Group Title: orthotropic circular disk subjected to its own weight when supported at a point
Title: An orthotropic circular disk subjected to its own weight when supported at a point
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Title: An orthotropic circular disk subjected to its own weight when supported at a point
Physical Description: vi, 97 leaves : illus. ; 28 cm.
Language: English
Creator: Wilson, James Blake, 1924-
Publication Date: 1957
Copyright Date: 1957
 Subjects
Subject: Elasticity   ( lcsh )
Calculus of tensors   ( lcsh )
Mathematics thesis Ph. D
Dissertations, Academic -- Mathematics -- UF
Genre: bibliography   ( marcgt )
non-fiction   ( marcgt )
 Notes
Thesis: Thesis--University of Florida, 1957.
Bibliography: Bibliography: leaves 95-96.
Additional Physical Form: Also available on World Wide Web
General Note: Manuscript copy.
General Note: Vita.
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Bibliographic ID: UF00098024
Volume ID: VID00001
Source Institution: University of Florida
Holding Location: University of Florida
Rights Management: All rights reserved by the source institution and holding location.
Resource Identifier: alephbibnum - 000568659
oclc - 13696249
notis - ACZ5398

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AN ORTHOTROPIC CIRCULAR DISK SUBJECTED

TO ITS OWN WEIGHT WHEN SUPPORTED

AT A POINT







By
JAMES BLAKE WILSON


A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY








UNIVERSITY OF FLORIDA
JUNE, 1957





















































UNIVERSITY OF FLORIDA


3 1262 08552 2869













ACKNOWLEDGEMENTS


The author is deeply grateful to Professor C. B. Smith
for his able and considerate direction of this work and to

all the members of his supervisory committee Professor
Smith, Professor F. W. Kokomoor, Dr. R. W. Cowan,

Dr. J. T. Moore, and Dr. W. A. Nash for their instruction,

counsel, and cooperation. To his wife, Nell, for her oon-

siderateness and encouragement, as well as her typing and

proofreading, and to young Lois and Gary, for their patience

in having a parttime father, the author also is very grateful,









TABLE OF CONTENTS


Page

ACKNOWLEDGEMENTS * * * * * * * ii

LIST OF FIGURES . . . . . . . . .

NOTATION . . . . . .. . v

INTRODUCTION . . . . . . . . 1

CHAPTER

I. EQUATIONS GOVERNING THE STRESS DISTRIBUTION
IN AN ORTHOTROPIC PLATE SUBJECTED TO ITS OWN
WEIGHT UNDER THE CONDITION OF PLANE STRESS 4

1.1. Derivation of the Differential
Equation Governing Stress Distri-
bution in an Orthotropio Plate
Under Plane Stress . . . . . 4

1.2. A General Solution of the Differ-
ential Equation .......... 9

1.3. Effeot of Boundary Foroes . . 13
II. STRESS DISTRIBUTION IN AN ORTHOTROPIC DISK
SUBJECTED TO ITS OWN WEIGHT WHEN SUPPORTED
AT A POINT ON ITS BOUNDARY . . . 14

2.1. Boundary Conditions . . . . . 14

2.2. Mapping Functions . . . . 15

2.3. The Stress Function; Conditions for
Single-Valuedness and Continuity . 18

2.4. The Constants Determined from the
Boundary Conditions. . . . 21

2.5. Displacements . . . . . . 40
2,6. The Isotropio Stress Function as the
Limiting Case of the Orthotropio
Stress Funtion . . . . 48


iii









TABLE OF CONTENTS (CONTD.)


Page


STRESS DISTRIBUTION IN AN ORTHOTROPIC DISK
SUBJECTED TO ITS OWN WEIGHT WHEN SUPPORTED
AT ITS CENTER . . . . . . .

3.1. Plan for Solution . . . . ..

3,2, Stresses in the Disk Due Only to the
Supporting Force P at Its Center .
3.3. Superposition of the Stresses Due to
the Weight of the Disk . . . .

3,4, Displaoements . . . . .


BIBLIOGRAPHY . . . . . . . .
BIOGRAPHICAL SKETCH .......


CHAPTER

III.









LIST OF FIGURES

Figure Page
1 . . . . . . . 4
2 . . 7

3 1 . . . . . . 14
4 . . . . 20










NOTATION


(The symbols are given for the x- and y-direotions, x, y,
and z being taken as rectangular coordinates; if the z-
direotion is to be considered, suitable changes in sub-
scripts and definitions are easily made.)

C7x, Oy Normal components of stress parallel to the x-
and y-axes.

TIxy Shearing stress component parallel to the y-axis.
EX, Ey Strain components in the x- and y-directions,

7_ Shearing strain component in a plane parallel to
the xy-plane.

Ex, By Moduli of elasticity (Young's module) for the x-
and y-direotions.

Gxy Modulus of elasticity in shear (modulus of
rigidity) associated with the directions of x and
Y.


up,


Poisson's ratio associated with the directions of
x and y; ixy is the ratio of the contraction in
the y-direction to the extension in the x-direotion
caused by a tensile force in the x-direction, and
Lyx is defined similarly.

Components of displacement in the x- and y-
direotions.









INTRODUCTION


A substance for which the elastic properties are inde-

pendent of direction is said to be isotropio. If the

substance is not isotropio but possesses three mutually

perpendicular planes of elastic symmetry, it is called

orthotropio. A plate of uniform thickness made of ortho-

tropic material in such a way that the faces of the plate

are parallel to a plane of elastic symmetry will have two
perpendicular axes of elastic symmetry in its own plane and

is called an orthotropio plate. The objective of this work

is the determination of the stress distribution in a circular

disk, i.e., a circular plate, of orthotropio material sub-

jected to its own weight when supported, with its plane

vertical, (1) at a point on its boundary and (2) at a point

at its center. Point support, as considered here, implies

uniform support along a line segment joining the faces of the

disk and perpendicular to each face at the point specified.

As a practical example, a board out from a log in such

a way that a transverse cross section is on a diameter of the

log (quarter-sawn board), or in such a way that, at its mid-

point, a transverse cross section is perpendicular to a

diameter of the log (plain-sawn board), may be considered as
an orthotropio plate, and a disk taken from such a board may
be considered as an orthotropio disk. The problem conaid-

ered here, then, is the mathematical idealization of the

-1-





-2-

problem of a wooden disk suspended in a vertical plane from
a point an its boundary or, in the second ease, from a slen-
der peg at its center.
The stress distribution in a oiroular isotropic disk
subjected to its own weight when supported, with its plane
vertical, at any point has been obtained by R. D. Mindlin.

The problem of the stress distribution in a oiroular ortho-
tropio disk under various edge loadings has been solved by

H. P. Cleaves,2 using a complex potential function, but the
weight of the disk was not considered.
This discussion is divided into three chapters.
Chapter I furnishes, in a sense, the tools for the investi-
gation, containing a derivation of the differential equation
governing the stress distribution in an orthotropio plate
in a state of plane stress, a discussion of a general solu-
tion of this equation, and a consideration of the effect of
boundary forces. The stress distribution in the disk sup-
ported at a boundary point is obtained in Chapter II, and
the problem of the disk supported at its center is solved
in Chapter III. In these treatments, the stresses are


1
R. D. Mindlin, "Stress in a Heavy Disk Suspended from an
Ecoentrio Peg," Journal of Apolied Physios, Vol. 9, Nov.,
1938, pp. 71-717.
2
H. F. Cleaves, "The Stresses in an Aeolotropio Circular
Disk, (rt rl t Jurpnl of .Me.hanion and Anolled Mathe-
matim Vol. VII, Part 1, March, 1955, PP. 59-o0.





-3-

obtained from a stress function, the form of which is assumed
and the constants of which are evaluated from boundary oandi-
tions and from considerations of continuity and single-

valuedness. In each of Chapters II and III the displacements

in the disk are obtained following the determination of the

stresses, and in Chapter II the stress function for the

orthotropio disk is shown to contain as a limiting case the

stress function for an isotropic disk under the same condi-

tions. Differences in coordinate systems and method of

attack make a comparison of this isotropic stress function

with that obtained by Mindlin unprofitable, but direct solu-

tion by the author of the problem for the isotropio disk has

verified that the limiting case referred to above is, in faot,

the isotropio stress function.










CHAPTER I


EQUATIONS GOVERNING THE STRESS DISTRIBUTION
IN AN ORTHOTROPIC PLATE SUBJECTED TO ITS OWN
WEIGHT UNDER THE CONDITION OF PLANE STRESS.

1.1. D rivation f thfa Differential EQuation govern Stress
Distribution ga AAU Orthotropic P3at Underu gJLao Stre a.
Let the plate be referred to a space rectangular coor-

dinate system (Fig. 1) with the axes taken parallel to the
planes of elastic symmetry, and let the displacements in the
directions of the x-, y-, and z-axes be u = u(x,y,z),
v = v(x,y,z), and w = w(x,y,z), respectively. Then the



z









Fig. 1

components of stress and strain are related as follows:3


3
H. W. March, Stress-Strain Relations in Wood and Plywood
Considered as OrthotroDio Materials, Forest Products
Laboratory Report No. R1503, February, 1944, p. 2.
-4.






-5-


( A -L .Lq -ix -r..tK)
x ax Ex x E y Ez z


E bv E- Z E
x y EU z




(1)
y = .i + i L= r
xy ax by G C x *

z = G.- + EX a + I
yZ ay az C) yz
zz


zx bx az ci zx
zx

In these equations Ex, Ey, and Ez are Young's moduli in
the x-, y-, and z-directions, respectively, and Gxy, Gyz,
and Gzx are the moduli of rigidity associated with the direc-

tions of x and y, y and z, and z and x, respectively. The
symbol 1/ represents Poisson's ratio; in particular, 1)x is
the ratio of the contraction in the y-direotion to the ex-
tension in the x-direction caused by a tensile force in the
x-direction.
If the plate is loaded only by forces parallel to the
plane of the plate and distributed uniformly over the thick-
ness of the plate, then the stress components z, Tyz, and

Tzx are zero on the faces of the plate and, for a thin plate,
it may be assumed that these components are zero within the





-6-
plate, also. The remaining stress components, Qy' (., and
T y, may then be assumed to be independent of z, that is,
assumed to be functions of x and y only. Such a state of
stress is called plane stress, and will be taken to repre-
sent the stress condition in the treatments of this and the
following chapters.
Under the condition of plane stress, equations (1)
become


E-Y x dyy G.






xy
ay EX E y Y


S= ^+ a T3W


The equations of equilibrium for plane stress,

x +X=O

(3)
++ +Y= o .


in which X and Y are the x- and y-components, respectively,
of the body force per unit volume, are obtained by applying
the conditions of static equilibrium to an elementary block
of the plate, shown with its applied stresses in Pig. 2. If
the body force acting on the plate is that due to gravity,
and if the plate be so oriented that the x-axis is downward,





-7-


O" -A

Fig. 2

then
X =pg and Y=0, (4)

where p is the mass density of the material and g is the
acceleration due to gravity. For this situation, equations

(3) become


ax + ay + pg =0
(5)

+ = .0


These equations will be satisfied if there is intro-
Aueed a stress function F(x,y) such that


0. B. Airy, Reort of the British Asoaaiatian for the
Adiancoement of SoIf--na 1862.









x y2 gx, dy (6)


The requirement that the displaoements u(x,y) and
v(x,y) be single-valued and continuous (class three) gives
rise, in the case of plane stress, to the following differ-
ential equation in the strains, known as the compatibility
equation for plane stress;


yL + j JB
SXy2 b2 ,xoy (7)

Equations (6) substituted into equations (2) yield




y2gx) x2
tVzlar I, L
i 2


xy
xy Gxy axby .


