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Analysis of elastic shells of revolution with membrane and flexure stresses under arbitrary loading using… Agrawal, Krishna Murari 1969

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ANALYSIS OF ELASTIC SHELLS OF REVOLUTION WITH MEMBRANE AND FLEXURE STRESSES UNDER ARBITRARY LOADING USING TRAPEZOIDAL FINITE ELEMENTS by KRISHNA MURARI AGRAWAL B.E. (Hons.) University of Jabalpur, India, 1961 M.E. (Struct.) University of Roorkee, India, 1963 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY in the Department of CIVIL ENGINEERING We accept this thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA A p r i l , 1968 In presenting t h i s thesis i n p a r t i a l fulfilment 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 freely available for reference and study. I further agree that permission for extensive copying of t h 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 publication of t h 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. K.M. Agrawal Department of C i v i l Engineering The University of B r i t i s h Columbia Vancouver 8, Canada Date A p r i l , 1968  ABSTRACT Analysis of a general s h e l l of revolution with a r b i t r a r y loading and boundary conditions using the F i n i t e Element approach, well-suited for use with the electronic computer, i s presented. The s h e l l i s approximated •by an assemblage of f l a t , e q u i l a t e r a l trapezoids and isosceles triangles connected to each other at the corners. The assumptions involved i n transforming a piece of plate into a f i n i t e element are defined. Uncoupled plane stress and flexure s t i f f n e s s matrices for the above-mentioned shapes of the f i n i t e elements are derived from considerations of ( i ) s t a t i c s , and ( i i ) v i r t u a l work (energy). Sta t i c s matrices are asymmetric with the exception of the t r i a n g l e plane stress s t i f f n e s s matrix. However, i t i s important to note that irrespective of the size of the trapezoid element, i n conditions of uniform stress the nodal forces s a t i s f y Betti's reciprocal theorem. When a trapezoid reduces to a rectangle, the asymmetry of plane stress and flexure s t i f f n e s s matrices disappears. Asymmetry of the Statics matrix i s removed by averaging the matrix and i t s transpose. This process corresponds to introducing s e l f - e q u i l i b r a t -ing nodal forces which disappear i n conditions of uniform stress. Suitable d i r e c t i o n cosine matrices are derived to transform the displacements and forces from the element coordinate system to the s h e l l coordinate system. The accuracy of the formulation i s demonstrated i n several examples by comparing the f i n i t e element solution v/ith the e l a s t i c i t y s o l ution. The comparison suggests convergence of the results to the correct solution on reduction of the element s i z e . DEDICATED TO MY PARENTS SHANTI DEVI AND GOVIND DAS AGRAWAL TABLE OF CONTENTS Page No. ABSTRACT TABLE OF CONTENTS i LIST OF TABLES v i LIST OF FIGURES x LIST OF SYMBOLS ' x i i i ACKNOWLEDGEMENT CHAPTER I INTRODUCTION 1 1.1 General 1 1.2 The F i n i t e Element Method of Solution 2 1.3 The Flat Polygonal F i n i t e Element 2 1.4 Shape of the Polygonal F i n i t e Element 3 1.5 Types of Stresses i n a Shell 4 1.6 The Nature of a F i n i t e Element 4 1.7 C r i t e r i o n for Selecting Displacement Modes 6 II DERIVATION OF PLANE STRESS STIFFNESS MATRICES 10 2.1 General 10 2.2 Selection of Stress Fields for a Quadrilateral Element 10 2.3 Selection of Displacement Functions to Provide the Assumed Stress Fields. 13 2.4 Evaluation of the Constant Parameters (t>i,...,bg) for an Equilateral Trapezoid F i n i t e Element 15 1 1 TABLE OF CONTENTS (Cont'd) CHAPTER ' Page No 2.5 Evaluation of Stresses at any Point i n a Plate Element. 18 2.6 Derivation of Stiffness Matrices 22 2.7 Derivation of Plane Stress Statics Stiffness Matrix For an Equi l a t e r a l Trapezoid 22 2.8 Sample Procedure to Determine the Equivalent Nodal Forces due to the Displacement Mode bi+^o, a l l other bj,s = 0 23 2.9 Addition of Equivalent Corner Forces from Individual Displacement Modes 26 2.10 Derivation of Plane Stress Energy Stiffness Matrix for an Equilateral Trapezoid 36 2.11 Derivation of Plane Stress Stiffness Matrix for an Isosceles Triangle 44 2.12 Derivation of Plane Stress Statics Stiffness Matrix for an Isosceles Triangle 46 2.13 Derivation of Plane Stress Energy Stiffness Matrix for an Isosceles Triangle 46 I I I DERIVATION OF FLEXURE STIFFNESS MATRICES 50 3.1 General _ 50 3.2 Equation of the Deflection Surface for a Quadri-l a t e r a l Plate Element i n Flexure 50 3.3 Evaluation of the Constant Parameters for an Equ i l a t e r a l Trapezoid Element i n Flexure 53 i i i TABLE OF CONTENTS (cont'd) CHAPTER Page No. 3.4 Moments and Transeverse Shears at a Point 57 3.5 Moments at a Point on the Trapezoid Element 58 3.6 Transeverse Shears at a Point (x,y) on the Trape-zoid Element 63 '3.7 Equivalent Nodal Forces ' 63 3.8 Statics Stiffness Matrix for an Equilateral Trapezoid Element i n Flexure 64 3.9 Forces on Inclined Edges 68 3.10 Sample Procedure to Determine the S t a t i c a l l y Equivalent Nodal Forces due to the Displacement Mode, w = bgxy 2 70 3.11 Equivalent Nodal Forces from a l l Displacement Modes 74 3.12 Energy Stiffness Matrix f o r a Trapezoid Element i n Flexure 81 3.13 Flexure Stiffness Matrix for a F i n i t e Element i n the Shape of an Isosceles Triangle - Selection of -Displacement F i e l d 88 3.14 Choice-1, Omission of bsxy term 90 3.15 Choice-2, Omission of bgxy 2 term 90 3.16 Choice-3, Combination term b (x 2y + xy 2) 90 3.17 Evaluation of the Constant Parameters 92 3.18 Curvatures and Moments at a Point 9 5 3.19 Statics Stiffness Matrix for an Isosceles Triangular Element i n Flexure 95 IV TABLE OF CONTENTS (cont'd) CHAPTER Page No. 3.20 Energy Stiffness Matrix for an Isosceles Triangular Element i n Flexure 95 IV TRANSFORMATION OF STIFFNESS MATRICES FROM ELEMENT COORDINATES TO SHELL COORDINATES 107 4.1 General 107 4.2 Transformation of Deformations and Forces 109 4.3 Transformation Matrix for a Plane Stress Shell Element 119 4.4 Transformation Matrix for a Shell Element having Membrane and Flexure Properties 120 V APPLICATION OF THE FINITE ELEMENT METHOD TO VARIOUS SHELL OF REVOLUTION PROBLEMS 123 5.1 General 123 5.2 A General Solution Procedure 124 5.3 Example-I. Ci r c u l a r Plate Subjected to the Action of a Pair of Diametrically Opposite Point Loads 138 5.4 Example-II. Hemispherical Dome Under Snow Load Uniformly Distributed over the Horizontal Projected Area 151 5.5 Example-Ill. Hemispherical Dome Subjected to Wind Load. 162 5.6 Example-IV. Hemispherical Dome Subjected to Radial Load Applied Along the Entire Edge 213 V TABLE OF CONTENTS (cont'd) CHAPTER Page No. VI DISCUSSION 257 VII CONCLUSION 262 REFERENCES 263 APPENDIX B r i e f Description of Computer Programs and Flow Diagrams 266 LIST OF TABLES Table No. Page No. I I - l Nodal Displacements of an Equilateral Trapezoid Pldte Element i n Terms of the Eight Constant Parameters 0 16 II-2 Parameters {B} i n Terms of the Eight Nodal Displacements 17 II-3 Strains at Any Point (x,y) i n the Trapezoid Element 19 II-4 Stress - Strain Relationship 19 II-5 Stresses at Any Point i n the Trapezoid Element i n Terms of the Nodal Displacements 21 II-6 Equivalent Corner Forces i n Terms of Constant Parameters (bj ,b2 ,.. . ,b8) (equilateral trapezoid element i n plane stress) 27 II-7,A Plane Stress Stiffness Matrix 29 II-7,B Equalities i n the Stiffness Coefficients due to the.Geometrical Symmetry of the Trapezoid about Y-Axis 29 II-7,C Entries of the Statics Stiffness Matrix f o r a Trapezoid i n Plane Stress 32 II-8 Integrated [H] Matrix for a Trapezoid i n Plane Stress 41 II-9 Entries of the Energy Stiffness Matrix for a Trapezoid i n Plane Stress 42 11-10 Nodal Displacements of an Isosceles Triangle i n Terms of Constant Parameters 47 11-11 Evaluation of Constant Parameters for an Isosceles Triangle i n Terms of Nodal Displacements 47 11-12 Equivalent Nodal Forces i n Terms of Constant Parameters for an Isosceles Triangle 48 11-13 Plane Stress Stiffness Matrix for an Isosceles Triangle 48 v i i LIST OF TABLES (cont'd) Table No. ' Page No, 11-14 Integrated [H ] Matrix (Isosceles Triangle i n Plane S t r e s s ) A 49 I I I - l Nodal Deformations for a Trapezoid Element i n Terms of Constant Parameters 54 III-2 Evaluation of Constant Parameters i n Terms of Nodal Displacements and Rotations for a Trapezoid i n Flexure 55 III-3 Curvatures and Twist at Any Point (x,y) i n Terms of Constant. Parameters for a Trapezoid i n Flexure 111-12 Evaluation of Parameters i n Terms of Nodal. Deformations (Isosceles Triangle i n Flexure) 59 III-4 Relation Between Internal Moments and Curvatures and Twist 59 III-5 Moment Matrix for a Trapezoid Element i n Flexure 60 III-6 Transeverse Shear at a Point on the Trapezoid Element 62 III-7 Entries of [C*] Matrix (Trapezoid Element) 75 * III-8 Integrated [H ] Matrix for a Trapezoid 84 III-9 Equalities i n the Stiffness Coefficients due to the Geometrical Symmetry of the Equi l a t e r a l Trapezoid i n Flexure 86 111-10 Equalities i n the Stiffness Coefficients due to the Geometrical Symmetry of the Isosceles Triangle 91 I I I - l l Nodal Deformations i n Terms of Constant Para-meters (Isosceles Triangle i n Flexure) 93 94 111-13 Curvatures and Twist at a Point on a Triangular Element ' 96 * 111-14 Entries of [C^] Matrix (Isosceles Triangular Element i n Flexure) 97 111-15 Statics Stiffness Matrix for an Isosceles Triangular Element i n Flexure 100 v i i i LIST OF TABLES (cont'd) Table No. Page No. * 111-16 Integrated [H^] Matrix for an Isosceles Triangle i n Flexure 103 111-17 Energy Stiffness Matrix for an Isosceles Triangular Element i n Flexure 104 IV-1 Transformation Submatrices for Various Nodes of a Trapezoid 112 IV-2 Transformation Submatrix for the Apex Node of a Triangular Element 116 IV-3 Transformation Matrices for Shell Elements Having Plane Stress Properties 116 IV-4 Transformation Matrices for Shell Elements Having Plane Stress and Flexure Properties 121 TABLES OF PERCENTAGE ERROR ANALYSIS V - I - l C i r c u l a r Plate-Two Point Loads; P-Displacement 141 V-I-2 C i r c u l a r Plate-Two Point Loads; M-Displacement 143 V-I-3 C i r c u l a r Plate-Two Point Loads; P-Force/Length, (N e) 145' V-I-4 C i r c u l a r Plate-Two Point Loads; M-Force/Length, V-I-5 Ci r c u l a r Plate-Two Point Loads; Shear Force/ Length, (N ) 149 V-II-1,2 Hemispherical Dome-Snow Load; M-Displacement 154,155 V-II-3,4 Hemispherical Dome-Snow Load; R-Displacement 156,157 V-II-5 Hemispherical Dome-Snow Load; P-Force/Length, (N Q) 159 V-II-6 Hemispherical Dome-Snow Load; M-Force/Length, (N ) 160 V-III-1 to 6 Hemispherical Dome-Wind Load; P-Displacement 169-174 V-III-7 to 12 Hemispherical Dome-Wind Load; M-Displacement 176-181 i x LIST OF TABi.ES (cont'd) Table No. Page No. V-III-13 to 18 Hemispherical Dome-Wind Load; R-Displacement 183-188 V-III-19 to 24 Hemispherical Dome-Wind Load; P-Force/Length, (N ) 190-195 V-III-25 to 30 Hemispherical Dome-Wind Load; M-Force/Length, (N ) 197-202 V-III-31 to 36 Hemispherical Dome-Wind Load; Membrane Shear, (N ) 203-208 1 6<f>J V-III-37 Hemispherical Dome-Wind Load; Bending Moments 209 V-III-38 Hemispherical Dome-Wind Load; R-Displacements from Membrane and Flexure F.E. Solutions 211 V-IV-1 to 4 Hemispherical Dome-Radial Edge Load; Horizontal 216,218, Displacement 220,222 V-IV-5 to 8 Hemispherical Dome-Radial Edge Load; P-Rotation 224,226, 228,230 V-IV-9 to 12 Hemispherical Dome-Radial Edge Load; P-Force/ 232,234, Length, ( N J 236,238 V-IV-13 to 16 Hemispherical Dome-Radial Edge Load; M-Force/ 240,241, Length, (N ) 243,244 V-IV-17 to 20 Hemispherical Dome-Radial Edge Load; P-Moment/ ' 246,247, Length, (M ) 249,250 V-IV-21 to 24 Hemispherical Dome-Radial Edge Load; M-Moment/ 252,253, Length, (M ) 255,256 LIST OF FICURES Normal and Shear Stresses i n a Plate E q u i l a t e r a l Trapezoid element i n Plane Stress (Positive directions of nodal forces and displacements) Force Di s t r i b u t i o n Due to the Displacement Mode, b 4 Equivalent Nodal Force contributions From Inclined Edges Equivalent Nodal Forces Corresponding to the Displacement Mode, b^ Isosceles Triangular Element i n Plane Stress (Positive directions of nodal forces and displacements) Eq u i l a t e r a l Trapezoid i n Flexure (Positive directions of nodal forces and deformations) Directions of Positive Transeverse Shears and Moments Conversion of Twisting Moment into Distributed Transeverse Shear Conversion of Edge Bending Moment into Equivalent Direct Forces n and t Directions for the Inclined Edges Calculation of Equivalent Nodal Forces on Edges 1-2 and 3-4 of the Trapezoid Element due to the Displacement Mode, w = bgxy 2 Equivalent Nodal Forces from Edge 1-3 (w = bgxy 2) Equivalent Nodal Forces from Edge 2-4 (w = bgxy 2) Isosceles Triangular Element i n Flexure (Positive directions of nodal forces and deformations) Shell Coordinate Directions; P, M and R Location of a F i n i t e Element on the Shell XI LIST OF FIGi'RES (cont'd) Figure No. Page No. IV-3 Relating y and e Angles to Angles <j> , 8 and 9 e 113 IV-4 Derivation of Direction Cosine Matrix at Node 1 117 V-0-1 Positive Directions of Stresses and Moments i n a Shell Element 130 V-0-2 Calculation of Stresses at an Internal Node 131 V-0-3 Calculation of Stresses at an Edge Node 133 V-0-4 Dis t r i b u t i o n of P-stress (N.) on Meridian 1-2 133 GRAPHS OF ELASTICITY SOLUTION V- I - l C i r c u l a r Plate Subjected to the Action of a Pai r of Diametrically Opposite Point Loads 137 V-I-2 C i r c u l a r Plate-Two Point Loads, E l a s t i c i t y Solution for P-Displacement 140 V-I-3 ~ Circular Plate-Two Point Loads, E l a s t i c i t y Solution for M-Displacement 142 V-I-4. C i r c u l a r Plate-Two Point Loads; E l a s t i c i t y Solution for P-Stress 144 V-I-5 C i r c u l a r Plate-Two Point Loads; E l a s t i c i t y Solution for M-Stress 146 V-I-6 C i r c u l a r Plate-Two Point Loads; E l a s t i c i t y Solution for Shear Stress 148 V-II-1 Hemispherical Dome Subjected to Snow Load 150 V-II-2 Hemispherical Dome Subjected to Snow Load -Calculation of Nodal Loads 152 V-II-3 Hemispherical Dome-Snow Load; E l a s t i c i t y Solu-t i o n for M-Displacement 153 V-II-4 ' Hemispherical Dome-Snow Load; E l a s t i c i t y Solu-tion for R-Displacement 153 LIST OF FIGURES (Cont'd) Figure No. Page No. V-II-5 Hemispherical Dome-Snow Load; E l a s t i c i t y Solution f o r P-Stress 158 V-II-6 Hemispherical Dome-Snow Load; E l a s t i c i t y Solution f o r M-Stress 158 V-III-1 Hemispherical Dome Subjected to Wind Load 161 V-III-2 Hemispherical Dome-Wind Load; E l a s t i c i t y Solution f o r P-Displacement 168 V-III-3 Hemispherical Dome-Wind Load; E l a s t i c i t y Solution f o r M-Displacement 175 V-III-4 Hemispherical Dome-Wind Load; E l a s t i c i t y Solution for R-Displacement 182 V-III-5 Hemispherical Dome-Wind Load; E l a s t i c i t y Solution f o r P-Stress 189 V-III-6 Hemispherical Dome-Wind Load; E l a s t i c i t y Solution for M-Stress 196 V-III-7 Hemispherical Dome-Wind Load; E l a s t i c i t y Solution f o r P-Stress 196 V-III-8 Hemispherical Dome-Wind Load; F.E. Solution for R-Displacements along 8 = 0 ° Meridian 210 V-IV-1 Hemispherical Dome Subjected to Radial Load Along the Entire Edge 212 V-IV-2 to 5 Hemispherical Dome-Radial Load Along the Edge; 215,217, E l a s t i c i t y Solution for Horizontal Displacement 219,221 V-IV-6 to 9 Hemispherical Dome-Radial Load Along the Edge; 223,225, E l a s t i c i t y Solution for Rotation about P-Axis 227,229 V-IV-10 to 13 Hemispherical Dome-Radial Load Along the Edge, 231,233, E l a s t i c i t y Solution for P-Stress 235,237 V-IV-14 to 17 Hemispherical Dome-Radial Load Along the Edge, E l a s t i c i t y Solution for M-Stress 239,242 V-IV-18 to 21 Hemispherical Dome-Radial Load Along the Edge, E l a s t i c i t y Solution for P-Moment 245,248 V-IV-22 to 25 Hemispherical Dome-Radial Load Along the Edge, E l a s t i c i t y Solution f o r M-Moment 251,254 L I S T O F S Y M B O L S Symbol a h a l f of the top side of a trapezoid & i , . . . , a 1 2 constant parameters b^ constant parameter associated with the i th displacement mode c projection of the i n c l i n e d side of a trapezoid on the X-axis, positive i f the bottom, side i s bigger than the top side; also, h a l f of the base of an isosceles triangle Dj = E t / ( l - y 2 ) , extensional r i g i d i t y per unit length of the plate D2 = Et 3/12 (1-y 2), fle x u r a l r i g i d i t y per unit length of the plate E modulus of e l a s t i c i t y f. nodal force i n the i th direction 1 h h a l f of the height of a trapezoid or a triangle K r a t i o of the bottom side to the top side of a trapezoid; also, r a t i o of height to base of a triangle K]^  r a t i o of height to top side of a trapezoid K. . s t i f f n e s s cbefficient belonging to i throw and j th column of 1 t h e s t i f f n e s s matrix M bending moment/length i n n-direction (perpendicular to the i n c l i n e d edge) M twisting moment/length on n-plane M , M bending moments/length i n X and Y directions x y M ' , M twisting moments/length on X and Y planes xy yx M Q , M , bending moments/length i n 6 and <j> ( p a r a l l e l and meridian) * directions' M ,M twisting moments/length on 9 and <f> planes n no. of corners i n a f i n i t e element; also, direction normal to the i n c l i n e d edge N t o t a l degrees of freedom of•corner movements of an element xiv LIST OF SYMBOLS (cont'd) Symbol N ,N membrane direct force/length i n 9 and tj> ( i . e . p a r a l l e l and meridian)directions membrane shear on 9 and <J> planes intensity of the distributed load s h e l l coordinate directions (tangent to the p a r a l l e l c i r c l e , tangent to the meridian c i r c l e and normal to the s h e l l surface) transeverse shears on X and Y planes radius of curvature of the s h e l l surface at a point radius of the p a r a l l e l c i r c l e containing lower nodes 1 and 2 of an element Ry radius of the p a r a l l e l c i r c l e containing upper nodes 3 and 4 of an element t thickness of a plate element (assumed same as shell) u,v displacements i n X and Y directions in the plane of the plate w displacement i n Z-direction We. external work done by the nodal forces of the f i n i t e element due to a v i r t u a l displacement {&} . which i s unity i n the j t h direction and zero i n a l l other-1 directions Wi. internal work done by the internal forces of the plate element due to a v i r t u a l displacement {^}j x,y coordinates of a point on the plate element XYZ element coordinate direction - right hand system X-Y-Z s h e l l coordinate directions at the apex a accute angle i n the plane of the f l a t element between Y-axis and the in c l i n e d side - positive Xf counter-clockwise ai,ot2,... constant coefficients g . at the lower node, angle between tangent to the meridian and the chord line joining the lower and the upper nodes of the element N ,N 8<)> <)>9 P P,M,R Q ,Q x x ' y LIST OF SYMBOLS (cont'd) xv Symbol 3 1 at the upper nodes, angle between tangent to the meridian and the chord l i n e j oining the lower and the upper nodes of the element ({3,= <£>u _ 4*^- £) Y angle of i n c l i n a t i o n of the element with the plane of a p a r a l l e l c i r c l e 6^ nodal displacement i n the i t h direction e h a l f the accute angle between the planes of two equal adjacent f l a t elements intersecting along a common inc l i n e d edge e ,e direct strains i n X and Y directions x y i e shear s t r a i n xy 0 angle between the planes of reference meridian and a meridian passing through a point on the s h e l l j looking from top - counter-clockwise angle positive 9 angle between the adjacent meridian planes which enclose the element 9 horizontal angle between the base of a triangle element and the X-direction at the apex y Pbisson's r a t i o £ horizontal displacement a ,o direct stresses i n X and Y directions x' y a shear stress xy $ . angle between the axis of rotation and tangent to the meridian at a point on the s h e l l $ difference of the angles made by tangents to the meridian at upper and lower nodes of an element with the axis of rotation (= ^  - <j>L) <j>k,<J>jj angle between the axis of rotation and tangent to a meridian at lower and upper nodes respectively of an element v rotation about P-axis XVI LIST OF SYMBOLS (cont'd) MATRIX SYMBOLS Note: (i) a subscript A refers to a triangle element ( i i ) a superscript * refers to a flexure element Symbol [A] N x N matrix i n terms of nodal coordinates {B} N x 1 matrix of constant parameters; b 1,b 2,...,b N [C] N x N matrix of s t a t i c a l l y equivalent corner forces. Entries of the j t h column of this matrix correspond to s t a t i c a l l y equivalent nodal forces due to a displacement mode b.. [D] 3 x 3 symmetric matrix containing e l a s t i c constants {f} N x 1 matrix of nodal forces; f 1 , f 2 , . . . , f N ^XYZ'^PMR f ° r c e s i - n element coordinates X-Y-Z and i n s h e l l coordinates 1 P-M-R respectively [G] 3 x N matrix i n terms of variables x and y T [H] . N x N symmetric matrix defined by ff [G] [D][G]dxdy [I] N x N iden t i t y matrix [K] N x N sti f f n e s s matrix of the f i n i t e element [ K ] X Y Z , [ K ] p M R s t i f f n e s s matrix i n element coordinates X-Y-Z and s h e l l coordinates P-M-R respectively 4M*V 3 X 1 moment matrix (M ,M and M ) \ J • x' y xy' [MT*]. 3 x N moment matrix i n terms of.variables x and y [Q*] 2 x 1 matrix of transeverse shears (Q x and Q^ ) [R*] 2 x N; matrix i n terms of variables x.and y [S] entire structure s t i f f n e s s matrix [ST] 3 x N stress matrix i n terms of variables x and y [T] transformation matrix [T 1] 3 x 3 matrix'containing cosines of the angles between P-M-R axes and the XYZ axes at node i LIST OF SYMBOLS (cont'd) x v i i Symbol [T 3] 3 x 3 matrix containing cosines of the angles between A X-T-E, directions of the s h e l l at the apex and X-Y-Z directions of a triangle {We}. N x 1 matrix of external work done by the nodal forces of the f i n i t e element due to a l l v i r t u a l displacement {&}., j = 1,2,.. .,N 3 {Wi} N x 1 matrix of internal work done by the internal forces of the plate element due to a l l v i r t u a l displacements {<§}. , j = 1,2,...,N 3 {&] N x 1 matrix of nodal displacements ;61,62 >••• »<$ N {<$} . N x 1 matrix o f . v i r t u a l displacements; here, the displace-3 ment i n the j t h direction i s equal to unity and zero i n a l l other directions [6] N x N matrix of a l l v i r t u a l displacements; i . e . {6}., i = 1,2,.,.,-N = N x N iden t i t y matrix, [I] ^XYZ'^PMR n o d a l displacements i n element coordinates X-Y-Z and s h e l l coordinates P-M-R respectively {a} structure deformation vector \e} 3 x 1 matrix of strains (e , e and e ) L ' ^ x y xy [c]j 3 x 1 matrix of strains due to a v i r t u a l displacement, {<$}.. {a} 3 x 1 matrix of stresses (a , a and a ) 1 J x' y xy r * i • ^ ^ /-32w 82w „92w -. {x } 3 x 1.matrix of c u r v a t u r e s > ~ > 2J£iy * {x*} • 3 x 1 matrix of curvatures due to v i r t u a l deformations, {<S*}. ACKNOWLEDGEMENTS The author i s greatly indebted to his thesis d i r e c t o r , Dr. A. Hrennikoff for recommending the thesis topic and for of f e r i n g suggestions and c r i t i c a l checks at various stages of the study. His keen insight into the physical behaviour of f i n i t e elements greatly helped the development of the subject. The author i s also grateful to Dr. S. Tezcan, Dr. S. Cherry, Dr. B. Ovunc and other members of his Ph.D. committee with whom he had helpful discussions. Thanks are also due the University of B r i t i s h Columbia Computing Centre for free use of f a c i l i t i e s , Diane Johnson for e f f i c i e n t typing of the thesis, and f i n a l l y my wife, Christine, whose constant encouragement made t h i s thesis become r e a l i t y . CHAPTER I INTRODUCTION 1.1 General Shells of revolution are structural forms whose middle surfaces are generated by rotating a plane curve about an axis i n i t s plane. Conical she l l s and domes are examples of shells of revolution. These st r u c t u r a l forms provide a unique fe e l i n g of s p a t i a l continuity and architectural beauty and are used to enclose large column-free spaces. Due to the curved nature of t h e i r surface, s h e l l s are capable of carrying external loads mainly by membrane action. In buil d i n g construction,they are as ancient as the Roman Pantheon of 120 A.D. The Pantheon, b u i l t by the Roman Emperor, Hadrian, for the worship of pagan gods, i s a massive dome structure over 4 f t . i n thickness.* Today, however, with better understand-ing of the inherent strength i n shape, shel l s are being b u i l t with thick-2 nesses as l i t t l e as 2.5 inches. Rigorous solution for deflections and stresses i n shell s of revolution are obtainable for very simple geometry and certain s p e c i f i c boundary and loading conditions. However, when complicated geometry and arbitrary boundary and loading conditions are involved, the solution of the governing d i f f e r e n t i a l equation becomes extremely complicated and often impossible. 2 1.2 The F i n i t e Element Method of Solution Approximate solution of a continuum problem can be obtained by recourse to the F i n i t e Element (F.E.) method. For axisymmetric shells,the f i n i t e element method has been suggested using:(1),elements i n the form of 3-5 a frustrum of a cone ,and ( 2),doubly curved elements i n the form of a frustrum of a s h e l l of revolution.^ The above methods consider axisymmetric deformations only. The extension to unsymmetric deformations, by means of 7 Fourier expansion, i s suggested by Percy et a l . An extensive survey of the e x i s t i n g methods of analysis of shells of revolution i s presented by 8 Jones and Strome. However, these methods are r e s t r i c t e d to axisymmetric boundary conditions. It i s therefore desirable to have a method of solution, which w i l l handle a completely general s h e l l of revolution problem with arbitrary loading and boundary conditions. 1.3 The Flat Polygonal F i n i t e Element In the present analysis, the r e s t r i c t i o n s of axisymmetric loading and boundary conditions involved i n the use of f i n i t e elements i n the shape of a frustrum of a cone are removed. The use of polygonal elements, connected to each other at the corners, allows the solution of a general s h e l l of revolution with completely arbitrary boundary and loading conditions. The matrix algebra method of structural analysis, best-suited for use with the high-speed electronic computer, i s applicable. I t has, at times, been suggested that the curved shape of poly-gonal s h e l l elements should be used to avoid d i f f i c u l t y of discontinuities i n geometry. This may increase the accuracy of the solution, but w i l l also undoubtedly increase the derivation complications and computational 3 d i f f i c u l t i e s of the force-displacement relations of the f i n i t e element. The need of curved elements may not arise i f i t can be shown that good approxima-t i o n to the true solution can be achieved using f l a t elements. In a s h e l l problem, the size of subdivision necessary to obtain reasonable accuracy i s not obvious, but i n the regions of rapidly-changing displacements and stresses, small sized elements must be used. I f the r a t i o of the radius of curvature of the curved element to i t s side i s large, the influence of curvature on the force-displacement properties of the element i s small,and i n such a case,a curved element e s s e n t i a l l y behaves as a f l a t element. Thus, the use of f l a t element seems to be adequate for the analysis of s h e l l s . 1.4 Shape of the Polygonal F i n i t e Element A model of a s h e l l of arbitrary shape can be made from an assemblage 9 10 of f l a t t r iangular elements. ' However, i t has been shown that i n general quadrilateral elements provide greater accuracy than the triangular 11 12 elements. ' Yet, i n special circumstances the use of triangular elements cannot be completely avoided. In a s h e l l of revolution, advantage of geometrical symmetry i s taken i n selecting the shape of quadrilateral f i n i t e elements. The s h e l l surface i s subdivided by a number of meridian lines and p a r a l l e l c i r c l e s . The curved portion of the s h e l l enclosed between two meridian lines and two p a r a l l e l c i r c l e s i s approximated by a f l a t surface i n the shape of an equ i l a t e r a l trapezoid. The elements j o i n i n g the apex are i n the shape of isosceles triangles. Therefore, the solution of a s h e l l of revolution w i l l involve f i n i t e elements i n the shape of e q u i l a t e r a l trapezoids and isosceles t r i a n g l e s . This approximation introduces some errors i n the results apart from those due to the f i n i t e size of the elements. However, as the element size decreases, the model approaches the 4 continuously curved shape sthereby reducing the errors. 1.5 Types of Stresses i n a Shell The scope of the study here i s limited to e l a s t i c shells of revolution subjected to small deflections. A s h e l l normally has both membrane and flexure stresses induced i n i t . It i s therefore necessary to assign both plane stress and flexure properties to the f i n i t e element used i n the analysis. For small deflections, the equations of s t a t i c s are written for an undeformed plate. This permits an independent derivation of the stress-deformation relations of an element i n plane stress and flexure which are f i n a l l y reduced to the combined re l a t i o n s . 1.6 The Nature of a F i n i t e Element The nature of a plane stress f i n i t e element may be explained i n the following manner. Consider a homogeneous plate of large dimensions i n X-Y plane subjected to a simple stress condition such as uniform unidimen-sional s t r a i n or shearless bending i n X-direction. Remove from t h i s plate a polygonal area of the shape of the assumed f i n i t e element. I t i s held i n equilibrium by the appropriate edge stresses. Assume now that the edge stresses are replaced by proper s t a t i c a l l y equivalent nodal forces, the corner displacements remaining unaffected by t h i s operation. This trans-forms the given piece of plate into the f i n i t e element proper which i s i n .effect a body whose corners move i n several simple s t r a i n conditions through the same distances as the same points on the plate when acted upon by corner-forces s t a t i c a l l y equivalent to the edge-stresses. Each s t r a i n condition may be viewed as an independent displacement mode. The f i n i t e element must act i n such manner under several simple s t r a i n conditions including uniform strains i n X and Y directions and uniform shearing s t r a i n . The number cf these independent displacement modes including the three r i g i d body moJes i s selected to match the t o t a l number of possible movements of the corners of the f i n i t e element. A displacement expression, obtained by combining these displacement modes i n certain proportions through coefficients or parameters, i s known as the displacement function. Proper selection of the values of these parameters allows to effect separate X or Y displacements of the corners of the f i n i t e element. By converting the combined edge stresses i n the element into s t a t i c a l l y equivalent corner forces, i t ' i s possible to formulate the force-displacement r e l a t i o n , i . e . , the s t i f f n e s s matrix of the f i n i t e element [K] r e l a t i n g the corner-forces {f} with the corner-displacements {6} by the equation^ The nature of a flexure f i n i t e element.whose corners are free to deflect and rotate about two axes i n i t s plane i s explained i n a s i m i l a r manner. Here, the simple s t r a i n conditions are replaced by the simple curvature conditions,and edge stresses by edge moments and transeverse shears. The selected independent curvature modes must include the uniform curvatures i n X and Y directions and uniform twist. There are two d i s t i n c t methods available for conversion of the edge stresses (or moments) into the nodal forces: the direct s t a t i c s method,and the energy method. The entries of the s t i f f n e s s matrix [K], known as the s t i f f n e s s - c o e f f i c i e n t s , obtained by these methods are usually different. i t i s curious that the energy derivation of the s t i f f n e s s matrix i n the preserce of a certain qualifying condition coincides exactly with the result of application of the Rayleigh-Ritz method to the structure as a whole, an approach completely unrelated to the f i n i t e element method. In the analysis of a problem by the f i n i t e element method, the results depend on the s t i f f n e s s matrix used. The s t i f f n e s s matrix depends on the displacement modes employed and also on the manner of derivation of the corner forces from the edge stresses. Thus, the selection of these dis-placement modes i s important and i s based on certain q u a l i f y i n g conditions. 1.7 Cri t e r i o n for Selecting Displacement Modes I f the displacement modes, related to an element, are so selected that continuity of deflections and slopes i s always preserved on the i n t e r -element boundaries, then the force-displacement relations can be derived by the Rayleigh-Ritz method. In such a case,the formulation always leads to an underestimate (iower bound) of s t r a i n energy of the structure. On successive reduction of the element s i z e , the s t r a i n energy of the f i n i t e element approximation approaches the true s t r a i n energy of the structure monotoni-13-19 c a l l y from below. However, i t should be noted that the c r i t e r i o n of continuity of deformations on the interelement boundaries i s a s u f f i c i e n t condition for monotonic convergence of results to the true solution with regard to the s t r a i n energy of the entire system but not a necessary condi-20 ti o n for monotonic convergence of stresses to t h e i r true values. Clough used displacement modes which do not preserve continuity of displacements along element boundaries for his plane stress rectangular f i n i t e element and demonstrated convergence of displacements and stresses to t h e i r true values. 