THE CONTRIBUTION OF AN ELASTIC WALL SUPPORT TO THE DEFLECTIONS 7 OF A THIN CIRCULAR PLATE Thesis for the *Degreerof PII; D. 7 MICHIGAN STATE UNIVERSITY WILLIAM v. BREWER ,, t 11968» ’ LIBRrSRY "In.“ Umvcrszty This is to certify that the thesis entitled THE CONTRIBUTION OF AN ELASTIC WALL SUPPORT TO THE DEFLECTIONS OF A THIN CIRCULAR PLATE presented by WILLIAM V. BREWER has been accepted towards fulfillment of the requirements for DoctoraI d . Mech. Engr. _ egree 1n. Thesis AdVTSOY‘ Major professor Date M 0.169 Michigan Stair: ABSTRACT THE CONTRIBUTION OF AN ELASTIC WALL SUPPORT TO THE DEFLECTIONS OF A THIN CIRCULAR PLATE by William V. Brewer The axially-symmetric problem examined is as follows. A thin circular plate (or disk) is attached (bonded, glued, welded) to the wall of a cylindrical shaft (or hole) in a massive or thick walled solid, The disk and shaft share the same axis of symmetryo The generality of the mathematical model of this problem is limited by the two component theories used in its solution: classical, three-dimensional, small-strain, theory for a homogeneous isotrOpic elastic solid used for the supporting wall; classical, small—strain, thin—plate theory without mid-plane forces used for the plate fixed to the supporting walla The other two limitations on the generality of the theory are: choice of axial-symmetry to obtain a non-plane- strain solution in two dimensions; selection of the stress profiles at the boundary shared by the two bodies. Thus the matching of the SIOpe and displacements at this common boundary is limited to a single apprOpriate point. William V. Brewer Within the above limitations the problem is solved for all plates without holes or inclusions and having uni- form thickness and material parameters. Any load q(r) is admissible. The solution for the wall would allow the other plate solutions (i.e., with a hole and/or inclusions and also having variable thickness and/or material parameters) to be develOped in a parallel fashion with comparative easeo The solution allows the wall and plate to have dif— ferent material parameters. The solutions for the shear and the moment applied to the cylindrical wall are easily used together or separately to solve a variety of other problems where such a wall is similarly loaded. Since the mathematical model for the wall is solved for all of space and since the solutions all decay with respect to increasing r and 2, these solutions can be used to approximate those of objects with sufficiently thick walls having cylindrical cavities. In the case of the self- equilibrating moment, where shear is small, the walls need not be very thick. The specific example used to illustrate the solution procedure sets the load q(r) = q(a constant) and lets the 10 5 2 100' 100' 100' ratios the ratio of the elastic moduli of plate and wall For these plate thickness/diameter ratio be is allowed to range from Ep/Ew = 0 to Ep/Ew = 10. For each of the thickness/diameter ratios, the deflections w(r) based William V. Brewer on the assumptions of a rigid wall and of a wall and plate of same material, never differ by less than 300% for any given r. Similarly the maximum radial tensile stress at the wall never varies by less than 200%. The effect of Poisson's ratio varying from .25 to .30 is relatively small but becomes more significant for thinner plates. The deflection at the wall is small for thinner plates, while rotation at the wall is still comparatively large. There is little reason to believe that the trends indicated by this example would change significantly for other usual loadings and geometries within the scope of the problem. THE CONTRIBUTION OF AN ELASTIC WALL SUPPORT TO THE DEFLECTIONS OF A THIN CIRCULAR PLATE by William V.VBrewer A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mechanical Engineering 1968 ACKNOWLEDGMENT S In this work, as in all human activity, little progress would have been made without support. Dr. R. W. Little suggested the area of study, pro- vided resource materials, answered essential questions, _gave encouragement and insisted on a thorough investiga- tion. Without his help this work would not have been started or completed. My thanks to the Guidance Committee, Dr. G. H. Martin, Dr. R. T. Hinkle, and Dr. E. A. Nordhaus, who gave generously of time and knowledge whenever called upon. My thanks also to Dr. I. N. Sneddon and Dr. J. S. Frame who contributed measurably to the outcome of the work. I am indebted to the staff members of the several engineer- ing and science departments for support of various kinds. I am grateful especially to my wife and to several others who helped translate the manuscript into a finished product. I would dedicate this effort to my parents who from the beginning to the end did everything they could to en- courage an advanced education. Finally, my thanks to God without whom nothing would be possible. ii TABLE OF CONTENTS Page LIST OF TABLES . . . . . . . . . . . . . . . . . . . . . iV LIST OF FIGURES. . . . . . . . . . . . . . . . . . . . . v LIST OF APPENDICES . . . . . . . . . . . . . . . . . . . Vi Section I INTRODUCTION . . . . . . . . . . . . . . . . . . 1 Design and the boundary value problem Area of interest II SELECTION AND STATEMENT OF THE PROBLEM . . . . . 3 Symmetry Coordinates and geometry Component theories Boundary conditions III MATHEMATIC DEVELOPMENT . . . . . . . o . . . . . 7 Axially-symmetric isotrOpic elastic solid Strain function Boundary problem and Fourier integral transforms The plate IV ILLUSTRATIVE EXAMPLE . . . . . . . . . . . . . . 