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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 Degree ( C i v i l E n g i n e e r i n g ) Swiss 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 i n the Department o f CIVIL ENGINEERING We accept t h i s t h e s i s as conforming to the r e q u i r e d standard The' U n i v e r s i t y o f B r i t i s h 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 for an advanced degree at the University of B r i t i s h Columbia, I agree that the Library s h a l l make i t f r e e l y available for reference and study. I further agree that permission for extensive copying of th i s thesis for scholarly purposes may be granted by the Head of my Department or by his representatives. It i s understood that copying or pu b l i c a t i o n of th i s thesis for f i n a n c i a l gain s h a l l not be allowed without my written permission. W. K. Thut Department of C i v i l Engineering The. University of B r i t i s h Columbia Vancouver 8 , Canada A p r i l 1 9 7 0 -ABSTRACT The stress d i s t r i b u t i o n near the centerline apex of pitch-cambered glulam beams i s analysed. The analysis i s done with an orthotrbpic trapezoidal 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. The element i s tested with various g r i d sizes against known solutions and shown to give : an o v e r a l l accuracy to within f i v e percent. Radial tension stresses perpendicular to grain due to moment, were calculated for various geometries and plotted as a design aid. These stresses can be several times those found i n a uniform curved beam. The e f f e c t of changing the e l a s t i c moduli of the whole beam or i n d i v i d u a l laminations was investigated and found to be unimportant. Radial stress due to shear forces was found to be unimportant. Radial stress for one t y p i c a l geometry due to moisture change was investigated i n d e t a i l . This showed that ten pounds per square inch t e n s i l e stress, was gene-rated f o r a one percent change i n moisture content. • Twenty two.tests were performed on tension per-pendicular to grain to indicate that the allowable stress would be .near t h i r t y f i v e pounds per square inch. A numerical example and design recommendations are included. i i . TABLE OF CONTENTS ABSTRACT i TABLE OF CONTENTS i i LIST OF FIGURES AND TABLES i v LIST OF MAJOR SYMBOLS • v i ACKNOWLEDGEMENT x 1. INTRODUCTION 1 2. DERIVATION OF A TRAPEZOIDAL FINITE ELEMENT IN PLANE STRESS 5 2. 1. GENERAL 5 .2. 2. DERIVATION OF THE FINITE ELEMENT 8 2. 3- TRANSFORMATION INTO POLAR COORDINATES 17 3. COMPUTER SOLUTION PROCEDURE ; 21 3- 1. BASIC STEPS IN THE PROCEDURE 21 3. 2. SIMPLIFIED FLOW DIAGRAM 24 4. CHECKING OF THE FINITE ELEMENT WITH VARIOUS . PROBLEMS 29 4. 1. ISOTROPIC CIRCULAR BEAMS AND RINGS . 2 9 ^ . 4. 2. ORTHOTROPIC CIRCULAR BEAMS . .'. 40 4. 3- PITCH-CAMBERED ORTHOTROPIC BEAM 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 . ' 6 2 5 . 4 . VARIATION IN ELASTIC PROPERTIES . . 7 1 SUMMARY . ; ' 7 3 6 . . EXPERIMENTAL TEST ON STRESS PERPENDICULAR TO GRAIN . 7 4 " 7 . NUMERICAL EXAMPLE 7 8 8 . CONCLUSIONS 8 3 LIST OF REFERENCES 8 5 APPENDIX: LISTING OF THE COMPUTER PROGRAM . . 8 7 i v LIST OF FIGURES AND TABLES Pig. 1. Pitch-cambered glulam beam. 4 Pig. 2. Arrangement of nodal points. 4 F i g : 3. Trapezoidal f i n i t e elements assumed with rectan-gular orthotropy. 5 F i g . 4a. Element parameters. . 9 Pig. 4b. Directions of p o s i t i v e nodal displacements and nodal forces. 9 F i g . 5.. S t i f f n e s s matrix [k]. 14 Pig. 6. Parameters and coordinates determining the t r a -pezoid. 18 Pig. 7- The two sets of nodal forces and displacements. 18 Pig. 8. Approximation of the upper straight edge. 22 Pig. 9a. Isotropic c i r c u l a r beam, shear loading. 31 F i g . 9b. Stresses at cj) = 0/2. 32 Pig. 9c. Tangential stresses along r = R + AR/2. 31 F i g . 10.. Isotropic c i r c u l a r beam, moment loading. 34 Pig. 11. Thick i s o t r o p i c cylinder under external pressure. 36 F i g . 12. Thick i s o t r o p i c cylinder, uniform temperature gradient. 39 F i g . 13- C i r c u l a r orthotropic beam under pure moment. 42 Fi g . 14. C i r c u l a r orthotropic beam under shear load. 44 F i g . 15. Geometric dimension of the tested beam. 46 Pig. 16. F i n i t e element g r i d , loading condition moment." 47 F i g . 17- Stresses at centerline of the tested beam. . 48 F i g . .18. Arrangement f o r d/R = .2 and tan a = .4. 52 F i g . 19. Three possible cases of pitch-cambered beams. 52 Pig. 20. Beams No. 2, 9, 24 and 27- 54 F i g . 21. C h a r a c t e r i s t i c stress d i s t r i b u t i o n f o r pure bending moment. 54 Pig. 22. Stress c o e f f i c i e n t Cp^ for maximum r a d i a l stress at centerline due to pure bending. 57 V. F i g . 2 3 . Stress c o e f f i c i e n t s Cijjjyj and CQJVJ for tangential stresses at centerline due to pure bending. 5 8 • F i g . 24. Antimetric load case for a pure shear load at centerline. . . • 6 . 1 F i g . 2 5 . Shear stresses at centerline from pure shear load. 6 l Fi g . 2 6 . Two cases of a group of strained elements. 6 3 Fig . 2 7 . Parameters for stress i n element n due to s t r a i n i n group of elements m. 6 3 Fig . 2 8 . Influence l i n e s for r a d i a l stresses due to . i n i t i a l r a d i a l s t r a i n of 1%. 6 6 Fi g . 2 9 . Influence l i n e s for r a d i a l stresses due to i n i t i a l tangential, s t r a i n of 1%. 6 7 Fi g . 3 0 . Influence l i n e s for r a d i a l stress at center-l i n e of element 9. 6 8 Fig . 3 1 . Assumed l i n e a r change of moisture content. 7 0 Fi g . 3 2 . Example of Testmember. 7 5 Fi g . 3 3 . Roof beam. 7 8 Table 1 . Terms of the s t i f f n e s s matrix. 1 6 Table 2 . Parameters and stress c o e f f i c i e n t s of the considered beams. 5 6 Table 3- 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 properties ' of the whole beam. 7 1 Table 4. Ef f e c t of v a r i a t i o n i n e l a s t i c properties i n one lam. 7 2 Table 5 . Results of tests on tension perpendicular to grain. 7 6 LIST OP MAJOR SYMBOLS x , y coordinates within the trapezoidal f in i te element. u , v displacement in x and y direct ion. T {a} = {a x , ay, T} stress vector. {e} = (e x , £ y 5 strain vector. a x stress in x direct ion. Gy stress in y direct ion. T shear stress. e x , £y , Y s imilar for strains. {a} , {e} i n i t i a l stress vector and i n i t i a l o o strain vector. °"Y » i T i n i t i a l stresses e X o , ey o , YO i n i t i a l strains. NOTE: x and y direct ion, in the f in i te element, correspond to the tangential and radia l direction in the beam. {6} displacement vector containing the nodal displacements of the f i n i t e element. '{f} force vector containing the nodal forces of the f i n i t e element. {f} Q force vector due to i n i t i a l s t r a i n . [S] stress matrix. [k] s t i f f n e s s matrix of the f i n i t e element. [X] transformation-matrix r e l a t i n g the polar coordinates to the xy coordinates. [k] 1,{f}',{f}^,{6} 1 matrix and vectors of the f i n i t e element r e f e r r i n g to polar coordinates. [K],{F},{F} Q,{A} s i m i l a r matrix and vectors for the structure. {B} structure load vector. [D] e l a s t i c i t y matrix. Ei = E x/(1 -v XyVy X) i n the tangential d i r e c t i o n of the structure. E 2 = E v / ( l - V y X v X y ) i n the r a d i a l d i r e c t i o n of the structure E v = v x y ' E x = v y x - E y modulus of e l a s t i c i t y i n x d i r e c t i o n (tangential). modulus of e l a s t i c i t y i n y d i r e c t i o n ( r a d i a l ) . shear modulus. Poisson's r a t i o ( s t r a i n i n y d i r e c t i o n f o r stress i n x d i r e c t i o n ) . s i m i l a r to v x y . bending moment. shear force. width of the beam. depth of the beam. radius of curvature of lower edge of the beam. distance i n r a d i a l d i r e c t i o n between two adjacent nodes of the f i n i t e element g r i d . angle between two adjacent r a d i i d e s c r i -bing the f i n i t e element g r i d . reference angle of the r a d i i . slope angle of upper s t r a i g t h edge of pitch-cambered beam w.r.t. a horizontal l i n e . a t tangential stress. .a '. r a d i a l stress, r x shear - stress. C"rM> C ^ M stress c o e f f i c i e n t for stresses i n r a d i a l and i n tangential d i r e c t i o n due to moment loading. Cry3 C^Y s i m i l a r due to shear loading. C"rpj Ctp s i m i l a r due to uniform d i s t r i b u t e d pressure. C r r j i j C ^ T s i m i l a r due to temperature gradient. A difference i n %. between f i n i t e element solution and known solution. o-n = C-n-M'r^r maximum r a d i a l stress at centerline of rmax wm bd • . a pitch-cambered beam due to pure bendin °"tens = C r p j y p j ^ p ^ maximum tangential t e n s i l e stress at centerline of a pitch-cambered beam due to pure bending. °"comp = CgjyTkflZ maximum tangential compression stress at centerline of a pitch-cambered beam due to pure bending. CRM' CTM» CCM stress c o e f f i c i e n t s for above maximum stresses. ACKNOWLEDGEMENTS The author wishes to thank his supervisor, Dr. R.F. Hooley, for his invaluable a s s i s t -ance and encouragement during the develop-ment of thi s work. Gratitude i s also ex-pressed to Le Conseil des Arts du Canada for f i n a n c i a l support, and to the U.B.C. Computing Centre for the use of the f a c i l i t i e s . A p r i l 1970 Vancouver, B.C. INTRODUCTION Pitch-cambered glulam beams ( f i g . l ) are used as roof beams i n many buildings. However the stresses within these beams have not been determined.in a rigorous manner. Since the abundance of recorded f a i l u r e s i n North America have not been s a t i s f a c t o r i l y explained, i t seems j u s t i f i e d to investigate the problem deeper. The pitch-cambered glulam beam i s manufactured i n a way that, at any cross section, the grain 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 laminations are curved i n the middle part of the span, but are usually straight towards the supports. For rectangular cross sections, such a beam can be considered to be i n a state of plane stress. In this two dimensional state, the axes of 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 perpendicular to the grain, coincide with polar coordinates. This i s c a l l e d polar anisotropy. In these pitch-cambered beams, there exist t e n s i l e stresses perpendicular to the grain. These stresses are caused by the v a r i a t i o n i n depth of the cross sections and by the curvature of the gravity axes. Present design procedures estimate these stresses from the formula where M i s the applied moment, b the width of the beam, d the depth o f the beam and R a the r a d i u s o f c u r v a t u r e o f the beam a x i s . T h i s f o r m u l a i s f o r i s o t r o p i c m a t e r i a l o f c o n s t a n t c u r v a t u r e , c o n s t a n t depth and c o n s t a n t moment, and so n e g l e c t s many parameters o f the problem. Most f a i l u r e s show a c r a c k p a r a l l e l t o the g r a i n i n t h e l o w e r h a l f o f the c r o s s s e c t i o n , c l o s e t o t h e midd l e , o f the span. At t h e s e s e c t i o n s t h e r e a re l a r g e moments and 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 