Using these expressions for the strains in equation (7)
and using the relation5

.iX=S .X (9)
E x


we obtain the differential equation


5Mroh, p 7
Maroh, OP. cit.9 p. 7.





-9-

L+- + 2 t 3 F (10)
4) bE1/3'7 2 =0)ax a
-6

Under the substitution6
= Ey (11)
with

,4 (12)
y

equation th becomes, upon multiplication by EonI


m+2 a(13)
ax 63c~a7e
where

PI ---E2 (t, .(14)


From the foregoing derivation of equation (13), it is
clear that any function F which satisfies that equation will,
by means of relations (6), satisfy the equilibrium equations










6
H. W. March, Flati Pltes of Plywood Under Uniform or Con-
centrated Loads. Forest Laboratory Report No. 1312,
March 1942, p. 40.






-10-


F = F(x + pj), we are led to the auxiliary equation


4+ 2K2 + = O ,


(15)


a quadratic equation in f 2 with roots


P2 K t VK2 -1
P


(16)


For wood, K, as defined by equation (14), is probably always
greater than 1.7
Let


K = cosh C .


Then
p2 cosh +. Voosh2 1
= cosh P +s sinh 9

= -,k + e-.a + a ._ e,
2 2

= ef .

Hence, the roots of equation (l*)may be written

p= + iei or


(17)


(19)


(19)


where


7
C. B. Smith, Effect of ElliDtic or Circular Holes on the
Stress Distribution i Plates of Wood er Plywood Considered
as Orthotronfo Materials. Forest ProdUcts Laboratory Report
No. 1510, 1944, p. 5.


P3 = iP P4 = i #


Pl = ia, P2 = I X





-11-
ia l = \[K +VK2 1
and (20)
P e' = VK -fK2 -1 .
It will be assumed throughout this work that a O /, unless
specifically stated otherwise.
A general solution of equation (13) is, then,

F = F1(x+lia) + F2(z-Ila + F3(x+ip) + F4(x-ij) (21)

where the functions Fj, j = 1, 2, 3, 4, are any analytic
functions of their respective arguments.
The requirement that the stresses obtained from this
stress function be real, leads one to consider the real part
of F, denoted by Rh{), as a possible solution of equation
(13). That BR{) is, indeed, a solution of that equation can
be seen by examining separately the terms R{Fj), J = I, 2, 3,
4, of R{F}. For example, substitution of R{Fl(x + i0o)} for
F in equation (13) gives


(1 2K12 + 4) d F (x + A iL7) (22)
d(x + ic-o)4

This equation is satisfied by virtue of equation (17) and the
first of equations (19) and (20). Hence, R{Fl(x + iav)} is a
solution of equation (13). Similar results obtain for
R{F2(x iax)), R{IF(x + ip)), and R{F4(x i()), whence it
follows that fRF), with F given by equation(21), is a solu-
tion of equation (13). And it is H{P), then, which will be






-12-

used in place of F in equations (6) to give the (real)
stresses.
Since, for f(z) analytic, the real part of f(z) is har-
monio, we can obtain an analytic function g(z) such that
R{g(z)) = R{f(E)}. Thus, the function R{P), from equation
(21), may be written

RB{) = R{G(x + + 2(x + ( i)} (23)

where G1 and G2 are analytic functions of their respective
arguments.
The problem of obtaining the stress distribution in
such a plate as is considered here is solved, then, when
there is found a function in the form of equation (23) such
that the corresponding stresses and strains satisfy the bound-
ary conditions on the plate.
It is apparent from equation (23) that it will be
useful to introduce two new systems of orthogonal Cartesian
coordinates (xl, yl) and (x2, y2) related to the coordinates
(x,N) by the formulas

xI + iyI = z = x + i~ and x2 + y2= z2 = x + i., (24)

Hence, to any point P in the region R lying in the z-plane
correspond points P1 and P2 inside some regions 81 and R2
which lie in the complex zl- and z2-planes, respectively. In
order to solve a problem of the proposed type, we shall make
use of two mapping functions zI a= zl(1) and z2 = Z2(T2)





-13-
which are analytic functions of the complex variables l and

T2, respectively, and which map the regions RI and B2 into
regions more easily discussed.

1.3. Effect of Bo undar Faorges.
If the force on the boundary per unit length and unit
thickness has x- and y-oomponents denoted by X, and Y, ,
respectively, then the following relations must be satisfied
on the boundary:.

v Ox ds -xy ds
(25)

V ixy ds y d *


where ds denotes an element of the boundary. With the
stresses given by equations (6), equations (25) may be re-
written in terms of the stress function to yield

X 2 = -/E) pgx Ax
da (bay) do
(26)

do +x do








S. Timoshenko and J. N. Goodier, Theory of Elasticity,
McGraw Hill Co., New York, 1951, p. 190.










CHAPTER II


STRESS DISTRIBUTION IN AN ORTHOTROPIC DISK

SUBJECTED TO ITS OWN WEIGHT WHEN SUPPORTED

AT A POINT ON ITS BOUNDARY.


2.1. B oudary Conitions.
Consider an orthotropio disk of radius a and, for con-

venience, of unit thickness. Let the center of the disk be

taken as the origin of a plane rectangular coordinate sys-

tem (Fig. 3) with x- and y-axes parallel to the axes of

elastic symmetry. The disk will be considered to be so

oriented that the force on it due to gravity has the same

direction as the positive x-axis. The case of point support

(at the point (a, 0)) will be considered as the limit of the

case in which the supporting force gna2 is distributed over









r y






Fig. 3
-14-






-15-
an arbitrarily small arc of the boundary, according to the
following boundary conditions (see Fig. 3):

2
Xv = for 9 ;
2at
X, = 0 for all other values of 9; (27)
Y, = 0 for all values of 8.

In these conditions, e is an angle measured from the posi-
tive x-axis, and q is an arbitrarily small (positive) angle.

2.2, MaPing Functions.
Referring to the discussion of mappings on page 12, we
observe that the regions RI and R2 are in this case bounded
by elliptical curves given, respectively, by the equations

Z1 = a(cos 9 + ioa sin 9)
and (28)

z2 = a(cos B + 1 9 sin 9)

Let t1 and 52 be functions of z1 and z2, respectively, which
map these boundaries into the unit circles 11 = eiG and

2 = eiO in the r1- and T2-planes. Then, in terms of fl,

cos 9 1= ( + -) and sin e = ( ) (29)


so that

o1 1 1 1

or

z= a +)2 t) (30)






-16-

Similarly, considering 12 and z2, it is found that

= A(U + 7 + a( l r) 1. 0 (31)
2 2 2 2 (2

Solution of equation (30) for 7l and equation (31) for r2
yields


1 = ad(lIo \1 + a( 1)
and (33)

2 = a(l + Z2 V + a2P 1)
If we let

rI ac = 13e (33)

W = t1 + a2( 1) and W = 2 + a2(r 1) (34)

then
07 + h
a 1 + 1l)


z2 + W
2 all + r2a

The function W1 has branch points z = l a\fl-7 ,

which are the foci of the first ellipse (2g). Let the line
segment joining these points be taken as a branch out. From
equations (12), (20), and (33), it may be seen that r > 0
and r2 > 0. If rl < 1, then, on the branch out

z = x, with Ix i a \II -r (36)






-17-


and

s1 ry \+ a2( -(j

x + 1a2(1 r) x
a(l + r) 1 -a(l + r) (37)

Representing T, by X + 1 with x and a real, we have

L2(1 r nT(x
S-a(l + r) and = a(l+ r s

whence

S++ 2 (39)

Thus, the image in the Sl-plane of the branch out in the
z,-plane is a circle of radius A If r3 > 1. then

the branch points are on the axis of imaginaries in the
zl-plane, and, on the branch out,
z1 = iy, with Iyl a r -1 (4o)

An analysis like the one above shows, however, that the image
in the 11-plane is again a oirole, this time with radius



r -
a = + (41)

In either case, then, the function 31 maps the region
h1 bounded by the first ellipse (28), with the cut Joining
its fool, into the region in the ~l-plane lying between the
concentric circles Il11 =- ol and Ill = 1. It can be shown






u18-
similarly that the function S2 maps the region R2 bounded by
the second ellipse (28), with a branch cut joining its fooi,
into the region in the T2-plane lying between the concentric
circles I21 = Idl and I21 = 1, where


d +- (42)
r2 + 1 "


2.3. Ma Stress Funot on; Conditions fe Sinrle-Valuednesa
IA GContiuty.
The stress distribution in the disk under consideration
is obtained by finding a solution of equation (13) in the
form of equation (23), i.e., the real part of the sum of an
analytic function of z1 and an analytic function of z2, which
satisfies equations (26) on the boundary of the disk, with
X, and YV given by equations (27). The ring-shaped regions
into which the regions B1 and R2 of section 2.2, with their
branch outs, are mapped by the functions l1 and 72 suggest
the use of Laurent series in 1 andT2 for this purpose; in
particular, they suggest the assumptions

= R ( an + bn
and (43)

HI Rf I ^Wni + 1 bfl
=y + ir2 2

where an and bn are complex coefficients to be determined by
the boundary conditions.
The requirement that the stresses obtained from this






-19-
stress function F be single-valued is tantamount to the
requirement that i1 and 2 be single-valued functions, i.e.,
that the variables z, and z2 be not allowed to cross the
branch outs in their planes. This, in turn, amounts to the
stipulation that the signs taken with W1 and W2 (equations
(34)) be such that
I I i + =I I 4r, 1
!1 aTl + rl)+ r1

and (44)
z, + W r -1
I21 = I S V r 7 K

i.e., suoh that
zi + W, e a \ r- 1
and (45)
1z2 + W21 1 a r \ 1

A further consideration, if the stresses are to be oon-
tinuous, is the behavior of 2 and P in the vicinity of
ax ay
the branch outs in the z1- and z2-planes. Assume for the
moment that r > 1, so that the branch out in the zl-plane
is on the imaginary axis. As a point on the out is approached
from below (see Fig. 4),
T1 cee (46)

and as the same point is approached from above,

h --. ojde, i (47)
where 81 depends on the particular point approached, The







-20-


z-plane
zl-Plane


il-plane


Fig. 4

values approached by 6 and y as TI approaches its limit
must be the same for both oases. It is seen upon reference
to equations (43) that this requirement will be met if

S neni = (-o)e-ne (4w)

Equating the coefficients of like powers of eig we are led
to the relations

a.n = (-o2)na n = 0, t 1, t 2, . .(9)

If r2 > 1, this same process, with subscripts 1 replaced by
subscripts 2 and o replaced by d, leads to the relation

b-n (d2)n n = 0, 1 1, 2, . (50)

If one of the constants rl and r2 is less than 1, then
the corresponding branch cut is on the real axis. The case
rI < 1, for example, implies, by equation (41), that o is






-21-
imaginary, so that Icl o,. As z1 approaches a point on
the branch out from the right,

S-- (-l)e1 (51)
and as z1 approaches the same point from the left,

T --1(-Io) e"le (52)

Now, the requirement that the values approached by and i

be the same for both cases can be shown to lead again to
relation (49). If r2 < I, this process, with subscripts 1
replaced by subscripts 2 and o replaced by d, yields relation
(50) again.