28 The plate loading Limiting cases Plate geometries Material parameters Results Other results V SUMMARY 0 O O O O O O O O O 0 c O 0 O 0 O O 0 O O 3 6 Limitations in theory Applications iia. TABLE OF CONTENTS (CONTINUED) Section Page REFERENCES 0 O O O O O O O O O O O O O O O O O O O O O 3 8 APPENDICES O O O O O O O O O O O O O O O O O 0 O O O O 40 iii LIST OF TABLES Table Page 1 Table of symbols . . . . . . . . . . . . . . . vii 2 Results of the numerical integrations used in the illustrative example . . . . . . 47 iv Figure LIST OF FIGURES Thin plate with thick wall support . . Coordinates and geometry . . . . . . . . Plate displacements for the illustrative example 0 O O O O O O O O O C O O O O O A sampling of wall displacements for the illustrative example . . . . . . . Untitled . . . . . . . . . . . . . . . . Untitled . . . . . . . . . . . . . . . . Untitled . . . . . . . . . . . . . . . . Untitled . . . . . . . . . . . . . . . . untitled O O O 0 O O O O O O O O G O O O Page 32 34 48 49 50 51 52 LIST OF APPENDICES Appendix Page I Numerical Integration . . . . . . . . . . . . 40 II Computer Programs . . . . . . . . . . . . . . 53 Vi TABLE OF SYMBOLS English Symbols AIL»). B(w) a,b c=2(l-v) D_ 2Eb3 _ 3(l-v2) arbitrary coefficient functions of solutions to V” E (r,z)=0 radius and half-thickness of the undeformed plate group of constants appearing frequently plate modulus D(x)=[x2(K2(x)—l)-c] group of terms appearing frequently in the displacement solutions elastic modulus in tension and compression or Young's modulus E for the circular plate E for the wall supporting the plate vector potential function defined by Galerkin Elastic modulus in shear G: the Lame‘ constant u HP: R/b or na/b half-period of functions to be integrated in HPS HPR expressions for displacements starting point for the numerical integrations expressed in half-period lengths range of the numerical integrations expressed in half-periods vii 6 B % f pq(p)dded6dY an integral appearing o Iw(r)= i OH Y I 6 f o o in the general solution for the plate displace— ment w(r) Iw=Iw(r) at r=a the edge of the plate Iwr(r) = 5%Iw(r) Iwr = Iwr(r) at r=a the edge of the plate 1 _d(rIwr(r)) Iwrr(r)=-E-dr Iwrr = Iwrr(r) at r=a Kn(x) Modified Bessel Functions of the Second Kind and order n K(x)=KO(x)/Kl(x) group of terms appearing frequently K=K(x) at x=wa Mr(r) moment resultant per unit circumferential length applied in the 0-direction (right-hand-rule) to the r-face of a differential element of plate material (dr)x(rd0)x(2b) z=b M(o)= f [or(a,z)]sz = f [0°z]z(l°dz) moment resultant per A z=_b unit circumferential length applied to the wall N(x) = [l+xK(x)] group of terms appearing frequently in the displacement solutions viii PHP partition of the half-period HP into PHP equal parts Qr(r) shear resultant per unit circumferential length applied in the z-direction to the r-face of a differential element of plate material (dr)x(rd0) x(2b) z=+b Z 2 0(1) = f [Trz(a,z)]dA = f T(l-Bf)(l°d2) shear resultant per A z=-b unit circumferential length applied to the wall q(r) plate load per unit area applied in the z-direc- tion to the z-face of a differential element of plate material (dr)x(rde)x(2b) R(r) one of the solutions to V2 §2(r,z)=0 where gZIrIZ)=R(r)Z (2) R =3R(r) r 3r =32R(r) rr 3r2 (r,6,z) cylindrical coordinates S(x) [§$2I§L - Cos(x)] group of terms appearing frequently X > . u=(ur, v6, w) displacement vector in the elastic solid ur(r,z,o,T) displacement in the r—direction ur(r,z) displacement in the r-direction g u (r,z,o,0) infinite integrals us(r,z)= r us(z) = us(a,z) infinite integrals us = us(a,0) constant ix G ut(r,z) =I? ur(r,z,0,T) infinite integrals ut(z) = ut(a,z) infinite integrals ut = ut(a,b) constant G aur uzs(r,z) = 3 FE— (r,z,c,0) infinite integrals uzs(z) = uzs(a,z) infinite integrals uzs = uzs(a,0) constant G aur uzt(r,z)= ? FE—'(r’z'0’T) infinite integrals uzt(z) = uzt(a,z) infinite integrals uzt = uzt (a,0) constant w(r) displacement of the plate in the z- direction w(r,z,o,T) displacement of the elastic solid in the z— direction w(r,z) displacement of the elastic solid in the z- direction ws(r,z)= g-w(r,z,o,0) infinite integrals ws(r) = ws(r,0) infinite integrals ws(z) = ws(a,z) infinite integrals ws = ws(a,0) constant wt(r,z) = g-w(r,z,0,T) infinite integrals wt(r) = wt(r,0) infinite integrals wt(z) = wt(a,z) infinite integrals wt = wt(a,0) constant g(r,z) component of Galerkin's vector potential F in the z-direction....also referred to as the Love strain function §4(r,z) satisfies V1+ §4(r,z)=0 §2(r,z) satisfies V2 §2(r,z)=0 Z(z) one of the solutions to V2 §2(r,z)=0 where 52(r,z)=R(r)Z(z) _ 3Z(2) ZZ— 32 Z = 822(2) 22 32 Greek Symbols v Poisson‘s ratio or, 06, oz, etc. tensile stress in the indicated directions (not partial derivatives) 0 maximum or on the common boundary between the plate and wall support Trz, Ire, T62 shear stress on the faces and in the direc— tions indicated where Tij= Tji for all cases considered T maximum Trz on the common boundary between the plate and wall support w separation constant in the solution of V2§2(r,z)=V2R(r)Z(z)=0 but considered a variable in the Fourier integral sine and cosine transformations Other Symbols 2_ 32 1 a 32 a . V -(§ET + 5-5? + 5;?) LaplaCian operator shown for the ax1ally symmetric case xi xii I. Introduction Design and the boundary value problem Design may be defined as the combination of two or more known techniques or solutions in a new application to achieve a desired result. Most design problems in mechanics are in some way concerned with specifications and/or results at boundaries and therefore may also be interpreted as boundary value problems. We will combine two solutions in such a manner as to both match and influence each other at their common boundaries. Area of interest In the area of elastostatics the interaction of bodies having a thin cross—section with those having massive, solid or thick cross-sections is often neglected. In such cases the solid is usually assumed to be rigid and