are almost i n v a r i a b l e d e t e r m i n a t e and s i m p l y s u p p o r t e d . T h i s l e a d s 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 on the s p e c i a l shaped m i d d l e p a r t - m i g h t cause h i g h r a d i a l s t r e s s e s w h i c h b r i n g t h e beam t o f a i l u r e . A l s o , m o i s t u r e change c o u l d c r e a t e r a d i a l s t r e s s e s a d d i t i v e t o t h o s e g e n e r a t e d by the moment. The t h e o r y o f e l a s t i c i t y g i v e s s o l u t i o n s f o r curved-beams o f c o n s t a n t s e c t i o n and c u r v a t u r e . A s o l u t i o n f o r a c u r v e d beam o f Douglas F i r , under pure b e n d i n g , i s shown by N o r r i s ( 1 ) . A g e n e r a l s o l u t i o n , g i v e n by C a r r i e r ( 2 ) , i s d i s c u s s e d and a p p l i e d by F o s c h i ( 3 ) j f o r t h e . c a s e o f c u r v e d Douglas F i r beams o f c o n s t a n t c r o s s s e c t i o n s under combined b e n d i n g moments, a x i a l and s h e a r f o r c e s . F o r p i t c h -cambered c u r v e d 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 a r e no e x a c t s o l u t i o n s a v a i l a b l e . However, t h e r e i s a p u b l i c a t i o n o f S.P. Fox ( 4 ) i n p r e p a r a t i o n , w h i c h w i l l t r e a t t h i s problem. A l s o , as f a r as i t i s known, t h e i n f l u e n c e o f m o i s t u r e change on s t r e s s e s i n beams o f t h i s t ype has not y e t been t r e a t e d . The f i n i t e element method e n a b l e s us today t o f i n d t h e s t r e s s and 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 continuum. I n t h i s p o w e r f u l method the r e a l c o n t i n e o u s body, i n t h i s case a p l a n e 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. These f i n i t e elements are intercon-nected by a discrete number of nodal points. The continuum .is now replaced by a type of structure which can be treated by means of normal s t r u c t u r a l analyses, usually the s t i f f n e s method. The analysis i s completely routine work and can be done by d i g i t a l computers. The r e l a t i v e l y large number of f i n i t e elements required to represent r e a l i s t i c a l l y the continuum, re s u l t s i n a large number of l i n e a r equations. To solve these, the computer i s an important t o o l . For t h i s s p e c i f i c problem an appropriate f i n i t e element was derived. Then a computer program was written to handle c y l i n d r i c a l anisotropic structures with ease. After t e s t i n g the f i n i t e element, pitch-cambered glulam beams, under external loads and under moisture- change, were studied. F i n a l l y recommendations are made to modify the e x i s t i n g design procedures. v I n addition, some laboratory tests on tension perpendicular to the grain were made and compared with published data, to show t h a t . r e v i s i o n was necessary i n this area as w e l l . P i g . l . Pitch-cambered glulam beam. Pig.2. Arrangement of nodal points. 2. DERIVATION OF A TRAPEZOIDAL FINITE ELEMENT IN PLANE STRESS 2.1. GENERAL Our interest is focused on plane stress problems in pitch-cambered glulam beams. The or ig inal plate is replaced by f in i te elements in an arrangement of nodal points on polar coordinates,as shown in 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 or ig inal .structure has cy l indr ica l ortho-tropy but this w i l l be approximated by rectangular ortho-tropy within the range of one element,as shown in f i g . 3 . The smaller the angle 0 , the better w i l l be this appro-ximation. I \ I F i g . 3 - Trapezoidal f in i te element assumed with rectangular orthotropy. The f i n i t e element must take into account i n i t i a l s trains due to change i n moisture content and temperature. I n i t i a l s trains occur i f the single element i s free to expand. I f the element i s constrained, then i n i t i a l s t r e s -ses occur under the action of a change i n moisture content or temperature. Again these strains w i l l be orthotropic on a rectangular system within an element, rather than on a polar system. e l a s t i c continuum i s by means of displacement functions,, as described i n several papers ( 5 * 6 ) . These displacement ^functions specify uniquely the state of s t r a i n within the element. I f , the i n i t i a l strains are included, together with the e l a s t i c properties, the stresses are also defined throughout the f i n i t e element, and, hence along the boundaries. Expressing these stresses i n terms of the nodal displacements w i l l lead to the stress matrix. In matrix notation t h i s i s : A p r a c t i c a l and convenient way to i d e a l i z e the {a} = [S].{6}+{a} o {a} = >- stress vector i n i t i a l stress vector 3x8 stress matrix { 6 } =X 6 1 6 2 6 3 65 6 8 - = displacement vector The boundary stresses are replaced by a set of generalized forces, corresponding to the generalized nodal displacement. These nodal forces can be expressed i n terms of the nodal displacements. This yields the s t i f f -ness matrix of the f i n i t e element or {f} = [k ]- {6}+{f} {f} =< f i f 2 f 3 f* fs fe f 7 fs * = nodal forces [k] 8x8 = s t i f f n e s s matrix of the f i n i t e element {f} °(8xD = nodal forces due to i n i t i a l s t r a i n 2 . 2 . DERIVATION OF THE FINITE ELEMENT The variables used i n the derivation are shown i n f i g . 4 . It indicates, the element parameters, coor-dinate axes, nodal coordinates and the numbering of the nodal forces and displacements. Following the standard procedure two displace-ment functions are assumed as - u = a i + a 2 x + a 3 y + a t, xy v = as + a 6 X + a7y + aexy This assumption trapezoids have for small a rectangle. The nodal displacements {6} i n terms of the constants a i , az, ........... as are obtained by placing the nodal coordinates into the displacement functions. This gives: {6} = [T] • {a} where {a}T= {ai, a 2 , a 3 , a^, a 5 j a 6 , a 7 , a 8} seems reasonable, since the angles 8, a shape close to Pig.4b. Directions of p o s i t i v e nodal displacements and nodal forces. and. [ T ] 1 1 1 1 b 2 b "2 a "2 a 2 h h 0 0 bh 2 bh "2 0 0 1 1 1 1 b 2 b "2 a "2 a 2 h h 0 0 bh 2 bh "2 0 0 For later use, equation 2a has to be solved for {a} as {a> = [ T ] _ 1 { 6 > where [T] \= 0 0 1 2 1 2 0 0 _1 a 1 a 1 2h 1 2h 1 2h 1 2h 1 bh 1 "bh 1 ah 1 "ah 0 0 1 2 1 2 0 0 _1 a 1 a 1 2h 1 2h 1 2h 1 "2h 1 bh 1 "bh 1 ah 1 ah {2b} The t o t a l s t r a i n s are found by appropriate d i f f e r e n t i a t i o n of the displacement f u n c t i o n s as: 9u • ^ 3x where [C]= 3y_ 3u , 8v 3y 8x y = [C] {a} 0 1 0 y 0 0 0 0 0 0 0 0 0 0 1 X 0 0 1 X 0 1 0 y (3.} The i n i t i a l s t r a i n s , due to change i n moisture content or temperature, are assumed independent of x and y w i t h i n the element and are described by the i n i t i a l s t r a i n v e c t o r . Wo'" 'Xo •yo Hookes law now gives {a} = < y {e} - {e> (5> Herein [D] is called the e las t ic i ty matrix and is [D] 1 v 0 0 0 0 G where and G E x / ( 1 - v x y - v y x ) E y / ( 1 " V X y ' V y X ) v x y * E x . = v y x * E y is the shear modulus It should be noted that x and y correspond to tangential and radia l direction so that E x and E y are E ^ a n g e n t ^ a 2 and E r a c ^ a ] _ for the wood considered. The stresses and strains may be expressed now in terms of nodal displacements by substituting equation 2b into equation 3> and equations 2b and 3 into equa-tion 5, to give iz} [C] CT]_ 1{6} {a} = [D]([C][T]"1{6}-{e} o] Equation 7a can be written in the form {a} = [S]{<5}+{a} where [S] = [D][C][T] - I i s the stress matrix and { 0 > o = -[D]{e}o i s the i n i t i a l stress vector. 13 Once the nodal displacements have been d e t e r -mined by. s o l v i n g the s t r u c t u r e s t i f f n e s s e q u a t i o n s , the s t r e s s e s and s t r a i n s at any p o i n t o f the f i n i t e element can be found by the above r e l a t i o n s . The s t r e s s matrix [S] o b t a i n e d from the chosen displacement f u n c t i o n s i s e x p l i c i t l y : [ S ] = [ D ] J_ bh y b H a ah i - i o a ah 0 — + — u 2h +bh •2h +bh 2h bh 2h ah 2h"ah bh 0 1 X 1 X 1 X 2h"bh ~2h +ah ~2h~ah ' y -A+JL 1-2-bh ah ah a ah The p r i n c i p l e o f v i r t u a l displacements i s now used t o d e r i v e the element s t i f f n e s s m atrix [k] as shown below d { 6 } T { f } = / d [ e ] T [ a ] dV where d{6} i s a v i r t u a l displacement of the nodes and d{e} i s the c o r r e s p o n d i n g change i n the s t r a i n s . D i f f e r e n t i a t i o n of .equation 6 y i e l d s a l i n k between d{6} and die} so t h a t d { 6 } T { f } = y^([C][T]~ 1d{6}) T[D]([C][T]- 1{6}-{e} 0)dV v wherein io} i s expressed as i n e q u a t i o n 7a. Expansion of t h i s g i v e s d{S} T{f} = d { 6 } T [ T ] " l T Since d{6} i s a r b i t r a r y y [ C ] T [ D ] C C ] d V [ T ] " 1 { 6 } - [ T ] " T / [ C ] T [ D ] { e } 0 d V {f} = [T ] " l T j T[C] T[D][C]dV[T]" 1{6}-[T ] " i y[C] T[D]{e} odV {8a} This r e l a t i o n can be rewritten as {f} = [k]{5> + {f}, {8b} where the s t i f f n e s s matrix i s [k] = [ T ] - l T o / L C ] T [ D ] [ C ] d V [ T ] - 1 (9> and the nodal forces due to i n i t i a l s t r a i n are {f} 0 = - [ T ] _ l T y [ C ] T [ D ] { e } 0 d V {10} In a plane stress element with a constant thick-ness t the volume dV i s t*dA or t-dx-dy. The s t i f f n e s s matrix [ k ] , established by the previous matrix operation, is.symmetric due to the r e c i p r o c a l theorem so that only the- 36 terms of the lower h a l f need to be stored. Because of element geometric symmetry, only twenty of 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]=t x D -F K - M -E M - L Q G - H N - 0 R H - G 0 - N U Fig.5. S t i f f n e s s matrix [k]. To reduce the lengthy expressions In [k] a set of abbrevatlons for the integr a l s over the range of the trapezoidal element i s introduced as follows: = j T l dxdy = ^/^ydxdy -*-xx = ^yx-dxdy Jyy = ^ y 2 dxdy The in t e g r a l s h' due to symmetry with respect to the Y-axis The twenty terms entering the s t i f f n e s s matrix are given e x p l i c i t e l y without presenting the intermediate calculations i n table No.l. The nodal forces due to i n i t i a l s t r a i n , deter-equation 10, are given as follows: *10 ht, 6"b( a+2b; ( E i e x o + E v £ : y o ) a+ b] ' GYxyo f 2 0 ht, ~6¥ ( a+2b; ( E i £ x o + E v e y o ) a+ b. ' GYxyo f 3 0 b] ( E i e x o + E v e y o ) a+ b ) G y x y o f40 < - = •< b] ) ( E i £ x o + E v e y o ) t ( a+ b > GYx y o f 5 0 a+ b] >(E ve x o+E 2e y o) + h t . a+2b ) GY Xyo f 6 0 a+ b] •ht, a+2b > GY Xyo * ( a+ b] ' ( E v e x o + E 2 e y o ) - | | ( 2 a +-b ' GYxyo **80 - i r ( a+ b. ) ( E v e x o + E 2 e y o ) +g!(2a+ b >GYxyo Ei E 2 G I i Iyy I i Ixx I i I i h Ixx ^ y A l / b 2 h 2 l/4h 2 l / b 2 h 2 B - l / b 2 h 2 l/4h 2 - l / b 2 h 2 C -1/abh 1/abh2 - l / 4 h 2 1/abh2 D 1/abh -1/abh 2 - l / 4 h 2 -1/abh 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 2 -2/a 2h l / a 2 h 2 l/4h 2 l / a 2 h 2 K -1/a 2 2/a 2h - l / a 2 h 2 l/4h 2 - l / a 2 h 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 1/2ah2 P, l/4h 2 l / b 2 h 2 l / b 2 h 2 Q: l/4h 2 - l / b 2 h 2 - l / b 2 h 2 R; - l / 4 h 2 1/abh2 -1/abh 1/abh2 s: T; - l / 4 h 2 l/4h 2 -1/abh 2 l / a 2 h 2 1/a 2 1/abh -2/a 2h -1/abh 2 l / a 2 h 2 l/4h 2 - l / a 2 h 2 -1/a 2 2/a 2h - l / a 2 h 2 2 . 