7 What interests an engineer i s the stresses and to a smaller extent deflec-tions. Energy as such i s of no interest to him unless i t i s a proof of the correctness of solution. In a general case, selection of displacement functions that w i l l s a t i s f y continuity of a l l deformations along element boundaries i s 18 2 1 ' impossible. Bazely, Cheung, Irons and Zienkiewcz assert that i t i s possible to achieve convergence without the conformity requirements. They state that the only necessary condition for convergence of the solution to i t s true value i s that of the displacement function being capable of representing r i g i d body movements and constant s t r a i n (curvature) states throughout a f i n i t e element irrespective of i t s size or shape. Thus, i f the selected displacement function having dis c o n t i n u i t i e s at the inter-element boundaries i s such that as the element size decreases continuity at the i n t e r -element boundaries tends to be restored, then the formulation must tend to the correct solution. I f the constant s t r a i n (curvature) c r i t e r i o n i s used i n the basic derivation of the displacement function then as the elements decrease i n d e f i n i t e l y i n s i z e , the continuity at the nodes w i l l require a constant s t r a i n (curvature) state to be reached i n the l i m i t within each element. This state of constant s t r a i n (curvature) w i l l automatically require that compatibility of the deformations e x i s t across the interelement 21 22 boundaries. ' 23 Hrennikoff demonstrates the necessity and sufficiency of the presence of constant stress states i n the formulation of plane stress s t i f f n e s s matrix, derived using the s t a t i c s approach, for convergence of the calculated results to i t s true value. He views the f i n i t e elements i n the l i m i t , as the molecules of a continuum and defines stresses as i n the theory o-f e l a s t i c i t y . The d i f f e r e n t i a l equations of the theory of e l a s t i c i t y 8 governing the conditions of plane stress are based on s t a t i c s , continuity of the material and i t s e l a s t i c i t y properties. He shows that these equations are equally applicable to the model which Jn the l i m i t becomes i n effect a s i m p l i f i e d representation of the molecular structure of the plate. This observation i s e a s i l y extended to f i n i t e elements i n flexure where the constant s t r a i n condition i s replaced by the constant curvature condition. In a s h e l l model, where the continuously curved surface i s approximated by f l a t polygonal elements meeting at an angle, a s h e l l dis-placement w i l l produce both inplane and out of plane displacements i n the f l a t plate elements and the continuity along the element boundaries would be even harder to s a t i s f y than either i n plane stress or flexure of a 13-17 plate. Thus, the Rayleigh-Ritz sufficiency condition appears a l l but impossible i n the application to s h e l l s . However, convergence of the results to the true solution can be s t i l l claimed by selecting displacement functions that w i l l s a t i s f y the constant s t r a i n (curvature) c r i t e r i o n of 21 22 Bazely et a l and Zienkiewicz. In the following study a piece of plate, whose displacement and stress properties are defined at any point within i t s boundaries, w i l l be referred to as a plate element whereas when the force and displacement properties are defined only at the nodes, i t w i l l be referred to as a f i n i t e element. In Chapter II plane stress and i n Chapter I I I flexure s t i f f n e s s matrices for e q u i l a t e r a l trapezoid and - isosceles triangular elements have been derived using s t a t i c s and v i r t u a l work approaches. These matrices refer to the element coordinate directions. In Chapter IV, transformation matrices are derived which transform the s t i f f n e s s matrices from element coordinates to s h e l l coordinates. Several examples demonstrating the s u i t a b i l i t y of both types s t i f f n e s s matrices and convergence of calculated results to the true solution on reduction of the element size are presented i n Chapter V. CHAPTER II DERIVATION OF PLANE STRESS STIFFNESS MATRICES 2.1 General The s t i f f n e s s matrix formulation may be based on the assumption of simple states of stresses or displacements of an element. In the following formulation of plane stress s t i f f n e s s matrices, assumptions are made on the states of stress i n the element. Displacement surfaces are then selected i n the form of polynomials i n x and y to provide the assumed stress states. 2.2 Selection of Stress Fields for a Quadrilateral Element In a plane stress problem, a corner or a node of a f i n i t e element i s allowed to displace independently i n two perpendicular directions. Thus, a polygonal plate element having n nodes has 2n degrees of freedom to d i s -place as these nodes move while the element deforms under stress. However, by r i g i d body movements three of these displacements may be made zero without i n t e r f e r i n g i n any way with the stresses imposed on the plate element. This leaves (2n-3) independent stress conditions required for the formulation of the force-displacement r e l a t i o n s , i . e . , s t i f f n e s s matrix of the f i n i t e element irrespective of i t s actual shape. For a quadrilateral element with four nodes, the number of these independent stress conditons i s (2)(4)-(3)=(5). Three of these stress 11 . , • conditions must be constant direct stresses i n two perpendicular directions and one constant shear. The two additional conditions of the stresses selected are, shearless bending i n the plane of the plate with the direct stresses vary-ing l i n e a r l y i n the two perpendicular directions. In x-y coordinate system, (Fig. ( I I - l ) ) , t h i s corresponds to the selection of stress states as (i ) Constant a ... ( i i ) Linear function of y ( i i i ) Constant ^ °y J (iv) Linear function of x a (v) Constant xy Here, o^, a - direct stresses i n x and y directions respectively. a - shear stress. xy Since the plate material obeys Hooke's law, the stresses through-24 out the element are related to the strains as, ( e v + i i e j f £ y + v*'y? (II-2) E (1 E (1 E xy 2(l+y) xy 12 FIG. (IE-1) NORMAL AND SHEAR STRESSES IN A PLATE „ . 2(a+c)  K " ~2o K, 2h 2a FIG.(II-2) EQUILATERAL TRAPEZOID ELEMENT IN PLANE STRESS. (Positive directions of nodal forces and displacements.) Here, E = Modulus of e l a s t i c i t y u = Poisson's r a t i o e ,e = direct s t r a i n i n x and y directions respectively x y e = shear s t r a i n The strains may be expressed as the p a r t i a l derivatives of the displacements. 9u 3v 3u 9v_ rn %-\ e x 9x ' E y 9y exy " 8y dx Ui~,i-' Here, u and v are displacements i n x and y directions respectively. 2.3 Selection of Displacement Functions to Provide the Assumed Stress Fields A quadratic polynomial i n x and y i s assumed to represent u and v displacements of a point i n the plate element. u = aj + a 2x + a3y + a^x 2 + asxy + agy 2 v = + aQX + a gy + a 1 0 x 2 + a l l x y + a l 2 y 2 (II-4,a) Here, a l s a 2, a 1 2 are constant parameters. Substituting the expressions for u and v into the Eqns. (II-3) and (II-2) i t i s found °x = ( l - y 2 ) ^ a 2 + + (2ai+ + y a n ) x + ( a 5 + 2ya 1 2)y] E O y = ( l - p 2 ) ^ y a 2 + a9^ + f 2 ^ + a l l ) x + (^a5 + 2 a 1 2 ) y ] (U-4,b) °xy = 2(1 + M ) [ ( a 3 + a e ) + ( a 5 + 2 a 1 0 ^ X + ( 2 a6 + a l l ) y l 14 For Eqn. (II-4,b) to agree wi ;-.h Eqn. (II -1) (II--4,b) must vanish. These are: (i) 2a 4 + y a n = °y or, a n = 2 y ait ( i i ) y a 5 + 2 a 1 2 = 0; or, a 1 2 = y 2 *5 ( i i i ) a 5 + 2 a 1 0 = 0; or, a10 = l 2 as and (iv) 2a 6 + a n = 0; or, »6 = l 2 a l l (H-4,c) — att y Thus, there are only eight independent constants i n the quadratic polynomials of Eqn. (II-4,a). Renaming these constants as b l = a l J b2 = a2> b 3 = a3> bh = a5 b 5 = b 6 = a 9 ; b 7 = a 8; b 8 = a 7 and rewriting Eqn. (II-4,a) i n terms of the eight new constants, one obtains u = hi + b 2x + b 3y - bi t(yx 2 + y 2) + 2b 5xy (II-5) v = b 8 + b 7x + b 6y - b 5 ( x 2 + yy 2) + 2bltxy Constants bj and b 8 correspond to the r i g i d body displacements and (b 3-b 7) corresponds to r i g i d body rotation of the plate element. The other constants simply provide fi v e independent stress conditions as given below. 15 o Y = ^- [(b 2 + yb 6) + 2 t 5 ( l - y 2 ) y ] Constant Linear function of y o v = ^- [(yb 2 + b 6) + 2b l t(l-y 2)x] Constant Linear function of x (II-6) Constant Et Here, = ^ Extentional r i g i d i t y per unit length of the plate t = thickness of the piece of plate (constant) The displacement functions of Eqn. (II-5) include constant s t r a i n 21 22 states - a c r i t e r i o n suggested by Bazely et a l and Zienkiewicz for the convergence of solution to the true r e s u l t . 2.4 Evaluation of the Constant Parameters (bi ,b2 ,. . . ,b8) for an Equilateral Trapezoid F i n i t e Element Equation (II-5) provides displacements at any point (x,y) of the plate element i n terms of the eight parameters (b^ ,b2,... ,bs) and the coordinates of that point. Eight independent equations for nodal displace-ments 6 I , 6 2 , 6 3 are obtained i n terms of the eight parameters. These are written i n the matrix form i n Table ( I I - l ) . In short, these can be written as W = [A]{B] (II-7) Here, {&} [A] •W = 8 x 1 column vector of nodal displacements 6^  , 6 2 , 6 Q (Fig. 11-2) = 8 x 8 matrix i n terms of nodal coordinates = 8 x 1 column vector of constant parameters, bi,...,bs. Table ( I I - l ) Nodal Displacements of an Equilateral Trapezoid Plate Element i n Terms of the Eight Constant Parameters Fig. (II-2) <*1 1 (a + c) -h -{y(a + c ) 2 + h 2} -2(a + c)h. 0 0 0 b l 6 2 0 0 0 -2(a + c)h -{(a + c ) 2 + yh 2} -h (a + c) 1 b 2 «3 1 -(a + c) -h -{u(a + c ) 2 + h 2} 2(a + c)h 0 0 0 b 3 64 = 0 0 0 2(a + c)h -{(a + c ) 2 + yh 2} -h -(a + c) 1 b 4 65 1 a h -{ya 2 + h 2} 2 ah 0 0 0 b 5 «6 0 0 0 2 ah -{a 2 + yh 2} h a 1 b 6 6 7 1 -a h -{ya 2 + h 2} -2ah 0 0 0 b 7 0 0 0 -2ah -{a 2 + yh 2} h -a 1 b 8 {&} = [A]{B} Eqn. (II-7) Table ( I I - 2 ) Parameters {B} i n Terms of the Eight Nodal Displacements b i 1/4 -013014 1/4 a 3a 1 + 1/4 <*3°<5 1/4 - « 3 a 5 b 2 <*6 0 -<*6 0 0 - a 7 0 b 3 -o 8 - y a ^ -o 8 ya 2a l t a 8 ya 2ot 5 a 8 . -ya 2a 5 bit = 0 0 0 « 5 0 -015 b 5 -uk 0 cu, 0 ' J a 5 0 -<*5 0 b 6 - 0 2 ^ . - a 8 - a 8 a 2 a 5 a 8 - a 2 a 5 <*8 b 7 0 C6 0 "«6 0 a 7 0 -a 7 b 8 -<*l«it 1/4 aiatt 1/4 0 4 0 : 5 . 1/4 ~ai<*5 1/4 {B} = Eqn. ( I I - 8 ) Here, «1 = i '+ (a + c )2 "2 + 2yh 2 h «2 - ;a2 - (a 1 2h + C ) 2 } ; a 3 = 2h2 + y{a 2 0 6 2 | 6 3 j 5. 66| >8 a 4 = 8h(a + c) ' <*5 1 . = 1 _ 1 1 8ha ' a6 4 ( a + c ) J «7 " 4^ ; a 8 = ^ -18 The parameters, {B}, are evaluated i n terms of the nodal displace-ments as {B} ="[A"1]{6} (II-8) The [A - 1] matrix i s given i n Table (II-2). Knowning the corner displacements {&} of a plate element, the displacements at any point of the element are obtained simply by substituting Eqn. (II-8) i n Eqn. (II-S) 2.5 Evaluation of Stresses at Any Point i n a Plate Element Using Eqn. (II-5) for displacements, strains at any point (x,y) of the plate are given by e x = 3u _ 3x ~ b 2 e = y 3v 3y ~ be 3u 3v xy " W 9* (II-9) b3 + b? Writing i n the matrix form, these are W = [G]{B} (11-10) [G] i s a 3 x 8 matrix and i s shown i n Table (II-3). Substituting Eqn. (II-8) for {B} i n Eqn. (11-10) one obtains W = [G][A"1]{6} (11-11) 19 Table (II-3) Strains at Any Point (x,y) i n the Trapezoid Element _ — e x 0 1 0 -2yx 2y 0 0 0 e y = 0 0 0 2x -2yy 1 0 0 exy 0 0 1 0 0 0 1 0 b 2 b 3 b 4 b 5 b 6 by b 8 {e} = [G]{B} Eqn. (11-10) xy Table (II-4) Stress-Strain Relationship E. yE (1 - y2) yE (i -y2) o (i - y2) E (i - y?) o 0 0 E 2(1 + y) xy {a} = [D]{e} Eqn. (11-12) 20 Stress-strain relationships of Eqn. (II-2) can be rewritten i n matrix form as [a] = [D]{e} (11-12) The (3 x 3) symmetrix matrix [D] contains the e l a s t i c constants and i s shown i n Table (II-4). Substituting Eqn. (11-11) for {e} i n Eqn. (11-12) one gets, {a} = [D][G][A-!]{6} (H-13) Letting, [ST] = [D][G][A" 1] (II-14,a) Eqn. (11-13) i s rewritten as {a} = [ST]{6} (II-14,b) (3 x 8) stress matrix [ST] i s shown i n Table (II-5). Thus, knowing the nodal displacements for any p a r t i c u l a r element, stresses at any point within the element may be obtained by simply substitut-ing the coordinates of that point i n the matrix Eqn. (11-14,b). Table (II-5) Stresses at Any Point i n the Trapezoid Element i n Terms of the Nodal Displacements 1 2 3 4 L x y J (2h-uot2) - 2(1 - y 2)y E 8h(a+c) 2uh-ct2 8h(a+c) (i-y) 8h y 4h f 2 ( l - y 2 ) x + 2(a+c)v i 8h(a+c) 1 (1-y)(2h-yg 2) 16h(a+c) f(2h-yg ?) - 2 ( l - y 2 ) y ] 1 Rhfa + r-V ' 8 ( c) r2yh-ct2 \ l8h(a+c) 1(i-y) 8h y 4h { 2 ( l - y 2 ) x - 2(a+c) 8h(a+c) •(l-y)(2h-yg ?) 16h(a+c) f (2hnict2) + 2 ( l - y 2 ) y i y_ t. 8ha r 4h 2yh+a2 8ha (i-y) 8h  4h 2a+2(l-^y2)x 8ha (l- u)(2h+ya 2) 16ha j(2h+ y a 2 ) + 2 ( l - y 2 ) y i _ 1 8ha 1 2yh+a2 8ha (i-y) 8h y 4h 2a-2(l-y 2)x 8ha (1-y)(2h+ya2) 16ha 6 l 62 63 64 65 «6 6 7 68 L J -c(2a+c) Here, a 2 = ^ — {a} = [ST]{6} Eqn. (II-14,b) 22 2.6 Derivation of Stiffness Matrices Force-displacement relations or "he s t i f f n e s s matrix of the f i n i t e element i s obtained by converting the edge stresses corresponding to individual displacement modes into equivalent nodal forces. There are two d i s t i n c t methods for conversion of the edge stresses into the equivalent nodal forces (i) The distributed edge stresses are transformed into equivalent nodal forces by the simple Taw of the lever. The s t i f f n e s s matrix so obtained w i l l be referred to as the Statics Matrix, and ( i i ) The p r i n c i p l e of v i r t u a l work i s used for obtaining the equivalent nodal forces. The s t i f f n e s s matrix so obtained w i l l be referred to as the Energy Matrix. 2.7 Derivation of Plane Stress Statics Stiffness Matrix for an Equilateral  Trapezoid. Each c f the component displacement modes i s considered separately. The edge stresses are apportioned to the connecting nodes by the simple law of the lever. In the present case, only l i n e a r l y distributed direct stresses and constant shear stress i n x and y directions are present. Collecting direct stresses by the law of the lever i s consistent with s t a t i c s . For uniformly distributed shear stress, i t i s only l o g i c a l to apportion h a l f of the t o t a l shear force on the edge to each of the connecting nodes. The stresses on the i n c l i n e d sides of the trapezoid act at an angle. I t i s convenient to replace the i n c l i n e d sides by i n f i n i t e s i m a l steps with horizontal and v e r t i c a l parts, the stresses on which are respectively v e r t i c a l and horizontal. The components of corner forces are then found by transferring the stresses acting on the projected sides i n accordance with s t a t i c s . 2.8 Sample Procedure to Determine the Equivalent Nodal Forces Due to the Displacement Mode b^O, A l l other b-j's = 0 The displacements for t h i s mode are given by u v = 2bi+xy Corresponding to these displacements, the strains are e = — = -2yb ux ; e = — = 2bhX x 9x 4 ' y dy 4 xy 8y 8x Using Eqn. ( I I - 2 ) , the forces/length of the plate are; Direct force/length i n x-direction = ta-' = 0 Direct force/length i n y-direction = ta = 2Etb4X Shear force/length = to = 0 ° xy Here, t = thickness of the plate element. The d i s t r i b u t i o n of these forces on the element i s shown i n Fig. (II-3). The equivalent nodal forces are collected for each edge separately using the law of the lever and are shown i n Fig. (II-4). F i n a l l y , the nodal forces from a l l edges are combined Fig. (II-5). = -b 4(yx 2 .+ y 2 ) ; 24 25 pa pa Edge 2-4 FIG(U-4) EQUIVALENT NODAL FORCE CONTRIBUTIONS FROM INDIVIDUAL EDGES - | ( a + c ) ( 2 a + c) ^ A -|(a+c)(2a+c) (Here,p = 2Etb 4 ) t pa(2a+c) t-pa(2a + c) ' 6 * 6 * FIGCLT-5) EQUIVALENT NODAL FORCES CORRESPONDING TO THE DISPLACEMENT MODE b 4 26 2.9 Additioi. of Equivalent Corner Forces from Individual Displacement Modes For o. general deformed state of the plate element given by Eqn. (II-5), the equivalent corner forces are simply the sums of equivalent corner forces from in d i v i d u a l displacement modes. Although there are only fiv e independent stress modes, for the purpose of compatible matrix m u l t i p l i -cation a l l the eight displacement modes are considered. The equivalent force contributions from modes corresponding to r i g i d body motion w i l l be zero. C o l l e c t i v e l y , the equivalent forces for a l l the displacement modes are written i n the matrix form as {f} = [ C ] { B } (11-15) Here, | f j = 8 x 1 column vector of equivalent corner forces i n the pos i t i v e directions of Fig. (II-2). [C] = 8 x 8 matrix. Jth column of t h i s matrix contains equivalent corner forces i n eight p o s i t i v e directions of Fig. (II-2) due to a displacement mode bj The matrix Eqn. (11-15) i s shown i n f u l l i n Table (II-6). Using Eqn. (II-8) to substitute for {B} i n Eqn. (11-15), the nodal forces are: {f} = [C][A-1]{5} (H-16) Thus, Eqn. (11-16) gives the relationship between nodal forces and nodal displacements. This relationship i s commonly known i n the form {f} = [K]{6} (11-17) Table (II-6) Equivalent Corner Forces i n Terms of Constant Parameters (b t,b 2,...be) (equilateral trapezoid element i n plane stress) V 0 h - fr: y) (2a+c) 0 - | ( l - y2 ) h 2 yh > -y) (2a+c) 0 bl f 2 0 - ^ -P(2a+c) y)h y 2) (2a+c)a. 0 - y(2a+c) -y)h 0 b 2 0 -h >-P)(2a+c) 0 §(l-y2)h2 -yh > -y) (2a+c) 0 b3 Di 0 - jH(2a+c) y)h y 2) (2a+c)a 0 - y(2a+c) -y)h 0 b4 0 h y)(2a+c) 0 §(i-y2)h2 yh -y) (2a+c). 0 b 5 f 6 0 jy(2a+c) y)h y2) (2a+c). (a+c) 0 y(2a+c) I d -y)h . 0 b 6 f 7 0 -h y)(2a+c) 0 - | ( i - y 2 ) h 2 -yh -y)(2a+c) 0 *8 0 iy(2a+c) - j d -y)h . - ± d - y 2)(2a+c)(a+c) 0 y(2a+c) 4 d -y)h 0 > {f} = [C]{B} Eqn. (11-15) Et Here, = = Extensional r i g i d i t y per unit length of the plate t = thickness of the plate (constant for the element) 28 Here, [K]. = s t i f f n e s s matrix of the f i n i t e element i n plane stress. See Table (II-7,a) Comparing Eqns. (11-16) and (11-17) [K] = [C][A _ 1] (11-18) I t may be noted that a force i n the i t h direction due to a general state of displacements i s referred to as f^ whereas, a force i n the i th direc-t i o n due to a unit displacement i n the j th .direction alone i s cal l e d K. . ( d e f i n i t i o n of s t i f f n e s s c o e f f i c i e n t ) . Thus, the following r e l a t i o n holds, •^ 1 = ^1 1^1 + ^1 2°~2 + . . . + ^ 1 ^ 3 6 3 f 2 = K 2 161 + K 2 2 6 2 + . . . + K 2 Q8Q * f i = K i , 1 < S l + K i , 2 6 2 + • • • + K i j 8 6 8 Due to the geometrical symmetry of the f i n i t e element about y-axis, s t i f f n e s s c o e f f i c i e n t s belonging to columns 3, 4, 7 and 8 (due to unit displacements at j o i n t s 2 and 4) may be obtained simply by inspection from those of columns 1, 2, 5 and 6 (due to unit displacements at j o i n t s 1 and 3). This i s shown i n Table (II-7,b). The e x p l i c i t expressions for 29 [ K ] = Table ( I I T7,A) Plane Stress Stiffriess Matrix K l l K12 K13 K 1 4 K 15, Kl6 K17 K18 K21 K2 2 K2 3 K2H K25 . K2 6 K 27 K2 8 K31 K32 K 3 3 K 3 4 K 3 5 . K36 K 3 7 K38 K m K 4 2 Ki+3 K*+*+ K4 5 Ki+6 K 4 7 Ki+8 K51 K52 " K53 K 5 4 K 5 5 K 5 6 K 5 7 • K58 K61 K62 K63 K6 4 K6 5 K66 K67 K68 K 7 1 K 7 2 K 7 3 K 7 5 K76 K 7 7 K78 K81 K82 K83 ^84 K85 K86 K87 K88 Table (II-7,B) Equalities in the Stiffness Coefficients due to the  Geometrical Symmetry of the Trapezoid about Y-Axis K l l K 1 2 K31 - K 3 2 Kl5 K16 K35 "K36 K21 K 2 2 ! Ki+2 K2 5 K2 6 _ K 4 5 K46 K31 K32 K l l - K 1 2 K35 K36 Kl5 _ K 1 6 Ki+1 Ki+2 - K 2 1 K 2 2 K1+5 Ki+6 _ K 2 5 K26 K51 K52 K 7 1 - K 7 2 K 5 5 K56 K 7 5. _ K 7 6 Kei K62 - K 8 1 K82 K 6 5 K66 - K e 5 K86 K 7 1 K72 K 5 1 _ K 52 K7 5 K76 K55 "K56 K81 K82 - K 61 K62 K85 K86 - K 6 5 K66 30 s t i f f n e s s coefficients belonging to columns 1, 2, 5 and 6 are given i n Table (11-7,0 . The c'imensionless ratios of the sides K and K^  are normally defined as (See Fig. ( I I - 2 ) ) 2a 1 2a In order to obtain s t i f f n e s s coefficients belonging to columns 5 and 6 from those of columns 1 and 2, the above dimensionless ratios would have to be written by analogy, as K n e w 2a j. new _ 2h = 2 (a + c) ' 1 " 2 (a + c) which are equal to l/K and K^/K respectively. Therefore, the values of the s t i f f n e s s coefficients of columns 5 and 6 may be obtained by inspection from those of columns 1 and 2 by replacing the dimensionless ratios of the sides K and Kj by l/K and Ki/K respectively. It may be noted that the s t i f f n e s s matrix [K] for an e q u i l a t e r a l trapezoid obtained from the s t a t i c s approach i s not symmetrical about the main diagonal. However, i f the trapezoid reduces to a rectangle, the s t a t i c s s t i f f n e s s matrix becomes symmetrical. The s t a t i c s plane stress 20 s t i f f n e s s matrix for a rectangle coincides with the one derived by Clough using Energy considerations and based on the same assumption of stress states. Apparently, i t s lack of symmetry r e f l e c t s the asymmetry of the element i t s e l f about X-axis. 31 It i s more important to note that the s t i f f n e s s coefficients for a trapezoid element s t i l l s a t i s f y r e c i p r o c i t y relations i n conditions of uniform stress. Thus when the element decreases i n size and the stresses i n the v i c i n i t y of the element become nearly uniform, the i r r e g u l a r i t y due to the asymmetry of the s t a t i c s s t i f f n e s s matrix w i l l tend to correct i t s e l f . r i i i i t i u i II U - - - - -uPi P 2 P 3 Under the uniform stress conditions, the edge displacements of the piece of plate become lin e a r . Since i t i s the lack of the edge l i n e a r i t y that causes 23 the difference between the Statics and the Energy s t i f f n e s s c o e f f i c i e n t s the nodal forces i n the f i n i t e element corresponding to the Statics and the Energy s t i f f n e s s matrices w i l l coincide and the conditions of recipro-c i t y w i l l be restored as i s always the case with the Energy matrix. 32 Table (II-7,C) Entries of the Statics Stiffness Matrix for a Trapezoid i n  Plane -Stress. (Fig. (II-2)) Et Let, =' ^ 2 j = Extensional r i g i d i t y per unit length of•the plate. Here, t = thickness of the plate element. Also, defining the dimensionless numbers £ _ 2(a•+ c) _ Bottom side 2a ~ Top side K = — Height  1 ~ 2a Top side The entries of the sta t i c s s t i f f n e s s matrix, defined by.Eqn. (11-18), are: Entries of the 1st Column, (6i = 1; a l l other displacements are zero) -1 1 » 4 (4 - M2)Ki + (1 - y)(K + 1) + y(K 2 - 1) 3K 4Ki 4KKX K 2 i -EL 8 y (K - 1)(K + 1) K" 4 l ? >3 1 2 i 4 Di (4 y 2 ) K i (1 - y)(K + 1) y(K 2 - 1) 3K Kif . l = i ^ -v 5 , l = ^ 3K 4Ki 4KKX + (K 2 - 1)(K + 1) 4KK12 (1 - y)(K + i ) . y(K 2 - 1) 4Ki 4KKi 6^.1 - •£] 8 y(2 + ±) (K 2 - 1)(K + 1) 4KK12 K 7 ! 4 (2 + y 2 ) ^ (1 - y)(K + 1) y(K 2 - 1) 3K " 4K: " 4KKX •Pj 8 , y (K 2 - i)(K + .1) K 4KKj2 Continued Table (II-7.C) 33 Entries of the 2nd Column (6? =1; a l l other displacements are zero) K l 2 4 E. rs h + 1 + K + y ( i - y)(K 2 - D(K + l) 16KK!2 2^ 2 2 l 4 (1 - y 2 + 3K)(K + 1) + (1 - y)Kt + y(l - y)(K 2 - 1) 6KK, 2K 8KKi k3, 2 4 y r^ + I i + (1 + K) + y(l -y)CK 2 - i)(K + l) 4 K 4K 16KKi2 C - 1 + .y2 + 3K)(K + 1) (1 - p)K] 2K 6KK 1 uCl - y)(K 2 - l) 8KK1 -Di K5,2 = (1 * K) y(l - M)(K2 - 1)(K + 1) 4K 16KKX2 - £L *6,2 -. 4 _ (4 - y 2)(K + 1) + (1 - y)Ki + y ( l , - y)(K 2 - 1) 6K, 2K 8KKi K7,2 •Di_ 4 y_ 1 ^ l + K y d - y)(K 2 - 1) (K, + l ) ~ 4 K J " 4K 16KKf K8,2 4 (2 + y2)(K + 1) 6K, [1 - y)Ki uCl - y)(K 2 - 1) 2K " 8KKi Entries of the 5th Column ( 6 5 = 1; a l l other displacements are zero) Kl,5 - ^ 4 (2 • y 2)Ki. (1 - y)(K + 1) y(K 2 - 1) 3 4K, 4KX K2,5 = ~ •Di 8 1 + .y(2 + K ) (K 2 - D(K + 1) 4KX2 Y - D l K3>5 - -&— (2- + y 2)Ki (1 - M ) C K + .1) . y(K2 - l) 3 • " 4KX 4Kj 34 Continued Table (11-7,0 ^ 5 ~ -Di 1 + yK (K2 - 1)(K + 1) 4KX2 ^5.5 Di (4 - yQKi (1 - P)(K + 1) y(K^ - 1) 3 4K: 4KX •Di - 1 - yK + ^ if - D l (4 - M 2 ) K I (1 - M ) ( K ,+ 1) y ( K 2 - 1) 3 4KX 4K: ^8.5 •Di 1 - y(2 + K) + (K2 - 1)(K + 1) 4Ki2 Entries of the 6th Column ( 6 6 = 1; a l l other displacements are zero) Kl>6 - — £(5 + K) + (1 + .K) y ( l - y)(K2 - 1)(K + 1) 4 ' 16KX2 K2,6 = ~ Dl 4 (4 - y 2)(K + 1) (1 - y)Ki • y ( l - y) (K 2 6KX 2 8K: - 1) ^3, 6 -EL 4 £(3 - K) + 1 * K y ( l - y)(K 2 - 1) (K + 1) 4 16Kj2 if - E L (2 + y2)(K f 1) (1 - y)Kl x y ( l - y) (K 2 - 1) 6 K 1 8K, K5,6 " — n C3 - K) - L±L + yd - y)C«2 - I K K +,i) 4 ^ N J 4 16K7T K6 .6 - — (K - Ky 2 + 3)(K + 1) ( 1 - M)Ki y ( l - y)(K 2 - 1) 6KX 2 " 8KX 35 Continued Table (II-7.C) K7>6 4 7 C5 + K) (1 + K) y ( l 4 y)(K 2 - 1)(K + 1) 16Ki2 8>6 (-K + KPZ + 3)(K + 1) (1 - y)K x y ( l - y) (K 2 - 1) 6Ki ' 2 8KX 30 2.10 Derivation of Plane Stress Energy Stiffness Matrix for an Equilateral  Trapezoid This method uses the v i r t u a l work p r i n c i p l e to define the equivalent nodal forces. The f i n i t e element has been defined as a body, the nodes of which undergo the same deformations as the plate and are acted upon by forces s t a t i c a l l y equivalent to the distributed edge stresses of the plate. The condition imposed oh the values of the equivalent nodal forces here i s that during a v i r t u a l deformation the external work done by the equivalent nodal forces of the f i n i t e element i s the same as the external work done by the edge forces of the plate element. During a v i r t u a l deformation, the external work done by the edge forces of the plate i s equal to the internal work done by the internal stresses i n the plate. The equality of the external work of the plate and the f i n i t e element requires the equality of t h e i r i n t e r n a l work. Let these nodal forces i n reverse directions be applied to the plate element. This action makes the plate element suffer additional deformations and stresses super-imposed on the ones carried o r i g i n a l l y . However, the edge forces are not affected because the new forces have been applied at the nodes only. The plate element i s now acted upon by two sets of forces external to the element; the edge forces and the nodal forces, each set being i n a state of equilibrium by i t s e l f . In this state the plate element of course continues to carry some in t e r n a l stresses. The error of present formulation l i e s i n disregarding the work of these in t e r n a l stresses within the plate element. This work i s not l i k e l y to be great because the o r i g i n a l l y present i n t e r i o r stresses were largely cancelled by the action of reversed nodal forces, but 37 s t i l l t h i s vvork i s not zero. Thus, the supposition of the equality of external work of the f i n i t e element and of the plate element i s hypothetical, I f the selected v i r t u a l displacement {<5}.. i s such that i t i s unity i n the direction of a p a r t i c u l a r external force f. and zero i n the direction of a l l other forces, then the external work i n the f i n i t e element We. w i l l be the same as the value of the nodal force, i . e . J We. = { d j / ^ f } = f. (11-19) Here, {<5}j = 8 x 1 column vector of v i r t u a l displacements. Its j t h entry i s unity and a l l other entries are zero, {f} - 8 x 1 column vector of nodal forces. Writing the external work expressions for a l l such v i r t u a l displacements {&}. (j = 1,2,...,8) due to {6}i; Wej = (fij = 1,0,0,...,0){f} = fl due to { s } 2 ; We2 = (0,6 2 = 1,0,...,0){f} = f 2 due to {I}.; We. = (0,...,6 = 1,0,...,0){f} = £ o . . • * • . . . . C o l l e c t i v e l y , these v i r t u a l displacements {s}^ (j = 1,2,...,8) can be written as [6] v/hich i s the same as an 8 x 8 Identity matrix [ I ] . The corresponding external works for a l l v i r t u a l displacements can be expressed 33 i n matrix form as {We} = [ 6 ] T { f } = [I]{f} = { f j (11-20) 8x1 8x8 8x1 . 8x1 During each of these v i r t u a l displacements { 6 } ^ , an int e r n a l work Wi.. i s done by the internal stresses over the area of the plate which i s the same as the external work We., of the edge stresses. For a plate of constant 24 thickness t , the int e r n a l work may be expressed as Wi.. = tff {e}.-T{o}dxdy (H-21) area Here {e}.. = 3 x 1 column vector of strains due to v i r t u a l displacements {&}^ The integration i s to be carried out over the entire area of the plate. Using Eqn. (11-11) , the strains due to v i r t u a l displacements {<$} . may be expressed as {ej-j = [G][A-1]{5}. Substituting for {e}j from above and for [a] from Eqn. (11-13) i n the Eqn. (11-21) one obtains, Wi. = t // { 6} j T[A"l] T[G] T[D][G][A"l]{ (S}dxdy (11-22) area l x l 1x8 8x8 8x3 3x3 3x8 8x8 8x1 Writing the internal work expressions for a l l such v i r t u a l displacements [S}y CJ = 1,2,...,8), 39 due to Wii = t // {6} 1 T[A" 1] T[G] T[D][G][A" 1]{6}dxdy area due to Wi 2 = t ff {6} 2 T[A" 1] T[G] T[D][G][A" 1]{6}dxdy area due to {fi}.-; Wi . = t ff {&} . T [ A " 1 ] T [ G ] T [ D ] [G] [A _ 1]{6}dxdy ^ area C o l l e c t i v e l y , i n t e r n a l works {wi} due to a l l such v i r t u a l displacements [<5] ( i . e . {6}j, j = 1,2,...,8) can be written i n the matrix form as {Wi} = t // [ 6 ] T [ A _ 1 ] T [ G ] T [ D ] [G][A _ 1]{6}dxdy (11-23) area 8x1 8x8 8x8 8x3 3x3 3x8 8x8 8x1 [A i s a matrix i n terms of constant corner coordinate and therefore can be taken out of the integration sign. Also, note that [ 6 ] T [ A " 1 ] T = [ I ] [ A _ 1 ] T = [A"1] T Only the [G] matrix contains variables x and y. Equation (11-23) can be rewritten as {Wi} = [ A _ 1 ] T t // [G] T[D][G]dxdy [A _ 1 ] { s } area (11-24) 40 Equating the external work done by the nodal forces of the f i n i t e element durii.g a l l v i r t u a l displacements (Eqn. (11-20)) to the corresponding internal work done by the internal stresses i n the plate element (Eqn. ( I I -24)), one obtains; {£} •= [ A " 1 ] 1 t // [G] T[D][G]dxdy [A"l]{s} (11-25) area Comparing Eqn. (11-25) with the force-displacement r e l a t i o n of Eqn. (11-17), the Energy s t i f f n e s s matrix may be written as [K] = [ A _ 1 ] T t // [G] T[D][G]dxdy [ A _ 1 ] (11-26) area Letting, [H] = t ff [G] T[D][G]dxdy area T The entries of the product matrix [G] [D][G] are simple polynomials of x and y which are e a s i l y integrated over the area of the element. Due to the symmetry of the element about the Y-axis, integration of the polynomial terms containing odd powers of x result i n zero value. The integrated [H] matrix i s given i n Table (II-8). Equation (11-26) i s rewritten as [K] = [A' 1] T[H][A' 1] (II-27,b) The congruent transformations of the symmetric matrix [D] i n Eqn. (11-26) 25 ensures the symmetry of the energy s t i f f n e s s matrix . Table (II-7,b) showing the equality of s t i f f n e s s c o e f f i c i e n t s due to the geometrical symmetry of the element about the Y-axis i s s t i l l v a l i d . Table (II-8) [H] = t j j j " [G] 1 [D] [G] dxdy area Eqn. (II-27;a) [H] = Et (1 - P z ) '1 Here, 0 0 0 0 0 0 0 0 2(1 - n 2)L 2 0 } i L 1 0 0 0 0 L 2 = 2(2a + c)h _L2 = - \ ch 2 0 ( l - y)Li 2 0 0 0 (1 - -JO-Li • 2 0 0 0 0 4(1 - y 2 ) L 3 0 0 0 0 2(1 - y 2)L 2 0 0 4(1 - y 2 ) ^ 0 0 0 0 o 0 0 Li 0 0 0 0 0 0 ( i - y)Li 2 0 0 0 (1 - y)Li 2 0 •201-h 3 = M 2 a j c) { c 2 + ( 2 a + c ) 2 } Li, = - (2a + c ) h 3 Table (II-9) 42 Entries of the Energy Stiffness Matrix for a.Trapezoid i n Plane Stress (Fig. ( I I - 2 ) ) [K]< Et (1 - V*) H + 6 6 1 2 - ^ 2 I 4 + 6 5 -H + 06 n 5+6i H+06 symmetrical 1 2 + 0 2 - n i + + B 5 - 1 2 + 0 2 11++05. ( K j i = K i j ) n 3 - 6 6 - n 5 - 6 3 -H3-66 1 5 - 6 3 l7 +06 - n 2 - 3 4 ne-65 n 2 - 6 i + . -16-05. 18+04 I 9 + 6 5 -03-66 -r)5 + 6 3 13-06 15 + 63 -17+06 -18+01+ 17+66 -i2+6i+. -n 6 - 6 5 12+04. 16-05 i8 - 6 i + -119 + 65 - I 8 " 6 i + Here, 61 = J4UKi 2 + (K 2 - 1)}(K + 1) 32KK22 ft J4Ki 2 + y(K 2 - 1)}(K + 1) (1 - y)  P 2 32KKT2 2 B, - J4yKi 2 - ( K 2 - 1 ) } ( K + 1) P 3 04 U K I2 - y ( K 2 - 1 ) } ( K + l ) ( l - y} 3 2 1 7 ^ 2 05 (K + 1) 8 k 1 6 SKI 2 11 1 r{4K1 2 + " 4 k | _ .-X.\l 8 k k 1 3 y(K 2 - 1)}(K + 1) + (1 - ^ 2 ) ( K - l ) K i 8 k k 1 6K (1 - y 2 ) K x 2 + J 4 y K i 2 ^ C K 2 - 1 ) } ( K + 1) ( K 2 ^ - 1) 8 k k 1 2Ki 43 Continued Table (II-9) n 2 = (K + 1) I 6 K K 1 1 1 3 = 4K { 4 K 1 2 - y(K 2 - 1)}(K + 1) (1 - y 2)(K - l)Ki 8K1 6 2 n . . 2 ^ 2 r 4 u K i 2 - (K2 - 1) 1 (K + 1) [K2 - 1) 8 K K T 3 C 1 " M ) K l " 1 ~ 8 K l 1 2KT~ Tilt = 4 K i 2 + y(K2 - 1) 8KK1 n 2 (K + 1) (1 - M) 2Ki 2 ^5 = 16 = 4 K j 2 .+ y(K2 - 1) 8KK1 4 K X 2 + y(K2 - 1) (K + 1) (1 - IQ _ ft 4Ki 2 0 2 8KKj { 4 K l2 - y(K 2 - 1)} i ^ L n i L ^ l i6Kr 1 (1 - P 2) (K + 1){(K - l ) 2 + (K + l ) 2 } 8KK1 12 ^7 = 8K1 {4Kl2 - y(K2 - 1)}SJLLH - I L ^ J i l L CK " DK!2 + I (1 - y 2 ) K l 2 - ( K + " {4UK!2 - (K2 - 1)} n e = {4yKi 2 - (K2 - 1)} (K + 1) = B, 32Ki2 P 3 n 9 = {4Ki 2 .- M(K2 - 1) }2(K + 1). (1 - U) 8K1 2Ki + _L_ CI - y 2) (K + 1){(K - l ) 2 + (K + l ) 2 } 8K1 12 4J The s t i f f n e s s coefficients belonging to the lower h a l f of the symmetrical energy s t i f f n e s s matrix are given i n Table (II-9). It i s of interest to note that i n conditions of uniform stress, the nodal forces obtained by using either Statics or Energy approach are 23 i d e n t i c a l . Thus, i n conditions of uniform stress, the Statics and Energy s t i f f n e s s matrices w i l l y i e l d i d e n t i c a l r e s u l t s . When the trapezoid reduces to a rectangle, the Energy s t i f f n e s s 20 matrix coincides with the one derived by Clough. It i s of further interest to note that the Statics and Energy s t i f f n e s s matrices for a rec-tangular element are i d e n t i c a l . 2.11 Derivation of Plane Stress Stiffness Matrix for an Isosceles Triangle Fig. (II-6) A t r i a n g l e has s i x degrees of freedom. Three r i g i d body move-ments are possible without i n any way affecting the state of stress. Thus, three independent stress conditions are required for the formation of the s t i f f n e s s matrix. The three necessary stress conditions are, constant a , 38 constant a and constant a . These stress states are conveniently y xy 1 achieved by selecting a displacement function of the type. u = bj + b 2x + b 3y (11-28) v = hh + b 5x + b 6y Nodal displacements i n terms of the s i x constant parameters are expressed through an equation s i m i l a r to Eqn. (II-7). The 6 x 6 matrix [A.] of t h i s FIG.(lT-6) ISOSCELES TRIANGULAR ELEMENT IN PLANE STRESS. (Positive directions of forces and displacements.) 46 case i s shown i n Table (11-10). The parameters b^ , b 2 , • • • ..bg are evaluated i n terms of the nodal displacements as i n Eqn. (II-8) and are shown i n Table (11-11). 2.12 Derivation of Plane Stress - Statics Stiffness Matrix for an Isosceles  Triangle Equivalent nodal forces corresponding to each displacement mode are expressed as i n Eqn. (11-15). The 6 x 6 matrix of equivalent corner forces, i s shown i n Table (11-12). The s t i f f n e s s matrix, i s obtained using Eqn. (11-18) and i s shown i n Table (11-13). Note that the s t a t i c s s t i f f n e s s matrix for a tr i a n g l e i n plane stress i s symmetrical. 2.13 Derivation of Plane Stress Energy Stiffness Matrix for an Isosceles  Triangle The plane stress energy s t i f f n e s s matrix for an isosceles t r i a n g l e element i s expressed as i n Eqn. (11-26) or Eqn. (11-27,b). The [A -*] matrix i s given i n Table (11-11) and [H^] matrix i s shown i n Table (11-14). Note that, f o r a tria n g l e i n plane stress, the energy s t i f f n e s s matrix coincides with the s t a t i c s s t i f f n e s s matrix shown i n Table (11-13). This matrix i s 20 38 i d e n t i c a l to the matrix derived by Clough and Turner et a l . 47 Table (11-10) Nodal.Displacements of an Isosceles Triangle i n Terms of Constant Parameters Fig. (II-6) «1 1 c -h 0 0 0 6 2 0 0 0 1 c -h b 2 «3 1 -c -h 0 0 0 b 3 6 4 0 0 0 1 -c -h b^ «5 1 0 h 0 0 0 b 5 «6 0 0 0 1 0 h b 6 Table (11-11) Evaluation of Constant Parameters for an Isosceles Triangle i n Terms of Nodal Displacements bl 1 4 0 1 4 0 1 2 0 b 2 1 2c 0 -1 2c 0 0 0 b 3 -1 4h 0 -1 4h 0 1 2h 0 bi* 0 1 4 0 1 4 0 1 2 b 5 0 1 2c 0 -1 2c 0 0 b 6 0 -1 4h 0 -1 4h 0 1 2h ] «1 6 2 S3 &k -Table (11-12) Equivalent Nodal Forces i n Terms of Constant Parameters for an Isosceles Triangle f l f 2 f 3 = f'k fs f6 h - d - y) c 2 2 -yc (1 -2 2 -h - d - y) C 2 2 -JlC -Ci -2 2 0 (1 -2 yc 0 0 -(1 2 y) C 2 0 (1 2 c 2 0 -(1 2 y) c 2 -yh 0 - ( i 2 c " 2 0 ( i 2 0 0 0 c bl b 2 b 3 b 5 b 5 Table (11-13) Plane Stress Stiffness Matrix for an Isosceles Triangle (Fig. (II-6)) [ K A 3 = ( i E - V ) . 