i l a r due t o u n i f o r m d i s t r i b u t e d pressure.  C rjij  s i m i l a r due t o temperature  C ^ T  r  A  gradient.  d i f f e r e n c e i n %. between f i n i t e  element  s o l u t i o n and known s o l u t i o n . o-n r  =  max  °"tens  =  C-n-M'r^r  wm bd  Crpjypj^p^  maximum r a d i a l s t r e s s at c e n t e r l i n e o f • . a pitch-cambered beam due t o pure bendin maximum t a n g e n t i a l t e n s i l e s t r e s s at c e n t e r l i n e o f a pitch-cambered beam due t o pure  °"comp  =  CgjyTkflZ  bending.  maximum t a n g e n t i a l  compression  stress  at c e n t e r l i n e o f a pitch-cambered beam due  C  RM' T M » CM C  C  t o pure  bending.  s t r e s s c o e f f i c i e n t s f o r above maximum stresses.  ACKNOWLEDGEMENTS  The  author wishes t o thank h i s  supervisor,  Dr. R.F. Hooley, f o r h i s i n v a l u a b l e  assist-  ance and encouragement d u r i n g the development o f t h i s work. pressed  G r a t i t u d e i s a l s o ex-  t o Le C o n s e i l des A r t s du Canada  f o r f i n a n c i a l support,  and t o the U.B.C.  Computing Centre f o r the use o f the facilities.  A p r i l 1970 Vancouver, B.C.  INTRODUCTION Pitch-cambered  glulam beams ( f i g . l ) are used as  r o o f beams i n many b u i l d i n g s .  However the s t r e s s e s w i t h i n  these beams have not been determined.in a r i g o r o u s manner. Since the abundance o f r e c o r d e d f a i l u r e s i n North have not been s a t i s f a c t o r i l y to i n v e s t i g a t e the problem The pitch-cambered  America  e x p l a i n e d , i t seems j u s t i f i e d  deeper. glulam beam i s manufactured  i n a way t h a t , a t any c r o s s s e c t i o n , the g r a i n i s p a r a l l e l to the lower edge.  The" i n d i v i d u a l l a m i n a t i o n s are curved  i n the middle p a r t o f the span, but are u s u a l l y  straight  towards the s u p p o r t s . F o r r e c t a n g u l a r cross s e c t i o n s , such a beam can be c o n s i d e r e d to be i n a s t a t e o f plane s t r e s s .  In  t h i s two d i m e n s i o n a l s t a t e , the axes o f the e l a s t i c  pro-  p e r t i e s , p a r a l l e l and p e r p e n d i c u l a r t o the g r a i n , c o i n c i d e with p o l a r coordinates.  This i s c a l l e d polar anisotropy.  In these pitch-cambered  beams, there e x i s t  s t r e s s e s p e r p e n d i c u l a r to the g r a i n .  tensile  These s t r e s s e s are  caused by the v a r i a t i o n i n depth o f the c r o s s s e c t i o n s and by the c u r v a t u r e o f the g r a v i t y axes. procedures  Present d e s i g n  estimate these s t r e s s e s from the formula  where M i s the a p p l i e d moment, b the width o f the beam,  d t h e d e p t h o f t h e beam a n d R o f t h e beam a x i s . of constant and  This  curvature,  so n e g l e c t s  formula constant  of the span. and  the radius of curvature i s forisotropic  material  d e p t h and c o n s t a n t  moment,  many p a r a m e t e r s o f t h e p r o b l e m .  Most f a i l u r e s i n the lower  a  show a c r a c k p a r a l l e l  h a l f o f the cross At these  to the grain  s e c t i o n , c l o s e t o t h e middle,  sections there  a r e l a r g e moments  s m a l l s h e a r f o r c e s , as p i t c h - c a m b e r e d beams a r e a l m o s t  invariable  determinate  and s i m p l y  supported.  This  leads  t o t h e c o n c l u s i o n , t h a t a moment a p p l i e d o n t h e s p e c i a l shaped middle p a r t - m i g h t  cause h i g h  b r i n g t h e beam t o f a i l u r e .  r a d i a l stresses which  A l s o , moisture  create r a d i a l s t r e s s e s a d d i t i v e t o those the  theory  of elasticity  curved-beams o f c o n s t a n t f o r a curved  g e n e r a t e d by  discussed  curved  solutions f o r  s e c t i o n and c u r v a t u r e .  (1).  A general  A solution  s o l u t i o n , g i v e n by C a r r i e r ( 2 ) ,  and a p p l i e d by F o s c h i  ( 3 ) j f o r the.case  D o u g l a s F i r beams o f c o n s t a n t  c o m b i n e d b e n d i n g moments, a x i a l cambered c u r v e d exact  gives  beam o f D o u g l a s F i r , u n d e r p u r e b e n d i n g , i s  shown by N o r r i s  S.P.  could  moment. The  is  change  cross  sections  and s h e a r f o r c e s .  beams w i t h p o l a r a n i s o t r o p y , t h e r e  solutions available.  of under  For pitcha r e no  However, t h e r e i s a p u b l i c a t i o n o f  Fox ( 4 ) i n p r e p a r a t i o n , which w i l l  treat this  problem.  A l s o , a s f a r as i t i s k n o w n , t h e i n f l u e n c e o f m o i s t u r e on  s t r e s s e s i n beams o f t h i s The  the  finite  type  has n o t y e t been t r e a t e d .  e l e m e n t m e t h o d e n a b l e s us t o d a y t o f i n d  s t r e s s a n d s t r a i n d i s t r i b u t i o n i n any e l a s t i c  In t h i s powerful case a plane  change  method t h e r e a l c o n t i n e o u s  continuum.  body, i n t h i s  s t r u c t u r e , i s r e p l a c e d by a number o f f i n i t e  s t r u c t u r a l elements. nected by .is now  These f i n i t e  elements are  a d i s c r e t e number of n o d a l p o i n t s .  replaced  by  intercon-  The  continuum  a type o f s t r u c t u r e which can be  by means o f normal s t r u c t u r a l a n a l y s e s ,  u s u a l l y the  treated stiffnes  method. The be  done by  of f i n i t e  a n a l y s i s i s completely r o u t i n e work and  digital  computers.  elements r e q u i r e d  The  relatively  to r e p r e s e n t  can  l a r g e number  realistically  the  continuum, r e s u l t s i n a l a r g e number o f l i n e a r e q u a t i o n s . To s o l v e t h e s e , the  computer i s an important  tool.  F o r t h i s s p e c i f i c problem an a p p r o p r i a t e element was  derived.  Then a computer program was  finite written  to handle c y l i n d r i c a l a n i s o t r o p i c s t r u c t u r e s with ease. A f t e r t e s t i n g the  finite  beams, under e x t e r n a l were s t u d i e d .  element, pitch-cambered glulam  loads  and  under moisture- change,  F i n a l l y recommendations are made to modify  the e x i s t i n g d e s i g n p r o c e d u r e s . v  In  perpendicular published  a d d i t i o n , some l a b o r a t o r y to the  t e s t s on  g r a i n were made and  data, to show t h a t . r e v i s i o n was  t h i s a r e a as w e l l .  tension  compared w i t h necessary i n  P i g . l . Pitch-cambered glulam beam.  P i g . 2 . Arrangement o f nodal p o i n t s .  2. DERIVATION OF A TRAPEZOIDAL FINITE ELEMENT IN PLANE STRESS  2.1.  GENERAL Our interest  i s focused on plane stress problems  i n pitch-cambered glulam beams. replaced by f i n i t e  The o r i g i n a l plate  is  elements i n an arrangement of nodal  points on polar coordinates,as  shown i n f i g . 2 .  Herein,  AR can vary with depth only and R and 3 can vary with '<j>. Trapezoidal elements are the most convenient for this arrangement of nodal points. The o r i g i n a l .structure has c y l i n d r i c a l orthotropy but this w i l l be approximated by rectangular orthotropy within the range of one element,as shown i n f i g . 3 . The smaller the angle 0 , the better w i l l be this approximation.  I \  Fig.3-  Trapezoidal f i n i t e element assumed with rectangular orthotropy.  I  The f i n i t e element  must take i n t o account  s t r a i n s due to change i n moisture content and I n i t i a l s t r a i n s occur i f the s i n g l e element expand.  I f the element  initial  temperature.  i s f r e e to  i s c o n s t r a i n e d , then i n i t i a l  stres-  ses occur under the a c t i o n o f a change i n moisture content or  temperature.  Again these s t r a i n s w i l l be  on a r e c t a n g u l a r system w i t h i n an element, a polar  orthotropic  r a t h e r than on  system. A p r a c t i c a l and convenient way  e l a s t i c continuum  to i d e a l i z e  i s by means o f displacement  as d e s c r i b e d i n s e v e r a l papers  (5*6).  the  functions,,  These displacement  ^functions s p e c i f y u n i q u e l y the s t a t e o f s t r a i n w i t h i n the element.  I f , the i n i t i a l s t r a i n s are i n c l u d e d , t o g e t h e r  w i t h the e l a s t i c p r o p e r t i e s , the s t r e s s e s are a l s o d e f i n e d throughout  the f i n i t e  element,  and, hence along the boundaries.  E x p r e s s i n g these s t r e s s e s i n terms o f the nodal displacements w i l l  lead to the s t r e s s matrix.  In matrix  notation this i s :  {a} = [ S ] . { 6 } + { a }  o  {a} =  >-  stress vector  initial  3x8  stress vector  s t r e s s matrix  61 62 63  {6}  - = displacement  =X  vector  65  6  8  The boundary s t r e s s e s are r e p l a c e d by a s e t o f generalized  forces, c o r r e s p o n d i n g t o the g e n e r a l i z e d  displacement.  nodal  These nodal f o r c e s can be expressed i n  terms o f the nodal d i s p l a c e m e n t s . ness matrix o f the f i n i t e  This y i e l d s  the s t i f f -  element or fi  {f} =  [k]-{6}+{f}  {f} =<  f  2  f  3  f* fs  * = nodal  forces  fe f  7  fs  [k]  {f}  8x8  °(8xD  = s t i f f n e s s matrix o f the f i n i t e  element  = n o d a l f o r c e s due t o initial  strain  2 . 2 . DERIVATION  OF THE FINITE ELEMENT  The v a r i a b l e s used i n the d e r i v a t i o n are shown in fig.4.  I t i n d i c a t e s , the element parameters,  coor-  d i n a t e axes, nodal c o o r d i n a t e s and the numbering of the nodal f o r c e s and  displacements.  F o l l o w i n g the standard procedure  two  displace-  ment f u n c t i o n s are assumed as  - u = ai  + a 2 x + a 3 y + a t, xy  v = as + a 6 X + a7y + aexy  This  assumption seems r e a s o n a b l e , s i n c e the  t r a p e z o i d s have f o r s m a l l angles 8 , a shape c l o s e to a rectangle. The nodal displacements constants a i , az,  {6} i n terms of the  ........... as are o b t a i n e d by  placing  the nodal c o o r d i n a t e s i n t o the displacement f u n c t i o n s . This gives:  {6} where  = [T] • {a}  {a} = T  { a i , a , a , a^, a 2  3  5 j  a , a , 6  7  a} 8  Pig.4b. D i r e c t i o n s of p o s i t i v e nodal displacements and nodal  forces.  and.  1 1 1 1 [ T ]  b  h  2  b  h  "2  a "2  a 2  bh 2  bh "2  0  0  0  0 1 1 1 1  b 2  b "2  a "2  a 2  h h  bh 2  bh "2  0  0  0  0  For l a t e r use, equation 2a has to be solved for {a} {a>  =  _  0  0  0  0  1 2h 1  where [ T ] \=  [T]  bh  1  {2b}  {6>  1 2  1 2  _1  1  a  a  1 2h  1 2h  1 2h  1  1  "bh  as  ah  1  "ah 1 2  1 2  0  0  0  0  1 2h  1 2h  1 2h  1 "2h  1  1  1  ah  ah  1  bh  "bh  _1  a  1  a  The t o t a l s t r a i n s a r e found by a p p r o p r i a t e d i f f e r e n t i a t i o n o f the d i s p l a c e m e n t f u n c t i o n s a s : 9u 3x  •  3y_  ^  = [C] {a}  y  3u 3y  ,  (3.}  8v 8x  where  [C]=  0  1  0  y  0  0  0  0  0  0  0  0  0  0  1  X  0  0  1  X  0  1  0  y  The i n i t i a l s t r a i n s , due t o change i n m o i s t u r e c o n t e n t o r t e m p e r a t u r e , are assumed independent o f x and y w i t h i n t h e element and a r e d e s c r i b e d by the i n i t i a l s t r a i n vector. 