2.4, %ha Constants Determined 4rom. BoundaUr Conditions.
For the boundary of the disk, where I, = 2 = eil
equations (43) may be written

= RE (a, +* b)enie

and
(53)
a jllR (ranr + rabn) }el

Now, x = a oos 9 and y = a sin 9 on the boundary, so that

= 1, (54)
di a
and

t( )) R ( ni(an + bn)e1 !
do ax n)
and (55)

SR.) ~"I n(rjan + r2bn)8ne 1
ds ay'






-22-
where the symbol Z* is used to indicate that the constant
term, corresponding to n = 0, is missing.
Substituting these expressions for () and a)
in equations (26), and noting that

pgx =o pga cos 0 a cos 9 a gaoos2

= MA (1+ oos 20) = 4-(2 + e210 + e-21)

and (56)
pgx = pga oos (- a sin ) = pga sin 9 cos 9

- sin 2n = '-(e219 e-219)

we obtain
x, = n(r a + r2b nIeniU -P 2 + e21e + e-21e

and (57)
= - RI 1n(a + bn)enie + (e2 i e-2ie)

The identity R{w) = 1(w + W), where W denotes the complex
conjugate of w, permits the rewriting of equations (57) in
the form
x,= s *tn(rl an + r2n)e + n(rii + r2 )e-"ni

P 2 + e210 + e-21)
and (5)
1*n(an + bn) nie n(a + ^Ee 3nie1

+ 1 e219 e-210) .

Let the function X1 given by the first two of equations






-23-
(27) be represented by a complex Fourier series as follows:

OD niQ
xv E '- Onen (59)

where

-n = xe-nie-1 -- f (- ga),-nied (60)

Integrating, we obtain


n = (- P.)ae- -iede
-nik
= e 1.n

Se-M, _i 1
2 nig
n = 0, 1, 2, . . (61)

These constants represent the only appearance of the quantity
g in this discussion. Their limits as k approaches zero,
representing the desired case of point support of the disk,
are found by 1'Hospital's rule to be

on = 2 n = t 1, 2,. . (62)

Hence, by equation (59),

x = E. ei (63)

The third of equations (27) states that

YV = 0 for all values of 9.
With the use of these last two equations, equations (58)
become






-24.


E* ,n(r an + r2ba)e!n + n('1in + 2I)1n

"g(2 + e21 + e-219) = i eO
and (64)
l [n(a + bn)e ni n(In + n)e-ni

+ A e21@ e-21) = 0 .
In each of these equations the coefficients of like powers
of ea in the lefthand and righthand members must be equal,
in order that the equations may hold for all values of .,
Equating the ooeffioients of enei in the two members of each
of equations (64), we obtain

t n(ran + r2b.)] + Jn(rl1 + r2S.n)) a

and (65)
lEn(an + b) 5n(5a + n)3 = o
a+ n+ -
n= 1 3, - * ,
and, for n = 1 2,

.n(r2an + r2n)+ + n(r 1n + r n) = -

and (66)

-Ln(a9 + bn)3 an(it + n o S

where the sign of the term fj in the last equation is the
same as that of n. In the expressions for M and given
ax yo
by equations (43), the terms involving ao and bo are constant






-25-
terms and hence do not affect the stresses, given by equa-
tions (6), There is no need, therefore, to consider the
case n = 0 in evaluating an and bn.
Equations (65) are equivalent to the following system
of equations, wherein n may assume the values 1, 3, 4,


2
rla, + r2bn rlan r2b n n


e- a r2 a-n + r + r2bn = *
(67)
n + bn + -n + 0-n =3

and
an + b-n + + bn 0

With the use of equations (49) and (50), this system may be
written
r1 an + r2 bn rl(-c2)nn r2( 2)n =

r (-o2n -1 (d2)nb + 1 .n r+ 2n = n

(68)
S+ bn + (-o2)nn + (-d22)n = 0,

and +
a (-o2)n + (-2)nb + n+ n =0,

n = 1, 3, 4, . .
It may be seen similarly that equations (66) are equivalent
to the system of equations






-26.


r2 bn r(-o2)Fn r2( d2n = 2.


.rl(-O2)an *- r2(-d2)nb +


1 1 +


bn + (-o2) an +


2
r2 bn =
(69)
(-da2)nrn 2


and
(-o2) na +


(-a2)bn +


2
n L.2n


n = 2.


The sum of the first two of equations (69) and the sum of the
last two of those equations are, respectively,
r
riCl (-o,2)11^ 'S) + r.Cl (-d2)2](b, + Sn) F n


and
E[1 + 2)( + (-o)n(a) +


(70)


l + (-d2)"n(bn + i) = o ,


n=1, 3, 4, .* *
The respective differences are

r1il + (-o2)n](an n) + r2[l + (_42)tn]b bn) = 0


(71)


Cl (-o2)n3(a l) +


[ (-a2)n](bn 0) =o


n = 1, 3, 4, 0


Let


%n = n + i%


and


bn = Bn + in ,


where A, an, Bn, and Pn are real constants. Then equations
(70) and (71) become, after division by 2,


an +


(72)







rll1 (-c2)n]A +


l[ + (-c2)n]^ +


-27-
2
r2Cl (-cd2)]Bn, = a


l[ + (-.2)n]B n o
[z +


r11l + (-c2)n]J + r2gl + (2)n]Pn = o

and (74)


[1 (-c2)n n +


[1 (-d2)n1n = 0 ,

n = 1 3, 4, . .


Corresponding equations for n = 2, obtained from
(69) by the same method as that applied above to
(68), are

r11l (-c2)n]A + r211 (-d2) nB =


[1 + (-c2n]An +
U


[ + (-2)"B n =
1l + (-d 2)n]Bn 2n


equations
equations


(75)


r1Pl + (-o2)n]n + r2El + (-d2)JnPn = 0 ,

and (76)


[i (-o2)n3]n +


[1 (-d2)n]n = 0 ,


n = 2.


It will be shown that the determinant of the system
(73), and hence of the system (75), can equal zero for
n = k, a natural number, only if Ak and Bk do not affect the
stress distribution in the disk and, similarly, that the
determinant of the system (74), and hence of the system (76),


(73)






-28-
can equal zero for n ; k only if ak and Pk do not affect the
stresses. Let these determinants be denoted, respectively,
by Dn and Dn, let

r1[l (o2)n] r2[l (-d2)]n

D 3

1 + (-02) + (d2)n

and (77)

r1El + (.-2)n] r2C1 + (-d2)n

Dt =
n
1- (-02)" (-a2) .


If D = 0 for n = k, and if we consider g = 0, then equations
(73) (or equations (75) if k = 2) permit arbitrary values
of Ak and Bk and hence, if Ak and Bk affect the stresses,
allow a stress distribution in the disk when no external
forces act. This situation cannot exist physically. There-
fore, Dn = 0 for n = k implies that Ak and B have no part
in determining the stresses, or, stated otherwise, if Ak
and Bk do affect the stresses, then Dk f 0. The same argu-
ment applied to Dn shows that DA = 0 implies that an and
@n have no part in determining the stresses, which is to say
that if a and 13 do affect the stresses, then D 0O
To determine which of the constants do affect the stress






-29-
distribution and which do not, consider equations (43), which
may be rewritten in the form

6F= R1 (aj +a n) + S(b n + bin,) + a + b o
)x 1 1 n n 2 n2
(7S)
and

=F cc n+ anccin) +
S= Irl (nT _n 1a ) + ir2 E (bnTn + b.gn-n)
+ irlao + ir2bo)


Since a and b0 do not enter into the stresses, given by
equations (6), they may be taken equal to zero. With the use
of equations (49) and (50), the equations above become (with
ao = bo = 0)


% = R, a rn + (_.2)nvn) + \b + -d2)n n)

and (79)
Rir n + 2n n+ b n 2n -n
aF ya + (. +o2' + ir2 + (-d2)n


Consider now the factor 7n + (-c2 )n 2 n It will be proved

that
n + (o2 = (zI) n ( 1, 2, . (80)

where Pn(zl) is a polynomial in z1 of degree n in which the
coefficient of zk, k = 0, 1, . n, depends on both n
and k. From equation (30), the first of equations (33), and
equation (41), we have






-30-
a(l + r) a( -


1 2l+r2I
l2 + 1 + -r- 11


a(t + --1



2 .. = + (-02) 1 ]

Mll + rl)


A similar equation can be seen to hold with subscripts 1
replaced by 2 and c replaced by d thus


al + 2) 2 2)2


Squaring both members of equation (81) and letting



we obtain
k2 z2 2 + 2(-o2) + (-o2)2 2
o 01 1
or
2 + k =22 + 2 ( 4.)
2k1+kZ1 1 01 '
where
k0 = (-o2) (s)

Thus, equation (80) holds for n = 1, 2. Assume that it holds
for n = m, a natural number. To demonstrate that it holds





-31-
for n = m + 1, we multiply the righthand and lefthand members
of equation (80), with n = m, by the lefthand and righthand
members, respectively, of equation (81) to obtain

EP(z)3 0* k = (m + (-o2 )mr1m 1) + (-o2) T-1]
S 1 1 1
SiM+ (.o2)m+l(lm+l)