8K ' ( i > y) 8 K ( i - y) , 8K (3y - l ) 8 ( i - y) 8K. £ + 2(1 - y)K "(3y - 1) 8 j - 2(1 - y)K "(1 - y) + - y) 8K ( i + y) 8 ( i - y) 8K symmetrical 1_ 8 £ + 2(1 - y)K 4 ( i - y) 4K 1_ 4K 1_ 4K 1_ 2K Table (11-14) Integrated [H A] Matrix 0 0 0 0 1 0 0 0 i±f> 0 0 0 o o i ^ l O p 0 0 0 0 M ^ 0 0 0 i l ^ l 0 0 1 CHAPTER III DERIVATION OF FLEXURE STIFFNESS MATRICES 3.1 General In a plate bending problem, a point on the plate i s allowed to displace i n a direction normal to the plate and to rotate about two mutually perpendicular axes i n the plane of the plate. Assuming that the deflections are small, stretching of the plate i n i t s plane due to the deflections i s neglected. In the following discussions, the term "deformation" w i l l include displacement or rotations, and the term "nodal force" w i l l include transeverse force or moments. The formulation of flexure s t i f f n e s s matrices follows a s i m i l a r systematic matrix procedure as i s adopted i n the derivation of plane stress s t i f f n e s s matrices. 3.2 Equation of the Deflection Surface for a.Quadrilateral Plate Element  i n Flexure A f l e x u r a l f i n i t e element having n nodes has (3n) degrees of freedom as these nodes move or rotate while the plate element deforms under the action of moments. However, there are three r i g i d body movements: a r i g i d body displacement i n the Z direction, and r i g i d body rotation about X and Y directions respectively. Thus, three of the corner displacements may be made zero by r i g i d body movements without i n t e r f e r r i n g i n any way with the shears and moments imposed on the plate element. This leaves (3n-3) independent deformation conditions for the formulation of the s t i f f n e s s matrix, irrespective of the shape of the f i n i t e element. For a quadrilateral with four nodes, this number i s (3)(4)-(3)=(9). Three of these conditions must be constant curvatures for bending i n two perpendicular directions and a constant twist. Lateral deflections are represented throughout the plate element 27 by a polynomial i n x and y as w = bj + b 2x + b3y.+ b 4 x 2 + b 5xy + b 6 y 2 + b 7 x 3 + b 8x 2y + b 9xy 2 + b 1 0 y 3 + b n x 3 y + b 1 2 x y 3 ( I I I - l ) Here, w = displacement normal to the plane of the plate and i s positive upwards bi ,b 2,...,bi2 = constant parameters Rotations are defined as x " 3y.' y " 3x and are positive i n the direction of right hand screw when viewed i n the positive directions of x and y axes. Note that each term of the displacement surface Eqn. ( I I I - l ) s a t i s f i e s the biharmonic equation governing the deflection of an unloaded plate. 52 fIO'SIO * fl2>Sl2 a >H Q — 0 X (a+c) H*-FIG(DI-l) EQUILATERAL TRAPEZOID IN FLEXURE (Positive directions of nodal forces and deformations) t = thickness FIG (HI-2) DIRECTIONS OF POSITIVE TRANSEVERSE SHEARS AND MOMENTS 53 3.3 Evaluation of the Constant Parameters for an Equilateral Trapezoid  Element i n Flexure For an equilateral trapezoid plate element, the twelve nodal displacements and rotations are'obtained i n terms of the twelve parameters, b 1 ,b2 ,. .. ,b 1 2. These displacements and rotations may be written i n a matrix form: {&*} = [A*]{B*| (III-2) Here, = 12 x 1 column vector of nodal deformations &i,&z> • • • >&\z Fig. (IH-1) [A*] = 12 x 12 matrix in terms of corner coordinates of the trapezoid. {B*} = 12 x 1 column vector of constant parameters b1,b2,...,b 1 2. The matrix Eqn. (III-2) i s shown i n f u l l i n Table ( I I I - l ) . The parameters, {B*} , are evaluated in terms of the corner deformations: {B*} = [A* _ 1]{5*} (III-3) The matrix [A* - 1] i s given i n Table (III-2). Knowing the corner deformations, {6*}, of a f i n i t e element, the displacement surface of the piece of plate having the same deformations as the f i n i t e element i s completely known. Table ( I I I - l ) Nodal Deformations for a Trapezoid Element i n Terms of Constant Parameters Fig. ( I I I - l ) w = b x + b 2 X + b 3 y + b^x 2 + bsxy + b 6 y 2 + t>7> 3 + b 8 x 2 y + bgxy 2 + b 1 0 y 3 + b n x 3 y + b 1 2 x y 3 6 = X e = -; y . • 3w 3 x -1 (a+c) -h (a+c)2 -(a+c)h h 2 (a+c)3 -(a+c) 2h (a+c)h2 -h 3 -(a+c) 3h -(a+c)h 3 b i 6 2 0 0 1 0 (a+c) -2h 0 (a+c) 2 -2(a+c)h 3h2 (a+c)3 3(a+c)h2 b 2 . 6 3 0 -1 0 -2(a+c) h 0 -3(a+c) 2 2(a+c)h -h 2 0 3(a+c) 2h h 3 b 3 6 if 1 -(a+c) -h (a+c) 2 (a+c)h h 2 -(a+c) 3 -(a+c) 2h -(a+c)h 2 -h 3 (a+c) 3h (a+c)h3 b 4 6 5 0 0 1 0 -(a+c) -2h 0 (a+c) 2 2(a+c)h 3h 2 -(a+c) 3 -3(a+c)h2 b 5 66 0 -1 0 2(a+c) h 0 -3(a+c) 2 -2(a+c)h -h 2 0 3(a+c) 2h h 3 b 6 67 - 1 a h a 2 ah h 2 a 3 a 2h ah 2 h 3 a 3h ah 3 b 7 6 8 0 0 1 0 a 2h 0 a 2 2 ah 3h 2 a 3 3ah 2 b 8 V 0 -1 0 -2a. -h 0 -3a 2 -2ah -h2 0 -3a 2h -h 3 b 9 1 -a h a 2 -ah h 2 -a 3 a 2h -ah 2 h 3 -a 3h -ah 3 b i o V 0 0 1 0 -a 2h 0 a 2 -2ah 3h2 -a 3 -3ah 2 b l l 6 1 2 0 -1 0 2a -h 0 [#l = -3a 2 [A*]{Bl 2 ah Eqn. -h 2 (III-2) 0 -3a2h -h 3 b 1 2 Table (III-2) Evaluation of Constant Parameters i n Terms of Nodal Displacements and Rotations for a Trapezoid i n Flexure b l 1/4 h/8 <*1 1/4 h/8 ~<*1 1/4 -h/8 a 2 1/4 -h 8 -oi 2 b 2 0 3 h aj " a 3 -h. - h / 8 a az h <*2 8(a+c) (a+c) 8(a+c) (a+c) a 8 a a 3 8 h -1/8 <*5 - 3 / 8 h -1/8 3 / 8 h .-1/8 <*6 3 / 8 h 1 8 0 0 1 0 0 1 0 0 -1 0 0 1 8 a 8(a+c) 8(a+c) 8 a ct7 1 <*5 - a 7 1 as «8 . - l / 8 a a - a 8 1 a 8(a+c) (a+c) 8(a+c) (a+c) 8 a b& 0 1 Otg 0 - l / 8 h -ag 0 l / 8 h -ag 0 1 a 9 8 h 8(a+c)h 8(a+c)h 8 a h 8 h 8 a h by 1 0 1 1 0 -1 -1 0 -1 1 0 -1 8(a+c) 3 8(a+c)2 8(a+c)3 8(a+c)2 8 a 3 8 a 2 8 a 3 8 a 7 b 8 0 0 1 0 0 -1 0 0 -1 0 0 1 8(a+c)h 8(a+c)h 8 a h 8 a h b 9 a 9 1 a 9 - a 9 1 a 9 -a 9 1 -ag a 9 -1 -a 9 8 h(a+c)3 8h(a+c) 8 h(a+c ) 2 8h(a+c)3 8h(a+c) 8 h(a+c) 2 8 h a J 8 h a . 8 h a 2 8 h a 3 8 h a 8 h a 2 b 1 0 1 1 a 9 1 1 -ag -1 1 a 9 -1 1 -ag 8 h ^ I P " 8(a+c)h2 8P" 8(a+c)h2 8 P ~ 8 h 2 8 a h 2 8 h 3 8 h 2 8 a h 2 b l l 1 0 1 -1 0 1 -1 0 -1 1 0 -1 8 h(a+c ) 3 8 h(a+c ) 2 8h(a+c)3 8 h(a+c ) 2 8 h a 3 8 h a 2 Sha^ 8 h a 2 b 1 2 1 a 9 015 -1 a 9 -a 6 1 a 9 a6 1 ag ( a + c ) ^ 8(a+c)h^ 8 h 2 ( a + c ) 2 (a+c) 2 h 2 8(a+c)h2 8(a+c)2h2 a*h 2 8 a h 2 8 h ^ a ^ aW 8 a h 2 8 h 2 a 2 {if} = [A*-L]{#\ Eqn. (III-3) 56 Continued Table (IH-2) Here, 0 4 a 2 a 3 °<5 <*6 ay a 8 a 9 3(a+c) 2 + a 2} 32(a+c) (a+c) 2 + 3a2} 32 a l l ( a + c ) 2 + a 2} 32(a+c)3 (a+c) 2 + 11a2} "32aT •5(a+c)2 + a 2} 32h(a+c) (a+c)- 5a 2} 32ha -17(a+c) 2 + a 2} 32(a+c)3h -(a+c) 2 + 17a2} 32a3h; (a+c) 2 - a 2} 4h 57 3.4 Moments and Shears at a Point Moments and transeverse shears at any point of the plate can be 26 calculated using e l a s t i c i t y formulae. The curvatures and twist at any 27 point of the plate are written i n terms of the twelve parameters; 92w - j£ = -(2b,, + 6b 7x + 2b 8y + 6b nxy) - l £ = -(2b 6 + 2b 9x + 6b 1 0y + 6b 1 2xy) 92w 9x9y = (2b 5 + 4b 8x + 4b 9y + 6 b n x2 + 6b 1 2y 2) In the matrix form, these are written as, {x*} = [G*]{B*} (III-4) Here, {x*} = 3 x 1 vector of curvatures and twist [G*] = 3 x 12 matrix i n terms of variables x and y Expression (III-4) i s shown i n f u l l i n Table (III-3). Substituting i n Eqn. (III-4) for {B*} from Eqn. ( I l l - 3 ) , one obtains {x*} = [G*][A* _ 1]{6*} (IH-5) 2 6 Moments and shears per unit length of the plate are given by 32w 92w Moments, M x = D 2[- - v—^ ] My = D 2[- w - y - ^ ] (III-6,a) M o -M = J 2 Z K L D 2 ( 2 | ^ _ ) xy yx 2 ^ 9x9y 1 58 and Transeverse shears, Q x " ~ ° 2 3 x W 9y2 J 3 .3 2w , 3 2w (III-6,b) + Q y = " D 2 3 y ^3x2 ' 3 y 2 E t 3 Here, D2 = 2^ = fle x u r a l r i g i d i t y per unit length of the plate The positive directions of these .moments and shears are shown i n Fig. (III-2). 3.5 Moments at a point :on the Trapezoid Element Internal moments, given by Eqn. (III-6,a), can be expressed i n the matrix form: {M*} • [D*]{x*} (III-7) Here, {M*} = 3 x 1 matrix of internal moment [D*] = 3 x 3 symmetrix matrix of e l a s t i c constants The matrix equation (111-7) i s shown i n Table (III-4) . Substituting for {x*} for a.trapezoid from Eqn. (111-5) i n Eqn. (III-7) one obtains {M*} = [D*][G*][A* _ 1]\6*} (III-8) Letting [MT*] = [D*][G*][A* - 1] ( I I I - 9 ) Eqn. (III-8) can be rewritten: {M*} = [MT*]{6*} (111-10) Here, [MT*] = 3 x 12 moment matrix i n terms of the variables x and y 59 Table (III-5) Curvatures and Twist at Any Point (x,y) in Terms of Constant 92w " ax 2 ( 92w 9y 2 = ( 2 3 2 w 9x9y i Parameters for a Trapezoid i n Flexure 0 0 0 -2 0 0 -6x -2y 0 0 -6xy 0 0 0 0 0 0 -2 0 0 -2x -6y 0 -6xy 0 0 0 0 2 0 0 4x 4y 0 6x 2 6y 2 b i b 2 b 3 b H b 5 b 6 by b 8 b 9 b 1 0 b l l j b 1 2 1X7 = [G*]{B*f Eqn. (III-4,b) Table (III-4) Relation Between Internal Moments and Curvatures and Twist Mx My = Mxy Do yD 2 yD2 D2 0 0 a2w ~ 9x 2 0 92w 9y 2 U)D2 2 9 2 w 2 9x9y_ (III-7) E t 3 Here, D2 = y2) f l e x u r a l r i g i d i t y per unit length of the plate 60 Table (II1-5) Moment Matrix for a Trapezoid^Element i n Flexure A l l entries to be multiplied by D2 = Et 3 / 1 2 ( l - y 2 ) MT*i,i = [x(6h 3-2ya 9h 2). - 6y(a+c) 3y - xy{6h 2-48ya 5h(a+c)}]/8(a+c) 3b 3 MT*2 MT*3 MT*i MT*2 MT*3 MT*i MT*2 MT*3 MT*i MT*2 MT*3 MT*i MT*2 MT*3 MT*i MT*? .1 = [x(6yh 3-2a 9h 2) - 6y(a+c) 3 - xy {6yh 2-48a 5h (a+c) } ]/8h 3(a+c) 3 _ a , n u , + { 3 h ( l - y ) x 2 + 2(l-y)a 9hy - 24(l-y)a 5(a+c)y 2} 1 - «7(1-W) + : 8h2(a+c)3 2 = {2y(a+c)h + 2yhx - 6y(a+c)y - 6yxy}/8h2(a+c) 2 = {2h(a+c) + 2hx - 6(a+c)y - 6xy}/8h2(a+c) 2 = {-(l-y)h 2 - 2(l-y)hy + 3(l-y)y 2}/8h 2(a+c) 3 = [2(a+c)h(h-ya 9) + x(6h 2-2ya 9h) - y(a+c)(2h+6ya9) -6xy(h+ya 9)]/8h 2(a+c) 2 3 = [2(a+c)h^lh-a9) + x(6yh 2-2a 9h) - y(a+c) (2yh+6a9) -6xy(yh+a 9)]/8h 2(a+c) 2 3 = [8h 2a 5(a+c)(1-y) + 2xh(a+c) (1-y) + 3h(l-y)x 2 + 2h(l-y)ya 9 + 3(l-y)a 9y 2]/8(a+c) 2h 2 i+ = [2xh2(-3h+ya9) - 6yy(a+c) 3 + 6xyh{h-8ya 5(a+c)}]/8h 3(a+c) 3 k • • [2xh2(-3yh+ag) - 6y(a+c) 3 + 6xyh{uh-8a 5(a+c)}]/8h 3(a+c) 3 u = -a 7 ( l - y ) - J3h(l-y)x 2 + 2(l-y)a 9hy - 24(l-y)a 5(a+c)y 2}. k a ? u M J 8h2(a+c)3 5 = {2yh(a+c) - 2yhx - 6y(a+c)y + 6yxy}/8h2(a+c) 5 = {2h(a+c) - 2hx - 6(a+c)y + 6xy}/8h2(a+c) 5 = { ( l - y ) h 2 + 2(l-y)hy - 3(l-y)y 2}/8h 2(a+c) 5 = [-2h(a+c)(h-yag) + 2hx(3h-ya9) + 2(a+c)y(h+3ya9) - 6xy(h+ya 9)]/8h 2(a+c) 2 6 = [-2h(a+c)(yh~ag) + 2hx(3yh-a9) + 2(a+c)y(yh+3a 9 ) - 6xy(yh +a 9)]/8h 2(a+c) 2 61 Continued Table (III-5) MT*3,6 = [8C5(l-M)h2(a+c) - 2h (1-y) (a+c)x + 2h(l-y)a 9y + 3h(l-y)x 2 + 3(l-M)a 9y 2}/8h 2(a+c) 2 MT*i,7 = [2h2x(3h+pa9) +6Pa 3y + 6hxy(h+8yaa 5)]/8h 3a 3 MT*2,7 = [3h 2x(3yh+a 9) + 6a3y.+ 6hxy (yh+8aQ6) ]/8h 3a 3 MT*3,7 = ( l - y ) a 8 - [3d-y)hx 2 + 2 ^ ~ ^ h / + 2 4 ( l - y ) a a R y 2 j MT* 1 ) 8 = -2y[ah + xh + 3ay + 3xy]/8ah 2 MT*2 ,8 = -2[ah + xh + 3ay + 3xy]/8ah 2 MT*3j8 = [-(l-y)h 2 + 2(l-y)hy + 3(1-y)y 2]/8ah 2 M T * l , 9 = [-2ah(h-ya9) + 2hx(3h+ya9) + 2ay(h-3ya 9) + 6xy(h-ya 9)]/8h 2a 2 MT*2 ,9 = [-2ah(yh-a9) + 2hx(3yh+a9) + 2ay(yh-3ag) + 6xy(yh-a 9)]/8h 2a 2 MT*3,9 = [8a6(l-y)h 2a - 2(l - y ) a h x - 2(l-y)h<x9y - 3(l-y)hx 2 + 3 ( l - y ) a 9 y 2 ] / 8 a 2 h 2 M T*1,10 = [-2h2x(3h+ya9) + 6ya 3y - 6hxy(h + 8ya 6a)]/8a 3h 3 M T*2,10 = [-2h?.x(3yh+a9) + 6a 3y - 6hxy (yh+8a6a) ]/8a 3h 3 MT*, ,„ = -(l-y)ct f t + [ 2 h ( l - y ) a g y * 3h(l-y)x 2 + 24a(l-y)a By 2] 3 ,10 u y j a 8 + 8 h 2 a 3 M T*1,11 = 2y[-ah + hx - 3ay + 3xy]/8h 2a MT*2 ,n = 2 [-ah + .hx - 3ay + 3xy]/8h 2a MT*3 n = [ ( l - y ) h 2 - 2(l-y)hy - 3(l-y)y 2]/8ah 2 ,IZ = [-2ah(h+ya9) + 2hx(3h+ya9)' - 2ay(h-3ya 9) + 6xy(h-ya 9) ]/8a 2h 2 MT*2,12 = [-2ah(yh+a9) + 2hx(3yh+a9) - 2ay(yh-3a 9) + 6xy(yh-a 9) ]/8a 2h 2 MT*3,12 = [8h 2(l-y)aa 6 + 2(l - y ) a h x - 2(l-y)a 9hx - 3(l-y)hx 2 + 3(l-y)a 9y 2]/8a Table (III-6) Transeverse Shear at a Point on the Trapezoid Element Qx 0 0 0 0 0 0 -6D2 0 -2D2 0 -6yD2 -6yD2 Qy 0 0 0 0 0 0 0 -2D2 0 -6D2 -6xD2 -6xD2 Ibi Ib 2 l b 1 2 | {Q*} = [R*]{B*} Eqn. (III-6,c) Once the deformation vector {&*} for an element i s known, the moments at any point within the plate can be obtained from Eqn. (111-10) by simply substituting the coordinates of that point. Entries of the moment matrix [MT*] are shown in Table (III-5). 3.6 Transeverse Shears at any Point (x,y) on the Trapezoid Element Substituting for w from Eqn. (III-1) i n the expression for transeverse shears, Eqn. (III-6,b) can be written i n the matrix form: {Q*} = [R*]{B*} (III-6.C) Here, {Q*} = 2 x 1 vector of transeverse shears at a point (x,y) i n the plate, and [R*] = 2 x 12 matrix i n terms of variables x and y. See Table (III-6) Substituting i n the above for {&*} from Eqn. ( I I I - 3 ) , one obtains {Q*} = [R*][A* - 1]{6*} (III-6,d) 3.7 Equivalent Nodal Forces The st i f f n e s s matrix depends upon the d e f i n i t i o n of equivalent nodal force. As i n plane stress, the plate edge transeverse shears and moments may be transferred as equivalent nodal forces to the corners of the f i n i t e element i n two different ways: 1. The edge shears and bending moments are transferred to the adjacent corners by the simple law of the lever - an operation completely consistent with s t a t i c s . The st i f f n e s s matrix so obtained w i l l be referred to as the S t a t i c s s t i f f n e s s matrix 64 of the f i n i t e element i n flexure, 2. The pr i n c i p l e of v i r t u a l work i s employed i n a special way to determine the equivalent nodal forces from the plate edge shears and moments. The matrix so obtained w i l l be referred to as the Energy st i f f n e s s matrix of the f i n i t e element i n flexure. 3.8 Statics Stiffness Matrix for an Equilateral Trapezoid Element i n Flexure Each term of the displacement function given by Eqn. ( I I I - l ) represents an independent simple displacement mode (or an independent curva-ture and twist state). Its f i r s t three terms (b 1,b 2x,and b 3y) correspond to r i g i d body motions and, as such, do not cause any shear or moments. The remaining displacement terms are considered one at a time. The corresponding edge moments and shears are transferred to the adjacent nodes es s e n t i a l l y i n accordance with s t a t i c s . With regard to edge shears this obviously means that they are carried to the nodes at the ends of the edge by the law of the lever. The situation with bending and torsional moments i s more complicated. The procedure for calculating equivalent nodal forces from edge bending and twisting moments i s explained as follows. Twisting Moment One might think that there i s no unique way i n which the twisting moments may be transferred to the nodes as moments since any proportion of the t o t a l edge twisting moment as node moment would s a t i s f y s t a t i c s . Figure (III-3,a) shows the d i s t r i b u t i o n of the torsional moment M along the edge 1-2 of the element. The elementary torques, M dx, are yx 1 yx made up of.couples of equal horizontal forces indicated i n Fig. (III-3,b). 65 (a) Distribution of Torsional Moment, M v x on Edge 1-2 ->|dx|dx ldx|<— (b) Couples of Equal Horizontal Forces forming M (T) t = thickness of the plate r yx - j^dx|dx |dx|<— ©I t l t i l l t l t l t l t l t i t l t l t l t© ^thickness of the plate _ 1 " T (c) Horizontal Couples replaced by Equivalent Couples of Vertical Forces © dM yx dx '/\ ength 1 ( M y x ) 2 t ® (d) Equivalent Shear and Corner Concentrated Shear FIG. (HI-3) CONVERSION OF TWISTING MOMENT INTO DISTRIBUTED TRANSVERSE SHEAR AND CORNER CONCENTRATED FORCE 66 These couples, however, may be replaced by the equivalent couples of v e r t i c a l forces. Fig. (III-3,c). The neighbouring members of these couples i n the two adjacent elements, dx, subtract from each other giving 3M CM +-~- dx)dx M dx 3M yx 3x _ yx = yx d x dx dx 3x 3M yx on every length dx which i s equivalent to the distributed shear ——— per o X unit length (Fig. (III-3,d)). The members of the torques at the corners of the edge have no partners to combine with and so they persist as the con-3M y x centrated corner shears (M ). The distributed edge shear, ' , i s ^ yx 6 3x carried uniquely to the corners by the law of the lever. The combination of these two effects ( i . e . , distributed edge 3M yx shear, — — , and concentrated corner shear, M ) amounts to equal and ' 3x ' ' yx 1 n opposite v e r t i c a l corner forces whose moment equals the sum of a l l torques on this p a r t i c u l a r edge. Edge Bending Moment The bending moment on the edge of the plate (Fig. I I I -4,a) can be viewed as a result of distributed direct stresses (Fig. III-4,b). These direct stresses over the thickness may be replaced by two equal and opposite lines of forces (Fig. III-4,c). Thus, the edge moments are equivalent to two equal and opposite lines of forces. These forces may now be collected at the corners uniquely by the simple law of the lever. This results i n two equal and opposite nodal forces at each corner which are equivalent to nodal moments. This procedure i s obviously equivalent to the transfer of the bending moments to the nodes i n accordance with the law of the lever i n the same way as i n the case of shears. 67 t= thickness of the plate i (a) Plate Edge Moment (b) Equivalent Flexure Stresses over the Thickness (c) Equivalent Concentrated Forces FIG. (ILT-4) CONVERSION OF EDGE BENDING MOMENT INTO EQUIVALENT DIRECT FORCES. 68 3.9 Forces on Inclined Edges Equations (II1-6,a and b) give expressions for the moments and shears on the x and y planes. Calculations of moments and shears for inclined edges must refer to the normal and p a r a l l e l directions of the par t i c u l a r edge. The directions of normal and p a r a l l e l axes, Fig. (III-5), are selected to be the same as the x and y directions (Fig. (III-2)) when the angle of i n c l i n a t i o n of the edge to the Y-axis would approach zero. 26 By simple direction transformation, the moments and shear in n and t directions of the incl i n e d edge are given by M = M Cos 2a + M Sin 2 a - M Sin2a n x y xy M . = I (M - M )Sin2a + M Cos2a (III-6,d) nt 2 v x y xy ' 1 and, = QxCosa + Q^Sina Here, a = the angle which the positive direction of n-axis makes with the positive direction of x-axis and i s considered positive i n the counter clockwise direction, looking from above. Fig. (III-5) The angle a i s positive for edge 1-3 and negative for edge 2-4 of the trapezoid The expressions for moments and shears on edge 2-4 are obtained from those given above by simply replacing the angle a with -a. These are NL = M Cos 2a + M S i n 2 a + M Sin.ioc' ^ 2 - 4 x y *y Mpt •'=:=- (M - M )Sin2a + M Cos2a ' (III-6,e) m o A 2 y x xy v ' J '2-4 1 y x XX and, Q n 2_ 4 = QxCosa - Q ySina 69 FIG.(HT-.5) n AND t DIRECTIONS FOR THE INCLINED EDGES 70 The procedure for calculating nodal forces i s i l l u s t r a t e d for the displacement mode bgxy 2. 3.10 Sample Procedure to Determine the S t a t i c a l l y Equivalent Nodal Forces Due to the Displacement Mode, . w = bgxy 2 The distributions of moments and shears on the X and Y planes are obtained from Eqns. (III-6,a and b) by subsituting w = bgxy 2. These are:: Mx = -2 D 2b 9x My = -2D 2b 9x Mxy = " Myx =:2(l-y)D 2b 9y Qx = -2D2b9 Q = 0 y Nodal Force Contributions from Edges 1-2 and 3-4 S t a t i c a l l y equivalent nodal force contributions are calculated from the distributed edge forces M^., M^ and and are shown i n Fig. (III-6) . Edge 1-3 Angle a for this edge i s positive. Fig. (III-5,a). The d i s t r i -butions of edge moments and shear are obtained from Eqn. (III-6,d). These are given by Mn = -2D 2b 9{x(Sin 2a + pCos2ct) + (l-y)ySin2X'| Mnt = (l-y)D 2b 9{xSin2a + 2yCos2a} = -2D2b9Cosa and are shown i n Fig. (I I I - 7 ) . The components of equivalent nodal forces are 71 Edge 1-2 2D2b9(o+c) 2D 2b g(a+c) Nodal Moments due to M, M yx o e e e e e e e e ©t —(a+c) >N (a+c)-V2(l-/*)D2b9h Nodal Reactions due to M 2 b 9 h J2(I-^)D, 2(l-//.)D2b9h yx E d g e 3 - 4 2D2b9a « | D 2 b 9 a 2 • | D 2 b 9 a 2 2 D 2 b 9 a Nodal Moments due to M, (4).k—- a H>k a —->i(3) Icecccc~y1)2(i-M)D,bah 1 M yx 2(l-^.)D 2b 9h 2(l-/i)D2b9h Nodal Reactions due to M yx FIG(m-6) CALCULATION OF EQUIVALENT NODAL FORCES FROM EDGES 1-2 AND 3-4 OF THE TRAPEZOID ELEMENT DUE TO DISPLACEMENT MODE W=b9xy2 72 3 w2 u9~cosa {(3a+c)(sin 2a 4-yu.cos2a)+(l-yu.)h sin2aj 13J 2 D 2 b 9 { a ( s i n 2 a " l "A i C o s 2 a ) +(r-/x)hsin2a} 2D 2b g{ (a+c)(sin2a+^cos2a)-(l-/x)h sin2a} Nodal Moments due to M, S D 2 b 9 - c ^ { ( 3 a + 2 c K s i n 2 a ; +/xcos2a)-(l-^)hsin2a| Nodal Forces due to Q n FIG(n-7) EQUIVALENT NODAL FORCES FROM EDGE 1-3 (W=bQxv2) 7 3 2D2b9{a(sin2a+yu.cos2a)+(l-/i)h sin2a 2D 2b g^a +c)(sin2a+-/zcos2a) -0-/i)h sin2aj> - ^ K ® / 3" D2 b9^sa {(3a+c)(sin2a y +/tcos2a)+(|-^.)h sin2d} 2h sec a Nodal Moments due to M. 3 2 9 cosa [ (3 a + 2 c) (s i n2a +/z cos2a) - (I -/i)h sin 2aj 2(l-^)D2bg [(a + c)sinacosa y ^ ^ j —h(cos2a - s in2a)}j 2(l-/z)D2b9|a sin a cos a + h(cos2a — sin2a)^ (l-^.)D2bg(2a+c)sinacosa 2°2 b g cosa -ft)Dib> (2a+c)sinacosa I 2D b h (up) c. y 2 9 2D 2b gcosa 2D2bgh(up) Nodal Forces due to M nt Nodal Forces due to Q FIG.(ILI-8) EQUIVALENT NODAL FORCES FROM EDGE 2-4( W = b gxy 2) 74 calculated using the law of the lever and are shown i n the same figure. Edge 2-4 Angle a, for this edge i s negative Fig. (III-5,b). The d i s t r i b u -tions of edge moments and shear are obtained from Eqn. (III-6,e). These are = -2D 2b 9{x(Sin 2a + yCos 2a) - (l-u)ySin2a) Mnt = (1-y)D 2b 9{-xSin2a + 2yCos2a} and, = -2D2bgCosa The d i s t r i b u t i o n and nodal force contributions are shown i n Fig. (I I I - 8 ) . The nodal components of moments from the incl i n e d edges, 1-3 and 2-4, are resolved into x and y directions. F i n a l l y , the contributions of nodal forces from a l l the edges are added. This results i n the equivalent nodal forces due to the displacement mode, w = bgxy 2. 3.11 Equivalent Nodal Forces from a l l Displacement Modes The equivalent nodal forces from a l l displacement modes b.. (j = 1,2,..., 12) can be written c o l l e c t i v e l y i n a matrix form {f*} = [C*]{B*} ( I I I - l l ) Here, [C*] = 12 x 12 matrix of equivalent nodal forces. The equivalent nodal forces corresponding to a displacement mode b.. constitute the j t h column of th i s matrix^ and {f*} = 12.x 1 column vector of nodal forces due to a general state of deformation of the f i n i t e element. Entries of [C*] matrix are shown - in Table (III-7) . 75 Table (III-7) Entries of [C*] Matrix (Trapezoid Element) Entries of the f i r s t three columns of [C*] matrix are zero as displacement surfaces b j , b 2x and b3y correspond to r i g i d body displacement and rotations and therefore do not produce any shears or moments i n the plate. Entries of the 4th Column (due to w = b^x 2) r* L 1 »4 = 2D2 (1-y) Sinct Cpsa C*2 ,h = D2{ (Gos2a + ySin 2a)c - 2y(a+c)} C*3,<* . = -2D 2(Cos 2a + MSin 2a)h = 2D 2(l-y)Sina Cosa 0*5,4. = D2{ (Cos 2a + ySin 2a)c - 2y(a+c)} C*6,4 ' = 2D 2(Cos 2a + ySin 2a)h C*7,4 = -2D 2(1-y)Sina Cosa C*8 ,k = D2{ (Cos 2a + y S i n 2 a ) c + 2ya} C*9 ,4 = -2D2(Cos2ct + ySin 2a)h C*10 ,4 -2D 2(1-y)Sina Cpsa C*U ,4 = D 2{(Cos 2a + y S i n 2 a ) c + 2ya} C*12,4 . = 2D 2(Cps 2a + ySin 2a)h Entries of the 5th Column (due to w = bsx; C*i,5 = -2D 2(l-y)Cos 2a C*2,5 = 2D 2(l-y)h Sin 2a C*3,5 = -2D 2(l-y)h Sina Cosa C*4,5 = 2D 2(l-y)C 0s 2a C*5,5 = -2D2(l-y)h Sih 2a C*6,5 = -2D 2(l-y)h Sina Cosa C*7,5 = 2D 2(l-y)Cos 2a C*8,5 = 2D 2(l-y)h Sin 2a C*9,5 = -2D2(l-y)h Sina Cosa 76 Continued Table (III-7) C*i 0>,5 = - 2 D 2 (l-M)C 0s 2a C * n , 5 = - 2 D 2 (l-y)h Sin 2a C* i 2 , s ' = - 2 D 2(T-y)h Sina Cosa Entries of the 6th Column (due to w = b6y 2) C*l, 6 = - 2 D 2(1-y)Sina Cosa C*2, 6 = D 2[(Sin 2a + yCos 2a)c - 2(a+c)] C*3, 6 = - 2 D 2 h ( S i n 2 a + PCos2a) C*4, 6 = - 2 D 2(1-y)Sina Cosa C*5, 6 = D 2{(Sin 2a + yCos 2a)c - 2(a+c)} C*6, 6 = 2 D 2 h ( S i n 2 a + yCos 2a) C*7, 6 = 2D2(1-y)Sina Cosa C*8, 6 = D 2 J ( S i n 2 a + yCos 2a)c + 2a} C*g, 6 = - 2 D 2 h ( S i n 2 a + yCos 2a) C * l 0 ».6 = 2D2(1-y)Sina Cosa C*n , 6 = D 2{(Sin 2a + yCos 2a)c + 2 a } C*12 . 6 = 2 D 2 h ( S i n 2 a + yCos 2a) Entries of the 7th Column (due to w = b7X ) C*!,7 = -6D 2[h{l + (l-y)Sin 2a} - (1-y)(a+c)Sina Cosa] C*2,7 = D 2[c(3a+2c){l - (l-y)Sin 2a} - 2y(a+c) 2] C*3,7 = -2D 2h(3a+2c){l - (l-y)Sin 2a} C*^,7 = 6D 2[h{l + (l-y)Sin 2a} - (1-y)(a+c)Sina Cosa] C*5i.7 = -D 2[c(3a+2c){l - (l-y)Sin 2a} - 2y(a+c) 2] C*6,7 = -2D 2h(3a+2c){l - (l-y)Sin 2a} C*7,7 = -6D 2[h{l + (l-y)Sin 2a} + (l-y) a Sina Cosa] C*8»7 = D 2[c(3a+c){l - (l-y)Sin 2a} + 2ya 2] Continued Table (III-7) C*g, y = -2D 2h(3a+c){l - (l-y)Sin 2a} C*10 7 y = 6D 2[h{l + (l-y).Sin2cx} + (l-y)a Sina Cosa] C*ll» 7 = -D 2[c(3a+c){l - (l-y)Sin 2a} + 2ya 2] C*12,7 , = -2D 2h(3a+c){l .- (l-y)Sin 2a} Entries of the 8th Column (due to w = b 8x 2y) • C*l,8 •= D2{y + 2(l-y)Sin 2a}(2a+c) C*2,8 = D2[2yh(a+c) + | (3(l-y)(4a+3c)Sin 2a - c}] C*3„8 = -|D 2h{2(l-y)(3a+2c)Sina Cosa - h(Cos 2 a + ySin 2a)} C\,8 = D2{y + 2(l-y)Sin2 a} (2a+c) L 5,8 = D2[2yh(a+c) + |{3(l-y)(4a+3c)Sin 2a -C*6,8 = |D 2h[2(l-y)(3a+2c)Sina Cosa - h(Cos 2 a + ySin 2a)] C*7,8 = -D2{y + 2(l-y)Sin 2a](2a+c) C*8,8 = D2[2yha + |{3(l-y)(4a+c)Sin 2a + c}] C*9,8 - -|o 2h[2(l-y)(3a+c)Sin a Cosa + h(Cos 2 a + ySin 2a)] C*10,8 = -D 2[y + 2(l-y)Sin 2a](2a+c) C*ll,8 = D 2[2yha + |{3(l-y)(4a+c)Sin 2a + c}] C*12,8 = |p 2h[2(l-y) (3a+c)Sin a Cosa •+ h(Cos 2a + ySin 2a)] Entries of the 9th Column (due to w = b qxy 2) C*l,9 ' = -D2[2yh + (1-ry) (2a+c)Sin a Cosa] C*2,9 •~2[a(2a+c) •+• 3(l-y)c(a+c)Cos 2a] C*3,9 = -|D2h[(3a+2c) - 3(l-y)(a+c)Cos 2a] C%,9. = D 2[2yh + (1-y)(2a+c)Sina Cosa] C*5,9 | 2[a(2a+c) '•+ 3(l-y)c(a+c)Cos 2a] C*6,9 = -|D2h[(3a+2c) -,3(l-y)(a+c)Cos 2a] C*7,9 = -D2[2yh - (1-y)(2a+c)Sina Cosa] 78 Continued Table (III-7) <- 8» 9 I2- [(a+c)(2a+c) - 3(l-y)ac C 0s 2a] L 9, 9 -|-D2h[(3a+c) - 3(l-y)a Cos 2a] C*10,9 = D2[2yh - (1-u)(2a+c)Sina Cosa] C*ll,9 = ~ [(a+c)(2a+c) -,3(1-U)a.c Cos 2a] C*12,9 = | D2h[(3a+c) - 3(l-y)a Cos 2a] Entries of the 10th Column (due to w =.bioy ) C*i > 10 = 3D2 (2a+c) C*2, 10 = D2h[6(a+c) - c(Sin 2a + yCos 2a)] C*3, 10 = 2D2h 2(Sin 2a + yCos 2a) 10 = 3D2(2a+c) C*5, 10 = D2h[6(a+c) - c(Sin 2a + yCos 2a)] C*6, 10 = -2D2h 2(Sin 2a + yCos 2a) C*7, 1.0 = -3D2(2a+c) C*8, 10 = D2h[6a + c(Sin 2a + yCos 2a)] C*9, 10 = -2D2h 2(Sin 2a + yCbs 2a) C*10 ,10 = -3D2(2a+c) C*u ,1.0 = D2h[6a + c(Sin 2a + yCos 2a)] C*i2 ,.10 = 2D2h 2(Sih 2a + yCds 2a) Entries of the 11th Column (due to w = b u x y) c * l , l l " D 2[-(l-y)(4a 2+5ac+ 3 c 2 ) + a(2a+c) + 2h2+ (1-y)(6a 2+6ac+3c 2+2h 2)Sin 2a] C*2,n a D 2h[2y(a+c) 2 + (1-y) ( 6 a 2 + 9 a c+4c 2)Sin 2a - c(a+c)] c * 3 , l l B - n 2h[(l-y ) ( 6 a 2 + 9 a c+4c2)Sinoi Cosa - 2h(a+c)] c V l l B -D 2[-(l-y)(4a 2+5ac+3c 2) + a(2a+c) + 2h2+ (1-y)(6a 2+6ac+3c 2+2h 2)Sin 2a] c*S»ll ° -D 2h[2 y(a+c)2 + (1-y)(6a 2+9ac+4c 2)Sin 2a - c(a+c)] c * 6 i U B -D 2h[(l-y) (6a 2+9ac+4e 2)Sin a Cosa - 2h(&+c)] 1 1 , y / — \ 3 CM 3 CM i — i XI CM CM to CM i i C M x: to O x; CM i — i C M o C J i — i x: ,—^ + I 1 i-1 C J i — i 3 3 ,—^ 3 o i — i + O 1—1 3 CM C N 3 3. CM /—•» nj X! cd •5 / N CM x: + 1 CM C M to o cd / — \ cd CO x: + ,—\ i—1 to 1 : O + + CM o + CM X / — \ 3. 3 ^—J O C J cd »—i + r — i X 3 3 1—1 CM CM (—1 CM C J v—' 3 CN 3 3 + nl 3 3 + CM CM CM f—\ 1 C , s •3. 1 1 C M ,—1 v 1 CM CM C N CM CM f—i 1 C 3 • H 3 1 1—\ + /—N t—> C C 3 v—> C C 3 f—( •H CM v ' co CM + 3 1 3O •H •H to t— •H •H co ^—^ C O to CM v ' to 3 CM 3 CM + C O C O o CJ co co O II CM >—' o f—^ o CM 3 to 3 CM to CTJ / — \ / — \ C J + f—% / — \ C J / — \ C J 1 o u to CM o CM C o v — v CM' CM cd CM CM s- 1 o 3 to 3. 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CM ^ ^ >—t C O v / C O cd v — ' cd o v—' cd 3 CM i—\ 1 O i—\ + CJ V I V J to + / N to cd + / — \ to 1 CJ o i—l o , — ^ + + r—^ + 3. + to CM 3 + i—1 to v — ' + to / — \ + O o r—\ CTS 1 CM + cd 1 CM 3 v—/ + + + CJ 1 I—\ CM • 1 cd MO 1—1 Cd vD •—1 cd - i—I cd + CM cd + CM cd CJ CJ cd i v—' > -* cd v ' v ' vD O + CM CM ,—^ CM cd CJ C M + C M I—< v ' /—y —' / — \ *—1 C J / — \ v ' / N CJ v ' CM v ' cd + t—l / — \ 3 + r \ \ — ' 3 + t \ / — \ O O + v—/ + CM 1—1 3. 1 3. / — \ 1 CM 3. xi (J + 1 + cd 1 , — , + * > CM + v — ' 1 r—i 1 3 t—1 cd 1 + J + cd cd ^ y CM t—\ cd i—t v ' cd •—1 1 — ' 3 i—1 CM cd v—' CJ CN o + cd o o o <D v — y 3. v ' t-i CM >—' i—1 CM I—1 • i I—I cd + l-llj 1 1 •—1 1 + CM CM cd CM V / CM cd CM •s 1 — • 1 JC 1 XI XX <D cd x; L 1 x:| x: 1—' X! v ' x: cd CM l_-J- CM CM x: x: 1 CM CM CMICN CM CM x: 1 CM t—-1 CM) C M CM H Q CM Q CM a a + J CM <M|CM a a Q 1 a a CM CM a CM Q| a I a I a i i a Q | i i 1 i i Q | i Q 1 X ) CD 4-1 II ii II n II II O II II II II II II II II II II II II i—i i—( to CM CM CM r—1 >—< <—i •—i i—< CD CM CM CM CM CM CM CM CM CM i—i .—< •H I 1 .—< .—i •* •H r-~i i—i ,—t .—* .—i i—i 1—1 i-H »\ • •M «\ o i—1 CM u o r—i CM c CD c n I 1 •—i 4-> r—* CM m J - IT) CD CO c n 1—1 t 1 o * * * * * * C * * * * * * * * * * * * C J C J C_> C J C J C J cj UJ C J C J C J C J U U C J C J U C J C J 80 Using Eqn. ( I l l - 3 ) to replace {B*} i n Eqn. ( I I I - l l ) one obtains {£*} = [C*] [A* _ 1]{6*} (111-12) Equation (111-12) relates the nodal forces with the nodal deformations,'. The force-deformation r e l a t i o n for a f i n i t e element i s commonly expressed i n the matrix form as {f*} = [K*]{(5*} (111-13) Here, [K*] = 12 x 12 s t i f f n e s s matrix for a trapezoid element i n flexure. Comparing matrix equations (111-12) and (III-13), one obtains [K*] = [C*][A* - 1] (111-14) The entries of [K*],, being very long expressions, are not l i s t e d . Note that the flexure s t i f f n e s s matrix of the trapezoid element obtained by using the s t a t i c s approach i s asymmetrix ( i . e . K*£ .. $ K*.. ^) . Note also that i'f the trapezoid reduces to a rectangle (top side = bottom side), the asymmetry of the s t a t i c s s t i f f n e s s matrix disappears and the resulting matrix 37 coincides with the one derived by the author e a r l i e r . Apparently i t s lack of symmetry re f l e c t s the asymmetry of the trapezoid element i t s e l f about the x-axis. However, i t i s more important to note that i n conditions of uniform curvatures or twist the nodal forces s a t i s f y the reciprocal relations 23 with the element remaining trapezoid. When the element size i s decreased i n the l i m i t j uniform curvature conditions w i l l i n fact p r e v a i l i n the v i c i n i t y of the element and the reciprocal relations w i l l be automatically s a t i s f i e d . 81 3.12 Energy Stiffness Matrix for a Trapezoid Element i n Flexure As i n plane stress, t h i s method uses the v i r t u a l work p r i n c i p l e to define the equivalent nodal forces. " F i n i t e element" has been defined as a body whose nodes undergo the same deformations as the plate whose nodes are acted upon by concentrated forces s t a t i c a l l y equivalent to the edge forces. Since during a v i r t u a l deformation i n a plate the same nodal deformations occur i n the f i n i t e element, the condition imposed on the values of the equivalent nodal forces then i s , that during v i r t u a l deformation, the external work done by the equivalent nodal forces of the f i n i t e element be the same as the external work done by the edge forces of the plate Note that during v i r t u a l deformation of a plate loaded along the edges only, the external work done by the edge forces of the plate i s equal to the internal work done by the internal moments i n the plate. However, as explained i n the case of plane stress, the supposition of equality of external work i n the f i n i t e element and i n the plate element i s hypothetical. In p a r t i c u l a r , i f the v i r t u a l deformation {6*}^ i s such that i t i s unity i n the direction of a selected nodal force f* and zero i n the direc-J * tion of a l l other forces, then the external work i n the f i n i t e element We. 3 * w i l l be the same as the value of th i s selected nodal force f., i . e . 3 We. = {6*}. T{f*} = f* (111-15) 3 3 3 • J Here, = 12 x 1 column vector of v i r t u a l deformations. Its j t h entry i s unity and a l l other entries are zero. A l l such v i r t u a l deformations {6*}j . (j = 1,2,...,12) can be c o l l e c t i v e l y written as [6*] which i s the same as a 12 x 12 Identity matrix [ I ] . 82 The corresponding external works for a l l such v i r t u a l deformations can be expressed i n the matrix form {wS} = [6*]{f*f = [I]{f*} = Jf*} (111-16) 12x1 12x12 12x1 12x1 Here, {We} = 12 x .1 column matrix of external work done by the nodal forces for a l l v i r t u a l deformations. {6*}.. (j = 1,2,..., 12) Work i s done corresponding to each of these v i r t u a l deformations {6*}^, by the edge moments and shears of the plate element which i s the same as the internal work Wi.. done by the internal moments over the area of the plate. 26 For a plate of constant thickness, the internal work may be expressed as, Wi. = // {x*},T{M*}dxdy (II1-17) 3 area 3 Here, {x*}j = curvatures due to a v i r t u a l deformations {6*}^. The integra-tio n i s carried out,over the entire area of the plate. Using Eqn. (I I I - 5 ) , the curvatures due to v i r t u a l deformations can be expressed as, •{x*}j = [G*][A*" l]{6*}j (HI-18) Substituting for {x*}^ from Eqn. (111-18) and for {M*} from Eqn. (III-8) into Eqn. (111-17) for internal work, one obtains Wi. = // {&*}. T[A* - 1] T[G*] T[D*][G*][A* _ 1]{6*}dxdy (111-19) J area J '1 ] T [ G * ] T | l x l 1x12 12x12 12x3 3x3 3x12.12x12 12x1 83 C o l l e c t i v e l y , internal works for a l l such v i r t u a l deformations {6*}.., j 1,2,..., 12 ( i . e . [6*]), can be written i n matrix form {Wi} = // [6*] T[A* 1] T[G*] T[D*] [G*] [A* 1]{6*}dxdy (111-20) 12x1 a r 6 a 1 2 x l 2 .12x12 12x3 3x3 3x12 12x12 12x1 * Here, {Wi} = 12 x 1 column matrix of internal work done by the internal forces i n the plate during a l l v i r t u a l deformations {<$*}.., j = 1,2,...,12 and [6*] =i [ I ] , 12 x 12 Identity matrix. The matrix [A* 1 ] i s expressed i n terms of constant nodal coordinates and, as such, can be taken out of the integral sign. Eqn. (III-20) can be rewritten as {Wi} = [ A * _ 1 ] T // [G*] T[D*][G*]dxdy[A* _ 1]{6*} (111-21) area Equating the external work done by the nodal forces of the f i n i t e element, Eqn. (III-16), to the corresponding internal work done by the internal moments i n the plate element, Eqn. (111-21), one obtains ;{f*} = [A* 1] T a^ a[ G*] T[D*|[G*]dxdy[A*- 1]{6*} (111-22) l e t , [H*] = If [G*] T[D*][G*]dxdy (111-23) area The integrated 12 x 12 [H*] matrix i s given i n Table (II1-8). Equation (III-22) can be rewritten as, {f*} o [A* _ 1] T[H*][A* _ 1]{6*} (111-24) Table (III-8) Integrated [H ] Matrix for a Trapezoid Et^ 12(l-tO * [H..J = 0 0 0 0 0 0 0 0 0 0.0 0 0 0 0 0 0 0 0 0 0 0 Q 0 Here, [H ] [G*] T[D*][G*] dxdy 4L.i 0 0 4L 2 0 0 0 12yL2 area 2(l-y)L! 0 0 0 4 L 2 4Li 0 4yL 2 0 0 0 0 6 ( l - u ) L 3 0 Symmetrical 36Lc 41^ + 8 ( l - u ) L 3 0 12pL3 12L 2 0 12uL4 36L 5 0 0 0 0 6(l-y)Lit 0 36uL5 Eqn. (111-23) L1 = 2(2a+c)h L 2 = -§ch 2 L3.*&pL{c*+ (2a +c) 2} L 4 = |(2a+c)h 3 L 5 =-§5-[c2 + 5(2a +c)2] L 6 = 4 ch« L 7 = ^ f V + ^ ( 2 a + c ) 2 + .