contri- butions to deflections therefrom are not investigated. That the solid may contribute to deflections is recognized even in cases where both bodies are made of the same mater- ial. A plane stress (strain) case of a beam intersecting a half-plane body at right angles was investigated inde- pendently by Weber [1] and Muskhelishvili [2] who employed different assumptions. Weber modeled the problem by applying to the half-plane the linear axial tensile stress distribu- -b$y$b tion of simple beam theory: 0X(0,y) = (Omax) y -b>y>b = O Muskhelishvili however applied a linear axial displacement distribution: u (0,y) = (u )y -by>b It is noted by O'Donnell [3], who compared these solutions, that even though the latter case results in an infinite value of stress Omax at y = 1-b the resultant displacements obtained due to rotation at the wall differ by only 15%. O‘Donnel chose a cubic stress distribution as a compromise between the first two models and obtained results between. He also investigated the effect of shear by applying a constant shear to the wall: I (0,y) = T constant -b$ys+b XY = O -b>y>+b O‘Donnell also investigated the plane stress case experi- mentally. The various results and comparisons are displayed in several graphs. More recently Cook [4] generalized this problem to many evenly spaced beams intersecting a half- plane. An investigation of a non-plane-stress problem was conducted by Brown and Hall [5]. In this case a shaft of circular cross-section intersects a half-space body at right angles. The problem is modeled by applying the axial tensile stress from simple beam theory to the half-space. Deflections are obtained both theoretically and experiment- ally. Shear stresses were not.considered. II. Selection and statement of the problem We will study the contribution of an elastic wall support to the deflection of a thin plate whose edge is at every point attached (bonded, glued, welded) to the surface or cast in one piece with the surface of the wall support. See figure 1. We will want to be able to specify the ma— terial parameters of plate and wall separately. Symmetry The axially—symmetric problem to be examined is as follows. A thin circular disk or plate is fixed at its edge to the wall of a cylindrical hole or shaft in a solid. The disk and shaft share the same axis of symmetry. Coordinates and geometry Cylindrical coordinates (r, 0, z) are used to des- cribe the problem with independency with respect to 0 due to symmetry. The axis of symmetry (0, z) is taken with "z" positive downward. The undeformed disk is described in the obvious way, occupying the space -b$zs+b and 0>b. Similarly the undeformed solid occu- pies the space -wa where an infinite solid is chosen to further simplify the problem. The positive direc- tions of important quantities listed in the table of symbols are shown in figure 2. Component theories The deformed solid is to be described by the classi— cal three-dimensional elasticity theory for homogeneous isotrOpic solids experiencing small strains. The deformed disk is to be described by classical elastic plate theory approximating the conditions to be prescribed at the common boundaries and compatible with the resulting displacements. Boundary conditions On the common boundary between the plate and solid let the radial tensile stress or(r,z) and the shear on the r" face Trz(r,z) be assumed to be or(r = a,-szSb) = 0:2 2 — .. = -31.. Trz(r — a, bszsb) T(l b2) where o and I are constants to be determined. Only the stress profiles are specified. It is not expected that any choice of o and T will produce identical displacement solu- tions at every point on the common boundary. If however 0 and T are determined by requiring displacements and slope be equal at the plate midplane (a,0) then displacements in each solution should be close for all coordinate values on the common boundary except perhaps near points of stress discontinuity (a,b) and (a,-b). In O'Donnel's [3] discus- sion of the analogous plane stress (straim case, where Figure 1. Thin plate with thick wall support 9 I I/ 7/ / \J \§§5§Q§S§§\ \Q§SS§§Q§5\ ‘EEEEEE; / / ? \s\\\\\\\\\ Figure 2. Coordinates and geometry /1§7Trz = Tzr /// 4;_i_( q / /L___. ‘”“fi a /’ O / r / / ‘::i::fi; jy/ /// r: \ /// A / /, u(r.2) \// I //»/’ w(r,z) in the solid // /// Iw(r) in the plate I L\ +4.--.” . _. 1..-... ,___.___,__.__._.. __.._. “IF“ Mm--.” N 1““'V T 3 H shear is neglected, he notes that if a linear displacement profile or a linear stress profile is assumed the variance of the general results from experimental evidence is approxi- mately 15% in either case even though stress becomes in— finite at (a,b) and (a,-b) for the linear displacement assumption. The stress profiles imposed are the major restrictions on the problem. First, solutions for the wall will be found and then the requirements of the plate will be examined in Section III. III. Mathematic development Axially-symmetric isotropic elastic solid In elastostatic problems without body forces the Navier equation [6, p. 88] becomes 1 > _ mVIVIIU~0o Galerkin defined a vector function [V2 + 265 s [2(1-v)v2 -V(V-)]F such that the Navier equation reduces [6, p. 119] to V“E'=o. In the case of symmetry with respect to the z-axis the z—component of Galerkin's vector is particularly important. Let F = (0,0,§(x,y,z)). If both the elastic body and its loadings are axially sym- metric the gbis a function of r and 2 only. Let §(x,y,z) = §(r,z). This last function is called the Love strain function be- cause it was developed earlier by Love [7, pp.274—277] using a different method [6, p. 130]. The displacements and stresses may be expressed in the following manner [6, pp. 129-130] as functions of §(r,z) where §(r,z) satisfies the biharmonic equation V“§[r,z) = 0 32 2Gw = [2(1—\))V2 — €57] 2(r,z) 2Gve 0 _ ’32 (r12) 2Gur _ 3r 32 2 o = 5%I<2-v)v2 - 37:] a _§ IvVZ - 32 ) Z 0r = 32 3rz — oe=_2.z 32 r 3r _ 3 _ 2 _ 32 Tzr - 3?