'Xo  Wo'"  •yo  Hookes law now g i v e s  {a}  =  <  y  {e}  -  {e>  (5>  Herein [D] i s c a l l e d the e l a s t i c i t y matrix and i s 1  0  v  [D]  0 0  where  E  x  Ey v  G  and  G  0  /  ( 1 -  v  x  y  /  (  1  V  X  y ' V y  xy*  E x  .  " =  v  - v  yx*E  y  x  )  X  )  y  i s the shear modulus  It should be noted that x and y correspond to tangential and r a d i a l d i r e c t i o n so that E and E x  are E ^  a n  g  e n t  ^ 2 and E a  r a c  y  ^ ] _ for the wood considered. a  The stresses and strains may be expressed now i n terms of nodal displacements by substituting equation 2b into equation 3> and equations 2b and 3 into equat i o n 5, to give [C]  iz}  {a}  =  CT] {6} _1  [D]([C][T]" {6}-{e} ] 1  o  Equation 7a can be written i n the form  where  {a}  =  [S]{<5}+{a}  [S]  =  [D][C][T] -  i s the stress matrix and  {0>  o  = -[D]{e}  o  i s the i n i t i a l stress vector.  I  13  Once t h e n o d a l d i s p l a c e m e n t s m i n e d by. s o l v i n g stresses  the s t r u c t u r e  and s t r a i n s  c a n be f o u n d  stiffness  equations, the  a t any p o i n t o f t h e f i n i t e e l e m e n t  by t h e above r e l a t i o n s .  [S] o b t a i n e d from  have b e e n d e t e r -  the chosen  The s t r e s s  displacement  matrix  functions i s  explicitly: y  J_ bh  a ah  b H  [S]=[D]  o  0  —+ — 2h bh  2h"ah  bh  u  •2h bh  2h bh  +  2h ah  The p r i n c i p l e used  i - i a ah  of virtual  0  1  +  '  y  bh  matrix  1  X +  displacements  t o d e r i v e the element s t i f f n e s s  1  X  2h"bh ~ 2 h a h  X  ~2h~ah  -A+JL ah ah  1-2a ah  i s now  [ k ] as shown  below d{6} {f}  =  T  where is  d{6}  / d [ e ] [ a ] dV T  is a virtual  displacement  o f t h e nodes and  t h e c o r r e s p o n d i n g change i n t h e s t r a i n s .  d{e}  Differentiation  o f . e q u a t i o n 6 y i e l d s a l i n k b e t w e e n d{6} and die} so t h a t d{6} {f} T  =y^([C][T]~ d{6}) [D]([C][T]- {6}-{e} )dV 1  T  1  0  v wherein this  io} i s e x p r e s s e d  7a.  Expansion  of  gives  d{S} {f} = d { 6 } [ T ] " T  T  S i n c e d{6}  {f}  as i n e q u a t i o n  =  l T  y[C] [D]CC]dV[T]" {6}-[T]" /[C] [D]{e} dV T  1  T  T  0  i s arbitrary  [T]" jT[C] [D][C]dV[T]" {6}-[T]" y[C] [D]{e} dV l T  T  1  i  T  o  {8a}  T h i s r e l a t i o n can be r e w r i t t e n as {8b}  {f} = [k]{5> + {f}, where the s t i f f n e s s m a t r i x i s [k]  =  [T]-  l T  / C] [D][C]dV[T]T  o  and the nodal f o r c e s due {f}  =  0  -[T]  to i n i t i a l  _ l T  (9>  1  L  strain  are  {10}  y[C] [D]{e} dV T  0  In a plane s t r e s s element w i t h a constant ness  t the volume dV i s t*dA or t-dx-dy.  The  thick-  stiffness  m a t r i x [ k ] , e s t a b l i s h e d by the p r e v i o u s matrix o p e r a t i o n , is.symmetric  due  to the r e c i p r o c a l theorem so that only  the- 36 terms o f the lower h a l f need t o be o f element geometric  stored.  symmetry, only twenty o f these 36  need c a l c u l a t i o n as shown i n f i g . 5 . A.  B  C  D K  D  [k]=t x  -F  G  H  -M  -E  M  - H  N  - G  0  Because  -L  -0 - N  Q R  U  F i g . 5 . S t i f f n e s s matrix [ k ] .  To reduce the lengthy e x p r e s s i o n s o f abbrevatlons  In [k] a set  f o r the i n t e g r a l s over the range o f the  t r a p e z o i d a l element i s i n t r o d u c e d as f o l l o w s : =jTl  dxdy  = ^/^ydxdy -*-xx = ^yx-dxdy  yy  = ^ y dxdy  The  integrals  J  due  h'  2  t o symmetry w i t h r e s p e c t t o the Y-axis The  twenty terms e n t e r i n g the s t i f f n e s s  are g i v e n e x p l i c i t e l y without  matrix  p r e s e n t i n g the i n t e r m e d i a t e  calculations i n table No.l. The  nodal f o r c e s due t o i n i t i a l  strain,  deter-  e q u a t i o n 10, are g i v e n as f o l l o w s : ht, a+2b; ( i x o v £ y o ) 6"b  *10  E  (  e  + E  f  20  ht, a+2b; ( i x o ~6¥  f  30  b] ( i x o  40  b]) ( E  f  < f  f  E  (  E  -  =  £  +  i £ x o  a+ b.' Yxyo G  e  E  v e  y  +E e v  a+ b ) G y  )  o  y o  )  t  (  x y o  a+ b > Y x G  yo  •<  50  a+ b]> ( E e + E e )  60  a+ b]  v  * **80  G  v yo)  + E  e  a+ b]' Y x y o  :  -ir  (  (  a+ b]'  x o  2  y o  +  h t . a+2b •ht,  ) G  Y yo X  a+2b > Y yo G  X  ( E  v xo e  a+ b.) ( E e v  x o  + E  e 2  yo  +E e 2  y o  )  - | | ( 2 a - b ' Yxyo +  G  ) +g!(2a+ b > Yxyo G  G  E2  Ei Ii  I  Ii  yy  h  Ii  Ii  Ixx  Ixx  ^y  A  l/b h  2  l/4h  2  l/b h  2  B  -l/b h  2  l/4h  2  -l/b h  2  2  -l/4h  2  1/abh  -l/4h  2  -1/abh  2  2  C  -1/abh  1/abh  D  1/abh  -1/abh  2  2  2  E  l/2bh  2  l/2bh  2  F  l/2bh  2  -l/2bh  2  G  -l/2bh  2  -1/2ah  l/2ah  2  H  -l/2bh  2  l/2ah  -l/2ah  2  I  1/a  K -1/a  2  2  2  2  -2/a h  l/a h  2  l/4h  2  l/a h  2  2/a h  -l/a h  2  l/4h  2  -l/a h  2  2  2  2  2  2  2  L  -l/2ah  l/2ah  2  -l/2bh  2  M  -l/2ah  l/2ah  2  l/2bh  2  N  l/2ah  -l/2ah  2  l/2ah  -l/2ah  2  0  l/2ah  -l/2ah  2  -l/2ah  1/2ah  2  P,  l/4h  2  l/b h  2  l/b h  2  Q:  l/4h  2  -l/b h  2  -l/b h  2  R;  -l/4h  2  1/abh  s:  -l/4h  2  -1/abh  T;  l/4h  2  l/a h  2  l/4h  2  -l/a h  2  2  2  2  2  2  2  2  2  1/a -1/a  2  2  -1/abh  1/abh  1/abh  -1/abh  -2/a h  l/a h  2  2/a h  -l/a h  2  2  2  2  2  2  2  2.3.  TRANSFORMATION INTO POLAR COORDINATES  B e f o r e t h e m e m b e r . s t i f f n e s s m a t r i x [ k ] can be added t o g e n e r a t e t h e s t r u c t u r e s t i f f n e s s m a t r i x [ K ] , t h e l o c a l c o o r d i n a t e s must be t r a n s f o r m e d t o a g l o b a l p o l a r system. The parameters d e t e r m i n i n g t h e t r a p e z o i d a r e r e l a t e d t o t h e p o l a r c o o r d i n a t e s by: h = (R  2  - R ) cos | x  a,= 2R  2  sin |  b = 2R  2  sin |  where R , R  2  and B a r e d e f i n e d i n f i g . 6 .  A f t e r c a l c u l a t i o n o f t h e s t i f f n e s s m a t r i x and the n o d a l f o r c e s due t o i n i t i a l s t r a i n i s c a r r i e d out i n r e c t a n g u l a r c o o r d i n a t e s , i t i s d e s i r a b l e t o convert the result i n t o polar coordinates. The new s e t o f a x e s , w i t h the p o s i t i v e d i r e c t i o n s o f t h e n o d a l f o r c e s and d i s p l a c e ments chosen r a d i a l l y and t a n g e n t i a l l y , w i l l be denoted w i t h a prime " ( f i g . 7) • The d i r e c t i o n c o s i n e s o f t h e p o l a r s e t r e f e r r e d t o t h e r e c t a n g u l a r s e t , w r i t t e n i n a square a r r a y , cons t i t u t e a transformation matrix [X] as:  {11}  Fig.6. Parameters  and  coordinates  determining  the  trapezoid.  3  [X]  =  0 3  0  cos|  0  0  0  0  0  0  B C O S ;  0  0  •sin|  0  0  0  0  -sinf-  0  0  . 3  0  sinf  cos|  sinf  0  -sin|  sinf  0  0  sinf  0  0  .0  0  -sinf  0  0  3  COSTJ-  0  C O S ;  0  0  0  0  3  0  COS75-  3 COSTJ  The components o f the v e c t o r s {f}' and {f}' are r e l a t e d t o o  the components r e f e r r e d to the r e c t a n g u l a r axes by: {f}'  = [X] {f}  {f}'  = [X] {f}  O  { 1 2 } o  and s i m i l a r l y f o r the components o f the nodal {«}'  displacements  = [X] {6}  The member m a t r i x i n p o l a r c o o r d i n a t e s i s then [k]'  = [X] [k] [X]-  { 1 3 }  {f}'  - {f}' = [ k ] \ { 6 } \  {14}  so t h a t  The matrix  [ k ] ' i s e a s i l y o b t a i n e d from [,k] by  A by A', B by B  f  e t c . i n f i g . 5 where  A'  A  -2E  P  B'  B  2P  -Q  C  C  G-L  -R  D'  D  -H+M  S  E'  E  A-P  -E  F'  P  -B-Q  -F  G'  G  -C-R  L  H'  H  D-S  M  I'  I  2N  T  K'  K  -20  -U  cos  L  L  C+R  G  sin  M'  M  -D+S  H  N'  N  -I+T  -.N  0'  0  K+U  -0  P'  P  2E  A  Q'  Q  2P  -B  R'  R  G-L  -C  S  S  H-M  D  1  1  m  i  T  U'  U  T  replacing  -2N -20  I -K  cos  2 z  6 ^ B 2  2  |  sin  3. COMPUTER SOLUTION PROCEDURE  3.1. BASIC STEPS IN THE PROCEDURE  The computer program.developed;Is  based on the  displacement method u s i n g a s t r u c t u r e s t i f f n e s s m a t r i x . The program, w r i t t e n f o r the IBM appendix.  7 0 4 4 , i s g i v e n i n the  The setup i s shown i n the s i m p l i f i e d flow d i a -  gram o f ( 3 - 2 ) . The b a s i c steps i n the procedure a r e : a) The pitch-cambered beam i s s u b d i v i d e d by l i n e s and many c i r c u l a r a r c s .  radial  The area e n c l o s e d  by the r a d i a l l i n e s and the a r c s , i s approximated by e q u i l a t e r a l t r a p e z o i d s connected at the corners only.  These t r a p e z o i d a l elements r e t a i n the appro-  p r i a t e m a t e r i a l and geometric p r o p e r t i e s o f the structure. An i n c l i n e d , s t r a i g h t upper edge cannot be r e p r e sented e x a c t l y i n p o l a r c o o r d i n a t e s .  Therefore i t  must be approximated by v a r y i n g the boundary wise.  step-  The r e s p e c t i v e t h i c k n e s s g i v e n to these boun  dary elements i s i n p r o p o r t i o n to the p l a t e a r e a t h e y have to r e p r e s e n t .  The r e s p e c t i v e  thickness  o f an element i s taken as • A  £  .  _  resD  *\  structure A  element  structure  Fig". 8. Approximation The area A  s  t  r  u  c  t  u  r  o f the upper s t r a i g h t  e  i s hatched i n f i g . 8 .  edge. A lement e  i s the area e n c l o s e d by the arcs and the r a d i a l lines.  This stepwise approximation i s checked .  i n chapter f o u r and shown t o be s u f f i c i e n t f o r the l i m i t s o f accuracy r e q u i r e d h e r e i n . b) The elements  and the nodes are numbered.  c ) The boundary c o n d i t i o n s are s i m u l a t e d by l o c k i n g or r e l e a s i n g the a p p r o p r i a t e nodal displacements. The unlocked nodal d i s p l a c e m e n t s , i . e . the components o f the s t r u c t u r e displacement v e c t o r , are numbered. d) The element s t i f f n e s s matrix [k] i s generated and transformed i n t o p o l a r c o o r d i n a t e s [ k ] ' . e) The s t r u c t u r e s t i f f n e s s matrix [K] i s generated [ k ] ' by u s i n g the code number technique.  from  This tech-  nique a u t o m a t i c a l l y  eliminates  the rows and  columns  o f the s t r u c t u r e s t i f f n e s s m a t r i x , c o r r e s p o n d i n g to the r e s t r a i n e d j o i n t s . f)  The  structure load vector The  loads  {B}  i s generated.  a c t i n g on the beam are a p p l i e d to  model at the nodal p o i n t s o n l y .  Therefore-,  d i s t r i b u t e d e x t e r n a l s t r e s s e s have to be by  statically  equivalent  The  nodal f o r c e s due  by e q u a t i o n 10. to p l a c e due  replaced  to i n i t i a l  s t r a i n are  given  code number technique i s used {P} , Q  strain.  structure load vector {B}  = {F}  -  {B}  {F}  i s then  0  g) Cholesky's method i s used to s o l v e the  system o f  simultaneous l i n e a r equations i n the s t i f f n e s s {B}  element displacement v e c t o r number technique and  7b.  {A},  { 6 } ' i s found by  s i n c e there  the  the. code  transformed i n t o { 6 } .  s t r e s s e s w i t h i n the element are found by They are  relation  = [K]{A}  h) Knowing the s t r u c t u r e displacement v e c t o r  i ) The  the  {P}.  them i n t o a s t r u c t u r e f o r c e v e c t o r  to i n i t i a l The  The  the  nodal f o r c e s to y i e l d  terms o f the s t r u c t u r e f o r c e v e c t o r  the  c a l c u l a t e d at the  center  the d i r e c t i o n s of the XY  c o i n c i d e w i t h the p o l a r  coordinates.  