(2a +c)^] „ L 8 - (2a +c)h3[^i + .I2g£2l] L 9 = |-(2a+c)h5 4L 3 +8(I-u)L 4 0 361^ 36L 8 12L 5 0 +18(l-u)L 7 12L 5 36L 8 + 12(l-u)L 6 0 18(l + y)L 8 + 18(l-y)L 9 85 Equation (II1-24) provides the force-deformation relationship for an equilateral trapezoid i n flexure. Comparing Eqn. (III-13)'with Eqn. (111-24), the Energy st i f f n e s s matrix for a trapezoid i n flexure i s [K*] = [ A * _ 1 ] T [ H * ] [ A * - 1 ] (111-25) Note that [D*] i s a symmetric matrix. Equation (III-22) represents a 25 congruent transformation of the symmetric matrix. Thus, the energy s t i f f -ness matrix [K*] i s symmetrical. The matrix multiplication of Eqn. (111-25) being very involved, i s performed preferably by computer. Due to the geometrical symmetry of the f i n i t e element about Y-axis, certain equalities i n the st i f f n e s s coefficients exist irrespective of the manner i n which these are calculated. These equalities i n s t i f f n e s s c o e f f i c i e n t s , shown i n Table (111-9) , provide an additional check on the computation of s t i f f n e s s matrices. When a trapezoid i s reduced to a rectangle, the Energy s t i f f n e s s 27 matrix becomes i d e n t i c a l to the one derived by Zienkiewicz and Cheung for an i s o t r o p i c plate. Table (III-9) Equalities i n the Stiffness Coefficients due to the Geometrical Symmetry of the Equilateral Trapezoid in,Flexure Fig. ( I I I - l ) 'K.1,2 * * Ki+,1 * K 4 , 2 * - K 4 , 3 K l , 7 * K 1 , 8 * K l , 9 * K 4 , 7 * K 4 , 8 * - K 4 Y 9 * K 2 , l * ^2 , 2 K 2 / 3 * K 5 , l * K 5 , 2 * " K 5 , 3 * K 2 , 7 * K 2 , 8 * K 2 , 9 * K 5 , 7 * K 5 , 8 * _ K 5 , 9 * K 3 , 1 * K 3 , 2 * K 3 , 3 * - K 6 , i * " K 6 , 2 * K 6 , 3 * K 3 , 7 * K 3 , 8 * K 3 , 9 * " K 6 , 7 * _ K 6 , 8 * K 6 , 9 * K 4 , 1 * K L + , 2 * K 4 , 3 * K l , l * K l , 2 * " K l , 3 * K 4 , 7 * K<t,8 * K 4 , 9 * K l , 7 * K l , 8 * _ K 1 , 9 * * 5 , 1 . * K 5 , 2 K 5 , 3 * K 2 , 1 * K 2 , 2 * - K 2 , 3 * K5,7 * K 5 , 8 * K 5 , 9 * K 2 , 7 * K 2 , 8 ' * _ K 2 , 9 K 6 , l * K 6 , 2 * • • K 6 , 3 * - K 3 , 1 * - K 3 , 2 * K 3 , 3 * K 6 , 7 * K 6 , 8 * K 6 , 9 * " K 3 , 7 * - K 3 , 8 * K 3 , 9 * K 7 , 1 * K 7 , 2 * K 7 , 3 * K 1 0 , 1 * K 1 0 , 2 * " K 1 0 , 3 * K 7 , 7 * K 7 , 8 * K 7 , 9 * K 1 0 , 7 * K 1 0 , 8 * Y 1 0 , 9 4,1 * K 8 , 2 * K 8 , 3 * K l l , 1 * K l - 1 , 2 * - K l l , 3 K 8 . , 7 * K 8 , 8 * K 8 , 9 * K l l , 7 * K l l , 8 * * K11,9 * K 9 , 1 * K 9 , 2 * K 9 , 3 * ~ K 1 2 , 1 * - K 1 2 , 2 * K 1 2 , 3 * K 9 , 7 * K 9 , 8 * K 9 , 9 * " K 1 2 , 7 * v " 1 2 , 8 * Y 1 2 , 9 * K 1 0 , 1 * K 1 0 , 2 * K 1 0 , 3 * K 7 , 1 * K 7 , 2 " * " K 7 , 3 * K 1 0 , 7 * K 1 0 , 8 * K 1 0 , 9 * K 7 , 7 * K 7 , 8 * " K 7 , 9 * K l l , 1 * K l l , 2 * K l l , 3 * K 8 , 1 * K 8 , 2 * - K 8 , 3 * K l l , 7 * K l l , 8 * K l 1 , 9 * K 8 , 7 * K 8 , 8 * _ K 8 , 9 * K 1 2 , 1 * K 1 2 , 2 * K 1 2 , 3 * - K 9 , 1 * - K 9 , 2 * K 9 , 3 * K 1 2 , 7 , * K 1 2 , 8 * K 1 2 , 9 * " K 9 , 7 . * " , K 9 , 8 * K 9 , 9 . 87 FIG.(rLT-9) ISOSCELES TRIANGULAR ELEMENT IN FLEXURE. (Positive directions of nodal forces and deformations.) 88 3.13 Flexure Stiffness Matrix for a F i n i t e Element i n the Shape of an Isosceles Triangle. Fig. ( I l l - 9 ) Selection of Displacement F i e l d An isosceles triangular element i n flexure has nine degrees of freedom of displacements and rotations at i t s j o i n t s . Thus a displacement f i e l d containing nine independent displacement modes i s required for the complete determination of the element s t i f f n e s s matrix. A polynomial expression i n x and y i s selected to represent various displacement modes of the element. The expression must include three displacement modes corresponding to r i g i d body movements of the element which do not affect the stresses i n any way. It must also contain three independent displace-ment modes corresponding to constant curvatures i n x and y directions and constant twist. Thus, the assumed general expression for w must necessarily contain the linear and quadretic terms, b 1 } b2X, b3y, b^x 2 ,b5xy^ and bgy 2. Restricting the choice of the remaining three terms to the cubic, the follow-ing four terms are available for selection; byx^, b 8 x 2 y , b 8xy 2 and bgy 3. In the present study, only isosceles triangles symmetrical about Y-axis are encountered (Fig. ( I l l - 9 ) . The symmetry of the element about Y-axis i s used to select the displacement modes. In the cubic polynomial containing ten terms, s i x terms marked S_produce symmetric and four terms marked A produce antisymmetric deflections with respect to Y-axis. bl» b2x> b3y, t"tx2» b 5 x y , b 6 y 2 , b 7 x 3 , b 8x 2y, b 8xy 2, b 9 y 3 S A S S A S A S . A S An arbitrary deflection f i e l d represented by a p a r t i c u l a r polynomial may be considered as made up of a symmetrical part composed only of the symmetrical 89 terms and an antisymmetrical part composed only of the antisymmetrical terms. Nine deflection f i e l d s are needed for determination of the s t i f f n e s s c o e f f i c i e n t s : 6 2 ' and 63 : for node 1; 64 •', 6 5 " and 6 e for node 2, and 67' ' 6 g ' a n d 6 9 ' for node 3 Fig. (III-9). The s i x displacement f i e l d s &l': to 6g*, related to nodes 1 and 2, may evidently be made up of three symmetric and three antisymmetric terms. As to the deflection f i e l d s of the node 3, two, 67 and &Q":, are symmetrical, and one, 6 9 - , i s a n t i -symmetrical. Thus, five symmetrical and four-antisymmetrical terms are needed i n the polynomial capable of describing the nine independent displace-ment f i e l d s required for the derivation of the s t i f f n e s s matrix of an element i n the shape of an isosceles tr i a n g l e . Thus, a l l four antisymmetric terms must be used. Out of the s i x symmetric terms, only five are required leaving the option of discarding either b 8x 2y or bgy 3 symmetric term. In the present case, b 8x 2y term has been omitted. Retaining b 8 x 2 y symmetric term i n place of the b 9 y 3 term i s expected to y i e l d an equally good s t i f f -ness, matrix. Note: Various choices have been suggested i n the past for selecting nine 1 1 1 4 2 8 2 9 independent displacement modes for a general triangle ' ' ' . The s u i t a b i l i t y of the resulting polynomials i s examined for the special case when the general triangle reduces to an isosceles triangle of the present study. 90 3.14 Choice 1. Omission of D5xy term. 1 1 This i s an antisymmetric term. A l l antisymmetric terms are required for the evaluation of the parameters b},. .. ,bg . ' Omission of this term makes i t impossible to evaluate these parameters as the resulting [A*] matrix w i l l be singular. Moreover, b 5xy term corresponds to constant twist in the X and Y planes. Omission of this 21 22 term w i l l violate the necessary condition for convergence. ' 3.15 Choice 2. Omission of bsxy 2 term. This i s an antisymmetric term and, as explained before, can not be omitted. 28 3.16 Choice 3. ,Combination term bs(x 2y + x y 2 ) . Here the symmetric term x 2y i s combined with the antisymmetric term xy 2 under a common parameter b 8. Using t h i s combination term, the s t a t i c s and energy s t i f f n e s s matrices for an isosceles triangle were evaluated. Due to the geometrical symmetry of the element about Y-axis, certain, equality relations among the s t i f f n e s s coefficients must hold. Table (111-10). These equality relations are independent of the size of the element and the manner i n which the s t i f f n e s s coefficients are derived. The s t i f f n e s s coefficients of the evaluated st i f f n e s s matrices f a i l e d to s a t i s f y these equality relations suggesting the use of the above combination term unsuitable for an isosceles t r i a n g l e . 25 For c l a r i f i c a t i o n of this point, assume that i t i s necessary to determine r * 1 the values of the parameters i n several independent antisymmetrical f i e l d s . 1, 8y 3; 2, Vl\ and -W2 meaning by this simultaneous equal and oppositf^leflections at the nodes 1 and 2 ( a l l other nodal deflections or rotations being zero); 3, simultaneous 6 X l and -8 X , and 4, simultaneous '8. and Q.. . These four independent antisymmetric displacement f i e l d s require definite values of the four antisymmetrical parameters. Among these Table (111-10) Equalities i n the Stiffness Coefficients due to the Geometrical Symmetry of the Isosceles 92 i s the combination term parameter b 8, whose value cannot be zero i n a l l four f i e l d s . The existence of b 8 introduces a symmetrical component of W associated with i t s antisymmetrical term. Other symmetrical terms w i l l also be found necessary to s a t i s f y the equations. This means that although the corner conditions i n the assumed f i e l d s are antisymmetrical the displacement W i s not, because i t contains also some symmetrical terms which contradicts structural theory. Thus, the symmetrical and a n t i -symmetrical terms must not be combined under the same parameter. 3.17 Evaluation of the Constant Parameters On the basis of the foregoing arguements, the following d i s -placement function containing nine independent displacement modes i s selected: W =-bi-+ b 2x + b 3y + b^x 2 + b 5xy + b 6 y 2 + b 7 x 3 + b 8xy 2 + bgy 3 (111-26) The displacements and rotations,{6*}, at the nodes of the isosceles triangle are expressed i n terms of the nine constant parameters, {B*}, as i n Eqn. (III-2) and are shown i n Table (III-11) . The parameters {B^} are evaluated i n terms of the nodal deforma-tions as i n Eqn. (III-3) and are shown i n Table (111-12) . Table ( I I I - l l ) Nodal Deformations i n Terms of Constant Parameters (Isosceles Triangle i n Flexure) (Fig. (III-9)) «1 1 c -h c 2 -ch h 2 C3 ch 2 -h 3" 6 2 0 0 1 0 c -2h 0 -2ch 3h 2 0 -1 0 -2c h 0 -3c 2 -h 2 0 6 4 1 -c -h c 2 ch h 2 -c-3 -ch 2 -h3 6 5 0 0 1 0 -c -2h 0 2ch 3h 2 -56 0 -1 0 2 c h 0 -3c 2 -h 2 0 6 7 1 0 h 0 0 h 2 0 0 h3 <58 0 0 1 0 0 2h 0 0 3h 2 6 9 0 -1 0 0 -h 0 0 -h 2 0 b l b 2 b 3 b^ b 5 b 6 b 7 b 8 b 9 r * 1 * r * , Table (111-12) Evaluation of Parameters i n Terms of Nodal Deformations (Isosceles Triangl e i n Flexure) b l 1 4 h 8 o |oo 1 4 h 8 -c 8 1 2 -h 4 0 «1 9 16c h 4c 3 16 -9 16c -h 4c 3 16 0 0 -1 4 6 2 b 3 3 8h -1 8 -3c 16h -3 8h 1 8 3c 16h 3 4h -1 4 0 «3 \ 0 0 -1 4c 0 0 1 4c 0 0 0 64 b 5 -3 8ch 0 -1 8h 3 8ch 0 -1 8h 0 0 -1 2h <55 b 6 0 -1 8h ! 0 0 -1 8h 0 0 1 4h 0 . «6 b 7 -1 4c"3" 0 -1 4 c 2 1 4cT 0 -1 4c2 0 0 0 6 7 b -3 -1 -1 3 1 . -1 0 0 -1 5 8 8 16ch2 4ch 16h2 16ch2 4ch 16h2 4h2 b 1 1 c 1 1 -c -1 1 0 6 9 9 8h3 8h2 16h3 8h3 8h2 16h3 4h3 4h2 95 3.18 Curvatures and Moments at a Point Curvatures and twist are expressed through a matrix equation s i m i l a r to Eqn. (III-4) and are shown i n Table (II1-13). Moments at any point i n the element are expressed through a matrix equation s i m i l a r to Eqn. (III-8) and are obtained by substituting proper values of the component matrices [D*], [G*],and [A* *] for an isosceles triangle given i n Tables (111-4, 13, and 12). 3.19 Statics Stiffness Matrix for an Isosceles Triangular Element i n Flexure Equivalent corner forces for each displacement mode b^ are obtained in a manner s i m i l a r to a trapezoid. C o l l e c t i v e l y , these forces can be expressed i n terms of constant parameters as i n Eqn. ( I l l - 1 1 ) . The equiva-lent corner forces for individual displacement mode are shown i n Table (III-14). F i n a l l y , the Statics s t i f f n e s s matrix for flexure i s obtained using Eqn. (III-14). The component matrices of this equation, i . e . [A* 1 ] and [C*] are given i n Tables (II1-12) arid (111-14). The entries of the Statics s t i f f n e s s matrix are shown i n Table (111-15). The Statics s t i f f n e s s matrix for a triangle in. flexure i s found to be asymmetric. 3.20 Energy Stiffness Matrix for an Isosceles Triangular Element i n Flexure The derivation of the flexure s t i f f n e s s matrix for an isosceles triangle by the Energy approach i s carried out i n a manner si m i l a r to that for a trapezoid and i s given by a matrix equation s i m i l a r to Eqn. (111-25). The component matrices of Eqn. (111-25) for an isosceles t r i a n g l e , i . e . , [A* - 1] and this central integrated matrix [H*] are given i n Tables (111-12) and (III-16). Theentries of the Energy s t i f f n e s s matrix are given i n Table (111-17). Table (111-13) Curvantures arid Twist at a Point on a Triangular Element - 32w 3x 2 - 32w ay 2 -2 3x3y 0 0 0 -2 0 0 -6x 0 0 0 0 0 0 0 -2 0 -2x -6y 0 0 0 0 2 0 0 4y 0 '1 »2 b 3 b 7 b 8 b 9 r * i * r * i 1XA) = [G A]{B A} * Table (111-14) Entries of [C A] Matrix (Isosceles Triangular Element i n Flexure). F i r s t three terms of the displacement function, (bj , b^x, b$ y) , correspond to r i g i d body displacement and rotations thus they do not produce any forces or moments. Entries of 4th Column (due to w = K x2 ) * = 2D 2(l-y)Sina Cosa * C ? 4 A = D2c{-2y + Cos 2a + ySin 2a} * = -2D 2h{Cos 2a + ySin 2a) = 2 D 2(1-y)Sina Cosa * C C u A ' = D2c{-2y + Cos2a + ySin 2a} * CA 6 , 4 = 2D 2h{Cos 2a + ySin 2a} = -4D 2(l-M)Sina Cosa C a k A ' = 2D 2c(Cos2a + ySin 2a) * C q k = 0 Entries of 5th Column (due to w * C 1 5 A ' = -2(l-y)D 2Cos2a * C 5 c A^' ' = 2D 2h(l-y)Sin 2a * C 3 5 A -2D 2h(l-y)Sina Cosa * C k s A ' = 2(l-y)D 2Cos 2a * CA 5 , 5 = -2D 2h(l-y)Sin 2a CA6,5 = -2D 2h(l-y)Sina Cosa * C 7 5 A • = 0 * C B s = 0 * CA9,5 = -4D 2h(l-y)Sina Cosa Continued Table (111-14) Entries of 6th Column (due to w = bg y2 ) * .6 = -2(l-y)D 2Sina Cosa * C A 2 .6 = D2c[-2 + {y + (l-y)Sin2a}] * .6 = -2D2h{y + (l-y)Sin 2a} * CA» . 6 = -2(1- y)D 2Sina Cos a * C A ^ 6 = D 2c[-2 + {y + (1- y)Sin2a}] * CA&< 6 = 2D2h {y + (1- y)Sin2a} * .6 = 4(1- y)D2Sina Cosa * CA* 6 = 2D2c{y + (l-y)Sin 2a} * C 9 6 = 0 Entries of 7th Column (due to w = b 7x ) * C A l .7 = -6D 2[h{l + ( l - y ) S i n2 a } - c ( l - y ) S i n a Cosa] * CA 2 7 = 2D 2c2[(l-y)Cos 2a] * CA3 .7 = -4D 2hc{l-(l-y)Sin 2a} * 7 = 6D 2[h{l + ( l - y ) S i n 2 a } - c(l-y)Sina Cosa] * 7 = -2D 2c 2(l-y)Cos 2a * 7 = -4D 2hc{l - ( l - y ) S i n 2 a } *. 7 = 0 * CA8, 7 = 0 * C A 9 ) 7 = -4D 2hc{l - (l-y)Sin2a} Entries of 8th Column (due to w = bftxy 2) * CAl> 8 = -D2{2yh + (l-y)c Sina Cosa} * CA2, 8 = -D 2c2(l-y)Cos 2a * CA3, 8 = -| D2hc{2 - 3(l-y)Cos2a} * A H > 8 = D2{2yh + (l-y)c Sina Cosa} 99 Continued Table (111-14) * CA5,8 = D 2 c 2 ( l - y ) C o s 2 a * CA6,8 = -~ D2hc{2 - 3(l-y)Cos 2a} * CA7,8 = 0 * CA8>8 0 * CA9,8 = 4 D2bc Entries of 9th Column (due to w = * CA1,9 = 3D2c * CA2,9 = D 2h c{6 - (Sin 2a•+ yCos2ct)} * CA3,9 = 2 D 2 h 2 ( S i n 2 a + yCos 2a) * CA4,9 = 3D2c * CA5,9 = D 2h c{6 - (Sin 2a + yCos 2a)} * CA6,9 - 2 D 2 h 2 ( S i n 2 a + yCos 2 a) * CA7,9 = -6D2c * CA8,9 = 2D 2hc(Sin 2 a + yCos 2 a) * CA 9 » 9 = 0 bg/ 3) 100 Table (III-15) Statics Stiffness Matrix for an Isosceles Triangular Element i n Flexure. (Fig. ( I l l - 9 ) ) Let, K = 2±= ^ f i i ^ l 2c base Et3 and D 2 = j 2 ( i - 2) = f l e x u r a l r i g i d i t y per unit length of the plate. y* _ ( 3 3 D2 V ' - l " W " 2K + 6 K ) 4cT * " 1 /&L-1»-24K 2 + 5 + 4yK2 (4K2-1) D2 V ' 1 7 '^ '^  ' 4K7(4KZ+1) " 2C A 3' 1 " [ 2 K K(4K2 +1)J 2C if* - r 3 3 fim 1 ) 2 * _ - 3 D 2 V ' 1 ' KT 4C2" * 1 + 4yK2. D2_ A 8' 1 " 2K2(4K^+1) 2C * 16K1* + 16K2 + 1 - 8MK2 D2_ A 9' 1 2K(4K2+1) 2C A 1' 2 L4K^ M 4K2+1 J 2C V , 2 - [|K * 5K . - 2yK] ^ 1 <3,2 • [§• " § K2 + 4 MK 2] ^ v* - r 3 i 22. ' V.' 2 " [4KT " y ] 2C 101 Continued Table (III-15) KA 5 ,2 = [ 4 T + 3 K ] 7K2+T K A 6 > 2 ~ _ [6 + 3 K ] 4K^T K*7 9 - r 2(1-P) i D2 A " L" 2K 7 " 4K*+1 J 2C K*8,2 = 0 K%,2 = l D 2 A v* r 3 j. 1 ... 7v 4K(l-u) , Do V ' 3 = [ 8 K T + 4K + 3 K - -4K2T1 3 2C KA*3»3= [ f^ + | + 6 1 ( 3 4 § n V ' 3 = " 4T • 3 K ] f c K K% o - T 3 4 K(l-y) 1 Dg_ A7,3 - L" 4 K 2 + i J 2C 1 „ „ 9 -l Do V ' 3 = [ 8 K 2 - " 2 K 2 1 4 K 2 ^ T V ' 3 • t w + r - + 4 K 3 i °> K 3 1 , 7 ' " W AC 102 Continued Table (111-15) K* - r L. + 1 + 4yK 2 D2 V ,7 l " K2 2K 5 r C 4 K 2TlJ 2C 2K*(4I0 K A 9 8 = 0 A3 , 7 LK(4K*+1.) J 2C A 1,9 - LK 4K2+ 1 J 2C K* - 3 D 2 . . * V , 9 = 0 * _ r 3 1 + 4yK^ D2 A 5 >7 L" 2K 2(4K 2>1) J 2G * 1 CA 3> 9 = 3K °2 * _ r l • 4 y K 2 D2 'A6 »7 " lK(4K*+l) J 2C A1* »9 " " lK 4K2 + 1 J 2C v * 6 D2 KA 7' 7 " K"3" 4C^ K A 5 I 9 - 0 A 8' 7 " " LK*(4K Z+1) J 2C KA 6> 9 = 3K ° 2 K A9,7= 0 K A 1 , 8 = [ 3 2(1-M) 1 D2 2K? ' 4K*+1 J 2C * KA 7> 9 * KA 8, 9 0 0 < 2 > B 4 ° 2 i f * _ r l . + 1 6 K2 - 12yK2 1  KA 9, 9 ' [ 3K(4K2 + 1) ] ° 2 K 3 8 = 0 A V ' 8 = [2T2" 2(1-y) -i P_2 4KZ+1 J 2C K*5,8 D2 = 0 - r 5 4(l-y) , D2 4K*+1 1 — 1 2C j.* _ 1 * 4yK z  KA8 > 8 " K(4K2+1) ° 2 Table (111-16) Integrated [H A] Matrix for an Isosceles Triangle i n Flexure p y = D2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8ch 0 8ych 0 0 -8ych2 area 0 0 0 0 4(l-p)ch 0 0 -§{l-y)ch2 *][G*I dxdy 0 0 0 8ych 0 8ch 0 0 0 0 0 0 0 0 12c 3h 4yc 3h -8ch 2 0 0 0 0 -§(l-y)ch2 0 4yc 3h | c 3 h +^-(l-y)ch 3 0 0 0 0 -8ych 2| .; 0 -8ch 2| 0 0 24ch 3 104 Table (111-17) Energy Stiffness Matrix for an Isosceles Triangular Element  in Flexure. (Fig. ( I l l - 9 ) ) KA1>1 = [ W " + IK + 3 K ] 4^" K* - r-L_ + L l v* _ r 1 3 1 3K 3 y A 3' 1 ~ 1 32KT + 4K + 2~ + 4 K ] 2C V ' - 1 = [ W " I K " 3KJ % . v* - r 7 y i Do V ' 1' ~ [8K2- ' 2 ] 2C K* - r + 1 + 3 K ^ V * 1 ' L" 32KT + 4K + 2~ • 4K ] 2C v* _ 1 A 8' 1 " K2- 2 C i f * r 1 . 1 y n D 2 A 9' 1 " L8K3 . K " 2K J 2C i f * _ r_JL . y i V ' 2 ~ [8P" + 2 ] 2C i f * _ r5_ . K yK . V ' 2 - " [6K + 3 " 3~ ] ° 2 If* - r 7 y i Do A1*'2 " [8K2" " 2 ] 2C Continued Table (111-17) 105 * K « * SL A?>2 F 2C V '3 = [48T3- + + I2K ] D 2 1 V , 2 = 2K ° 2 V > 2 = 12P" ° 2 * r 13 1 3 „ 3 u n D? V ' 3 = [m? + r-> ° 2 KA3 ' 3 = [192KT + 24K + 4 K + 3K ] °2 Y* - r 3 * v i 22_ V > - 3 = [ 4 W + J ] ° 2 17 V ' 3 = [" W P " + 24K + 4 " 6K ] ° 2 A ' L4K^ K J 2c K * 8 , 3 = ^ D 2 1 1 A 1' 7 K 3 4 C 2 K * _ 2 Do V ' 7 _ " K2 2 C KA3 ' 7 = " [ 4 l T + K ] I c K* _ 3 Dz_ A1*' 7 " " KT 472 106 Continued Table (111-17) * 2 D2 * 1 1 y KA 5' 7 = " KT 2C K A 3 , 9 = [ 4 W + 6K + 12K ] ° 2 * KA 6 , 7 = r 3 + V I 22.. L4K3 K J 2C * V > 9 "[8K3 •1 + -y 1 D2 2K J 2C * V ' 7 6 ' K3 * KA 5, 9 1 12K2 D2 * 2 D2 " W 2C * KA 5, 9 1 + 6K + 12T] D ' KA 9' 7 = 0 * V , 9 = 0 * V ' 8 1 = W 22. 2C * KA 8, 9 = 0 * V > 8 1 2K D2 >• * KA 9, 9 1^2K3 + I -3K " I K ] °2 * KA3,8 1 " 4P" D2 * . V ' 8 1 = w 22_ 2C * V ' 8 2K ° 2 * V ' 8 4KT D 2 V ' 8 2 _D2L IcT 2C KA8,8 = K °2 ! K*9,8 = 0 V ' 9 = L8KT + K " 2K ] 2C * 1 A2 ' 9 = W ° 2 CHAPTER IV TRANSFORMATION OF STIFFNESS MATRICES FROM ELEMENT COORDINATES TO SHELL COORDINATES 4.1 General In the f i n i t e element model of a s h e l l of revolution, the adjacent f l a t elements meet at an angle. The s t i f f n e s s matrices for these elements, derived i n chapters II and I I I , are i n the l o c a l element coordinate direc-tions. Thus, deformations and forces of a l l the elements meeting at a node have different directions. In order to be able to d i r e c t l y superimpose the st i f f n e s s coefficients from the adjoining elements to obtain the force-displacement relations at any node of the assemblage, i t i s necessary to transform the deformations and forces from the element coordinate directions into the common s h e l l directions. For the present analysis, i t i s most suitable to select at any point of the s h e l l surface directions defined by; (see Fig. (IV-1)) (i) P - along tangent to the p a r a l l e l c i r c l e , positive to the right looking from outside of the s h e l l ( i i ) M - along tangent to the meridian, po s i t i v e towards the apex and ( i i i ) R - along the direction of the normal to the surface, p o s i t i v e outwards. 108 Axis of Revoiution (Vertical) . , axis of revolution P — Tangent to the parallel circle in the plane of the parallel, positive to the right-looking from outside. M — Tangent to the meridian in the plane of the meridian , positive towards the apex. R - Normal to the surface, positive outwards. £ - Angle between M-direction and the axis of revolution. 9 - Angle between the planes of the meridian and the reference meridian. FIG. ( EM ) SHELL CO-ORDINATE DIRECTIONS = P, M and R. 10? Nodal displacements of a s h e l l element are defined as; (i) P-displacement, ( i i ) M-displacement, ( i i i ) R-displacement, (iv) P-rotation ( i . e . , rotation about P-axis, positive i f clockwise looking i n the positive direction of the a x i s ) , and (v) M-rotation. No rotation about R-axis need be considered since i t i s already covered by the displacements i n P and M directions. S i m i l a r l y , nodal forces of a shell-element are defined as; (i) P-force, ( i i ) M-force, ( i i i ) R-force, (iv) P-moment ( i . e . , moment about P-axis, considered positive i f clockwise looking i n the positive direction of the a x i s ) , and (v) M-moment. No rotation or moment about the R-axis may be present i n an i n f i n i t e -simal element of the s h e l l i t s e l f and for t h i s reason no R-rotation need be contemplated at any node of the f i n i t e element of the s h e l l model, even though some unbalanced R-moment w i l l remain at the nodes. 4.2 Transformation of Deformations and Forces Let { <5 1}^YZ a n c* ^^XYZ ^ e ^ e t n r e e displacements and three forces i n the XYZ (element) directions at a node i of the element. Also, l e t {(51}p^pv {^Jp^R ^ e t n e corresponding displacements and forces i n the PMR (shell) directions at the same node. The displacements and forces of the two coordinate systems may be related to each other through an orthogonal transformation matrix [T 1] as follows: and s i m i l a r l y , {^}yCiz = P^H^PMR 110 i Here, [T ] = 3 x 3 matrix containing the direction cosines of the element axes XYZ with respect to the s h e l l axes PMR at node i . Writing the abcve relations c o l l e c t i v e l y for an n cornered element, , 1, = . { 6 n }. [T 1] 3x3 [T 2] 3x3 0 XYZ [T n] 3x3 JPMR or, s i m i l a r l y , {«5}XYZ = [T]{6} { f } X Y Z = [T]{f} PMR PMR (IV-1) (IV-2) Here, { i } m " a n d {f} 'XYZ { 5 } P M R and {f} PMR [T] = displacements and forces at a l l nodes of the element i n XYZ (element)directions displacements and forces at a l l nodes of the element i n PMR (shell) directions, transformation matrix containing diagonally the direction cosine matrices of a l l corners The (3x3) direction cosine matrices [ T 1 ] , [ T 2 ] , [T 3] and [T^] belonging to nodes 1, 2, 3,and 4 respectively of a trapezoid element are shown i n Table (IV-1). The various angles used i n t h i s table are defined as follows; (see Fig. (IV-2)) 0 e = angle between the meridian planes enclosing the element FIG.(JSr-2) LOCATION OF A FINITE ELEMENT ON THE SHELL.. Table (IV-1) Trails format ion Submatrices f o r Various Nodes of a Trapezoid [T l ] 11 = [T 2] = [T J] r 9 e Cos^-9e Sin—y Cosy ee -Sin—2 Siny 8 e C o s — ee -Sin—2" Cosy S i n — ^ Siny 3i = Cos-Si n — j Cosy -Sin—y Siny 6e Cos—2 [T*],= ee Sin—2 Siny (Sing Sing Cos a -Cosg S i n a ) (Sing Sine Sina + Cosg Cosa) Sing Cose (Cosg Sin€ Cosa + Sing Sina) (Cosg Sine Sina - SinB Cosa) Cosg Cose for node 1 (Sing Sine Cosa - Cosg Sina) -(Cosg Sine Cosa + Sing Sina) (Sing Sine Sina + Cosg Cosa) Sing Cose (Cosg Sine Sina - Sing Cosa) Cos3 Cose for node 2 -(Singi Sine Cosa + Cosgi Sina) (Cosgi Sine Cosa - Sin8i Sina) -(Singi Sine Sina - Cosgi Cosa) (CosBi Sine Sina + Singi Cosa) -SinBi Cose Cosgi Cose for node 3 (Singi Sine Cosa + Cosgi Sina) -(Cosgi Sine Cosa - Singi Sina) 0 -Sin—2 Cosy -(Sing! Sine Sina - Cosgi Cosa) -Singi Cose (Cosgi Sine Sina + Singi Cosa) Cosgi Cose for node 4 113 FlG.CEZ:-3) RELATING yancU ANGLES TO <£L>/3and 0e ANGLES 114 cf>k = angle between tangent to the meridian at the lower node.% of the element and the axis of revolution <(>^  = angle between tangent to the meridian at the upper nodes of the element and the axis of revolution 8 = angle between the tangent to the meridian at the lower node of the element and the chord l i n e j o i n i n g the lower and the upper nodes of the element $ 1 = angle between the tangent to the meridian at upper node and the chord j o i n i n g the lower and the upper nodes [g^ = <J> -(+ L + &) ] a = angle i n the plane of the element which the i n c l i n e d side makes with the Y-axis of the element Y = angle of i n c l i n a t i o n of the element with the plane of a p a r a l l e l c i r c l e e = h a l f the angle between two equal elements intersecting along a common i n c l i n e d edge (e.g. AB) Angles y and e are related to the other angles as follows: ee Tany = Cot(cJ>L + g)sec -y 9e Tane = Tan -y Cos(<J>L + 8) For derivation of these r e l a t i o n s , refer to Figs. (IV-2 and 3). The apex node of a tri a n g l e must be treated i n a special manner. A new set of coordinate system X, "Y~, % i s defined at the apex, (see Figs. (IV-1 and 2)). X and T axes l i e i n a plane tangent to the s h e l l at the apex and Z. axis i s normal to the s h e l l surface, positive outwards. The Direction of X-axis may be selected a r b i t r a r i l y . The direction of Y-axis i s selected so as to form a righthand system of coordinates. An independent U S transformation submatrix [T 3] i s define! for the apex node of a triangular element and i s given i n Table (IV-2). Here, 0^ = angle i n the horizontal plane (assuming the axis of revolution to be v e r t i c a l ) between the X- d i r e c t i o n at the apex and X-axis of the tri a n g l e element. See Fig. (IV-2). Derivation of the direction cosine matrix [T 1] at node 1 of an equilateral trapezoid element i s explained as below; vectors (6P), (<5M), and (<5R) along P, M and R directions are i n d i v i d u a l l y resolved into components along X, Y and Z direction (see Fig. (IV-4)). (a) Components of (6P) along X, Y and Z directions 6e '' (6.X) = Cos(-|) (6P) e (6Y) = Sin(-|)Cosy (6P) See Fig. (IV-4,a,b,d) 9e (6Z) = - S i n ( - | ) S i n Y (6P) (b) Components of (SM) along X, Y and Z directions (6X) = (SinBSineCosa-CosBSina)(6M) (6Y) = (SinBSineSina+CosgCosa) (6M) R e f e r t 0 F l g * c,a,ej (6Z) = (SinBCose) (<SM) (c) Components of (<5R) along X, Y and Z directions (<5X) = (CosBSineCosa + SinBSinct) (SR) rx-v-i tn oc- c- c- nn w * r ^ Refer to Fig. (IV-4, (6Y) = (CosBSmeSma - SmBCosa) (6R) 6 v. ' c,d,e) (6Z) = (CosBCose)(6R) The force-displacement r e l a t i o n i n the XYZ (element) direction i s given by < f W = WXYZ^XYZ 116 Table (IV-2) Transformation Submatrix for the Apex Node of a Triangular Element. [T A 3] = Cos(e x) Sin(e x) 0 -Sin(6 x)CosY Cos(9 X)Cosy Siny S i n ( 6 j S i n y -Cos(6 v)Siny Cosy A A Table (IV-3) Transformation Matrices for Shell Elements Having Plane-Stress  Properties (a) Transformation Matrix for an Equilateral Trapezoid in Plane Stress (membrane shell) [T] 12x12 [Tl] [T 2] [T3] Each block [T ] i s a 3 x 3 submatrix, i t contains direction cosines at i * * 1 corner. See Table (IV-1) (b) Transformation Matrix for an Isosceles Triangle i n Plane Stress (membrane;she 11) [T] 9x9 [T 1] 0 [T 2] 0 [T 3] A Here [T 1] and [T 2] are the same as for a trapezoid and are given i n Table (IV-1). [T 3] i s a 3 x 3 submatrix containing the direction cosines at the apex; i t i s given i n Table (IV-2) SP)cos(-^-) 117 Z-Axis (SPJsin^sin^ (SP )s in&cos y Y-Axis Plane of element ,<w / 6*" v Plane of the parallel Plane of the Parallel (SP)sm(-^) circle through A,C /IT/ftr* • , ^ C L C i r c l e t n r o u g n lower nodes. / v (8P)sin(-j£) . ^ T v P - A x i s Section through Y - Z p l n n g of the element. R-Axis Axis of Revolution (8M)sin/3 (SM)cosyS (SR) \ jSR)cos/3 (c) Components of (SM) and (SR) along and normal to AB in the plane of  the meridian AB. Z-Axis (8R)cos/3 (SM) sin/3 A '(SM) cos/3 (d) Plane of the element ( SM ) sin/3 sine ^ / (8R)cos/3sine ^ J • &#(8R)sin / 3 \ ^ / ( S P ) c o s ( % ) Plane of the element Normal to AB " *^^ (8M)s in/3s in ^ ( S M ) sin/3 cos € and (SM)cos/3cos e \and(SR)cos/3sine (e) Resolving components of (SM) and  (SR) normal to AB in the plane of the element and in Z direction. FIG.(EZ-4) DERIVATION OF DIRECTION COSINE MATRIX AT NODE I. 118 Here, [^]^y T ^ s t n e s t i f f n e s s matrix of the element i n XYZ (element) coordinates. -Substituting Eqn. (IV-1) for {d} and Eqn. (IV-2) for { f } Y V 7 i n Eqn. (IV-XYZ XYZ 3), one obtains Premultiplying each side of the Eqn. (IV-4) by [T] 1 {f}PMR= m " 1 m X Y Z [ T ] { 6 } P M R dv-5) Since, the matrix [T] transforms the deformations and forces from one set of orthogonal directions to the other, i t i s an orthogonal matrix. Thus, [T]" 1 = [ T ] T (IV-6) Thus, Eqn. (IV-5) may be rewritten as { f } p M R = [ T l ^ ^ x Y Z ^ H ^ P M R t I V - 7 ^ Equation (IV-7) provides the force-deformation r e l a t i o n of the element i n the PMR (shell) coordinate system. The transformed s t i f f n e s s matrix [ K j p ^ of the element i n the s h e l l coordinate system i s given by W PMR = m T [ K ] X Y Z [ T ] (IV-8) 119 It should be noted that rotations and moments at any node ma/ be transformed from XYZ coordinate system to PMR coordinate system by means of the same direction cosine matrix as used for the transformations of d i s -placements and forces. 4.3 Transformation Matrix for a Plane Stress Shell Element When analysing a s h e l l devoid of flexure stresses, the f i n i t e elements used are the ones possessing only plane stress properties. A node of a s h e l l element i s free to displace along P, M and R directions while stresses are induced i n the s h e l l . Thus, the transformation matrix i s (12 x 12) for a trapezoid and (9 x 9) for a triangular element. These matrices are shown i n Table (IV-3). The element s t i f f n e s s matrices i n t h e i r l o c a l coordinates for a trapezoid and a triangle are (8 x 8) and (6x6) respectively. To make these matrices compatible for matrix m u l t i p l i c a t i o n of Eqn. (IV-8), rows and columns containing zero are introduced to correspond to the non-existing Z-direction s t i f f n e s s e s . The re s u l t i n g s t i f f n e s s matrix i n the loc a l XYZ coordinate system i s (12 x 12) for a trapezoid and (9 x 9) for a triangle element. The 3rd, 6th, 9th and 12th rows and colunns containing zero, correspond to zero force due to the Z-direction displacements. 120 •4.4 Transformation Matrix for a Shell Element having Membrane and Flexure  Properties In the analysis of a s h e l l having membrane and flexure stresses, f i n i t e elements possessing both plane stress and flexure properties have to be used. The combined plane-stress and flexure s t i f f n e s s matrices are (20 x 20) for a trapezoid and (15 x 15) for a triangular element. A node of a s h e l l element i s free to displace i n P, M and R directions and rotate about P and M axes while stresses and moments are induced i n the s h e l l . No rotation about R axis need be considered since i t i s covered by the displacements i n P and M directions. The transforma-t i o n matrices are (24 x 24) for a trapezoid and (18 x 18) for a tria n g l e and are given i n Table (IV-4). The element s t i f f n e s s matrices i n t h e i r l o c a l element-coordinates are made compatible for matrix m u l t i p l i c a t i o n of Eqn. (IV-8) by introducing rows and columns containing zero to correspond to the rotation about Z axis. The resu l t i n g s t i f f n e s s matrix i n the lo c a l XYZ coordinate system i s (24 x 24) for a trapezoid and (18 x 18) for a t r i -angle element. The 6th, 12th, 18th and 24th rows and columns containing zero forces due to rotation about Z-axis. The nodes of the f i n i t e elements are not permitted to rotate about the R-axis because no such rotation takes place i n the s h e l l i t s e l f as i t s points f i n d t h e i r deformed positions by undergoing two components of displacements i n the plane of the s h e l l . However, i n the shell-model, prevention of rotation about R-axis results i n some moment unbalance about the same axis brought about by the conversion of X and Y f i n i t e element moments into P, M, R s h e l l moments. The R-moments so produced contain as factors sines of very small angles 8 and e, which cause the R-moments to 121 Table (IV-4) Transformation Matrices for Shell Elements Having Plane Stress  and Flexure Properties (a) Transformation Matrix for an Equilateral Trapezoid i n Plane Stress and Flexure. [T] 2^X2^ [T 1 ] [T 1] [ T 2 ] 0 [ T 2 ] [ T 3 ] o [T3; [T 4] Each [T 1] i s a (3 x 3) sub-matrix of direction cosines at the i t b corner. Entries of [T 1] matrix are shown in Table (IV-1) (b) Transformation Matrix for an Isosceles Triangle in Plane Stress and Flexure [ T ] l 8 X l 8 [Tl] [T 1] [T 2] [T 2] [T|] [T>] Entries of [T 3] are shown i n Table (IV-2) 122 disappear i n the l i m i t . With f i n i t e size elements, the R-moments are l e f t unbalanced because they are purely f i c t i t i o u s from the viewpoint of the s h e l l i t s e l f . With t h i s i n mind, R-rotations are made zero and the unbal-anced R-moments are neglected i n the analysis. CHAPTER V APPLICATION OF THE FINITE ELEMENT METHOD ^ 0 VARIOUS SHELL OF REVOLUTION PROBLEMS 5.1 General In the following study, s u i t a b i l i t y of the formulation i s established by comparing the f i n i t e element solution with the theoretical solution of several problems. Convergence trends are also demonstrated by reducing the element s i z e . The s t a t i c s s t i f f n e s s matrices derived i n chapters II and I I I for trapezoidal and triangular elements i n plane stress and flexure are asymmetric with the exception of a triangular element i n plane stress. Asymmetry of the s t i f f n e s s matrix means v i o l a t i o n of Betti's reciprocal theorem. Although the evaluation of equivalent nodal forces from distributed edge forces i s consistent with the laws of s t a t i c s , yet, the asymmetry of the s t i f f n e s s matrix raises some doubts with regard to the method. The asymmetry of the s t i f f n e s s matrices of a trapezoid element i n plane stress and i n flexure disappears when the element reduces to a rectangle. I t i s even more important that as the states of stress reduce to uniform, the reciprocal theorem i s s a t i s f i e d i n the l i m i t even with the trapezoidal ele-ment. The asymmetric s t i f f n e s s matrices were used and accuracy comparable to that of symmetrical energy s t i f f n e s s matrices was obtained. However, the asymmetry of matrices i s a disadvantage i n the computer work i n view of 124 the need for a double storage capacity. Asymmetric s t i f f n e s s matrices, a r i s i n g from the s t a t i c s approach, have been derived and used successfully 12 13 30 by Antebi et a.\, Gallagher and Shah. In the present study, the asymmetric s t i f f n e s s matrices are made symmetric by averaging the terms of the matrix and i t s transpose. The s t i f f n e s s coefficients belonging i n each column of an asymmetric matrix s a t i s f y the equilibrium requirements. I t i s observed that the terms belonging to each column of the average matrix also s a t i s f y the equilibrium requirements. Thus, the process of averaging corresponds to introduction of additional s e l f e q u i l i b r a t i n g nodal forces making the asymmetric s t i f f n e s s matrix symmetric. A s i m i l a r procedure was independently adopted 12 by Antebi et a l . They observed that the errors introduced by the averaging process are minor and that they are more pronounced i n stresses than i n displacements. 