+ (-o2C-E + (-c2 m-l] 1

mm+1 + (2)D1q-( 1) + kP1(z) *

or
SM+l + (2 m+1 -m+1 ( )

where
P (zl) = ko zPmz(l) k1m1(z1) (s)

It follows, by the principle of mathematical induction, that
equation (80) holds for the indicated range of values for n.
n (2)n -n
A similar result obtains for the factors 7n + ( T2 and

polynomials P'(z2). Therefore, each of the terms in equa-

tions (79) for i and is equal to a polynomial In z- or

z2 of degree equal to the index of the term. Since, by
equations (6), the stresses are obtained by differentiation

of l and ., and since = 1 and r = irj, j = 1, 2, it
follows that all terms with indices greater than 1 will have






-32-
a part in determining the stress distribution. That is, all
terms containing one of the constants An an, Bn,3 n with
n = 2, 3, . do affect the stress distribution. The
contributions to 9- and ae of the terms with index 1 are
ax 6y
seen, from equations (79), (91), and (82), to be, respeo-
tively,

RB(A i + ial)1 4 z1 + (B1 + i )i)a(i + ) z21

and (88)
R(irl(A1 + lal)ai +r) z + ira2(B + al+ r2) +2

Hence, for the terms with index 1, the contribution to

is
418

the ontrbuton to is
2A, 2B (9)
;( a+ +r1 =a + r2)

the contribution to L2 ? is
ay2

2r *A, 2r2B21
ail + r) al + r2) (90)


and the contribution to 2 is
xay

2r-pL+ 2r2 9
-all + ri) a(l + r2T 1






-33-
Referring to equations (6), we see that A1 and B1 do affect
the stresses GX and c-, whereas al and pi do not, and that
the contribution of the terms containing a1 and P$ to the
stress TXY is


+ (92)
a 1 + r 1 + r2

It will be shown later (after solution of equations (74))
that this expression is equal to zero, so that a1 and P0 do
not affect the stresses.
Since An and Bn affect the stresses for n = 1, 2, . .
we conclude from the argument on page a that

D # 0, n = 1 2, . (93)
and, since an and p. affect the stresses for n = 2, 3, ,
we conclude that
D# o, n 2, 3, . . (94)

Calculation shows that D' = 0. This in itself implies, by
the argument on page 28, that L1 and P1 do not affect the
stresses, but, as stated below expression (92), this conolu-
sion will be verified later, as a matter of interest.
Simultaneous solution of equations (73) yields

A ia i + (-d2)"
"n n Dn (95)
and
Bn (96)
B~~ F'~ n


n = 1. 3, 4, a .






-35-


where D is the determinant given by the first of equations
(77) whioh, upon expansion, may be written

Dn = (r r2)1l (o2d2)n] (, + r2)[(_o2)n (2)nl,
n 1, 2,. . (97)
The values of A2 and B2 obtained by simultaneous solution
of equations (75) are






i.e.,




A o'2 1 + d4) r (1+ d )
2 1D









BA2 = bn, equation (96)1n=2 go ; D)
B2 2 + D





It follows from equations (74) and (76) and statement
(94) that
n "= n = 0, n = 2, 3, . . (100)

As stated earlier, Di = 0, which implies that equations (74)






-35-
are dependent for n = 1. Solution of either of these equa-
tions for al, utilizing the definitions (41) and (42) of o
and d, yields
r (r, + 1)
al -(r2 +) 1 1 (101)

where the constant/~ is arbitrary. Use of equation (101)
in expression (92) shows that that expression vanishes
identically, which completes the demonstration that al and
P1 do not affect the stresses. It is convenient to set
oal l 0 o. (102)

As a convenient summary of the results of the two pre-
ceding paragraphs, the formulas for the values of an and bn,
n = 1, 2, . obtained from equations (72), (95) through
(100), and (102), are given below.

an n n = ,3, 4. . (103)


bn =- + 12 n = 1, 3, 4, . (104)

r2 + 1 (r2 1)d
a2 [a, equation (O13)n2 l- 2 C(105)

and
and r0 + 1 (r, 1 )o 4
b2 [ba, equation (04)] + p -106)

where
qDn = (r r2)[1 (o2d2)n] (rl r2)1(-o22)n (-d2)n],
n = 1, 2, . . (97)






-36-
With the constants an and b1 evaluated, attention is

returned to equations (79) for and Z. The differentia-
tion of those equations required by equations (6) in order
to obtain the stresses is facilitated by considering first
the derivatives of 1, with respect to zi, i = I, 2. Differ-
entiating equations (32) with respect to their respective
independent variables, making use of equations (33), we
obtain










=. i= 1, 2;

i

i.e.,


dP. and d -2 (107)
dzl si dz2 2


Then, since a i,
zt: +=






-37-


- + 1 -2)n = [l-1
ax 1


= En (-_2)fn n]


and, sinoe az i1,


(los)


S+ (-2)n ] I [nf-1


- n(o2)n 1-n-1] 3 *
1 Ir1


irn T (.-o2 2)n


Similarly, it is found that

B n+ (-d2) n n 2 n


S7+ (-a2)n.-n, -(-a2) n -1 -
ty-1 ,2 W 2 2 2

Now, by the first of equations (6),


-hene, by PX, ad
whenoe, by equations (79), (108), and (109),


2- a(Tn (_-02n -,n)
x i~ni w
ITX ,"*IWl


r+ (n (_2)n
+ 2n 2 9. ; 2 .1


-pa'.


(110)


Similarly, by means of the second and third of equations (6)
and by equations (79), (108), and (109), we obtain


(109)


- n(-o2)1n -n-l]








cr, pgx
g dx ax I t


" R n[] ,2)


n ( 4 (a2)n


- pgx


and


n 12 n ,

?+~ Wi


(11)


r an Tn 2
(11)
(112)


Substitution of the values of an and bn given by equa-
tions (103) through (106) into equations (110) through (112)
leads to the following formulas for the stresses:

S= (g r^ (1 +D J2) n 2n e)

r2(l + (~-2)n) n
t- (-- ( -d2)fnl i)]

r2r1 + 1 (r2 1)4] 2 4 -

w --- 1 s 4 2 (113))
r + 1 = 2 ) 12.






-39-


+ 2n E 2n 2n
i n -l" n1



r + 1 (r2 1)d4
42 2 1 2


+ 2-1 o2 2 ) pg, (1n4)


and

r ( + (-d2)n) n2





tr- ID2 b --a"- )"






= [- s[1 (o2d2)] (r + r)[ (-o2)n- (2)

n = 1, 2, . . (97)
The siil ty of these three fo as the riti of
(115)

where
D = (r, p[l (o~d2)n] (rl + r2)[(-o2)11, (.d2)ni,
a I 2p .(97)

The similarity of these three formulas allows the writing of


the composite equation










s= p2k R E (1 JlnL IK (j 12


2- (-d' r 22)


- 1 -


-"k


r t J. ( -1/j


r +1 (r, 04 1)o


,IJ


- oi)

-


- Vk pgx ,


k = 1, 2, 3,


(116)


where Sk, Tk, Uk, Vk are given by the table below.


Sk Tk Uk V


0. -r. -?
1 12


1 1 1


'Ty irl ir2


This form is more convenient for computation than are the

forms of equations (11 ) through (115).


2.5. PDis ementa.
The displacements u and v in the disk are obtained from
the first two of equations (2),


x



i=E7 --
ay y S


y


and


I


i





-41-
by integrating the first with respect to x and the second
with respect to y. In view of relation (9) among the con-
stants, these equations may be written



and (117)



Substitution of the righthand members of equations (110) and
(i11) for (OT and CTy, respectively, leads to


is - Vn z") (2 (-e-)n ii
+- n 2 qn) (118)



and

AX= R En r y) n C-)2 -n

2
+ ( 1)- (2) )] ( E pgx.( 119)

The required integration is facilitated by observing the
implications of equations (108) and (109) that

S- = + (1_2)n nn)] + constant,

,n 2)n (120)
f 1 dy - ? + (-o2 _; _02 n ] + constant,








and that similar equations hold for subscripts 2, with o
replaced by d. With the use of these relations, equations
(118) and (119) are readily integrated to yield


U = an (T + (2-o i1)

+ + Ex b b + (d 2)n -n


1 -.... Ape- + f(Y)
2-- z + f(y) (121)


and

v = E an(n + (-,2)n T-n)

2
+ r2 1v'2 + n 2 n


+ pgzy + g(x) (122)
B y

where f(y) and g(x) are as yet undetermined functions of
their arguments.
It will be shown that comparison of the expression for
'y which may be obtained from the last one of equations (2)
and equations (121) and (122) with that given by equation
(112) leads to the determination of f(y) and g(x), Differ-
entiation of equations (121) and (122) with respect to y and
x, respectively, with the aid of equations (108) and (109),
leads to









( 1


r 2 n- (. 2)n
2


2
r= +1 [
= -ai H n[ +I
L I n X


+ 1 +
,2Ey


S(..o2) n )


b
2


pgy + A *


With the use of these expressions for the derivatives, the
last one of equations (2) yields


= o ai n
xy i "ai


2


2
Ez +1
E y


S (. (02)" ?n)
1~


(td2)" ]


r2E 2
2 7


L+ 1
E Y Pgy
xlg


+ ,[ T


Now, from equation (14),


1 (126)


= ..- ln[
niL +_


ay


+2
+ 2E X -


and


(-o2)n ~


.2-n] +fl


(123)


+
E


(124)


,r2 + i_
+ (2+ "
3


(125)


x


2 n -n)


Ut w VSjy-


+ ] w








where, by equations (17), (20), and (33),
r2 2
o2K nr2 + 3a2 2 +r,


or, alternatively,


2K= o


With E2 =

become


and


by equation (12), these last two equations

by equation (12), these last two equations


2K r2
2 JX-I


*^~i


(12:)


Insertion of these expressions, in turn, for 2K in equation
(126) gives


-X =
a -


and


L- =
XY


2
r'2


2
r


+ I +
cj7


21)
X-
X


(129)


If now, on the authority of equation (9),


X


2i V
g?1 be written
X


+ t then these equations become
-y


(127)


2
+ Myf
Xr


e2"Y





-45-


r2 + I. r +1


and (130)




These equations allow the rewriting of equation (125) as

S R{i 1n2 [ ( 2 n)

r+ bf 2n n)a2
iCn (-d 11 2


x[1 pIrgy + U ) + z(] (131)

Comparison of this equation with equation (112) indicates
that
[ P Y i+ + ] o. ] (132)

i.e., since GOy 0 o,


SPg& + d z (133)

Since the lefthand member here is a function of y only,
while the righthand member is a function of x only, the fact
that the equation must hold for all points (x,y) within the
disk leads to the oonolusion that both members must be equal
to some constant, oall it k; then





-46-

ay ) 3.ir + k (1354)

whence

f(y) E -k+ ky+ m (135)

where m is an arbitrary constant, and

=- k (136)

whence
g(x) = kx + n (137)

where n is an arbitrary constant. Now, the constants m and
n represent constant contributions to u and v, as is seen
from equations (121) and (122), and hence correspond to a
rigid displacement of the whole disk; it follows that, if
the support is to be stationary, then m = n = 0. Also, the
fact that the displacement u (in the x-direction) is, physi-
oally, symmetrio with respect to the x-axis requires mathe-
matically, considering equation (121), that f(y) be an even
function; it follows, considering equation (135), that k = 0.
With k = m = n = 0,
(y) = -

and (138)
g(x) ,

whence equations (121) and (122) become







2
U = R{ g


x+
lux


S)n Tn )-n


t + (-d2)n ;n)]
2 n -


S2 + y y2)
EXi E y


and

= -R (1 + 1 +
v rr1.r - an (T 1


(-02)n 1)


2
S- b(+ n (-d2)n )]
r2E -bn72+


(14o)


+ = 1
+ p3x .
y r


Inserting the expressions given by equations (103) through
(106) for an and b M, we obtain


U2 2n + V (1 + (-d42)n) (n
U - peg2ai C x I


+ (-,2)n -n-)
1


(1 + (-2)n) + (d2)n ;n)
xnD .