3 . TRANSFORMATION INTO POLAR COORDINATES Before the member.stiffness matrix [k] can be added to generate the s t r u c t u r e s t i f f n e s s matrix [K], the l o c a l coordinates must be transformed to a g l o b a l p o l a r system. The parameters determining the t r a p e z o i d are r e l a t e d to the p o l a r coordinates by: h = (R 2 - R x) cos | a,= 2R 2 s i n | {11} b = 2R 2 s i n | where R , R 2 and B are defined i n f i g . 6 . A f t e r c a l c u l a t i o n of the s t i f f n e s s matrix and the nodal forces due to 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 re 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 to convert the r e s u l t i n t o p o l a r coordinates. The new set of axes, w i t h the p o s i t i v e d i r e c t i o n s of the nodal forces 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 cosines of the p o l a r set r e f e r r e d to the 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 , con-s t i t u t e a t r a n s f o r m a t i o n matrix [X] as: F i g . 6 . P a r a m e t e r s a n d c o o r d i n a t e s d e t e r m i n i n g t h e t r a p e z o i d . [ X ] = 3 0 0 0 s i n f 0 0 0 3 c o s | 0 0 . 3 • s i n | 0 0 0 0 B c o s | 0 0 -sinf-0 - s i n | 0 0 C O S ; 0 0 s i n f 3 C O S T J -0 0 s i n f 0 .0 0 C O S ; 0 0 s i n f 0 0 3 COS75-0 0 - s i n f 0 0 3 COSTJ The components of the vectors {f}' and {f}' are rel a t e d to o the components referred to the rectangular axes by: {f}' = [X] {f} {f}' = [X] {f} O o { 1 2 } and s i m i l a r l y f o r the components of the nodal displacements {«}' = [X] {6} The member matrix i n polar coordinates i s then [ k ] ' = [X] [k] [X]- { 1 3 } so that {f}' - {f}' = [ k ] \ {6}\ {14} The matrix [k]' i s e a s i l y obtained from [,k] by replacing A by A', B by B f etc. i n f i g . 5 where A' B' C D' E ' F' G' H' I' K' L 1 M' N' 0' P' Q' R' S 1  T i U' A B C D E P G H I K L M N 0 P Q R S T U -2E 2P G-L -H+M A-P -B-Q -C-R D-S 2N -20 C+R -D+S -I+T K+U 2E 2P G-L H-M m -2N -20 P -Q -R S -E -F L M T -U G H -.N -0 A -B -C D I -K 2 6 cos z ^ B 2 cos s i n s i n 2 | 3. COMPUTER SOLUTION PROCEDURE 3.1. BASIC STEPS IN THE PROCEDURE The computer program.developed;Is based on the displacement method using a structure s t i f f n e s s matrix. The program, written for the IBM 7 0 4 4 , i s given i n the appendix. The setup i s shown i n the s i m p l i f i e d flow d i a -gram of ( 3 - 2 ) . The basic steps i n the procedure are: a) The pitch-cambered beam i s subdivided by r a d i a l l i n e s and many c i r c u l a r arcs. The area enclosed by the r a d i a l lines and the arcs, i s approximated by e q u i l a t e r a l trapezoids connected at the corners onl y . These trapezoidal elements r e t a i n the appro-p r i a t e material and geometric properties of the structure. An i n c l i n e d , straight upper edge cannot be repre sented exactly i n polar coordinates. Therefore i t must be approximated by varying the boundary step-wise. The respective thickness given to these boun dary elements i s i n proportion to the plate area they have to represent. The respective thickness o f an element i s taken as • A . £ _ structure resD A *\ element structure Fig". 8. Approximation of the upper straight edge. The area A s t r u c t u r e i s hatched i n f i g . 8 . A element i s the area enclosed by the arcs and the r a d i a l l i n e s . This stepwise approximation i s checked . i n chapter four and shown to be s u f f i c i e n t for the l i m i t s of accuracy required herein. b) The elements and the nodes are numbered. c ) The boundary conditions are simulated by locking or releasing the appropriate nodal displacements. The unlocked nodal displacements, i . e . the compo-nents of the structure displacement vector, are numbered. d) The element s t i f f n e s s matrix [k] i s generated and transformed into polar coordinates [ k ] ' . e) The structure s t i f f n e s s matrix [K] i s generated from [ k ] ' by using the code number technique. This tech-nique automatically eliminates the rows and columns of the structure s t i f f n e s s matrix, corresponding to the restrained j o i n t s . f) The structure load vector {B} i s generated. The loads acting on the beam are applied to the model at the nodal points only. Therefore-, the d i s t r i b u t e d external stresses have to be replaced by s t a t i c a l l y equivalent nodal forces to y i e l d the terms of the structure force vector {P}. The nodal forces due to i n i t i a l s t r a i n are given by equation 10. The code number technique i s used to place them into a structure force vector {P} Q, due to i n i t i a l s t r a i n . The structure load vector {B} i s then {B} = {F} - {F} 0 g) Cholesky's method i s used to solve the system of simultaneous l i n e a r equations i n the s t i f f n e s s r e l a t i o n {B} = [K]{A} h) Knowing the structure displacement vector {A}, the element displacement vector { 6 } ' i s found by the. code number technique and transformed into { 6 } . i ) The stresses within the element are found by equation 7b. They are calculated at the center of the elements, since there the directions of the XY coordinates coincide with the polar coordinates. 3.2. SIMPLIFIED FLOW DIAGRAM Start Read and Print the general geometric structure data: Max. No. of AR, Values of AR, No. of r a d i a l groups, No. of elements i n each group, innermost radius R and angle $ of each group. Divide the structure into the elements. Number the elements and the nodes. Assign the nodal numbers to the elements [Done i n subroutine DIVIDE]. Generate for each element the appropriate data. [Done i n subroutine DISTRI] Read: The thickness of the normal elements No. of boundary elements. Thickness of the boundary elements. V continued : _\k • Print the: No. of elements. No. of boundary elements. No. of nodes. Geometric data of a l l elements Nodal No.s of the elements. V Read and Pri n t : The data of the restrained nodes. Generate the nodal displacement numbers w  Print the nodal displacement numbers Generate the: Code numbers. [using subroutine CODE] Bandwidth. V P r i n t the: Code numbers Bandwidth. continued : ± Read and Print the: Moduli of e l a s t i c i t y of the structure, i . e . of the normal elements. No. of s p e c i a l elements. Moduli of e l a s t i c i t y of 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 into [ k ] ' . [Done i n subroutine SMEM] \ f Place the element s t i the structure s t i ffness matrix [k]' into ffness matrix [K]. Is i t done for a l l elements? YES Y N O A Read the number of loadcases V Read the type of loading. V continued Are there s t a t i c a l l y equivalent nodal forces? N O YES JL Read the nodal forces.P. JL Is there i n i t i a l strain? YES Read the: I n i t i a l strains of the structure, No. of s p e c i a l strained elements, I n i t i a l s t r a i n of the s p e c i a l strained elements. N O Generate the nodal forces {f} Q due to i n i t i a l s t r a i n . Transform them into polar coordinates {f}' o [Done i n subroutine EPSINI] f B u i l d up the load,vector [B]. continued V Solve f o r the displacement vector {A}. [Done i n UBC-subroutine BAND] f F i l l the element displacement vector { 6 } ' Transform { 6 } ' into' { 6 } . Calculate the element stresses. [Done with subroutine SIGMA] 3L. Print stresses. Is i t done for a l l elements? YES IL Are a l l loadcases considered? YES JL More structures? A NO YES NO End, 29. 4. CHECKING OF THE FINITE ELEMENT WITH VARIOUS PROBLEMS I t i s the o b j e c t of t h i s chapter to see i f the f i n i t e element p r e v i o u s l y developed w i l l p r e d i c t , w i t h s u f f i c i e n t accuracy, the s t r e s s d i s t r i b u t i o n i n pitch-cambered beams. To do t h i s , f i n i t e element so-l u t i o n s are compared a g a i n s t e l a s t i c i t y s o l u t i o n s f o r v a r i o u s problems. A l l percentage e r r o r s g i v e n i n comparison w i t h known s o l u t i o n s are c a l c u l a t e d c o r r e s p o n d i n g to the f o l l o w i n g r e l a t i o n based on absolute values l e r r o r % = I c a l c u l a t e d value) - | known value [ ^ -^QQ |known v a l u e | 4.1. ISOTROPIC CIRCULAR BEAMS AND RINGS The m a t e r i a l i n the f i n i t e element s o l u t i o n i s determined by the [D]-matrix. To r e p r e s e n t i s o t r o p i c m a t e r i a l [D] has to be the corres p o n d i n g e l a s t i c i t y m a t r i x , i . e . or E> E v 0 1 V 0 CD] = E V E 2 0 E V 1 0 0 0 G 0 0 1-v 2 E 1 ^ > E v VE 1-V and G = 2(l+v) EXAMPLE 1 ISOTROPIC CIRCULAR BEAM, SHEAR LOADING P i g . 9 a shows p r o p e r t i e s o f a c u r v e d beam u n d e r s h e a r l o a 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 s o l u t i o n i s g i v e n i n T imoshenko ( 5 ) on page 7 3 - P i g . 9 c g i v e s a p l o t o f t h e t a n g e n t i a l s t r e s s e s n e a r t h e i n s i d e edge. H e r e i n , two f i n i t e e l e m e n t s o l u t i o n s a r e shown: One f o r a c o n -s t a n t d i s t r i b u t i o n o f s h e a r l o a d a t <J> = TT/2, and t h e o t h e r f o r t h e s h e a r d i s t r i b u t i o n a c c o r d i n g t o t h e e l a s t i c i t y s o l u t i o n . I n f i g . 9 b a r e shown t a n g e n t i a l and r a d i a l s t r e s s e s a t <j> = 3/2. The two f i n i t e e l e m e n t s o l u t i o n s p r o d u c e d t h e same r e s u l t a t t h i s s e c t i o n remote f r o m t h e a p p l i e d l o a d . E v e n f o r t h i s c o a r s e g r i d s i z e , t h e e l e m e n t p r e d i c t s t h e s t r e s s d i s t r i b u t i o n w i t h s u f f i c i e n t a c c u r a c y . — > F i g .9a. Isotropic c ircular beam, shear loading.. F i g .9c. 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 MOMENT LOADING. Fig. 1 0 shows the p r o p e r t i e s and s t r e s s e s o f a curved beam under pure bending 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 g i v e n i n Timoshenko (5) on page 6 l . I t i s the same beam as i n example 1, except t h a t the l o a d i s pure moment. The nodal f o r c e s were d i s t r i b u t e d l i n e a r l y at <j> = TT/2 to r e p r e s e n t the a p p l i e d moment. Again, the element p r e d i c t s the s t r e s s e s w i t h s u f f i c i e n t accuracy. E X A M P L E 3 THICK I S O T R O P I C C Y L I N D E R UNDER UNIFORM E X T E R N A L P R E S S U R E •• The e l a s t i c i t y solution i s given i n Wang (6) on page 5 4 . Due to r a d i a l symmetry the problem i s solved with one r a d i a l group,(fig.11). However, to check the program, a quarter of the cylinder has been used too, with the same element s i z e , and the same resu l t was obtained. The depth of the beam i s divided into three elements and the group angle B i s taken as 6 ° . The stresses are calculated for a r a t i o ^ = . 3 . The r a d i a l nodal forces, due to external pressure, are found by the formula - _ p • (R + d) • B  f i ~ 2 The remarkable accuracy found here with only three ele-ments i s due, of course, t o the uniform stress set up within the element. However, previous examples have con-tained a l l modes of deformation and s t i l l showed good accuracy. 