[(l V)V 322]Z Ire - 0 I02 = O (3.1) Strain function A function §4(r,z) must be found such that V“§4(r,z) = 0 . . . . . . [6, p. 122] is satisfied. The solution begins with V2§2(r,z) = 0 where §2(r,z) is of the form §2(r,z) = R(r)Z(z). In cylindrical coordinates the Operator V2 is (32+_1__§_1 32+32) 3r2 r 3r F7 362 az'2 ' 2 — l‘. 2 Then V (R(r)Z(2)) — Rrrz + rRrZ + R222 0 and (R +lR)z+Rz =0. rr r r 22 This equation is solved by separating the variables. Choice of constant w2 and the minus Zzz gives results suit- able for the application to be made. The resulting equations ZZ _ 2 Z (.0 Rrr ERr become Z + wZZ = O 22 rZR + rR - rzsz = 0 respectively. rr r They are well known and possess the solutions Z = AlSin(w2) + BlCos(wz) R = A2I0(wr) + B2K0(wr) where Al’ A2, B1’ and B are arbitrary constants and I0 2 and K are Modified Bessel Functions of the first and second 0 kind and order zero. Boundary problem and Fourier integral transforms From section II the conditions required on the two boundaries of the elastic solid are these: 1. The stresses must be bounded at infinity. 2. The displacement in the radial direction ur must be zero at (r,z) = (a,o) where the midplane of the plate intersects the cylindri— cal wall. lO 3. The stresses at r = a on the interior surface of the hollow cylindrical shaft are 0 a. or(a,z) = 52 -bz>b _ _ E 2 _ b. Trz(a,z) — TIl (b) ] bszsb = 0 -b>2>b where o and T are the maximum stress values and (plate thickness "t") b = 2 (3.2) Consider condition (1) above. It requires that A2 = 0 since IO + w as r + w and would lead to unbounded stress solutions at infinity. Note that solution Z(2) may be separated into even and odd parts. Condition (3) above requires only the even part. Therefore let A1 = O. The desired solution is 22 = R(r)Z(2) = K0(wr)Cos (LUZ) It can be verified by substitution that Z4 satisfies V“Z = 0 when _ 3 £4 ‘Aéz + Brsi'i’lz and since —2K (wr) = -wK (wr) 3r 0 1 then Z _4 = [AK0(wr) + Berl(wr)] Cos (wz) (3.3) also satisfies V“Z = 0. 11 If equation (3.3) is a solution, then any finite or integral sum of such functions, having different arbi- trary coefficients A(w) and B(w), is a solution. The A(w) and B(w) are independent of r and 2 but will be considered functions of w in the Fourier transformations to be used in the solution of the boundary problem. The desired strain function is given below. £4 = [JgLa%[A(w)Ko(wr) + B(w) erl(wr)]Cos(w2) dw o (3.4) Substitution of this function into equations (3.1) yields the following expressions (3.5). 2Gw(r,z) = f7; i—[A(w)K0(wr) o + B(w)(erl(wr)-4(l—v)K0(wr))] Cos(wz)dw 2 _°°—1 2Gur(r,z) — /;\—5[A(w)Kl(wr) f o + B(w)er0(wr)] Sin(wz)dw At 2 = 0 the displacement in the "r" direction ur(r,0) is zero. Condition (2) is satisfied in equations (3.2), 3u (r,z) w 2G r32 = £¢§TA(M)Kl(wr) + B(w)erO(wr)] Cos (mz)dw Or(r,2) = éI/glAm) (K0(wr) + UEKIIer) -B(w)([l-2v]K0(wr) — erl(wr))]Sin(wz)dw _°°2_ Trz(r,z) — é%;‘[ A(w)Kl(wr) + B(w)(2[1-v]Kl(wr)-er0(wr))] Cos(wz)dw (3.5) 12 To satisfy the last boundary condition "3" (3.2) let the general expressions (3.5) for or and Trz equal "3" at r = a. —bz>b 0 o N “a -B(w)([l-2v]K0(wa)-waKl(wa))]Sin(wz)dw z 2 -b€Z€b T (l—[B] ) = f/Z[-A(w)Kl(wa) -b>z>b o n O +B(w)(2[l-v]Kl(wa)-waK0(wa»] Cos(wz)dw (3.6) There are now two equations which may be solved for the two unknown functions A(w) and B(w). Equations (3.6) may be expressed in the form or(a,z) = J; f[f(w)] Sin(wz)dw o T (a,2) = I/ZHgMH Cos(wz)dw r2 ITO (3.7) which are already the Fourier Integral Sine and Cosine Transforms of f(w) and g(w). These transforms possess the property of being the same as their inverse transformation, hence if the above equations (3.6) are transformed jngor(a,z)] Sin(wz)dz J§f[J%IIf(w)I Sin(wz)dw]Sin(wz)dz 0 o o f(w) l3 J%}[Trz(a,z)] Cos(wz)d2 J%T[J§I[g(w)l Cos(wz)dw]Cos(wz)dz o o o = q(w) A(w) and B(w) can be determined explicitly by substituting expressions (3.2) for or and Trz in the above equations. b 00 J%W[%.z] Sin(wz)dz + )ZU[0]Sin(wz)dz = f(w) o Trb b C!) X§I[T(l-(%)2)] Cos(wz)dz + JEUIOI Cos(wz)dz = g(w) o b Integration yields the following results [8,pp. 80-81] b I)??? [%-2-Sin(wz)—Z-];—2 Cos(wz)] + 0 = f(uo) o fiql Sin(w2) - l (22 c ( )+{ZZ — 2}s‘ ( ))]b R w 37- 57 05 OZ 5— ET in wz o + 0 = g(w) @%I§%HEEL - Cos (wb)I = f(w) Vng§;[Slgéwb) - Cos(wb)] = g(w) Groups of terms that appear frequently will be given a symbol for brevity. Sin(wb) Let S(wb) = [ wb - Cos(wb)] then 2 o _ 463(wb) - f(w) 5-533:- smb) = 9(3) If the expressions (3.6) and (3.7) are used, substituting for f(w) and g(w), a matrix form of the result becomes l4 Ifl g-S(wb) R w _ : wzbs(w wb) = E{K0(wa) + afixlman {-[l-2v1KO(wa) + waKl(wa)} A(w) _ {-Kl(wa)} {2[l-v]Kl(wa) — waKo(wa)} B(w)_ Let K(wa) = K0(wa)/Kl(wa) and divide by the factors Kl(wa), S(wb), and J3, yielding the form -I [9. w 2T -5731 {K(wa) + l—} {-[l-2v]K(wa) + waII J§"A(“)K1(wa) wa S(wb) {—1} {2[l-v]-waK(wa) JZ'B(R)K1 (Pa) - _ . S(wb) — The functions A(w) and B(w) become _ 2 A(w) — I);S(wb) [oa(waK(wa) - c) D(wa)Kl(wa) 6§%%(w2a2 -(c—l)waK(wa))] -J§S(wb) B(w) = 2Ta D(wa)Kl(wa) [Ga + waTbwaK) K(wr) - (1 +waK(wa))(wr — 2cK(wr))}ICos(wz)dw R -3Gur(r,z) = g S (wb) K1 (wr) wD(wa)Kl(wa) [o{MaK(wa)-c)—er(wr)} + 2T wawb{(w2a2-(c-l)waK(wa)) -(l+waK(wa))er(wr)}]Sin(wz)dw l6 fl G311r w S(wb)Kl(wr) -§--§;—(r.z) = g D(wa)Kl(wa)[0{(waK(wa)-C)-er(wr)} + 7:59 (mZaZ- (c-l)maK (wan -(l+waK(wa))er(wr)}]Cos(w2)dw n w S(wb)Kl(wr) 1 530r(r’z) = g DTUgTEITEETIOLOaK(wa)(K(wr) + a?) -(K(wr)+wr+6%q} 21 - _i + wawb {(wzaZ-(c-l)waK(wa))(K(wr)+ wr) + (l+waK(wa))((c-1)K(wr)-wr)}] °Sin(w2)dw --11 (r 2) 2a r2 ’ w S(wb)K1(wr) = g D(wa)Kl>a or r< a or 26 2 Dw(r) = Iw(r) + A: + C dw(r) _ D—dE—— — Iwr(r) + Ar (3.21) Two limiting cases are of interest for purposes of compari- son later 1) Rigid wall support: Gwall + 2) Simply supported plate: Gwa11+ O and terms involving T are neglected. 