Er + 1- (r d1)d C[T
2 2 3-


+ o0


rl + (r1 1)o0 4c72 + d64 ]
----- ~ U-E~ ---- -


+ x+I 2)
P+? ^x
x2*-^-y2


(139)


(141)


m


-47-


A +
x
---^



r + +z_
r2 +

1-
2 E x


[.







and

v - pea2Ri El4 -----E--- 3
4 +(1 + -)n2)n) n + (, 2)nl -n )



S- 21y D n


+rx + 1 [r1 + 1 ( l)oa4[E + o4 2]


2 y n


+2 pxsy (142)


2.6. TAl Isotroai Stress Punction aL t.e& Z mitinL Case .EOf
te Orthotropio Stress Function.
It is of interest to obtain from the stress function for
this problem the stress function for the corresponding prob-
lem for isotropio materials. The elastic properties of
isotropic materials are independent of direction; thus,
EX = E = E, G = G and 1Vx = V7 = 1). By equation (12),

then, E= ,1 Further, for suoh materials, the elastio con-
stants are related by the equation9


21 + V) = G i i(143)

9 p
AL P. 9.






-49-
It follows from this, by equation (14), that

K = 1) = 1 (144)
(124)
whereupon, by equation (17), c = 0. With E = 1 and 4p = 0,
equations (11), (20), (24), (33), (34), (35), (41), (42), and
(97) show that

21 = z2 = X + iy = z,

'1 = r2 = 1,

o = d = o, (145)
W1 = W2 = z,

1 q= 2 a I
and
Dn = 0, n = 1, 2, . ,

where, in the fourth and fifth equations the positive sign
is taken with W1 and W2 in order that %1 and V2 be functions

of z, instead of vanishing identically.
Insertion of the expressions for an and bn given by
equations (103) through (106) into the first of equations
(79) leads to


x 11+(-d2)n n _2 n ] 11+(-o2)n ]a 2)n? 1nT
pga2a E 2

[r2+l-(r2-1)d4j[T +(-o2)2j2]3 1[r+l-(rI-1)02][2 2- 2 -2


(146)
For the values of the constants given by equations (145), it





-50-
may be seen that the major terms (fractions) in equation
(146) are indeterminate. To obtain E (and hence, by inte-
ax
gratin, F) for the isotropic ease from equation (146),
therefore, we shall set E = 1 in that equation and find,
using l'Hospital's rule, the limit of the resulting equa-
tion as


r1 = e and r2 = e, (147)

by equations (20) and (33). The roles of (p in the quantities
or d, Zi, z2, WI, W2, Tl, T2, and Dn (with E = 1) are estab-
lished by the equations defining these quantities, together
with relations (147) above. Equations (145) indicate the
limits of these quantities as q< approaches zero.
It will prove convenient to take the preliminary step
of finding the limits as qc approaches zero of |,

d~ and O With r = e1 with z = x + iey, and

with W1 and T7 given by the first of equations (34) and (35),

differentiation yields


UqP zl '2i p a?3 dT
ac= az1q a 1

i. e Y+ ar+l) + 51+W) a(r +1)2 e

,( a(lg)

whence, by equations (145), construed as limit equations as






-51-


cp approaches zero,


lim
(9--0


2a + [2z a 2a
2a 2z 2a


i.e.,

UrnO ,y a2 t Xa 21-o
Similar it- ay be found that
Similarly, It may be found that


lim
)--w0


With ri = ev, -o2 is given by

2 1+ae
1+ e'


whenoe

d-) (1 + e)(-eT) (1 e)(et)
d( (1 + ev)2


- (152)
(1 + e)2


and


lim
4--0


With r2 = e ,


-d2 is given by


-d2
1+e


1+ e


whenoe


lima
<9-&0


(-ad2-
d4 4


Now,


1 it


(149)


(15o)


(151)


(153)


o2 .


(154)


(155)


Sa2 2 + 21yz
4az


dyliL 1
d(-o2 = -





-52-


S 2^^ U- rn-^1 2 2-ni3 2a
Tn + (-. ) l- = En (-"8 n3> )q,

+ n(-o2)n-1 -n (156)

Henoe, by means of equations (145), construed as limit equa-
tions as cp approaches zero, and equations (149) and (153).

Iao--,. 0 2n

= n(f)n-1 o-+ 2 + i -

i.e.,
l., 2)n
10 -n. + (-o20 -

*+ r (157)

where the star (*) beside a term is used to indicate that
that term is present for n = 1 only, vanishing for all other
values of n. Similarly, it can be shown that



.(2 -2 + +-1) (.1 5)

Let the numerator of the nth term of the summation In
equation (146) be denoted by Mn, i.e., let

an = C1 + (-d2)][n + (-_2)n q-n

C1 + (-o2)2 + (-d2n (159)
el2





-53-


Then

N CL + (-2)n] [In + (.-o2)'nhf


+ [C + (.02)n nl n(t-d2-1 (2 ]d(dfl)
1dp

1c + (-o2))n] I [T + (_-d2)n ]n


a- + b (-d 5), (5n), (-2)1 a(17)] ,n (16(o)

and, by equations (145), (153), (155), (157), and (158),


1im 2 n(a2zun2 n + Ri2 n_-1)
q~-'-O acp 4a

n(a2zn2 z n + 2vYPn-1)
4an


tz' a(\


i.e.,


lim 5 L


2? aa


From equation (97), we obtain, by differentiation with
respect to p,

S- t r23 S (o2d2)J] + 1 (o2d2)n][e + ]

1r, + rdn(-o )n-l ql _n(-d2)n-1 Ad&2L


- [(-o.)n (-2)n][ d" ]
-'- d. J7


(162)


wherein equations (147), r1 = st and r2 = e have been used

to obtain L-(rl r2) for the second term. Hence, by the secmd
d~p


n(a2zn-2 n + 2 1vzn-l)
2a"


(161)


z-9?- (-






Sthi of equations (1 an by equations (153) an (155)
and third of equations (l45) and by equations (153) and (155).


lim d(.o)
Q-o -0 d1 2(- )


i.e.,


lim d(,) 2 for n = 1
P-00 dp 11, for n = 2, 3,. . .


(163)


By 1'Hospital's rule, now, the limit, as cp approaches
zero, of the nth term of the summation in equation (146) is,
provided all the limits exist,


lia S&
n lim d(
q-U0 64)
(164)


hence, by equations (161) and (163),


lim N for n
~-2,-o ,p2f,. n + 2ivA,,1l
S, for n 2, 3
^ ^ ^ - ^ e T ^ , ,n


, g


(165)


Consider now the last term in equation (146), which may
be denoted by ,-|. where

NO = Er + 1 (r2 l)a[ + (-o2)2 2

Cr +1 (rl 1)o 2 + (-a 2)2?].

(166)


lim n li-O 11 O
T-'* O n p9O $ n -OW


2a",






-55-


Then


Sm- Ir2 + I (r2 i)-] d [2 + (-o2) 22]


22 222
+ [1 + (-u2) ~- e- (i2 1)2d2 g




Cr + 1 ( 1)o] [T2 + (-d2)22


12 (422 2)2 3[e (Cr 1)2o2 da

0 1] (167)

wherein equations (147), rl = e2 and r2 e-, have been
used as needed for differentiation. It follows, by equa-
tions (145), (153), (155), (157) and (158), that


0o a. -23Ca2 "2a+ 21Ts 3 [ C()2]c.
2a2 a 2

2. 2][-3 2 2 + 2iv] ()2 12

2( 2 _2 + 2r1) (y,)2
a2 a )
i.e.,


1 4 + 41 (16)
(P-.. ay &2 a2


Henoe, by I'Hospital's rule






-56-


(lim
p -9- 1
0-


2


d(D )


S --, 0

, lim (169)


(169)


equation (163) being used, with equation (16) in the last
step.
Now, from equations (146), (159), (165), (166), and
(169), it can be stated that


= Pga2 R( (lirn
"?1 -.0.


-ps2 H 25-2


c-~0


N2


nnn


2 +
2an


+ s2 + 21yr
4az


(170)


This may be written


= pg2 *. Rj 2,-2 zn + 2iyz1,-]


(171)


a2


Integration with respect to x yields


F pga2


HB(t .La2
2San


+3
+Wd


lim
cp-o


-N8


lim
cp --1.


lim
Cp-*o


lim
(p-10


-n+l
n+1


+ 2i ]3
a


(172)


M


-- a


2ivzn-l]


+ a


+x-p


*izz + l)
2.=- 2a 21


= 11 0n
tO>p0


+A+ I
eL2|+Z


-a-2


Oft t.






-57-
where the constant of integration, which does not affect the
stresses, has been taken to be zero. This result may be
simplified as follows:

lim F pga2 B f - + 2iy E] + + +
p- 0 2an n+1 n+1 n 2


2a
p2- + )ga


"^jUP +f 4a2



li (Z )+1 + 2 (173)




-p R- 0 2 (n + 1)an 2a 2

This function, lia F, is, in fact, the stress function for
cp- 0
an isotropio disk under the same loading as that considered
for the orthotropio disk.