3 6 . EXAMPLE 4 THICK ISOTROPIC CYLINDER UNDER UNIFORM TEMPERATURE GRADIENT The e l a s t i c i t y solution i s developed according to Timoshenko (5) on page 407. By using the boundary conditions crr = 0 at r = a (=R) and at r = b (=R+d) and by introducing an assumed l i n e a r d i s t r i b u t i o n of the temperature T = T Q r , the following e l a s t i c i t y solution for the stresses i s obtained aET, 3r"^ ( r2 - a 2 ) ( ^ f z - ) - ( r 3 - a 3 ) aET ( r 2 + a 2 ) ( j ^ f p - ) - (2r 3+a 3) ] In the f i n i t e element solution, the respective i n i t i a l s t r a i n i s given by the formula r ^ aT E aT E • 0 where Tg i s the temperature i n the center of the element as given by T E = T q (R + | (n - 1/2)) N = t o t a l number of elements r a d i a l l y n = element number from in s i d e . Again, due to r a d i a l symmetry, only one r a d i a l group i s required to solve the problem. By taking f i v e elements i n the group, as shown i n f i g . 1 2 , the error i n the maximum ar i s With 10 elements i n the group th i s i s reduced to 1%. F i g . 1 2 . T h i c k i s o t r o p i c c y l i n d e r , u n i f o r m t e m p e r a t u r e g r a d i e n t . 4 . 2 . O R T H O T R O P I C C I R C U L A R BEAMS The remaining examples consider orthotropic material and the [D]-matrix i s given with the example EXAMPLE 5 CIRCULAR ORTHOTROPIC BEAM UNDER PURE MOMENT Fi g . 1 3 shows properties of a curved ortho-t r o p i c beam of Douglas F i r under pure moment for which the e l a s t i c i t y s olution i s given by Foschi ( 3 ) . Foschi used the following e l a s t i c properties of Douglas F i r given by Hearmon ( 7 ) . 2 . 2 7 6 5 ' • 1 0 6 p s i E y 0 . 1 5 3 7 ' • 1 0 6 p s i vxy 0 . 2 9 0 vyx 0 . 0 2 0 G 0 . 1 2 7 6 ' • 1 0 6 p s i From these E x , E 2 , E v and G of the CD] tained as E i 2 . 2 9 • 1 0 s p s i E2 = 0 . 1 5 5 • 1 0 6 p s i E v 0.045 • 1 0 6 p s i G 0 . 1 2 8 • 1 0 s p s i To demonstrate the convergence trend by reducing the element s i z e , the depth of the beam i s divided once into three, once into f i v e and once into ten elements, while 3 i s kept constant. The nodal forces at the ends were d i s t r i b u t e d l i n e a r l y to represent the applied moments. The^ r e s u l t obtained with ten elements comes very close to the exact sol u t i o n . Fig.13 shows the three f i n i t e element TT approximations and the exact solution at <J> = g". Also a part of the i s o t r o p i c case i s plotted for 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 In this 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 is represented by a set of-equal nodal forces at the ends of the beam. In f i g.l4 the f in i te element approximation with ten elements over the depth is compared to Foschi's exact solution. 4 5 . 4 . 3 . .PITCH-CAMBERED ORTHOTROPIC BEAM A favourable opportunity to check the f in i te element is offered by Fox. In his PROGRESS REPORT No. l ( 4 ) , on DOUBLE-TAPERED PITCHED BEAMS, a theoretical solution for the problem is obtained by using Fourier series and a point-matching method to satisfy the upper boundary conditions at several points. The other boun-dary conditions are sat isf ied everywhere. This theory has been veri f ied by testing experimentally, a glued laminated beam with the geometric properties given in f i g . 1 5 . This test beam under pure moment is chosen as an example to compare the two theoretical solutions. The stress distribution given by S.P. Fox is based on the material properties " Ex " 2 . 4 5 • 1 0 6 psi v - 0 . 1 6 9 ' • 1 0 6 psi V = xy 0 . 0 3 3 v y x = 0 . 4 7 3 G 0 . 1 3 7 .' • 1 0 6 psi The equivalent [D]-matrix in the f in i te element solution has the terms = 2.46 1 0 6 psi = 0 . 1 6 9 * 1 0 6 psi E = 0 . 0 7 9 9 * 1 0 6 psi G = 0 . 1 3 7 ' 1 0 6 psi The f i n i t e element g r i d , -used to represent the 0 beam, i s shown i n f i g . l 6 . The boundary elements are approx-imated as desc r i b e d i n chapter 3 . 1 . page 2 1 . The s t r e s s e s at c e n t e r l i n e are obtained by qua 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 are c l o s e together. The d i f f e r e n c e i n maxi-mum 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 $ . Laminations 1 / 2 " t h i c k Beam 4" wide 9 = 1 6.-646° a = 1 8 . 4 3 0 ° (tana = . 3 3 3 ) symmetric about £ F i g . 1 5 . Geometric dimension of the t e s t e d beam. 4 7 . P i g . 1 7 . Stresses at c e n t e r l i n e of the t e s t e d beam. The chosen length 9.-.R i n the f i n i t e element approximation should be s u f f i c i e n t by St. Venant's p r i n -c i p l e to produce accurate r e s u l t s at the centerline. To check t h i s , the beam under the same loading condition has been extended to a length of 1.5 8R. The change i n maxi-mum stresses at centerline was less than 1/2%, well with-i n the desired accuracy. In the extended case, where a part of the straight beam has to be used to get enough length, this straight part has to be simulated by a curved beam with a large radius R = 10 R, since the computer program i s written for curved beams. A change from R = 10 R to R = 20 R varies the maxi-mum stresses at centerline not more than .07%. The l a s t thing to be checked i s the influence of the stepwise approximation of the upper boundary. To do t h i s , stresses i n thi s beam are calculated with upper boundary elements of zero thickness and also with f u l l thickness. The maximum tangential stress changes by 8% between those two extremes, while the maximum r a d i a l stress changes by 5%. The approximation, using a thickness bet-ween the two extremes, can produce an error w.r.t. the o r i g i n a l beam which i s only a part of the t o t a l difference. Since this error has the same magnitude as i n the examples 1 to 6 for curved beams, where no stepwise approximation i s required, the desired accuracy i s obtained. The actual accuracy i s well within 5%> quite s u f f i c i e n t for the prac-t i c a l purpose herein. The good re s u l t s i n the examples of thi s chapter, obtained by the f i n i t e element so l u t i o n , j u s t i f y the a p p l i -cations made i n the next chapter. There the parameters d, R and a i n the pitch-cambered beam are varied to cover the p r a c t i c a l range. 5 . STRESSES IN PITCH-CAMBERED BEAMS PREAMBLE In the following are shown stresses due to moments, shear loads, change i n moisture content and change i n e l a s t i c properties, i n an attempt to evaluate the factors which might create high stresses. The beams have been divided into 1 0 or more r a d i a l groups, with about 1 5 elements at centerline, to provide good r e s u l t s . The results are presented i n terms of the parameters d/R and tana. F i g . 1 8 shows the parameters and the g r i d of an example which i s used several times i n this chapter to demonstrate various e f f e c t s . I f not mentioned s p e c i f i c a l l y , the following e l a s t i c i t y constants are used E t a n E y = E t a n / 2 0 G E t a n / 1 5 vxy . 0 1 8 5 v y x = . 3 7 0 This gives E i E v 0 [D] = E v E a 0 = 0 0 G 1 E /E x 0 E/E 1 E 2 / E 1 0 0 0 G E a = 1 . 0 1 8 5 0 . 0 1 8 5 . 0 5 0 0 0 0 0 . 6 6 2 1 where E1 = E x / ( l - v X y - v y x ) can have any value, as this constant factor does not effect the distribution or magni-tude of the stresses, except in the case of moisture chang where E1 i s taken as 1810. k s i . This is reasonable for Douglas F i r . The effect bf ratio changes in this [D]-matrix is checked in section 5.4. It should be noted that x and y correspond with the tangential and radia l direction of the beam, that is with the grain and perpendicular to the grain direction of the. wood. 52. 5.1. STRESSES FROM MOMENTS The peak at midspan of the pitch-cambered beam Induces a stress concentration. This stress concentration i s l o c a l by St. Venant's p r i n c i p l e . At some distance from the centerline where the beam i s st r a i g h t , uniform tapered or curved (see fig.19), the e x i s t i n g formulas are v a l i d . The stress concentration at centerline i n beams under pure bending moments, i s investigated i n this section. To cover the p r a c t i c a l range a group of 26 beams, with d/R varying from ."01 to .8 and tana from .1 to .6, has been chosen for the f i n i t e element solu t i o n . A sketch of four beams i n the upper and lower parameter range i s given i n fig.20 to aid i n v i s u a l i s i n g the parameters. A l l beams showed at centerline the c h a r a c t e r i s t i c stress d i s t r i b u t i o n sketched i n fig.21 with zero r a d i a l and tangential stress at the top of the beam. For cross sections adjacent to the centerline the stresses at the top edge are d i f f e r e n t from zero. The stress concentration produces a l o c a l maximum i n the r a d i a l stresses. The maximum and minimum stresses at centerline, as indicated i n fig.21, are rela t e d to the value M = 6 M Z . , . bd2" centerline by the dimensionless c o e f f i c i e n t s Cjyyj, dpjyi and CQJ^ as follows: 54. P i g . 2 1 . 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. maximum r a d i a l stress a r m a x = Sd^ maximum tangential _ _ 6 M t e n s i l e stress <*tens - ^TM bd7". maximum tangential _ 6 M compression stress acomp " CCM bd2" Due to symmetry,, only one h a l f of the beam had •to.be solved by f i n i t e elements and a quadratic interpo-l a t i o n had to be used to get the values at centerline, since the computer program produces stresses at the centers of the elements. Table 2 shows for each considered beam the para-meters d/R and tana, the number of elements NE at center-l i n e , the number of r a d i a l groups NG and the stress c o e f f i -cients Gjyyj, Crpjyj and G Q ^ . A graphical presentation of the r a d i a l stress c o e f f i c i e n t s i s given i n f i g . 2 2 . The c o e f f i c i e n t s are p l o t t e d against (d/R). A set of curves i s obtained by connecting the points of constant tana. The c o e f f i c i e n t s GRM °f r a d i a l stresses can r e a d i l y be taken out of t h i s graph f o r any parameters. The values f o r r a d i a l stresses obtained by the c l a s s i c a l formula f o r curved beams rt - i M _ 1 d 6 M °r 2 bd(R+d/2) ¥*(R+d/2) bcF are" plotted 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 stress c o e f f i c i e n t s .of the considered beams No tana d/R NG .NE CTM C.CM CRM 1 - 0 . 0 . 5 9 1 0 1 . 0 6 7 - 1 . 3 2 3 . 0 5 9 9 2 0 . 1 . 0 1 1 2 2 2 1 . 1 5 1 - 0 . 7 1 0 .0243 3 0 . 1 . 0 5 1 5 1 6 1.124 - 0 . 7 5 1 . 0 2 5 3 4 0 . 1 . 1 1 6 1 9 1 . 1 2 7 - 0 . 7 6 3 . 0 3 0 2 5 - 0 . 2 . 0 1 1 1 2 3 1 . 3 7 1 - 0 . 6 8 7 ' .0455 . 6 0 . 2 . 0 5 1 0 1 5 1 . 3 2 4 - 0 . 7 0 2 5 . 0 4 6 3 7 0 . 2 . 1 1 0 1 6 1 . 2 8 2 6 - 0 . 7 1 6 5 . 0 4 8 7 8 0 . 2 . 2 1 3 1 5 1 . 2 7 4 3 - 0 . 7 3 2 0 . 0 6 0 1 9 0 . 2 . 4 14 14- 1 . 3 8 6 9 - 0 . 7 2 8 1 . 0 8 2 5 10 0 . 3 . 1 1 0 1 5 1 . 5 5 3 8 -0.7341 . 0 7 3 1 1 1 0 . 3 . 2 1 0 1 5 1 . 4 8 8 1 - 0 . 7 4 3 4 . 0 8 1 7 1 2 0 . 3 . 4 1 0 1 3 1 . 5 2 4 1 - 0 . 7 2 8 1 . 1 0 5 9 13 0 . 3 . 6 1 2 1 3 1 . 6 3 4 3 -O . 7 1 8 2 . 1 2 5 2 14 0 . 4 . 0 1 1 6 2 0 2.042 - 0 . 8 1 8 . 0 9 4 6 1 5 0 . 4 . 1 1 0 1 1 1 . 9 1 7 4 - 0 . 7 7 7 5 . 1 0 0 0 1 6 0 . 