of the  equation elements,  coordinates  3.2. SIMPLIFIED FLOW DIAGRAM  Start  Read and P r i n t the g e n e r a l geometric s t r u c t u r e data: Max. No. o f AR, Values o f AR, No. o f r a d i a l groups, No. o f elements i n each group, innermost r a d i u s R and angle $ o f each group.  D i v i d e the s t r u c t u r e i n t o the elements. Number the elements and the nodes. A s s i g n the nodal numbers to the elements [Done i n s u b r o u t i n e DIVIDE].  Generate f o r each element the a p p r o p r i a t e data. [Done i n s u b r o u t i n e DISTRI]  Read:  The t h i c k n e s s o f the normal elements No. o f boundary elements. Thickness o f the boundary elements.  V continued  :  _\k  •  Print the: No. o f elements. No. o f boundary elements. No. o f nodes. Geometric data o f a l l elements Nodal No.s o f the elements.  V Read and P r i n t : The data o f the r e s t r a i n e d  nodes.  Generate the n o d a l displacement numbers  w  Print  the nodal displacement numbers  Generate t h e : Code numbers. [ u s i n g s u b r o u t i n e CODE] Bandwidth.  V  P r i n t the: Code numbers Bandwidth.  continued  :  ±  Read and P r i n t t h e : Moduli o f e l a s t i c i t y o f the s t r u c t u r e , i . e . o f the normal elements. No. o f s p e c i a l elements. Moduli o f e l a s t i c i t y o f the s p e c i a l elements.  Generate  the element  s t i f f n e s s matrix [ k ] .  Transform i t i n t o [ k ] ' . [Done i n s u b r o u t i n e SMEM]  \  A  f  P l a c e the element s t i f f n e s s matrix [ k ] ' i n t o the s t r u c t u r e s t i f f n e s s m a t r i x [ K ] .  Is i t done f o r a l l elements?  YES Y Read the number o f loadcases  V Read the type o f l o a d i n g .  V continued  NO  Are t h e r e s t a t i c a l l y  e q u i v a l e n t nodal f o r c e s ?  YES JL Read the nodal f o r c e s . P .  JL Is  there i n i t i a l  strain?  NO  YES Read t h e : I n i t i a l s t r a i n s o f the s t r u c t u r e , No. o f s p e c i a l s t r a i n e d elements, I n i t i a l s t r a i n o f the s p e c i a l s t r a i n e d elements.  Generate the nodal f o r c e s { f } i n i t i a l strain. Transform  Q  due to  them i n t o p o l a r c o o r d i n a t e s {f}'  [Done i n s u b r o u t i n e EPSINI]  f  B u i l d up the l o a d , v e c t o r [ B ] .  continued  o  NO  V  Solve f o r the displacement v e c t o r [Done i n UBC-subroutine  {A}.  BAND]  f  Fill  the element displacement v e c t o r { 6 } '  Transform  { 6 } ' into'{6}.  C a l c u l a t e the element  stresses.  [Done w i t h s u b r o u t i n e SIGMA]  A 3L.  Print  stresses.  NO  Is i t done f o r a l l elements?  YES IL  Are a l l loadcases  considered?  YES JL More s t r u c t u r e s ?  NO  End,  YES  29.  4. CHECKING OF THE F I N I T E ELEMENT WITH VARIOUS  It the  finite  with  i s the o b j e c t o f t h i s  chapter  element p r e v i o u s l y developed  sufficient  pitch-cambered  accuracy, beams.  the s t r e s s  To do t h i s ,  various  problems.  w i t h known s o l u t i o n s the  l  4.1.  % =  solutions f o r  corresponding to  on a b s o l u t e  value)  -  values  | known v a l u e [ ^  ISOTROPIC CIRCULAR BEAMS AND  determined  material  element s o -  -^QQ  |known v a l u e |  The is  are c a l c u l a t e d  Icalculated  predict,  e r r o r s given i n comparison  f o l l o w i n g r e l a t i o n based  error  will  finite  a r e compared a g a i n s t e l a s t i c i t y  percentage  t o see i f  distribution i n  lutions  All  PROBLEMS  RINGS  m a t e r i a l i n the f i n i t e by t h e [ D ] - m a t r i x .  element  solution  To r e p r e s e n t  [D] h a s t o be t h e c o r r e s p o n d i n g  isotropic  elasticity  matrix, i . e .  CD]  =  E>  E  v  V  E  2  E  0 or  0 E  0  0  G E  1^  > E  v  VE 1-V  1  V  0  V  1  0  0  0  1-v  and  2  G  =  2(l+v)  EXAMPLE  1  ISOTROPIC  C I R C U L A R BEAM,  Pig.9a shear in  loading  Timoshenko  the  tangential  two  finite  stant for  f o r which (5)  stresses  73-  a t <j> =  produced  t h e same  element  gives  solutions  a r e shown:  One  result Even  finite  at this for this  the stress  under given of  Herein, for a  con-  a t <J> = TT/2, a n d t h e o t h e r to the  a r e shown t a n g e n t i a l T h e two  i s  a plot  edge.  load  beam  solution  the inside  3/2.  load.  predicts  accuracy.  Pig.9c  d i s t r i b u t i o n according  stresses  curved  near  In fig.9b  applied  of a  the e l a s t i c i t y  on page  element  the shear  LOADING  shows p r o p e r t i e s  d i s t r i b u t i o n o f shear  solution.  the  SHEAR  elasticity  and  element  section coarse  radial solutions  remote grid  d i s t r i b u t i o n with  from  s i z e , the sufficient  —  F i g . 9 a . Isotropic  >  c i r c u l a r beam, shear loading..  F i g . 9 c . Tangential stresses along  r = R+ — .  P i g . 9b. S t r e s s e s  at  <j> = B/2.  •EXAMPLE  2  ISOTROPIC CIRCULAR BEAMj Fig.10  MOMENT LOADING.  shows t h e p r o p e r t i e s  and s t r e s s e s  of a  c u r v e d beam u n d e r p u r e b e n d i n g f o r w h i c h t h e e l a s t i c i t y solution  i s given It  that  (5)  on page  6l.  i s t h e same beam as i n example 1 , e x c e p t  the load  distributed  i n Timoshenko  i s p u r e moment.  linearly  The n o d a l  forces  a t <j> = TT/2 t o r e p r e s e n t  were  the a p p l i e d  moment. Again, sufficient  the element p r e d i c t s  accuracy.  the stresses  with  EXAMPLE  3  THICK I S O T R O P I C C Y L I N D E R UNDER U N I F O R M E X T E R N A L •• The e l a s t i c i t y  PRESSURE  s o l u t i o n i s g i v e n i n Wang ( 6 )  on page 5 4 . Due t o r a d i a l symmetry the problem i s s o l v e d w i t h one r a d i a l g r o u p , ( f i g . 1 1 ) .  However, t o check the  program, a q u a r t e r o f the c y l i n d e r has been used t o o , w i t h the same element s i z e , and the same r e s u l t was obtained. The  depth o f the beam i s d i v i d e d i n t o three  elements and the group angle  B i s taken as 6 ° .  s t r e s s e s are c a l c u l a t e d f o r a r a t i o ^ = . 3 .  The  The r a d i a l  nodal f o r c e s , due t o e x t e r n a l p r e s s u r e , are found by the formula i  f  _ ~  The remarkable  p • (R + d) • B  2  accuracy  found here w i t h only three e l e -  ments i s due, o f course, t o the uniform s t r e s s s e t up w i t h i n the element.  However, p r e v i o u s examples have  t a i n e d a l l modes o f deformation accuracy.  and s t i l l  showed good  con-  3 6 .  EXAMPLE  4  THICK ISOTROPIC CYLINDER UNDER UNIFORM TEMPERATURE GRADIENT The e l a s t i c i t y s o l u t i o n i s developed a c c o r d i n g to Timoshenko ( 5 ) on page 407. By u s i n g the boundary  c o n d i t i o n s cr = 0 at r  r = a (=R) and at r = b (=R+d) and by i n t r o d u c i n g an assumed l i n e a r d i s t r i b u t i o n o f the temperature T = T r , Q  the f o l l o w i n g e l a s t i c i t y  s o l u t i o n f o r the s t r e s s e s i s  obtained aET, 3r"^  (r -a )  (r +a )( j ^ f p - ) -  (2r +a )  2  aET  2  In the f i n i t e initial  ]  (r -a )( ^ f z -) 2  2  3  3  3  3  element s o l u t i o n , the r e s p e c t i v e  s t r a i n i s g i v e n by the formula  r  ^ aT  E  aT • E  0 where Tg i s the temperature i n the c e n t e r o f the element =  T  N  =  t o t a l number o f elements  n  =  element number from i n s i d e .  as g i v e n by T  E  q  (R + | (n -  1/2)) radially  Again, due t o r a d i a l  symmetry, only one  r a d i a l group i s r e q u i r e d t o s o l v e the problem.  By  t a k i n g f i v e elements i n the group, as shown i n f i g . 1 2 , the e r r o r i n the maximum a  r  is  With  10 elements i n the group t h i s i s reduced to 1%.  Fig.12.  Thick  uniform  isotropic  temperature  cylinder,  gradient.  4.2.  ORTHOTROPIC  The  CIRCULAR  remaining examples c o n s i d e r  m a t e r i a l and the [D]-matrix  EXAMPLE  BEAMS  orthotropic  i s g i v e n w i t h the example  5  CIRCULAR ORTHOTROPIC BEAM UNDER PURE MOMENT Fig.  1 3 shows p r o p e r t i e s o f a curved  ortho-  t r o p i c beam o f Douglas F i r under pure moment f o r which the e l a s t i c i t y  s o l u t i o n i s given by F o s c h i  (3).  Foschi  used the f o l l o w i n g e l a s t i c p r o p e r t i e s o f Douglas F i r given by Hearmon ( 7 ) . 2 . 2 7 6 5• 1 '0  6  psi  6  psi  6  psi  E  y  0 . 1 5 3 7• 1 '0  v  xy  0 . 2 9 0  v  yx  0 . 0 2 0  G  0 . 1 2 7 6• 1 '0  From these E , E , x  2  E  v  and G of the CD]  t a i n e d as E  s  psi  • 1 0  6  psi  0.045  • 1 0  6  psi  0 . 1 2 8  • 1 0  s  psi  = 0 . 1 5 5  E 2  E  • 1 0  2 . 2 9  i  v  G  To demonstrate the convergence t r e n d by r e d u c i n g the element s i z e , the depth o f the beam i s d i v i d e d once i n t o t h r e e , once i n t o f i v e and once i n t o t e n elements, w h i l e 3 i s kept  constant.  The nodal f o r c e s at the ends  were d i s t r i b u t e d l i n e a r l y to r e p r e s e n t the a p p l i e d moments. The^ r e s u l t o b t a i n e d w i t h t e n elements comes very c l o s e to the exact s o l u t i o n .  Fig.13 shows the three f i n i t e  element  TT  approximations  and the exact s o l u t i o n at <J> = g".  a p a r t o f the i s o t r o p i c  Also  case i s p l o t t e d f o r comparison.  P i g . 1 3 . C i r c u l a r o r t h o t r o p i c beam under pure moment.  EXAMPLE  6  CIRCULAR ORTHOTROPIC BEAM UNDER SHEAR LOAD I n t h i s example shear loads, instead of moments, are applied on the same beam with the same [D]-matrix as i n example 5. The applied shear load i s represented by a set of-equal nodal forces at the ends of the beam. In f i g . l 4 the f i n i t e element approximation with ten elements over the depth i s compared to Foschi's exact s o l u t i o n .  4 5 .  4 . 3 . .PITCH-CAMBERED ORTHOTROPIC BEAM A favourable opportunity to check the element i s offered by Fox.  finite  In his PROGRESS REPORT N o . l  ( 4 ) , on DOUBLE-TAPERED PITCHED BEAMS, a t h e o r e t i c a l solution for the problem i s obtained by using Fourier series and a point-matching method to satisfy boundary conditions at several points.  the upper  The other boun-  dary conditions are s a t i s f i e d everywhere.  This theory  has been v e r i f i e d by testing experimentally, a glued laminated beam with the geometric properties given i n fig. 1 5 . This test beam under pure moment i s chosen as an example to compare the two t h e o r e t i c a l solutions. The stress d i s t r i b u t i o n given by S.P. Fox i s based on the material properties " Ex  "  v V y  x  6  psi  1 0 0 . 1 6 9• '  6  psi  6  psi  =  0 . 0 3 3  =  0 . 4 7 3  xy v  • 1 0  2 . 4 5  0 . 1 3 7 • .'1 0  G  The equivalent [D]-matrix i n the f i n i t e element solution has the terms =  2.46  =  0.169  E G  1 0  6  psi  *  1 0  6  psi  =  0.0799 *  1 0  6  psi  =  0.137  1 0  6  psi  '  The f i n i t e element g r i d , -used t o r e p r e s e n t the beam, i s shown i n f i g . l 6 . The boundary elements a r e approxi m a t e d as d e s c r i b e d i n c h a p t e r 3 . 