CHAPTER III


STRESS DISTRIBUTION IN AN ORTHOTROPIC DISK
SUBJECTED TO ITS OWN WEIGHT WHEN
SUPPORTED AT ITS CENTER

3.1, l eU L= Solution.
Consider an orthotropio disk of radius a, referred to
a rectangular coordinate system as described in section 2,1.
The stress distribution in the disk subjected to its own
weight and supported at its center will be found by (1)
finding the stresses due only to the supporting force,
P = pgMa2, with no gravity (weight) force, and (2) superim-

posing on this stress distribution the stresses due to the
weight of the disk, with the requirement that the resultant
stresses be zero on the boundary of the disk. Section 3.2
will be concerned with the first ease, section 3.3 with the
second. Each of these oases, like the problem of Chapter I,
consists essentially of finding a solution to equation (13)
in the form of equation (23) which satisfies equations (26)
on the boundary of the disk. In the first ease, of course,
the equations referred to above must be modified by setting
g = 0.

3j.2, Streasseai Xa MS Mak aA llz Rf JAga SunI9ortn9r Eame
Z at It Centera


-59-






-59-
Let the stress function, the stresses, and the displace-
ments corresponding to the single force P at the center be
distinguished by primes ('). With g = 0, then, equations (6)
become
3'2p' z a2p_


and equations (26) become

x' = )

and (175)


A suitable stress function for this case is given by

P' = B(Azl log zI + Bz2 log z2) (176)

where zl and z2 are defined by equations (24) and A and B are
complex coefficients. The constants A and B are determined
by the following conditions:
(1) the stress o iss symmetric with respect to the
z-axis,
(2) the Integral / Xds, taken around the boundary

C of the disk, representing the resultant of
the x-components of the boundary stresses, is
equal to the magnitude P of the applied load,
and
(3) the displacements u' and v, obtained by Integra-
tion from equations (2), are single-valued.





-6o0-
These conditions will be applied in succession in the para-
graphs which follow.
Differentiation of F' gives

= BaA(1 + log z1) + B(l + log z2)}
ax=

and (177)

-L= R{ irA(1 + log zi) + ir2B(1 + log 2))

and, again,

3x2 sl Z2 *

a~ r hA + rLB
-ml = Ri- 1 9
C y21 (2 (17g)
and
=t ) + ir B '

Let
A=A' + ia' and B =B' + ip (179)
where A', B', ', and p' are real constants. Then, by the
first of equations (174) and the second of equations (178),
with and replaced by their respective equals x ~-
I1 Z2 Z2 + 2
x ir-y
and 2 o ,
x2 + ry2

(A + ia') (x iy) (B + 2)(x ir2y)
I x2+ +rjy2 + + x2+ 2


i.e.,






-61-

S 2 x + rI'y 2Bx+ r3'y
Ox I r 2 2 2 y2 (1+0)

Condition (1), that 0( be symmetrio with respect to the
x-axie, is seen now to require that (since we have assumed

3 ( ri2)
= p' 0 (181)

Consider next condition (2), /X', ds = P, where C is the
boundary of the disk. By the first of equations (175), this
condition allows us to write

C)d Td a 2 ) 1dzl + ')d2 = P (182)

whence, by the second of equations (177),

Rfirolk^ % + Ir2B f (3-93)

where C1 and C2 are the ellipses in the z1- and z2-planes,

respectively, corresponding to C in the z-plane (the plane
of the disk). The residues of the integrands I- and I- at
z1 z2

z = 0 and Z2 m 0, respectively, are both equal to 1, so
that application of Cauchy's residue theorem to the integral
of equation (183) yields

RirA 2ni + ir2B 2ni) = P (lg4)

or since A = A' (real) and B = B' (real) by equations (179)
and (181),
r2A + r2B 2 -i (195)






-62-


The last equality here results from the observation that
P = pgwa2
By the first two of equations (2) and the first two of
equations (174),

ai. 32' a2F
ax xa

and (316)

ay -- + a x2

Substitution of the expressions given by the first two of
A2,V* a2p'
equations (178) for and in equations (186) leads to


= RB(( + ) B( + )}
ba ^xZ l x y 22

and (187)

R= B( + ).+ B( 2+ .+

Integration of the first of these equations with respect to
x and the second with respect to y yields

u' R{ B-A( + (

B(I+ )(log 6z + ')
x y
and (18g)

v' r A B(( + g ) t(log 111 + i )

+ B(-Y- + iz)5 (log l, + iy!)






-63-
where 8' and cp' are the (multiple-valued) arguments of zl
and z2, respectively. Since A and B are real, by equations
(179) and (1i1), equations (188) may be written

ui M A + uZ) log z1 B(g + ) log I z21

and (189)
T' A(1T + --1)0' + B( + 4)-pI )

The displacement u' as given here is clearly single-valued.
Since, e' = Aro tan MlL + 2nr and (p' = Are tan r2 + 2nn,
X x
for n I 0, t 1, t 2, . the single-valuedness of v'
requires that

A( rN+ 1) + B( + o) = 0 (190)

Equations (185) and (190), solved simultaneously, yield
expressions for A and B which may be put into the form

pga2rr(E 2 + gC)
A = CEX + 2 .v..- y)
2EX(r1 r2)

and (191)
B pga2r2(E, + r 7Ey)
B=
B 2E,(r2 r*)

Replacement of ,Ecyy by VyXEx, permissible by equation (9),
leads to the simpler forms
pa2r (1 +
2(r r)
and (192)
pga2r2(1 + r14 )
B = 2(r )






-64-
With A and B determined, the stresses due to the load
P may be found from equations (174) and (178), Thus, since
x iry
A and B are real and since x =y and

L x ir- y
2 "x2+ rj 2

r2Ax 2Bx
+ry2 x2 +r2y2 '


+ B (193)

andAy B
r Ay i|By
Lxy - x2 + ry2 X2 + rjy2


where A and B are the constants given by equations (192),
The boundary stress components XI and YT may be obtained
from equations (175) and (177). There is no need for ex-
plicit expressions for these quantities at this point,
however, and their derivation will not be carried out here.

3.3. Superpositionof pt h Stresses Pue I to e Weiaht of,
the Disk.
Inasmuch as this phase of the problem differs essentially
from the problem of Chapter II only in the boundary condi-
tions, the same stress function as that of Chapter II, except
for the values of the constants, will be assumed here; i.e.,
the stresses due to the weight of the disk may be obtained






-65-
from equations (6) and equations (43) once the constants an
and bn have been evaluated. In fact, with an and bn evalu-
ated, these stresses may be obtained directly by insertion
of the new values for an and bn in equations (110) through
(112). The determining condition for the oonstants here is
that the resultant of this stress distribution and that due
to the force P at the center of the disk must be such that
Xv = Yv = 0 on the boundary, where, from equations (26) and

(175),
xv M 4-4a + pgm Ax
dsoy ay ) pz d
and (194)



With Xv = YV 0, these equations may be written


L(M) = (') + pgx A
dso y ds ay ds
and (195)

e( = drF +a
de ax dse x s

or, with the aid of equations (54) and (56),

e) - L( ) + -a 1 + cos 2e)
ds ay de ay 2 da
and (196)

dsE = ) a 2se 2
do ax da ax 2 do


Integration gives now






-66-


=i + -(2 + sin 29)
ay ay
and (197)

+- +0 20
-l + cos 29 ,
where the constants of integration, which do not affect the
stresses, have been taken to be zero.
It will prove convenient to consider, from equations
(197), the equation

+x J t + cos 29)

+ 1[- y + 2 +(2 sin 29)] (198)

If in equations (177) the constant terms be neglected, inas-
much as they do not affect the stresses obtained from F',
then those equations become


C R{A log zI + B log Z2}
and (199)

= y Blrl( A log zi + irgB log z2)

Since A and B are real coefficients, by equations (192), and
since, on the boundary
V 4 r- sin 0
log z1 = log a oos2' + r sin% + i aro tan ( os
and (200)
log X2 = log a o/os2a + r sin2 + i are tan ( sei ooi
equations (199) may be written, for the boundary of the disk,








i- A log (a Voo2 9 + sin2 )

+ B log (aVoos2 0 + r aa2 )

and (201)
ap, '1 sin 9 Pr sin 9
y -- r arc tan ( cos ) r2 aro tan ( oos 9 ).

Substitution of these expressions for 2aE and in equa-
ax ay
tions (198) gives

ax + 1 ay = f() + ig() ,(202)
where

f(e) + ig(9) [-A log (aVoos2 9 + r2 sin2 )

-B log (a oos2 9 + r2 sin2 9 )

2 rl sin 9
+ c' os 20] + I[CrA arc tan ( oos )

r2 sin9 2
+ r2B arc tan ( osn ) + -(29 + sin 200,
(203)
or, since cos 29 + i sin 20 = e21i

f(e) + ig(C) = A log (a Vos2 9 + r n2 )

B log (a Voos2 g + r sin2 )

ri sin 9
+ irAA arc tan ( oos 0
r2 sin 9
+ ir2B arc tan ( 'o 0 )

+ m e 210 + l 2 9 (204)








The function f(S) + ig(Q) can be represented by a complex
Fourier series; then equation (202) becomes


+0 (In enie (205)
where

on ji /n Cf(e) + ig()e-nie a& ,
n = o, 1, 2, . (206)

For n = 0, 1, 2, . let

II = Clog (a Voos2 9 + r sin2 9 ) ]e-nae 9 ,

12 = 1log (a Voos2 9 + j sin2 9 )e-nie d ,
r1 sin 9
3 = /' Care tan ( cos ) e-ne d0

r sin (207)'
14 = I lare tan ( g co )]e d

15 = /T -(n-2)i ,d
--Tt

and

16g / e"-nie .