1) Let G + w from (3.19) 0- 2b3 A = ( -§-9Iwr + O _ ”EEE 3ab3 + o a 3 from (3.20) _ a2 -Iwr _ a2 C—O 7(a)-Iw—-2—5Iwr-Iw from (3.21) * Dw(r) = [Iw(r)-Iw(a)] + l¥§(a2-r2) since Iw = Iw(r) at r = a dw(r) _ _ E D —aE—— — Iwr(r) aIwr(a) since Iwr =Iwr(r) at r = a (3.22) 2) First let T 0 then multiply numerator and denominator of (3.18) and (3.19) by g and let G + o 27 A = -(il:§LEE§ - 0) Iwr + (+uzs Iwrr + O) -(l + v)uzs = (1-v)Iwr _ Iwrr (1+v)a (1+v) 2 - C = [O + 0] _ a [(1 v)Iwr Iwrr —7' (l+v)a — (1+v)] ' IW from (3.21) - 2_ 2 Dw(r) = Iw(r)-Iw(a)+[W .. avg“ (r 2a ) dw(r) dr = Iwr(r) + [(1-v) Iwr _ Iwrr D (l+v) a _(1+v)] (3.23) IV. Illustrative Example Further investigations and comparisons of interest would be awkward to carry out in general. The remaining text will deal with specific loading, geometry and material parameters. The plate loading Let q(r) = q a constant -3 a Recall T = 153-: rq(r)dr then T = 2.3.53. 8 b q Recall r l B Iwrr(r) = f ? f pq(p)dpd8 o 0 then r 2 2 Iwrr(r) = q f l%§§0d8 = 3%_. o B 2 and Iwrr = Iwrr(a) = 9%— _ l r _ g l r 3 _ qr3 Iwr(r) — E £ 6 Iwrr(6)d6 — 4 Ebé 6 d6 — —l€ 3 = 9__ and Iwr 16 m 28 29 r r Iw(r) = f Iwr(y)dy = 3 f ysdy = fir” 16 64 o o = .9. k Iw 64 a . y . dw(r) With the above formuli DW(r) and D—aE—- may be found by algebraic substitution. The load q in the examples is a multiplying factor in the expressions for w(r) and is set at 100 for all numerical results. First the two limiting cases in the previous chapter will be examined. Limiting cases 1) Rigid wall support: equations (3.22) become Dw(r) = 3% [r”-a”] - §%:«r2-a2) = €%{r“-2r2a2+a“] dw(r) ...—.33- 2 D dr 16[r r a ] which are the classical plate equations for this problem. 2) Simply supported plate: equations (3.23) become _ _ qaz (l-V) ._ ...—.12... _ Dw(r) — €%[r” a”] + 2 [16(l+v) 4(l+v)](r2 a2) = €%(r”-a”) - q§:[%$%l(r2-a2) dw(r) 3+v D dr T%[ 3‘(I¥U)ra2] and again these are the expected solutions. 30 Plate geometries For purposes of illustration three plate geometries will be used. (a) Thickness to diameter ratio b/a = 1/10 will be used to explore the maximum differences be- tween the rigid and elastic wall assumptions even though it is recognized that "thin" plate theory is probably not a very good model for this ratio. (b) Thickness to diameter ratio b/a = 2/100 will be used to explore the differences between the rigid and elastic wall assumptions for a plate that "thin" plate theory should describe quite well. (C) Thickness to diameter ratio b/a = 5/100. With the geometry specified the quantities ws, wt, uzs, uzt from (3.12) can be found using the integration techniques as explained in Appendix 1. Material parameters For the given loading and thickness to diameter ratios the ratio of the elastic moduli of the plate to that of the wall Ep/Ew will be allowed to range over eight values from zero to ten where zero represents the plate built into an inflexible wall. Ep/EW = w does not repre- sent the simple support because w(r) # O at r = a. 31 Each of the above cases will be done for v = .25 in both wall and plate and for v = .30 in both wall and plate though v in the wall and the plate need not be the same. Results The deflections Dw(r) of the midplane of the plate together with corresponding maximum stresses 0 and T at the wall are displayed in Figure 3. It is apparent that for each of these thickness/diameter ratios, the deflec— tions w(r) based on the assumptions of a rigid wall and of a wall and plate of same material, never differ by less than 300% for any given r. Similarly the maximum radial tensile stress 0 at the wall never varies by less than 200%. From a practical viewpoint the latter variation is on the safe side. The effect of Poisson's ratio varying from .25 to .30 is relatively small but becomes more sig- nificant for thinner plates. As expected the deflection at the wall is small for thinner plates, while rotation at the wall is still com- paratively large. There is little reason to believe that the trends indicated by this example would change significantly for other usual loadings and geometries. A3339 AMESBQ A .n \\\$m. AM\HV3Q .\“v\\lmm I \x.\ mNO”? |.|.| N. \.\ OH \HNOM' .II I . \ . aim... m."9|l.ll NI OH .\HHHWoMI \.\ .m\\\““mfl.on 3.0:» on #3: .\\\.\._H . u \ .\\.\Hu\\lam.vu mm.u> no... >93 3 \.\. T \x. \me mm x .\ ..\ ._ Hmsvm wanmmc mum .u> .\\H\H._Mnmmvu \H\ .“H‘IVMN6MI \\ .\\n\\m mmél czonm uoc wwmwmmnz...\\wm..\wmmu .\\ .\\\H\\\\u\m_‘ma.mmu .\\ \..\\\ .\\ludlmafin .HOmu mGDH R.VUMHHHHHow¢mI \\ \H\\\.N \meomml \. \. 0%..4” mm.OHl \ \. \ul .l n u \. "\\ .\\.\\ .\\ 9.3 \H\\\”& H #03? \ \u\\ a). L153- \.\Qx u\\q..a \ momT \\\\ \ \.\YJmmmm: \ \\u\"\ .\\x. .\ m \\\m\\ .\ .\\“Wnfllmmmal \.w\\\\\.\m..\\.\\.$ m " \.\. ..\\mm\ ..\\ \. \mfl .\. m3: \ \. \ .\ T} m \ x." .\ .\mm 3: \.\\.\ \.\ .\.\. m \.\\ \.\. .. \ \\..\ \.\..‘m u . o . o . c »\ c | u . . 0 MM .\\\\.\\\.\\ WM ml‘ \ n\\\ \\.\ .\.\I\om.mNH ..\ A.\\\.\ .\ H .ull\mm. .mHl .\ .a . .\\ A. l \. .\ .\ \.\ \M\\. ..\. \. \. .\ 1100.03: .\ \. \ .\.o.o 1 \ \.\ .\ \.\A N.— \ .\\\ \ \.\ . .\\.\ . .\ H + u\ .. . . ..II.\.wvmmi . . . \. 0.0 l ..\ .\ \ 3 ”a . \.x .\ \...\\ ”.\\\ .\ \. 9 a 9 \..\\\ m m moa\o \. \ .\ O O I. H\. .\.\. 3 AH H\..O m\..\..\. _ p :\\ \.\\ >nm 4/ \u\..\\ m\ m we \“.ML 0 .t. w ._ N ._ O p\\\\m\\. 3M\ m oa\o.hWWfiHW\\ on v.“ m» o m L“..\ l ._1 Mn 0 _ «.