1 . page 2 1 . The s t r e s s e s a t c e n t e r l i n e a r e o b t a i n e d by q u a d r a t i c i n t e r p o l a t i o n and are- p l o t t e d i n f i g . 1 7 * t o g e t h e r -with Pox's s o l u t i o n . The two s o l u t i o n s a r e c l o s e t o g e t h e r . The d i f f e r e n c e i n maximum t a n g e n t i a l s t r e s s i s 1.9%, i n maximum r a d i a l s t r e s s 4 . 6 $ and i n minimum t a n g e n t i a l s t r e s s 3 . 0 $ . 0  symmetric  about £  Laminations 1 / 2 " t h i c k Beam 4 " wide 9 = 16.-646° a = 1 8 . 4 3 0 ° (tana = . 3 3 3 ) F i g . 1 5 . Geometric d i m e n s i o n o f t h e t e s t e d beam.  47.  Pig.17.  Stresses  at c e n t e r l i n e o f the t e s t e d beam.  The  chosen l e n g t h  approximation should  9.-.R i n the f i n i t e  element  be s u f f i c i e n t by S t . Venant's p r i n -  c i p l e t o produce accurate  r e s u l t s at the c e n t e r l i n e .  To  check t h i s , the beam under the same l o a d i n g c o n d i t i o n has been extended t o a l e n g t h o f 1.5 8R.  The change i n maxi-  mum s t r e s s e s at c e n t e r l i n e was l e s s than 1/2%, w e l l i n the d e s i r e d  with-  accuracy.  In the extended case, where a p a r t o f the s t r a i g h t beam has to be used t o get enough l e n g t h , t h i s s t r a i g h t p a r t has  t o be s i m u l a t e d  by a curved beam with a l a r g e  radius  R = 10 R, s i n c e the computer program i s w r i t t e n f o r curved beams.  A change from R = 10 R to R = 20 R v a r i e s the maxi-  mum s t r e s s e s at c e n t e r l i n e not more than .07%. The  l a s t t h i n g to be checked i s the i n f l u e n c e  o f the stepwise approximation o f the upper boundary.  To  do t h i s , s t r e s s e s i n t h i s beam are c a l c u l a t e d w i t h upper boundary elements o f zero thickness.  thickness  and a l s o w i t h  full  The maximum t a n g e n t i a l s t r e s s changes by 8%  between those two extremes, w h i l e the maximum r a d i a l  stress  changes by 5%.  bet-  The approximation, u s i n g a t h i c k n e s s  ween the two extremes, can produce an e r r o r w.r.t. the o r i g i n a l beam which i s only a p a r t o f the t o t a l d i f f e r e n c e . Since  t h i s e r r o r has the same magnitude as i n the examples  1 to 6 f o r curved beams, where no stepwise approximation i s r e q u i r e d , the d e s i r e d accuracy i s obtained. accuracy i s w e l l w i t h i n  5%> q u i t e s u f f i c i e n t  The a c t u a l  f o r the p r a c -  t i c a l purpose h e r e i n . The obtained  good r e s u l t s i n the examples o f t h i s  by the f i n i t e  element s o l u t i o n , j u s t i f y  c a t i o n s made i n the next chapter.  chapter,  the a p p l i -  There the parameters d,  R and a i n the pitch-cambered beam are v a r i e d to cover the p r a c t i c a l range.  5 . STRESSES IN PITCH-CAMBERED BEAMS  PREAMBLE In the f o l l o w i n g are shown s t r e s s e s due to moments, shear l o a d s , change i n moisture content and change i n e l a s t i c p r o p e r t i e s , i n an attempt to e v a l u a t e the f a c t o r s which might create h i g h  stresses.  The beams have been d i v i d e d i n t o 1 0 o r more r a d i a l groups, w i t h about 1 5 elements at c e n t e r l i n e , to p r o v i d e good r e s u l t s .  The r e s u l t s are p r e s e n t e d i n terms o f the  parameters d/R and tana.  Fig.18  shows the parameters and  the g r i d o f an example which i s used s e v e r a l times i n this  chapter to demonstrate v a r i o u s  effects.  I f not mentioned s p e c i f i c a l l y , the f o l l o w i n g elasticity  constants are used E  E  y  =  G v  This  x  tan / 2 0  E  tan / 1 5  =  . 3 7 0  gives E  [D] =  y  E  . 0 1 8 5  xy  v  tan  i  E 0  v  E  v  E  a  0  0 0 G  1 = E/E  0  1  E /E  x  0  E /E  1  0  2  0  G  1 E  a  .0185  =  0  .0185  0  .0500  0  0  .6621  where E = E / ( l - v y v ) can have any value, as this constant factor does not effect the d i s t r i b u t i o n or magnitude of the stresses, except i n the case of moisture chang where E i s taken as 1810. k s i . -  1  x  X  y x  1  This i s reasonable for Douglas F i r . The effect bf r a t i o changes i n this [D]-matrix i s checked i n section 5.4. It should be noted that x and y correspond with the tangential and r a d i a l d i r e c t i o n of the beam, that i s with the grain and perpendicular to the grain d i r e c t i o n of the. wood.  52.  5.1.  STRESSES FROM MOMENTS The  peak at midspan o f the pitch-cambered beam  Induces a s t r e s s is the  concentration.  This  l o c a l by S t . Venant's p r i n c i p l e .  stress  concentration  At some d i s t a n c e  from  c e n t e r l i n e where the beam i s s t r a i g h t , uniform tapered  or curved (see f i g . 1 9 ) , the e x i s t i n g formulas are v a l i d . The  stress concentration  at c e n t e r l i n e i n beams  under pure bending moments, i s i n v e s t i g a t e d To  i n this  section.  cover the p r a c t i c a l range a group o f 26 beams, with d/R  varying  from ."01 t o .8 and tana from .1 to .6, has been  chosen f o r the f i n i t e element s o l u t i o n .  A sketch o f f o u r  beams i n the upper and lower parameter range i s g i v e n i n fig.20  to a i d i n v i s u a l i s i n g the parameters. A l l beams showed at c e n t e r l i n e the c h a r a c t e r i s t i c  s t r e s s d i s t r i b u t i o n sketched i n f i g . 2 1 w i t h zero r a d i a l and tangential  s t r e s s at the top o f the beam.  adjacent to the c e n t e r l i n e the s t r e s s e s different  from zero.  F o r cross  sections  at the top edge are  The s t r e s s c o n c e n t r a t i o n  produces  a l o c a l maximum i n the r a d i a l s t r e s s e s . The  maximum and minimum s t r e s s e s  at c e n t e r l i n e ,  as i n d i c a t e d i n f i g . 2 1 , are r e l a t e d t o the value M  Z . ,. centerline by  =  6 M bd " 2  the dimensionless c o e f f i c i e n t s Cjyyj, dpjyi and CQJ^ as  follows:  54.  Pig.21.  Characteristic  stress distribution  f o r pure b e n d i n g moment.  maximum r a d i a l s t r e s s maximum t a n g e n t i a l tensile stress  l a t i o n had s i n c e the o f the  a  to symmetry,, only one  •to.be s o l v e d by  a  7  6 M bd "  C  a quadratic  the values  2  interpo-  at c e n t e r l i n e ,  computer program produces s t r e s s e s at the  centers  elements.  and  c i e n t s Gjyyj, Crpjyj and  coefficients  (d/R).  The  coeffi-  r a d i a l stresses  stress  c o e f f i c i e n t s are  A s e t of curves i s obtained  c o n n e c t i n g the p o i n t s o f constant tana.  The  center-  the s t r e s s  of the r a d i a l  i s given i n f i g . 2 2 .  p l o t t e d against  and  at  para-  GQ^.  A graphical presentation  graph f o r any  beam the  t a n a , the number o f elements NE  l i n e , the number o f r a d i a l groups NG  °f  6 M bd ".  h a l f o f the beam had  f i n i t e elements and  to be used to get  meters d/R  RM  x  _ comp " CM  Table 2 shows f o r each considered  G  Sd^  = m  _ _ <*tens - ^TM  maximum t a n g e n t i a l compression s t r e s s Due  r  a  by  The c o e f f i c i e n t s  can r e a d i l y be taken out o f t h i s  parameters. values  f o r r a d i a l s t r e s s e s obtained  by  the  c l a s s i c a l formula f o r curved beams rt  °r  - i M _ 1 d 6 M 2 bd(R+d/2) ¥*(R+d/2) bcF  are" p l o t t e d by a dashed l i n e .  Prom the comparison i t can  be seen that the whole p r a c t i c a l range produces up to 4 time  TABLE 2 Parameters and s t r e s s  coefficients  .of the c o n s i d e r e d beams  No  tana  1  -0.0  2 3 4  0.1 0.1 0.1  5-  . 6 7 8  9 10 11 12  13 14 15 16 17 18 19  20 21 22 23  24 25 26 27  •  d/R  NG  .NE  C  TM  .CM  C  C  RM  9  10  1.067  -1.323  .0599  .01 .05 .1  12 15 16  22 16  1.151  .0243  19  1.127  -0.710 -0.751 -0.763  0.2 0.2 0.2 0.2 0.2  .01 .05 .1 .2 .4  11 10 10 13  23 15 16 15  -0.687' -0.7025 -0.7165 -0.7320 -0.7281  .0455  14-  1.371 1.324 1.2826 1.2743 1.3869  0.3 0.3 0.3 0.3  .1 .2 .4 .6  10 10 10 12  15 15 13 13  1.5538 1.4881 1.5241 1.6343  -0.7341 -0.7434 -0.7281 -O.7182  .0731 .0817 .1059 .1252  0.4 0.4 0.4 0.4  .01 .1 .2 .4 .6  16 10 10 16 11  20 11  2.042  -0.818  1.9174 1.7805 1.7244 1.8015  -0.7775 -0.7937 -0.7794 -O.7672  .0946 .1000 .1076 .1312 .1550  0.5 0.5 0.5 0.5  .2 .4. .6 .8  15 13 15 14  15 15  2.2418 2.0202 2.0202 2.1001  -0.8964 -0.8569 -0.8271 -0.7999  .1438 .1608 .1851 .2071  0.6 0.6 0.6 0.6 -0.6  .01 . 2 .4 .6 .8  16 15 13 13  20 15 13 13 12  3.016 2.6702 2.4215 2 . 3208 2.3561  -1.066 -0.9870 -0.9502 -0.9030 -0.8764  .1600 .1788 -1975 .2189. .2431  M  .5  14  14  14 15 16 13  14  1.124  .0253 .0302  .0463 .0487 .0601 .0825  .30i  Pig.22.  Stress c o e f f i c i e n t  C  R M  f o r maximum r a d i a l  at c e n t e r l i n e due to pure  bending.  stress  .4  .2  ;0I  .1  .05  F i g . 23-  Stress  .2  .3  c o e f f i c i e n t s .C at  centerline  T M  .4 and due  C to  C M  .5  .6  f o r tangential pure  bending.  .7 stresses  d  R  .8  higher stresses Even i f the the  than given by  depth at the p o i n t  formula, the  are s t i l l  stresses  formula.  of tangency i s used i n  o b t a i n e d by  finite  remarkably h i g h e r f o r s m a l l  d/R's.  The  elements  calculated stress coefficients for  tangential stresses are  this classical  are p l o t t e d i n f i g . 2 3 .  The  the points  connected by hand, to produce curves from which  stress c o e f f i c i e n t s for tangential stresses o b t a i n e d f o r any  parameter.  w i t h deepest cross  sections  ficant tangential stresses c e n t e r l i n e cross  sections.  For the  observed that  due  to pure bending, was  line.  considered  at c e n t e r l i n e , the f o r design do not  beams signi-  occur at  the s m a l l e s t  Even though dj.^ i s g r e a t e r  at the  than 1 ,  the  depth at c e n t e r keeps the t a n g e n t i a l s t r e s s sections.  the  computer output, i t  the t a n g e n t i a l bottom f i b r e  than i n adjacent cross  be  From an i n s p e c t i o n of  s t r e s s e s i n each element i n the was  can  the  stress, centerextra  smaller  5.2.  STRESSES FROM SHEAR LOAD  A pure shear l o a d at c e n t e r l i n e i s produced by the  loading  conditions  shown i n f i g . 2 4 .  Due t o a n t i m e t r y , there i s , at c e n t e r l i n e , no tangential (e ). r  stress  (a^) and a l s o there i s no r a d i a l  Introducing  a  = e  t  t  it  can be seen t h a t To  = 0 i n t o Hooke's law  ' V  a  a  r  t  = E e  r  strain  v  t  +  E  v r e  + E e 2  r  the r a d i a l stress•must be zero.  check the magnitude o f the c e n t e r l i n e  shear  s t r e s s e s , the beam w i t h tana = .4 and d/R = .2 o f f i g . l 8 is investigated. bution  Fig.25  at c e n t e r l i n e .  shows the shear s t r e s s  distri-  The maximum shear s t r e s s at center-  line is: x  max  =  ( | L ) 2 bd  Even though the maximum shear s t r e s s at c e n t e r l i n e i s 40$ more than normal, t h i s w i l l  not govern d e s i g n s i n c e the  depth here i s l a r g e r than at the supports and the maximum shear f o r c e i s about one q u a r t e r the support  shear.  P i g . 25.