Then, from equations (204) and (206),

On = t(- A 11 B 12 + irlA 13 + ir2B 1

2 2
+ a I5 + i 16} ,

n = 0, + 1, 2, . . (20)
The integrals I1 through 16 will be evaluated successively in







-69-


the following paragraphs.
In the evaluation of 11 it will be necessary to have

the value of O a-~ oa dq, where k = 0, 1, 2, ..

and a and b are real constants suoh that a > 0 and a2 > b2
To this end, consider


k


(209)


where C is the circle Izl = 1. The integrand -d +- l
b

C+L- 2 b nZ=b
has simple poles at z = -- and z2 b

of which points only z2 is inside the contour C. The residue

of k at z = z2 is
z Z + 1
b


e[(z k
(z zl( z2 z=z2


= lim (z
Z -E2
k
2
z2 E1
Z2 Z-


S z z~1z z2


I (210)


whioh may be written
k
Ree s2--#z --- 2) 22) .(211)
=T b


Hence, by Cauohy's residue theorem,






-70-


2 + 1 d2 2i a2 -

Now, on the oirole C, z = eiP and

k z i
2 dz ... do
z2 Wb z + 1 elp + e-4
S- 1

= kioscknL dq, (213)
2 a b oos cp

whence it follows that

Ik L dz E 1Tt =B Aa d(p
C Z2 kA'z +1 2 0 a-b oosq)

Si k /Tr gooa 102
2 a b osp d *

(21.)
The equality of the righthand members, and henoe of the
imaginary parts of the righthand members, of equations (212)
and (214) leads to the conclusion

12T cosg 4(9 2w a a2
0 a b oosp / b (215)

Consider now

~I = Clog (a Voos2 9 + i sin2 9 )3e ni' d
-wT

(page 68) and assume that n # 0. The radioand in 11 may be
rewritten as follows:






-71-


os2 0 + 2 sin.2 O 1 + (r2- 1 .in2 9

= 1 + (1 cos 29)
2
r2 + 1 2 1
2 2 -oo8 20

i.e.,
oos2 9 + r sin2 9 = a b oos 2 (216)

where

a = 2 and b = 2 (217)

It may be noted that a > 0 and a2 > b2. With the use of
equation (216), 13 may be written

I =- log a /" e-nie d9


-WI
(218)
or, since
/" e"- de = 0, for n # 0
-w

I1 f [log (a b oos 29)]e-ne de (219)
2 ,"

Beplaoement of e"a by oos ne 2 sin ne leads to

I /I [log (a b oos 20)]cos n 609

I Clog (a b 0oo 20)]sin no .

(220)
But the integrand of the second integral here is an odd






-72-
funotion; hence, / [log (a b cos 20)]sin no d9 = 0, and

I 1 log (a b oos 29)oos n de (221)
1 2 -T
Now, integration by parts gives

II = log (a b oos 29)] a inA


2_ T sin a sin 2
n a b oos 29

s sin n- sin2 de (222)
n a b cos 2

Use of the identity
sin ne sin 29 = kcoos(n 2) 0 oos(n + 2) e] (223)

in equation (222) gives

Soa(n + bo 2)9 dO -f / o(anI 2)a de (224)
1 2n na b cos 2 a b oos 2

In these integral, let cp = 29, d(9= 2de; then, for example,

coa(n + 2)9 de 2 /0T cos(n + 9)0 a
Sa b cos 29 0 a b oos 20

S/2TT 0co(8 + -L)CC
P p d (225)
o a b (225)
and equation (224) becomes


S 2n oa co( (226
o, t va s of t ntras n ts euaton ar 226)

Now, the values of the tntegrals in this equation are given





-73-


by equation (215), with k L n- 2 and k E~ respectively.
Thus, +2
i -h -2- Jfa -a 2 2
1 n b


(a )b 2 .] (227)

This expression for 11 simplifies when it is observed from
equations (217) that

)a -b2 = and ba + 1 = 02 (228)

where a is the quantity defined by equation (41). The use
of these relations and equations (217) in equation (227)
leads to
w(r 1)(.a ) 229)
2nr *

A further simplification follows from the observation that,

asnoe 62 Ir 1 and d2 -
Sand a r2 + 1

(r1 i)(o4 1) (r ) + 1)(r 1)(o2 + 1)(o2 1)

S- .4rIa2

and similarly, (230)
2 4 2
(r2 l)(d^ 1 - 1rg2 .

With the use of these relations, equation (229) becomes,
finally,
11 2f (231)






-74-
Examination of the first two of equations (207) reveals
that 12 differs from II only in the replacement of rl by zr2
It follows that, under the assumption n # 0,

a2 =- 0 (232)

where

a r2+l

Consider next
r sain
1 = I'T Caro tan (r. ooa e)e-nie d4

(page 68), and assume again that n # 0. Integration by
parts, with the observation that
r' sin 6 rile
d Care tan ( -os e a (23 )
os2 e + r2 sin2 *

yields
fr i sin 0, T
1 { Caro tan ( 0os 9e 3] e'ni



n cos2 + sin 9

i.e.,

2 -1. nni 2oos a -
3 n n -1 cos +r sin 0


S r,. sin ne -- d2 (234)
In -00oo2 + 2 sin2 e






-75-


But /" arn nto
-T cos 9 + sjin

odd function, Hence,


de 0, sinoe the integrand is an


S L-)n -2d oi2 fT c n 2
3 n n 0082 + 21 sin2 0


The application of equations (216), (225), (215), and (228),
suoessively, to the integral here gives


-n o0os0 + r sin
1


_C a b coso de

e oos '
a b cos 9




aron
^ '


i.e.,


SU


The use of this result in equation (235) leads to

2Tnr[(-1)n onl
3 n

Since it may be seen from equations (207) that I1
differs from I3 only in the replacement of rl by r2, it
follows, by equations (41), (42), and (237), that

I 2 n .


d0 (235)


(236)




(237)


(238)


/T," oo- c Oaa 3
-Tn 0o8 a + r sin-
1






-76-
As was the case for 13, it is assumed here that n 0.
The integral
S= /I e-(n-2)9 e

(page 68) can be evaluated directly to yield

I= 2r, for n 2 (29)
0, for n = + 1, 2, 3, 4,

Consider lastly

16 = I1 e-nie d

(page 68), and assume again that n # 0, Integration by
parts gives

16 = [e -n -e-n s

or
ig (24o)

sinoe
/ e" de = 0 for n ? 0.
-IT

Attention is returned now to oa, given by equation (208).
With I1 through 16 given, respectively, by equations (231),
(232), and (237) through (240), equation (208) becomes, for
n = 2, t 3, + 4, ,

on + (.l)n on] r2B[(-l)n dn
=- d1 2
na n n n

2 n (241)






-77-
or, with like terms in n oolleoted,

n= [A(rl + 1)on + B(r2 + l)dn

+ (-l)n+lrlA + r2B + 2)]

n = + 1, 2, 3, 4, . . (242)

Substitution of the values given by equations (192) for A
and B in the expression rlA + r2B + ga2- shows that this

expression is identically equal to zero, Henoe, equation
(242) may be written
A(r, + l)on + B(r2 + l)dn
On = n

n = 1, 2, t+ 3, 4 . (243)

For n = 2, the same equations oited for on with n $ 2 lead to
2
02 = C%, equation (243)]n + Pa- (244)

The value of on for n = 0 will be simply denoted by Oo not
evaluated.
The constants on, n = t 1, 2, . have been
evaluated to serve in equation (205), from which the oan-
stants an and bn oan be evaluated. For oonvenienoe, however,
the values of on will not be used in the following analysis
until explicit expressions for an and bn have been obtained.
With






-78-
aF n ,b .
a3 =- RL an; + -z b

and (43)

= R{ rlanZ +Z ir2bn

and with = '2 =. eion the boundary of the disk, equation
(205) becomes

RBz aneie + 0 b enI

+ iBR iraen e + Ir2bnel = one (245)

With a, and En denoting the complex conjugates of an and bn,

respectively, this equation may be written

E (aene~t + ie-nie) + E (b ene + 6ne-n~e)

-O (r nene r1nei ) r(r2ba r2 -e-i )

= 2ane1i (246)

or, rearranging the terms,

(1 r) c anene + (1 r2) n ben

+ (1 + ) e- + (1 + e?) $ ane bn

= 2 oneie (247)

The equating now of the coefficients of are in the two
members of this equation gives






-79-
(1 rli) + (1 + rl)a, + (1 r2)bn + (1 + r2)bn = 2on

n 0, 1, 1 2 . ,
(241)
i.e.,

(1 ra) + (1 + P)_-n + (1 r2)bn + (1 + r2)b-n 2n

and (249)
(1 r). + (1 + rl)i + (1 r2)bn + (1 + r2)S 2o.,

n = 0O 1, 2, . .

Now, by the same argument as that given on pages 19 and
20, equations (49) and (50) apply here; i.e.,

a-n = (-o2)na, and b-n =-2)bn

Hence, equations (249) may be written,

(1 rQ)a% + (1 + rl)(-o2)n~ + (1 r2)bn

+ (1 + r2)(-d2)n 2%,

and (250)
(1- r)(-o2)an + (1 + rl)

+ (1 r2)(-2)b n + (1 + 2) n 2o-

n = 0, 1, 2, @. .

As has been stated earlier, the constant terms in
dx
and which, in equations (43), are the only terms involv-

ing ao and bo, have no effect on the corresponding stresses,






-go-
given by equations (6). There is no need, therefore, to
evaluate ao and bo, and equations (250) will be considered
only for the oases n = 1, 2, . . It is seen from equa-
tions (192), (243) and (244) that, for these values of n,
an is real.
Let

an = + i and bn Bn + iFn (72)

where An, an, Bn, and (n are real constants. Then oonsidera-

tion of the real parts of equations (250), noting that an is
real for n # 0, leads to

C(1 ri) + (1 + r)(-02)n ]A

+ [(1 r2) + (1 + r2)(-d2A]n = 2%o

and (251)

C(1 r)(-o2)n + (1 + rl)]A

+ [(1 r2)(-d2)n + (1 + r2)]B3 = 2c. ,

n = 1, 2, . ;

and consideration of the imaginary parts leads to
[(1 r1) (1 + rl)(-o2)n]

+ [(1 r2) (1 + r2)(-d2)na3p =
and (252)
[(1 ri)(-o2)n (1 + r)]an

+ E(1 r2)(-2)n (1 + r.)]n = o ,


n = 1, 2, . .






-81-


From equations (251), now,

20n (1-r2)+(l+r2)("_2)n

2o0n (1-r2) (-d2)+(1+r2)


(l-r2)+(l+r2) (.2)n2






2on

20o
-n


(253)


Evaluation of the determinants and subsequent simplification
yield
rp(o,+o-n)E[il-(-.d2)n]+(,on-O)-n[ (+d2)n]


A M-
n


(254)


and


Bn -


where

Dn (.-r2)[l --22)n-(rl+2)[(-c2)n- (-.2)n]
n = 1, 2, . .


(97)


An m





and





Ba


Y


rI(n+o-n) 1-(-o2)n]+(e-0o)11+(-o2)n]


(l-ri)+(l+ri) (-o2)n

(1-i) (-o2)4n+(1+ri)


(1.rl)(..2)n+(,lrl)


(1-r2)+(l+r2) (d2)n

(1-r2) (_2)n+(1+r2)


3.-rl)+(l+rl) (..o2)n








It was shown in Chapter II, pages 28 through 33, that Dn # 0
for n = 1, 2, . From equation (243), which gives the
values of en for n = + 1, 2, t 3, + 4, . it follows
that
an + e-n = [A(l + 1)(on e-n)
+ B(r2 + 1)(an -a'f)] (255)
and
n o-n = ( A(rI + 1)(0 + e-n)
+ B(r2 + 1)(d~ + 4-n)]
n l 1 3, 4, * (256)

Use of these equations in equations (254), followed by some
straightforward manipulation, yields the results

A(r + i)(o" + (-2)n)
=- -(r2 + +) 1 on

+ (1 + (-l)n)B(r2 + 1)e]

A((r + 1)(1 + (-0o22)n)
(r )1 -
a on

B(pr + !)(1 + (- 4)n
+ d)a} (257)

and
Bh = (rz + 1)[(1 + (-1)n)A(r + 1)on


B(ra + 1)((-o2)n + 21)
+








A(r + 1)(1 + (-i)
(r1 1)- --51 -

B(r2 + 1)(1 + (-o2d2)n)
+ 2 ]
+ ---- 5---- *

n = 1, 3, 4,. . (25)

From equation (244), which gives the value of 02, it follows
that
02 + o_2 = [o, + ca, equation (255)]n= + DPt

and (259)
2
02 0-2 = [n .-n, equation (256)]n= + 2 .