4 . 2 1 0 14 1 . 7 8 0 5 - 0 . 7 9 3 7 . 1 0 7 6 1 7 0 . 4 . 4 1 6 1 5 1 . 7 2 4 4 - 0 . 7 7 9 4 . 1 3 1 2 1 8 • M . 6 1 1 1 6 1 . 8 0 1 5 -O . 7 6 7 2 . 1 5 5 0 1 9 0 . 5 . 2 1 5 1 3 2 . 2 4 1 8 - 0 . 8 9 6 4 . 1 4 3 8 20 0 . 5 . 4 . 1 3 14 2 . 0 2 0 2 - 0 . 8 5 6 9 . 1 6 0 8 21 0 . 5 . 6 1 5 1 5 2 . 0 2 0 2 - 0 . 8 2 7 1 . 1 8 5 1 2 2 0 . 5 . 8 1 4 1 5 2 . 1 0 0 1 - 0 . 7 9 9 9 . 2 0 7 1 2 3 0 . 6 . 0 1 1 6 2 0 3 . 0 1 6 - 1 . 0 6 6 . 1 6 0 0 24 0 . 6 . 2 1 5 1 5 2 . 6 7 0 2 - 0 . 9 8 7 0 . 1 7 8 8 25 0 . 6 . 4 1 3 1 3 2 . 4 2 1 5 - 0 . 9 5 0 2 - 1 9 7 5 2 6 0 . 6 . 6 1 3 1 3 2 . 3 2 0 8 - 0 . 9 0 3 0 . 2 1 8 9 . 2 7 - 0 . 6 . 8 14 1 2 2 . 3 5 6 1 - 0 . 8 7 6 4 . 2 4 3 1 .30i P i g . 2 2 . Stress c o e f f i c i e n t C R M for maximum r a d i a l stress at centerline due to pure bending. .4 .2 ;0I .05 .1 .2 .3 .4 . 5 .6 .7 d .8 R F i g . 23- S t r e s s c o e f f i c i e n t s . C T M a n d C C M f o r t a n g e n t i a l s t r e s s e s a t c e n t e r l i n e d u e t o p u r e b e n d i n g . higher stresses than given by this c l a s s i c a l formula. Even i f the depth at the point of tangency i s used i n the formula, the stresses obtained by f i n i t e elements are s t i l l remarkably higher for small d/R's. The calculated stress c o e f f i c i e n t s for the tangential stresses are plotted i n f i g . 2 3 . The points are connected by hand, to produce curves from which the stress c o e f f i c i e n t s for tangential stresses can be obtained for any parameter. For the considered beams with deepest cross sections at centerline, the s i g n i -f i c a n t tangential stresses for design do not occur at centerline cross sections. From an inspection of the stresses i n each element i n the computer output, i t was observed that the tangential bottom f i b r e s t r e s s , due to pure bending, was the smallest at the center-l i n e . Even though dj.^ i s greater than 1 , the extra depth at center keeps the tangential stress smaller than i n adjacent cross sections. 5.2. STRESSES FROM SHEAR LOAD A pure shear load at centerline i s produced by the loading conditions shown i n fig.24. Due to antimetry, there i s , at centerline, no tangential stress (a^) and also there i s no r a d i a l s t r a i n ( e r ) . Introducing a t = e r = 0 into Hooke's law at ' V t + E v e r a r = E v e t + E 2 e r i t can be seen that the r a d i a l stress•must be zero. To check the magnitude of the centerline shear stresses, the beam with tana = .4 and d/R = .2 of f i g . l 8 i s investigated. Fig.25 shows the shear stress d i s t r i -bution at centerline. The maximum shear stress at center-l i n e i s : x = ( | L ) max 2 bd Even though the maximum shear stress at centerline i s 40$ more than normal, th i s w i l l not govern design since the depth here i s larger than at the supports and the maximum shear force i s about one quarter the support shear. P i g . 25.-' Shear* s t r e s s e s at c e n t e r l i n e from pure shear load. 5 . 3 . STRESSES DUE TO CHANGE IN MOISTURE CONTENT Change in moisture content causes the i n i t i a l strains e r o and e t 0 , i . e . {e} = -I 0 If the distribution of strains {e) 0 satisf ies the compa-t i b i l i t y equation 9r r I F x 2 z t ~ V T F 2" 36 30 r 3r36 then no stresses are produced, i f the structure Is suppor-ted in a determinate way. With {e} = {e}' the equation is 3r* - *~ r 3r {2£ t o X o p 32e 30 = 0 This equation is sat isf ied i f £ r o and e ^- 0 are indepen-dent of r and 0, or i f e to = I / 2 e r o a n d varies l inearly with r and 0. The latter case is unlikely. The f i r s t case shows that a uniform moisture change causes no stres-ses, but nonuniform moisture change w i l l induce stresses. Since in a pitch-cambered beam i t is not obvious what strain distr ibution {e} cfauses maximum stresses, a o * closer look into a representative example is indicated. The beam with tana = .4 and d/R = .2 shown in fig.l8 is investigated. As shown in the figure, 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 total of 14 pairs of load cases were run. For each pair one group of elements of constant radius were given 6 V 0> a. f i r s t and then In f i g . 2 6 two d i f f e r e n t cases of a group^of strained elements are shown. Group a has elements along the whole beam, while group b reaches the top edge before the support. The stresses of a l l these load cases can be com-bined to represent other states of i n i t i a l s t r a i n by super-p o s i t i o n . S i m i l a r l y influence l i n e s can be obtained. The l a t t e r was done for r a d i a l stresses. As the r a d i a l stresses at centerline are the largest, influence l i n e s are of i n t e r e s t there. The stresses Ao"r at the center of an element n adjacent to the centerline, due to s t r a i n change i n one group of elements at l e v e l y, i s a function of seven parameters. Aa r = f[Ay, y, d, R, a , e , E] o The parameters are shown i n f i g . 2 7 . A dimensionless re-presentation i s i n the form E 1 L d 3 d ' R 3 o' a J which requires only f i v e parameters. Here ^ and tana are set at .2 and .4 by the chosen geometry of the- example; the modulus E i s fixed as given on page 50; the r a d i a l or tangential s t r a i n e Q i s taken as .01. Since there exists a l i n e a r system, the stress varies d i r e c t l y with e Q so that results for other s t r a i n magnitudes are i n dir e c t r a t i o to the one chosen. The stresses can now be written as . AO- = f [ £ ] u o r 1 1 d ' d J since the other three parameters are now constant. i t can be assumed that that Ay g^ d ; and °r • • - . / 8 . C 5 ) ^ ' O For an influence l i n e at one point, due to i n i -t i a l s t r a i n i n the element groups 1 to 14, the stresses at this point are required from a l l load cases. The sum of the stresses of a l l load cases i s zero, since i t represents uniform s t r a i n over the whole beam. From these stresses the values g are calculated and plotted at the correspon-ding l e v e l (y/d) of the strained group. Now the t o t a l area, enclosed by an influence l i n e , corresponding to uni-form s t r a i n over the whole beam, i s zero. Influence lines, were calculated for the centers of a l l 14 elements at centerline. A few of them are shown i n fig.28 and fig.29. • It should be noted that these plots are for r a d i a l or tangential s t r a i n e =.01, so that r e s u l t s for other strains can be obtained by d i r e c t r a t i o . As well they are calculated using the modulus E'l = 1810 k s i of page 51 so that results for other E are obtained by d i r e c t r a t i o again. The influence l i n e s for r a d i a l stress snow that the largest possible stress, due to pure r a d i a l or tangential s t r a i n , i s obtained at the center of element 9, however, with 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 for each. The two influence l i n e s for t h i s point are shown i n figures 30a and 30b together with the two s t r a i n loads providing maximum av. From the figure i t can be seen, that each s t r a i n load produces a considerable stress only i n one influence l i n e , while the other influence l i n e does not contribute. Because ^ i s small, Aa r varies l i n e a r l y with Ay so F i g . 2 8 . Influence lines for r a d i a l stresses due to i n i t i a l r a d i a l strain-of 1%. top 2000 0 2000 0 2 0 0 0 -1000 0 2000 - 2000 0 2000 y N= 3 N =6 N =9 N= 12 g ( d " a at £ of element N Pig.29. In f l u e n c e l i n e s f o r r a d i a l stresses, due to i n i t i a l t a n g e n t i a l s t r a i n , o f 1%. top •o O O •a o o c o (A W C E o pos. Area !,82lpsX%< bottom I 1 I L J 1 I. -80 - 4 0 40 (a) STRESS <TX DUE TO INITIAL RADIAL STRAIN e OF 1% ro •to bisture change -2000 0 2000 (b) STRESS Of DUE TO INITIAL (c) STRESS o~r DUE TO MOISTURE TANGENTIAL STRAIN € f 0 OF 1% CHANGE OF I % Pig.30. Influence lines for r a d i a l stress at centerline of element 9. OO . . • . 69. These two influence lines for radia l stress due to i n i t i a l strain can be combined to one for radia l stress due to moisture change i f the relat ion between moisture 'content and shrinkage of the wood is known. Wood shrinks most in the direction of the annual growth rings (tangentially to the tree) , somewhat less across these rings (radia l ly ) , and very l i t t l e along the grain ( longitudinally) . The Wood Handbook ( 8 ) , f i g . 7 1 on page 3 1 9 , gives for Douglas F i r a f lat grain (tangential) shrinkage of about 1 . 5 $ when the moisture changes from 10% to 15%. F. Kollmann in TECHNOLOGIE DES HOL'ZES ( 9 ) gives the longitudinal strain due to moisture change as approxi-mately 1 / 2 3 of the f lat grain shrinkage. Herewith, the following i n i t i a l strains, corres-ponding to 1% moisture change are assumed to be reasonable e r o = ( 1 . 5 $ ) / 5 = .3% for., a 1% moisture change e to = ( e r o ) / 2 3 = (.3%)/23 = . 0 1 3 $ for a 1% moisture change. With these relat ions, the influence l ine , due to moisture change, as shown in f i g . 3 0 c , is obtained by- super-position of the adjusted influence lines due to strain change from figures 3 0 a and 3 0 b . With the maximum area of 1 0 . 1 psi/$ m.c. under the influence l ine , a moisture change of 5%, distributed as shown in f i g . 3 0 b , produces the following maximum stress a r = 5 * 1 0 . 1 = 5 0 . 5 ps i This w i l l be tension perpendicular to grain when the moisture in the central region decreases. However, i t has to be noted that these influence l i n e s represent moisture or s t r a i n d i s t r i b u t i o n s , as shown i n f i g . 2 6 , which means the moisture contents within the groups of elements are constant. This occurs, for example, i f a d i f f e r e n t amount of moisture content i s present i n the laminations at the time of glueing, and subsequent drying shrinkage produces stresses. The influence lines cannot be used for a moisture d i s t r i b u t i o n as shown i n f i g . 3 1 j since the moisture content there varies along one group of elements. Influence lines could be made by an analogous procedure to treat t h i s type of s t r a i n d i s t r i b u t i o n s . Instead the case was solved by applying to a l l elements the corresponding i n i t i a l s t r a i n as shown i n f i g . 31. The maximum r a d i a l stress o"r at center l i n e was then a r = 4 6 . 8 p s i tension For the increase i n moisture content the r a d i a l i n i t i a l s t r a i n e r o contributes the most, namely 4 5 - 5 p s i , while e-to with 1.3 p s i produces almost no r a d i a l stress i n thi s case. I 5% Fig.31. Assumed l i n e a r change of moisture content. 5 . 4 . VARIATION IN ELASTIC PROPERTIES Up to here, the [D]-matrix given on page 5 0 , has been used as the 'basic'matrix for the investigations. •Table 3 shows how the stresses due to pure moment are affected 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 of v a r i a t i o n i n e l a s t i c properties of the whole beam. D-matrix E 2 / E 2 b a s i c E v / E v b a s i c G/Gbasic Aa i n •? I tang e n t i a l r a d i a l tens. comp. basic 1 1 1 - - -E 2 varied 2 1 • • 1 - 2 . 4 + 4 . 9 E v varied 1 2 1 - 0 . 9 - 1 . 3 + 1 . 1 G varied 1 1 1 / 2 + 1 0 . 1 + 1 3 . 0 - 1 1 . 