-' Shear* s t r e s s e s a t c e n t e r l i n e from pure s h e a r l o a d .  5.3.  STRESSES DUE TO CHANGE I N MOISTURE CONTENT Change i n moisture content causes the i n i t i a l  strains e  and t ,  i.e.  e  r o  0  = -I  {e}  0 I f the d i s t r i b u t i o n of strains t i b i l i t y equation  9r  r I F  x 2 z  t  ~ V  T  {e)  s a t i s f i e s the compa-  0  F " 36  30  2  r 3r36  then no stresses are produced, i f the structure Is supported i n a determinate way. With {e} = {e}' the equation i s 3e 30 2  3r* This dent with case ses,  {2£ t o r- *~ 3r  X  o  p  = 0  equation i s s a t i s f i e d i f and ^ - are indepenof r and 0, or i f t o I/ ro varies l i n e a r l y r and 0. The l a t t e r case i s u n l i k e l y . The f i r s t shows that a uniform moisture change causes no s t r e s but nonuniform moisture change w i l l induce stresses. £  e  r  e  =  2 e  o  0  a  n  d  Since i n a pitch-cambered beam i t i s not obvious what s t r a i n d i s t r i b u t i o n {e} cfauses maximum stresses, a o  *  closer look into a representative example i s indicated. The beam with tana = .4 and d/R = .2 shown i n fig.l8 i s investigated. As shown i n the f i g u r e , the elements 1 to 8 are spaced equally while the depth of the elements 9 to lk are .fixed by the approximation of the upper edge. A t o t a l of 14 pairs of load cases were run. For each pair one group of elements of constant radius were given  6 V  0>  a.  first and In  f i g . 2 6 two  then different  elements are shown.  cases o f a group^of  Group a has elements a l o n g the whole  beam, while group b reaches The  strained  the top edge before the  support.  s t r e s s e s of a l l these l o a d cases can be com-  bined to r e p r e s e n t other s t a t e s o f i n i t i a l position.  Similarly influence lines  l a t t e r was  done f o r r a d i a l  s t r a i n by  super-  can be o b t a i n e d .  The  stresses.  As the r a d i a l s t r e s s e s at c e n t e r l i n e are the l a r g e s t , i n f l u e n c e l i n e s are o f i n t e r e s t t h e r e . s t r e s s e s Ao"  at the c e n t e r o f an element n adjacent to  r  the c e n t e r l i n e , due  to s t r a i n change i n one  elements at l e v e l y, i s a f u n c t i o n of seven Aa  The  r  =  group of parameters.  f [ A y , y, d, R, a , e , E] o  The parameters are shown i n f i g . 2 7 . p r e s e n t a t i o n i s i n the  E  1  d  L  A dimensionless r e -  form  3  d  ' R  3  o'  a  which r e q u i r e s only f i v e parameters. set  at .2 and  J  Here ^ and tana are  .4 by the chosen geometry of the- example;  the modulus E i s f i x e d as g i v e n on page 5 0 ; tangential strain e  i s taken as  Q  .01.  the r a d i a l or  Since there e x i s t s  a l i n e a r system, the s t r e s s v a r i e s d i r e c t l y w i t h e  Q  results  ratio  to  f o r o t h e r s t r a i n magnitudes are i n d i r e c t  the one .  chosen. AO-  u o  r  =  The f[  1 1  d  s t r e s s e s can now  be w r i t t e n as  £ ] ' d J  s i n c e the o t h e r three parameters are now  so that  constant.  Because ^ Aa  r  i s small, i t  v a r i e s l i n e a r l y w i t h Ay  Ay  g  ^  d  can be  assumed that  so that  ;  and  °r  ••-./ .C5)^ 8  '  O  For an i n f l u e n c e l i n e at one tial  p o i n t , due  s t r a i n i n the element groups 1 to 14,  t h i s p o i n t are r e q u i r e d  to  the s t r e s s e s at  from a l l l o a d cases.  The  sum  the s t r e s s e s of a l l l o a d cases i s zero, s i n c e i t u n i f o r m s t r a i n over the whole beam. the values  g are  c a l c u l a t e d and  ding l e v e l  (y/d)  o f the  area, enclosed  by  From these  p l o t t e d at the  s t r a i n e d group.  Now  were c a l c u l a t e d f o r the A few  centers  by  l  other E The  correspon-  the  total  Influence  lines,  of a l l 14 elements at fig.29.  so that r e s u l t s f o r other s t r a i n s  direct ratio.  u s i n g the modulus E' for  stresses  be noted that these p l o t s are f o r r a d i a l  or t a n g e n t i a l s t r a i n e =.01, obtained  represents  o f them are shown i n f i g . 2 8 and  • I t should can be  of  an i n f l u e n c e l i n e , corresponding to u n i -  form s t r a i n over the whole beam, i s zero. centerline.  ini-  = 1810  are obtained  As w e l l they are c a l c u l a t e d  k s i o f page 51 so that r e s u l t s  by  direct ratio  again.  i n f l u e n c e l i n e s f o r r a d i a l s t r e s s snow that  the l a r g e s t p o s s i b l e s t r e s s , due s t r a i n , i s obtained  at the  to pure r a d i a l or t a n g e n t i a l  center of element 9, however,  w i t h a d i f f e r e n t s t r a i n d i s t r i b u t i o n f o r each.  The  two  i n f l u e n c e l i n e s f o r t h i s p o i n t are shown i n f i g u r e s 30a 30b  together  From the  w i t h the two  f i g u r e i t can be  produces a c o n s i d e r a b l e w h i l e the other  and  s t r a i n loads p r o v i d i n g maximum seen, that each s t r a i n  s t r e s s only i n one  i n f l u e n c e l i n e does not  load  influence  contribute.  line,  av.  Fig.28.  I n f l u e n c e l i n e s f o r r a d i a l s t r e s s e s due  to i n i t i a l r a d i a l s t r a i n - o f  1%.  top  2000  0  2000  N= 3  0  0  0  0  -1000  N =6 a  Pig.29. I n f l u e n c e  2  lines  2000  - 2000  N =9  at £ of element  for radial  0  0  2000  N= 12  N  s t r e s s e s , due t o i n i t i a l  tangential  s t r a i n , o f 1%.  g (  y  d"  •o  •a o o  O O  top  co  W C  (A  E o  bisture change  •to  pos. Area  X%<  !,82lps  bottom I  -80  1  I L  -40  J  1  40  I.  (a) STRESS <T DUE TO INITIAL RADIAL STRAIN e OF 1% ro X  -2000  0  2000  (b) STRESS Of  DUE TO INITIAL (c) TANGENTIAL STRAIN € OF 1%  Pig.30. I n f l u e n c e l i n e s  f 0  for radial  stress  STRESS o~ DUE TO CHANGE OF I %  at centerline  r  o f element  MOISTURE  9.  OO  . .  •  .  69.  These two influence lines for r a d i a l stress due to i n i t i a l s t r a i n can be combined to one for r a d i a l stress due to moisture change i f the r e l a t i o n between moisture 'content and shrinkage of the wood i s known. Wood shrinks most i n the d i r e c t i o n of the annual growth rings (tangentially to the t r e e ) , somewhat less across these rings ( r a d i a l l y ) , and very l i t t l e along the grain ( l o n g i t u d i n a l l y ) . The Wood Handbook ( 8 ) , f i g . 7 1 on page 3 1 9 , gives for Douglas F i r a f l a t grain (tangential) shrinkage of about 1 . 5 $ when the moisture changes from 10% to 15%. F . Kollmann i n TECHNOLOGIE DES HOL'ZES ( 9 ) gives the l o n g i t u d i n a l s t r a i n due to moisture change as approximately 1 / 2 3 of the f l a t grain shrinkage. Herewith, the following i n i t i a l s t r a i n s , corresponding to 1% moisture change are assumed to be reasonable e e  ro  to  = (1.5$)/5 = =  .3%  for., a 1% moisture change  ( r o ) / 3 = (.3%)/23 = . 0 1 3 $ for a 1% moisture change. e  2  With these r e l a t i o n s , the influence l i n e , due to moisture change, as shown i n f i g . 3 0 c , i s obtained by- superp o s i t i o n of the adjusted influence lines due to s t r a i n change from figures 3 0 a and 3 0 b . With the maximum area of 1 0 . 1 p s i / $ m.c. under the influence l i n e , a moisture change of 5%, d i s t r i b u t e d as shown i n f i g . 3 0 b , produces the following maximum stress a  r  =  5 * 1 0 . 1  = 5 0 . 5 psi  This w i l l be tension perpendicular to grain when the moisture i n the central region decreases.  However, i t has to be noted t h a t these i n f l u e n c e l i n e s r e p r e s e n t moisture in  or s t r a i n d i s t r i b u t i o n s ,  f i g . 2 6 , which means the moisture  as shown  contents w i t h i n the  groups o f elements are constant.  This o c c u r s , f o r example,  i f a different  content i s present i n  amount o f moisture  the l a m i n a t i o n s at the time o f g l u e i n g , and subsequent d r y i n g shrinkage produces The distribution  stresses.  influence lines  cannot be used f o r a moisture  as shown i n f i g . 3 1 j  s i n c e the moisture  there v a r i e s a l o n g one group o f elements. c o u l d be made by an analogous of s t r a i n d i s t r i b u t i o n s .  procedure  Influence l i n e s  to t r e a t  this  type  Instead the case was s o l v e d by  a p p l y i n g t o a l l elements the c o r r e s p o n d i n g i n i t i a l as shown i n f i g . 3 1 .  content  strain  The maximum r a d i a l s t r e s s o" at c e n t e r r  l i n e was then a  r  =  4 6 . 8 p s i tension  For the i n c r e a s e i n moisture strain e  r o  content the r a d i a l  initial  c o n t r i b u t e s the most, namely 4 5 - 5 p s i , w h i l e  e-to w i t h 1.3 p s i produces almost no r a d i a l s t r e s s i n t h i s case.  I 5%  F i g . 3 1 . Assumed l i n e a r change o f moisture  content.  5 . 4 . VARIATION IN ELASTIC  PROPERTIES  Up t o here, the [D]-matrix  g i v e n on page 5 0 , has  been used as the ' b a s i c ' m a t r i x f o r the i n v e s t i g a t i o n s . •Table 3 shows how  the s t r e s s e s due to pure moment  are a f f e c t e d by changes i n the moduli o f ' e l a s t i c i t y . TABLE 3 E f f e c t o f v a r i a t i o n i n e l a s t i c p r o p e r t i e s o f the whole beam. Aa i n •?I D-matrix  E /E basic 2  E /E basic  2  v  v  -  1  - 2 . 4  2  1  -0 . 9  1  1 / 2  + 1 0 . 1  1  1  E  2  varied  2  1  E  v  varied  1  varied  1  A doubling of E  2  or E  v  ••  amount o f change i s s t i l l  lam to lam.  4.9  -1 . 3  +  1.1  + 1 3 . 0  - 1 1 . 7  tolerable. vary from  T h i s i n f l u e n c e has been checked f o r bending  moment, by assuming a v a r i a t i o n i n one modulus o f e l a s t i c i t y i n one lam at a time.  Table 4 shows the changes i n  maximum r a d i a l s t r e s s , due to changes i n the e l a s t i c  +  However, the  the e l a s t i c p r o p e r t i e s may  radial  -  has a s m a l l e f f e c t , but the s t r e s s e s  are somewhat s e n s i t i v e t o a change o f G.  More l i k e l y  tang e n t i a l tens. comp.  1  basic  G  G/Gbasic  pro-  p e r t i e s o f the bottom l a y e r o f elements with h i g h t a n g e n t i a l  t e n s i l e stresses  (see f i g . 2 6 ) ;  high r a d i a l stresses;  o f the seventh l a y e r w i t h  o f the tenth  l a y e r with high  tan-  g e n t i a l compressive s t r e s s .  TABLE 4 Effect of v a r i a t i o n i n e l a s t i c properties Change o f modulus i n one lam to  Aa Var.  i n layer 1 - 8.82 0 . 0 0  E G  v  -»• 2 E  V  *1/2G  +  0.16  - 0.22  r  Var.  i n one lam.  in %  i n layer 1  Var.  i n layer 1 0  -  4.58.  - 2.14  +  0.72  + 0.84  +  0.61  +  1.03  .0.00 -  As b e f o r e , the e f f e c t o f these changes i s t o l e r a b l e .  1.86  SUMMARY T h i s chapter has shown, that the maximum r a d i a l s t r e s s e s at c e n t e r l i n e due t o shear and v a r i a t i o n  in'elastic  p r o p e r t i e s are n e g l i g i b l e , but that the s t r e s s e s due t o moment and moisture change are important. The p o i n t e d peak induces c o n c e n t r a t i o n s o f r a d i a l s t r e s s e s from moments at the c e n t e r l i n e , and these s t r e s s e s must be c a l c u l a t e d by the graphs o f f i g . 