5.2 A General Solution Procedure A computer program for use with the I.B.M. 7044 electronic computer, based on the general matrix algebra method of s t r u c t u r a l analysis, has been developed and i s given i n the Appendix. The basic steps i n the solution procedure are: (i) The s h e l l surface i s subdivided by meridian lines and p a r a l l e l c i r c l e s . The surfaces enclosed between meridian lines and p a r a l l e l c i r c l e s are approximated by f l a t elements connected to each other at the corners only. These elements are i n the shape of equ i l a t e r a l trapezoids with the exception of the ones at the apex which are isosceles t r i a n g l e s . Different shape elements are normally not needed. I f the meridians are spaced at a fixed horizontal angle, 125 then the elements inscribed between any two adjacent p a r a l l e l c i r c l e s are i d e n t i c a l . These elements w i l l have the same s t i f f n e s s and transformation matrices which, thus, need to be generated only once. This i s a big advantage as i t results i n a large saving of computer time and storage space. The transformation matrix for each triangular element i s different and has to be generated for every element independently. ( i i ) The loads acting on the s h e l l are applied to the model i n the same re l a t i v e locations - at the nodes only. These loads are specified i n the P, M,and R directions. In most cases the loads acting on the s h e l l are distributed over certain areas of the s h e l l and so nodal concentrations are simply parts of the distributed load tributary to the nodes and computed, with some exercise of judgement, i n conformity with the requirement of s t a t i c equivalence. An alternative procedure i s to assign the nodal loads consistent with the v i r t u a l work formulation. The d i s t r i b u t i o n of the load applied within each element becomes dependent on the movements of i t s nodes. Both procedures suggested are i n agreement with s t a t i c s . The second one i s unique, the f i r s t one i s not. However, due to complexity of the second procedure, the f i r s t procedure i s adopted i n the present study. ( i i i ) The model i s endowed with the same boundary conditions as the prototype. These are specified i n the s h e l l directions. Boundary conditions for displacements do not present any problem. However, rotation boundary conditions involve some special considerations. The rotation boundary conditions for a s h e l l are 126 prov:ded s i m i l a r to those of a plate. For a plate model, these are 23 as follows; (a) Simply Supported Edge Suppose th i s i s an edge y=c perpendicular to the Y-axis. The two obvious boundary conditions are w=0 and My=0. This means that the edge nodes have no deflection and are free to rotate about the edge. In the plate prototype the edge y=c may develop no torsional 9^ moment Myx. Since M^x = - (1 - y) D2r^^ , t h i s may appear to mean that, 92w _ _9_ 9w 9x9y " 9x <-9yJ On integration with regard to x, th i s r e l a t i o n gives 9w/9y = constant. This conclusion i s obviously incorrect, since the slope of the deflection surface does vary along the simply supported edge. The explanation of the seeming contradiction l i e s i n the fact that the tor s i o n a l moment along the simply supported edge i s transformed into the edge shear 9M V Y / 9 x which i s resisted yx by the additional v e r t i c a l reactions rather than the reactive torques. Thus, although Myx=0, 92w/9x9y ? 0, and i n order to comply with these two, the edge j o i n t s must be allowed to rotate about the Y-axis to bring into balance the moments acting about t h i s axis. (b) B u i l t - i n Edge Assume again t h i s edge to be y=c. Along t h i s edge w=0 and 9W/ay=0. From these two r e l a t i o n s , i t follows that 127 and at the same time 9w/3x=0 along the edge. This means that i n vhe prototype, the edge develops no M and the edge points do yx no^ rotate about the axis Y, perpendicular to the edge. An absence of the edge torque accompanied by an absence of rotation about the Y-axis i s however, impossible i n the model since i f the edge nodes do not rotate about the Y-axis some torsional moments w i l l be present there being brought about by the rotations and displacements of the nodes one ele-ment away from the edge. To eliminate these the edge j o i n t s must be allowed to rotate about the Y-axis, but t h e i r displace-ments must s a t i s f y the r e l a t i o n w=0 and 9w/3y=0. (c) Free Edge Nodes are permitted to deflect and rotate freely about both axes. Simi l a r , force considerations are used to decide the rotation boundary conditons of a corner, of the plate model at a right angle intersection of i t s two edges. From the supplied geometry, the element s t i f f n e s s matrix [^l^yz a n c * the transformation matrix [T] are calculated. The s t i f f n e s s matrix [ K ] p ^ i n the s h e l l coordinates i s obtained from the matrix equation [ K ] p M R = [ T ] T [ K ] X Y Z [ T ] Stiffness matrix [S] for the entire model of the s h e l l i s generated from the transformed element s t i f f n e s s matrices [K]p^ R using the 31 code number technique. This technique automatically eliminates the rows and columns of the structure s t i f f n e s s matrix [S] corres-ponding to the re s t r a i n t directions. 128 Load vector {F} i s generated fro.n the supplied j o i n t loads. 32 Choleski's method of solution (also called the square root method) i s employed to solve the system of simultaneous l i n e a r equations, {F} = [S ] { A } Here, {A} i s a column matrix of unknown nodal deformations of the structure. ( v i i i ) Knowing the structure deformations {A}, the s h e l l element deformation vectors {^Ip^ a r e found by using code numbers. (ix) Following the calculation of nodal displacements and rotations, the s h e l l stresses may be found by one of the two methods; the method of nodal displacements or the method of nodal force concentrations. In the former method, once a l l nodal displacements and rotations of a trapezoid element are known i n the element coordinates, the in-plane and flexure stresses may be found using Eqns. (11-14,b) and (111-10). Similar expressions may be written for a triangular element. The in-plane and flexure stresses, calculated at the nodes, are i n the X-Y-Z (element coordinate) directions and need to be further transformed into PMR (shell) directions. At an i n t e r i o r node where four elements meet, i t i s necessary to compute and average the stress values at the node i n question i n a l l four elements. At the edge node there are only two elements involved and at the corner-only one. On account of the involved conversion of stresses from X-Y-Z directions into PMR directions, the method of determining stresses from nodal displacements i s inconvenient to use when applied to s h e l l s . (vi) ( v i i ) 129 In the present study, the method of nodal force concentra-tion i s used to calculate stresses at various node points. For each element, the nodal forces i n s h e l l d i r e c t i o n s , {f}p^R» a r e calculated from the equation ^f}pMR = ^PMR^PMR The average stresses and moments (Fig. (V-0-1) are found by spreading 23 33 the nodal forces over the appropriate areas of the section. ' The procedure i s i l l u s t r a t e d by reference to Fig. (V-0-2) represent-ing four elements surrounding the common node 1 of the model. In the absence of an external load at the node 1, the internal forces acting on the four elements are mutually balanced. Should an external load be present, the nodal forces w i l l balance i t . The following membrane and flexure forces per unit length may be assumed to exis t i n the s h e l l i n P, M and R-directions at the node 1 i n the absence of the external load. The positive directions of stresses and moments are shown i n Fig. (V-0-1). P-force/length, (N ) 2(N p 4 + N p 2) 2 ( N p 3 + N p i) (a 2 + a 3) ( a 2 + a 3) (N M_ +' NM.) (N + N,. ) M2 Ml _ M4 M3 M-force/length, (N,) <P a l a l Membrane shear force/length e<J>J ( a 2 + a 3) " (a 2 + a 3) ( N p 2 + N p i) ( N p 4 + N p 3) also, (N ) <{>e ai &i 130 FIGURE Yr0-i. POSITIVE DIRECTIONS OF STRESSES AND MOMENTS IN A SHELL ELEMENT FIGURE 3Z-0-2. CALCULATION OF STRESSES AT AN INTERNAL NODE 132 Ng^ and N^^  should t h e o r e t i c a l l y be equal and since t h e i r calculated values nearly always disagree somewhat, they should be averaged. P-moment/length, (M ) (M p 4 + M p 3) (M p 2 + M p i) <f> ai aj 2 ( MM4 + V - 2 ( M M3 + V M-moment/length, (M0) = - ^ + ^ = Transeverse shear/length, CQJ = 2(NR4 + NR2) _2(NR3 + N R 1) ( a 2 + a 3) ( a 2 + a 3) Transeverse shear/length, CQJ = <NR2 + NR1> ' <NR4 + NR3> ai a! Twisting moment/length, R M - " 2(MP4 + MP2) 2<MP3 + V © 4> ( a 2 + 33) ( a 2 + a 3) „ , (MM2 + »W &m  + »W and, ( M + e) - = Twisting moments M^ ^ and M^g should be numerically equal and opposite i n sign. I f they are not, the two values should be averaged. I f concentrated forces P p, P^ and P R are present at the node 1, they should be properly apportioned between the four elements into the parts P p i, P M 1, P R 1; P p 2, P M 2, P R 2; P p 3, P M 3, P R 3; and Pp^, PM4, PR^. The nodal forces s a t i s f y the r e l a t i o n s , ( N p i " PPP + (NP2 - PP2> + <NP3 " PP3> + CNP4 - PP4> = 0 <NM1 - PMJL> + <NM2 " <W + <NM3 " PM3> + ^NM4 - PM4> = 0 <NR1 - PRP + <NR2 " PR2> + <NR3 " PRP + (NR4 " PR4> = 0 132,a Thus, the stresses are obtained from the previous equations by simply replacing N p l, N p 2, ... by ( N p i - P p l) , ( N p 2 - P p 2 ) , ... respectively. Additional explanation i s needed with regard to the stresses at the edges of the s h e l l . Some decisions regarding these stresses must be based on judgement p a r t l y i n view of the d i f f i c u l t y of deciding to which of the two planes some of the nodal forces should be attributed, (a) Free Edge Nodal forces acting on elements having a common edge node are shown i n Fig. (V-0-3). I f the d i s t r i b u t i o n of P-stress (N ) i s free edge nonuniform on the meridian 1-2 (Fig. (V-0-4)) then i t s calculation on the basis of the edge nodal concentration alone results i n high error. In such a case, the following modification i s suggested for calculation of P-stress at the edge: P-force/length (N ) at the i n t e r i o r node 2 i s f i r s t 02 calculated. Assuming the d i s t r i b u t i o n of P-stress to be li n e a r on the meridian 1-2, the equivalent P-force at node 1 due to P-stress alone (Fig. (V-0-4)) i s 4 - r- (2Nei + V Thus, P-force/length at node 1 (Ng^).may be expressed as N = 1 6 N P 6 1 2 C-iJ " N e 2) 133 > FIGURE 1-0-3. CALCULATION OF STRESSES AT AN EDGE NODE FIGURE(3T-0-4) DISTRIBUTION OF P-STRESS (N« ) ON MERIDIAN 1-2 134 Absence of N J n on the free edge results i n 9 8 Thus, and N p = N p l 1 6 NP1 61 2 L a 2 62 j 2 NR2 2 NR1 c0 a 2 M 2MM2 2 M a 2 Ml 6 a 2 a 2 Stresses N j 5 N^. M, , M,_ and Q, may be assumed zero. 9 9 8 9 96 9 (b) Fixed Edge - Supported i n P, M and R Directions (Figs. (V-0-3 and 4)) The forces/length are; 1 6 N P N e i = 2 - ^ - V Here, the equivalent force at node 1 due to P-stresses alone i s ( N P 1 + N ) N 1 = N + • N a i = N - — — - = — or, Nj = ( N p 2 - N p i)/2 Thus, from above, M 1 r 3 ( N P 2 - N P P % " 2 t a 2 - N 9 2 ^ N 9 = " (NM1 + N M 2 ^ / a l N , 0 = - ( N P 1 + N P 2 ^ a l 135 In view of the edge node be.vng allowed to rotate about M-axis, M6 = 2 MM2 / a 2 = - 2 M M l / a 2 = t M P l + M P 2 : ) / a l = M^ g = 0 (from considerations of plate theory) % = 2Kl + Q,Cai/2)}/a 2 = ( N R 2 - N R 1 ) / a 2 and QA = - ( N R 1 + N R 2)/a (c) Simply Supported Edge - Displacements Restricted i n P, M and  R Directions (Figs. (V-0-3 and 4)) Here, N Q 1 ,N^, N^, QQ and are the same as calculated for a fixed edge, (case b) In view of the edge nodes being allowed to rotate about both axes, M p i + M p 2 = 0 and + = 0. i s zero and MQ may be assumed zero from general considerations of plate theory. M p 2 may be assumed to correspond to the torsion moment i n 2M, 0-plane, M P2 09 a 2 The torsion moment M,„ has been transformed 90 into transeverse edge shear Q^ -(d) Edge with Displacements Restricted i n the P a r a l l e l and V e r t i c a l Directions. (Fig. (V-0-3)) V e r t i c a l component of the reaction at node 1 i s resolved i n M and R direc-tions; M = V Cos 9 ; R = V Sin 9 . 136 The components of nodal forces i n P, M and R directions are shown i n Fig. (V-0-3). Here, calculations for N Q 1, N,, N,n, o i <p 9 8 Q and Q are the same as for a fixed edge (case b). In view of the edge nodes being allowed to rotate about both P and M axes, M p i + Mp2. = 0 and + = 0. Thus, i s zero and MQ = 2 ^ / 3 2 = - 2 ^ / 3 2 -Mp2 may be assumed to correspond to the torsion moment i n 9 plane, M = 2MDO/a2> Th e torsion moment M,0 has been transformed 0 Cp r /. Cp D into transeverse edge shear Q,. Calculation of stresses by the above suggested formulae i s not exact but for small sized elements i t provides a good approximation. Large errors are expected i n areas where the stress gradients change suddenly. These errors are expected to reduce on reduction of the element s i z e . (xi) Percentage errors i n displacements and stresses are calculated using the following r e l a t i o n based on absolute values of displace-ments and stresses o _ ,1 calculated value - e l a s t i c i t y valueL n n^ % Error = (-1 .—-—' . . ^  -.—^ ')(100) [. e l a s t i c i t y value | v J Several s h e l l of revolution problems for which e l a s t i c i t y solutions are known, are solved to demonstrate the s u i t a b i l i t y of the f i n i t e element method employing both s t a t i c s and energy s t i f f n e s s matrices of the trapezoid and isosceles t r i a n g l e elements. Several values of Poisson's r a t i o . (y = 1/3 for c i r c u l a r plate problem and y = 0.0 and 0.2 for spherical s h e l l problems) are used. Convergence of the calculated results to the true solution i s demonstrated by reducing the element s i z e . 137 f R \ > _____^^^e=3o° e = 90° = 60' FIGURE 3T-I-I. CIRCULAR PLATE SUBJECTED TO THE ACTION OF A PAIR OF DIAMETRICALLY OPPOSITE POINT LOADS 138 5• 3 Example-I Ci r c u l a r Plate Subjected to the Action of a Pair of Diametri- c a l l y Opposite Point Loads. Fig. (V-I-l) A c i r c u l a r plate may be considered as a s h e l l of revolution having zero curvatures. Here, the generating curve i s a straight l i n e , perpendicular to the axis of rotation. The f i n i t e element method using equi l a t e r a l trapezoids and isosceles triangles i s well suited for the analysis. 34 The result of e l a s t i c i t y solution . for displacements (u = 1/3) and stresses (independent of y) at various sections i s shown i n Figs. (V-I-2 to 6).^ The e l a s t i c i t y solution i s compared with the f i n i t e element solution using Sta t i c s and Energy s t i f f n e s s matrices. Due to the symmetry of the problem about two axes, only one quarter of the plate needs to be .analysed. The application of point load produces very steep displacement and stress gradients requiring use of small elements to give a reasonable representation of displacement and stress d i s t r i b u t i o n i n the v i c i n i t y of the load. The convergence trends are obtained by reducing the element size which i s defined by 0 e and AR. Here, 6 g i s the angle between two adjacent meridian lines and AR i s the difference of r a d i i of the two p a r a l l e l c i r c l e s enclosing the element. The percentage errors are calculated at various points and are shown i n Tables ( V - I - l to 5). Greater error i n calculation of stresses than i n displacements i s noted. Calculation of stress by evenly d i s t r i b u t i n g the nodal force on the appropriate cross section i s reasonable where stress-gradients are small. At points where the absolute value of the function i s small com-pared to i t s maximum value elsewhere, higher percentage error i s noted. Such points may be disregarded as they are of l i t t l e s ignificance. 139 Stresses produced i n the plate by a concentrated force are i n f i n i t e at the point of application of the force according to the theory of e l a s t i c i t y . The f i n i t e element solution results i n f i n i t e value of stresses. Thus, i t i s not r e a l i s t i c to expect the f i n i t e element results to agree with the e l a s t i c i t y at the point of application of the concentrated force. Also, due to the presence of high displacement and stress gradients i n the v i c i n i t y of the concentrated force, large errors i n the f i n i t e element solution near the load are inevitable. Reduction of the size of elements improves the precision of the results i n the neighbourhood of the load. Coarser mesh may be used away from the load with no i l l effects on precision. The results presented i n Tables (V-I- l to 5) reveal that for 6 g = 7.5°, AR = 0.0625 R element size and y= 1/3, the f i n i t e element solution d i f f e r s from the e l a s t i c i t y solution i n displacements by 2% and i n stresses by 4%. The error analysis c l e a r l y shows the rapid convergence of displace-34 meats and stresses to t h e i r e l a s t i c i t y values on reduction of the element size . 141 Table (V-I-l) Circular Plate - Two Point Loads At ijr/R 30' 45 60 .25 .50 ' .75 1.00 .25 .50 ' .75 1.0.0 .25 .50 ' .75 1.0 P-Displacement = Poisson's Ratio E l a s t i c i t y Solution c i .1813 .3332 .3480 .1305 .2007 .3250 .3010 ..1611 .1671 .2476 .2203 .1362 O! (P/Et) Fig. (V-I-2) 1/3 Percentage Error -Ge=15° AR=.125R 0.66 2.49 6.23 21.23 1.44 2.37 • 5.12 -10.55 1.91 1.33 - 1.18 - 1.17 Statics Matrix 6e=10° AR=.0833R 0.33 1.14 1.06 -10.50 0.84 0.52 -1.04 -1.32 6e=7.5° AR=0.0625R 0.22 0.66 0.29 -0.08 0.40 0.52 -0.40 -0.99 6.47 0.28 -0.50 -0.81 6e=15° AR=.125R - 0.72-0.93 4.71 13.10 0.05 1.02 -5.32 - 8.26 0.48 0.12 - 2.09 - 2.28 Energy Matrix 6e=10° AR=.0833R -0.28 0.45 0.60 -8.05 0.24 -0.04 -1.36 -1.54 6e=7.5° AR=.0625R -0.11 0.30 0.03 -0.23 0.05 0.22 -0.60 -0.99 0.18 0.00 -0.68 -0.88 143 Table (V-I-2) Circular Plate - Two Point Loads M-Displacement = Poisson's Ratio At t r/R .25 .50 0° .75 1.00 .25 .50 30° .75 1.00 .25 .50 45° .75 1.00 .25 .50 60° .75 1.00 .25 .50 90° .75 1.00 Elasticity-Solution a 2 .2721 .5933 1.0797 . 1533 .2635 .2791 .2561. .0445 .0463 .0123 .0042 .0558 .1169 .1608 .1731 .1488 .2489 .2935 .3033 6e=15° AR=. 125R -0.44 1.15 2.94 1.37 2.20 -5.34 -9.88 5.17 -2.81 -98.50 93.0 -1.79 2.91 1.80 3.58 0.47 0.68 0.07 0.33 a2'(P/Et) Fig. (V-I-3) 1/3 Percentage Error Statics Matrix 6e=10° • AR=.0833R -0.18 0.49 1.69 0.59 0.91 -2. 72 -2.22 -0.72 1.20 1.18 1.16 0.13 0.32 0.07 0.06 ee=7.5° AR=.0625R 0.07 0.27 0.93 0.33 0.49 - 1.47 - 2.05 1.12 -0.86 -18.70 59.0 -0.36 0.68 0.62 0.58 0.p7 0.16 0.03 0.03 9e=15° AR=.125R 1.58 0.79 2.18 0.52 1.86 - 5.37-- 5.27 5.62 - 1.73 -84.5 • 93.0 -4.84 1.20 - O.44 - 0.12 - 1.68 -0.72 - 1.57 - 1.68 Energy Matrix 9e=10° AR=.0833R -0.70 0.30 1.31 0.20 0.80 -2.72 -2.62 -2.15 0.43 0.19 -0.17 -0.87 -0.32 -0.68 -0.79 9e=7.5° AR=.0625R - 0.37 0.19 0.72 0.13 0.46 -1.47 - 2.03 1.35 -0.65 -14.63 33.4 - 1.08 0.26 0.12 0.12 - 0.34 -0.16 -0.34 - 0.43 145 Table (V-I-3) Circular Plate - Two Point Loads P-Force/Length, (N ) = a3(P/R) - 6 At j r/R .25 0° .50 .75 .25 30° .50 .75 45' .25 .50 .75 .25 60° .50 .75 90e .25 .50 .75 E l a s t i c i t y Solution °<3 .3183 .3183 .3183 -.0829 -.3447 -.5174. -.3899 -.4923 -.3461 -.6264 -.5132 -.2565 -.8095 -.4966 -.2032 A6=15° AR=1.25R 1.51 8.58 26.86 - 3.02 4.12 22.24 0.62 . 4.87 -23.98 0.59 0.74. - 2.61 - 0.02 -1.95 -5.46 Fig. (V-I-4) Percentage Error Statics Matrix A6=10° AR=.0833R 1.22 3.23 19.76 2.05 2.41 4.93 0.45 0.15 4.01 0.22 0.86 2.66 A0 = AR= 7.5° . .0625R 0.69 1.70 16.18 • 1.21 1.45 1.12 0.00 1.06 • 1.73 0.25 0.08 • 1.87 0.13 •0.48 • 1.52 A9=15° AR=.125R 0.16 7.79 28.68 - 4.22 2.23 19.87 - 0.62 3.60 -21.76 -0.70 - 0.23 - 3.08 1.46 2.90 5.12 Energy Matrix Ae=10° AR=.0833R 1.04 2.80 20.58 3.26 1.59 4.56 0.08 0.23 3.82 - 0.16 - !1.27 - 2.51 A6=7.5° AR=.0625R 0.60 1.45 16.24 • 1.81 0.99 0.93 0.00 0.77 -1.76 0.05 • 0.16 •1.87 • 0.07 • 0.72 • 1.48 147 Table (V-I-4) Circular Plate - Two Point Loads M-Force/Length, (N ) = c^CP/R) Fig. (V-I-5) Percentage Error Statics Matrix Energy Matrix At E l a s t i c i t y A6=15° A6= 10° A9=7.50 A9=15° A6=10° A9=7.5° e.5 r/R Solution AR=.125R AR= .0833R AR=.0625R AR=. L25R AR=.0833R AR=.062. .25 -1 0398 - 2.86 - 1 .43 - 0.85 - 3 .05 - 1 .62 - 0.97 0° .50 -1 3793 - 6.02 - 3 .21 - 1.95 - 5 .94 - 3 .23 - 1.97 .75 -2 5920 -19.79 -11 .60 - 7.85 -20 .08 -11 .89 - 8.07 .25 _ 5907 4.28 2 .10 1.22 4 .72 2 .34 1.34 30° .50 - .3899 14.21 6 .33 3.46 15 .72 7 .10 3.87 .75 - .0599 35.56 47 .91 26.04 18 .53 42 .40 23.87 .25 _ 2418 11.37 2.85 12 .90 3.31 45° .50 - .0684 8.50 - 0.14 13 .83 1.01 .75 0155 13.55 14.19 .25 0317 -12.93 - 3 .78 - 1.89 -16 .40 - 5 .99 - 2.84 60° .50 0585 22.39 8 .89 4.79 22 .56 9 .23 4.96 .75 0249 -28.51 6 .42 1.61 -20 .08 8 .03 2.41 .25 2478 10.77 4 .80 2.70 ' 12 63 5 57 3.11 90° ,5p .1146 6.54 3 .23 1.74 9 .24 4 .36 2.44 .75 0250 -4.80 - 0 . 40 . - 0.40 1 .20 : 2 .00 1.20 149 Table (V-I-5) Circular Plate - Two Point Loads Shear Force/Length, (N ) :—: 8(p a 5(P/R) Fig. (V-I-6) Percentage Error Statics Matrix At E l a s t i c i t y A9=15° A9= = 10° A9 = 7.5° A0= 15° A6=10° A9 = 7.5° r/R Solution AR=. 125R AR= =.0833R AR= =.0625R AR=.125R AR=.0833R AR= =.0625R «5 .25 .5809 - 1 .98 0.91 _ 0.50 - 5.15 - 2.39 _ 1.31 30° .50 .5872 3 .03 1.62 0.90 0.29 0.48 0.34 .75 .2901 23 .34 0.72 - 1.55 23.72 1.58 - 1.21 .25 .5899 - 0 .64 _ 0.19 - 3.64 _ 0.92 45° .50 .3965 3 .35 0.60 1.56 0.20 .75 .1099 ^50 .86 - 2.91 -48.13' - 2.27 .25 .4527 - 0 .04 _ 0.22 0.13 - 2.83 - 1.48 0.82 60° .50 .2250 -. 1 .07 - 0.71 - 0.40 - 2.18 - 1.15 - 0.62 .75 .0467 8 .35 8.99 - 3.43 12.63 - 8.35 - 2.78 Energy Matrix 150 FIGURE ' E - H - l . HEMISPHERICAL DOME SUBJECTED TO SNOW LOAD 151 5.4 Example II - Hemispherical Dome Under Snow Load Uniformly Distributed over the Horizontal Projected Area. Fig. (V-II-1) In t h i s problem, the dome i s loaded and supported axisymmetrically. At the base, the dome i s free to deflect i n the r a d i a l directions. The 35 36 e l a s t i c i t y solution based on the membrane theory ' i s obtained for d i s -placements (y = 0.0 and 0.2) and stresses (independent of y) and i s shown graphically i n Figs. (V-II-3 to 6). In view of the axisymmetry of the boundary and loading conditions of the s h e l l , i t i s s u f f i c i e n t i n the f i n i t e element analysis to consider elements bounded by a p a i r of adjacent meridian l i n e s . The snow load on the horizontal projected area of a s h e l l element i s apportioned to the corners according to the laws of s t a t i c s , Fig. (V-II-2). The v e r t i c a l load at each node i s f i n a l l y resolved into M and R directions. Percentage errors i n the F.E. solution employing Statics and Energy matrices and using two values of Poisson's r a t i o (y = 0.0 and 0.2) . are shown i n Tables (V-II-1 to 6). For 0 g ='<(> = 2.5° element s i z e , excellent accuracy i n the calculation of displacements and stresses (error less than a f r a c t i o n of 1%) i s obtained. Higher percentage errors i n displacements (3% from Statics matrix and 8% from Energy matrix) and stresses (5% from Stati c s matrix and 6% from Energy matrix) are noted at the nodes of the triangle element. These errors are attributed to the behaviour of the F.E. model as a cone rather than as a smooth spherical surface at the apex. Rapid convergence of the F.E. results to the e l a s t i c i t y solution i s cl e a r l y seen on reducing the element siz e . 152 Total vertical load on the element, V =-g-0e IRi^-R^ ) p For small angle 8e , distance of C.6. of the load from 0 , R . 2 f R L + R L R U + RLV1 c " 3 1 • * L ^ U j From Statics- 2 V y + 2V,_ = V and 2 V U - R U + 2 V L - R L = V R c Here, and Vy are nodal components of the vertical load on the element Solving for and Vy - V( R L -R c ) /2 (R L-Ru) and V L = V ( R C - R U ) / 2 ( R L - R U ) FIG. ( l T - n - 2 ) HEMISPHERICAL DOME SUBJECTED TO SNOW LOAD. CALCULATION OF NODAL LOADS. 154 Table (V-II-1) Hemispherical Dome - Snow Load M-Displacement = a!(pR2/Et) Fig. (V-II-3) Poisson.'s Ratio = 0.0 Percentage Error E l a s t i c i t y ee= <t>e- 5° <J>e= 2.5° j * Solution Statics Energy Statics Energy ' Matrix Mat r i x Mat r i x Mat r i x 0.0 0 00000 0 00 0 00 0 00 0.00 2.5 -0 04358 -0 07 -0.04 5.0 -0 08682 -0 24 -0 16 -0 06 -0.04 7.5 -0 12941 -0 06 -0.04 10.0 -0 17101 -0 24 -0 16 -0 06 -0.04 12.5 -0 21131 -0 06 -0.04 15.0 -0 25000 -0 25 -0 17 -0 06 -0.04 17.5 -0 28679 -0 06 -0.04 20.0 -0 32139 -0 26 -0 18 -0 06 -0.05 22.5 -0 35355 -0 06 -0.05 25.0 -0 38302 -0 27 -0 19 -0 07 -0.05 27.5 -d 40958 , -0 07 -0.05 30.0 -d 43301 -0 28 -0 20 -0 0.7 • -0.05 32.5 -d 45315 -d 07 -0.p5 35.0 -d 46985 -0 30 -d 22 -d 08 -o.bs 37.5 -0 48296 -0 08 -0.06 40.0 -0 492-40 -0 32 -0 23 -0 08 -0.06 42.5 -0 49810 -0 08 -0.06 45.0 -0 50000 -0 33 -0 25 -0 08 -0.06 47.5 -0 49810 > -0 09 -0.07 50.0 -0.49240 -o. 35 -P 28 -Q 09 -0.07 52.5 -0 48296 -P -0 09 -0.07 55.0 -0 46985 -0 37 3d -0 d9 -0.07 57.5 -0 45315 -0 09 -0.08 60.0 -d 43301 -0 38 -o 32 -0 d9 -0.08 62.5 -0 40958 -0 10 -0.08 65.0 -0 38302 -0 38 -0 34 -0 09 -0.08 67.5 -0 35355 -0 09 -0.08 70.0 -0, 32139 -0 36 -d 33 -0 09 -0.08 72.5 -0 28679 -0 08 -O.07 75.0 -Q 25000 -0 26 -d 24 -0 06 -0.06 77.5 -0 21131 -0 02 -d.03 80.0 -d 17101 +0 22 -d d4 +0 05 +0.04 82.5 -d 12941 0.22 0.22 • 85.0 -d 08682 +2 0d + 3.0d 0 79 0.63 87.5 -d d4358 1 82 2.89 90.0 d ddddd 0 dd d do o.dd .o .dd 155 Table (V-II-2) Hemispherical Dome - Snow Load, M-Displacement = c4(pR 2/Et) Fig. (V-II-3) Poisson's Ratio = 0.2 Percentage Error E l a s t i c i t y 9e = 5° ee= 4>e= 2.5° * Solution Statics Energy Statics Energy Matrix Matrix . Matrix Matrix 0 .0 0 00000 0 00 0 00 . 0 .00 ,0.00 2 .5 -0 05229 -0 .06 -0.04 5 .0 -0 10419 -0 24 -0 17 -0 .06 -0.04 7 .5 -0 15529 -0 .06 -0.04 10 .0 -0 20521 -0 25 -0 17 -0 .06 -0.04 12 .5 -0 25357 -0 06 -0.04 15 .0 -0 30000 -0 25 -0 18 -0 06 -0.05 17 .5 -0 34415 -0 07 -0.05 20 .0 -0 38567 -0 26 -0 19 -0 06 -0.05 22 .5 -0 42426 -0 06 -d.05 25 .0 -0 45963 -0 27 -0 20 -0 07 -d.05 27 .5 -0 49149 -0 07 -0.05 30 .0 -P 51962 -0 28 -0 21 -0 07 rd.05 32 .5 -0 54379 -0 07 -p. 05 35 0 -P 56382 -0 29 -0 22 -0 07 t-d.06 37 .5 -0 57956 -0 08 -d.06 40 .0 , -d 59089 -0 31 -0 24 -0 08 -d.06 42 .5 i 59772 -0 08 -0.06 45 0 -d 60000 -0 32 -0. 26 -Q 08 -0.07 47 5 -0 59772 -0 08 -d.07 50 0 -0 59089 -0 34 -0. 29 -0 09 -0-07 52 5 -0 57956 -0 09 -0.08 55 0 -0 56382 -0 35 -0. 31 -0 09, -0.08 57 5 -0 54379 -0 09 -0.08 60 0 -0 51962 -0 36 -0. 33 -0 09 -0.08 62 5 -b 49149 -0 09 -0.09 65 0 , -0 45963 -0 36 -0. 35 -0 09 -0.09 67 5 -0 42426 -0 09 -0.09 70 0 -d 38567 -0 33 -P. 35 -0 08 -0.09 72 5 -0 34415 -d 07 -0.08 '• 75 0 -0 30000 -0 22 -0. 25 -d 05 -0.06 77 5 -0 25357 -d 01 -0.03 80 0 -0 20521 +0 27 +d. 01 +d d7 +d.d3 82 5 -0 15529 0 23 0.22 85 0 -0 10419 2 12 3. 14 d 81 d.61 87 5 -0 05229 l 91 3.05 90 0 0 00000 0 00 d. 00 d dd d.dd 156 Table (V-II-3) Hemispherical Dome - Snow Load R-Displacement = a 2(pR 2/Et) Fig. (V-II-4) ! Poisson's Ratio =0.0 Percentage Error * E l a s t i c i t y Solution a 2 9e= $e= Statics Matrix 5° Energy Matrix e e = <f>e= 2 Statics Matrix 5° Energy Matrix 0,0 0.50000 0 09 0 25 0 02 0.06 2.5 0.49620 0 02 0.06 5.0 0.48481 0 i o ; 0 26 0 02 0.07 7.5 0.46593 0 02 0.07 10.0 0.43969 0 n 0 27 0 03 0.07 12.5 0.40631 0 03 0.07 15.0 0.36603 0 13 ' 0 29 0 .03 0.07 17.5 0.31915 0 04 0.08 20.0 0.26605 0 18 0 35 0 04 0.09 22.5 0.20711 0 05 0.10 25.0 0.14279 0 34 0 53 0 08 0.13 27.5 0.07358 0 16 0.23 30.0 0.00000 -- -- -- --32.5 -0.07738 -0 17 -0.16 35.0 -0.15798 -0 36 -0 26 -0 09 -0.07 37.5 -0.24118 -0 06 -0.03 40.0 -0.32635 -0 20 -0 08 -0 05 -0.02 42.5 -0.41284 -0 04 -0.01 45.0 -0.50000 -0 16 -0 03 -0 04 -0.01 47.5 -0.58716 -0 04 : -0.01' 50.0 -0.67365 -0 14 -0 02 -0 04 -0.01 52.5 -0.75882 -0 04 -0.00 55.0 -0.84202 -0 15 -0 02 -0 04 -0.01 57.5 -0.92262 -0 .04 -0.01 60.0 -1.00000 -0 16 -0 04 -0 04 -0.01 62.5 -1.07358 -0 04 -0.01 65.0 -1.14279 -0 18 -0 07 -0 05 -0.02 67.5 -1.20711 -b 05 -0.02 70.0 -1.26604 -0 23 -0 09 -0 05 -0.03 72.5 -1.31915 -0 06 -0.04 75.0 -1.36603 -0 19 -0 25 -0 07 -0.04 77.5 -1.40631 -0 07 -0.06 80.0 -1.43969 -0 54 +0 08 -0 09 -0.05 82.5 -1.46593 -0 04 -0.16 85.0 -1.48481 , +o 87 -1 53 -0 30 +0.25 87.5 -1.49620 +1 .11 ' -1.29 90.0 -1.50000 +1 68 +5 37 +1 67 +5.38 157 Table (V-II-4) Hemispherical Dome - Snow Load R-Displacement = a 2(pR 2/Et) Fig. (V-II-4) Poisson's Ratio =0.2 Percentage Error E l a s t i c i t y *e= 5° *e= 2.5° Solution Statics Energy Statics Energy a 2 Matrix Matrix Mat r i x Matri x 0 .0 0 60000 0 04 0 17 . 0 .01 0.04 2 .5 0 .59581 0 .01 0.04 5 .0 0 58329 0 04 0 .17 0 .01 0.04 . 7 .5 0 56252 0 .01 0.04 10 .0 0 53366 0 05 0 18 0 .01 0.04 12 .5 0 49694 0 .01 0.04 15 .0 0 45263 0 06 0 19 0 .02 0.05 17 .5 0 40107 0 .02 0.05 20 .0 0 34265 0 09 0 21 0 .02 0.05 22 .5 0 27782 0 .03 0.06 25 .0 0 20707 0 17 0 28 0 .04 0.07 27 .5 0 13093 0 08 0.10 30 .0 0 05000 +0 .84 +0 89 0 .20 0.22 32 .5 -0 03512 -0 34 -0.27 35 .0 -0 12378 -0 42 -0 26 -0 11 -0.06 37 .5 -0 21530 -0 07 -0.03 40 .0 -0 30899 -0 21 -0 08 -0 06 -0.02 42 .5 -0 40413 -0 05 -0.01 45 .0 -0 50000 -0 17 -0 04 -0 04 -0.01 47 .5 -0 59587 -0 04 -0.01 50 .0 -0 69101 -0 16 -0 04 -0 04 -0.01 52 .5 -0 78470 -0 04 -0.01 55 .0 -0 87622 -0 16 -0 05 -0 04 -0.01 57 .5 -0 96488 -0 04 -0.02 60 .0 -1 05000 -0 18 -0 07 -0 05 -0.02 62 .5 -1 13093 -0.05 -0.02 65 0 -1 20707 -0 20 -0 12 -0 05 -0.03 67 5 -1 27782 -0 05 -0.04 70 .0 -1 34265 -0 25 -0 14 -0 06 -0.04; 72 .5 -1 40107 -0 06 -0.05 75 .0 -1 45263 -0 22 -0 33 -0 07 -0.06 77 5 -1 49694 -0 08 -0.08. 80 0 -1 53366 -0 54 +0 04 -0 10 -0.07 82 5 -1 56252 -0 05 -0.19 85 .0 -1 58329 +0 74 -1 67 -0 29 +0.26 87 .5 -1 59581 + 1.00 -1.35 90 0 -1.60000 +2 82 + 7 96 2 80 + 7.90 159 Table (V-II-5) Hemispherical Dome.- Snow Load P-Force/.Length, .(N .) = a 3(pR/2) Fig. (V-II-5) Percentage Error t-i = 0.0 1-j = 0.2 E l a s t i c i t y 6e= 9e= 5° ee= 9 e= 2 5° ee= 9 = 5° e 3 ee= 9 e= 2 . 5 ° Solution Statics Energy Statics Energy Statics Energy Statics Energy a 3 Mat r i x Matrix Mat r i x Mat r i x Matrix Matrix Mat r i x Matrix 0.0 1.0000 -0 .48 -0 .48 -0 12 -0 12 . -0 .48 -o 48 -0.12 -0.12 2.5 0.9962 -0 12 -0 12 -0.12 -0.12 5.0 0.9848 -0 .48 -0 .48 -0 12 -0 12 -0 .47 -0 48 -0.12 -0.12 7.5 0.9659 -0 12 -0 12 -0.12 -0.12 10.0 0.9397 -0 .47 -0 .47 -0. 12 -0 12 -0 47 -0 47 -0.12 -0.12 12.5 0.9063 -0 12 -0 12 -0.12 -o.i2 : 15.0 0.8660 -0 .46 -0 .46 -0 12 -0 12 -0 .46 -0 46 -0.12 -0.12 17.5 0.8192 -0 11 -0 11 -0.11 -0.11 20.0 0.7660 . -0 .45 -o .45 -0. 11 -0 11 -0 45 -0 45 -0.11 -0.11 22.5 0.7071 -0 11 -o 11 -0.11 -0.11 25.0 0.6428 -0 .42 -0 .42 -0 11 -0 11 -0 .42 -0 42 -0.11 -0.11 27.5 0.5736 -0 10 -0 10 -0.10 ; -0.10 30.0 0.5000 -0 .37 -0 :37 -0. 09 -0 09 -0 .37 -0 37 -0.10 -0.10 32.5 0.4226 -0 09 -0 09 -0.09 -0.09 35.0 0.3420 -0 .27 -0 .27 -0 07 -0 07 -0 .27 -0 27 -0.07 -0.07 37.5 0.2588 -0. 04 -0 04 i -0.04 -0.04 40.0 0.1736 +0 .07 +0 .07 . +0. 02 +0 02 +0 .07 +0 07 +0.02 +0.02 42.5 0.0872 +0. 18 +0 18 +0.18 +0.18 45.0 0.0000 -- • - -- -- --47.5 -0.0872 -0. 53 -0 53 -0.53 -0.53 50.0 -0.1736 -1 •41 . -1 .41 -0. 36 -0 36 -1 .41 -1. 41 -0.35 -0.35 52.5 -0.2588 -0 30 -0 30 -0.30 -0.30 55.0 -0.3420 -1 .09 -1 .09 -0. 27 -0 27 -1 .09 -1 09 -0.27 -0.27 57.5 -0.4226 -0. 27 -0 27 -0.27 -0.27 60.0 -0.5000 -1 .04 -1 .04 -0. 26 -b 26 -1 04 -1. 04 -0.26 -0.26 62.5 -0.5736 -0. 27 -0 27 -0.27 . -0.27 65.0 -0.6428 -1 .08 -1 .08 -0. 27 -0 27 -1 08 -1. 08 -0.27 !-0.27 67.5 -0.7071 -0. 29 -0 29 -0.29 -0.29 70.0 -0.7660 -1 .24 -1 .24 -0. 31 -0 31 -1 24 -1. 24 -0.31 -0.31 72.5 -0.8192 -0. 35 -0 35 -0.35 -0.35 • 75.0 -0.8660 -1.64 -1 .64 -0. 41 -0 41 -1 64 -1. 64 -0.41 -0.41 77.5 -0.9063 -0. 51 -0 51 -0.51 -0.51 80.0 -0.9397 -2 .87 -2 .87 -0. 70 -0 70 -2 87 -2. 87 -0.70 -0.70 82.5 -0.9659 -1. 12 -1 12 -1.12 -1.12 85.0 -0.9848 +5 .24 +5 .24 -2. 39 -2 39 +5 24 +5. 24 -2.39 =2.39 87.5 -0.9962 +5. 02 +5.02 +5.02 +5.02 . 90.0 -1.0000 160 Table (V-II-6) Hemispherical Dome - Snow Load M-Force/Length, (N ) = a4(pR/2) Fig. (V-II-6) Percentage Error = 0.0 M = 0.2 E l a s t i c i t y ee= <f>e = 5° 9e= <j)e = 2.5° ee = <j>e = 5 0 6e= = 2.5° Solution Statics Energy Statics Energy Statics Energy Statics Energy Matrix Matrix Matrix Matrix Matrix Mat r i x Matrix Matrix 0.0 -1.0000 -0 13 -0.13 -0.03 -0.03 -0. 13 -0 .13 -0 03 2.5 -0.03 -0.03 -0 .03 5.0 -0 13 -0.13 -0.03 -0.03 -0. 13 -0 .13 -0 03 7.5 -0,03 -0.02 -0 03 10.0 -0 13 -0.13 -0.03 . -0.03 -0. 13 -0 .13 -0 03 12.5 -0.03 -0.03 -0 .03 15.0 -0. 13 -0.12 -0.03 -0.03 -0. 14 -0 .13 -0 03 17.5 -0.03 -0,03 -0 04 20.0 -0 14 -0.12 -0.04 -0.03 -0. 14 -0 .12 -0 04 22.5 -0.04 -0.03 -0 04 • 25.0 -0 15 -0.12 -0.04 -0.03 -o. 15 -0 .12 -0 04 27.5 -6.04 -0.03 -0 04 30.0 -0. 15 -0.12 -0.04 -0.03 -0. 15 -0 .12 -0 04 32.5 -0.04 -0.03 -0 04 35.0 -0. 16 -0.12 -0.04, -0.03 -0. 16 -0 .12 -0 04 37.5 -0.04 -0.03 -0 .04 40.0 -0. 17 -0.12 -0.04 i-0.03 -d. 17 -0 .13 -0 04 42.5 -0.04 -0.03 -0 05 45.0 -0. 18 -0.13 -0.05 -0.03 -0. 18 -0 .14 -0 05 47.5 -0.04 -0.03 -0 05 50.0 -0. 19 -0.15 -0.04 -0.04 -d. 19 -0 .16 -0 05 52.5 -0.05 -0.04 -0 05 55.0 -0. 19 -0.17 -0.04 -0.04 -0. 21 -0 .20 -0 05 57.5 -ID. 05 -0.05 -0 05 60.0 -0. 20 -0.23 -0.05 -0.06 -0. 22 -0 .26 -0 05 62.5 -0.05 -0.07 -0 06 65.0 -0. 21 -0.32 -0.05 -0.08 -0. 24 -0 .38 -0 06 67.5 -0.05 -0.10 i -0 06 70.0 -0. 21 -0.51 -0.06 -0.12 -0. 26 -0 .60 -0 07 72.5 -0.05 -0.16 -0 06 75.0 -0!24 -0;.88 -0.06 -0.22 -0. 33 -1 .04 -0 08 77.5 -0.06 -0.32 -0 09 80.0 -0. 13 -2.22 -0.04 -0.51 -0. 32 -2 .66 -0 09 82.5 j -0.08 -0.88 -0 17 85.0 -1.65 -6.18 -0.07 -2.23 -2. 36 ' -7 .37 -0 13 87.5 rl.36 -5.97 i • -2 08 90.0 -0. 67 -0,67 -0.14 -0.14 -0. 67 j-0 .67 -0 14 161 WIND LOAD p/area Axis of | Revolution p cos y< ELEVATION p/area 9=0 symmetrical pcos G 9 = 180° 9=90° PLAN VIEW Antisymmetrical FIGURE 3Z:-UI-I. HEMISPHERICAL DOME SUBJECTED TO WIND LOAD 162 5.5 Example III - Hemispherical Dome Subjected to Wind Load Fig. (V-III-1) 35 36 The available e l a s t i c i t y solution ' i s based on the following assumption of the wind pressure on the dome surface. PP = °' PM = ° ; PR = " P C 0 S * C o s 8 " Here, p i s the int e n s i t y of the wind load on a surface normal to the direction of the wind, and Pp, p^, p R are components of the distributed pressures on the s h e l l surface i n P, M and R directions respectively. .Corresponding to the above d i s t r i b u t i o n the r a d i a l load i s symmetrical about 6= 0° and G = 180° meridian planes and antisymmetrical about 9 = ±90° meridian planes. The s h e l l i s supported i n such a way that the deflections at the base are permitted only i n r a d i a l directions. By assuming proper boundary conditions along 8 = 0 ° and 9 = 90° meridians, i t i s s u f f i c i e n t to solve 35 36 only one quarter of the dome. E l a s t i c i t y solution ' for displacements (y = 0.0 and 0.2) and stresses (independent of y), based on the membrane theory which neglects flexure, i s plotted (Figs. (V-III-2 to 7)) at 6 = 0°, 30°, 60° and 90° meridians. For the f i n i t e element analysis, nodal loads i n R-directions are calculated by multiplying the intensity of the normal pressure p at the R node by 1/4 of the areas of a l l s h e l l elements meeting at that node. Both Statics and Energy plane stress s t i f f n e s s matrices are employed to analyse the s h e l l model. Different values of Poisson's r a t i o and various element sizes are used i n the analysis. Percentage errors are calculated and are shown i n Tables (V-III-1 to 36). Higher percentage errors i n R-displacement and membrane shear, noticed at the nodes of the 163 -triangles, are probably due to the manner of calculation of nodal loads and due to the behaviour of the model as a cone rather than as a spherical surface near the apex. However, here the values of R-displacement and shear are small as compared to t h e i r maximum values elsewhere and as such these errors are considered n e g l i g i b l e . The stresses and the displacements i n the upper part of the s h e l l are obtained with f a i r accuracy. Larger deviation i n R-displacements near the base nodes i s noticed. The accuracy of the solution i s improved by using smaller sized elements. In the f i n i t e element analysis, errors are introduced due to ( i ) approximating the continuously curved surface by f l a t surfaces connected only at the corners, ( i i ) replacing the continuously varying load by approximate concentrated loads applied at the nodes and ( i i i ) replacing the complicated deflected shape of an element by a simple deflected shape. However, as the element size decreases, the geometry, the loading, the support conditions,and the deflected shape of the model approach closer those of the s h e l l . The errors of the F.E. solution, therefore, are expected to reduce. This i s confirmed by the error analysis presented i n Tables (V-III-1 to 36). A further subdivision of the entire s h e l l increases the size of the problem beyond the scope of the present computer program and i s unnecessary. In view of the f a i r l y accurate solution obtained for the upper part of the s h e l l , smaller sized elements are only necessary i n the base zone. This i s conveniently achieved by providing displacement boundary conditions along a p a r a l l e l c i r c l e at which displacements have been calculated with good accuracy. This p a r a l l e l c i r c l e i s selected high enough so that the base de-flectio n s are unaffected by the small errors i n displacements provided along i t . 