\m m m SHE QM wJ w v omens. ooaxoacw H\mu0\n ooau mnmaup oo lg ooa\mHM\n coal coauv .m musmflm 33 Other results Also of interest is the behavior of the wall. For the case where b/a = 1/10 the displacement Dw(r,z) where z = 0, r>a is displayed in Figure 4. These results are obtained by using the technique described in Appendix I to integrate the first of the five equations (3.10). This equation is shown below. . e-wr{eer (wr)} ( ) a Iw(Cos(wb) Sinéwb)) wa ma 1 G w r,z = — - — n o e {e Kl(wa)} {ewaK0(wafl w((wa)2[(——+-————4)2-l]-2(l-v)) {ewaKl(wafl wa wr .{%{(wa{e K0(wa)} + 2(l—v))({e K0(wr)})_ wr} {ewaKl(wa)} {eerl(wr)} {ewaK0(wa)} {eer0(mr)} \ + {(wzaZ—(2(l—v)—l)wa T (wa)(wb> {ewaKl(wa)}l {eerl(wr)} {emaK0(wa)} {eer0(wr)} —(l+wa——BE———————Q(wr —2(2(1—v))——zfi?——————4} {e Kl(wa)} {e Kl(wr)} °Cos(wz)dw where o and T are obtained from the previous results and the exponentials are required in the computational algorithm for the Bessel functions. Having obtained Gw(r,0) for various r it is now necessary to apply the scaling factor 34 .HAIIAI Ham3 mumam may ca map CH I s .H u \ 0-10 us a muoz .-roa. Ao.nvsir Amos 3”: NA... g LHM :> mamfimxw m>fluwnpmsaafl may now mpcmamomammflp Hams mo mcflamfimm 4 .v musmflm 35 g in each case to obtain Dw(r,0). 3 2E b E 4b3(l+vw) 2_ 3(1-v) =<> _ G — E 2 EW 3 1 vp 1+vp) __W__ 2(l+vw) - _ _ .1— In thlS case vp — Vw — .25 and b — l l E 4(-———) G E 3 W 3"4‘) 9__fi§'E—w In a variety of related problems it would be useful to know the behavior of specified points in or on the wall responding to a unit load 0 or T. These are essentially the integrals in (3.10, 3.11, or 3.12) and were used to obtain the previous results. At the end of Appendix I, figures 5, 6, 7, 8, and 9 display these results for b/a = %0’ v = .25 together with the output stresses (3.10, 3.11) which were used as a computational check. It will be noted in Appendix I that the apparent uncertainty of convergence for some of the values is exaggerated in the display which gives the bounds on the value of the infinite integrals. V. Summary Limitations in theory The generality of the mathematical model of this pro- blem is limited by the two component theories used in its solution: classical, three—dimensional, small-strain, theory for a homogeneous isotropic elastic solid used for the sup— porting wall; classical, small-strain, thin-plate theory without mid—plane forces used for the plate fixed to the supporting wall. The other two limitations on the generality of the theory are: choice of axialsymmetry to obtain a non-plane— strain solution in two dimensions; selection of the stress profiles at the boundary shared by the two bodies, thus the matching of the slope and displacements at this common boundary is limited to a single appropriate point. Applications Within the above limitations the problem is solved for all plates without holes or inclusions and having uni— form thickness and material parameters. Any load q(r) is admissible. The solution for the wall would allow the other plate solutions [i.e., with a hole and/or inclusions and 36 37 also having variable thickness and/or material parameters] to be developed in a parallel fashion with comparative ease. The solution allows the wall and plate to have dif— ferent material parameters. The solutions for the shear and the moment applied to the cylindrical wall are easily used together or separately to solve a variety of other problems where such a wall is similarly loaded. Since the mathematical model for the wall is solved for all of space and since the solutions all decay with respect to increasing r and z, these solutions can be used to approximate those of objects with sufficiently thick walls having cylindrical cavities. In the case of the self— equilibrating moment, where shear is small, the walls need not be very thick. REFERENCES 10. REFERENCES C. Weber, “The Deflection of Loaded Gears and the Effect on Their Load Carrying Capacity," Department of Scientific and Industrial Research, Sponsored Re- search, Germany. Report No. 3, part 1, England, 1949. N. I. Muskhelishvili, "Some Basic Problems of the Mathe- matical Theory of Elasticity," P. Noordhoff Limited, Groningen-Holland, third edition, 1953, p. 471. W. J. O'Donnell. "The Additional Deflection of a Canti- lever Due to the Elasticity of the Support," Journal of Applied Mechanics, Vol. 27, No. 3, Trans. ASME, Series E, Vol. 82, Sept. 1960, pp. 461-464. R. D. Cook. l“Deflections of a Series of Cantilevers Due to Elasticity of Support," Journal of Applied Mechanics, Vol. 34, No. 3, Trans. ASME, Series E, Sept. 1967, pp. 760-761. J. M. Brown and A. S. Hall. "Bending of a Circular Shaft Terminating in a Semi-Infinite Body," Journal of Applied Mechanics, Vol. 29, No. 1, Trans. ASME, Series E, Vol 84, Mar, 1962, pp. 86-90. H. M. Westergaard. “Theory of Elasticity and Plasticity," Harvard University Press, New York: John Wiley & Sons, Inc., 1952. A. E. H. Love. "The Mathematical Theory of Elasticity," Cambridge University Press, fourth edition, 1927. Richard Stevens Burington. "Handbook of Mathematical Tables and Formulas," Handbook Publishers, Inc., Sandusky, Ohio, third edition, 1949. S. Timoshenko and S. Woinowsky-Krieger. "Theory of Plates and Shells," McGraw-Hill Book Co., Inc., second edition, 1959. J. S. Frame. "Numerical Integration and the Euler- MacLauren Summation Formula," Michigan State Uni- versity, 1967. 39 APPENDIX I APPENDIX I Numerical Integration The integrated quantities in equations (3.10, 3.11, or 3.12) must be computed for the illustrative example of Section IV. Examination of the functions to be integrated re- veals several things. 1. They are reasonably well behaved at the origin even though some of the component functions are not. 2. They have a periodic behavior for large values of w. 3. They decay for large values of w. The partial sums of a good numerical integration procedure should therefore converge within any specified bounds on the desired value "I" for sufficiently large w and suffi- ciently small increments h. Summation formula A suitable numerical integration procedure is ob— tained from the Euler—Maclaurin summation formula and is developed as follows [10]. 