2 2 . . on the beam, one o r two depths  At o t h e r p o i n t s  away from t h i s s t r e s s  riser,  the r e g u l a r curved or t a p e r e d beam formulas may be used f o r stress  calculations. T h i s p o i n t e d peak a l s o generates s t r e s s e s due to  non uniform moisture change, whether the beam i s loaded o r not.  It i s difficult  to s e t the magnitude, as f i e l d  testing  i s necessary to e s t a b l i s h some r e a l i s t i c moisture g r a d i e n t s . In the meantime, however, i t might be a d v i s a b l e t o add 20 p s i r a d i a l t e n s i o n t o any c a l c u l a t e d s t r e s s from moments at the peak.  T h i s 20 p s i i s about h a l f the a b s o l u t e magni-  tude determined h e r e i n f o r one geometry.  6.  EXPERIMENTAL TESTS ON STRESS PERPENDICULAR TO GRAIN Average s t r e n g t h values  o f Douglas F i r are  listed  i n the p u b l i c a t i o n STRENGTH AND RELATED PROPERTIES OF WOODS GROWN IN CANADA by the F o r e s t Products Research Branch o f the Department o f F o r e s t r y , Canada  (10).  For a i r - d r y c o n d i t i o n the u l t i m a t e perpendicular value  t o g r a i n i s g i v e n there  tensile stress  as 440.  r e s u l t s from short time t e s t s .  psi.  This  To compare i t with  the a l l o w a b l e working s t r e s s under l o n g time l o a d i n g , i t may  be m u l t i p l i e d by the f a c t o r 9/l6.  For Douglas F i r , the  maximum t e n s i l e s t r e n g t h p e r p e n d i c u l a r time l o a d i n g , now becomes 9/l6 The allowable  x 440.  CSA-Code 086 (11)  t o g r a i n under l o n g = 248. p s i .  s p e c i f i e s , on t a b l e 8, the  working t e n s i l e s t r e s s p e r p e n d i c u l a r  glulam Douglas F i r 24 f s t r e n g t h grade i n dry 65 p s i , o r 1/3.8 The  c o n d i t i o n s as  times the l o n g term u l t i m a t e .  calculated r a d i a l stresses i n section 5 . 1 - ,  are about twice But w i t h  to grain f o r  the values  g i v e n by the curved  the u l t i m a t e s t r e n g t h o f 3 . 8 times the  s t r e s s , a f a c t o r o f two does not  beam  theory.  allowable  explain f a i l u r e , especially  when many f a i l u r e s occured under l i t t l e more than dead l o a d . Other i n f l u e n c e s must be s i g n i f i c a n t . allowable  stress perpendicular  i s too h i g h .  Therefore  the a l l o w a b l e  stress.  t o g r a i n , given i n the  code,  a s e r i e s o f t e s t s were done t o check  Samples from two d i f f e r e n t Vancouver area were t e s t e d . given i n f i g . 3 2 ,  I t could be that the  glulam p l a n t s i n the  The t e s t p i e c e s , o f the  shape  had a cross s e c t i o n o f 4 . 5 " x 4 . 5 " o r 20.2 i n  Pig.32. Example o f Testmember Two  s t e e l p l a t e s , with c o n c e n t r i c  glued  threaded h o l e s , were  on w i t h Epoxy f o r a p p l i c a t i o n of t h e Twenty two  The  =128.  average value  failure  psi ^  T h i s value  i s l e s s than 1/3  .,  forces.  members have been t e s t e d with the  s u l t shown i n t a b l e 5a„  tensile  is  o f the  advertised  ultimate  and  w i t h the  c l a s s i c a l formula, under pure bending o n l y , might  be already  i t must be r e c o g n i z e d  re-  i n the  critical  t h a t beams designed,  range.  Although the number o f t e s t s i s s m a l l , the  ten-  dency i s c l e a r and.the c o n c l u s i o n must be drawn, that ultimate  r a d i a l s t r e s s e s must be  Such i n v e s t i g a t i o n s should study should  not  the  further investigated.  use  c l e a r samples and  i n c l u d e s i z e e f f e c t s and  drying  cracks.  the  TABLE .5 •RESULTS OF TESTS ON TENSION PERPENDICULAR TO GRAIN Loading time about Test member  5 minutes;  U l t . load lb.  moisture content about 10%  Ult. stress psi  vv  V  1  2890  143  - 15  . 2 2 5  2  1685  83  + '45  2025  3 4  2415  119  2000  5  8 1  99  9 + 29  841  1740  86  + 42  1764  6  1940  9 6  + 32-  1024  7  2130  105  +  "23  8  2120  1 0 5  '+  23  9  2 3 0 0  114  + 14  1 9 6  1 0  2020  100  + 28  784  11  2310  114  + 14  1 9 6  12  2090  1 0 3  +  25  6 2 5  1 3  3300  •163  - 35  1 2 2 5  14  1955 4290  + 31 - 84  9 6 1  1 5  97 212  7056  1 6  2340  1 1 6  + 12  . 144  17 1 8  3100  1 5 3  - 25  625  2 3 7 5  + .11  121  + 18  324 10000  +  .529 529  1 9 2 0  2 2 3 5  117 110  4625  228  -100  21  2725  1 3 5  -  7  . 49  22  4350  215  -'87  7569  2813  + 356  36893  128  -353  0  Sum Average  =  Standard d e v i a t i o n Coefficient  of v a r i a t i o n  y  3  g  = ± 4l. psi  TTS"  =  32 % ,  .. T h i s d i f f e r e n c e o f three between the u l t i m a t e s t r e s s o f Douglas F i r i n r a d i a l t e n s i o n and the values found h e r e i n are probably  due t o d r y i n g cracks which  appeared i n many o f t h e 22 samples.  I f a clear,  g r a i n e d sample i s used, d r y i n g w i l l not induce as many cracks  as i f ' r e a l ' wood i s used.  straight  nearly  Those  cracks  normal to the d i r e c t i o n o f r a d i a l t e n s i o n induce  excee-  dingly large stress concentration the u l t i m a t e The  f a c t o r s which reduce  strength. l o n g term u l t i m a t e s t r e s s taken as 9_/l6  o f the s h o r t term gives 12 p s i . g i v e s an a l l o w a b l e  A safety factor of 2  s t r e s s o f 36 p s i .  Since t h i s i s  &Q% o f the a l l o w a b l e now used, and the peak about doubles the a c t u a l s t r e s s c a l c u l a t e d by the c l a s s i c a l curved beam formula, occur.  i t i s not s u r p r i s i n g that  failures  7.  NUMERICAL The  EXAMPLE geometry o f a r o o f beam f o r use i n wet  'conditions i s shown i n f i g . 3 3 with a spacing o f 1 6 f e e t .  The beam spans 4 0 f e e t  The l a m i n a t i o n s  o f the s i z e  1 5 / 8 " x 7 " are out o f Douglas F i r 2 4 f s t r e n g t h The  c e n t e r l i n e cross s e c t i o n o f 3 9 inches  grade.  requires 2 4  lams.  DL = 20 psf LL = 40psf • • j / V & ' i / ' i / i / ' v i t  \|/v]/\|/vl/\l/>J/>l/\|/\|/\l/\|/>l/V  F i g . 3 3 . Roof beam. .  The .20  (normal d u r a t i o n o f l o a d ) i s  p s f and the l i v e l o a d ( 2 months d u r a t i o n b f l o a d ,  i.e. be  dead l o a d  snow d u r a t i o n )  i s 4 0 psf.  Herewith, the beam has t o  designed f o r a t o t a l u n i f o r m l y  (20. + 40.)  x 16  =  d i s t r i b u t e d load o f  9 6 0 . 0 . 9 6  lb/ft k/ft,  The  allowable  working s t r e s s e s  Douglas F i r 2 4 f s t r e n g t h given  f o r glulamed  grade f o r wet c o n d i t i o n s are  i n t a b l e 8 o f the. CSA-code 0 8 6 ( 1 1 ) to 1 9 0 0 p s i f o r bending 1 4 5 p s i f o r shear and 55 p s i f o r tension perpendicular  to g r a i n .  These values are f o r normal d u r a t i o n Since the governing s t r e s s e s over a two months d u r a t i o n , increased  f o r the considered the a l l o w a b l e  beam occur  s t r e s s e s may be  by the f a c t o r 1 . 1 5 t o  °t il  *  x  1 . 1 5  =  2 1 8 5 . psi  -  55.  x  1 . 1 5  =  6 3 .psi  =  145.  x  1 . 1 5  =  1 6 7 . psi  =  1 9 0 0  a  o-r  a l l  fall The and  of loads.  d e s i g n i s made by checking the t a n g e n t i a l  r a d i a l stresses  at c e n t e r l i n e o f the beam and at the  p o i n t o f tangency and by checking the shear s t r e s s at the support. The  stresses  at c e n t e r l i n e are c a l c u l a t e d f o r  comparison by the proposed as w e l l as by the e x i s t i n g formulas.  The parameters d/R = . 1 and t a n a = . 2 give  the s t r e s s c o e f f i c i e n t s  (see t a b l e 2 or f i g u r e s 2 2 and  2 3 ) CRM Crpjyj c  C M  = 0 . 0 4 8 7 =  1  .  2  8  3  = - 0 . 7 1 7  The  moment at c e n t e r l i n e i s  M  The  =  g i l =  .96 x 4 o  2  =  1  9  2  j  Q  k  maximum bending s t r e s s e s obtained  t  t  by the formula  M/Z are _ 6 M _ 192 x 12000 x 6 _ , „ o . ~ b d " " 7 x 39 x 39 1298. p s i Q  a  2  B e t t e r d e s i g n p r a c t i c e uses the s m a l l e r at  the p o i n t o f tangency  rr o-  =  The  6  M _ 192 x 12000 x 6 _ 30.6 x 30.6 "  2  7  proposed formula gives  C  tens  =  C  TM  It the present  '  x  S d ^  ^ all n  p  s  l  K  a  =  1  '  2  8  3 x  "  2  section  _ , . „ , •P  1  8  ^  5  the bending s t r e s s e s  = - 0 . 7 1 7 x 1298  Ocomp = C M x  a  9 i n Q 1 0 9  x  cross  1  2  9  8  =  l  =  6  6  5  '  -931. p s i  P  s l  can be seen t h a t the s t r e s s e s obtained by design  method give very  little  information  about the a c t u a l t a n g e n t i a l s t r e s s e s , although the s t r e s s e s are on the safe s i d e , i f the cross s e c t i o n at the p o i n t o f tangency i s used.  The r a d i a l t e n s i l e s t r e s s e s by the curved beam formula are  =  °r x  3M  3 x 192 x 12000  2 b d (R+d/2)  m a  2 x 7 x 39 x (390 + 19-5)  - on'Q 5  y  p  s  i  Here, t o o , b e t t e r d e s i g n p r a c t i c e would use the depth at the p o i n t o f tangency  3M _ —  =  a  max  2_  bd  The proposed ar  max  =  C  RM  x  to give  3 x 192 x 12000 ^ 2  =  ( R + d / 2  )  o n  p  =39.8 psi  2 x 7 x 30.6x (390+15-3)  formula g i v e s Ed^  =  '°  4  8  7  x  1  2  9  8  =  6  3  '  2  p  The curved beam formula underestimates stress considerably.  s  l  the maximum  radial  The a c t u a l s t r e s s i s about twice as  much as g i v e n by the curved beam formula. the present a l l o w a b l e s t r e s s s l i g h t l y .  I t even  exceeds  I f 2 0 p s i are  added f o r coverage o f s t r e s s e s due t o change i n moisture content, the present a l l o w a b l e s t r e s s i s exceeded  by an  i n t o l e r a b l e amount. For the s t r e s s e s at p o i n t o f tangency formulas can be used.  existing  The bending moment t h e r e , at a  distance, x = 6 . 6 3 f t from c e n t e r l i n e , i s  M = q( j j j - - f - ) = - 9 6 ( ^  ) = . 9 6 ( 2 0 0 - 2 2 ) = 1 7 0 . 9 kft  Herewith, the t a n g e n t i a l s t r e s s i s  6 M _ 170-9 x 12000 x 6 _ 170.9 x 12000 " b d ^ " 7 x 30.6 x 30.6 1092.7 "  The  , ' *  l P i 7 1  J  P  maximum r a d i a l s t r e s s i s  „  2 L  .  2bd (R+d/2)  m d X  Both s t r e s s e s  3 x 170.9 x 12000  „  p s ±  2 x 7 x . 3 0 . 6 x (390+15-3)  are s m a l l e r  The  3 5  than the allowable  shear s t r e s s e s  ones.  are checked at the support,  where the maximum shear f o r c e occurs.  The v e r t i c a l r e -  a c t i o n at the support i s  F  =  | i  =  -96 x 4o  =  1  9  >  2  k  Herewith, the maximum shear s t r e s s i s o b t a i n e d as  T  max  3 V  3 Fcos a  2 bd  2  bd  n = 3 19. .2 x 1000 x c o s ( l l ° 2 0 ' ) _ • • - 13^-3 p s i 2 7  x  3 0  m6  T h i s example shows t h a t the present d e s i g n method gives but  the s i g n i f i c a n t s t r e s s e s  does not s a t i s f y  i n tangential direction  for radial stresses.  F o r a safe  design the a c t u a l s t r e s s e s have t o be known at c e n t e r l i n e . They are r e a d i l y o b t a i n e d by the proposed method u s i n g thestress and  c o e f f i c i e n t s CQ^, Crpjyi and Cp^ as shown i n f i g . 22  f i g . 23 - -  8.  