164 I t i s observed that the differences i n R-displacement at tht base nodes between the F.E. and the e l a s t i c i t y solutions are reduced on reduction of the element size but the pattern of deviation, i.e.,the base deflections being too large and the ones one node above being too small, remains unchanged (Fig. (V-III-8)). The cause for t h i s deviation pattern may be explained as follows; loads at the base nodes are determined using the wind in t e n s i t y at the base, which i s somewhat greater than the one over most of the area tributary to the base node. Furthermore, the base loads are shift e d v e r t i c a l l y down compared to t h e i r actual location. Both these causes tend to make the R-deflection at the base i n the F.E. solution greater compared to the e l a s t i c i t y solution. The assumed load action i n the F.E. solution agrees closely with the e l a s t i c i t y solution with regard to Y P condition ( i . e . sum of the forces i n r a d i a l direction) but d i f f e r s from i t substantially i n £ Mp condition ( i . e . sum of the moments about P-axis) i n view of the v e r t i c a l s h i f t of the load. This inconsistency between the two solutions might be removed by the application of an equivalent balancing moment made up of two equal forces applied i n opposite directions at the base node and at the node one step up. Such couple of forces would evidently decrease the R-deflection at the base and increase i t at the next node up. Presumably the action of these forces would restore the deflections of the F.E. solution to the e l a s t i c i t y l e v e l . Without such action, the base R-deflections w i l l be greater and the deflections one step up smaller than the deflections found by the e l a s t i c i t y . It i s reminded that the comparison of the two deflections just concluded, refers to the membrane solution. The results presented i n Tables (V-III-1 to 36) show that for 0 = cp = 10° element size (u = 0.0 and 0.2), the membrane F.E. solution d i f f e r s from the e l a s t i c i t y solution by 2% i n P and M displacements by 4% i n 165 R-displacen.ents and i n stress by 3.5% using S t a t i c s matrix and by 5.5% using Energy matriy. In the above, the base nodes and the nodes at which values of displacements or stresses are small are excluded. 35 36 The supposedly exact e l a s t i c i t y solution ' used i n the preced-ing discussion assumes the absence of f l e x u r a l stresses and th i s i s not the case i n the structure under consideration. In view of the bottom edge of the s h e l l carrying r a d i a l l y directed loads and not being restrained i n R-direction, some f l e x u r a l stresses, are induced near the base zone and the 39 load there i s p a r t i a l l y carried by flexure. The e l a s t i c i t y membrane 35 36 solution ' i s therefore somewhat erroneous i n the base zone. Away from the base the f l e x u r a l stresses quickly die out as i s a general condition i n the shells of revolution having positive gaussian curvatures i n the two p r i n c i p a l directions. To study the effect of flexure i n the s h e l l , the f i n i t e elements were endowed with both plane stress and flexure properties. Due to the increased degrees of freedom at each node, the size of the problem i s increased. To be within the scope of the present computer program, only the base zone of the s h e l l i s analysed i n two cases. The displacement and rotation boundary conditons are provided at <f> = 30° for 6 g = <j>e = 5° element size i n one case and at <f> = 15° and 6 = 45° for 0 = A> = 2.5° T e re element size i n the other. I t i s assumed that near the new boundary, the displacements found i n the membrane solution and the rotations implied i n them are es s e n t i a l l y correct. The flexure solution i s carried out i n three stages. In the f i r s t stage, displacements at a l l nodes are calculated using the membrane theory alone. In the second stage which includes flexure, P and M rotations of various nodes are found which would effect t h e i r moment equilibrium i n the presence of deflections determined i n the f i r s t 166 stage. F i n a l l y , i n the t h i r d stage, the displacements and rotations i n the base zone are calculated using the combined membrane and flexure theory while providing at the new boundary the displacements and rotations from the second stage of the solution. The percentage deviations i n displacements and membrane stresses of the flexure F.E. solution from the e l a s t i c i t y 35 36 membrane solution ' are shorn i n Tables (V-III-1 to 36). These results are for a thin s h e l l (radius R = 100.0 i n . ; thickness t = 1.0 i n . ) . The differences i n results of displacements and stresses calculated by membrane and by flexure solutions show the effect of flexure. These differences are small. The flexure stresses are then largely the p a r t i c i p a t i o n stresses having v i r t u a l l y no effect on the load carrying capacity of the s h e l l except near the base where there i s small effect present. In spite of thi s the unit f l e x u r a l stresses (#/in 2) may be s i g n i f i c a n t and may cause the s h e l l to crack. The bending moments/length along 9 = 0 ° meridian are shown i n Table (V-III-37). In order to study the effect of flexure i n a thicker s h e l l (radius = 100.0 i n . ; thickness =5.0 i n . ) , the deviations i n R-displacements and bending moments (at 9 = 0° meridian; for y = 0.2 and 9 g = <pe = 5° element size) are calculated. Table (V-III-37) shows that on increasing the thickness of the s h e l l , larger bending moments/length are induced. The small differences i n the R-displacements i n the membrane and flexure solutions shown i n Table (V-III-38) suggest that the effect of flexure on the load carrying capacity of the s h e l l on increasing the thickness (from t = 1.0 i n . to t = 5.0 in.) i s small. Furthermore, the flexure effects are r e s t r i c t e d at the base zone ( cp = 0° to 9 = 35°) above which the s h e l l action i s ess e n t i a l l y membrane. 167 The ov e r a l l results presented i n the Tables (V-III-1 to 36) show that the solution i s obtained with good accuracy for both values of Poisson' ratio.(0.0 and 0.2) by using either Statics or Energy s t i f f n e s s matrices. However, better accuracy i s noticed from the use of Statics matrices. 169 Table (V-III-1) Hemispherical Dome - Wind Load P-Displacement = a 1(pR 2/Ei) Fig. (V-III - 2 ) y = 0.0; R/t =i 100.0; 9 = 30° meridian (Fig. (V-III-1)) Percentage Error - < — — Membrane —y Flexure E l a s t i c i t y 9 e=9e 0e=9e 9 Solution 9 e=9 e=15° 9'e=<f> e=10° 9e=9 e~ 5° 9 e-9 e = 2 .5° =5° =2 .5° a, Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0 .0 0 .0000 0 .00 0.00 0 .00 0.00 0.00 0 .00 0 .00 .0 .00 0 .00 0.00 2 .5 0 .0286 0 .00 0 .00 0.00 5 .0 0 .0563 0.36 0 .36 0 .18 0 .18 0 .35 0.18 7 .5 0 .0832 0 .00 0 .00 0.00 10 .0 0 1092 1 .83 1.83 0.18 0 .18 0 .00 0 .00 0 .09 0.00 12 .5 0 1345 0 .00 0 .00 0.00 15 .0 0 1589 4 .03 4.15 0.06 0 .06 0 .00 0 .00 0 .06 0.00 17 .5 0 .1826 20 .0 0 .2055 1 .07 1.07 0.05 0 .05 0 .05 22 .5 0 2277 25 .0 0 .2491 0.04 0 .04 0 .04 27 .5 0 2698 30 .0 0 2898 2 .24 2.31 0 .93 0.97 0.00 0 .00 0 .00 35 0 0 3275 40 0 0 3622 0 .86 0.83 45 0 0 3939 1 .98 1.98 50 0 0 4224 0 .78 0.76 55 0 0 4478 60 0 0 4700 1 .64 1.53 0 .72 0.66 65 0 o 4889 70 0 0 5044 j 0 .67 0.56 75 0 0 '5165 1 .55 i 1. 36 80 0 0 5252 0 67 0.51 85 0 0 5304 90 0 Stat. - Statics matrix Engy. - Energy matrix 170 Table (V-III-2) Hemispherical Dome - Wind Load P-Displacement = a i(pR 2/Et) Fig. (V-III-2) y = 0.2; R/t = 100.0; 6 = 30° meridian (Fig. (V-III-1)) Percentage Error < • •  M e m b r a n e - - • > Flexure E l a s t i c i t y 6 e-<f>e 9e=<t)e Solution Q Q=15° e e-9 s=10° 6e-<j> e - - 0 a e-<l>< ==2.5° =5° =2.5° Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0 .0° 0 0000 0 .00 0.00 0 .00 0.00 0.00 0 .00 0 .00 0.00 0 .00 0.00 2 .5° 0 0343 0 .29 0.29 0.29 5 .0° 0 0676 0.30 0 .30 0 .00 0.00 0 .15 0.00 7 .5° 0 0998 0 .00 0.00 0.10 10 .0° 0 1311 1 .45 1.45 0.08 0 .08 0 .00 0.00 0 .08 0.00 12 .5° 0 1613 0 .06 0.06 0.06 15 .0° 0 1907 3 .30 3.46 0.05 0 .05 .0 .00 0.00 0 .05 0.00 17 .5° 0 2191 20 .0° 0 2466 0 .93 0.93 0.04 0 .04 0 .04 22 .5° 0 2732 25 .0° 0 2990 0.00 0 .00 0 .00 27 .5° 0 3238 30 .0° 0 3478 1 .93 2.01 0 .81 0.83 0.00 0 .00 0 .00 35 .0° 0 3930 40 .0° 0 4346 0 .74 0.76 45 .0° Q. 4726 1 .69 1.74 50 .0° 0 5069 0 .67 0.67 55 .0° 0 5374 60 .0° 0 5640 1 .37 1.35 0 .62 0.60 65 .0° 0 5866 70 0° 0 6053 0 .56 0.50 75 0° 0 6198 1 .24 1.10 80 .0° 0 6303 0 .54 0.40 85 0° 0 6365 90 0° Stat. - Statics matrix Engy. - Energy matrix 171 Table (V-III-3) Hemispherical Dome - Wind Load P-Displacement = a1(pR2/Et) Fig. (V-III-2) M = 0.0; R/t = 100.0; 9 = 60° meridian (Fig. (V-III-1)) Percentage Error Membrane E l a s t i c i t y 9e=<f>e 9e=4>e Solution 9 e = <(> =,= 15° 9e=<P e-10° 9e=<P =>=5 0 ee=<p s=2.5° =5° =2.5° • <*l • Stat Engy Stat Engy Stat En Stat Engy Stat Stat 0.0 0 0000 0. 00 0.00 0. 00 0 .00 0.00 0. 00 0. 00 0.00 0 .00 2.5 0 0496 0. 00 0.00 5.0 0 0976 j 0.31 0. 31 0. 00 0.00 0 .21 7.5 0 1441' 0. 00 0.00 10.0 0 1892 1. 74 1 .74 0.11 0. 11 0. 00 0.00' 0 .11 12.5 0 2329 0 00 0.00 15.0. 0 2752 3. 96 4.11 0.11 0. 11 0. 00 0.00 0 .07 17.5 0 3162 20.0 0 3560 1. 01 1 .01 0.03 0. 03 0 .03 22.5 . 0 3944. 25.0 0 4315 0.02 0. 02 0 .02 27.5 0 4674 30.0 0 5019 2. 19 2.23 0. 92 0 .92 0.00 0. 00 0 .00 35.0 0 5672 40.0 0 6274 0 81 0 .80 45.0 0 6822 1. 92 1.91 50.0 0 7317 0 75 0 .72 55.0 0 7757 60.0 0 8141 1. 60 1.49 0 70 0 .64 65.0 0 8467 70.0 0 8736 0 66 0 .56 75.0 b 8946 1. 53 1.34 80.0 0 9097 0. 66 0 .51 85.0 0 .9188 90.0 Flexure Stat. - Statics matrix Engy. - Energy matrix 172 Table (V-III-4) Hemispherical Dome - Wind Load P-Displacement = a^pR^/Et) Fig. (V-III-2) M = 0.2; R/t = 100.0; 0 = 60° meridian (Fig. (V-III-1)) Percentage Error CI MemDrane r i e x u r e E l a s t i c i t y e s=9e 0e =9e 9 Solution ee=9 e=15° ee=9 e=10° 0e=9e=5 0 5=9 2.5° =5° =2.5° Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0.0° 0 0000 0. 00 0.00 0 .00 0.00 0.00 0.00 0 .00 0 .00 0 .00 2.5° 0 0595 0 .00 0 .00 . 5.0° 0 1171 0.26 0.26 0 .00 0 .00 0 .17 7.5° 0 1729 0 .00 0 .00 10.0° 0 2270 1 .45 1.45 0.09 0.09 0 .00 0 .00 0 .09 12.5° 0 2795 0 .00 0 .00 15.0° 0 3303 3. 24 3.36 0.06 0.06 0 .00 0 .00 0 .06 17.5° 0 3795 J 20.0° 0 4271 0 .87 0.91 0.05 0.05 0 .05 22.5° 0 4733 25.0° 0 5178 0.02 0.02 i 0 .02 27.5° 0 5608 30.0° 0 6023 1. 86 1.94 0 .78 0.81 0.00 0.00 0 .00 35.0° 0 6807 40.0° 0 7528 0 .70 0.72 45.0° 0 8186 1. 63 1.67 50.0° 0 8780 0 .65 0.65 55.0° 0 9308 60.0° 0 9769 1. 33 1,30 0 .60 0.57 65.0° 1 0161 70.0° 1 0484 0 .55 0.49 75.0° 1 0736 1. 21 1.08 80.0° 1 0916 0 54 0.41 85.0° 1 1025 90.0° Stat. - Statics matrix Engy. - Energy matrix 173 Table (V-III-5) Hemispherical Dome - Wind Load P-Displacement =cti (pR 2/Et) (Fig. (V-III-2) y = 0.0; R/t = 100.0; 9 = 90° meridian (Fig. (V-III-1)) Percentage Error -< Membrane • Flexure E l a s t i c i t y 5° .5° 9 e=?e 9'e=<J>e Solution 6 e=<f> s=15° 9 e=<f> e=10° ee=<t> e= e e=9 e=2 =5° = 2.5° . _ a l . . Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0.0 0 0000 0 .00 0.00 0 .00 0 .00 0.00 0 .00 0 .00 0. 00 0 .00 2.5 0 0572 0 .18 0. 18 5.0 0 1127 0.27 0 .27 0 .00 0. 00 0 .18 7.5 0 1664 0 .00 0. 00 10.0 0 2185 1 .56 1 .60 0.05 0 .05 0 .00 0. 00 0 .05 12.5 0 2689 0 .00 0. 00 15.0 0 3178 3 .65 3.78 0.06 0 .06 0 .00 0. 00 0 .03 17.5 0 3682 20.0 0 4110 0 .90 0 .93 0.02 0 .02 0 .02 22.5 . o 4554 25.0 0 4983 0.00 0 .00 0 .00 27.5 0 5397 30.0 0 .5796 2 .02 2.07 0 .83 0 .83 0.00 0 .00 35.0 0 .6550 40.0 0 7244 0 .76 0 .75 45.0 0 .7877 1 .83 1.80 50.0 0 .8449 0 .72 0 .69 55.0 0 8957 60.0 0 .9400 1 .55 1.44 0 .68 0 .63 65.0 0 .9777 70.0 1.0088 0 .64 0 .55 75.0 1.0330 1 .52 1.34 80.0 1 .0504- 0 .67 0 .51 85.0 1 .0609 90.0 Stat. - Statics matrix Engy. - Energy matrix 174 Table (V-^III-6) Hemispherical Dome - Wind Load P-Displacement = g^p^/Et) Fig. (V-III-2) p = 0.2; R/t = 100.0; 9 = 90° meridian (Fig. (V-III-1)) Percentage Error Membrane >- Flexure E l a s t i c i t y 9 e = 9 e 9e-<j>e Solution 6 =.= <)> e=15° 9 e = 9 e - L0° 9 e = 9 9 2=9 e=2.5° 5'° =2.5° CM. . Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0 0° 0 0000 0 .00 0.00 0. 00 0 .00 0.00 0 .00 0 .00 0.00 0. 00 2 5° 0 0687 0 .00 0.00 5 0° 0 1352 0.22 0 .22 0 .00 0.00 0. 15 7 5° 0 1997 0 .00 0.00 10 0° 0 2621 1. 30 1 .30 0.08 0 .08 0 .04 0.00 0. 08 12 5° 0 3227 0 .00 0.00 15 0° 0 3814 2 .94 3.07 0.03 0 .03 0 .00 0.00 0. 03 17 5° 0 4382 20 0° 0 4932 0. 77 0 .79 0.02 0 .02 0. 02 22 5° 0 5465 25 0° 0 5979 0.02 0 .02 0. 02 27 5° 0 6476 30 0° 0 6955 1 .70 1.77 0. 71 0 .73 0.00 0 .00 0 00 35 0° 0 7860 40 0° 0 .8693 0 64 0 .67 45 0° 0 .9453 1 .51 1.57 50 0° 1 0138 0 61 0 .61 55 0° 1.0748 60 0° 1 1280 1 .29 1.25 0 58 0 .55 65 0° 1 1733 70 0° 1 2105 0 55 0 .48 75 0° 1.2397 1 .20 1.07 80 0° 1 2605 0 54 0 .41 85 0° 1.2731 90 0° Stat. - Statics matrix Engy. - Energy matrix 176 Table (V-III-7) Hemispherical Dome - Wind Load M-Displacement = g ?fpR 2/Et) Fig. (V-III-3) p = 0 . 0 ; R/t = 1 0 0 . 0 ; 6 = 0° meridian (Fig. (V-III - 1 ) ) Percentage Error < Membrane ' Flexure E l a s t i c i t y 6e = <t>e 9 e = 4 i e * Solution eft=cj> 15° 9R = 9 10° 9e -<j> 5 = 5° ee=e|>e= 2.5° 5° = 2.5° 0-2 Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0 .0° 0 0000 0 .00 0 .00 0 .00 0 .00 0. 00 0. 00 0 .00 0 .00 0. 00 0.00 2 .5° 0 0436 1 .38 1 .38 0.92 5 .0° 0 0873 2. 52 2. 52 0 .34 0 .34 2. 29 0.46 7 .5° 0 1308 0 .23 0 .23 0.31 10 .0° 0 1744 5 .33 5 .62 0. 63 0. 63 0 .06 0 .06 0. 92 0.06 12 .5° 0 2177 0 .05 0 .05 0.05 15 .0° 0 2609 7 .44 •8 .20 0. 38 0. 38 0 .00 0 .00 0. 38 0.00 17 .5° 0 3039 20 .0° 0 3465 1 .65 1 .91 0. 17 0. 17 0. 17 22 .5° 0 3887 25 .0° 0 4304 0. 07 0. 07 0. 07 27 .5° 0 4716 30 .0° 0 5120 2 . 12 2 .79 1 . 15 1 .41 0. 00 0. 00 0. 00 35 .0° 0 5906 40 .0° 0 6653 0 .77 •1 .04 45 .0° 0 7355 1 .51 2 .14 50 .0° 0 8005 0 .60 0 .86 55 .0° 0 8596 60 .0° ;o 9121 1 . 12 1 .71 0 .48 0 .72 65 .0° 0 95 76 70 .0° 0 9955 0 .44 0 .64 75 .0° I 0254 0 .61 0 .93 80 .0° I 0470 0 .28 0 .33 85 .0° I 0600 90 .0° •;i 0644 0 .97 0 .89 '0 .46 0 .34 Stat. - Statics matrix Engy. - Energy matrix 177 Table (V-III-8) Hemispherical Dome - Wind Load M-Displacement = ct2(pR2/Et) V = 0.2; R/t = 100.0; 8 = 0° Fig. (V-III-3) meridian (Fig. (V-III-1)) Percentage Error -Membrane E l a s t i c i t y ee=9e 9 e-9 e ( Solution = <j> e-15° 9e-9e- 10° 6e=9e=5 9e = cf> 3=2 .5° =5° =2.5° • 012 Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0 0° 0 0000 0 .00 0 .00 0. 00 0 .00 0.00 0.00 0. 00 0. 00 0.00 0.00 2 5° 0 0524 1. 15 1. 15 0.76 5 0° 0 1047 2.48 2.48 0. 38 0. 38 2.29 0.48 7 5° 0 1570 0. 19 0. 19 0.25 10 0° 0 2092 5. 12 5 .40 0.62 0.62 0. 10 0. 10 0.91 0.14 12 5° 0 2613 6. 04 0. 04 0.04 15 0° 0 3131 7 .09 7 .79 0.35 0.35 0. 00 0. 00 0.38 0.00 17 5° 0 3647 20 0° 0 4158 1. 54 1 .78 0.17 0.17 0.17 22 5° 0 4665 25 0° 0 5165 0.06 0.06 0.06 27 5° 0 5659 30 0° 0 6144 1 .99 2 .56 1. 06 1 .30 0.00 0.00 0.00 35 0° 0 7087 40 0° 0 7984 0. 70 0 .94 45 0° '• 0 8826 1 .34 1 .92 50 0° 0 9606 0. 53 0 .78 55 0° 1.0315 60 0° 1 0946 0 .95 1 .49 0. 41 0 .64 65 0° 1 1491 70 0° 1 1946 0. 39 0 .57 75 0° 1 2304 0 .50 0 .73 80 0° 1 2564 0. 24 0 .26 85 0° 1 2720 90 0° 1 2773 0 .78 0 .66 0. 38 0 .25 Flexure Stat. - Statics matrix Engy. - Energy matrix 178 Table (V-III-9) Hemispherical Dome - Wind Load M-Displacement = g 2(pR 2/Et) Fig. (V-III-3) p = 0.0; R/t = 100.0; 0 = 30° meridian (Fig. (V-III-1)) Percentage Error < Membrane E l a s t i c i t y e e = 9 e 6 e - 9 e ( P Solution 8 e- 9 e-15° e<e=<pe=10° _ S'e-tP'e21-O e e-9'e-2.5° =5° =2.5° a 2 Stat Engy Stat Engy ' Stat Engy Stat Engy Stat Stat 0 0° 0 0000 0 00 0 .00. 0 00 0.00 0.00 0. 00 0 .00 0 .00 0 .00 0.00 2 5° 0 0378 1 .32 1 .32 0.79 5 0° 0 0756 2.51 2. 51 0 .27 .0 .27 : 2 .25 0.53 7 5° 0 1133 0 .18 0 .18 0.26 10 0° 0 1510, 5 30 5.63 0.66 0. 66 0 .07 0 .07 0 .93 0.13 12 5° 0 1886 0 .00 0 .00 0.00 15 0° 0 2260 7 39 8 .14 0.35 0. 35 0 .00 0 .00 0 .35 0.00 17 5° 0 2632 20 0° 0 3001 1 63 1.87 0.17 0. 17 0 .17 22 5° 0 3366 25 0° 0 3728 0.05 0. 05 0 .05 27 5° 0 4084 30 0° 0 4434 2 16 2 .77 1 13 1.40 0.00 0. 00 0 .00 35 0° 0 5114 40 0° 0 5762 0 76 1.02 45 0° 0 6370 1 48 2 .10 50 0° 0 6933 0 58 0.85 55 0° 0 7444 60 0° 0 7899 1 10 1 .70 0 47 0.71 65 0° 0 8293 70 0° 0 8621 0 .44 0.65 75 0° 0 8880 0 61 0 .92 80 0° 0 9067 0 29 0.34 85 0° 0 9180 90 0° Flexure Stat, Statks matrix Engy. - Energy matrix 179 Table (V-III-10) Hemispherical Dome - Wind Load  M-Displacement = ct 2(pR 2/Et) Fig. (V-III-3) y = 0.2; R/t = 100.0; 9 = 30° Meridian (Fig. (V-III-1)) Percentage Error Membrane Flexure E l a s t i c i t y Solution 8e=<i>e=15' 9e=<j)e=10c 9e=ie=5' 6e-<}>e 9e=le=2.5' 9e=<t>e =2.5' a2 Stat Engy Stat Engy Stat Engy Stat 1 Engy Stat Stat 0.0° 2.5° 5.0° 7.5° 10.0° 12,5° 15.0° 17.5° 20.0° 22.5° 25.0° 27.5° 30.0° 35.0° 40.0° 45.0° 50.0° 55.0° 60.0° 65:0° 70.0° 75.0° 80.0° 85.0° 90.0° 0.0000 0.0453 0.0907 0.1360 0.1812 0.2263 0.2712 0.3158 p.3601 0.4040 0.4473 0.4901 0.5321 0.6137 0.6914 0.7644 0.8319 0.8933 0.9479 0.9952 1.0345 1.0656 1.0880 1.1016 0.00 7.04 1.96 1.32 0.94 0.50 0.00 7.74 2.54 1.90 1.48 0.72 0. 00 5.08 1. (53 1.05 0.69 0.53 0.41 0.39 0.25 0.00 5.41 1.78 1.30 0.94 0.77 0.63 0.57 0.27 0.00 2.43 0.61 0.33 0!.17 0.07 0.00 0.00 2.43 0.61 0.33 0.17 0.07 0.00 0.00 .1.33 0.33 0.15 0.11 0.04 0.00 0.00 1.33 0.33 0.15 0.11 0.04 0.00 0.00 2.01 0.94 0.37 0.17 0.07 0.00 0.00 0.88 0.44 0.22 0.11 0.04 0.00 Stat. - Statics matrix Engy. - Energy matrix 180 Table (V-III-11) Hemispherical Dome - Wind Load  M-Displacement = ct2(pR2/Et) Fig. (V-III-3) p= 0.0; R/t 100.0; 9 = 60° meridian (Fig. (V-III-1)) Percentage Error < Membrane Flexure E l a s t i c i t y 9e=4>e 9e=ej>e <p ' Solution e £=cpe-15° 9 e = 9 3=10° 9e=<)> 5°. 9 e = 9 e = 2.5° =5° = 2.5° Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0.0° 0 0000 0 .00 •0 .00 0 00 0.00 0.00 0 .00 0. 00 0 .00 0 00 2.5° 0 0218 1. 38 1 .38 5.0° 0 0436 2.52 2 .52 0. 46 0 .46 2. 29 , 7.5° 0 0654 0. 31 0 .31 10.0° 0 0872 5 23 5.62 0.57 0 .57 0. 12 0 .12 0. 92 12.5° 0 1089 0. 00 0 .00 15.0° 0 1305 7 .28 8 .05 0.31 0 .31 0. 00 •0 .00 0. 31 17.5° 0 1519 20.0° 0 1733 1 56 1.79 0.12 0 .12 0 12 22.5° 0 1944 25.0° 0 2152 0.05 0 .05 0. 05 27.5° ; 0 2358 30.0° 0 2560 2 .07 •2 .70 1 09 1.37 0.00 0 .00 0 00 35.0° 0 2953 40.0° 0 3327 0. 72 '0.99 45.0° 0 3678 1 .41 2 .04 50.0° 0 4003 0 55 0.82 55.0° 0 4298 60.0° 0 4561 1 .05 1 .64 0 44 0.68 65.0° 0 4788 70.0° 0 4977 0 44 0.64 75.0° 0 5127 0 .61 0 .92 80.0° 0 5235 0. 29 0.34 85.0° 0 5300 90.0° Stat. - Statics matrix Engy. - Energy matrix 181 Table (V-III-12) Hemispherical Dome - Wind Load M-Displacement = g ?(pR 2/Et) Fig. (V-III-3) y =0.2; R/t = 100.0; 6 = 60° meridian (Fig. (V-III-1)) Percentage Error Membrane Flexure Elasticity-Solution ee=<J>e=15' 9e=<i>e=10' 6e=4>e=50 6e=cf>e=2.5' ee=<f>e =5° 3e-<Pe =2.5C Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0.0° 2.5° 5.0° 7.5° 10.0° 12.5° 15.0° 17.5° 20.0° 22.5° 25.0° 27.5° 30.0° 35.0° 40.0° 45.0° 50.0° 55.0° 60.0° 65,0° 70.0° 75L0° 80.0° 85.0° 9 0 . 0 ° 0.0000 0.0262 0.0524 0.0785 0.1046 0.1306 0.1566 0.1823 0.2079 0.2332 0.2583 0.2829 0.3072 0.3543 0.3992 0.4413 0.4803 0.5158 0.5473 0.5746 0.5973 0.6152 0.6282 0.6360 0.00 6.90 1.89 1.25 0.90 0.49 0.00 7.60 2.44 1.84 1.43 0.72 0.00 j i 5.07 i 1 1.49 1.01 0.65 0.50 0.38 0.37 0.24 0.00 5.35 1.73 1.27 0.90 0.75 0.60 0.55 0.26 0.00 2.29 0.67 0.32 0.14 o'. 04 0.00 0.00 2.29 0.67 0.32 0.14 0.04 0.00 0.00 1.15 0.19 0.25 0.10 0.08 0.00 0.00 1,15 0.19 0.26 0.10 0.08 0.00 0.00 2.10 0.96 0.32 0.14 0.04 0.00 Stat. - St at its matrix Engy. - Energy matrix 183 Table (V-III-13) Hemispherical Dome - Wind Load R-Displacement = - a 3(pR 2/Et) Fig. (V-III-4) y =0.0; R/t = 100.0; 6 = 0 ° meridian (Fig. (V-III-1)) Percentage Error « Membi rane — 1 Flexure E l a s t i c i t y = <Pe ee =?e Solution = cpe-15 O e e -cj>< , = 1C 0 9 e - 9 e - 5 : 9 e : = 2.5° =5° =2.5° a 3 Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0.0° 1 0000 23. 96 24. 87 16. 28 16. 64 8.06 8 .10 4 08 4 .09 4 .84 1.46 2.5° 1 0272 -1 20 -1 .20 0.52 5.0° 1 0506 -2.33 -2 .31 0 19 0 .19 • 0 .27 -0.02 7.5° 1.0701 -0 17 -0 .16 -0.21 10.0° 1.0859 -4. 23 -4. 13 0.41 0 .42 -0 06 -0 .06 -0 .70 -0.19 12.5° 1 0980 -0 12 -0 .11 -0.11 15.0° 1 1062 -5. 93 -5. 81 -0.29 -0 .28 0 00 0 .00 -0 .33 0.00 17.5° 1 1108 20.0° 1 1117 1. 23 1. 36 -0.04 -0 .04 -0 .13 22.5° 1 1089 25.0° 1 1025 -0.10 -0 .11 -0 .15 27.5° 1 0927 30.0° 1 0793 2. 48 2. 73 -0. 10 -0. 04 0.00 0 .00 0 .00 35.0° 1 0425 40.0° 0 9927 0. 36 • 0. 38 45.0° 0 9308 -0. 07 -0. 22 50.0° 0 8576 0. 32 0. 21 55.0° 0 7741 60.0° 0 6813 2 29 1. 67 0. 19 -0. 32 65.0° 0 5805 70.0° 0 .4728 1. 44 0 30 75.0° 0 .3595 -8 23 -16. 63 80.0° 0 2420 -8. 21 -16. 58 85.0° 0 1217 90.0° 0 0000 0 00 0. 00 0. 00 0. 00 Stat. '- Statics matrix Engy. - Energy matrix 184 Table (V-III-14) Hemispherical Dome - Wind Load R-Displacement = -a 3(pR 2/Et) Fig. (V-III-4) M = 0.2; R/t = 100.0; 6 = 0 ° meridian (Fig. (V-III-1)) Percentage Error CI Fiemurane r xexure E l a s t i c i t y ee= 9e Qe-^e Solution efi=<f)P=is O 6e=<»>e=10 ee=9e=5 ee= :<f,e = 2. - O =5 o =2.5° a 3 Stat Engy Stat Enj IY Stat Engy Stat Engy Stat Stat 0 .0° 1 0000 22 .8 23. 75 15 55 15 91 7.70 7 .74 3 90 3 .90 4. 64 1.44 2 .5° 1 0328 -1 16 -1 . 16 0.48 5 .0° 1 0614 -2.24 -2 .23 0 17 0 . 17 0. 24 -0.05 7 .5° 1 0859 -0 18 -0 .18 -0.24 10 .0° 1 1062 -4 07 -3 95 0.35 0 .36 -0 08 -0 .07 -0. 70 -0.23 12 .5° 1 1223 -0 13 -0 . 13 -0.12 15 .0° 1 1343 -5 . 73 -5. 58 -0.30 -0 .30 0 00 0 .00 -0. 34 0.00 17 .5° 1 1422 20 .0° 1 1461 1 11 1 25 -0.06 -0 .06 -0. 15 22 .5° 1 1459 25 .0° 1 1418 -0.13 -0 .13 -0. 17 27 .5° 1 .1338 30 .0° 1 1220 2 .22 2. 52 -0 13 -0 05 0.00 0 .00 0. 00 35 .0° 1 0871 40 .0° 1 0380 0 33 0 38 45 .0° 0 9755 -0 . 10 -0. 16 50 .0° 0 9005 0 29 0 22 55 .0° 0 8142 60 .0° 0 7176 1 .98 1. 53 0 20 -0 25 65 .0° 0 6121 70 .0° 0 4989 1 24 0 28 75 .0° 0 3797 -6 .79 •i4. 35 80 .0° 0 2557 -6 76 -14 29 85 .0° 0 1287 90 .0° 0 0000 0 .00 0. 00 0 00 0 00 Stat. - Static*, matrix Engy. - Energy matrix Table (V-III-15) Hemispherical Dome 185 - Wind Load R-Displacement = -a 3(pR 2/Et) Fig. (V-III-4) M = 0.0; R/t = 100.0; 9 = 30° meridian (Fig. (V-III-1)) Percentage Error -<— - Membrane — *- Flexure E l a s t i c i t y 0e: = <Pe 9e-9e > Solution 9 g=<pe-15° 9e -<j> , = 10° 9e = <J>e-5 :<Pe =2.5° =5° = 2.5° °<3 Stat Engy Stat Engy Stat Engy Stat En gy Stat Stat 0 0° 0. 8660 23 .89 24.79 16. 26 16.62 8.05 8 .10 4 08 4 .09 4 .83 1.51 2. 5° 0. 8896 -1 20 -1 .20 0.52 5 0° 0 9098 -2.34 -2 .33 0 20 0 .20 0 .26 -0.03 7 5° • 0 9268 -0 17 -0 . 17 -0.23 10 0° 0 9405 -4. 24 -4.15 0.42 0 .43 -0 06 -0 .06 -0 .71 -0.20 12 5° 0 9509 -0 12 -0 . 12 -0.08 15 0° 0 9580 -5 .98 -5,87 -0.28 -0 .27 0 00 0 .00 -0 .32 0.00 17 5° 0 9620 2o:o° 0 9627 . 1. 22 1.34 -0.04 -0 .04 -0 . 11 22 5° 0 9603 25 0° 0 9548 -0113 -0 .13. -0 .15 27 5° 0 9463 30 0° 0 9347 2 .42 2.67 -0. 13 -0.06 0.00 0 .00 0 .00 35 0° 0 9028 40 0° 0 8597 0. 34 0.36 45 0° 0 8061 -0 .14 -0.29 50 0° 0 7427 0. 30 0.19 55 0° 0 6704 60 0° 0 5900 2 .24 1.63 0. 19 -0.34 65 0° 0 5027 70- 0° 0 4094 1. 44 0.29 75 0° 0 3114 -8 .28 -16.70 80 0° 0 2096 -8. 19 -16.65 85 0 1054 90 0° 0 0000 0 .00 0.00 0. 00 0.00 Stat. - Statics matrix Engy. - Energy matrix 186 Table (V-III-16) Hemispherical Dome - Wind Load R-Displacement = -a 3(pR 2/Et) Fig. (V-III-4) y = Q.2; R/t = 100.0; 0 = 30° meridian (Fig. (V-III-1)) Percentage Error -<— - Membrane — Flexure • E l a s t i c i t y ee= <j>e 6e=9e <l Solution ee=<t>e=15° ee=<j>e=io 6e=<t>e=5 ee= :9e= 2.5 0 =5 O = 2.5' Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0 0° d 8660 22 .78 23.68 15. 53 15.90 7.69 7. 74 3 89 3. 90 4. 63 1.45 2 5° 0. 8944 -1 16 -1. 16 0.49 5 0° 0 9192 -2.26 -2. 25 0 16 0. 17 0. 24 -0.05 7 5° 0 9404 -0 18 -0. 18 -0.23 10 0° 0 9580 -4 08 -3.97 0.36 0. 37 -0 07 -0. 07 -0. 70 -0.22 12 5° 0 9719 -0 12 -0. 12 -0.10 15 0° 0 9823 -5 .77 -5.63 -0.30 -0. 29 0 00 0. 00 -0. 34 0.00 17 5° 0 9892 20 0° 0 9925 1 09 1.23 -0.07 -0. 07 -0. 14 22 5° 0 9924 25 0° 0 9888 -0.14 -0. 15 -0. 17 27 5° 0 9819 30 0° o 9716 2 .17 2.47 -0 14 -0.06 0.00 0. 00 0. 00 35 0° 0 9415 40. 0° 0 8990 0 29 0.35 45 0° 0 8448 -0 .15 -0.23 50 0° 0 7799 0 27 0.21 55 0° 0 7051 60 0° 0 6214 . 1 .95 1.50 0 19 -0.26 65 0° 0 5301 70 0° 0 4321 1 23 0.28 75 .0° 0 .3288 -6 .84 -14.42 80 .0° 0 2215 -6 80 -14,43 85 .0° 0 .1114 90 .0° 0 .0000 0 .00 0.00 0 00 0.00 Stat, - Statics matrix Engy. - Energy matrix i 187 Table (V-III-17) Hemispherical Dome - Wind Load R-Displacement = -a 3(pR 2/Et) Fig. (V-III-4) y = 0.0; R/t = 100.0; 6 = 60° meridian (Fig. (V-III-1)} Percentage Error - Membrane — 4- Flexure E l a s t i c i t y .= 10° 5° 6e= 9e ee=<pe •<l > Solution 6e=9< .= 15° ee=<p« ee=<pe=5 ee= :9e= 2. = 5 O =2.5° a 3 Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0 0° 0 5000 23. 54 24.44 16.08 16.44 8.04 8 .08 4 08 4 .08 4. 80 2 5° 0 5136 -1 21 -1 .21 5 0° 0 5253 -2.36 -2 .34 0 19 0 .19 0. 25 7 5° 0 5351 -0 17 -0 .17 10 0° 0 5430 -4.37 -4.27 0.39 0 41 -0 07 -0 .07 -0. 74 12 5° 0 5490 -0 11 -0 .11 15 0° 0 5531 -6. 16 -6.06 -0.33 -0 .31i 0 00 0 .00 -0. 36 17 5° 0 5554 20 0° 0 5558 1.06 1.19 -0.05 -0 .05 -0. 13 22 5° 0 5544 25 0° 0 5513 -0.15 -0 .16 -0. 16 27 5° 0 5463 30 0° 0 5396 2. 15 2.41 -0.28 -0.20 0.00 0 .00 0. 00 35 0° 0 5212 40 0° 0 4964 0.18 0.20 45 0° 0 4654 -0. 39 -0.56 50 0° 0 4288 0.14 0.02 55 0° 0 3870 60 0° 0 3407 2. 00 1.32 0.03 -0.50 65 0° . 0 2902 70 0° 0 2364 1.23 0.13 75 0° 0 1798 -8. 56 -17.02 80 0° 0 1210 -8.51 -17.03 85 0° 0 0609 90 0° 0 0000 0. 00 0.00 0.00 0.00 Stat. - Statics matrix Engy. - Energy matrix Table (V-III-18) Hemispherical Dome 188 • Wind Load R-Displacement = -a 3(pR 2/Et) Fig. (V-III-4) y - 0.2; R/t = 100.0; 9 = 60° meridian (Fig. (V-III-1)) Percentage Error Membrane Flexure E l a s t i c i t y Solution 8eB<fre=15 9e=<j>e=10C 9e=ie=5' 9 e =9e=2.5 c 9 e = 9 e =5° 9 e = 9 e =2.5' <*3 Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0.0° 2.5° 5.0° 7.5° 10.0° 12.5° 15.0° 17.5° 20.0° 22.5° 25.0° 27.5° 30.0° 35.0° 40.0° 45.0° 50.0° 55.0° 60.0° 65.0° 70.0° 75.0° 80.0° 85.0° 90.0° 0.5000 0.5164 0.5307 0.5429 0.5531 0.5611 0.5671 0.5711 0.5730 0.5730 0.5709 0.5669 0.5610 0.5436 0.5190 0.4878 0.4503 0.4071 0.3588 0.3060 0.2495 0.1898 0.1279 0.0643 0.0000 22.48 -5.94 1.91 -0.41 1.70 -7.06 0.00 23.38 -5.78 2.19 -0.49 1.23 -14;65 0.00 15.38 -4.19 0.94 -0.30 0.15 0.11 0.06 1.04 -6.96 0.00 15.74 -4.09 1.08 -0.21 0.21 0.04 -0.39 0.12 -14.54 0.00 7.68 -2.26 0.34 -0.32 -0.07 -0.18 0.00 7.74 -2.26 0.34 -0.32 -0.07 -0.18 0.00 3.90 -1.16 0.17 -0.17 -0.07 -0.11 0.00 3.90 -1.16 0.17 -0.17 -0.07 -0.11 0.00 4.62 0.21 -0.72 -0.35 -0.16 -0.19 0.00 Stat. - Statics matrix Engy. - Energy matrix 1 9 0 Table (V-III-19) Hemispherical Dome - Wind Load P-Force/length, (N )• = -a 4(pR] Fig. (V-III-5) y =0.0; R/t = 100.0; 6 = 0 ° meridian (Fig. (V-III-1)) Percentage Error Membrane : : ' *- Flexure E l a s t i c i t y = 10° ee= 9e 9 e=9 e l> Solution ee= :4>e: = 1 5 ° ee= = <f>e: 6e=<|>e= = 5 ° ee=cj>e= = 2 . 5 O =5 o = 2 . 5 ° Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0 0 ° 1 . 0 0 0 0 - 1 7 6 - 1 . 7 6 - 0 7 2 - 0 . 7 2 - 0 . 3 5 -0 3 8 - 0 . 1 3 - 0 . 1 3 - 1 . 5 1 - 1 . 2 7 2 5 ° 0 . 9 7 2 • - 0 . 1 3 - 0 . 1 4 0 . 6 1 5 0 ° 0 . 9 4 5 - 0 . 3 2 -0 3 5 - 0 . 1 3 - 0 . 1 4 0. 9 1 0 . 0 4 7 5 ° 0 . 9 2 0 - 0 . 1 3 - 0 . 1 4 - 0 . 2 1 1 0 0 ° 0 . 8 9 5 -0 7 9 - 0 . 7 9 - 0 . 3 0 -0 3 4 - 0 . 1 3 - 0 . 1 4 - 0 . 6 3 - 0 . 2 4 1 2 5 ° 0 . 8 7 1 - 0 . 1 4 - 0 . 1 5 - 0 . 1 4 1 5 0 ° 0 . 8 4 7 - 1 9 1 - 1 . 9 1 - 0 . 3 1 -0 35 - 0 . 5 3 1 7 5 ° 0 . 8 2 4 2 0 0 ° 0 . 8 0 0 -0 8 4 - 0 . 8 4 - 0 . 2 8 -0 3 2 - 0 . 2 9 2 2 5 ° 0 . 7 7 7 2 5 0 ° 0 . 7 5 3 - 0 . 2 8 - 0 3 3 - 0 . 2 5 2 7 5 ° 0 . 7 3 0 3 0 0 ° 0 . 7 0 6 - 1 9 6 - 1 . 9 6 -0 8 4 - 0 . 8 4 3 5 0 ° 0 . 6 5 6 . 4 0 0 ° 0 . 6 0 5 -0 8 4 - 0 . 8 4 4 5 0 ° . 0 . 5 5 2 - 1 9 0 - 1 . 9 0 5 0 0 ° 0 . 4 9 7 -0 8 0 - 0 . 8 0 5 5 0 ° 0 . 4 4 0 6 0 0 ° 0 . 3 8 1 - 1 5 8 - 1 . 5 8 -0 6 9 - 0 . 6 9 6 5 0 ° 0 . 3 2 1 7 0 0° 0 . 2 5 8 -0 3 2 - 0 . 32 7 5 0 ° 0 . 1 9 5 1 1 5 1 . 1 5 8 0 0 ° 0 . 1 3 0 2 5 3 . 2 . 5 3 8 5 0° 0 . 0 6 5 9 0 0 ° 0 . 0 0 0 Stat. - Statics matrix Engy. - Energy matrix 191 Table (V-III-20) Hemispherical Dome - Wind Load P-Force/Length, (N ) = -o^(pR) Fig. (V-III-5) u = 0.2; R/t = 100.0; 6 =0° meridian (Fig. (V-III-1)) Percentage Error Membrane E l a s t i c i t y 8e=<f>e 6e=<J>e Solution 6e-<J>e= = 15 0 ee= =<f>e= = 10c 6e-<|>e =5° 4>e: =2. 5° =5° = 2.5° Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0.0° 1 000 -1.76 -1 .76 -0 72 -0. 72 -0.35 -0 38 -0. 12 -0 .13 -1.45 -1.18 2.5° 0 972 -0. 12 -0 .13 0.60 5.0° 0 945 -0.32 -0 35 -0. 13 -0 .13 +0.87 0.02 7.5° 0 920 . -0. 13 -0 .14 -0.23 10.0° 0 895 -0 79 -0. 79 -0.31 -0 34 -0. 13 -0 .14 -0.62 -0.25 12.5° 0 871 -0. 13 -0 .14 -0.15 15.0° 0 847 -1.91 -1 .91 -0.31 -0 35 -0.53 17.5° 0 824 20.0° 0 800 -0 84 -0. 84 -0.28 -0 32 -0.32 22.5° 0 111 25.0° 0 753 -0.28 -0 33 -0.28 27.5° 0 730 30.0° 0 706 -1.96 -1 .96 -0 84 -0. 84 35.0° 0 656 40.0° 0 605 -0 84 -0. 84 45.0° 0 552 -1.90 -1 .90 50.0° 0 497 -0 80 -0. 80 55.0° 0 440 60.0° 0 381 -1.58 -1 .58 -o 69 -0. 69 65.0° 0 321 70.0° 0 258 -0 32 -0. 31 75.0° 0 195 1.15 1 .1.5 80". 0° 0 130 2 53 2. 53 85.0° 0 065 90.0° 0 000 Flexure Stat. - Statics matrix Engy. - Energy matrix 192 Table (V-III-21) Hemispherical Dome - Wind Load P-Force/Length, (N ) = -ai+(pR) Fig. (V-III-5) y = 0.0; R/t = 100.0; 6 =30° meridian (Fig. (V-III-1)) Percentage Error : Membrane • Flexure E l a s t i c i t y 5° 5° ee=<pe ee=<pe c P Solution = <pe= 15 0 ee= = <j)e = = 10° ee=<pe= :9e= 2. = 5° = 2.5° Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0 0° 0 866 2 5° 0 842 -0 13 -0 .14 0.62 5 0° 0 818 -0.33 -0 36 -0 13 -0 .14 0.90 0.02 7 5° 0 796 -0 13 -0 .14 -0.23 10 0° 0 775 • -0 81 -0. 81 -0.30 -0 34 -0 13 -0 .14 -0.63 -0.24 12 5° 0 754 -0 14 -0 .15 -0.14 15 0° 0 734 -1 95 -1 .95 -0.30 -0 34 -0.53 17 5° 0 713 20 0° 0 693 -0 86 -0. 86 -0.29 -0 33 -0.31 22 5° 0 673 25 0° 0 653 -0.28 -0 33 -0.27 27 5° 0 632 30 0° 0 611 -1 99 -1 .99 -0 86 -0. 86 35 0° 0 568 40 0° 0 524 -0 85 -0. 85 45 0° 0 478 -1 91 -1 .91 50 0° 0 431 -0 80 -0. 80 55 0° 0 381 60 0° 0 330 -1 54 -1 .54 -0 67 -0. 67 65 0° 0 278 70 0° 0 224 -0 27 -0. 27 75 0° 0 169 1 30 • 1 .30 80 0° . o. 113 2 64 2. 64 85 0° 0 057 90 0° Stat. - Statics matrix Engy. - Energy matrix 193 Table (V-III-22) Hemispherical Dome - Wind Load P-Force/Length, (N ) = -ct^pR) Fig. (V-III-5) y = 0.2; R/t = 100.0; 8 = 30° meridian (Fig. (V-III-1)) Percentage Error 1 wemDianc > Flexure E l a s t i c i t y 6e_(Pe 9e=<Pe Solution 6e=4>e= = 15° 6e=(Pe = 10° 9e_<Pe =5° 9e-<Pe= =2.5° = 5° =2.5° Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0.0° 0.866 2.5° 0.842 -0.13 -0.13 0.60 5.0° 0.818 -0.33 -0.36 -0.13 -0.13 0.86 0.01 7.5° 0.796 i -0.13 -0.14 -0.24 10.0° 0.775 -0.81 -0.81 -0.31 -0.34 -0.13 -0.14 -0.62 -0.25 12.5° 0.754 -0.13 -0.14 -0.14 15.0° 0.734 -1.95 -1.95 -0.31 -0.34 -0.53 17.5° 0.713 20.0° 0.693 -0.86 -0.86 -0.29 -0.33 -0.32 22.5°r 0.673 25.0° 0.653 -0.29 -0.34 -0.28 27.5° 0.632 30.0° 0.611 -1.99 -1.99 -0.86 -0.86 35.0° 0.568 40.0° 0.524 -0.85 -0.85 45.0° 0.478 -1.91 -1.91 50.0° 0.431 -0.80 -0.80 55.0° 0.381 60.0° 0.330 -1.54 -1.54 -0.67 -0,67 65.0° 0.278 70.0° 0.224 -0.27 -0.27 75.0° 0.169 1.30 1.30 80.0° 0.113 2.64 2,64 85.0° 0.057 90.0° 0.000 Stat. Engy. - Statics matrix - Energy matrix 194 Table (V-III-23) Hemispherical Dome - Wind Load P-Force/Length, (NQ) = -ait(pR) Fig. (V-III-5) y = 0.0; R/t = 100.0; 0 = 60° meridian (Fig. (V-III-1)) Percentage Error Membrane '• E l a s t i c i t y ee= 9e Se-9e 9 Solution = 9e = 15 0 e.e= = 9e: = 10° 6e =9e : =5° ee= 9e= =2.5 o =5 0 = 2.5° ait Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0.0° • 0 500 2.5° 0 486 -0. 14 -0. 14 5.0° 0 473 -0.35 -0 38 -0. 13 -0. 14 0. 88 7.5° 0 460 -0. 14 -0. 15 10.0° 0 447 -0 90 .-0.90 -0.33 -0 36 -0. 14 -0. 15 -o. 65 12.5° 0 435 I -0. 15 -0. 16 15.0° 0 424 -2 12 -2.12 -0.34 -0 38 -0. 56 17.5° 0 412 20.0° 0 400 -0 93 -0.93 -0.31 -0 35 -0. 33 22.5° 0 389 25.0° 0 377 -0.31 -0. 36 -0. 29 27.5° 0 365 30.0° 0-353 -2 08 -2 .08 -o 90 -0.90 35.0° 0 328 40.0° 0 303 -0 87 -0.87 45.0° 0 276 -1 90 -1 .90 50.0° 0 249 -0 78 -0.78 55.0° 0 220 60.0° 0 191 -1 38 -1 .38 -0 59 -0.59 65.0° 0 160 70.0° ' 0 129 -0 12 -0.13 75.0° 0. 097 1 77 1 .77 80.0° 0. 065 2 96 2.94 85.0° 0. 033 90.0° Flexure Stat. - Statics matrix Engy. - Energy matrix 195 Table (V-III-24) Hemispherical Dome - Wind Load P-Force/Length, (NJ =t-al+(pR) Fig. (V-III-5) _ — — — — — o Y-= 0.2; R/t = 100.0; 9= 60° meridian (Fig. (V-III-1)) Percentage Error Membrane Flexure E l a s t i c i t y Solution ep=<pe=15 66=96=10' 6e=9e=5< 6e=9e=2.5< 6e=9e =5° Qe=9e = 2.5C AIT Stat Engy Stat Engy Stat |Engy Stat |Engy Stat Stat 0.0° 2.5° 5.0° 7.5° 10.0° 12.5° 15.0° 17.5° 20.0° 22.5° 25.0° 27.5° 30.0° 35.. 0° 40.0° 45.0° 50.0° 55.0° 60.0° 65.0° 70.0° 75 80 85 90 ,0° ,0° ,0° ,0° 0.500 0.486 0.473 0.460 0.447 0.435 0.424 0.412 0.400 0.389 0.377 0.365 0.353 0.328 0.303 0.276 0.249 0.220 0.191 0.160 0.129 0.097 0.065 0.033 0.000 -2.12 -2.12 -2.08 •1.91 •1.38 1.77 -2.08 -1.91 -1.38 1.77 -0.90 -0.93 -0.90 -0.87 -0.78 •0.59 •0.13 2.96 -0.90 -0.93 -0.90 -0.87 -0.78 -0.59 -0.12 2.94 -0.35 -0.33 -0.34 -0.31 -0.32 -0.38 -0.36 -0.38 -0.35 -0.36 -0.13 -0.13 -0.14 -0.14 -0.15 -0.14 -0.14 -0.15 -0.15 -0.16 0.84 -0.65 -0.56 -0.35 -0.31 Stat. Engy. - Statics matrix - Energy matrix 197 Table (V-III-25) Hemispherical Dome - Wind Load M-Force/Length, (N^) = -a5(pR) Fig. (V-III-6) y = 0.0; R/t = 100.0; 9 = 0 ° meridian (Fig. (V-III-1)) Percentage Error Memurane ricxurc E l a s t i c i t y 9e= 4>e 9e=<t>e 4 > Solution 0e=<t>e= 15° 9e= <|>e = 10° 9e=<t>e: =5° 0e= = <j>e= 2. 5° =5 o =2.5° Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0 0° 0 000 0.00 0. 00 0. 00 0.00 0.00 0 00 0 00 0 .00 0. 00 0.00 2 5° 0 027 2 61 2 .90 3.04 5 0° 0 051 0.51 1 08 1 35 1 .49 -0. 14 1.51 7 5° 0 072 0 92 1 .02 1.01 10 0° 0 090 -1 89 -1.96 0.07 0 39 0 72 0 .80 -0. 42 0.79 12 5° 0 106 0 61 0 .67 0.69 15 0° 0 119 -4.39 -4. 62 -0.17 0 08 0 51 0 .57 -0. 45 0.56 17 5° 0 130 20 0° 0 139 -1 40 -1.55 -0.15 0 07 -0. 45 22 5° 0 147 25 0° 0 153 -0.23 -0 03 -0. 54 27 5° 0 157 30 0° 0 160 -3.33 -3. 95 -1 79 -2.06 -0.37 -0 17 -0. 47 35 0° 0 163 40 0° 0 161 -1 93 -2.44 45 0° 0 155 -4.95 -6. 46 50 0° 0 146 -2 40 -3.39 55 0° 0 133 60 0° 0 119 -8.00 -13. 