7 A doubling of E 2 or E v has a small e f f e c t , but the stresses are somewhat sensitive to a change of G. However, the amount of change i s s t i l l t o l e r a b l e . More l i k e l y the e l a s t i c properties may vary from lam to lam. This influence has been checked for bending moment, by assuming a v a r i a t i o n i n one modulus of 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 stress, due to changes i n the e l a s t i c pro-perties of the bottom layer of elements with high tangential t e n s i l e stresses (see fig.26); of the seventh layer with high r a d i a l stresses; of the tenth layer with high tan-g e n t i a l compressive stress. 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 properties i n one lam. Change of modulus i n one lam to Aa r i n % Var. i n layer 1 Var. i n layer 1 Var. i n layer 1 0 - 8.82 - 4.58. - 2.14 0 . 0 0 + 0 . 7 2 + 0.84 E v -»• 2E V + 0 . 1 6 + 0 . 6 1 . 0 . 0 0 G *1/2G - 0.22 + 1 . 0 3 - 1.86 As before, the e f f e c t of these changes i s to l e r a b l e . SUMMARY This chapter has shown, that the maximum r a d i a l stresses at centerline due to shear and v a r i a t i o n i n ' e l a s t i c properties are n e g l i g i b l e , but that the stresses due to moment and moisture change are important. The pointed peak induces concentrations of r a d i a l stresses from moments at the centerline, and these stresses must be calculated by the graphs of f i g . 2 2 . . At other points on the beam, one or two depths away from this stress r i s e r , the regular curved or tapered beam formulas may be used for stress ca l c u l a t i o n s . This pointed peak also generates stresses due to non uniform moisture change, whether the beam i s loaded or not. It i s d i f f i c u l t to set the magnitude, as f i e l d t e s t i n g i s necessary to es t a b l i s h some r e a l i s t i c moisture gradients. In the meantime, however, i t might be advisable to add 20 p s i r a d i a l tension to any calculated stress from moments at the peak. This 20 p s i i s about h a l f the absolute magni-tude determined herein for one geometry. 6 . EXPERIMENTAL TESTS ON STRESS PERPENDICULAR TO GRAIN Average strength values of Douglas F i r are l i s t e d 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 Forest Products Research Branch of the Department of Forestry, Canada (10). For air-dry condition the ultimate t e n s i l e stress perpendicular to grain i s given there as 440. p s i . This value r e s u l t s from short time t e s t s . To compare i t with the allowable working stress under long time loading, i t may be mu l t i p l i e d by the factor 9/l6. For Douglas F i r , the maximum t e n s i l e strength perpendicular to grain under long time loading, now becomes 9/l6 x 440. = 248. p s i . The CSA-Code 086 (11) s p e c i f i e s , on table 8, the allowable working t e n s i l e stress perpendicular to grain for glulam Douglas F i r 24 f strength grade i n dry conditions as 65 p s i , or 1/3.8 times the long term ultimate. The calculated r a d i a l stresses i n section 5 . 1 - , are about twice the values given by the curved beam theory. But with the ultimate strength of 3 . 8 times the allowable s t r e s s , a factor of two does not explain f a i l u r e , e s p e c i a l l y when many f a i l u r e s occured under l i t t l e more than dead load. Other influences must be s i g n i f i c a n t . It could be that the allowable stress perpendicular to grain, given i n the code, i s too high. Therefore a series of tests were done to check the allowable s t r e s s . Samples from two d i f f e r e n t glulam plants i n the Vancouver area were tested. The test pieces, of the shape given i n f i g . 3 2 , had a cross section of 4 . 5"x4 . 5 " or 20.2 i n Pig.32. Example of Testmember Two s t e e l plates, with concentric threaded holes, were glued on with Epoxy for a p p l i c a t i o n of the t e n s i l e forces. Twenty two members have been tested with the re-s u l t shown i n table 5- The average value i s a „ . , =128. p s i f a i l u r e ^ This value i s less than 1/3 of the advertised ultimate and i t must be recognized that beams designed, with the c l a s s i c a l formula, under pure bending only, might be already i n the c r i t i c a l range. Although the number of tests i s small, the ten-dency i s clear and.the conclusion must be drawn, that the ultimate r a d i a l stresses must be further investigated. Such investigations should not use clear samples and the study should include s i z e effects and drying cracks. TABLE .5 •RESULTS OF TESTS ON TENSION PERPENDICULAR TO GRAIN Loading time about 5 minutes; moisture content about 10% Test member Ult. load l b . U l t . stress p s i V vv 1 2890 143 - 15 . 2 2 5 2 1685 83 + '45 2025 3 2415 119 + 9 8 1 4 2000 99 + 29 841 5 1740 86 + 42 1764 6 1940 9 6 + 32- 1024 7 2130 105 + " 2 3 . 5 2 9 8 2120 1 0 5 ' + 2 3 529 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 + 2 5 6 2 5 1 3 3300 • 1 6 3 - 35 1 2 2 5 14 1955 97 + 3 1 9 6 1 1 5 4290 212 - 84 7056 1 6 2340 1 1 6 + 12 . 144 17 3100 1 5 3 - 25 625 1 8 2 3 7 5 117 + .11 121 1 9 2 2 3 5 110 + 18 324 2 0 4625 228 -100 10000 21 2725 1 3 5 - 7 . 49 22 4350 215 -'87 7569 Sum Average = 0 2813 128 + 356 -353 36893 Standard deviation y 3g = ± 4 l . p s i Coe f f i c i e n t of v a r i a t i o n TTS" = 3 2 % , .. This difference of three between the ultimate stress of Douglas F i r i n r a d i a l tension and the values found herein are probably due to drying cracks which appeared i n many of the 22 samples. I f a clear, straight grained sample i s used, drying w i l l not induce nearly as many cracks as i f 're a l ' wood i s used. Those cracks normal to the d i r e c t i o n of r a d i a l tension induce excee-dingly large stress concentration factors which reduce the ultimate strength. The long term ultimate stress taken as 9_/l6 of the short term gives 12 p s i . A safety factor of 2 gives an allowable stress of 36 p s i . Since t h i s i s &Q% of the allowable now used, and the peak about doubles the actual stress calculated by the c l a s s i c a l curved beam formula, i t i s not su r p r i s i n g that f a i l u r e s occur. 7. NUMERICAL EXAMPLE The geometry of a roof beam for use i n wet 'conditions i s shown i n f i g . 3 3 - The beam spans 40 feet with a spacing of 1 6 feet. The laminations of the size 1 5 / 8 " x 7" are out of Douglas F i r 24 f strength grade. The centerline cross section of 39 inches requires 24 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 dead load (normal duration of load) i s .20 psf and the l i v e load ( 2 months duration bf load, i . e . snow duration) i s 40 psf. Herewith, the beam has to be designed for a t o t a l uniformly d i s t r i b u t e d load of ( 2 0 . + 4 0 . ) x 1 6 = 9 6 0 . l b / f t 0 . 9 6 k/ft, The allowable working stresses for glulamed Douglas F i r 24 f strength grade for wet conditions are given i n table 8 of the. CSA-code 0 8 6 ( 1 1 ) to 1900 p s i for bending 145 p s i for shear and 55 p s i for tension perpendicular to grain. These values are for normal duration of loads. Since the governing stresses for the considered beam occur over a two months duration, the allowable stresses may be increased by the factor 1 . 1 5 to x 1 . 1 5 = 2 1 8 5 . p s i x 1 . 1 5 = 6 3 . p s i x 1 . 1 5 = 1 6 7 . p s i The design i s made by checking the tangential and r a d i a l stresses at centerline of the beam and at the point of tangency and by checking the shear stress at the support. The stresses at centerline are calculated for comparison by the proposed as well as by the e x i s t i n g formulas. The parameters d/R = . 1 and tan a = . 2 give the stress c o e f f i c i e n t s (see table 2 or figures 2 2 and 2 3 ) CRM = 0 . 0 4 8 7 Crpjyj = 1 . 2 8 3 c C M = - 0 . 7 1 7 ° t a i l = 1 9 0 0 * o - r a l l - 5 5 . f a l l = 1 4 5 . The moment at centerline i s M = g i l = .96 x 4o 2 = 1 9 2 j Q k t t The maximum bending stresses obtained by the formula M/Z are _ 6 M _ 192 x 12000 x 6 _ ,„ Qo . a ~ b d 2 " " 7 x 39 x 39 1298. p s i Better design practice uses the smaller cross section at the point of tangency rr - 6 M _ 192 x 12000 x 6 _ 9 i n Q ^ n _ , . „ , ^ o- = - 7 x 30.6 x 30.6 " 2 1 0 9 ' p s l K a a l l " 2 1 8 5 • P The proposed formula gives the bending stresses Ocomp = CCM x = -0 .717 x 1298 = - 9 3 1 . p s i atens = CTM x S d ^ = 1 ' 2 8 3 x 1 2 9 8 = l 6 6 5 ' P s l It can be seen that the stresses obtained by the present design method give very l i t t l e information about the actual tangential stresses, although the stresses are on the safe side, i f the cross section at the point of tangency i s used. The r a d i a l t e n s i l e stresses by the curved beam formula are = 3M 3 x 1 9 2 x 1 2 0 0 0 - o n ' Q ° r m a x 2 b d (R+d/2) 2 x 7 x 3 9 x ( 3 9 0 + 1 9 - 5 ) 5 y p s i Here, too, better design practice would use the depth at the point of tangency to give 3M 3 x 1 9 2 x 1 2 0 0 0 o n p a = _ — = ^ 2 = 3 9 . 8 p s i max 2_bd ( R + d / 2 ) 2 x 7 x 3 0 . 6 x ( 3 9 0 + 1 5 - 3 ) The proposed formula gives a rmax = CRM x E d ^ = ' ° 4 8 7 x 1 2 9 8 = 6 3 ' 2 p s l The curved beam formula underestimates the maximum r a d i a l stress considerably. The actual stress i s about twice as much as given by the curved beam formula. It even exceeds the present allowable stress s l i g h t l y . I f 2 0 p s i are added for coverage of stresses due to change i n moisture content, the present allowable stress i s exceeded by an int o l e r a b l e amount. For the stresses at point of tangency e x i s t i n g formulas can be used. The bending moment there, at a distance, x = 6 . 6 3 f t from centerline, i s M = q( j j j - - f - ) = - 9 6 ( ^  ) = . 9 6 ( 2 0 0 - 2 2 ) = 1 7 0 . 9 k f t Herewith, the tangential stress i s 6 M _ 170-9 x 12000 x 6 _ 170.9 x 12000 l P i 7 , " bd^ " 7 x 30.6 x 30.6 1092.7 " 1 ' J * P The maximum r a d i a l stress i s „ 2 L . 3 x 170.9 x 12000 3 5 „ p s ± m d X 2bd (R+d/2) 2 x 7 x . 3 0 . 6 x (390+15-3) Both stresses are smaller than the allowable ones. The shear stresses are checked at the support, where the maximum shear force occurs. The v e r t i c a l re-action at the support i s F = | i = -96 x 4o = 1 9 > 2 k Herewith, the maximum shear stress i s obtained as 3 V 3 Fcos a Tmax 2 bd 2 bd = 3 19. 2 .2 x 1000 x cos( l l ° 2 0 ' ) _ n 7 x 3 0- m6 • - 13^-3 p s i This example shows that the present design method gives the s i g n i f i c a n t stresses i n tangential d i r e c t i o n but does not s a t i s f y for r a d i a l stresses. For a safe design the actual stresses have to be known at centerline. They are re a d i l y obtained by the proposed method using t h e s t r e s s c o e f f i c i e n t s CQ^, Crpjyi and Cp^ as shown i n f i g . 22 and f i g . 