CONCLUSIONS  The  t r a p e z o i d a l element proposed does p r e d i c t  a s t r e s s d i s t r i b u t i o n c l o s e enough f o r a l l p r a c t i c a l purposes on these problems. determinations  A l s o , i t allows  from changes i n moisture  stress  content and  elastic properties. The peak at the c e n t e r l i n e o f beams induces formulas  pitch-cambered  a s t r e s s c o n c e n t r a t i o n so that curved beam  are i n v a l i d near the t o p . Therefore  coeffi-  c i e n t s from f i g . 2 2 and f i g . 2 3 should be used to c a l c u l a t e the maximum r a d i a l and t a n g e n t i a l s t r e s s e s , due'to bending  moment.  <  R a d i a l s t r e s s e s induced by shear and v a r i a t i o n i n e l a s t i c p r o p e r t i e s are not s i g n i f i c a n t , but those c r e a t e d by change i n moisture  content  are i n the order  o f magnitude o f 20 p s i and should be c o n s i d e r e d . The be reduced  a l l o w a b l e s t r e s s e s , used at p r e s e n t ,  should  t o about h a l f t h e i r v a l u e , but t e s t i n g needs  to be done to f i n d a more r e a l i s t i c r e d u c t i o n . With the i n c r e a s e d c a l c u l a t e d r a d i a l s t r e s s e s and the reduced difficult  a l l o w a b l e r a d i a l s t r e s s e s , i t w i l l be  to produce an economic d e s i g n i n some cases.  When such o c c u r s , the d e s i g n e r c o u l d c o n s i d e r r e i n f o r c i n g the beam w i t h s t e e l dowells. say 3/4" diameter  These dowells  from 3/8" t o  are p l a c e d i n s l i g h t l y o v e r s i z e d h o l e s ,  d r i l l e d almost the f u l l  depth o f the beam, and p a r t i a l l y  ' f i l l e d w i t h an adhesive such as 'Epoxy'. serve the same purpose It  Such dowells  as r e i n f o r c i n g s t e e l i n c o n c r e t e .  i s a l s o p o s s i b l e to c a r r y the t o t a l f o r c e a r i s i n g from  the r a d i a l s t r e s s e s by two v e r t i c a l s t e e l p l a t e s f a s t e n e d w i t h glulam r i v e t s to e i t h e r s i d e o f the beam at i t s centerline.  REFERENCES  NORRIS, C B . 1 9 6 3 . S t r e s s e s w i t h i n curved  lami-  nated beams o f Douglas F i r . ^ FPL - 0 2 0 . F o r e s t Prod. Lab., U.S.. Dep. A g r i . Madison, Wise.  CARRIER, G.F. 1 9 4 3 - S t r e s s d i s t r i b u t i o n i n c y l i n drically  aelotropic plates.  ' V o l . 1 0 , Trans,  ASME  J . Appl. Mech.  Vol.65.  A - 1 1 7 - 122.  FOSCHI, R.O. 1 9 6 8 . Plane S t r e s s problem i n a body with  cylindrical  reference  a n i s o t r o p y , with  t o curved  special  Douglas F i r beams.  Department o f F o r e s t r y and Rural Development, Departmental P u b l i c a t i o n No. 1244, Ottawa, Ont.  FOX, S.P. 1 9 6 8 . S t r e s s e s i n glued-laminated double-tapered  p i t c h e d beams.  timbers:  F o r e s t Prod. Lab,  Dept. F o r e s t . R u r a l Develop. Vancouver, Canada.  TIMOSHENKO, S. and GOODIER, J .  WANG, Chi-Teh,  '  Theory o f E l a s t i c i t y , McGraw-Hill. 1 9 5 7 -  A p p l i e d E l a s t i c i t y , McGraw-Hill. 1 9 5 3 -  HEARMON, R.F.S. 1 9 4 8 . The e l a s t i c i t y o f wood and plywood.  F o r e s t Prod. Res. Spec. Rep. No.1.  London, England.  WOOD HANDBOOK,  Handbook No. 7 2 , U.S. Department  of Agriculture, 1 9 5 5 .  KOLLMANN, F. Dr. Ing.,  Technologie des Holzes und  der W e r k s t o f f e , Volume I , S p r i n g e r B e r l i n 1 9 5 1 .  KENNEDY, E . I . 1 9 6 5 . S t r e n g t h and r e l a t e d P r o p e r t i e s o f Woods grown i n Canada, Department o f F o r e s t r y , F o r e s t Products, Research Branch, Dep. o f Forest'. P u b l i c a t i o n No. 1104, Ottawa, Ont.  CANADIAN STANDARDS ASSOCIATION, Ottawa, 0 8 6 - 1 9 5 9 , E n g i n e e r i n g Design i n Timber. August 1 9 6 3 .  Revised r e p r i n t -  87  APPENDIX Listing  of the  described  computer i n chapter  program 3-  ;IV  G COMPILER  t i ;  MAIN  09-08-69  11:47:38  PAGE 0 0 0 1  D I M E N S I O N R G ( 3 0 ) , M E G ( 3 0 ) , J O ( 2 0 0 , 4 ) , D E L R ( 2 0 ) , B E T A G ( 3 0 ) , 3F.TAM(200), IR 11 2 0 0 ) , R 2 ( 2 0 0 ) , THM(2 0 0 ) , N D ( 2 5 0 , 2 ) , N C O D E ( 2 0 0 , 8 ) , F I ( 8 ) , E P S I J I 3 ) DIMENSION SMM{36) , St 1 2 9 5 0 ), 3( 500 ) , DP ( 8 ) , EP SIM I 200 , 2 ) , E ( 2 0 0 , 4 ) DOUBLE P R E C I S I O N START,WORD MAXMEM=200 MAXJ0=250 C .. MAXMEM AND MAXJO ARE USED TO DEE INE THE D I M E N S I O N S IN T H E . S U B R O U T I N E S DATA START /5HS TART/ \00 P R I N T 20 READ 10 PRINT 10 P R I N T 20 10 .FORMAT ( 7 2 H __ _ _ _ _ _ 1 " J" ' "•" • 20 FORMAT ( / / 132H to************************* _ * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * x: * i * *•**_:** * * * ~ * * * * * * * ******* * v. ** ** *  :  '.  :  :  (  \ : I f i__  i |  2*** ***** * * * * * * / /) 2 101 102 103  201  .20 2 203  .204 208 210  212  214 216 220 222 224  READ 2, M XD EL R,MA ST R,MATE L,NP S FORMAT (18 14) _ __ P R I N T 1 0 1 , MXDELR I DELR) FORMAT (--'/25H NUMBER OF D E L T A RADIUS = , I 4 / 1 2 H . READ 1 0 2 , (DELP. ( I ), 1 = 1, MXDELR) FORMAT ( F 1 0 . 6 ) PRINT 1 0 3 , ( I , D E L R ( I ) , 1 = 1,MXDELR) FORMAT __( I 4 , F 12 ._6 ) READ 2,NRG P R I N T 2 0 1 , NRG FORMAT (//26H NUMBER OF R A D I A L GROUPS = , I 4 / 2 5 H I R G MEG RG 1BETAG) « , ' READ 2 0 2 , ' (MEG( I R G ) , R G ( I R G ) , B E T A G ( I R G ) , I R G = 1 , N R G ) F O R M A T . i J f t » F I Q . 3 , F 1 0 . 6.) ..„ PRINT 2C3,. ( I R G , M E G ( I R G ) , R G ( I R G ) , B E T A G ( I R G ) ,I RG=1,NRG) FORMAT (2I4,F10.3,F10.6) C A L L D I V I D E (MAXMEM,NPG,RG,MEG,JO,ME,MJ) C A L L D I S T R I (NRG,RG,MEG,DELR,BET AG,BETAM,ME,R1,R2) READ 2 0 4 , TH,NSM F0RMAT__.{F12. 6,14) _____ .__ _ DO 208 1=1,ME T H M ( I ) = TH P R I N T 210,MF,NSM,NJ FORMAT <//20H NUMBER OE MEMBERS = , I 4 / 2 0 H N OF SPEC.MEMBERS = , 1 4 / 2 0 IH NUMBER OF J O I N T S =,14) _IF_CNSM.EQ.0_ GO TO 2 2 2 _ P R I N T 212 FORMAT (//16H S P E C I A L MEMBERS/12H I THM) 00 2 2 0 1=1,NSM READ 2 1 4 , I S P , T H M ISM) FORMAT {-1.4, F I 0.6) ...PRINT 2.16',. ISM, THM( ISM.) FORMAT ( I 4 , 2 X , F 1 0 . 6 ) CONTINUE PRINT 224 FORMAT (//22H MEMBER SPEC I F I C A T I 0 N S / 5 9 H I J0INTNRS(I , J) I R l " R2 BETAM THM) PRINT. 2 2 6 , .( I , J 0 ( 1, 1 )_»JO ( 1 , 2 ) , J 0 ( 1,3) , J 0 ( 1,4) , R 1 (I ) , R 2 ( B IE )TAM( , I )  V  G COMPILER  MAIN  09-C8-69  11:47:38  PAGE  0002  1,THM(I)»I=1rME) FORMAT U4,2X,4I4,2F10.3,2F10.6) MS=8 NC.0 = 4 MST=MS/NCO R E A D 2, NR J PRINT 305 m J _ . _ '._ 305 . F O R M A T ( / / 3 C H N U M B E R OF R E S T R A I N E D J O I N T S = , T 4 / 1 2 H J R ND 1 ND2) DO 3 1 0 1 = 1 , N J 00 311 K=1,MST ' ND(I,K)=1 ' 311 CONTINUE 310 CONTINUE DO 3 1 2 I=1,NRJ READ 2 , J R , ND ( J R , 1 ) , ND ( J R , 2 ) PRINT 2 , J R , . M P ( J R , 1 ) , N D t J R , 2 ) 312 CONTINUE NUM=0 003.13 . I - l , N J DO 3 1 4 K = 1 , M S T IF ( N D ( I , K ) . E Q . 0 ) G O TO 3 1 4 NUM=NUM+1 N 0 ( I ,K.)=NUH 314 CONTINUE 313 CONTINUE PRINT 320 320 FORMAT C / / 1 9 H J O I N T CODE NUMBERS) P R I N T 3 2 1 , ( J , ( N O ( I , J ) , J = 1 , M ST) , 1 = 1 , N J ) 321 FORMAT (5(3I4,2X)) DO 3 1 6 1 = 1 , M E D0... 3 1 7 J = l . » MS _. C A L L CODE {MAXM E M , M A X J 0 » J 0 » N D , I > J » I COM) NCODEII,J)=ICDM 317 CONTINUE 316 CONTINUE PRINT 325 3 2 5 . . F O R M A T ( / / . ? 1 H .. E L E M E N T CODE. N'JMBf R S / P 4 H _I NCODEJ 1 , 1 T 0 3 ) . ) P R I N T 3 2 6 , I I , ( N C O D E ( I, J ) , J = l , M S ) , 1 = 1 , M E > ,, 326 FORMAT I I 4 , 2 X , 8 I 4 ) NU=NUM NB = 1 DO 3 5 0 1 = 1 , M E MSM.= M S - 1 DO 3 5 1 J = 1 , M S M ICDJ=NCODE(I,J) IF ( ICDJ ) 3 5 ' 353 JP=J+1 DO 3 5 2 K = J P , M S I C D K = N C O D E (.1 ,.K_) IF IICOK) 3 5 4 , 3 5 2 , 3 54 3 54 N8T=IASS((IARS( ICOKJ-IABSt ICDJ)))+1 IF (NST-NB) 352,352 355 355 N0=NB T 352 CONTINUE 351 CONTINUE  226 300  f  ! IVf  G COMPILER 350 362 360  370 _  372 373 376 380 400 402 401 403 404  405  40 8  ? 420  .421  ..  09-08-69  11:47:38  CONTINUE R E A D 3 6 2 , E 1 , E2 , E M U , G , N E S P F O R M A T I A X , 4 F 16 . 8 , I 4 ) PRINT 3 6 0 , E l , E 2 , E M U , G , N E S P . • FORMAT ( / / 5 H EI = , F 1 2 . 4 / 5 H E2 = , F 1 2 . 4 / 5 H E M U = , F 1 2 . 4 / 5 H 1 / 4 2 H N U M B E R OF M E M B E R S W I T H S P E C I A L F - M O D U L I = , 1 4 ) D C  S  MAIN  3 7 0  I = I,MF.  _  _ _ .  ...  _  PAGE  G  0003  =,F12.4/  _  E ( I , 1) =F1 El I ,2)=E2 . - '• E ( I t 3 ) = EMU . ' E(I,4)=G IF ( N E S P . E Q . 0 ) GOTO 3 0 0 ._ DO 3 7 6 1 = 1 , N E S P . : . _ READ 3 7 2 , IE S P , E ( I E S P , 1 ) , E ( I E SP , 2 ) , E { I E S P , 3 ) , E I I E S P » 4 ) PRINT 3 7 3 , I E S P , E ( I E S P , 1 ) ,E(I E S P , 2 ) ,E( I E S P , 3 ) , E { I E S P . 4 J FORMAT ( I 4 , 4 F 1 6 . 3 ) F O R M A T {I 4 , 4 E 1 6 - 3 ) CONTINUE CONTINUE IF (MATED 400,401,400 PRINT 402 .- F O R M A T ( 1 H 1 , 2 5 H E L E M E N T S T I F F N E S S MATRIX//) GO TO 4 0 4 PRINT 403 F O R M AT .( /. /.. 3 7 H _ ELE_.ME.NT_. ST I F F N E S S M A T R I X .NOT.__PR I NT E D ) CONTINUE NS - N U * N 3 00 405 I = 1 , N S S(I)=0.0 NB1=NB-1 ; _ _ D 0 407.L_=1.,.ME DO 4 0 8 1 = 1 , 3 6 SMMl I) = 0 . 0 R 1 L = B1 i I.) R2L=R2(L) BETAL=3ETAM(L) THL=THMIL> E1L=E(L,1) E2 L = E ( L , 2 ) E MU L= E (L.3) G L = E ( L.» 4 ) C A L L SMEM ( R 1 L , R 2 L , S E T A L , T H L , F 1 L , E 2 L , E M U L , G L , S M M ) IF ( M A TT EEL L . E Q . <0 ) GO TO 4 2 2 .... PRINT 4 2 0 , L FORMAT (/I4) P R I N T 4 2 1 , SMM( 1) P R I N T 4 2 1 , S M M( 2) , S M M ( 9 ) PRINT 4 2 1 , SMM(3) ,SMM{10) , S M M ( 1 6 ) P R I N T 4 2 1 , . . S M M l . 4 ) _ _ S M M l 1.1) , S M M ( 1 7 ) , S M M { 2 2 ) „ P R I N T 4 2 1 , SMM( 5 ) , S M M ( 12 ) , S M H ( 1 8 ) , S M M ( 2 3 ) , S M M I 2 7 ) P R I N T 4 2 1 , SMMl 6) , S M M ( 1 3 ) , S M M l 1 9 ) , S M M 1 2 4 ) , S M M l 2 8 ) , S M M l 3 1 ) P R I N T 4 2 1 , SMM( 7 ) , S M M ( 1 4 ) , S M M l 2 0 ) , S MM ( ? 5 ) , S M M l 2 9 ) , SMM( 3 2 ) , S M M { 3 4 ) PRINT 4 2 1 , S M M ( 0 ) , S M M l 1 5 ) , S M M l 2 1 ) , S M M l 2 6 ) , S M M l 3 0 ) , S M M { 3 3 ) , S M M 1 3 5 ) , 1 SMMl36) j .FORM*.]'. J . J . X , 8 G 1 6 . 6 . ) . _ _ _ __.._' _  IV  G COMPILER 422  412  _4C6 450 451  .4.1 3 414 415 410 .4.09 407  1000 460 462  464 465  MAIN  09-08-69  11:47:38  PAGE  CONTINUE DO 4 0 9 J = l , 8 IF ( N C O D E ( L , J ) ) 4 0 9 , 4 0 9 , 4 1 2 J l = ( J - l )-M 1 6 - J ) /2 DO 4 1 0 I = J , 6 IF ( N C O D E ( L , I ) ) 4 1 0 , 4 1 0 , 4 0 6 I F J N C O D E J L, J ) r NCODE.( L , I ).)_. 413 ,4.5 0,4 1.4 I F ( I - J ) 451,413,451 K =(MCOD E ( L »I ) - I )* NB1+ NC 00 £ I L , J ) N= J 1 + I S(K)=S(K)+(2.0*SMM(N)) GO TO 410 K M NCODE.( L , J ) — I.).J:.N.5 1+NCODEJ L , I } GO TO 4 1 5 K=(NCODE(L,I)-l)*NB1+NCODE1L.J) N=J1+I S(K)=S(K)+SMM{N) CONTINUE CONTINUE CONTINUE , IF {MASTR .EO. 0) GOTO 4 6 0 PRINT 1 0 0 0 FORMAT ( I H 1 ) CONTINUE P R I N T . 4 62_,.._NU,NB FORMAT (//22H NUMBER OF UNKNOWNS = , I 4 / / / 2 2 H BANDWIDTH" 1,14) IF (MASTR ) 4 6 4 , 4 6 6 , 4 6 4 PRINT 465 FORMAT {////27H S T R U C T U R E S T I F F M E S S M A T R I X / ) P R I N T 4 2 5 , ( S ( K ) , K= 1 , NS ) _ _ .  42 5 466 467 500 501  502 . 503  506  50 4 505 700 701 702 _ ...  