41 42 The Euler-Maclaurin formula may be stated thus Ch=h[%f(a) + f(a+h)+ f(a+2h)+ . . . +f(b-h)+ %f(b)] _ 2 4 6 Ch— A + b2h D1 + b4h D3 + b6h D5 + . . . 2313 — R (h) I O Q + b 25-1 ZS h 25 (I-l) Ch is the approximation to A where b A= f f(x) dx a h= 2&3 m is an integer to be specified k b 13k = ——d ff“ 1 = fk(b)—fk(a). dx a . . 25+2 R25(h) is the remainder of order h for small h and _ _ -1 b0 ‘ 1' b1 ' 2 n-l Z br/(n—r)l = 0 for n>2 r=0 _ l _ _ —1 b2 ‘ 1_2' b3 ‘ 0' b4 ‘ —720 b2k+l = 0 k>0. "A variety of formulas for numerical integration are obtained by taking weighted averages of Ch’c(2h)’c(3h)’c(4h)’ . . . With weights chosen to eliminate certain terms in Dl,D3,D5, . . . ." In this case the elimination of D1 and D3 terms keep the computation in manageable proportions. 43 From (I-l): _ 15 2 _ 15 4 15 15C(h)‘15A+ I7‘h) D1 736(h) D3+ 3024o(h) 6D5 = -6A+ T6(2h)2D- —(2h) 4D3 *(2h) 6D5 C(2h) 12 l 720 +30240 c = A+~l(3h)2D- w(Bh) 4D3 + 1 m(3h) 6D (3h) 12 1 720 30240 Adding these equations (15C -6C +C )- 10A+ h6D5 + R (h25 D ) for s>3 h 2h 3h _ _lé-T ZS 28-]. l . . . . Then I0(15Ch-6C2h+ C3h) IS a good approx1mation to A If h and D -l are sufficiently small. Note that h and D Zs 2s-l are functions of intervals (bi—ai) for sufficiently smooth f(x) where the required infinite integral I is broken into n a sum of integrals 2 Ai . Also note that the minimum value i=1 for m is six for this method and it may therefore be con- sidered a seven point formula. To obtain this formula sub- stitute the first of equations (I-l) into the estimate A = 1 _ I0(15Ch 6C2h + C3h) and let m — 6. 15C =h[i% f(a)+15f(a+h)+15f(a+2h)+15f(a+3h)+15f(a+4h)... ( h) 66(2h)=h[; 2 2f(a) -12f(a+2h) -12f(a+4h)... _ 3 . C(3h)_h[ §f(a) +3f(a+3h) 44 N l _ A — lo(15Ch 6C + h 2h c3h)= TO—[3f(a)+15f(a+h)+3f(a+2h) +18f(a+3h)+3f(a+4h)+15f(a+5h)+3f(a+6h)] (I-2) If m>6 it must be a multiple 6n and the formula is the same except that the terms f(a+6ih)are multiplied by 6 instead of 3 when i # 0 or n. This latter form reduces re- dundant computation where partial sums are not required and will be referred to as (I-3). Computation and limitations The f(x) in this computation has a decaying periodic behavior owing to the dominance of sinusoidal elements for large values of x (w in 3.10 and 3.11). It is expected that the successive partial sums approximating "I" (i.e., I 2 A1+A2+A3....+An) will exhibit maxima and minima for intervalsl (Bn - an) sufficiently small and that these values bound the desired value of infinite integrals "I." We choose to con- trol (Bn—dn) by partitioning the sinusoidal half—period (HP) of equations (3.10), HP = D/b, and (3.11), HP = Da/b. For b/a = I% and 5% the summation began at x = HP and ran in both directions. We wish to approach limiting values of component functions K0(x) and Kl(x) at x = 0 from the right. Summation formula (I—3) was used for x = HP summing toward 1The symbols a and b of the discussion "Summation formula" are changed to a and B to avoid confusion with plate dimensions (r=a) and (-b0 A O -...06 A o +Z - / _._.0<9 A \A --./0 o A J Z 050:) UT/Z) /. —.06058 -.006 77 2. —.000// -.0035 /O. JKOOO3 . -.OOOZ3 J'l . F/GUEE 8 A 025M) 0 UZHZ) ‘4 13 —.2 " -./ 0 + . r I J ) )' M o UZ5(—'Z) = 025(2) \ UZ7'(-Z) = UZT(Z) __.OZ . . \o \\ __.04 A O \ __.06‘ A A G 1-08 z 0250:) 02 7(2) \\ 09 {+.// 704.0 . ) ./O :3? (3 ‘H {-1244 /Z ‘ £0 +.00/ +005 ' " 2.0 +000/ +002 /0.0 -.0002 +0000?~ Q 6\ 52 30.1.8. [\.\xqi J./. 03 mm\ 6N4 3; NN.\ owe m: m: 1.: N: 92 m3 / “.3 N3 3 l. _ _ _ . r P b p _)))_ _ _ _ G c _ O .11 a _ d u q d H q q _ _ q 4 d / «o A . win. .40, 0,0 ,. o// orig... ill/II, Ollie //)o a // 4: 0/ a. O / a. “350: APPENDIX II 54 If .3n_._ . 11/06/67 _,70.0050047777270047/0) Fae/70721 I .3 COM/30755 PROGRAM 3 T N U ))z___,.. [004770.05 ?.// me 2%» X>0 . . ,? C FRAMES 7 PT DIMEUSlON AUOOO) {/OOCLssons I'CLtLO‘ _y_ READ 100.HPS.HPR.PHP.AQ RT. 8. T WSa-JT U S. U T, UZS.UZT: PR 2 __LQLO_.E.01IIIALL5F10 0),,.,:\_ ML thickness. “£21211?“ 012.... 1,, ___,___,_,,.., __ R33T/AR [0204.6 #0.de (3324,12 .,'3R,__1___ ,.,,. artcttoiux: (Acrflaér QkLOC£ _,,_,__- ___.,_,...l_..,,.__.,_._._,il_. PI=3.14159265358 f f ’ fl ______11_X3fl85tallR___fl____ m¢(__ ectocilKQJu c _ H=PI/(R*PHP) ‘/ F j K=HPRwHP \h..(,£_lpL)-L.L(;,_.st,aht, H06=h/5. CALL FNC~J_C.LRIAK1.ZL,, .Hzxo,._SlTol-'ISIHIIU S:U_._I1UZSIUZTI,,,,,AIM,),,, $0=S TQFT W50:NS NTinT U SO:U S ___,_.__,,-.U,,,T0=U_ T...,.,.,__.,..,_-,,.,.._...,.i - - -- .- - _ UZSO=UZS _"UZIO=UZTL_L_ __ DO 900 N=1.K __._,___ _X= X- H05, ‘ ____ _.-,_.,.,_._l_, CALL FNC (C.R,AH, Z, H,X, S I, HS: NT U S. U I, UZS, UZT. A,I’.) __,__W,Sl.=31+3 .,.._,,,_,,,__,,, ,._,.l_ 1., - , WW . T1=T1+T .___1"___U51:081+DS NI1=HT1+wT _nn_mm__flU_Sl=U Sl+U_S, _L._1_ U T1=U T1+U T _____u__-U151=UZSIvUZS._W_. ---”,N1,__m,-l 11- -1 r , "J 1 l , 1 , UZT1= UZT1+UZT _._.__-_ __.,,X_;_X'HOC,1,_.1,_______ _______ _ l______ _,,_,___ _‘.__,. _ ,,,,__-__ CALL FNC (C Q, AH. Z) HoX)— S, I. NSDwT, T) S, U IgUZSHJZIa AOM, _“__1___1_52=82+S, m. '._l,ltllml.l,_ T2=T2+T ..lll___1MSZ=NSQ+.S _WLWVMMMM,.lLWLU__l_W ,wu_m-_,ni,ll"11,ll,,,-,_._ ., - HI2= HT2+WI __,U 52: U SZ+U_ S,,,.______._,_,_,._,, ,,_,_,.,_ 1---, __1 U T2=U T2+U T _h___-11U232=U232+UZS,_L_,_____ ll_1, UZI2=UZT2+UZT ___._._____._X=.X.-.,‘:i06 .,__-,__,,-.---.,._._L,. -,---1- .-. -.-,“ W- 11., .. .,....” ...... CALL FNC (C.R:AR. Z. H:X. SJToNSaWIoU SoU T,UZS,UZT, A,M) __ _Sé:83‘s ____4_wv - 1 I5: I30T —__.. H85=k53+MIS _.-“,_m_"__n.__fl-ul_ _ --- _ HTS: wT3+WT —__«11."U153=U 83+U S 1_-_ - 1 l.” -- -1 _ -1 1 1- l- 1- L 1 - . U T3=U T3+U T _ZUZS3=U753+UZS_111“”U11__1-.___- .1111 UZT3=UZTS+UZT ~— -.-X3X2H05,l. ,L_1 .___,.. __._, ,,_-,.,-.__ -,.