23 - -8. CONCLUSIONS The trapezoidal element proposed does predict a stress d i s t r i b u t i o n close enough for a l l p r a c t i c a l purposes on these problems. Also, i t allows stress determinations from changes i n moisture content and e l a s t i c properties. The peak at the centerline of pitch-cambered beams induces a stress concentration so that curved beam formulas are i n v a l i d near the top. Therefore c o e f f i -cients from fig.22 and fig.23 should be used to calcu-late the maximum r a d i a l and tangential stresses, due'to bending moment. < Radial stresses induced by shear and v a r i a t i o n i n e l a s t i c properties are not s i g n i f i c a n t , but those created by change i n moisture content are i n the order of magnitude of 20 p s i and should be considered. The allowable stresses, used at present, should be reduced to about h a l f t h e i r value, but tes t i n g needs to be done to f i n d a more r e a l i s t i c reduction. With the increased calculated r a d i a l stresses and the reduced allowable r a d i a l stresses, i t w i l l be d i f f i c u l t to produce an economic design i n some cases. When such occurs, the designer could consider r e i n f o r c i n g the beam with s t e e l dowells. These dowells from 3/8" to say 3/4" diameter are placed i n s l i g h t l y oversized holes, d r i l l e d almost the f u l l depth of the beam, and p a r t i a l l y ' f i l l e d with an adhesive such as 'Epoxy'. Such dowells serve the same purpose as r e i n f o r c i n g s t e e l i n concrete. It i s also possible to carry the t o t a l force a r i s i n g from the r a d i a l stresses by two v e r t i c a l s t e e l plates fastened with glulam r i v e t s to either side of the beam at i t s centerline. REFERENCES NORRIS, CB. 1 9 6 3 . Stresses within curved lami-nated beams of Douglas F i r . ^ FPL - 0 2 0 . Forest Prod. Lab., U.S.. Dep. Agri. Madison, Wise. CARRIER, G.F. 1 9 4 3 - Stress d i s t r i b u t i o n i n c y l i n -d r i c a l l y aelotropic plates. J . Appl. Mech. ' Vol. 1 0 , Trans, ASME V o l . 6 5 . A - 1 1 7 - 122. FOSCHI, R.O. 1 9 6 8 . Plane Stress problem i n a body with c y l i n d r i c a l anisotropy, with s p e c i a l reference to curved Douglas F i r beams. Department of Forestry and Rural Development, Departmental Publication No. 1244, Ottawa, Ont. FOX, S.P. 1 9 6 8 . Stresses i n glued-laminated timbers: double-tapered pitched beams. Forest Prod. Lab, Dept. Forest. Rural Develop. Vancouver, Canada. TIMOSHENKO, S. and ' GOODIER, J. Theory of E l a s t i c i t y , McGraw-Hill. 1 9 5 7 -WANG, Chi-Teh, Applied 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 of wood and plywood. Forest 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 Werkstoffe, Volume I, Springer B e r l i n 1 9 5 1 . KENNEDY, E.I. 1 9 6 5 . Strength and related Properties of Woods grown i n Canada, Department of Forestry, Forest Products, Research Branch, Dep. of Forest'. Publication No. 1104, Ottawa, Ont. CANADIAN STANDARDS ASSOCIATION, Ottawa, 0 8 6 - 1 9 5 9 , Engineering Design i n Timber. Revised reprint -August 1 9 6 3 . 87 APPENDIX L i s t i n g o f t h e c o m p u t e r p r o g r a m d e s c r i b e d i n c h a p t e r 3-;IV G COMPILER MAIN 09-08-69 11:47:38 PAGE 0001 t i DIMENSION RG(30),MEG(30),JO( 2 0 0 , 4 ) , D E L R ( 2 0 ) , B E TAG(30), 3F.TAM(200), ; IR 11 200) ,R2(200) , 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 IJI 3) DIMENSION SMM{36) , St 12950 ), 3( 500 ) , DP ( 8 ) , EP SIM I 200 , 2 ) , E ( 200 , 4 ) : DOUBLE PRECISION START,WORD MAXMEM=200 MAXJ0=250 C . . MAXMEM AND MAXJO ARE USED TO DEE INE THE DIMENSIONS IN THE.SUBROUTINES DATA START /5HS TART/ '. \00 PRINT 20 READ 10 : - PRINT 10 PRINT 20 .FORMAT (72H __ _ _ _ _ _ 1 " J " ' "•" • FORMAT ( // 132H to************************* _ * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * x: * i: * *•**_:** * * * ~ * * * * * * * ******* * v. ** ** * ( i 2*** ***** ****** / /) \ READ 2, M XD EL R,MA ST R,MATE L,NP S : | 2 FORMAT (18 14) _ __ I PRINT 101, MXDELR f 101 FORMAT (--'/25H NUMBER OF DELTA RADIUS i _ _ . READ 102, (DELP. ( I ), 1 = 1, MXDELR)  FORMAT (F10.6) PRINT 103, ( I ,DELR(I),1 = 1,MXDELR) FORMAT __( I 4 , F 12 ._6 ) READ 2,NRG PRINT 201 , NRG FORMAT (//26H NUMBER OF RADIAL GROUPS =,I4/25H IRG MEG RG  1BETAG) « , ' READ 202, ' (MEG( IRG),RG(IRG),BETAG(IRG),IRG=1,NRG) FORMAT.iJft » FIQ. 3 , F 1 0 . 6.) ..„ PRINT 2C3,. (IRG,M E G(IRG),RG( IRG),BE TAG(IRG) ,I RG=1,NRG) FORMAT ( 2 I 4 , F 1 0 . 3 , F 1 0 . 6 ) CALL DIVIDE (MAXMEM,NPG,RG,MEG,JO,ME,MJ)  CALL DISTRI (NRG,RG,MEG,DELR,BET AG,BETAM,ME,R1,R2) READ 204, TH,NSM F0RMAT__.{F12. 6,14) _____ . _ _ _ DO 208 1=1,ME THM(I)= TH PRINT 210,MF,NSM,NJ  FORMAT <//20H NUMBER OE MEMBERS =,I4/20H N OF SPEC.MEMBERS =,14/20 IH NUMBER OF JOINTS =,14) _IF_CNSM.EQ.0_ GO TO 222 _ PRINT 212 FORMAT (//16H SPECIAL MEMBERS/12H I THM) 00 220 1=1,NSM READ 214, ISP, THM ISM) 214 FORMAT {-1.4, FI 0.6) ...PRINT 2.16',. ISM, THM( ISM.) 216 FORMAT (I4, 2 X , F 1 0 . 6 ) 220 CONTINUE 222 PRINT 224 224 FORMAT (//22H MEMBER SPEC I F I CATI0NS/59H I I R l " R2 BETAM THM) J0INTNRS(I , J) PRINT. 226, .( I , J 0 ( 1, 1 )_»JO (1,2) ,J0( 1,3) ,J0( 1,4) , R 1 (I ) , R 2 ( I ) , B E TAM( I ) 10 20 = , I4/12H I DELR) 102 103 201 .20 2 203 .204 208 210 212 V G C O M P I L E R M A I N 0 9 - C 8 - 6 9 1 1 : 4 7 : 3 8 P A G E 0 0 0 2 1 , T H M ( I ) » I = 1 r M E ) 2 2 6 FORMAT U 4 , 2 X , 4 I 4 , 2 F 1 0 . 3 , 2 F 1 0 . 6 ) 3 0 0 MS=8 NC.0 = 4  M S T = M S / N C O R E A D 2, NR J P R I N T 3 0 5 m J _ . _ '._ 3 0 5 . F O R M A T ( / / 3 C H NUMBER 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 0 0 3 1 1 K = 1 , M S T ' N D ( I , K ) = 1 ' 3 1 1 C O N T I N U E 3 1 0 C O N T I N U E DO 3 1 2 I = 1 , N R J R E A D 2 , J R , N D ( J R , 1 ) , N D ( J R , 2 ) P R I N T 2 , J R , . M P ( J R , 1 ) , N D t J R , 2 ) 3 1 2 C O N T I N U E NUM=0 0 0 3 . 1 3 . I - l , N J DO 3 1 4 K = 1 , M S T I F ( N D ( I , K ) . E Q . 0 ) G O TO 3 1 4 NUM=NUM+1 N 0 ( I , K . ) = N U H 3 1 4 C O N T I N U E 3 1 3 C O N T I N U E P R I N T 3 2 0 3 2 0 FORMAT C / / 1 9 H J O I N T CODE N U M B E R S ) P R I N T 3 2 1 , ( J , ( N O ( I , J ) , J = 1 , M S T ) , 1 = 1 , N J ) 3 2 1 FORMAT ( 5 ( 3 I 4 , 2 X ) ) 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) N C O D E I I , J ) = I C D M 3 1 7 C O N T I N U E 3 1 6 C O N T I N U E P R I N T 3 2 5 3 2 5 . . FORMAT ( / / . ? 1 H .. E L E M E N T CODE. N'JMBf R S / P 4 H _ I N C O D E J 1 , 1 T 0 3 ) . ) P R I N T 3 2 6 , I I , ( NCODE ( I, J ) , J = l , M S ) , 1 = 1 ,ME> ,, 3 2 6 F O R M A T 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,MSM I C D J = N C O D E ( I , J ) I F ( I C D J ) 3 5 ' 3 5 3 J P = J + 1 DO 3 5 2 K = J P , M S I CDK = NCODE (.1 ,.K_) IF I I C O K ) 3 5 4 , 3 5 2 , 3 5 4 3 5 4 N 8 T = I A S S ( ( I A R S ( I C O K J - I A B S t I C D J ) ) ) + 1 I F ( N S T - N B ) 3 5 2 , 3 5 2 f 3 5 5 3 5 5 N 0 = N B T 3 5 2 C O N T I N U E 3 5 1 C O N T I N U E ! IVf G C O M P I L E R M A I N .. 0 9 - 0 8 - 6 9 1 1 : 4 7 : 3 8 P A G E 0 0 0 3 3 5 0 C O N T I N U E R E A D 3 6 2 , E 1 , E2 , E M U , G , N E S P 3 6 2 FORMAT I A X , 4 F 16 . 8 , I 4 ) P R I N T 3 6 0 , E l , E 2 , E M U , G , N E S P . • 3 6 0 FORMAT ( / / 5 H E I = , F 1 2 . 4 / 5 H E2 = , F 1 2 . 4 / 5 H E M U = , F 1 2 . 4 / 5 H G = , F 1 2 . 4 / 1 / 4 2 H NUMBER 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 3 7 0 I = I , M F . _ _ _ . ... _ _ E ( I , 1 ) =F1 S E l I , 2 ) = E 2 . - '• E ( I t 3 ) = EMU . ' 3 7 0 E ( I , 4 ) = G I F ( N E S P . E Q . 0 ) GOTO 3 0 0 _ ._ DO 3 7 6 1 = 1 , N E S P . : . _ R E A D 3 7 2 , I E S P , E ( I E S P , 1 ) , E ( I E S P , 2 ) , E { I E S P , 3 ) , E I I E S P » 4 ) P R I N T 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 3 7 2 FORMAT ( I 4 , 4 F 1 6 . 3 )  3 7 3 FORMAT {I 4 , 4 E 1 6 - 3 ) 3 7 6 C O N T I N U E 3 8 0 C O N T I N U E I F ( M A T E D 4 0 0 , 4 0 1 , 4 0 0 4 0 0 P R I N T 4 0 2 4 0 2 .- FORMAT ( 1 H 1 , 2 5 H E L E M E N T S T I F F N E S S M A T R I X / / ) GO TO 4 0 4 4 0 1 P R I N T 4 0 3 4 0 3 F O R M AT .( /. /.. 3 7 H _ ELE_.ME.NT_. ST I F FN E SS M A T RI X .NOT.__PR I NT E D ) 4 0 4 C O N T I N U E NS - N U * N 3 0 0 4 0 5 I = 1 , N S 4 0 5 S ( I ) = 0 . 0 N B 1 = N B - 1 ; _ _ D 0 4 0 7 . L _ = 1 . , . M E DO 4 0 8 1 = 1 , 3 6 4 0 8 S M M l I ) = 0 . 0 R 1 L = B1 i I.) R 2 L = R 2 ( L ) B E T A L = 3 E T A M ( L ) T H L = T H M I L > E 1 L = E ( L , 1 ) E 2 L = E ( L , 2 ) E M U 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 ) I F ( M A T E L . E Q . 0 ) GO TO 4 2 2 ... .  EL < ? P R I N T 4 2 0 , L 4 2 0 F O R M A T ( / I 4 ) 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 ) P R I N T 4 2 1 , S M M ( 3 ) , S M M { 1 0 ) , 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 , S M M l 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( 32 ) , S M M { 3 4 ) P R I N T 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 S M M l 3 6 ) . 4 2 1 j .FORM*.] ' . J .J .X, 8 G 1 6 . 6 . ) . _ _ _ __.._' _ IV G COMPILER MAIN 09-08-69 11:47:38 PAGE 0004 422 CONTINUE DO 409 J = l , 8 IF (NCODE(L,J)) 409,409,412 412 J l = ( J - l )-M 16-J) /2  DO 410 I=J,6 IF (NCODE(L,I)) 410,410,406 _ 4 C 6 IFJNCODEJ L, J ) r NCODE.( L , I ).)_. 413 ,4.5 0,4 1.4 450 I F ( I - J ) 451,413,451 451 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 .4.1 3 K M NCODE.( L , J ) — I.).J:.N.5 1+NCODEJ L , I } GO TO 415 414 K = ( N C O D E ( L , I ) - l ) * N B 1 + N C O D E 1 L . J ) 415 N=J1+I S(K)=S(K)+SMM{N) 410 CONTINUE .4.09 CONTINUE 407 CONTINUE , IF {MASTR .EO. 0) GOTO 460 PRINT 1000 1000 FORMAT (IH1) 460 CONTINUE PRINT. 4 62_,.._NU,NB 462 FORMAT (//22H NUMBER OF UNKNOWNS =,I4///22H BANDWIDTH" 1,14) IF (MASTR ) 464,466,464  464 PRINT 465 465 FORMAT {////27H STRUCTURE STIFFMESS MATRIX/) PRINT 425 , ( S ( K ) , K= 1 , NS ) _ _ . 42 5 GO TO 500 466 PRINT 467 467 FORMAT {//39H STRUCTURE STIFFNESS MATRIX NOT PRINTED) 500 READ 2,NLC PR I.NT_..O.l ,. _NLC 501 FORMAT (////22H NUMBER OF LOADCASES = , 14) , DO 590 L=1,NLC 00 502 502 8( I ) = 0.0 READ 2, NBL,NEI . PRINT 503 , .L 503 FORMAT (LH1,9H LOADCASE,14) IF (NBL .EO. 0) GO TO 700 PRINT 506, NBL 506 FORMAT (//24H NUMBER OF LOADS GIVEN =,I4/12H K B{K)) DO 504 1=1,NBL READ. 50.5, ,K,B(K) 50 4 PRINT 5 05, K,3(K) 505 FORMAT ( 14 ,F10 .3) 700 IF ( N E D 701,722,701  701 READ 702, EPS I 1,EPS I 2,EPS I 3,NSSM 702 FORMAT ( 3 F 1 0 . 6 . I 4 ) _ ... PRINT .704, EPS. l.,.EP SI 2., EPS I 3 , NSSM I V G C O M P I L E R M A I N 0 9 - 0 8 - 6 9 1 1 : 4 7 : 3 8 P A G E 0 0 0 5 ! 7 0 4 F O R M A T ( / / 2 3 H I N I T I A L S T R A I N E T A N G . = , F 1 0 . 6 / 2 . 3 H E R A D I = , F 1 0 . 6 / 2 3 H E S H E A R = , F 1 0 . 6 / / 3 7 H NUM3ER OF S P E C I A L 2 S T R A I M E D M E M B E R S = , 1 4 ) DO 7 0 6 J = l , M E E P S I M ( J , 1 ) = E P S I 1 7 0 6 E P S I M U , 2 ) = E P S I 2 I F ( N S S M . E 0 . . 0 )_ G O „ T O . 7 0 9 . DO 7 0 8 1 = 1 , N S S M v R E A D 7 1 0 , I S S M , E P S I M { I S S M . l ) , E P S I M < I S S M , 2 ) 7 1 0 F O R M A T ( I 4 , 2 F 1 0 . 6 ) P R I N T 7 1 1 , I S S M , 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 7 1 1 F O R M A T U 4 , 3 ( 2 X , F 1 0 . 6 ) > 7 0 8 C O N T I N U E _ _ _ _ 7 0 9 0 0 7 2 0 j = l , M E R U = R1 ( J ) R 2 J = R 2 ( J ) B E T A J = B c T A M { J ) T H J = T H M ( J ) E P S I J ( 1 ) = E P S I M ( J , 1 ) E P S I J < 2 ) = E P S I M { J , 2 ) E P S I J I 3 ) - E P S I 3 E U = E ( J , l ) 1 E 2 J = E ( J , 2 ) E M U J = E ( J , 3 ) G J = E( J C A L L E P S I N I DO 7 1 4 N C = 1 , 8 I F { N C O D E ( J , M C ) ) ( 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 , 7 1 2 , 7 1 4 , 7 1 2 F I ) 7 1 2 7 1 4 7 2 0 7 2 2 N C M = N C O P E ( J , N C ) 8 ( NCM ) ~ B ( NCM ) + F I ( N O C O N T I M U E _ C O N T I N U E GO TO 7 5 0 DO 7 2 4 1 = 1 , M E DO 7 2 4 J = l , 2 7 2 4 E P S I M ( I , J ) = 0 . E P S I 3 = 0 . . 