GO TO 500 PRINT 467 FORMAT {//39H S T R U C T U R E S T I F F N E S S MATRIX NOT READ 2,NLC PR I.NT_..O.l ,. _NLC FORMAT (////22H NUMBER OF L O A D C A S E S = , 14) , DO 590 L=1,NLC 0 0 502 8( I ) = 0.0 READ 2, N B L , N E I P R I N T 503 , .L FORMAT ( L H 1 , 9 H L O A D C A S E , 1 4 ) I F (NBL .EO. 0) GO TO 700 P R I N T 5 0 6 , NBL FORMAT (//24H NUMBER OF LOADS GIVEN = , I 4 / 1 2 H DO 504 1=1,NBL READ. 50.5, ,K,B(K) PRINT 5 05, K , 3 ( K ) FORMAT ( 14 ,F10 .3) IF ( N E D 7 0 1 , 7 2 2 , 7 0 1 READ 7 0 2 , EPS I 1,EPS I 2,EPS I 3,NSSM FORMAT ( 3 F 1 0 . 6 . I 4 ) PRINT .704, E P S . l.,.EP SI 2., E P S I 3 , NSSM  PRINTED)  K  B{K))  0004  IV  G  COMPILER  !  704  706  v  710 711 708 709  MAIN  09-08-69  11:47:38  FORMAT ( / / 2 3 H I N I T I A L S T R A I N ETANG. = , F 1 0 . 6 / 2 . 3 H I =,F10.6/23H E S H E A R = , F 1 0 . 6 / / 3 7 H NUM3ER 2STRAIMED MEMBERS = , 1 4 ) DO 7 0 6 J = l , M E E P S I M ( J , 1 ) = E P SI 1 E P S I M U , 2 ) = EPSI2 IF ( N S S M . E 0 . . 0 )_ G O „ T O . 7 0 9 . DO 7 0 8 1 = 1 , N S S M READ 7 1 0 , I S S M , E P S I M { I S S M . l ) ,EPSIM< I S S M , 2 ) FORMAT (I4,2F10.6) PRINT 7 1 1 , ISSM , E P S I M ( I S S M , 1 ) , E P S I M I I S S M , 2 ) , E P S I 3 FORMAT U 4 , 3 ( 2 X , F 1 0 . 6 ) > CONTINUE _ _ __ 00 720 j = l , M E R U = R1 ( J ) R2J=R2(J)  PAGE  OF  ERAD SPECIAL  B ETAJ=BcTAM{J)  1  712 714 720 722 724 750 507 -t<.  508 511 509 512 r  >  510 514 513  0005  THJ=THM(J) EPSIJ(1 )=EPSIM(J,1) EPSIJ<2)=EPSIM{J,2) E P S I J I 3 ) - EP S I 3 E U = E( J , l ) E2J=E(J ,2 ) EMUJ=E(J,3) G J = E( J CALL E P S I N I ( R 1 J , R _ J , B E T A J , T H J , E 1 j » E2 J , E M U J , G J , E P S I J , F I ) DO 7 1 4 N C = 1 , 8 IF {NCODE(J,MC)) 712,714,712 NCM=NCOPE(J,NC) 8 ( NCM ) ~ B ( NCM ) + F I ( N O CONTIMUE_ CONTINUE GO TO 7 5 0 DO 7 2 4 1 = 1 , M E DO 7 2 4 J=l,2 E P S I M ( I , J ) =0 . E P S I 3=0.. CONTINUE PRINT 507 FORMAT (//11H LOAOVECTOR) PRINT 5 1 3 , (3(1),1=1,NU) DE T = 1 . E — 8 C A L L BAND (S,B,NU,NB,L,DET) IF(DET) 508,509,510 PRINT 511,DET F O R M A T ( 3 4 H 0 MAT R I X A I S NOT P O S I T I V E - D E F I N I T E / 7 H 0 D E T = ,E15.7) GO TO 5 9 0 PRINT 512,DET F O R M A T '(_2.0H0DET ER M IJMA.N.T _ J S Z E R 0 / 7 H 0 0 E T = , E 1 5 . 7 ) GO TO 5 9 0 PRINT 514 F O R M A T ( / / 1 9 H D I S P L A C E M i NTV E C T O R ) PR I N T 5 1 . 3 , ( B ( I ) , 1 = 1 NU.) FORMAT ( 1 X , 8 G 1 5 . 6 ) PRINT 555  IV  G  COMPILER 555 550  552 553 554  556 558  560 590 610  J  i  620  MAIN  09-08-69  11:47:38  PAGE  0006  TANGEN FORMAT {/////9H STRESSES/51 H RAOIAL IT TAU) 00 560 J=1,ME R I J = R1 ( J ) R 2J =R 2 ( J ) BETAJ =BETAMIJ ) DO 5 5 4 NC = 1 , 8 _ ___ IF (NCODE ( J . N £ ) ) 5 5 3 , 5 5 2 / 5 5 3 DP(NC)=0.0 GO TO 5 5 4 MCM=NCODE(J,NC) DP(NC)=8(NCM) ^CONTINUE E P S I J ( 1 ) = EPSIM( "j, 1) EPSIJ(2)=EPSIM(J,2) EPSIJ(3)=EPSI3 E U = E( J , l ) E2J=E(J,2) EMUJ^El J,3> GJ=E(J,4) C A L L S I O M ( R I J , R 2 J , BE T A J , E 1 J , E2 J , E MU J , G J , DP , E P S I J , S T H , SR. A D , SR TH ) PRINT 5 5 8 , J , S R A D , S T H , S R T H FORMAT ( 1 4 , 3 X , 3 ( 1 X , F 1 5 . 6 ) ) IF (J.GT.NPS) GO TO 5 6 0 WR I J F. . ( 7 5 5 8 ) J , S R A D , S T H , S R T H . CONTINUE CONTINUE R E A D 6 1 0 , WORD FORMAT (A5) IF (WORD . N E . S T A R T ) CO TO 6 2 0 PR I NT 1 0 0 0 GO TO 1 0 0 CONTINUE STOP END  IV  G COMPILER  . 11 N  13  12... •>  i  ~14  10  09-08-69  •  11:48:16  SUBROUTINE D I V I D E (MAXMEM,NRG,RC,MEG,JO,IMUS,NJ) DIMENSION R G ( l > , MEG(1),JO(MAXMEM,4} I8AS=0 IMUS=0 DO 10 IRG=1,NRG IF ( I P G . E Q . 1 ) GO TO 11 I F { M E G ( IRG-1 )-MEG(_IRG) ) . 1 1 , 1 1 , 12 MM = MF G ( I RG ) DO 13 JRG = 1,M I=IMUS + J'RG J 0 ( I ,1) = I B A S + J R G + M E G ( I R G ) + 2 J 0 ( I ,2) = I B A S + J R G + l J0(I,3)=I3AS+JRG J C ( I ,4]l =1BAS + J R G + M E G l I R G 3 + 1 CONTINUE IRAS=IBAS+MEG(IRG)+l IMUS=IMUS+MEGlIRG) GO TO 10 _MM = MEGJ.I RG) . DO 14 JRG.= 1,MM I=IMUS+JRG J 0 ( I ,1 )=T8AS+JPG*-MEG( I R G - D + 2 J 0 ( I ,2) = I B A S + JRG+1 J 0 ( I » 3 ) = I 3 A S +JRG _ JO { I ,4) = I B A S + J R G + MEG.(.IRG-1J_+1 CONTINUE IBAS=IBAS+MEG(IRG-1J+l IMUS=JMUS+MEGlIRG) CONTINUE NJ=IBAS+MEG(NRG)+1 RE T URN . END M  'V  1  DIVIDE  PAGE  0001  IV  G COMPILER  1  -V  'A  V  •v  21 20  . •»  * 5  I i  0 ISTR I  0 9 - 0 8-6 9  11:48:21  PAGE 0 0 0 1  S U B R O U T I N E O I S T R I (NRG ,RG, MEG,DELR,BETA G , B E T A M , I M U S f R l i R2 ) D I M E N S I O N R G ( 1 ) , M E G ( 1 ) , 0 E L R ( 1 ) , B E T A G <1) ,B E T A MI 1) , R l ( l ) ,R2( 1 ) IMUS=0 DO 20 I RG'=1 » NRG R1M=RG(IRG) MM=MEG(IRG) DO 21 JRG=1,MM I=IMUS+JRG R1(I)=R1M R2(I)=R1M+DELR(JRG) R1M=R21I) BETA M(I ) = B E T A G ( IRG) CONTINUE IMUS=IMUS+MEG(IRG) CONTINUE RETURN END  IV  G  COMPILER  2  >-  4  10 _5_ .  6 •<  y V. *'  e I  7 8 11 20  CODE  09-08-69  11:43:22  SUBPOUT INE C O D E ( M A X M E M , M A X J O , J O , I C D , I , J , I COM) DIMENSION J 0 ( M A X M E M , 4 ) , I C D ( M A X J 0 , 2 ) GO TO ( 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 ) , J JNU=1 GO TO 1 0 JNU = 2 GO TO 1 0 JNli =3 GO TO 1 0 JNU=4 JD=1 GO TO 2 0 JNU=_1 GO TO i l JNU=2 GO TO 11 JNU = 3 GO TO 1 1 JNU=4 JD=2 I I = J 0 ( I ,'JNU ) I C D H = I C D ( 11 , J D ) RETURN END  PAGE 0 0 0 1  IV  G COMPILER  S.VEM  09-08-69  11:43:25  S U B R O U T I N E SMEM I R I , R 2 , B E T A T H , E 1 , E 2 , E M U , G , S ) T R A P E Z O I D A L ELEMENT TRANSFORMED INTO PGLARCOORDINATES DIMENSION S K ( 3 6 ) , S ( 3 6 ) H=(R2-R1)*.CGS( BETA/2. ) A= 2 . ' R 1 S I N ( B E T A / 2 . ) B=2.*R2*SIN(BETA/2.) _ F l _ . = {A+B) ^ H / 2 . FY =(A+2.*B)*H*H/6. FXX=((A*A+3*B)*(A+B)) H/36. FYY=(A+3.*B)*H*H*H/12 • SA1-1./(A*A) SA2=1./(A H) . „ _ S A 3 = 1 . / ( A*A_*H)_ _ .__ SA4=1./IA*A*H*H> S.A5 = 1 . / ( A * H * H ) SA6=l./(A*B*H) SA7=1./(A~8"H*H) SB1=1./(H*H) _ S 8 2 . = 1_../H _____ _  PAGE 0 0 0 1  f  C  A  •_,  y  .  :  i ;  A  *>•  A  v ' 7  .  •  ...  _  _  SB3=1./(B.*H*H) ^ !_ '\ ^ j I » I * H i \ 1  "*  A  1  }  t  - — '1  >. >  A:  ;  y  A— J  SB4 = l./(B*B*H*-H) . S C l = l . / ( 4 . *H*-H)*iFl':-G) SC'2=(FYY*E1) + (FXX*G) SD1=FY*G S D 2 = FY*F_1 ' S D 3 = FY*--EMU SEI=F1*G SE2=F1*E1 " SE.3 = F 1 « E M U SFl=l./(4.tH*H)*(FlvE2) _ .. S F 2 = ( F X . X _ E 2 ).+ (. F Y Y * G ) S K ( 1 ) = SC1+SB4*SC2 S K ( 2 ) = SC1-S84'- SC2 SK(3) =-SCI-SA6~SD2+SA7~SC2 SK(4) =-SCl+SAo*SD2-SA7*SC2 SK(5) = SB3/2.*(SD1+SD3) SK ( 6 ) . _ = ' . S B 3 / . 2 . - : ( . - S D l + S D 3 ) _ ...... . _ SK(7) --SA2/2.*SEl+(SA5/2.*-SDi)-(S33/2.*Sn3J SK{8) = ( S A 2 / 2 . * S E 1 ) - ( S A 5 / 2 . * S D l ) - ( S B 3 / 2 . " S D 3 ) SK( 16)= ( S A l * S E 2 ) - ( S A 3 * 2 . * S D 2 ) + ( S A 4 * S C 2 )+ S C l SK(17)=-(SAl*SE2J+(SA3*2.*SD2 l-(SA4+SC2)+SCI SK(18)=-ISA2/2.*SE3) +( S A 5 / 2 . * S D 3 ) - ( S B3/2.*SD1) S K ( 1 9 ) = - ( S A 2 / 2 . * S E 3 ) +( S A 5 1 2 . * S 0 3 ) +(SB 3 / 2 . * S D 1 ) SK ( 2 0 ) = ( S A . 2 / 2 . * S E 3 ) - M S A 2 / 2 . < S E 1 ) - ( S A 5 / 2 . * S D 3 ) - ( S A 5 / 2 . *soT7~ S K ( 2 1 ) = ( S A 2 / 2 . - S E 3 > - ( S A 2 / 2 . * S E l ) - ( S A 5 / 2 . - S D 3 ) + { S A 5 / 2 .*SD1 J SKI 2 7 ) = S F l + ( S B 4 * S F 2 ) S K I 2 8 ) = SF 1 - ( S B 4 * S F 2 ) SK(29) = - S F l + ( S A 7 * S F 2 J - ( - S A 6 * S 0 1 ) SK(301=-SFl-{SA7-=SF2)+<SA6*SD1) — SKI 34)= S F l +( S A 1 * S E 1 ) - ( S A 3 * 2 . * S D l ) M S A 4 * S F 2 ) SK(35)= SFl-(SAl*SEl)+(SA3*2.*SD1)-(SA4*SF2) CC=COS(BETA/2.)*COS(BETA/2.) CS=COS(BETA/2.)*SINIBETA/2.) SS=SIN(BETA/2.)*S IN(BETA/2.) SJ l.).-:=-.CCr.SK (.1.) r . C S ? 2 . y.SK ( 5 ) + S S * S K ( 2.7) ._ :  •  —  IV  G COMPILER  SMEM  09-08-69  11:48:25  S(2) = C C * S K ( 2 ) + C S « 2 . * S K ( 6 ) - S S * S K 1 2 8 ) S(3) = CC*SK(3)+CS*(SK(7)-SK( 1B))-SS*SKI?9-) S(4) = CC*SK(4)+CS*(-SK(B)+S.K(19) )+SS*SK(30) S(5) = C C * S K ( 5 ) + C S » ( S K I 1 ) - S K ( 2 7 ) ) - S S * S K ( 5 ) S(6) = CC*SK{6)+CS M-SK(2)-SK(28) )-SS*SK(6) S(7) = CC*SK(7)+CS*(-SK(3)-SK(29)}+SS*SK{18) S ( 3 ) _=_ C C * S K _ 8 ) + C S = M S K ( 4 J - S K ( 3 0 ) ) + S S * S K ( 1 9 )  PAGE  •  !  si 9 )  s  I  _  =  sin  SI 1 0 ) = S(4) S ( 11)= S( 3) S(12)=-SI6) S(13)=-S15) S(14)=-S(8) S(15)=-S(7) S( 16)= C C * S K ( 1 6 H - C S * 2 . * S K ( 2 0 ) + S S * S K < 3 4 ) S U 7 ) = C C * S K ( 1 7 ) - C S * _ . * SKI 2 1 ) - S S * S K ( 3 5 ) S( 1 8 ) = C C * S K ( 1 8 ) + C S * ( S K ( 3 ) + S K ( 2 9 ) )+SS*SK(7) S(19)= CC*SK(19)+CS*(-SK(4)+SK(30))+SS*SK(8) _ S ( 2 0 ) - C C * S K ( 2 0 ) + C S * ( - S K ( 16) + S K ( 3 4 ) ) - S S * S K ( 2 0 ) S(21)= CC''SK(21 )+CS*lSK(L7)+SK(35))-SS*SK(21> S I 2 2 ) = S( 1 6 ) S(23) =- S ( 1 9 ) S(24)=-S(18) S(25)=-S(21) _S(26)=-SJ.20.) S ( 2 7)= CC*SK(27)+CS*2.*SK(5)*SS*SK(1) S( 2 8 ) = C C * S K l 28 ) + C S * 2 . * S ' K ( 6 ) - S S * S K ( 2 ) S ( 2 9 ) = C C * S K ( 2 9 ) +CS ( S K ( 7 ) - S K ( 1 8 ) ) - S S * S K { 3 ) S( 3 0 ) = C C * S K ( 3 0 ) - + C S * l S K { 8 ) - S K M 9 ) ) + S S * S K ( 4 ) S(31)= S(27 ) S ( 3 2 ) = _ S( 3 0 ) __ S ( 33 )= S( 29 ) S ( 3 4 ) = C C * S K [ 3 4 ) - C S * 2 . * SKI 2 0 ) + S S * S K { 16) S(35)= CC*SK(35)-CS*2.* SK<21)-SS*SK(17) SI 3 6 ) = S(34) DO 1 0 1 = 1 , 3 6 S ( I ) = S(.I_) T H RE T U R N END T  4  .10  0002  •  IV  G COMPILER  '_ r  *v  "  rr  -  X !  4  \  EPSIMI  '09-08-69  •  1 1 : 4 8 : 38  SUBROUTINE E P S I N I (RI,R2,BETA,TH,E1,E2,EMU,G,EPS D I M E N S I O N F T { B) , F I ( 1) , E P S K 1 ) C=C0S(BETA/2.) S=SIN<BETA/2.) H=C*(K2-R1) A=2.*Rl*S B=2. „R2*S A A = f H*'( A + 3 ) / 4 . A B = T H - H M 2 . *A + 3 ) / { 6 . - A ) AC = T H * H * ( A + 2 . - ' 8 ) /( 6 . * - B ) AE=E1*EPSI(1)+EMU*EPSI(2) AF=EMU*EPSI(1) + E2vEPSli2 ) _ ... A G = G * F p S I 13 F T t 1 ) = A C * A E +A A* AG FT(2)=-AC*AE +AA*AG FT(3)=-A3*AE -AA*AG F T ( 4 ) = A3 * A E — 4 A * A G F T ( 5 ) = A A * A F +AC*AG FT (6.)_=_AA*AF - A C * A G FT(7)=-AA*AF -AB*AG FT ( 8 ) - = - A A ' * A F + A 8 * A G : FI ( 1 ) = C-FT { 1 )-S*FT ( 5 ) F I ( 2 ) = C---FT(2) + S * F T ( 6 ) FI(3)= C*FT(3)+S*FT(7) F 1(4) = C*FT{4)-S*FT{3)_ F I ( 5 ) = O F T (5)+S*FT( 1) FI(6)= C*FT(6)-S*FT(2) FI(7)= C*FT(7)-'S-'«FT(3) FI(8)= C*FT(8)+S*FT(4) RETURN END  I,FI}  PAGE 0 0 0 1  IV  G COMPILER  SIGM  09-08-69  11:48:43  PAGE 0 0 0 1  S O B R O U T IM E S I G M ( R l , R 2 , B E T A , E l , E 2 , E M U , G , O P , E P S I , S I G X , S I G Y , T A U ) D I M E N S I O N D P I 1) , D K { 3 ) , C T I ( 3 , 3 ) , E P S ( 3 ) , E ° S I ( 1 ) C=COS(BETA/2 . ) S=SIN(BETA/2 . ) ' DKll)=C*DP(l)+S*E)P(5) nK(2)=C*DP(2)-S*0P<6) ___DK('3)=C*DP(3)-S*DP(7) . _ D K ( 4 ) = C * D P ( 4 ) + S - D P ( 8) DK(5)=C«DPI5)-S~DP(1) DK(6)=C*--DP(6)' + S * D P ( 2 ) ^ _ _ _ _ DK17)=C*DP(7)+S*DP(3) DX(8)=C*0P( 8 ) - S * 0 P ( 4 ) _ H = C * l _ R 2 - R 1) _ ._ _ ._ _ A=2.*R1*S B=2.*R2*S X=0.  S  Y=H/2. CTl (1,1)= Y/(B*H) CTl(l,2)=-CTI(l,l)  CT I ( 1 , 3 ) = - 1 . 6 / A  +  Y/(A*H)  CT n 1 , 4 ) = - C T 1 ( 1 , 3 ) CTl (1,5)'= 0. CTl  ( 1 , 6 ) =  0  CTI(1,7)= 0 C T l ( I »8__= . 0 , CTl (2,1)= 0 C T I ( 2 , 2 ) = 0, C TI ( 2 , 3 ) = 0 , CTl ( 2 , 4 ) =.0... C T I ( 2 , 5 ) = l . 0 / ( 2 . 0 * H ) + X/(B*H) CTI(2,6)=_1.0/(2.0*H) X/(B*H) CTl(2,7)=-l,0/(2.0*H) + X/(A*H) CTI(2,_)=-l.0/(2.0*H) X/(A*H) C T I ( 3 , 1) • CTl(3,2)= CTl(2,6) C T I ( 3 , 3 ) = C T H 2 ,7) CTl(3,4)= CTl(2,8) CT 1 ( 3 , 5 ) = C T I ( 1 , 1 ) CTl (3,6)= CTl(1,2) CTl (3,7)= C T K 1 ,3) CTl (3,8)= CTl (I,4) DO 2 0 K = l , 3 EP_S(K.)=-F.PST. ( K ) . DO 2 0 J = l , 8 EPS(K. ) = E P S ( K ) + C T I ( K , J ) ' D K ( J ) CONTINUE SIGX= EPS( 1)*E1+EPS(2)*EMU SIGY= E P S ( 1 ) " E M U + E P S ( 2 ) * E 2 TAU = EPS(3J»=G RETURN END T  20  

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