Use of these equations in equations (254) leads to

pea 2r2 + 1 (r2 1)d4]
A2= [A,, equation (257),n=2 4D2

and (260)
Pga2[rl + 1 (r 1)o ]
B2 [B., equation (25)]n. + -(---1- -


Consider next equations (252) in an and Pn The deter-

minant of these homogeneous equations, considered simultane-
ously, is

("-r l)-+ri) -o2)n (1-r2)-(l+r2) -d2)n
An = (261)
(1-r1)(-o)n-(1+r ) (l-r2) (g2n-( l+r2)

Upon addition of the second row to the first, removal of the







factor 2 from the new first row, and subtraction of the re-
sulting first row from the second, it is seen that

n 2Dn (262)

where DA is the determinant given by the second of equations
(77). It was shown in Chapter II, pages 2S through 33, that
D' f 0 for n = 2, 3, . and, as stated on page 33, oal-
oulation shows that D' = 0. Hence,

A1 = 0
but (263)

4n O n= 2, 3, . .

It follows from this last statement and equations (252) that

oLn = n = 0, n 1 2, 3, * (264)

but that, from the second of equations (252) and equations
(41) and (42),
(1 r2)(-d2) (1 + r2)
1 ( (l-r)(-o2) (1 + r)

1+ ri (1 !a)2 .(1 + r)2
1 + l r2 (l- )2 -(1+ r)21


"- r r +, (265)

where the constant ~ is arbitrary. But this relation be-
tween ox and 3l1, with p1 arbitrary, is the same as equation
(101), Chapter II, Since the same stress function exceptt





-85-
for constants) as that of Chapter II is being used here, it
follows that the contributions to the stresses here of terms
involving il and P, are the same as the contributions of the
terms involving.al and P, to the stresses in the problem of
Chapter II, whioh contributions were shown an pages 32 and
35 to be zero. It is convenient to set (as in Chapter II)
Oa = 0 1 o (102)

For conveneinoe, the formulas for the values of an and
b, n = 1, 2, . obtained from equations (72), (257),
(258), (260), (264), and (102), are given below.

a(+i () (o(+(-d2)n)
%a o- I 1 -P+ (1+(-)n)(r2+1dn1


- (r2-1)


n = 1, 3, 4, ,


(266)


Sn n B(r+1)((-a)nd22)


A(r+l1)(l+(-o )n) B(r2+1)(l+(-o2d2)n)
- (r-1) +- ?,
on dn


(267)


a2 = [an, equation (266))n=2 -


pga2[Cr+1-(r2-l)d4]


, (268)


and
P'ga2 Z+l-(r-1)- o ]
b2 = [h.n, equation (267)]n.2 + hn


* (269)


A(rl+1)(+(-o2d2)n) B(r2+1)(l+(-d4)n)
+ ,3),
on n


n = 1, 3. 4, . ,






-96-


where
Dn = (r-r)1[1-(o2d2)n] (rl+r2)(-c2)n-..m42):n]
n =1 2, . . (97)

With these constants evaluated, the solution of the problem
is essentially complete.
As stated at the outset of this chapter, the actual
stresses in the disk considered here are obtainable as the
resultants of the stresses due to the supporting force,
given by equations (192) and (193), and the stresses due to
the weight of the disk, given by equations (110) through
(112) and equations (266) through (269). Thus, insertion of
the values of an and bn given by equations (266) through
(269) into equation (110) and addition of the result to the
first of equations (193) lead to

G R)(r +l)(o22 +(-d2)n)-(r2-1)((+(-o2d2)n )
G-x = Rj, [A(r +L).. ... n ...


+(r2+)(r2+1)(l+(-l)n) (r2) (+(. )
an

2z n


_. (1ri+1)(1+(-1)n)o2n-(r -1)(l+(-o ) )


(r,+l)((-c2)n2n)-(r -.1)(l+(-od2)n)
+ B(r2+l) -

r--# -2D n j
"2D







2 22 -4 2






S" P (270)

where
pg 2r (l+32. ) pga2 2(1+ 2 V
A = and B = (192)
2( -r2 ) 2(r2-r)

Similarly, with equations (111) and the second of equations
(193) playing the roles of equations (110) and the first of
equations (193), respectively,

(r 2+) ( o2n+(- 2)n)-(r-1) (1+(-0o22)1)
-. RB- [A()j:-+l)
"2 1 A(+

(r+i) (1+(-1)n)da-(r-1) (1+(-a4)n)
+ B(r2+) !2+n1]
on


(+1)(+(1)n)o2n( 1)(+(.-o")n

+Z CA(rj+1)

) (r1+1) ((o2 )n+2n)-(r-l) (+(-od2)n)
+ B(r2+1)-



2n^


-87-







2 _-o4 2
- pga2[r+l-(r2-1)a4] 3 ..--..
2W1D2

+ pg,,a2, -2I
+-pga1ri+l-(-"zr o 2W2D2


+ Ax
xry


(271)


+ .
x2+r2y2 x
2


and, with equations (112) and the third of equations (193)
playing these respective roles,
00 (r2+1)(o 2n+(2)n)(r21) (1+(-o2d2)n)
Z-y = Rai, [A(r+l) ..


+ B(r2+1


1 I Z (-o2nn I
w 2 J


00
v4' rAt- .fi,


al------


+ Brpg2+) .. n

r2_-(-d.2)nq-=n



+ pgai[r2+l-(r2-1)ad4]a'
2WIPw2


4;2
- pgaiC20+1. (-l)0l 2' r2 'I


(r,+l)(l*(~l)n) a2P-( F~s~l)(l+(~LaL)n)







r2Ay
- X2+r2y2


r By
-?^


where, as for a-, A and B are given by equations (192). The
similarities of these stress formulas permit the writing of
the composite formula
R (r[+1)(o2( (d2)n)-(rp-.l)(l+(-o2d2 )n)
Sk- IkRTi[A(rl +l) n


+ B(r2+l


Go
4+ v r&.A +1


)- -- .. __ --- _--- ]
0 r i


1Dn

(r +1)(+-l)" ) 2-( -1)(.+-o 4)n
(r I r (+_)nn r-)(+-


(r+ ((-c2))d i- l -1) (,+(-o.2d2)2n)
+ B(r2+l) dn ]



W2Dn

pga2r2+1-( r2-l)d [.-0 c42
+ Tk 2W1D2

SUrs +l- W-lDo 4]2T 2-d42]
+ Uk -ga21+1- (- -)[2 -.
2W2D2


- 'V Pgz + A +


B


, k = 1, 2, 3,


(273)


where Sk, Tk, Uk Vk, Xk, and Yk are given by the table


-.9-


(272)






-90-


below.
k k Tk Uk k Xk Yk

1 T r -r2 1 -r2x -rx
2 -(T -1 1 1 x x

3 TZ ir -ir2 0 -ri y rJy



3.4. DiAlacements.
Those components of the displacements resulting from
the stresses due to the weight of the disk have been obtained,
except for the new values of the constants, in section 2.5
and are given by equations (139) and (140), together with
equations (266) through (269) for the values of an and bh,
n = i, 2, . To find the contributions, call them u'
and v', of the stresses due to the supporting force at the
center of the disk, let the values of the stresses c0 and c-'
given by the first two of equations (193) be substituted
into the first two of equations (2) to yield, after simpli-
fication,
= x Ax Bx
ax E x2+r2 2 x2+r y
x x?

and (274)
t),,+1 Ax_22v Br
by Ey By E2 xry2

Equation (9) has been used here to simplify the coefficients.
Integration of the first of equations (274) with respect to






-91-
x and the second with respect to y gives

Ut log(rx2 y2) log( p+4ry2)+f(y)
x 2 x 2 2
and (275)

o tn r- Ar tan y+ g(x)
SAr tan E r2 AxOt8fl

where f(y) and g(x) are as yet undetermined functions of
their arguments. These functions will be determined by the
method used in section 2.5.
The derivatives of u' and v' with respect to y and x,
respectively, obtained from the equations above, are
-^2 ^.2 2
r r ;+V r2By



and (276)


ax x yy F+r2 Y *


With these expressions for the derivatives, the last of
equations (2), with ZZY replaced by T yields

r +V r2 +1 Ay
zxy xy E r2 2+r2y2


r+ 1) z+1 r22By
+x dy dx
(277)






-92-
But, by equations (130), the quantities in parentheses are
each equal to Hence, the equation above may be written


iAy r2By ri+ G[ ) + (27z)
S=------+oG Ct l + .]* (27w)


Comparison of this equation with the third of equations (193)
indicates that
GxE + L = ) (279)

i.e., since Gy f 0,



Since the lefthand member of this equation is a function of
y only, while the righthand member is a function of x only,
the fact that the equation must hold for all points (x, y)
within the disk leads to the conclusion that

W.UX d- = = k, where k is some constant. Integration
dy dx

of these functions gives
f(y) = ky + m
and (291)
g(x) = kx + n,

where m and n are arbitrary constants. These constants m
and n represent constant contributions to u' and v', as is
seen from equations (275), and hence represent a rigid dis-
placement of the whole disk; it follows that, if the support






-93-
is to be stationary, then m = n = 0. Also, the fact that the
displacement u' (in the x-direotion) is symmetrio with respect
to the x-axis requires, considering the first of equations
(275), that f(y) be an even function, i.e., that k = 0.
With k = m = n = 0,
f(y) g(z) 0 o (282)
and equations (275) become

u A( ) log(z2 + xjy2)

2
B(r 2 v )
ati. s' log(x2 + r2 2) (283)

and A
a A (r^ @ +) +i
ay = +.E Aro tan

B(r 2v + 1)
+ r2Ey Arc tan (2&4)
r2Ey x

Now the actual displacements in the disk are given by
the sum of equations (139) and (283) and the sum of equa-
tions (14o) and (284); thus,




x I)


2S log(x2+ri2)- 2 X log( +r22) (285)
A(5\ + )




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