7 5 0 C O N T I N U E P R I N T 5 0 7 5 0 7 F O R M A T ( / / 1 1 H L O A O V E C T O R ) -t<. 5 0 8 5 1 1 P R I N T 5 1 3 , ( 3 ( 1 ) , 1 = 1 , N U ) DE T = 1 . E — 8 C A L L BAND ( S , B , N U , N B , L , D E T ) I F ( D E T ) 5 0 8 , 5 0 9 , 5 1 0 P R I N T 5 1 1 , D E T FORMAT ( 3 4 H 0 MAT RI X A IS NOT P O S I T I V E - D E F I N I T E / 7 H 0 D E T = , E 1 5 . 7 ) r > GO TO 5 9 0 5 0 9 P R I N T 5 1 2 , D E T 5 1 2 FORMAT '(_2.0H0DET ER M IJMA.N.T _ GO TO 5 9 0 5 1 0 P R I N T 5 1 4 5 1 4 F O R M A T ( / / 1 9 H D I S P L A C E M i J S Z E R 0 / 7 H 0 0 E T = , E 1 5 . 7 ) NTV E C T O R ) PR INT 5 1.3, ( B( I ) , 1 = 1 5 1 3 FORMAT ( 1 X , 8 G 1 5 . 6 ) P R I N T 5 5 5 NU.) I V G C O M P I L E R M A I N 0 9 - 0 8 - 6 9 5 5 5 F O R M A T { / / / / / 9 H I T T A U ) 5 5 0 0 0 5 6 0 J = 1 , M E R I J = R1 ( J ) S T R E S S E S / 5 1 H 1 1 : 4 7 : 3 8 R A O I A L P A G E 0 0 0 6 T A N G E N 5 5 2 R 2 J = R 2 ( J ) B E T A J = B E T A M I J ) DO 5 5 4 NC = 1 , 8 _ ___ I F ( N C O D E ( J . N £ ) ) 5 5 3 , 5 5 2 / 5 5 3 D P ( N C ) = 0 . 0 GO TO 5 5 4 5 5 3 M C M = N C O D E ( J , N C ) D P ( N C ) = 8 ( N C M ) 5 5 4 ^ C O N T I N U E E P S I J ( 1 ) = E P S I M ( " j , 1 ) E P S I J ( 2 ) = E P S I M ( J , 2 ) E P S I J ( 3 ) = E P S I 3 5 5 6 E U = E ( J , l ) E 2 J = E ( J , 2 ) E M U J ^ E l J , 3 > G J = E ( J , 4 ) C A L L S I O M ( R I J , R2 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 ) P R I N T 5 5 8 , J , S R A D , S T H , S R T H 5 5 8 5 6 0 5 9 0 F O R M A T ( 1 4 , 3 X , 3 ( 1 X , F 1 5 . 6 ) ) I F ( J . G T . N P S ) GO TO 5 6 0 WR I J F. . ( 7 5 5 8 ) J , SRAD , STH , SRTH . C O N T I N U E C O N T I N U E R E A D 6 1 0 , WORD J i 6 1 0 F O R M A T ( A 5 ) I F (WORD . N E . PR I NT 1 0 0 0 GO TO 1 0 0 6 2 0 C O N T I N U E S T O P  END S T A R T ) CO TO 6 2 0 IV G COMPILER DIVIDE 09-08-69 • 11:48:16 PAGE 0001 SUBROUTINE DIVIDE (MAXMEM,NRG,RC,MEG,JO,IMUS,NJ) DIMENSION R G ( l > , MEG(1),JO(MAXMEM,4} I8AS=0 IMUS=0 'V DO 10 IRG=1,NRG IF (IPG.EQ.1) GO TO 11 I F { M E G ( IRG-1 )-MEG(_IRG) ).11,11, 12 . 11 MM = MF G ( I RG ) N DO 13 JRG = 1,MM I=IMUS + J'RG J0( I ,1) = IBAS+JRG + MEG(IRG)+2 J0( I ,2) = IBAS + JRG + l J 0 ( I , 3 ) = I 3 A S + J R G JC( I ,4]l =1BAS + JRG + MEGlIRG3+1 13 CONTINUE IRAS=IBAS+MEG(IRG)+l IMUS=IMUS+MEGlIRG) GO TO 10 12... _MM = MEGJ.I RG) . • > DO 14 JRG.= 1,MM I=IMUS+JRG J0( I ,1 )=T8AS+JPG*-MEG( IRG-D+2 1 J0( I ,2) = IBAS + JRG+1 J0(I» 3 ) = I 3 A S +JRG _ JO { I ,4) = IBAS + JRG + MEG.(.IRG-1J_+1 ~14 CONTINUE IBAS=IBAS+MEG(IRG-1J+l IMUS=JMUS+MEGlIRG) 10 CONTINUE NJ=IBAS+MEG(NRG)+1 R E T U R N . END i IV G COMPILER 0 ISTR I 09-0 8-6 9 11:48:21 PAGE 0001 1 SUBROUTINE OISTRI (NRG ,RG, MEG,DELR,BETA G,BETAM,IMUSfRli R2 ) -V DIMENSION RG(1),MEG(1) ,0ELR(1),B E T A G <1) ,B E T A MI 1) , R l ( l ) , R2( 1 ) IMUS=0 ' A DO 20 I RG'=1 » NRG R1M=RG(IRG) MM=MEG(IRG) DO 21 JRG=1,MM V I=IMUS+JRG •v R 1(I)=R1M R2(I)=R1M+DELR(JRG) R1M=R21I) BETA M(I ) = BETAG( IRG) 21 CONTINUE IMUS=IMUS+MEG(IRG) 20 CONTINUE RETURN . •» END * 5 I i I V G C O M P I L E R CODE 0 9 - 0 8 - 6 9 1 1 : 4 3 : 2 2 P A G E 0 0 0 1 S U B P O U T I N E C O D E ( M A X M E M , M A X J O , J O , I C D , I , J , I COM) D I M E N S I O N 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 J N U = 1 2 GO TO J N U = 2 GO TO 10 10 >-4 J N l i = 3 GO TO J N U = 4 10 10 _5_ . 6 J D = 1 GO TO 2 0 JNU=_1 GO TO i l J N U = 2 GO TO 11 •< 7 J N U = 3 GO TO 11 8 J N U = 4 y 11 J D = 2 V. 2 0 I I = J 0 ( I , ' JNU ) *' I C D H = I C D ( 11 , J D ) R E T U R N END e I IV G C O M P I L E R S.VEM 0 9 - 0 8 - 6 9 1 1 : 4 3 : 2 5 P A G E 0 0 0 1 S U B R O U T I N E SMEM I R I , R 2 , B E T A f T H , E 1 , E 2 , E M U , G , S ) -A C T R A P E Z O I D A L E L E M E N T T R A N S F O R M E D I N T O P G L A R C O O R D I N A T E S •_, D I M E N S I O N S K ( 3 6 ) , S ( 3 6 ) H = ( R 2 - R 1 ) * . C G S ( B E T A / 2 . )  A= 2 . ' y R 1 i ; S IN ( B E T A / 2 . ) B = 2 . * R 2 * S I N ( B E T A / 2 . ) . : _ F l _ . = {A+B) ^ H / 2 . FY = ( A + 2 . * B ) * H * H / 6 . F X X = ( ( A * A + 3 * B ) * ( A + B ) ) A H / 3 6 . F Y Y = ( A + 3 . * B ) * H * H * H / 1 2 • • S A 1 - 1 . / ( A * A ) *>• S A 2 = 1 . / ( A A H ) . „ _ S A 3 = 1 . / ( A*A_*H)_ _ . _ _ _ S A 4 = 1 . / I A * A * H * H > v S.A5 = 1 . / ( A * H * H ) ' S A 6 = l . / ( A * B * H )  7 S A 7 = 1 . / ( A ~ 8 " H * H ) S B 1 = 1 . / ( H * H ) . ... _S82 .= 1_../H _____ _ _ SB3=1./(B.*H*H) ^ S B 4 = l . / ( B * B * H * - H ) !_ : . S C l = l . / ( 4 . * H * - H ) * i F l ' : - G )  '\ S C ' 2 = ( F Y Y * E 1 ) + ( F X X * G ) ^ S D 1 = F Y * G j S D 2 = FY*F_1 ' I S D 3 = FY*--EMU » S E I = F 1 * G I S E 2 = F 1 * E 1 "  * SE.3 = F 1 « E M U H S F l = l . / ( 4 . t H * H ) * ( F l v E 2 ) i _ .. S F 2 = (F X . X _ E 2 ).+ (. F Y Y * G ) \ S K ( 1 ) = S C 1 + S B 4 * S C 2 1 S K ( 2 ) = S C 1 - S 8 4 ' - ; S C 2 S K ( 3 ) = - S C I - S A 6 ~ S D 2 + S A 7 ~ S C 2 A "* S K ( 4 ) = - S C l + S A o * S D 2 - S A 7 * S C 2 S K ( 5 ) = S B 3 / 2 . * ( S D 1 + S D 3 ) 1 A— SK ( 6 ) . _ = ' . S B 3 / . 2 . y - : ( . - S D l + S D 3 ) _ . . . . . . . _ S K ( 7 ) - - S A 2 / 2 . * S E l + ( S A 5 / 2 . * - S D i ) - ( S 3 3 / 2 . * S n 3 J S K { 8 ) = ( S A 2 / 2 . * S E 1 ) - ( S A 5 / 2 . * S D l ) - ( S B 3 / 2 . " S D 3 ) S K ( 1 6 ) = ( S A l * S E 2 ) - ( S A 3 * 2 . * S D 2 ) + ( S A 4 * S C 2 ) + S C l S K ( 1 7 ) = - ( S A l * S E 2 J + ( S A 3 * 2 . * S D 2 l - ( S A 4 + S C 2 ) + S C I } J t S K ( 1 8 ) = - I S A 2 / 2 . * S E 3 ) + ( S A 5 / 2 . * S D 3 ) - ( S B 3 / 2 . * S D 1 ) - — S K ( 1 9 ) = - ( S A 2 / 2 . * S E 3 ) + ( S A 5 1 2 . * S 0 3 ) + ( S B 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~ ' 1 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 . S K I 2 7 ) = S F l + ( S B 4 * S F 2 ) * S D 1 J 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= - S F l - { S A 7 - = S F 2 ) + < S A 6 * S D 1 ) S K I 3 4 ) = S F l + ( S A 1 * S E 1 ) - ( S A 3 * 2 . * S D l ) M S A 4 * S F 2 ) — > S K ( 3 5 ) = S F l - ( S A l * S E l ) + ( S A 3 * 2 . * S D 1 ) - ( S A 4 * S F 2 ) C C = C O S ( B E T A / 2 . ) * C O S ( B E T A / 2 . ) C S = C O S ( B E T A / 2 . ) * S I N I B E T A / 2 . ) A : S S = S I N ( B E T A / 2 . ) * S I N ( B E T A / 2 . ) • SJ l.).-:=-.CCr.SK (.1.) r . C S ? 2 . y.SK ( 5 ) + S S * S K ( 2.7) ._ — I V G C O M P I L E R SMEM 0 9 - 0 8 - 6 9 1 1 : 4 8 : 2 5 P A G E 0 0 0 2 S ( 2 ) = C C * S K ( 2 ) + C S « 2 . * S K ( 6 ) - S S * S K 1 2 8 ) S ( 3 ) = C C * S K ( 3 ) + C S * ( S K ( 7 ) - S K ( 1 B ) ) - S S * S K I ? 9 - ) S ( 4 ) = C C * S K ( 4 ) + C S * ( - S K ( B ) + S . K ( 1 9 ) ) + S S * S K ( 3 0 ) S ( 5 ) = C C * S K ( 5 ) + C S » ( S K I 1 ) - S K ( 2 7 ) ) - S S * S K ( 5 ) • S ( 6 ) = C C * S K { 6 ) + C S ! M - S K ( 2 ) - S K ( 2 8 ) ) - S S * S K ( 6 ) S ( 7 ) = C C * S K ( 7 ) + C S * ( - S K ( 3 ) - S K ( 2 9 ) } + S S * S K { 1 8 ) S( 3 ) _=_ C C * S K _ 8 ) + C S = M S K ( 4 J - S K ( 3 0 ) ) + S S * S K ( 1 9 ) • si 9 ) = sin SI 1 0 ) = S ( 4 ) S ( 1 1 ) = S ( 3 ) s S ( 1 2 ) = - S I 6 ) S ( 1 3 ) = - S 1 5 ) I _ S ( 1 4 ) = - S ( 8 ) S ( 1 5 ) = - S ( 7 ) S ( 1 6 ) = 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 * _ . * S K I 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 ) ) + S S * S K ( 7 ) S ( 1 9 ) = C C * S K ( 1 9 ) + C S * ( - S K ( 4 ) + S K ( 3 0 ) ) + S S * S K ( 8 ) _ S ( 2 0 ) - C C * S K ( 2 0 ) + C S * ( - S K ( 1 6 ) + S K ( 3 4 ) ) - S S * S K ( 2 0 ) S ( 2 1 ) = C C ' ' S K ( 2 1 ) + C S * l S K ( L 7 ) + S K ( 3 5 ) ) - S S * S K ( 2 1 > SI 2 2 ) = S( 1 6 ) S ( 2 3 ) = - S ( 1 9 )  S ( 2 4 ) = - S ( 1 8 ) S ( 2 5 ) = - S ( 2 1 ) _ S ( 2 6 ) = - S J . 2 0 . ) S ( 2 7 ) = C C * S K ( 2 7 ) + C S * 2 . * S K ( 5 ) * S S * S K ( 1 ) S ( 2 8 ) = C C * S K l 2 8 ) + C S * 2 . * S ' K ( 6 ) - S S * S K ( 2 ) S ( 2 9 ) = C C * S K ( 2 9 ) + C S T ( 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 ( 3 1 ) = S ( 2 7 ) S ( 3 2 ) = _ S( 3 0 ) __ 4 S ( 3 3 ) = S ( 2 9 ) S ( 3 4 ) = C C * S K [ 3 4 ) - C S * 2 . * S K I 2 0 ) + S S * S K { 1 6 ) S ( 3 5 ) = C C * S K ( 3 5 ) - C S * 2 . * S K < 2 1 ) - S S * S K ( 1 7 ) S I 3 6 ) = S ( 3 4 ) DO 10 1 = 1 , 3 6 . 1 0 S( I ) = S(.I_) TH RE TURN END I V G C O M P I L E R E P S I M I ' 0 9 - 0 8 - 6 9 • 1 1 : 4 8 : 3 8 P A G E 0 0 0 1 S U B R O U T I N E E P S I N I ( R I , R 2 , B E T A , T H , E 1 , E 2 , E M U , G , E P S I , F I } D I M E N S I O N F T { B) , F I ( 1) , E P S K 1) C = C 0 S ( B E T A / 2 . ) S = S I N < B E T A / 2 . ) H=C* ( K 2 - R 1 ) A = 2 . * R l * S B = 2 . „ R 2 * S AA= f H*'( A+3 ) / 4 . AB = T H - H M 2 . *A + 3 ) / { 6 . - A ) AC = TH*H*( A + 2 . - ' 8 ) /( 6 . * - B ) A E = E 1 * E P S I ( 1 ) + E M U * E P S I ( 2 ) A F = E M U * E P S I ( 1 ) + E 2 v E P S l i 2 ) 'r_ _ ... A G = G * F p S I 13 F T t 1 ) = A C * A E +A A* AG F T ( 2 ) = - A C * A E + A A * A G *v F T ( 3 ) = - A 3 * A E - A A * A G F T ( 4 ) = A3 * A E — 4 A * A G F T ( 5 ) = A A * A F + A C * A G " r r \ FT (6 . )_=_AA*AF - A C * A G F T ( 7 ) = - A A * A F - A B * A G - FT ( 8 ) -= -AA '*AF + A 8 * A G : F I ( 1 ) = C - F T { 1 ) - S * F T ( 5 ) X ! F I ( 2 ) = C - - - F T ( 2 ) + S * F T ( 6 ) F I ( 3 ) = C * F T ( 3 ) + S * F T ( 7 ) F 1 ( 4 ) = C * F T { 4 ) - S * F T { 3 ) _ F I ( 5 ) = O F T ( 5 ) + S * F T ( 1 ) F I ( 6 ) = C * F T ( 6 ) - S * F T ( 2 ) F I ( 7 ) = C * F T ( 7 ) - ' S - ' « F T ( 3 ) F I ( 8 ) = C * F T ( 8 ) + S * F T ( 4 ) R E T U R N END 4 I V G C O M P I L E R S I G M 0 9 - 0 8 - 6 9 1 1 : 4 8 : 4 3 P A G E 0 0 0 1 SOBROUT 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 = C O S ( B E T A / 2 . ) S = S I N ( B E T A / 2 . ) '  D K l l ) = C * D P ( l ) + S * E ) P ( 5 ) n K ( 2 ) = C * D P ( 2 ) - S * 0 P < 6 ) _ _ _ D K ( ' 3 ) = C * D P ( 3 ) - S * D P ( 7 ) . _ D K ( 4 ) = C * D P ( 4 ) + S - D P ( 8 ) S D K ( 5 ) = C « D P I 5 ) - S ~ D P ( 1 ) D K ( 6 ) = C * - - D P ( 6 ) ' + S * D P ( 2 ) ^ _ _ _ _ D K 1 7 ) = C * D P ( 7 ) + S * D P ( 3 ) D X ( 8 ) = C * 0 P ( 8 ) - S * 0 P ( 4 ) _ H = C * l _ R 2 - R 1) _ ._ _ . _ _ A = 2 . * R 1 * S B = 2 . * R 2 * S X = 0 . Y=H/2. C T l ( 1 , 1 ) = Y / ( B * H ) C T l ( l , 2 ) = - C T I ( l , l ) CT I ( 1 , 3 ) = - 1 . 6 / A + Y / ( A * H ) CT n 1 , 4 ) = - C T 1 ( 1 , 3 ) C T l ( 1 , 5 ) ' = 0. C T l ( 1 , 6 ) = 0 C T I ( 1 , 7 ) = 0 C T l ( I »8__= . 0 , C T l ( 2 , 1 ) = 0 C T I ( 2 , 2 ) = 0 , C T I ( 2 , 3 ) = 0 , C T l ( 2 , 4 ) = . 0 . . . C T I ( 2 , 5 ) = l . 0 / ( 2 . 0 * H ) + X / ( B * H ) C T I ( 2 , 6 ) = _ 1 . 0 / ( 2 . 0 * H ) - X / ( B * H ) C T l ( 2 , 7 ) = - l , 0 / ( 2 . 0 * H ) + X / ( A * H ) C T I ( 2 , _ ) = - l . 0 / ( 2 . 0 * H ) - X / ( A * H ) C T I ( 3 , 1) • C T l ( 3 , 2 ) = C T l ( 2 , 6 ) C T I ( 3 , 3 ) = C T H 2 , 7 ) C T l ( 3 , 4 ) = C T l ( 2 , 8 ) CT 1 ( 3 , 5 ) = CT I ( 1 , 1) C T l ( 3 , 6 ) = C T l ( 1 , 2 ) C T l ( 3 , 7 ) = C T K 1 , 3 ) C T l ( 3 , 8 ) = C T l ( I , 4 ) DO 2 0 K = l , 3 EP_S(K . )= -F .PST . ( K ) . DO 2 0 J = l , 8 E P S ( K . ) = E P S ( K ) + C T I ( K , J ) T ' D K ( J ) 2 0 C O N T I N U E S I G X = E P S ( 1 ) * E 1 + E P S ( 2 ) * E M U S I G Y = E P S ( 1 ) " E M U + E P S ( 2 ) * E 2 T A U = E P S ( 3 J » = G R E T U R N END 

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