12 -3 10 -5.18 65 0° 0 102 70 0° 0 084 -6 67 -12.50 75 0° 0 064 -4.44 -12. 63 80 0° 0 043 -2 76 -11.60 85 0° 0 022 90 0° 0 000 Stat. - Statics matrix Engy. - Energy matrix 0 198 Table (V-III-26) Hemispherical Dome - Wind Load M-Force/Length, (NJ = -ct5(pR) Fig. (V-III-6) \i = 0.2; R/t = 100.0; 6 =0° meridian (Fig. (V-III-1)) Percentage Error -< Membrane Flexure Elasticity- e e -9e 6e = 9e 9 Solution 6e=9e= = 15° ee= :9e = 10° 0e = 9e: =5° ee= = 9e = 2. - o 5° =2.5° Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0 .0° 0 000 0.00 0.00 0 00 0.00 0.00 0 00 0 00 0 .00 0 .00 0.00 2 .5° 0 027 2 53 2 .79 2.86 5 .0° 0 051 0.55 1 08 1 31 1 .45 -0 .12 1.41 7 .5° 0 072 0 88 0 .99 0.93 10 .0° 0 090 -1 83 -1.90 0.10 0 40 0 69 0 .77 -0 .40 0.73 12 .5° 0 106 0 57 0 .64 0.62 15 0° 0 119 -4.29 -4.53 -0.14 0 09 • 0 44 0 .51 -0 .41 0.47 17 .5° 0 130 20 .0° 0 139 -1 39 -1.54 -0.12 0 08 -0 .41 22 .5° 0 147 25 .0° 0 153 -0.20 -0 01 -0 .38 27 .5° 0 157 30 .0° 0 160 -3.35 -4.00 -1 79 -2.07 -0.34 -0 16 -0 .35 35 .0° 0 163 40 0° 0 161 -1 96 . -2.49 45 .0° 0 155 -5.08 -6.65 50 .0° 0 146 -2 49 -3.52 55 .0° 0 133 60 .0° 0 119 -8.37 -13.70 -3 32 -5.47 65 .0° 0 102 70 .0° 0 084 -7 12 -13.15 75 .0° 0 064 -8.79 -17.78 80 .0° 0 043 -7 27 -16.95 85 .0° 0 022 90 .0° 0 000 Stat. - Statics matrix Engy. - Energy matrix 199 Table (V-III-27) Hemispherical Dome - Wind Load M-Force/Length, (N^ ,) = -a 5(pR) Fig. (V-III-6) y = 0.0; R/t = 100.0; 6 =30° meridian (Fig. (V-III-1)) Percentage Error Membrane Flexure E l a s t i c i t y Solution 9e=<f>e=15 6e=<j>e=10c 9e=4>e=5' 6e=4>e=2.5< 6e=<J>e =5° 6e=<f>e = 2.5C c*5 Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0.0° 2.5° 5.0° 7.5° 10.0° 12.5° 15.0° 17.5° 20.0° 22.5° 25.0° 27.5° 30.0° 35.0° 40.0° 45'. 0° 50.0° 55.0° 60.0° 65.0° 70.0° 75.0° 80.0° 85.0° 90.0° 0.000 0.024 .0.044 0.062 0.078 0.091 0.103 0.113 0.121 0.127 0.132 0.136 0.139 0.141 0.139 0.134 0.126 0.116 0.103 0.088 0.072 0.055 0.037 0.019 0.00 •4.75 •3.64 -5.30 •8.46 -5.16 0.00 -4.98 •4.i26 -6.82 •13.60 •13.38 0.00 -2.05 -1.54 -1.92 -2.08 -2,55 -3.29 -6.87 -3.42 0.00 -2-12 -1.69 -2.20 -2.59 -3.54 -5.37 •12.71 •12.30 0.00 0.41 0.06 -0.15 -0.24 -0.30 -0.39 0.00 0.63 0.04 -0.17 -0.19 -0.25 -0.30 0.00 2.58 1.33 0.92 0.71 0.59 0.53 0.00 2.84 1.47 1.01 0.78 0.67 0.58 0.00 -0.20 -0.44 -0.43 -0.46 -0.49 -0.39 0.00 2.97 1.67 1.20 0.90 0.72 0.59 Stat. - Statics matrix Engy. - Energy matrix 200 Table (V-III-28) Hemispherical Dome - Wind Load M-Force/Length, (N^) = -a 5(pR) Fig. (V-III-6) y = 0.2; R/t = 100.0; 0 = 30° meridian (Fig. (V-III-1)) Percentage Error < Membrane >- Flexure E l a s t i c i t y e e = 9e ee=<pe ( P Solution ee=ct>e= = 15° ee=<pe: = 10° ee=(pe =5° ee= = <pe= 2.5° 5° = 2.5° a5 Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0 0° 0 000 0.00 0.00 0.00 0.00 0.00 0 00 0 00 0 .00 0 .00 0.00 2 5° 0 024 2 50 2 .75 2.75 5 0° 0 044 0.45 0 97 1 27 1 .42 -0 .20 1.54 7 5° 0 062 0 87 0 .96 1.08 10 0° 0 078 -2.00 -2.07 0.09 0 39 0 67 0 .76 -0 .44 0.89 12 5° 0 091 0 55 0 .62 0.63 15 0° 0 103 -4.66 -4.89 -0.11 0 12 0 46 0 .53 -0 .43 0.50 17 5° 0 113 20 0° 0 121 -1.53 -1.68 -0.21 -0 01 -0 .45 22 5° 0 127 25 0° 0 132 -0.26 -0 08 -0 .45 27 5° 0 136 30 0° 0 139 -3.66 -4.31 -1.92 . -2.20 -0.35 -0 17 -0 .38 35 0° 0 141 40 0° 0 139 -2.11 -2.64 45 0° 0 134 -5.43 -7.00 50 0° 0 126 -2,63 -3.67 55 0° 0 116 60 0° 0 103 -8.84 -14.71 -3.51 -5.67 65 0° 0 088 • 70 0° 0 072 -7.33 -13.37 75 0° 0 055 -9.48 -18.51 80 0° 0 037 -7.97 -17.65 85 0° 0 019 90 0° 0 000 Stat. - Statics matrix Engy. - Energy matrix Table (V-III-29) Hemispherical Dome 201 - Wind Load M-Force/Length, (N^) = -a 5(pR) Fig. (V-III-6) p = 0.0; R/t = 100.0; 0 = 60° meridian (Fig. (V-III-1)) Percentage Error Membrane Flexure E l a s t i c i t y $* ee=*e Solution 9e=(f)e: = 15° ;?e = = 10° 0e=<j>e =5° = <J>e= =2.S o =5 0 =2,5° Stat Engy Stat Engy Stat Engy Stat Engy Stat Stat 0.0° 0 000 0.00 0.00 0 00 0.00 0.00 0 00 0 00 0. 00 0. 00 2.5° 0 014 2 42 2. 64 5.0° 0 026 0.08 0 63 1 25 1. 41 -0. 66 7.5° 0 036 0 86 0. 97 10.0° 0 045 -3 25 -3.31 -0.27 0 04 0 67 0. 76 -0. 73 12.5° 0 053 0 59 0. 65 15.0° 0 059 -6.95 -7.17 -0.40 -0 17 0 52 0. 59 -0. 64 17.5° 0 065 20.0° 0 070 -2 44 -2.58 -0.40 -0 19 -0. 62 22.5° 0 073 25.0° 0 076 -0.45 -0 25 -0. 61 27.5° 0 079 30.0° 0 080 -5.47 -6.11 -2 76 -3.03 -0.51 -0 30 -0. 61 35.0° 0 081 40.0° 0 080 -2 91 -3.42 45.0° 0 077 -7.19 -8.72 50.0° 0 073 -3 44 -4.44 55.0° 0 067 60.0° 0 059 -10.71 -15.87 -4 31 -6.38 65.0° 0 051 70.0° 0 042 -8 17 -14.03 75.0° 0 032 -8.19 -16.48 80.0° 0 022 -5 00 -13.89 85.0° 0 011 90.0° 0 000 Stat. - Statics matrix Engy. - Energy matrix 202 Table (V-III-30) Hemispherical Dome - Wind Load M-Force/Length, (N^J = -a 5(pR) Fig. (V-III-6) U -0.2; R/t = 100.0; 9 = 60° meridian (Fig. (V-III-1)) Percentage Error -<•—— : Membrane • Flexure E l a s t i c i t y 5° 5° 9e= <t>e ee-<j>e 4 > Solution 9e=<Pe=15° 9e= =<pe= 10° 6e=(pe= 0 e = =<Pe= = 2. = 5 O =2.5° <*5 Stat Engy. Stat Engy Stat Engy Stat Engy Stat Stat -0. 0° 0 000 0.00, 0.00 0. 00 0.00 0.00 0 00 0 00 0 .00 0. 00 2 5° 0 014 2 35 2 .57 5. 0° 0 026 0.08 0 63 1 21 1 .37 -0. 63 7 5° 0 036 0 83 0 .95 10. 0° 0 045 -3 18 -3.25 -0.25 0 07 0 65 0 ,73 -o. 71 12 5° 0 053 0 57 0 .65 15 0° 0 059 -6.84 -7.06 -0.39 -0 15 0 56 0 .61 -0. 62 17 5° 0 065 20 0° 0 070 -2 43 -2.58 -0.37 -0 17 -0. 60 22 5° 0 073 25 0° 0 076 -0.41 -0 24 -0. 63 27 5° 0 079 30 0° 0 080 -5.50 -6.15 -2 76 -3.03 -0.45 -0 27 -0. 61 35 0° 0 081 40 0° 0 080 -2 94 -3.47 45 0° 0 077 -7.31 -8.90 50 0° 0 073 -3 52 -4.56 -55 0° 0 067 60 0° 0 059 -11.08 -16.43 -4 53 -6.68 65 0° 0 051 70 0° 0 .042 -8 63 -14.67 75 0° 0 032 -12.38 -21.45 80 0° 0 022 -9 58 -19.26 85 0° 0 011 90 0° 0 .000 Stat. - Statics matrix Engy. - Energy matrix 203 Table (V-III-31) Hemispherical Dome - Wind Load Membrane Shear, (N ) yep a 6( PR) Fig. (V-III-7) y = 0.0; R/t = 100.0; 9 = 30° meridian (Fig. (V-III-1)) Percentage Error Membrane Flexure E l a s t i c i t y Solution 6e=<pe=10 6e=9e=5' 6e=9ef2.5 6e=4>e =5° 6e=<t>e =2.5C Stat | Engy Stat Engy Stat Engy Stat Engy Stat Stat 0.0° 2.5° 5.0° 7.5° 10.0° 12.5° 15.0° 17.5° 20.0° 22.5° 25.0° 27.5° 30.0° 35.0° 40.0° 45.0° 50.0° 55.0° 60.0° 65.0° 70.0° 75.0° 80.0° 85.0° 90.0° 0.333 0.312 0.293 0.275 0.259 0.244 0.229 0.216 0.204 0.192 0.181 0.170 0.160 0.142 0.125 0.109 0.095 0.081 0.069 0.056 0.045 0.033 0.022 0.011 0.000 -0.78 -0.31 -0.68 -0.70 0.03 •13.11 -0.80 -0.93 •1.15 -1.14 -0.57 -0.21 -0.31 -0.02 -0.17 -0.26 -0.29 -0.24 0.02 1.01 •11.99 -0.31 -0.32 -0.41 -0.47 -0.47 -0.41 -0.19 0.61 1.14 -0.24 -0.14 -0.17 -0.20 -0.25 -0.28 -0.24 -0.22 -0.24 -0.25 -0.28 -0.29 -0.06 -0.03 -0.03 -0.04 -0.04 -0.05 -0.06 -0.05 -0.06 -0.05 -0.05 -0.06 -0.22 -0.12 -0.22 -0.21 -0.24 -0.32 -0.06 0.05 0.03 -0.01 -0.04 -0.07 Stat. - Statics matrix Engy. - Energy matrix Table (V-III-32) Hemispherical Dome 204 Wind Load Membrane Shear, (N ) =a6(pR) Fig. (V-III-7) y = 0.2; R/t = 100.0; Q = 30° meridian (Fig. (V-III-1)) Percentage Error Membrane Flexure E l a s t i c i t y Solution 0e=<pe=15< 6e=cpe=10c 9e=(pe=5c 0e=<pe=2.5o 6e=<pe 0e=<Pe =2.5° Stat Engy Stat Engy Stat Engy Stat | Engy Stat Stat 0.0° ' 2.5° 5.0° 7.5° 10.0° 12.5° 15.0° 17.5° 20.0° 22.5° 25.0° 27.5° 30.0° 35.0° 40.0° 45.0° 50.0° 55.0° 60.0° 65.0° 70.0° 75.0° 80.0° 85.0° 90.0° 0.333 0.312 0.293 0.275 0.259 0.244 0.229 0.216 0.204 0.192 0.181 0.170 0.160 0.142 0.125 0.109 0.095 0.081 0.069 0.056 0.045 0.033 0.022 0.011 0.000 -0.79 -0.62 -0.93 -1.11 -0.54 -7.95 -0.79 -1.16 -1.42 -1.66 •1.30 3.87 -0.31 -0.15 -0.26 -0.39 -0.45 -0.43 -0.22 0.67 -6.52 -0.31 -0.40 -0.48 -0.61 -0.67 -0.67 -0.50 0.23 5.65 -0.23 -0.15 -0.17 -0.21 -0.24 -0.27 -0.23 -0.22 -0.23 -0.25 -0.28 -0.29 -0.06 -0.04 -0.03 -0.04 -0.04 -0.03 -0.06 -0.05 -0.05 -0.05 -0.05 -0.05 -0.21 -0.13 -0.22 -0.21 -0.23 -0.30 0.00 0.03 0.00 -0.03 -0.05 -0.06 -0.19 Stat. - Statics matrix Engy. - Energy matrix 205 Table (V-III-33) Hemispherical Dome - Wind Load Membrane Shear, (N f i A) = ag(pR) 60" y 0.0; R/t = 100.0; 6 Percentage Error Membrane Fig. (V-III-7) meridian [Fig. (V-III-1)) Flexure E l a s t i c i t y * Solution 0 .0° 0 577 2 .5° 0 541 5 .0° 0 508 7 .5° 0 477 10 .0° 0 449 12 .5° 0 422 15 0° 0 398 17 5° 0 374 20 0° 0 353 22 5° 0 332 25 .0° . 0 313 27 .5° 0 295 30 0° 0 278 35 .0° 0 246 40 0° 0 217 45 0° 0 190 50 0° 0 165 55 .0° 0 141 60 0° 0 119 65 .0° 0 098 70 0° 0 077 , 75 0° 0 057 80 0° 0 038 85 0° 0 019 90 0° 0 000 0e=<t>e=15 ;=9e=5< ee=(p( = 5° 9e=9e =2.5' Stat Engy Stat Engy Stat Engy Stat | Engy Stat Stat -0.88 -0.39 -0.68 -0.55 0.47 -11.77 -0.88 •1.01 -1.15 -0.99 -0.15 1.20 -0.36 -0.07 -0.19 -0.25 -0.24 -0.12 0.24 1.43 •11.32 -0.36 -0.37 -0.44 -0.46 -0.43 -0.30 0.03 1.01 1.87 -0.24 -0.15 -0.18 -0.20 -0.23 -0.26 -0.24 -0.33 -0.25 -0.25 -0.26 -0.27 -0.06 -0.03 -0.04 -0.03 -0.03 -0.03 -0.07 -0.06 -0.06 -0.05 -0.05 -0.04 -0.21 -0.12 -0.22 -0.20 -0.20 -0.25 Stat. - Statics matrix Engy. - Energy matrix 206 Table (V-III-34) Hemispherical Dome - Wind Load Membrane Shear, (N ) . 8(p a 5(pR) Fig. (V-III-7) y = 0.2; R/t = 100.0; 8 = 60° meridian (Fig. (V-III-1)) Percentage Error Membrane E l a s t i c i t y = 15° 9e 9e=9e Solution 8e=9e= 8e= = <pe= 10° 8e=<pe =5° 8e= :<pe; =2.5° = 5 O =2.5° .. a6 . Stat Engy Stat Engy Stat • Engy Stat Engy Stat Stat 0.0° 0 577 -0.88 -o 88 -0 36 -0 36 -0.23 -0 23 -0 06 -0.. 06 -0. 21 2.5° 0 541 -0 04 -0. 06 5.0° 0 508 . -0.17 -0 23 -0 04 -0. 05 -0. 14 7.5° 0 477 -0 03 -0. 05 10.0° 0 449 -0 19 -0 44 -0.18 -0 24 -0 03 -0. 05 -0. 23 12.5° 0 422 -0 03 -0. 04 15.0° 0 397 -0.70 -1 24 -0.21 -0 25 -0. 21 17.5° 0 374 20.0° 0 353 -0 28 -0 50 -0.23 -0 27 -0. 21 22.5° 0 332 25.0° 0 313 -0.26 -0 28 -0. 24 27.5° 0 295 30.0° 0 278 -0.93 -1 41 -0 39 -0 60 35.0° 0 246 40.0° 0 217 -0 40 -0 62 45.0° 0 190 -0.95 -1 51 50.0° 0 165 -0 31 -0 56 55.0° 0 141 60.0° 0 119 -0.09 -0 88 0 01 -0 28 65.0° 0 098 70.0° 0 077 1 09 0 64 75.0° 0 057 -6.61 5 28 80.0° 0 038 -5 87 6 37 85.0° 0 019 90.0° 0 000 Flexure Stat. - Statics matrix Engy. - Energy matrix 207 Table (V-III-35) Hemispherical Dome - Wind Load Membrane Shear, (N.J = a 6 (pR) Fig. (V-III-7) — — — — — — y (J) y = 0.0; R/t = 100.0; 6 = 90° meridian (Fig. (V-III-1)) Percentage Error Membrane Flexure E l a s t i c i t y =5° 6e= <j>e 0e=<f>e .Solution 9e=<f>e= 15° ee=<pe=10° .6e=<pe= 6e= :<pe= = 2.5 O =5 O = 2.5° Stat Engy, Stat Engy Stat Engy Stat Engy Stat Stat 0.0° 0 667 -1.28 -1.28 -0 57 -0. 57 -0.31 -0 31 -0 08 -0. 09 -0. 27 2.5° 0. 625 -0 04 -0. 06 5.0° 0 586 -0.18 -0 26 -0 04 -0. 06 -0. 15 7.5° 0 551 -0 03 -0. 05 10.0° 0 518 -0 .15 -0 45 -0.20 -0 26 -0 03 -0. 04 -0. 23 12.5° 0 487 -0 01 -0. 03 15.0° 0 459 -0.47 -1.10 -0.20 -0 25 -0. 19 17.5° 0 432 20.0° 0 407 -0 .19 -0 44 -0.20 -0 23 -0. 16 22.5° 0 384 25.0° 0 362 -0.19 -0 20 -0. 14 27.5° -0 341 30.0° 0 321 -0.52 -1.00 -0 .17 -0 38 35.0° 0 284 40.0° 0 250 -0 .07 -0 26 45.0° 0 219 -0.11 -0.56 50.0° 0 190 0 .15 -0 03 55.0° 0 163 60.0° 0 137 1.28 0.66 0 .63 0 44 65.0° 0 113 70.0° 0 089 1 .98 1 57 75.0° 0 066 -9.88 3.14 80.0° 0 044 -10 .81 2 42 85.0° 0 022 90.0° 0 000 Stat. - Statics matrix Engy. - Energy matrix 208 Table (V-III-56) Hemispherical Dome - Wind Load Membrane Shear, (NQ,) \x = 0.2; R/t a 6(pR) Fig. (V-III-7) 100.0; 6 =90° meridian (Fig. (V-III-1)) Percentage Error Membrane E l a s t i c i t y = 15° 10° =5° 0e= <(>e 0e = <pe P Solution = <Pe = =9e = e e E9e : e e = = 9e' =2. 5 ° =5 0 - -2.5° Stat Engy Stat Engy Stat •Engy Stat Engy Stat Stat 0 0° 0 667 -1 .30 -1.30 -0 58 -0.58 -0.30 -0 30 -0 08 -0 .08 -0. 26 2 .5° 0 625 -0 05 -0 .06 5 .0° 0 586 -0.20 -0 27 -0 04 -0 .06 -0. 16 7 5° 0 551 -0 03 -0 .05 10 0° 0 518 -0 27 -0.53 -0.20 -0 25 -0 03 -0 .04 -0. 24 12 5° 0 487 -0 02 -0 .04 15 0° 0 459 -0 78 -1.32 -0.20 -0 25 -0. 21 17 5° 0 432 20 0° 0 407 -0 29 -0.51 -0.20 -o 23 -0. 18 22 5° 0 384 25 0° 0 362 -0.18 -0 20 -0. 20 27 5° .0 341 30 0° 0 321 -0 78 -1/27 -0 31 -0.52 35 0° 0 284 40 0° 0 250 -0 23 -0.45 45 0° 0 219 -0 51 -1.08 50 0° 0 190 -0 04 -0.28 55 0° 0 163 60 0° 0 137 0 71 -0.09 0 41 0.12 65 0° 0 113 70 0° 0. 089 1 66 1.19 75 0° 0. 066 -4 73 7.22 80 0° 0 044 -5 49 6.79 85 0° 0 022 90 0° 0 000 Flexure Stat. - Statics matrix Engy. - Energy matrix 209 Table (V-IIi-37) Hemispherical Dome - Wind Load Bending Moments; M^ , M^  (Kip-inch/inch) At 6 = 0° Meridian; 6 = <J> =5° Element Size e Ye Radius = 100.0 inch; p = 0.001 Ksi E = 1000.0 K s i ; y = 0.2 Thickness = 1,0 i n . Thickness = 5.0 i n . 9° Me v M6 M4> 0° -0 .00006 0 .00000 -0 .00211 0 .00000 5° -0 .00002 0 .00034 -0 .00203 0 .00036 10° -0 .00007 0 .00001 -0 .00210 -0 .00039 15° -0 .00010 -0 .00014 -0 .00223 -0 .00131 20° -0 .00010 -0 .00013 -0 .00233 -0 .00198 25° -0 .00010 -0 .00010 -0 .00234 -0 .00232 30° -0 .0009 -0 .00009 -0 .00229 -0 .00239 35° -0 .00008 -0 .00008 -0 .00218 -0 .00229 40° -0 .00008 -0 .00008 -0 .00204 -0 .00212 45° -0 .00007 -0 .00007 -0 .00187 -0 .00192 50° -0 .00007 -0 .00007 -0 .00169 -0 .00171 55° -0 .00006 -0 .00006 -0 .00151 -0 .00151 60° -0 .00005 -0 .00005 -0 .00132 -0 .00131 65° -0 .00004 -0 .00004 -0 .00111 -0 .00093 70° -0 .00003 -0 .00004 -0 .00090 -0 .00086 75° -0 .00003 -0 .00002 -0 .00068 -0 .00063 80° -0 .00002 -0 .00000 -0 .00020 -0 .00017 85° -0 .00000 -0 .00000 -0 .00000 -0 .00000 90° I l l s gTC^irx- 8"), S P H E R I C AL'-DOMEE - W I N D , L O A D ifP'i+Ti+-i*!Ji^:U.|^;.!i!x-.: ;xj±!±ta MM '4±ff±l: spl: ^ a- 3-{-pR^Et -Table,! 3T-IE-38 -T-t-l-H. 211 Table (V-III-38) Hemispherical Dome - Wind Load R-Diaplacement = -a3(.pR / E t ) ; Fig. (V-III-8) At 6 = 0° Meridian; 6 = 9 =5° Element Size. e e y = 0.2; Radius = 100.0 i n . ; Thickness = t i n . F.E. Solution 013 E l a s t i c i t y Sol. Membrane Sol. Flexure Sol. «3 t=1.0, 5.0 i n t=1.0 i n . t= 5.0 i n . 0 1.0000 1.0786 1.0475 1.034 5 1.0614 1.0385 1.0646 1.0735 10 1.1062 1.1111 1.0990 1.1055 15 1.1343 1.1321 1.1309 1.129 20 1.1461 1.1461 1.1454 1.140. 25 1.1418 1.1414 1.1414 1.137 30 1.1220 1.1218 1.1215 1.119 35 1.0871 1.0871 1.0867 1.0855 40 1.0380 1.0381 1.0378 1.0375 45 0.9755 0.9756 0.9754 0.9755 50 0.09005 0.9006 0.9005 0.9005 55 0.8142 0.8142 0.8142 0.8143 60 0.7176 0.7176 0.7176 0.7175 65 0.6121 0.6120 0.6121 0.612 70 0.4989 0.4987 0.4989 0.499 75 0.3797 0.3794 0.3793 0.3795 80 0.2557 0.2552 0.2551 0.255 85 0.1287 0.1286 0.1287 0.1285 90 0.0000 0.0000 0.0000 0.0000 212 Plan - View FIGURE E-EZ-I. HEMISPHERICAL DOME SUBJECTED TO RADIAL LOAD ALONG THE ENTIRE EDGE. 213 5.6 Example IV Hemispherical Dome Subjected to Radial Load Applied Along  the Entire Edge. Fig. (V-IV-1) 35 In the e l a s t i c i t y solution of ttas problem allowing for flexure, the displacements, rotations, stresses, and moments are calculated for various values of (i) radius to thickness r a t i o (100.0 and 500.0)^and ( i i ) Poisson's r a t i o (0.0 and 0.2). These are shorn i n Figs. (V-IV-2 to 25). Severe displacement and stress gradients are noticed i n the base zone. These functions o s c i l l a t e exponentially and die out at some distance from the base. As the R/t r a t i o increases, the height of the zone gets smaller and gradients steeper. The s h e l l i s axisymmetrically loaded and i s s e l f supported. In view of the axisymmetry of the problem, i t i s s u f f i c i e n t i n the f i n i t e element analysis to consider only the elements bounded by a p a i r of adjacent meridian l i n e s . Statics type plane stress and flexure element s t i f f n e s s matrices are used i n the computation of re s u l t s . Due to the presence of high displacement and stress gradients, i t i s necessary to use small sized elements to obtain a f a i r representation of the displacements and stresses. Since the membrane and flexure stresses die out at a com-paratively short distance from the base, i t i s s u f f i c i e n t to analyse a model extending somewhat beyond the distance at which force-free boundary conditions may be applied. Percentage error i n the f i n i t e element solution using various element sizes for u = 0.0 and 0.2 and R/t = 100.0 and 500.0 i s calculated and shown i n Tables (V-IV-1 to 24). Excellent results are obtained i n the region of interest and remarkable convergence to the true solution i s noticed on reduction of the element s i z e . Higher percentage errors occur 214 at points where either the function changes sign or the value of the i i n c t i o n i s small compared to i t s value elsewhere. As expected, a higher percentage error i s observed i n stresses than i n displacements. For 6 = cp = 0.5° element size and radius to thickness r a t i o e e (R/t) of 100.0 the displacements, rotations, and stresses calculated by the F.E. solution d i f f e r from the e l a s t i c i t y solution by 2%. M. moment i s calculated with an^error of 5%. For R/t = 500.0, higher errors i n the order of 5% are observed. Table (V-IV-1) Hemispherical Dome - Radial Load.Along the Edge Horizontal Displacement, (£ ) = cx^pR/Et) U = 0.0; R/t = 100.0 (Fig. (V-IV-2)) E l a s t i c i t y Solution Percentage Error E l a s t i c t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9° 0 =cp e 0 e=cJ) e 9e_9e 9 • 0e=9e 8,e=9e 9e=9e 9 • 0e=9e 0e=9e 0e=9e "1 e =2° =1° =0:5° -<*1, =2° = 1° =0.5° =2° = 1° =0.5° 0.0 26.3215 - 1.67 -0.37 -0.11 8.5 -1.3761 0.79 17.0- -0.3664 1.93 1.92 0.5 23,3101 -0.12 9.0 -1.5669 2.41 0,79 17.5 -0.2879 1.96 1.0 20.3671 -0.30 -0.11 9.5 -1.6808 0.82 • 18.0 -0.2189 -2.88 1.32 1.97 1.5 17,5477 -0.11 10.0 -1.7309 7.38 2.19 0.86 18.5 -0.1591 1.89 2.-0 14,8935 -1.21 -0.26 -0.10. 10.5 -1.7293 0.91 19.0 -0.1080 -0r42 1.65 2.5 12,4331 -0.09 11.0 -1.6868 2.14 0.97 19.5 -0.0651 17-01 3.0 10,1850 -0.26 -0.08 11.5 -1.6128 1.04 20.0 -0.0295 -58.3 -9.95 -1.31 3,-5 8.1589 -0.07 12.0 -1,5156 6.11 2.16 1.10 20,5 -0:0008 -32.10 ' 4.0 6.3570 -1.91 -0.36 -0,07 12,5 -1.4024 1.19 21.0 0.0219 21,10 9.91 4,5 4,7761 -0,07 13.0 -1.2792 2.20 1.27 21.5 0.0393 7.39 5,0 3,4081 -0.70 -0.09 13.5 -1.1510 1.35 22.0 0.0521 36.54 10.72 6.80 5.-5 2,2416 -0.14 14.0 -1.0219 5.05 2.23 1.43 22.5 0.0610 6.69 6,0 1,2629 -10r52 -2.18 -0,31 14,5 -0.8953 1.52 23.0 0.0664 8.95 6.79 6.-5 0,4564 -1.09 15.0 -0.7736 • 2 ,24 1.61 23,5 0.0692 7.03 7.0 -0.1941 16.50 3.25 15.5 -0.6588 1.70 ' 24.0 0.0696 22.44 8.39 7.33 7.5 -0.7053 1.10 16.0 -0.5521 3.03 2.-16 1.78 24.5 0.0683 7.23 8.0 -1.0940 13.05 3.26 0.85 16.5 -0.4545 1.86 25.0 0.0656 8.24 8.18 Table (V-IV-2) Hemispherical Dome - Radial Load Along the Edge Horizontal Displacement, (5) = ai(pR/Et) U = 0.2; R/t = 100.0 (Fig. (V-IV-3)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9 ° c*i ee = <fe V * e 9° 9e=*e V * e 9 ° a l V ^ e V * e 6e =*e =2° =1° = 0.5° =2° = 1° =0.5° =2° = 1° =0.5° 0.0 26,0542 -1.69 -0.42 -0.16 8.5 -1.3149 0.49 17.0 -0.3912 2.04 1.88 0.5 23,1021 -0.15 9.0 -1.5167 2.23 0.54 17.5 -0.3113 1.94 1,0 20,2162 -0.34 -0.14 9.5 -1.6415 0.61 18.0 -0.2405 -1.72 1.60 1.97 1,5 17,4500 - -0.13 10.0 -1.7022 7.22 2.05 0.67 18.5 -0.1789 1.94 2,0 14,8431 -1.22 -0.28 -0.11 10.5 -1.7105 0.75 19.0 -0.1258 0.36 1.79 2,5 12,4238 -0.09 11.0 -1.6771 2.02 0.82 19.5 -0.0808 1.37 3,0 10,2099 -0.26 -0.06 11.5 -1.6112 0.90 20.0 -0.0432 -36.40 -4.99 0.11 3.5 8,2110 -0,04 12,0 -1,5211 5.99 2.06 0.98 20.5 -0.0125 -7.41 4,0 6,4298 -1.81 -0.32 -0,01 12.5 -1.4139 1.07 21.0 0.0121 32.96 14.87 4,5 4.8633 -- 0,02 13.0 -1.2956 2.12 1.16 21.5 0.0313 8.26 5,0 3,5041 -0.57 0.05 13.5 -1.1713 1.25 22.0 0.0458 40.12 11.22 7.14 5,5 2,3416 0,08 14.0 -1.0452 5.04 2.18 1.35 22.5 0.0561 6.86 6,0 1,-3627 -9.16 -1.69 0.10 14.5 -0.9206 1.43 23,0 0.0629 9.01 6.89 6.5 0.5526 0.05 15.0 -0.8001 2.22 1.53 23.5 0.0668 7.08 7,0 -0.1043 26.2 1.07 15.5 -0.6857 1.62 24.0 0.0683 22.87 8.38 7.37 7.5 -0.6239 0.44 16.0 -0.5788 3.27 2.19 1.71 24.5 0.0679 7.73 8.0 -1.0223 13.23 3.05 0.45 16.5 -0.4805 1.80 25.0 0.0659 8.23 7.90 Table (V-IV-3) Hemispherical Dome - Radial Load Along the Edge  Horizontal Displacement, (5) = aj.(pR/Et) V = 0.0; R/t = 500.0 (Fig. (V-IV-4)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E El a s t i c i t y Solution Percentage Error V<0e ee=*e ee =*e • a_ ee"*e 6e =*e 8e'*e . O 9e=4>e e e = f > e al =2° = 1° =0.5° =2° = 1° =0.5° ctl =2° = 1° =0.5° o-:-o 58,-8566 -8.36 -2.16 -0.53 8.5 -0.2512 4.24 17.0 -0.0070 38.17 9.59 0,-5 44,0317 -0,45 - 9,0 -0,0512 -93.9 -25,6 17.5 -0.0063 8.10 1,0 30,6669 -1.58 -0.42 9,5 0.0733 18,-71 18.0 -0.0053 25.82 6.83 1.5 19.5407 -0,52 10,0 0,1402 36.22 9,30 18.5 -0.0042 5.48 2,-0 10,8905 -9.39 -3.31 -0,88 10,5 0,1656 6.93 19,0 -0.0030 14.21 3.87 2,5 4,-6052 -2.15 11,0 0.1638 21.27 5,79 19,5 -0.0021 1.52 3,-0 0,-3768 -94.3 -2-7,35 11,5 0,1460 5,03 20,0 -0.0013 -6.53 -2.36 3,5 -2,1894 -- 4,-70 12,-0 0,1203 69.20 14.85 4,40 20.5 -0.0007 4,0 -3.4993 40.46 10.85 2.78 12,5 0.0927 3,77 21,0 -0.0002 4.5 -3.9242 2,21 13,0 0,0667 8.00 3.00 21.5 0.0000 5.0 -3,-7767 7.39 1.92 13,5 0.0444 1.92 22.0 0.0002 5,5 -3.3021 -- - • 1.70 14.0 0.0265 -35.5 -5.99 0.11 22,5 0.0003 6,0 -2.6807 25,13 5.51 1.49 14.5 0.0131 -3.83 23.0 0.0003 6.5 -2,0352 1.21 15.0 0.0038 -94.1 -20.9 23.5 0.0003 7,0 -1.4414 2.56 0.81 15.5 -0.0021 43.26 24.0 0.0002 7,5 '• -0,9392 0.16 16.0 -0.0054 71.48 16.61 24.5 0.0002 8.0 -0.5432 -17.09 -5.50 -1.09 16.5 -0,0068 11.81 25.0 0.0001 Table (V-IV-4) Hemispherical Dome - Radial Load Along the Edge Horizontal Displacement, (£) = c*i (pR/Et) j y = 0.2; R/t = 500.0 (Fig. (V-IV-5)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9 ° e =<i) e ve e Te 6 =<j> e re 9 ° 6 =<f> e Te 0 =d> e ye e =cb e Te 9 ° e Te 9 =9 e Ye 9 =9 e Te <*l =2° = 1° =0.5° a l =2° : = 1° • =0.5° =2° =1° =0.5° 0.0 58,2590 -8.22 -2.13 -0.54 8.5 -0.2918 -3.28 17.0 -0.0070 40.32 9.83 0.5 43,7273 -0,45 9,0 -0.0811 -63.8 -15,18 17.5 -0.0066 8.16 1,-0 30,-6028 -1.55 -0.42 9.-5 0.0535 24,71 18.0 -0.0056 27.16 6.79 1,-5 19,-6416 -0,51 10,0 0,1288 39.0 9.94 18,5 -0.0045 5.38 2.0 11,-0812 -9.02 -3.15 -0,84 10,5 0.1610 - - 7,13 19.0 -0.0034 16.37 3.71 2.5 4,8237 -1,96 11.0 0,1640 21.75 5,88 19.5 -0.0024 1.22 3,0 0,5789 -66.35 -16,99 11,5 0,1493 5.10 20.0 -0.0015 -33.12 -2.69 3,5 -2.0298 - - 4,85 12,0 0.1255 70.63 15.27 4.48 20.5 -0.0009 4.0 -3,3931 40.61 10.89 2,-77 12,5 0,0985 3,88 21.0 -0.0004 4.5 -3.8712 2.18 13.0 0.0724 8.95 3.20 •21.5 -0.0001 5.0 -3.7693 7.32 1.89 13.5 0.0495 2.27 22.0 0.0001 5.5 -3,3292 1.68 14.0 0.0308 -19.15 -2.79 0.80 22.5 0.0002 6,0 -2,7303 24.91 5.52 1.48 14.5 0.0164 -2.09 23.0 0.0003 6,5 -2,0962 - - . 1,2.3 15.0 0.0062 -60.24 -11.16 23.5 0.0003 7,0 -1,5048 2.80 0.87 15.5 -0.0005 24.0 0.0002 7,5 • -0.9984 -- 0.29 16.0 -0.0045 85.75 19.40 24.5 0.0002 8.0 -0.5942 -12.37 -4.24 -0.77 16.5 -0.0065 12.52 25.0 0.0001 Table (V-IV-5) Hemispherical Dome - Radial Load Along the Edge  P-Rbtation, (x) = -ot2(p/Et) y = 0.0; R/t = 100.0 (Fig. (V-IV-6)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9° ee=<fe V * e V<Pe 9° V * e V<f>e 0e=*e 9° ee=*e 8e=*e ee=*e ct2 ct2 =0.5° a 2 - =2° =1° =0,5° - =2° =1° =2° =1° =0.5° 0.0 346.409 -3.30 -0.73 -0.12 8.5 27.317 -0.10 17.0 -9.866 3.95 2.88 0,5 342,180 -0.08 9.0 17.660 -1.02 -0.47 17.5 -8.760 2.97 1,0 330,771 -0.40 -0,04 9.5 9.570 -1.43 18.0 -7.675 7.63 3.94 3.05 1,5 313.924 0,00 10.0 2.915 -37.5 -12.14 -6.42 18.5 -6.632 3.15 2,0 293.152 -1.04 -0.15 0,03 10,5 -2,442 9.54 19.0 -5.647 3.85 3.24 2,5 269,748 - - 0,06 11.0 -6.643 7.29 4.09 19.5 -4.731 3.35 3-.-0 244,803 0.05 0,10 11.5 -9.825 3.11 20.0 -3.892 4.93 3.60 3.45 3,5 219.221 - - - - 0,13 12,0 -12.124 12.55 4.65 2.73 20.5 -3.134 3.54 4,0 193,732 0.23 0.21 0.15 12,5 -13.665 2.57 21.0 -2.458 2.95 3.62 4,5 168.919 0.18 13.0 -14.569 4.09 2.50 21.5 -1.862 3.73 5, 0 145,225 0.33 0.20 13.5 -14.944 2.49 22.0 -1.345 -4.77 1.40 3.81 5.5 122,978 0.22 14.0 -14.891 9.68 3.93 2.51 22.5 -0.903 3.76 6.0 102.405 0.75 0.39 0.23 14.5 -14.499 2.54 23.0 -0.530 -4.28 3.60 6.5 83.645 0.23 15.0 -13.846 3.91 2.60 23.5 -0.221 2.67 7.0 66.-768 0.34 0.22 15.5 -13.002 2.65 24.0 0.030 19.67 7.5 51.785 0.17 16.0 -12.025 8.76 3.93 2.73 24.5 0.228 7.54 8.0 38.659 -0.43 0.06 0.08 16.5 -10.966 2.80 25.0 0.381 17.82 6.98 Table (V-IV-6) Hemispherical Dome - Radial Load Along the Edge  P-Rdtation, (x) = -ct 2(p/Et) JJ = 0.2; R/t = 100.0 (Fig. (V-IV-7)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9 ° a 2 V4>e ee =*e ee = <Pe 9 ° . 0e=*e ee =*e e e B *e 9 ° a 2 ee =*e V * e e e s *e - =2° = 1° =0.5° - =2° =1° =0.5° =2° = 1° =0.5° 0.0 339.410 -3.29 -0.77 -0.17 8.5 28.562 -0.06 17.0 -10.041 3.91 2.82 0.5 335.347 -0.13 9.0 18.914 -0.88 -0.36 17.5 -8.967 2.91 1.0 324.373 -0.44 -0.09 9.5 10.792 -1.10 18.0 -7.904 7.67 3.93 3.01 1.5 308,148 - - -0.06 10.0 4.073 -25.1 -8.12 -4.10 18.5 -6.876 3.09 2.0 288.114 -1.09 -0.20 -0.02 10,5 -1.373 15.35 19.0 -5.898 3.86 3.19 2.5 265.507 0,01 11.0 -5.681 8.04 4.38 19.5 -4.982 3.30 3.0 241.372 -0.01 0.04 11.5 -8.981 3.15 20.0 -4.138 5.30 3.67 3.40 3.5 216.578 0.08 12.0 -11.403 12.75 4.72 2.72 20,5 -3.370 3.50 4.0 191,829 0.15 0.15 0.10 12.5 -13.069 2.54 21.0 -2.681 3.16 3.58 4,5 167,688 0.13 13.0 -14.095 4.07 2.46 21.5 -2.070 3.66 5.0 144.588 0.27 0.16 13.5 -14.587 -- • • 2,44 22.0 -1.535 -2.61 1.95 3.76 5,5 122.850 0,18 14.0 -14.643 9.58 3.88 2.45 22.5 -1.074 3.76 6.0 102,699 0.68 0.33 0,19 14.5 -14.349 2.49 23.0 -0.682 -1.92 3.61 6,5 84,277 0.20 15.0 -13.784 3.86 2.54 23.5 -0.354 2.94 7,0 67,657 0.29 0,19 15.5 -13.016 2.60 24.0 -0.084 0.95 -51.4 -2.38 7,5 52,856 - - 0.16 16.0 -12.104 8.67 3.88 2.66 24.5 0.133 10.00 8.0 39.846 -0.34 0.05 0.08 16.5 -11.098 2.74 25.0 0.302 20.8 8.15 Table (V-IV-7) Hemispherical Dome - Radial Load Along the Edge  P-Rdtation, (x) = -ct 2(p/Et) u = 0.0; R/t = 500.0 (Fig. (V-IV-8))) E l a s t i c i t ) Solution Percentage Error E l a s t i c i t ) Solution Percentage Error E l a s t i c i t ) Solution Percentage Error 9° ct2 ee-9.e ee=?e ee~9e 9° a 2. ee-9e 6e-9.e 0e~9e 9° a 2 6e-9e 9e =9e 6e=9e =2° =1° =0.5° - =2° - = 1° =0.5° =2° = 1° =0.5° o.O 1732.046 -16.02 -4.27 -1.06 8.5 -28.172 1.59 17.0 -0.036 55.8 0.5 1636.122 -0.62 9.0 -18.479 2.60 0.97 17.5- -0.105 19.91 1.0 1411.794 -1.24 -0.31 9.5 -10.794 -0.23 18.0 -0:135 53.00 14.27 1.5 1133.527 -0.09 10.0 -5.095 -46.70 -14.70 -3.21 18.5 -0.139 11.77 2.0 851.378 3.18 0.30 0.06 10.5 -1.165 -21.01 19.0 -0.127 35.24 10.19 2.5 595.696 0.13 11.0 1.305 82.43 21.40 19.5 -0.107 8.99 3.0 381.935 0.55 0.08 11.5 2.650 10.52 20.0 -0.085 ,24.25 7.64 3.5 215.068 -0.19 12.0 3.184 29.06 7.98 20.5 -0.062 6.33 4.0 93.284 -2.26 -4.12 -1.27 12.5 3.179 6.79 21.0 -0.042 10.80 4.72 4.5 10.899 -16.30 13.0 2.852 20.66 6.00 21.5 -0.026 1.59 5.0 -39.506 19.35 5.38 13.5 2.365 5.36 22.0 -0.014 -24.14 -3.29 5.5 -65.633 3.40 14.0 1.832 74.15 14.72 4.69 22.5 -0.005 6.0 -74.569 42.17 10.40 2.86 14.5 1.327 3.89 23.0 0.001 6.5 -72.308 • 2.59 15.0 0.889 5.94 2.78 23.5 0.004 7.0 -63.577 8.64 2.41 15.5 0.536 0.88 24.0 0.006 7.5 -51.866 2.22 16.0 0.270 -90.54 -20.44 -3.26 24.5 0.006 8.0 -39.568 34.89 6.77 1.97 16.5 0.084 -19.93 25.0 0.006 Table (V-IV-8) Hemispherical Dome - Radial Load Along the Edge P-Rotation, (x) = -a 2(p/Et) y = 0.2; R/t = 500.0 (Fig. (V-IV-9)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9° ct2 ee-*e 9e=*e ee=+e a 2 ee =+e w ee=4>e 9° cx2 ' ee a+e e e a *e ee=0>e =2° = 1° =0.5° - =2° =1° =0.5° =2° = 1° =0.5° 0.0 1697.052 -15.74 -4.20 -1.06 8.5 -29.426 1.64 17.0 -0.001 0.5 1604.798 -0.62 9.0 -19.698 3.12 1.10 17.5 -0.084 24.66 1.0 1388,463 -1.24 -0,32 9.5 -11.873 0.09 18.0 -0.125 58,16 15;70 1.5 1119.182 -0.11 10.0 -5.977 -31.75 -10.33 -2,24 18.5 -0.137 12.78 2-.-0 845.046 • 2.94 0.26 0.04 10.5 -1.832 -12,15 19.0 -0.131 36.95 10.90 2,-5 595,467. 0,11 11.0 0.845 31.28 19.5 -0.114 9.74 3,0 385,674 0.55 0.08 11.5 2.371 11.43 20.0 -0.092 26.15 8.65 3.5 220,837 -0.17 12.0 3.050 30.20 8.28 20.5 -0.070 7.42 4,0 99.560 -1.26 -3.52 -1.10 12.5 3,154 6.92 21.0 -0,049 14.17 5.87 4,5 16.639 -10,03 13,0 2.899 21.0.7 6.11 21.5 -0.032 3.29 5.0 -34,902 20.78 5,79 13.5 2.456 5.46 22.0 -0.019 -12.13 -2.66 5,5 -62.403 3.44 14.0. 1.943 76.16 15.33 4.83 22.5 -0.009 6,0 -72,689 41.60 10.35 2.85 14.5 1.440 4.11 23.0 -0.002 6,5 -71,590 2.57 15,0 0.993 7.60 3.13 23.5 0.003 7,-0 -63,754 8.57 2,39 15.5 0.625 1.57 24.0 0.005 7,5 -52,650 2.21 16.0 0.340 -66.54 -12.36 -1.46 24.5 0.006 8.0 -40.695 -34.49 6.83 1.98 16.5 0.135 -10.55 25.0 0.006 Table (V-IV-9) Hemispherical Dome - Radial Load Along the Edge  P-Fbrce/Length, (NQ) = 0.3 (p) y = 0.0; R/t = 100.0 (Fig. (V-IV-10)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9° ee=0»e ee=<f>e 6e =*e 9° ee =*e 9e =*e ee=*e - * ° w 9e =*e V * e a3.. =2° = 1° =0 .5° a3 =2° = 1° = 0 , 5 ° =2° = 1° = 0 . 5 ° 0.0 26.321 -2.05 -0.48 -0.15 8.5 -1.391 -0.31 17.0 -0.383 3.67 2.35 0.5 23.311 -0.06 9.0 -1.586 -0.88 -0.02 .17.5 -0.302 , 2.51 1,0 20.370 : 0,09 -0.01 9.5 -1.704 0.19 18.0 -0.230 8.79 4.12 2.65 1,5 17.554 0.05 10.0 -1, 758 -0.98 0.18 0.36 18.5 -0.168 2.80 2,0 14.903 1.82 0.59 0.12 10.5 -1.759 0.52 19.0 -0.114 4,63 2.90 2.5 12.445 0.19 i i . o -1.718 0.87 0.66 19.5 -0.069 • ' 2.91 3.0 10.199 1.17 0.28 11,5 -1.646 0.80 20.0 -0.031 6.54 5.49 2.49 3.5 8.174 0.38 12.0 -1.549 2.99 1.43 0.93 20.5 -0.001 -40.70 4.0 6.373 6.93 1.93 0.51 12.5 -1.436 1.06 21.0 0.023 4.60 5.84 4.5 4.791 0.67 13.0 -1.313 1.91 1.20 21.5 0.042 . 5.43 5.0 3.421 3.21 0.89 13.5 -1.184 1.33 22.0 0.056 14.59 5.50 5.50 5.5 2.252 1.24 14.0 -1.053 5.50 2.37 1.47 22.5 0.066 5.76 6.0 1.270 25.50 6.86 1.95 14.5 -0.925 1.61 23.0 0.072 6.06 6.07 6.5 0.459 4.57 15.0 -0.801 2.80 1.75 23.5 0.075 6.46 7,0 -0.196 -31.2 -8.64 15.5 -0.684 1.90 24.0 0.076 14.93 6.60 6.90 7.5 -0.711 -1.78 16,0 -0.574 7.42 3.23 2.05 24.5 0.075 7.38 8.0 -1.105 -13.39 -3.23 -0.76 16.5 -0.474 2.20 25.0 0.072 7.13 7.90 Table (V-IV-10) Hemispherical Dome - Radial Load Along the Edge P-Force/Length, (NJ = 0 3 (p) y = 0.2; R/t = 100.0 (Fig. (V-IV-11J E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9 ° V * e V O e ee=*e 9 ° 8e=*e V ^ e 6e =*e 9° ee=*e V * e 9e=*e <*3 =2° =0.5° °<3 =2° = 1° =0.5° °<3 =2° = 1° =0.5° 0.0 26.054 -2.11 -0.58 -0.25 8.5 -1.335 -0.65 17.0 -0.409 3.60 2.27 0.5 23.104 -0.10 9.0 -1.541 -1.15 -0.29 17.5 -0.326 2.43 1.0 20.221 0.04 -0.04 9.5 -1.670 -0.04 18.0 -0.253 8.65 4.09 2.58 1.5 17.458 0.03 10.0 -1.734 -1.34 -0.02 0.16 18.5 -0.188 2.73 2,-0 14,-854 1.72 0.54 0.10 10,5 -1,-745 0.34 19.0 -0.133 4.66 2.85 2.-5 12.437 0.19 11.0 -1.713 0.71 0.50 19.5 -0.085 2.91 3.0 10.225 1.12 0.29 11.5 -1.648 0.65 20.0 -0.045 8.12 5.63 2.73 3,-5 8.226 0.40 12.0 -1.559 2.70 1.29 0.79 20.5 -0.013 0.60 4.0 6,-445 6.68 1.88 0.54 12.5 -1.451 0.93 21.0 0.014 2.73 7.19 4,5 4,876 - 0,73 13.0 -1.333 1.79 1.07 21.5 0.034 5.67 5,0 3,515 3.14 0.98 13.5 -1.207 1.22 22.0 0.050 14.25 5.09 5.59 5.5 2, 349 1.38 •14.0 -1.079 5.23 2.25 1.36 22.5 0.061 5.78 6.0 1.366 23.6 6.54 2.16 14.5 -0.952 1.50 23.0 0.069 5.79 6.06 6.5 0.551 4.68 15.0 -0.830 2.70 1.65 23.5 0.073 6.42 7.0 -0.110 -58.4 -19.7 15.5 -0.712 1.80 24.0 0.075 14.54 6.38 6.83 7.5 -0,635 -2.75 16.0 -0.603 7.18 3.15 1.95 24.5 0.075 7.29 8.0 -1.038 -14.7 -3.81 -1.26 16.5 -0.501- 2.11 25.0 0.073 6.96 7.78 1 1 i i n 1 i I 1 1 i nr - 1 1 1 i 1 " P 1 I _i i ; 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I 1 1 1 1 i I 1 I i I 1 L 1 1 1 | i 1 l _ _ U _ U I 1 i 1 1 i i 1 _J 1 1 Table (V-IV-11) Hemispherical Dome - Radial Load Along the Edge  P-Force/Length, ( N J = ct3(p) y = 0.0; R/t = 500.0 (Fig. (V-IV-12)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t ) Solution Percentage Error E l a s t i c i t y Solution Percentage Error <j>0 a 3 . ee = <Pe ee=*e V * e 8e=*e ee=*e 6e=<Pe V * e V * e = 2 ° = 1° = 0 . 5 ° = 2 ° = 1 ° = 0 . 5 ° = 2 ° = 1 ° = 0 . 5 ° 0.0 5 8 , 8 5 6 6 - 9 . 2 5 - 2 . 5 8 - 0 , 7 1 8,5 - 0 , 2 5 4 0 1.9.9 1 7 . 0 - 0 . 0 0 7 3 3 0 . 2 3 7 . 6 6 0 , 5 4 4 , 0 3 3 4 0 . 0 9 9.0 - 0 . 0 5 1 9 0 . 5 3 - 0 . 5 2 1 7 . 5 - 0 . 0 0 6 7 7 . 0 7 1.0 3 0 . 6 7 1 6 2 . 6 4 0 , 7 7 9.5 0 , 0 7 4 4 5 , 3 2 1 8 . 0 - 0 . 0 0 5 6 2 5 . 0 1 6 . 4 5 1,5 1 9 . 5 4 7 4 1 , 5 7 1 0 , 0 0 . 1 4 2 3 4 3 . 9 8 1 4 . 8 5 4 . 2 9 1 8 , 5 - 0 . 0 0 4 4 5 . 8 4 2 . 0 1 0 , 8 9 7 1 3 5 . 4 2 1 0 . 6 2 2 , 7 2 1 0 . 5 0 , 1 6 8 4 4 . 1 5 1 9 , 0 - 0 . 0 0 3 2 1 9 . 3 8 4 . 8 9 2 , 5 4 . 6 0 9 6 5 , 2 8 1 1 . 0 0 , 1 6 6 9 1 4 . 3 3 4 , 1 4 1 9 . 5 - 0 . 0 0 2 2 3 . 4 4 3,0 0,-3773 4 6 , 3 1 1 1 , 5 0 , 1 4 8 9 4 . 1 5 2 0 . 0 - 0 . 0 0 1 3 5 2 . 3 7 7 . 1 8 0 . 2 2 . 3 ,5 - 2 . 1 9 3 5 - 4 . 8 9 1 2 , 0 0 . 1 2 3 0 6 5 . 4 5 1 4 . 0 5 4 . 1 6 2 0 , 5 - 0 . 0 0 0 7 4 , 0 - 3 , 5 0 7 8 - 3 7 . 3 - 6 . 6 6 - 1 . 4 5 1 2 , 5 . 0 . 0 9 4 9 4 , 1 1 2 1 . 0 - 0 . 0 0 0 2 4,-5 - 3 , 9 3 6 4 - 0 , 2 3 1 3 . 0 0 , 0 6 8 4 1 2 . 7 1 3 , 9 9 2 1 . 5 0 . 0 0 0 0 5 , 0 - 3 , 7 9 1 1 1 , 2 0 0,-47 1 3 . 5 0 , 0 4 5 6 3 , 7 1 5 . 5 - 3 , 3 1 7 4 0 , 9 6 1 4 . 0 0 v 0 2 7 3 5 4 . 2 8 7 . 7 4 3 , 1 0 6,0 - 2 , 6 9 5 5 1 6 . 7 7 4 . 7 4 1 . 3 4 1 4 . 5 0 . 0 1 3 6 1 . 6 1 6.5 - 2 . 0 4 8 4 - 1 . 6 5 1 5 . 0 0 , 0 0 4 0 - 3 6 . 3 6 - 5 . 5 8 7.0 - 1 . 4 5 2 2 7 . 0 0 1 , 9 0 1 5 , 5 - 0 , 0 0 2 2 2 1 . 4 5 7,5 - 0 , 9 4 7 3 2 , 0 9 1 6 . 0 - 0 , 0 0 5 6 4 4 . 8 0 1 0 . 6 2 8.0 - 0 . 5 4 8 6 4 6 . 8 0 8 . 2 1 2 . 1 7 1 6 . 5 - 0 . 0 0 7 1 8 . 6 5 Table (V-IV-12) Hemispherical Dome - Radial Load Along the Edge  P-Fbrce/Lerigth, (N Q) = ct 3(p) y = 0.2; R/t = 500.0 (Fig. (V-IV-13)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error A 3 8 e - *e- fle=+e 6e=*e •° ee = <Ce 8e =*e 8e=*e A 3 8e =*e ee=*e 8e=*e =2° =i° =0~5° =2° = 1° =0.5° =2° =1° =0.5° 0,0 58,2590 -9,09 -2.54 -0.71 8.5 -0.2948 2.07 17.0 -0.0073 30.96 7.83 0.5 43,7299 0,07 9,0 -0,0818 4.09 0.83 17.5 -0.0069 7.04 1.0 30.6083 2.53 0,74 9,5 0,0545 5.84 18.0 -0.0059 25.34 6.50 1,5 19,6484 1,51 10,0 0,1311 41.15 14.75 4,-24 18.5 -0.0048 5.80 2,0 11,0871 33.91 10.17 2.61 10,5 0.1640 4,06 19.0 -0.0036 20.33 4.90 2,5 4,8267 4.98 11,0 . 0.1673 14.05 4.04 19,5 -0.0025 3.35 3,C 0,5775 30,30 11.5 0.1525 - 4,07 20.0 -0.0016 67.15 10.42 0.12 3.5 -2,0360 -5.39 12.0 0.1284 63.27 13.83 4.08 20.5 -0.0009 4,0 -3,4039 -38.91 -7.14 -1.60 12,5 0.1010 4,05 21.0 -0.0004 4,-S -3,8854 -0,33 13.0 0.0744 12.75 3.97 21.5 -0.0001 5,0 -3,7856 0.93 0.39 13.5 0,0509 3.74 22.0 0.0001 5.5 -3.3461 0,88 14.0 0,0317 55.91 8.77 3.25 22.5 0.0003 6,0 -2,7464 15.57 4.48 1,27 14.5 0.0170 2.18 23.0 0.0003 6.5 -2.1105 1,58 15.0 0,0064 -16.29 -1.67 23.5 0.0003 7.0 -1.5164 6.73 1.84 15.5 -0.0006 69.48 24.0 0.0003 7,5 -1.0072 2.03 16.0 -0.0047 50.76 11.80 24.5 0.0002 8.0 -0.5999 44.12 8.01 2.15 16.5 -0.0067 8.95 25.0 0.0002 Table (V-IV-13) Hemispherical Dome - Radial Load Along the Edge M-Fbrce/Lerigth, (N^) = cti+Cp) V = 0.0; R/t = 100.0 (Fig. (V-IV-14)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9 ° ee =*e 8 e = * e 9 ° 6e=*e ee =*e V * e 9 C ee=4>e V * e e e = * e =2° = 1° =0.5° - =2° = 1° =0.5° =2° = 1° - =0.5° 0,0 0.0000 0.00 0.00 0,00 8.5 -0.0276 1.05 17 0 -0.0002 -64.7 -9.5 0,-5 0,0068 -1.05 9,0 -0,0272 1.55 1.11 17 5 0.0007 6.22 1,0 0.0103 0.07 -0,68 9.5 -0.0263 1.17 18 0 0.0015 36.91 11.84 3.82 1.5 0,-0112 -0,65 10,0 -0,0250 3.68 1.68 1.23 18 5 0.0021 3.64 2,-0 0,-0100 1.79 0.19 -0,-71 10,5 -0,0235 1.28 19 0 0.0026 8.63 3.63 2,5 0,0073 -1,14 11,0 -0.0216 1.76 1.34 19 5 0.0030 3.42 3,0 0,0036 -0.11 -2.65 11.5 -0 ,0197 1.41 20 0 0.0032 19.32 7.45 3.42 3.5 -0,0007 17,75 12,0 -0,0176 3.37 1.75 1.44 20 5 0.0034 3.57 4.0 -0,0052 -0.32 1.07 2,21 12.5 -0.0154 1.52 21 0 0.0034 7.13 3.49 4,5 -0,0097 1.37 13.0 -0.0133 1.71 1.58 21 5 0.0034 3.55 5.0 -0,0139 1.02 1.06 13.5 -0.0112 1.59 22 0 0.0033 15.73 6.82 3.24 5,5 -0.0177 0.98 14.0 -0.0092 1.23 1.36 1.66 22 5 0.0032 3.13 6.0 -0.0209 2.35 1.15 0.94 14.5 -0.0073 1.71 23 0 0.0030 6.88 3.15 6.5 -0.0235 0.95 15.0 -0.0056 0.52 1.68 23 5 0.0028 2.78 7.0 -0.0255 1.29 0.95 15.5 -0.0040 1.52 24 0 0.0025 13.62 6.89 2.24 7.5 -0.0268 0.99 16.0 -0.0026 -12.03 -2.07 1.29 24 5 0/0023 1.65 8.0 -0.0275 3.26 1.43 1.04 16.5 -0.0013 0.85 25 0 0.0020 6.78 0.83 Table (V-IV-14) Hemispherical Dome - Radial Load Along the Edge M-Force/Length, (N^) = (p) 0.2; R/t = 100.0 (Fig. (V-IV-15)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9° ee=*e ee =*e 9° 9e =*e 9e =*e 9 ° 9e =*e 9e =*e 9e =*e =2'° =1° =0.5° =2° = 1° =0.5° =2° = 1° =0,5° 0.0 0.0000 0.00 0.00 0.00 8.5 -0.0279 1.02 17.0 -0.0006 -20.89 -0.71 0,5 0.0069 -0.76 9.0 -0.0276 1.48 1.07 17.5 0.0004 6.74 1.0 0.0104 0.16 -0.40 9.5 -0.0268 1.12 18.0 0.0013 43.5 14.29 4.29 1,5 0,0113 -0,39 10.0 -0.0256 3.57 1.59 1.20 18.5 0.0019 3.56 2,0 0,-0102 1.78 0.26 -0.43 10,5 -0,0241 1,27 19.0 0.0025 8.75 3.35 2.5 0.0075 -0,64 11.0 -0.0223 1.66 1,31 19.5 0.0029 3.65 3,0 0,0039 0.54 -1.27 11.5 -0.0204 1.40 20.0 0.0032 19.56 7.57 3.37 3.5 -0,0004 14.15 12.0 -0.0183 3.39 1-74 1,45 20.5 0.0033 3.46 4,0 -0,0049 -0.93 0.77 1,48 12.5 -0.0162 1,55 21,0 0,0034 7.01 3.28 4.5 -0.0094 1.02 13,-0 -0,-0140 1.68 1,56 21,5 0,0034 3.55 5,-0 -0.0137 0.82 0.81 13.5 -0,-0119 1.55 22.0 0.0034 15.86 6.95 3.40 5,5 -0,0175 0,80 14.0 -0.0099 1.53 1.45 1.63 22.5 0.0033 3.22 6.0 -0.0208 2.16 1.01 0,80 14.5 -0,0080 1,82 23.0 0.0031 6.90 3.10 6,5 -0,-0235 0.80 15.0 -0.0062 0.83 1.78 23.5 0.0029 3.00 7,0 -0,0255 1.18 0,87 15,5 -0.0045 1.85 24.0 0.0027 13.99 6.98 2.69 7.5 • -0.0269 0.89 16,0 -0.0030 -8.98 -1.25 1.84 24.5 0.0024 1.84 8.0 -0.0277 3.09 1.30 0.95 16.5 -0.0017 1.28 25.0 0.0022 6.73 1.54 i j -H— l._ i -i- - H i ' i i _ I l "TIT"' I - - • \-A-_! .L_ \ 1 l j ... i I i I I - 1 i i 1 i i l I ! 