-..Wwi, ,, CALL ch (LAMAR, z. H.x. S. T Us WT U S. U T" 028. UZT. AM) ~—_.~__.1_SZ= 52+S --..“ 1, _U-l,wwllwil_-1 _ .,.m ”_lm,_ll_.11 ..1 T2: T2+T ,3». 11/06/67 _-.-..-.,..1_-- W523 RS 231» S 11 ,_ 11.11 1 1,, _- . NT2=WT2¢WT -——~——U—.$2=l.1’§2+l} Q U T2=U T2+U T __1_____JJZSZ=UZS2+UZQ ' UZT2=UZT2+UZT XQXanfi _ -WM1WW”11111- CALL FNC (C;9;AR3 Z. HpX) 81T,NSJNT1U S5U T,UZS.UZTI AIM) _Sl=51+9 ‘ T1=T1*T ___-___H51=«S1+fi8 NT1=WT1YKT _11__.__JJ.-51:U-SJ_+U s ...... 1,1. 1.1111-1.1.1.111..-.-.-1.1- U T1=U Tl¢U T -_U251:U7ST+UZSh# - 1-1 UZT1=UZT1+UZT ————-~—X=X-H067—~— CALL FNC ST~15. NT1=UT1*15. ...-.,-..___-1_U , 51 =U 31 . 1 5 .,, _,.-,.-__1. -1. 1-11111. 111- .1. 11 ,, 111.1111. _-1.- - -, ....-.--1 .. ..1 .-. ....1-1- 1 . _. U T1=U Ti*15. ' ~~~~————»UZ s 1 = U z s 1 t 1 5 . ~- — ——-»- ,,_.___-1--- ~ ~—r ~77- --,_W1 W1- ~~~~~—~—-~- — -——- -; —- '~- UZT1=UZT1v15. ““”""‘ “” “"“'""' " " " ‘ 111111111152=(-152+ .30).3r_1_11-, x ——————— — T2=< T2+ TU)'5. “”“ ' --—~~u~—mH32=(7N52+ w30)*3.~ 1. , ~-~~ HT2=< ~12+ wT0)*3. ~“—-— ' ———-—-~—U~52=(U saw so; 5.1-1.1... -- .1 111-.- u T2=/A<2) ~ FA=1r+xg¢K - -Mndq-- w- “_MV "w” w __.fl_-"~r. ' F8=x2th<~v2-*.)-C - .... CAOB:C§F'A/FB . w ....-- , __-_....._..__-.__--—---.~-. .- ‘,-- ---.-m. SCZ=FS*CZ -m..-__.__.,SSZ:F-S tSZ . -..-V..._.,,-. ”-...“ , ---m, , - , -V . , ,---.-...._ .. - .--- ., -i _. .---.-~-_.... - S = S SZ/X ---mw.__.._.__'._1_:scz/-x2 -.w_--_-.-,.--_-_-- _W_ -,...... w. -... «-..».-i -._,-__-.- US :T *(1.+CAOS) ~W_Mwwm___.u1m=sczt<1,+CA08*FA)/X4.w.pr~w%_m__rmm”WWwwnwwnwmwwmwmwwwwmmmmwwuw.fim_ US=SIFR Mm__.ul,:ssz.(_1. +C,—mr-3 ) /x3 . .H--.-.....__...._.. ”4-...“ ....“ .. .. - .--- --." UZS : SCZ/FB ~._ 921;;ng -.- -. __._.-.--_.-___.,_ -..- _.-__-._..-.-_. -,.,-__-, ~ ....-. m"-.- _,_ -H.__--.. _‘ h__ ...- 7. r RETURN -_J_H_"___m_.5¢u1___“m_mfl_”m_.____._________m”_mmm__,m“_wq_- _a“__~_,_~_m.m_*n_mu__An "m,,n,w,_ {D ,- l - z A " :3 :94; :1? Y DECK. t 1. 5’. l1 .‘4 "xx .4.'Q-€-_§_/<_,::5.5.5kl“(J .."/.5 V! "I /.) . u.5.no.9uo.7.w ‘“ ~_.—- ...— ....._.-- - —_ - __.-. --.-- . _. -. -.,‘-_- , ._ _-.___._.-_.-.‘_.-.._..V-. _. ...-.. . _ “-f K7 . ‘2’ , {HEQLJ’UMEZCAAJ/KEGQA EOA/IO/Q/VUM [‘2 . (OM/307225 ,_1_110_é/_§7_____ -...-__ _ “PROGRAM 9 5 T “EAL“??? EQUAT/O/MS 43” FOE fl€<2§<°°_ ' ”' c FRAMES T DIMENSION A(1000) t ~N+@~q-mv -fi‘ v“ 'READ 100.HPS.HPR.PHP,AR.BT. s.T,wS.wr.u s.u T;uzs.uzr. PR.Z 100 FORMAT<5F10,0) ,¥»-._’__ 1- R=BTIAR C=2!12x'p8_ .,W Plé3.14159265358 X=HPSiPI/R' H=Rxxca~PHp) K=HPRtPHP H06=H/5, M“ iHoéllElfilQ;) F1=2,*r F8=F1IR “.,...“...w 7..” --...., .-., .. .7 ..A, ‘ A. r.~._..-r-,».....‘~..r ~wa « "4'... w .... ¥w V.-......_.r._..‘ --.. < . CALL FNC w__fin 0232: (UZ$2”“”"“‘)§3:“‘””’”””““ ”"‘””“*" ‘ ‘ UZT2=(LSZT2 "j:§i 7”__ $3= 83*18, A ‘ ‘“' ‘”_24" T5= T3*18. # h - ‘_ “W _ ,- - _ A .__w__ _“_____.___ WSSin3N18JWMGT _ UT3:VT3*18. A U U U ___”—_ ’"flwafi335UrSSt18gfl*umw2"‘w- N- _ 'TN5,3A U T3= U T3¢18. 11/06/67 qu3E u233~ia ” uzr3= uzr3.1s W900 P S=F1¢( 81* 82+ S3)*PS P T: F2¢¢ T1+ T2+ T3X+PT PWS: F3.(131+ w82+ LSS)+PWS PNT= F4o o. _ . READ 100.HPS.HDR.PHP,AR,BT. us.wT. PR.Z.R mains” Ania. 0‘. 0-)--. m - ___-___m-__._-mm-,--,_.,_i,..-._._ -_ -_ _- _. C22.02.*pR :, P153114159765358mawumwmnmw,mumu_1*Mm_mumwmwwwmmememm xéHPStPI/BT ‘ H.381 [(BDPHP) _v ___fi____,.----_-_____-V_____w__._»-, _ ‘ ______--___. . K=HPRtPHP ML. _ mic ec.n‘:z§‘;‘a‘i:2:‘ x, “ism?” “m5“ .WHWHSO=HSWm ' HTflawT 5 DD 90.0.__NE-11-K _______.__---.“__.___________.__-_-‘__ .______.--___,,,--____-__,.-__.____.,,7...”,,“_-____--_-_7. X=XpH06‘ _. QALKM_MWWWMEQCMJEJBLAB:8:111“ _____ )VL_N_WS!HIIMMW_MALM’M__HNM+" v N31SN31*NS ----k‘liz-‘HHJ 138.1 ---- X=XeH06 "mu__u_w__wCALL”“Jm__~FNCL(C.R.AR.BT;Z._“MX.._-”5.HTa”~m_uAoH)mmwMW_.L.L”- H82=U82+NS --- __ .___,H,.__.-,._.-,-HI2=N12{Kiwawwuhwflm_. ..- _-_. -_ Wm.“ _-____. -___.-_-___.. _ _ _. _. _ _- _ x=XsH06 "WEALLkmmwummFNCJ(C.R.ARgBT:Za_ “X. _ W$,HT.W_MV_A.H)JU_ u53=w53¢ws -.- __._~____,_.___.. “HESS-€191.13?“ _ .-._________‘__-_______ ___- _______. -. ____ --V-.-__-___,._ X=XHH06 - _-__.____.._.CALL ---.V-‘m_._FNC ‘ClptAROBT!ZJ X. ___N$.NT; V ADM,” NS?=WS?*WS .mmfi_mfir MH_HHT2=wTe+wI- ~_ “um” w_h-d _...HH _N..w .w” .Vfih---t- M,N' :XRH06 ' ‘ ”MMWW_m__m"m§ALLWM““_MMENCWCCs”:ARgsTIZI “X, _wsawT:u_n AgH)UM H31EW51*NS ---”wm_~_“mnwfl_mkt[1=krfi+?£[.HM_ "h_wm__uu,--.Mme_L_H_w"_,n_.__-__mwu M,__m__qn_-m..fl~~_,_vffif X=X9H06 ”CALL WWVWMMFNC_(C:9:ARgBT:Za_ ‘X. Hspr; . A.H)_ HS4$NS4fiNS _HHMW_WMQQQMNT43HT4fHL“WWALWWMWM,M HH,MJm‘v,H.I- 7,MW_,wmrmwmwyfiiq,ka+_ N313W81*15. ' ,WWH_M_*__~J[11=W11*1I5.__v”mmm_m_n‘,”_wwwwwflw ,, *TWUW_WMWF, 7 vHNHW__Mw~m_ «__,,H, H523! NSZ‘ HSfl)*3. _y72=(“H?2#_HIq)*3. HS3$HSS*18. .HI3:NIS*1S. _ “Mm ___¢____ ,U_LMWHHML W“H_J_w‘,fiwfl,m H84: WS¢*6.*3.* HS -~_uMJumH_..m_HI4:"HT4tb.93.t HT ‘ _ WT '_ Mm,w Wfi F=H06/(P!*10.) -*53=AP*F_HW F4=2.*F/BT “_._fiflSaF3¢ b Rx: RtX L_ ,_ _ X2=X0X ; AX=AR*Y ELLLWHMM__WH__Q”flu~-H _ ,__ AXZSAXoAX _.-” Bx_=8TtX ”H _«_mm.. m “_ -m.“. C... ‘ ‘“*_*b* _ zx= ZKX ____ __CZ§C0$E(ZX_ L ML~L__-_ LWLNW_L ”_~_- _ Fs=SINFtBX)/BX.COSF(BX3 ;_ V304 ”MW"L~.LLL__ CW” .L-LNW N31 .Uw_*w_m_wmm_CALL BESKCAXpV;N;A§H3 mwmmm_xfldm_gufimwm”JNWWWLMMwWWHM_me_Hh K:A(1)/A(2) ____,_W_L_WWA2%A(2LA “w.. _ m “LLB-“ AXK= AX¢AK "a AA: AXKAC “w_ _M___"_____‘_"_-_L,____«”M_«_ __ _ A8: Ax2+<1.-C)~AXK Mm‘-mmmuww_ AC§1.*KXK-. ,M, -MWMMMWUH,,_ “ - , “A” , _.L, L”- m AD=AX2*(AK**2-1.)-C ------..-.___.-._____CALL BESKmx v.N,A.M) --- --._,,-________----____._-_.h___.-.___--_C . _- w“ RK: A!1)/A(2) _qfl,qn_.._W_~R2=A(2) CWCMWLW _“m mwm-.u “Mp mwvwwhw,wn_-wn_ -wC HI=SCX3*(AB*RKeAC*RC)__MMMWEM_JCC“-MW_MMCJ,_MJQWH“MHCCWWLMU-uw-flfi,hr Z. G t.- -- _m-..-_~____-.____----- - .. FTLL15A_-W[CALflt/I£6M7ZOMEOIQHULA I «Z --.---_-___o.1z.o.9/98_-- -- .___._.__.--W.WPRPGRAM “AFORAZ .....COMPUTES THE 1 st OF 3. /0 C“ FRAHES 7 PT . = DIMENSION_ A(_1°M0_02.__-.. _ a“- FOE HP