1 " i 1 | i - < (1 •\ r~ • L±Q-_L i—rj i. i 1 ^ 1 i I i 1 1 ! 1 - l 1 j ..ii I - i- I j 1 1 i n 1 L . 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K M . 1 i U <T> i • i o i • • UJ I -i O ) 11 1 rr 1 1 i 5, V i l ko I i j—7^  i 1' 1 c L 1 i 1 1 F rn 1 i 1 II i • (|) 1 I r 1 c i i \ • i * / v. 1 - LU 01 ! i I - Pi- h+ i-jO 1 1 I 1 1 o 1 t/)-l l] 1 1 l .{£>- £.[.> >-| II 1 1 i ! 1 ^ N ! 1 1 1 1 1 \\\ L t> iTi d I i i 1 1 1 -1 1 I I i i 1 | J. - i 1 I 1 / \ r i 1 1 1 1 / 1 1 1 1 I t 1 1 / I i 1 1 1 i i i ! 1 i 1 ( i i 1 J._!.. s.i I! ! L ! i 1 J i" -}-, n i V -r > - u r D I 1 L b V \ I I i M r i | \ ! 1 V I , i a .H - J \ r A V r11 I \ j 7i i ! 1 i i \ I - n 1 i \ 1 1 i n k 1 1 n J 1 I j \ V. ! i* . ! ! 1 i 1 1 I 1 I I ! i 1 1 -- --i 1 1 I 1 I i i j 1 i 1 i - i j 1 | i 1 i i i i I 1 | 1 | A i 1 - i i i i I I i i i ! 1 1 . I 1 i i 1 i i t i 1 1 1 1 . - i 1 1 ...Li 1 -i 1 1 1 1 1 -I i ...L - 1 1 i i - i i i i 1 • .I. 1 l i i 1 1 1 i l i i i I i ! ! 1 1 Table (V-IV-15) Hemispherical Dome - Radial Load Along the Edge M-Force/Length, (N^) = a^ Cp) u = 0.0; R/t = 500-0 (Fig. (V-IV-16)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9° 6e-9e 9e~9e 6e~9e 9° ee~9e- 9e=9e 6e=9e =2° =1° =0.5° =2° =1° =0.5° 0.0 0.0000 0.00 0.00 0.00 8.5 0.0011 -2.10 0,5 0.0048 -1.66 9.0 0.0014 16.95 -2.27 1,0 0.0040 2.40 -1.74 9.5 0.0015 -3.13 1,-5 0,0003 -34.23 10.0 0.0014 67.23 12.19 -3.58 2,0 -0,0042 -5.93 0.97 2.72 10.5 0.0012 -5.55 2.5 -0.0082 T.70 11.0 0.0010 6.98 -8.02 3.-0 -0.0109 2.97 1.69 11.5 0.0007 -11.81 3,5 -0.0121 1.72 12.0 0.0005 23.80 -2.00 -18.60 4.0 -0.0121 15.78 3.74 1.75 12,5 0.0003 4.5 -0.0110 1.83 13.0 0.0002 5,0 -0,0093 3.66 1.89 13.5 0.0000 5,5 -0.0072 2.07 14.0 0.0000 6,-0 -0,0051 14.22 2.12 2.37 6,-5 -0,0032 2.81 7, 0 - -0.0015 -7.55 4.06 7.5 -0.0003 12.67 8.0 0.0006 40.17 -4.75 Table (V-IV-16) Hemispherical Dome - Radial Load Along the Edge  M-Force/Length, (N^) = (^(p) y = 0.2; R/t = 500.0 (Fig. (V-IV-17) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error an ee=*e ee=4>e ee = <fe <f>° V * e 9e =*e °e=*e - =2° =1° =0.5° - = 2° = 1° =0.5° 0.0 0.0000 O.00 0.00 0.00 8.5 0.0011 -1.02 0.5 0.0048 -1.24 9.0 0.0014 18.13 0.07 1,0 0.0041 2.39 -1.21 9.5 0.-0015 -0.20 1.5 0.0004 -13.33 10.0 0.0014 67.66 13.33 -1.21 2.0 -0.0041 -6.80 0.41 2.12 10.5 0.0012 -2.26 2,-5 -0.0081 1,56 11.0 0.0010 9.02 -4.71 3.0 -0.0109 2.74 1.54 11.5 0.O008 -7.05 3.5 -0.0122 1.59 12.0 0.0005 30.73 3.09 -10.36 4.0 -0.0123 15.17 • 3.56 1.63 12.5 0.0004 4.5 -0.0113 1.71 13.0 0.0002 5.0 -0.0096 3.57 1.84 13.5 0.0001 5.5 -0.0075 2.01 14.0 0.0000 6.0 -0.0054 14.03 1.96 2.09 6.5 -0.0034 2.32 7.0 -0.0018 -6.40 2.58 7.5 -0.0005 4.04 8.0 0.0005 48.12 -2.50 Table (V-IV-17) Hemispherical Dome - Radial Load Along the Edge P-M6merit/Lerigth, (M^ = a5(pR/100) u = 0.0; R/t = 100.0 (Fig, (V-IV-18)) E l a s t i c i t y Solution Perc .entage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9° 6e=*e 8e-*e fle=*e 9° as ee=*e ee =*e ee=*e 9° 6e = <Pe ee=<Pe ee = (Pe =2° =1° =0.5° - =2° =1° =0.5° = 2° = 1° =0.5° O.C 0.0000 0.00 0.00 0.00 8.5 1.0008 0.46 17.0 -0.1058 4.82 2.37 0.5 0.7763 0.05 9.0 0.8453 0.59 0.51 17.5 -0.1050 2.45 1,0 1,3750 0.07 0,02 9.5 0,7019 0.56 18.0 -0.1019 12.80 4.67 2.55 1.5 1.8185 - 0.03 10,0 0.5713 -0.32 0.48 0.60 18,5 -0.0970 2.65 2,0 2,1282 0.62 0.16 0.04 10,5 0.4541 0,64 19,0 -0.0909 4.64 2.76 2,5 2.3243 0.05 11,0 0.3503 0.16 0.68 19.5 -0.0839 2.87 3.C 2,-4255 0.25 0,07 11,5 0,2596 0.69 20.0 -0.0763 10.54 4.65 2.99 3.5 2.4488 0.09 12.0 0.1814 -7.19 -0.76 0.68 20.5 -0.0685 3.12 4,G 2.4098 1.15 0.34. 0.11 12,5 0.1149 0.58 21.0 -0.0607 4.66 3.25 4,5 2.3222 0.14 13.0 0.0594 -5.01 0.21 21.5 -0.0531 3.39 5.C 2,1979 0.43 0,17 13.5 0.0139 -2.74 22.0 -0.0457 7.93 4.62 3.56 5,5 2.0474 0.20 14.0 -0.0226 78.81 18.28 3.73 22.5 -0.0388 3.76 6.0 1,8796 1.41 0.50 0.24 14.5 -0.0511 2.49 23.0 -0.0325 4.47 4.02 6.5 1.7021 0.28 15.0 -0.0725 6.69 2.25 23.5 -0.0266 4.36 7.0 1.5210 0.57 0.32 15.5 -0.0878 2.22 24.0 -0.0213 2.51 4.06 4.86 7,-5 1.3413 0.36 16.0 -0.0979 17.41 5.26 2.24 24.5 -0.0167 5.62 8.0 1.1669 1.16 0.60 0.41 16.5 -0.1037 2.29 25.0 -0.0125 3.09 6.85 Table (V-IV-18) Hemispherical Dome - Radial Load Along the Edge P-Moment/Length, (MJ = a5(pR/100) M = 0.2; R/t = 100.0 (Fig. (V-IV-19)} E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9° ee =*e e ve 9° e =<*> e ye e =d> e Te e =<j> e ve 9° 9e=*e e ve 6e =*e -. =2° =i° =0.5° =2° =1°. =0.5° =2° = 1° =0.5° 0..0 0.0000 0.00 0.00 . 0.00 . 8.5 1.0393 0.48 17.0 -0.1065 5.29 2.78 0.5 0,-7772 0,04 9.0 0.8818 0.61 0.52 17.5 -0.1067 2.83 1.0 1,3784 0.08 0.02 9.5 0.7360 0.56 18.0 -0.1043 13.27 5.06 2.90 1.5 1.8256 0.03 10.0 0.6029 -0.20 0.48 0.58 18.5 -0.1000 2.97 2.0 2.1397 0.63 0.18 0.05 10.5 0.4829 0.61 19.0 -0.0943 4.96 3.06 2.5 2.3405 0.06 • 11.0 0.3762 0.14 0.61 19.5 -0.0876 3.16 ! 3.0 2,4465 0.28 0,08 11,5 0,2825 0.59 20.0 -0.0802 10.93 4.94 3.25 3.5 2,4745 0.11 12.0 0.2013 -6.40 -0.79 0.50 20.5 -0.0725 3.35 4.0 2,4398 1.16 0.37 0,14 12.5 0.1321 0,28 21.0 -0.0647 4.91 3.46 4.5 2.3558 0.17 • 13.0 0.0738 -4.50 -0.32 21.5 -0.0569 3.59 5,0 2,-2345 0.46 0.20 13.5 0.0257 -3.10 22.0 -0.0495 8.41 4.83 3.72 5,5 2,0864 0.24 14.0 -0.0132 34.73 9.90 22.5 -0.0424 3.91 6,0 1.9202 1.43 0.54 0.28 14.5 -0.0439 3.98 23.0 -0.0357 4.66 4.13 6.5 1,7436 0, 32 15.0 -0,0673 7.97 3.17 23.5 -0.0296 4.41 7,0 1.5626 0.60 0.36 15.5 -0.0845 2.90 24.0 -0.0241 3.54 4.23 4.83 7.5- 1.3824 0.40 : 16.0 -0.0962 18.44 5.92 2.79 24.5 -0.0191 5.43 8.0 1.2070 1.20 0.63 0.44 16.5 -0.1032 2.77 25.0 -0.0147 3.29 6.44 Table (V-IV-19) Hemispherical Dome - Radial Load Along the Edge P-Moment/Length, (MJ = cx5(pR/100) y = 0.0; R/t = 500.0 (Fig. (V-IV-20)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9 ° e =<t> e Te 6 =<b e e 9 = 9 e e 9 ° a 5 0 =4> e Ye 0 =* e Ye 9 =9 e Ye 9 ° 9 =d> e e 9 . =9 e e 9 =9 e e = 2° = 1° =0.5° - =2° = 1° =0.5° =2° =1° =0.5° 0.0 0,0000 0.00 0.00 0.00 8.5 -0.0406 3.24 17.0 0.0003 -5.71 0.86 0.5 0.6676 0.16 9.0 -0.0333 9.94 2.80 17.5 0.0002 1.0 0.9990 0.85 0.24 9.5 -0.0255 2.35 18.0 0.0001 1.5 1,0953 0.32 10.0 -0.0182 25.46 5.11 1.75 18.5 0.0000 2.0 1.0412 5.65 1.46 0.37 10.5 -0.0120 0.85 2.5 0.9025 0.40 11.0 -0.0071 -6.04 -0.83 3.0 0.7275 1.48 0.38 11.5 -0.0034 -4.53 3.5 0,5486 0.29 12.0 -0.0008 -93.8 -23.90 4.0 0.3855 4.53 0.33 0.10 12.5 0.0008 29.10 4.5 0,2486 -0.29 13.0 0.0016 52.7 13.41 5.0 0.1414 -4.63 -1.11 13.5 0.0020 9.40 5.5 0.0628 -3.44 14.0 0.0020 29.21 7.87 6.0 0.0093 -94.2 -26.4 14.5 0.0018 6.87 6.5 -0.0236 10.61 15.0 0.0015 20.60 6.42 7.0 -0.0409 22.49 5.76 15.5 0.0012 5.08 7.5 -0.0471 4.40 16.0 0.0009 46.16 10.70 3.84 8.0 -0.0460 60.62 13.94 3.69 16.5 0.0006 2.24 Table (V-IV-20) Hemispherical Dome - Radial Load Along the Edge P-Moment/Length, (MJ = a5(p.R/100) u = 0.2; R/t = 500.0 (Fig. (V-IV-21)) Elasticity-Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9 ° V*e ee=V 8e=*e 9 ° 0-5 ee=*e 8e=*e ee=+e 9 ee=*e 8e = (Pe 8e =*e = 2° = 1° =0.5° . =2° = 1° = 0.5° = 2° = 1° =0.5° 0.0 0.0000 0.00 0.00 0.00 8.5 -0.0421 3.32 17 .0 0.0004 -2.32 0.70 0.5 0.6695 0.16 9.0 -0.0350 10.15 2.89 17 .5 0.0002 1.0 1.0051 0.83 0.23 9.5 -0.0272 2.46 18 .0 0.0001 1.5 1.1062 0.31 10.0 -0.0198 27.43 5.67 1.91 18 .5 0.0000 2.0 1.0561 5.50 1.43 0.37 10.5 -0.0134 1.05 19 .0 0.0000 2.5 0,9202 0. 39 11,0 -0.0082 -3.91 -0.31 3.0 0.7461 1.47 0.38 11.5 -0.0042 -3.15 3,5 0,5667 0.30 12,0 -0.0014 -90.43 -62.9 -14.03 4,0 0.4020 4.61 0.41 0,12 12,5 0,0004 52.33 4,5 0.2626 -0,24 13.0 0,0015 59.73 15.34 5.0 0,1525 -4.08 -0.99 13.5 0.0020 10.25 5,-5 0.0709 - -3.00 14.0 0.0021 30.82 8.54 6,0 0,0148 38.69 -65.9 -16.63 14,5 0.0019 7.38 6.5 -0.0205 12.42 15.0 0.0016 21.23 6.38 7,0 -0,-0396 23.48 6.10 15,5 0.0013 5.31 7,5/ -0,0472 4.54 16.0 0.0010 54.43 12.47 4.33 8.0 -0.0470 60.77 14.15 3.81 16.5 0.0007 3.58 1 1 T i T - h - J 1 1 | i F j I — •- r i 1 1 P R R 1 r - 1 - 1 1 1 1 1 1 1 - 1 . J _ _ . J _ 1 1 1 1 1 1 i _ • 1 1 i 1 1 1 1 1 1 I 1  1 1 1 I 1 -- 1 "T - 1 1 1 1 I '-. 1 - „ CVJ _!_ 1 1 1 i i 1 — i • c C sJL 1 1 1 1 I -v Q C > . 1 1 > . > 1 l i - i 1 1 - i T r --- 'to 1 L - 1 1 -- • T - r 0 Ri 1 1 i - 1 1 D - - 1 1 - I •* t - 1 • ~c 1 1 1 1 "I I . V I "1 n 1 1 1 1 1 1 i y r i i - u _>-?__ - is T 1 1 _ P i 1 - 1 i i i i \h Zfe>_ i 1 1 1 pa 1 1 ? i ^ _ ' —V - 1 - j r , l 1 1 H i t -1 - _ -- Is - - - 1 1 1 R -C 1 1 1 1 r y -;_ v. i V L L r 1 i E ? - 1 A i 1 1 1 t ll © 1 1 - i j 1 J. 1 1 1 b -- 1 1 ' 1 1 1 i / 1 1 1 1 ii - .f- 1 1 1 - 'i 1 1 - -i 1 1 L ,> •T - I 1 1 1 1 1 1 1 1 i i 1 1 1 1 -1 1 1 l \ 1 1 1 i i 1 i \ ^1,. 1 - 1 i •> J ' 1 c If i f. 3 1 1 1 i 1 1 , _ f e V 1 0.1 ? K 1 1 -T 1 I T i i 1 1 - 1 1 ~] c - rn l\ L i r) 1 1 1 - - j 1 1 R I 1 1 1 1 ] 1 1 1 i t 1 1 0 - - .1 1 .1 1 - - „ -- A 1 - - 7 c 1 1 1 i - 1 ! -1 1 i 1 1 1 1 ! 1 1 1 , 1 - | 1 1 1 ---- - 1 -1 1 • A - A . 1 i I 1 1 1 1 1 1 1 - _ _ . 1 i r ! -1 1 ! v i 1 1 1 Lr 1 1 1 1 1 1 1 1 D nil I t L i 1 1 I ! 1 1 1 K 1 ! 1 • 1 >s l_ ir i H 1 1 I 1 1 1 i i 1 1 I i - CO I£ J X •15 I R 1 1 1 i r Y R I _ 1 R 1 1 w fi> V J\ 1 1 1 t I -a ~i~4—j-- , c r j -1 i O 1 1 -f- __ 1 1 1 1 1 1 L U 1 1 1 1 I L— i > 1 _ L. 1 1 1 i i - 1 1 1 N. 1 1 i 1 1 - I T t X T _ - ^ T 1 1 ... 1 1 ' 1 i _ J " U J 1 1 1 — i _ z 1 f i n l I j r 1 11 1 1 1 1 I t * R i c O , ^ r R 11 1 i 1 I •1 1 1 1 r v 1 1 1 1 1 T - 11 <— I 1 1 1 C P 1 'i N _ l 1 1 j -LJL.1 I L L .> r r 1 1 ! i_ 1 /f 1 1 1 1 1 1 | | 1 ~\ - 1 1 i 1 1 1 | l 1 1 1 1 i -< S i .__ 11 -»! 1 , 1 T! 1 1 t > —1 L \JS — => ! ! ! M l J 1 L -1 1 C 1 1 1 I : ! 1 - I 1 1 I 1 1 V l 1 a 1 - .1. 1 I 1 1 HI d 1 I I e 1 1 1 1 1 1 1 ! 1 - 1 1 1 - 1 -1 i 1 1 i lO 1 1 1 - 1 1 i r 01 1.. 1 1 , 1 1 1 i 1 i i 1 1 1 r --- 1 t ! - 1 -i 1 1 1 I 1 1 - - -- L i L 1 1 1 1 i 1 1 1 ! 1 1 I L 1 1 1 1 i 1 1 1 1 1 r\ Table (V-IV-21) Hemispherical Dome - Radial Load Along the Edge M-Moment/Length, (MJ = a6(pR/100) y = 0.0; R/t = 100.0 (Fig. (V-IV-22)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9 ° 9e= <Pe ee =*e ee =*e 9 ° ee=<Pe ee =*e ee=*e 9 ° ee=*e 9e = (Pe 6e=*e =2° = 1° =0.5° =2° = 1° =0.5° =2° =1° =0.5° 0,0 0.0000 0.00 0.00 0.00 8.5 -0.0034 1.73 17.0 0.0025 3.59 2.91 0,5 -0.0025 -1.36 9.0 -0.0023 10.00 2.23 17.5 0.0023 3.04 1,0 -0,0048 -5.59 -0.87 9.5 -0.0013 3.53 18.0 0.0021 8.99 4.28 3.12 1,5 -0,0069 -0.67 10.0 -0.0004 47.21 7.91 18.5 0.0018 3.35 2.0 -0.0085 -15.8 -4.06 -0.76 10.5 0.0004 -7.10 19.0 0.0016 5.31 3.70 2,5 -0.0098 -0.64 11.0 0.0011 -15.09 -1.48 19.5 0.0014 3.57 { 3.0 -0.0107 -2.66 -0.53 11.5 0.0017 -0.24 20.0 0.0012 16.27 6.36 4.24 ' 3.5 -0.0112 -0.25 12.0 0.0021 -26.37 -4.93 0.42 20.5 0.0010 4.08 4.0 -0.0113 -6.58 -1.40 -0.19 12.5 0.0025 1.03 21.0 0.0008 7.09 4.30 4.5 -0.0111 -0.07 13.0 0.0028 -1.61 1.18 21.5 0.0006 5.41 5.0 -0.0106 -0.18 0.11 13.5 0.0030 1.37 22.0 0.0004 29.56 10.00 6.44 5.5 -0.0099 0.26 14.0 0.0031 -6.31 0.23 1.68 22.5 0.0003 7.10 6.0 -0.0090 3.08 1.14 0.45 14.5 0.0031 1.95 23.0 0.0002 12.10 6.84 6.5 -0.0079 0.64 15.0 0.0031 1.46 2.07 23.5 0.0001 12.50 7.0 -0.0068 2.77 0.85 15.5 0.0030 2.35 24.0 0.0000 0.00 0.00 0.00 7.5 -0.0057 1.07 16.0 0.0029 .2.51 2.58 2.51 24.5' -0.0001 8.0 -0.0045 18.67 4.99 1.26 16.5 0.0027 2.47 25.0 -0,0001 Table (V-IV-22) Hemispherical Dome - Radial Load Along the Edge  M-M6ment/Length, (M ) = a6(pR/100) y = 0.2; R/t = 100.0 (Fig. (V-IV-23)) E l a s t i c i t ) Solution Percentage Error E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9° a6 ee-*e VOe •ee=*e 9° ee=*e 9e=*e ee =*e 9° ee=*e ee=*e V % = 2° = 1° =0.5° =2° = 1° =0.5° = 2° = 1° =0 5° 0.0 0.0000 0.00 0.00 0.00 8.5 0.2042 4.25 17.0 -0.0186 32.2 31 0 0.5 0.1529 -0.58 9.0 0.1738 4.58 3.72 17.5 -0.0189 28 6 1.0 0,2708 -4.04 1.67 9.5 0.1456 2.99 18.0 -0.0186 33.28 28.13 26 7 1.5 0,3581 2.61 10.0 0.1199 7.38 3.15 2.01 18.5 -0.0180 25 1 2,-0 0,4192 -10.22 0.57 3.21 10.5 0.0968 0.65 19.0 -0.0171 25.40 23 76 2,5 0,4580 3.67 11.0 0.0762 0.19 -1.26 19.5 -0.0160 22 44 3,-0 0,-4783 2.53 4.03 11.5 0.0581 -4.05 20.0 -0.0147 29.27 22.95 21 18 3.-5 0.4834 4.33 12.0 0.0424 0.98 -6.42 -8.32 20.5 -0.0134 19 90 4.0 0.4763 -0.03 3.74 4.58 12.5 0;0289 -15.6 21.0 -0.0120 20.47 18 64 4.5 0.4597 . 4.79 13.0 0.0176 -27.11 -27.9 21.5 -0.0107 17 23 5.0 0.4359 4.56 4.94 13.5 0.0082 -71.4 22.0 -0.0093 24.27 17.54 15 74 5.5 0.4070 5.05 14.0 0.0005 22.5 -0.0081 13 92 6.0 0.3747 4.61 5.09 5.10 14.5 -0:0056 23.0 -0.0069 13.43 11 90 6.5 0.3404 5.10 15.0 -0.0103 63.6 62.9 23.5 -0.0058 9 67 7.0- 0.3053 5,32 5.02 15.5 -0.0138 47.3 24.0 -0.0048 13.53 7.52- 6 68 7.5 0.2704 4.87 16.0 -0.0162 42.4 39.9 39.2 24.5 -0.0039 3 20 8.0 0.2365 7.15 5.20 4.61 16.5 -0.0178 34.3 25.0 -0.0031 -2.21 -1 24 Table (V-IV-23) Hemispherical Dome - Radial Load Along the Edge M-Momerit/Lerigth, (M ) = ct6(pR/l00) M = 0.0; R/t = 500.0 [Fig. (V-IV-24)) E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error 9° ee =*e e e = % V * e 9° ee=*e ee=*e V * e =2° = 1° =0.5° - = 2° = 1° =0.5° 0.0 0.0000 0.00 0.00 0.00 8.5 0.0001 3.60 0.5 -0.0005 -7.50 9.0 0.0001 15.00 2.00 1,0 -0.0008 -17.9 -4.51 9.5 0.0001 6.67 1.5 -0,0010 -2.93 10.0 0.0000 2,0 -0,0010 -29.2 -5.76 -1.11 2,-5 -0.0009 0.00 3,0 -0,0007 8.21 1.94 3.5 -0,0004 5.00 4,0 -0,0002 45.00 10,45 4,5 +0.0000 76.7 5.0 0,0001 -73.3 -20.0 5,5 0.0002 -6.19 6.0 0.0003 -82.3 -15.0 -2.69 6.5 0.0003 0.37 7.0 0,0003 -1.54 0.00 7,5 0,0002 0,00 8.0 0.0002 24.74 5.26 0.00 Table (V-IV-24) Hemispherical Dome - Radial Load Along the Edge M-M6ment/Length, (M ) = a6(pR/100) ]i = 0.2; R/t 500.0 (Fig. (V-IV-25)) E l a s t i c i t y Solution Percentage Error ! E l a s t i c i t y Solution Percentage Error E l a s t i c i t y Solution Percentage Error *° ee=<f>e 9e =*e ee=*e ee=*e ee=*e ee=*e ee=*e ee=*e = 2° = 1° =0.5° =2° = 1° = 0.5° =2° = 1° =0.5° 0.0 0.0000 0.00 0.00 0.00 8.5 -0.0083 6.10 17 .0 0.0001 4.44 -1.11 0.5 0.1334 -7.50 9.0 -0.0069 12.36 5.74 17 .5 0.0000 1.0 0.2002 -14.33 -2.87 9.5 -0.0054 5.30 18 .0 0.0000 1.5 0.2202 -1.09 10.0 -0.0039 50.41 11.65 4.61 2.0 0.2102 -22.33 -3.53 -0.05 10.5 -0.0027 3.72 2.5 0.1831 0.66 11.0 -0.0016 7.38 1.52 r L 3.0 0.1485 1.56 1.19 11.5 -0.0009 -2.79 c 3.5 0.1129 1.57 12.0 -0.0003 4.67 -23.00 -16.33 4.0 0.0802 16.24 5.15 1.81 12.5 0.0001 4.5 0.0525 1.85 13.0 0.0003 52.59 27.78 5.0 0.0306 8.22 1.55 13.5 0.0004 19.19 5.5 0.0144 0.24 14.0 0.0004 30.75 14.00 6.0 0.0032 15.25 -9.10 14.5 0.0004 13.24 6.5 -0.0038 13.46 15.0 0.0003 25.31 10.94 7.0 -0.0077 9.71 8.16 15.5 0.0002 12.00 7.5 -0.0092 7.00 16.0 0.0002 95.8 21.05 8.95 8.0 -0.0092 28.56 11.63 6.42 16.5 0.0001 9.23 CHAPTER VI DISCUSSION In deriving the element s t i f f n e s s matrix by the s t a t i c s approach, prorating of boundary stresses into equivalent forces at the immediately adjacent corners results i n an asymmetric s t i f f n e s s matrix. Asymmetry of the s t i f f n e s s matrix contradicts Betti's reciprocal theorem - v a l i d for a l l l i n e a r l y e l a s t i c bodies subject to superposition. However, the approach has the following advantages: (i) The procedure of prorating the edge stresses to y i e l d nodal forces i s c l e a r l y defined. ( i i ) The process i s formulated as a sequence of operations on matrices composed of r e l a t i v e l y simple expressions. This makes i t possible to derive the s t i f f n e s s matrix i n an e x p l i c i t algebric form. and ( i i i ) The significance of a nodal force i s more than being merely a number of computation. The nodal force i s s t a t i c a l l y equivalent to the adjoining edge stresses. The edges which are not adjacent to the corner point do not contribute towards the nodal force at the corner i n question. However, i t i s important to note that i n conditons of uniform stress or bending, the nodal forces do s a t i s f y Betti's reciprocal theorem 258 irrespective of the size and exact shapo of the trapezoidal element. As the element size decreases, the state of .".tress i n the v i c i n i t y of an element becomes more and more uniform. Thu.>, i n the l i m i t the reciprocal theorem i s i n fact s a t i s f i e d . It i s also worth noting that when a trapezoid i s reduced to a rectangle, the asymmetry of the plane stress and flexure matrices disappears. Apparently, the asymmetry of the s t i f f n e s s matrices r e f l e c t s the lack of symmetry of the trapezoid element i t s e l f about one of i t s axes.. In order to simplify the computation work, the asymmetric matrix i s made symmetrical by averaging the matrix and i t s transpose. The terms of the resultant symmetric matrix s a t i s f y the equilibrium requirements. The process of averaging i s thus equivalent to introducing additional s e l f e q u i l i b r a t i n g nodal forces i n each mode. The energy approach uses a systematic procedure for s t i f f n e s s matrix computation and i n a l l cases yields a symmetric matrix. However, the approach suffers from the following drawbacks: (i) For complicated shapes of a f i n i t e element, evaluation of the component matrix [ H ] = a ^ a [ G ] T [ D ] [ G ] d x d y required i n the calculation of the s t i f f n e s s matrix (Eqns. (11-27,b) and (111-25), i s not always possible without using the computer. and ( i i ) Matrix m u l t i p l i c a t i o n to obtain element s t i f f n e s s matrix (Eqns. ( I I -27,b) and (111-25)) , [K] = [A- 1] T[H][A- 1] 2S9 involves long expressions and thus, invariably needs to be done i n the computer. In such a case, the s t i f f n e s s c o e f f i c i e n t s are merely numbers and as such have no physical significance. Using the s t a t i c s approach, a symmetric s t i f f n e s s matrix for a rectangular element i n flexure was independently derived elsewhere^^ It was observed that, i n general, the S t a t i c s matrix w i l l y i e l d better results 17 27 than the Energy s t i f f n e s s matrices. ' Errors i n the f i n i t e element analysis of a s h e l l are due to approximating the curved s h e l l surface by f l a t elements, replacing the dist r i b u t e d load by approximately calculated nodal loads and preserving the continuity of the model at the nodes only. However, as the element size, decreases, the model imitates the s h e l l conditions more closely thereby reducing the errors. In s h e l l s with positive curvatures i n both p r i n c i p a l directions and with the load of d i s t r i b u t e d nature, the load carrying stresses are the membrane stresses except near the supports whose reactions develop components i n R-direction. In t h i s case f l e x u r a l stresses may take on some load carry-ing function, even though i t may be comparatively minor. Thus, normally the flexure stresses are simply the p a r t i c i p a t i o n stresses having l i t t l e or no effect on the load carrying capacity of the structure. In spite of t h i s minor role of flexure, the unit f l e x u r a l stresses (#/in^) may be quite s i g n i f i c a n t and may cause a nonreinforced or reinforced concrete or stone s h e l l to crack. The s h e l l also i s l i k e l y to develop s i g n i f i c a n t flexure stresses near concentrated or sharply l o c a l i s e d loads. The thinner the s h e l l i s ( i . e . the greater the radius to thickness r a t i o i s ) the smaller the region i s near the support or near the l o c a l i z e d load within which the I 260 flexure stress i s s i g n i f i c a n t . In t h i s region, smaller sized elements possess-ing both plane stress and flexure properties must be used i n order to obtain a reasonably gcod representation of displacements and stresses. In regions father away from t?ie boundaries or lo c a l i z e d loads, the s h e l l action i s largely membrane. In th i s region elements of larger size are f u l l y acceptable. The s h e l l i s f i r s t analysed with elements of f a i r l y large s i z e . The region which needs to be f i n e l y subdivided i s separated from the rest of the s h e l l by the boundaries whose displacements at the nodes have already been calculated f a i r l y accurately. The f i n e l y subdivided region i s analysed by assigning the calculated displacements at the boundaries. The d i s -placements at the intermediate nodes of the boundaries are interpolated. In cases when flexure properties of the elements must be used a three stage solution, described e a r l i e r , i s suggested. The size of mesh necessary to obtain a reasonable representation of displacements and stresses i s not d i r e c t l y obvious. It depends on the problem and i s determined largely by judgement. • Calculation of s h e l l stresses by the displacement method i s cumbersome. Calculation based on s t a t i c s and consisting i n d i s t r i b u t i n g the nodal forces over the tributary areas provides f a i r l y accurate results i n regions where stress gradients are not very steep. Where steep stress gradients are present, small element size improves the r e s u l t s . In general, higher percentage error i n stresses than i n displacements i s expected. Results of the numerical analysis on a variety of problems con-firm the s u i t a b i l i t y of f l a t elements for solving problems involving shells of revolution with positive curvatures. Stresses and displacements are 261 calculated with good accuracy by using either Sta t i c s or Energy s t i f f n e s s matrices. From the error analysis of the various problems presented, convergence of displacements and stresses to t h e i r exact e l a s t i c i t y values i s indicated on reduction of the element s i z e . It i s expected that the method i s equally applicable to the general case of shells of revolution which includes structures with reversed curvature. CONCLUSION A c i r c u l a r plate subjected to a p a i r of diametrically opposite point loads and a spherical s h e l l subjected to snow load, wind load, and ra d i a l edge load are analysed with good accuracy for both membrane and flexure stresses using f l a t e q u i l a t e r a l trapezoids and isosceles t r i a n g l e elements. The method may be applied to a s h e l l of revolution of an arbitrary shape subjected to any loading and provided with any boundary conditions, with an expectation of results of comparable precision. Three values of Poisson's r a t i o (0.0, 0.2 and 1/3) are used i n solution involving Statics and Energy s t i f f n e s s matrices. There i s no reason to doubt the a p p l i c a b i l i t y of the method with other values of Poisson's r a t i o . In the spherical s h e l l supporting a load of di s t r i b u t e d nature, the load carrying stresses are the membrane stresses except near sharply l o c a l i s e d loads or near the supports having reaction i n R-direction where the f l e x u r a l stresses become to some extent the load carrying stresses. Similar load carrying behaviour may be expected from a s h e l l of revolution having positive curvatures i n both p r i n c i p a l directions. In the problems analysed, the Statics and Energy s t i f f n e s s matrices provide, on the whole, comparable accuracy. In regions of rapidly changing displacements and stresses, small sized elements are necessary to obtain a f a i r accuracy of the r e s u l t s . On reduction of the element s i z e , the errors i n stresses and i n displacements decrease consistently. Convergence to the exact e l a s t i c i t y values i s expected on further decrease of fineness of mesh without l i m i t , apart from possible effects- of the rounding o f f errors of the computer. REFERENCES 1. Fletcher, B., "A History of Architecture," Seventeenth e d i t i o n , Athlone Press, 1961, pp 197-199. 2. Anon, "Dome Shell Roofs," AIA f i l e : 4-a, Published by Portland Cement Association, Chicago, I l l i n o i s . 3. Meyer, R.R., and Harmon, M.B., "Conical Segment Method for Analysing Open Crown Shells of Revolution for Edge Loadings," AIAA Journal, Vol. 1, No. 4, A p r i l 1963, pp 886-891. 4. Grafton, P.E., and Strome, D.R., "Analysis of Axisymmetric Shells by the Direct Stiffness Method," AIAA Journal, Vol. 1, No. 10, October 1963, pp 2342-2347. 5. Popov, E.P., Penzien, J . , and Lu, Z.A., "Finite-Element Solution for Axisymmetrical S h e l l s , " ASCE Eng. Mech. Div. Journal, October 1964, pp 119-145. 6. Jones, R.E., and Strome, D.R., "Direct Stiffness Method of Analysis of Shells of Revolution U t i l i z i n g Curved Elements," AIAA Journal, Vol. 4, No. 9, September 1966, pp 1519-1525. 7. Percy, J.H., Pian, T.H.H., K l e i n , S., and Navaratna, D.R., "Application of Matrix Displacement Method to Linear E l a s t i c Analysis of Shells of Revolution," AIAA paper No. 65-142, January 1965. 8. Jones, R.E., and Strome, D.R., "A Survey of Analysis of Shells by Displacement Method," AFFDL-TR-66-80, 'Matrix Methods i n Structural Mechanics', November 1966, pp 205-219. 9. Turner, M.J., Clough, R.W., Martin, H.C., and Topp, L.J., "Stiffness and Deflection Analysis of Complex Structures," Journal of Aeronautical Sciences, Vol. 23, No. 9, September 1956, pp 805-854. 10. Greene, B.E., Strome, D.R., and Weikel, R.C., "Application of the Stiffness Method to the Analysis of Shell Structures," ASME paper No. 61-AV-58, March 1961. 11. Adi n i , A., "Analysis of Shell Structures by the F i n i t e Element Method," Ph.D. Dissertation, University of C a l i f o r n i a , Berkeley, C a l i f o r n i a , 1961, 12. Antebi, J . , Rossow, E.C., Shah, I.K., and Shah, J.M., "Computer Analysis of the F - l l l Wing Assembly," ASCE Struct. Div. Journal, December 1966, pp 405-425. 13. Gallagher, R.H., "A Correlation Study of Methods of Matrix Structural Analysis," The MacMillan Co. 1964. 264 14. Clougl., R.W., "The F i n i t e Element Method i n Structural Mechanics," Chapter 7, 'Stress Analysis' edited by Zienkiewicz and H o l i s t e r , John WiJey and Sons Ltd., 1965. 15. Jones, R.".., "A Generalization of the Direct S t i f f n e s s Method of Structural Analysis," AIAA Journal, Vol. 2, No. 5, May 1964, pp 821-826. 16. Melosh, R.J., "Development of the Stiffness Method to Define Bounds on E l a s t i c Behaviour of Structures," Ph.D. Thesis, Department of C i v i l Engineering, Univ. of Washington, Seattle, June 1962. 17. Melosh, R.J., "Basis for Derivation of Matrices for the Direct S t i f f -ness Method," AIAA Journal, July 1963, pp 1631-1637. 18. Irons, B.M., and Draper, K.J., "Inadequacy of Nodal Connections i n a Stiffness Solution for Plate Bending," AIAA Journal, Technical Note, May 1965, page 961. 19. DeVeubeke, B.F., "Duality between displacement and equilibrium methods with a view to obtaining upper and lower bounds to s t a t i c influence c o e f f i c i e n t s , " Proc. 14th Meeting of AGARD 'Structures and Material Panel'. P a r i s , July 1962, AGARDograph 72. . 20. Clough, R.W., "The F i n i t e Element Method i n Plane Stress Analysis," Proc. 2nd ASCE Conference on Electronic Computation, Pittsburg, Pa., September 1960, pp 345-377. 21. Bazeley, G.P., Cheung, Y.K., Irons, R.M., and Zienkiewicz, O.C., "Triangular Elements i n Plate Bending - Conforming and Nonconforming Solutions," AFFDL-TR-66-80, November 1966, pp 547-576. 22. Zienkiewicz, O.C., "The F i n i t e Element Method i n Structural and  Continuum Mechanics," McGraw H i l l , 1967. pp 21-25. 23. Hrennikoff, A., "The F i n i t e Element Method Applied to Plane Stress and Bending of Plates," Lecture notes. 23a. Hrennikoff, A., "The F i n i t e Element Method i n Application to Plane Stress," Vol. 28, I.A.B.S.E., to be Published i n 2nd h a l f of 1968 volume. 24. . Timoshenko, S., and Goodier, J.N., "Theory of E l a s t i c i t y , " 2nd e d i t i o n , McGraw H i l l , 1951. 25. P e r i l s , S., "Theory of Matrices," Addison-Wesley Publishing Co., 1958 page 86. 26. Timoshenko, S., and Woinowsky-Kreiger, S., "Theory of Plates and Sh e l l s . " McGraw H i l l Book Co., second e d i t i o n , 1959. 27. Zienkiewicz, O.C., and Cheung, Y.K., "The f i n i t e element method for analysis of e l a s t i c , i s o t r o p i c and orthotropic slabs," Proc. Inst. C i v i l Engrs. (London), 28, A p r i l 1964, pp 471-487. 265 28. Tocher, J.L., "Analysis of Plate tending Using Triangular Elements," Ph.D. Dissertation, Univ. of C a l i f o r n i a , Berkeley, 1962. 29. Gallagher, R.H., and Huff, R.D., "Derivation of the Force-Displacement Properties of Triangular and Quadrilaveral Orthotropic Plates i n Plane Stress and Bending," B e l l Aerosystems Co., Report No. D2114-950005, 1964. 30. Shah, J.M., "A F i n i t e Element Technique for the Analysis of Thin Shells and Plates," Simpson-Gumpertz and Heger Inc., Cambridge Massachusetts, 1966. 31. Tezcan, S.S., discussion of "Simplified Formulation of Stiffness Matrices," by P.M. Wright, Journal of the Struct. Div. ASCE, Vol. 89, No. ST.6, December 1963, pp 445-449. 32. Faddeeva, V.N., "Computational Methods of Linear Algebra," Dover Publication, 1959, pp 81-85. 33. Hrennikoff, A., and Tezcan, S.S., "Analysis of C y l i n d r i c a l Shells by the F i n i t e Element Method," 'Symposium on Problems of Interdependence of Design and Construction of Large Span Shells for Industrial and C i v i c Buildings,' USSR, Lenningrad, 6-9 September 1966. 34. Muskhelishvili, N.I., "Some Basic Problems of the Mathematical Theory  of E l a s t i c i t y , " Translated from the Russian by J.R.M. Radok, P. Noorclhoff Ltd., 1963, pp 330-338. 35. "BETON=KALENDER", Taschenbuck fur Beton und Stahlbetonbau sowie die Verwandten Facher, Zweiter T e i l , B e r l i n 1958, pp 56-96. 36. Flugge, W., "Stresses i n Sh e l l s, " Second P r i n t i n g , Springer-Verlag, B e r l i n , 1962. Agrawal, K.M. and Hooley, R.F., " F i n i t e Element Solution of a Plate Bending Problem," Publication Pending. Turner, M.J., Clough, R.W., Martin, H.C., and Topp, L.J., "Stiffness and Deflection Analysis of Complex Structures," Journal of the Aeronautical Sciences, Vol. 23, No. 9, September 1956, pp 805-823. Pfluger, A., "Elementary Statics of Sh e l l s, " 2nd E d i t i o n , McGraw-H i l l Book Co., 1961. 37. 38. 39. APPENDIX BRIEF DESCRIPTION OF COMPUTER PROGRAMS AND FLOW DIAGRAMS Description of PR-I: This program i s intended to read the structure data from cards, to assign deformation numbers to the j o i n t s , to assign corner numbers to the elements, to calculate code nos. of each element, and to calculate s t i f f n e s s matrix i n element coordinates, transformation matrix and transformed s t i f f n e s s matrix of each t y p i c a l element. The above information i s stored i n Tapes 3 and 4. PR-I c a l l s two subroutines; SM3 and SM4. These subroutine's calculate element s t i f f n e s s matrix i n element coordinates, transformation matrix and transformed s t i f f n e s s matrix for triangular and trapezoidal elements respectively. These subroutines employ either Statics or Energy s t i f f n e s s matrices. For analysing a membrane s h e l l , •subroutines using only the plane stress properties of the f i n i t e element are used. For a s h e l l having flexure properties these subroutines are simply externally replaced by those which include both plane stress and flexure properties of the f i n i t e element. 266,a SIMPLIFIED FLOW DIAGRAM Start PR-I Read, Print and Store on Tape 3 Structure Data ( Read and Print Joint Data ) Generate and Store on Tape 3 Node Deformation Nos. , 1 , ( Print Node Deformation Nos.) Read, Print and Store on Tape 3 > Known Joint Deformations ( Generate Corner Data for a l l Elements ) Read, Print and Store on Tape 3 ' Geometry Data for a Typical Element For a Typical Element Generate and Store on Tape 3 Element Stiffness Matrix, Transformation Matrix Transformed Stiffness Matrix ± Generate and Store on Tape 3 Element Corner Nos., and Code Nos, I Element has Known Deformations Yes Generate the Force Vector Due to Known Deformation Calculate Band Width to Date Yes i ^ Identical Element y wo A l l Elements Considered Prin t and Store on Tape 4 ' \ Total Degrees of Freedom, Band Width / ( Read and Print Joint Load Data ) Generate and Store on Tape 3 Force Vector Y E S I v A l l Load Cases Considered \ 3_ End of PR-I 267 Description of PR-II: This program i s intended to read the structure l a t a from tapes 3 and 4, generate the structure s t i f f n e s s matrix and solve th^ system of simultaneous l i n e a r equations by Choleski's method. The Choleski's method of solution was o r i g i n a l l y programmed by Dr. R. F. Hooley for ALP II -a program for solving two dimensional frames on IBM 1620. The system of simultaneous equations are solved for a l l load cases and the deformation vectors are written on tape 4. No data i s required to execute the program. Start of PR-II ( Read Tape 3 Structure Data ) • \  ( Read Tape 4; Nos. of Unknowns, Band Width ) :—t. . / Read Tape 3; Typical Element Transformed Stiffness Matrix  —. «i— ( Read Tape 3 Element Corner Data, Code Nos.) Generate Structure Stiffness Matrix Identical Element ) *• A l l Elements Conversion of Band Stiffness Matrix into Lower Triangle Considered ( Read Tape 3, Load Vector") Forward Substitution to Obtain Deformation Vector Store on Tape 4 Deformation Vector More Load Cases y , \ ) End of PR-II 268 Description of PR-III: This program i s intended to read structure data and .shell deformation data from tapes 3 and 4, assign deformations to various j o i n t s and calculate nodal forces i n various elements. Deformations at various j o i n t s and element nodal forces are obtained as printed out-put and are also stored on disk 10. Start PR-III ("Read Tape 3 and Print Structure Data ) ( Read Tape 4 Structure Deformation Vector" -) ( Read Tape 3, Struct. Joint Deformation Nos.) Assign Joint Deformations ( Print Joint Deformations ) Read Tape 3; Typical Element Transformed Stiffness Matrix ( Read Tape 3; Element Corner Nos., Code Nos.) Generate Element Nodal Deformation Vector and Calculate Nodal Forces -c( Identical Element y A l l Elements Considered A l l Load Cases Considered~y YES End of PR-III 269 Description of PR-IV This program i s intended to read the nodal forc.^ concentrations from tape 10 (calculated and stored there by PR-III). Elements connecting a p a r t i c u l a r j o i n t are supplied as data to calculate stress resultants at that j o i n t . Start PR-IV Read Tape 3 and Print Structure Data Read Disk Id, for a l l Elements Element Nodal Forces ._ I Read Joint Information at which Stress desired YES Calculate Stress [ Print Stress at the Joint ]  I Stress Calculation at Desired Joints Completed Y E T S A l l Load Cases Considered YES End of PR-IV NO HO 

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