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On the Stresses in a Plate Containing Two Circular HolesChihBing Ling

Citation: J. Appl. Phys. 19, 77 (1948); doi: 10.1063/1.1697875View online: http://dx.doi.org/10.1063/1.1697875

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On the Stresses in a Plate Containing Two Circular HolesCHIH-BING LINGGuggenheim Laboratory, California Institute of Technology, Pasadena 4, California

(Received May 29, 1947)This note gives a theoretical solution to a plate containing two circular holes of equal size.

The method of solution is to add to the given stress system a suitable biharmonic functionwhich gives no stress at infinity. The parametric coefficients involved in the solution areadjusted so as to' satisfy the boundary conditions at the edges of the holes. Bipolar coordinatesare used in the solution, by means of which explicit expressions are obtained for the parametriccoefficients. Three fundamental stress systems are discussed in some detail, namely, the all-around tension case, the longitudinal tension case, and the transverse tension case. Formulasof the stress along the edges of holes are derived and, in particular, values of maximum stressare calculated. Th e limiting case in which the holes are tangential is also discussed.

T HE general solution of the stresses in aperforated plate, of infinite size, containinga group of circular holes was obtained by How-land and Knight,l and Green.2 Naturally, suchgeneral solutions do not give much informationas to the numerical aspect of the problem unlessparticular cases are worked out separately. Inhoth solutions, it may be mentioned that theparametric coefficients involved in the stressfunction are solved by the method of successiveapproximations, which is, however, a lengthlyprocess.

Th e simplest case of a perforated plate is theone containing a single circular hole. This prob-lem has been discussed in great detail by variousinvestigators. The next less simple case is aplate containing a group of two circular holes ofequal size. Owing to its importance in engineeringapplications, it is thought that it is also necessaryto discuss this problem in some detail.

A plate problem involving two circular boundaries is best treated by means of bipolar coordi-nates. A general discussion of stress and strainin bipolar coordinates was given in a paper byJeffery sometime ago.3 Nevertheless, the problemof a plate containing two equal holes has not beenworked out as fully as it deserves. In this note, asolution of this problem, differed slightly from

1 R. C. J. Howland and R. C. Knight, "Stress functionsfor a plate containing groups of circular holes," Phil.Trans. A 238, 357 (1939).2 A. E. Green, "General biharmonic analysis for a platecontaining circular holes," Proc. Roy. Soc. A 176, 121(1940).3 G. B. Jeffery, "Plane stress and plane strain in bipolarcoordinates," Phil. Trans. A 221, 265 (1921).

VOLUME 19, JANUARY, 1948

Jeffery's, wiII be given, corresponding to severalfundamental stress systems acting in the plate.The parametric coefficients involved in the so-lution are obtained in explicit forms in terms ofbipolar coordinates. Finally, the maximum stressesin the plate are calculated and plotted againstthe distance between the holes.

The bipolar coordinates ( ~ , 71) will be definedby the equation of transformation

x+iy= -a c o t h ! i ( ~ + i T / ) , (1)such that the two poles of the coordinates arelocated on the x axis at the points (a, 0), and

x=JsinhT/, y=Js i n t (2)where(3)

The biharmonic equation V4X = 0, which must besatisfied by the stress function x, then transformsto

(4)

Consider a solution, even in both and 71, ofthe type

(5)The transformed biharmonic equation then re-duces on substitution to an ordinary differentialequation

(6)

77




Consequently, the solution even in 1/ is

where An and En are parametric coefficients.This solution holds true for any n except n = O.The solution in the latter case, however, will berejected partially on the ground that it isredundant to stresses and partially that it givesrise to multi-valued displacements. There arealso some even solutions of Eq. (4), which arenot included in (7), for instance,

xlJ = K(cosh1/ - c o s ~ ) log (cosh!) - c o s ~ ) , (8)K being a parametric coefficient. I t is foundconvenient to include this solution also in thesolution concerned presently. A more generalsolution is, therefore,

00+ L f/ln(1/) cosnf (9)n ~ 1

Th e stresses derived from XI are as follows:

a }XI- s i n ~ - + c o s h 1 / -J

= -!K(cosh21/-2 cosh1/ c o s ~ - c o s 2 ~ ) 00+f/lI(!)+! L {(n-1)(n-2)f/ln_I(!)n=1

FIG. 1. Two equal holes.78



derived from xo at 'Y/ = a be denoted by00 00

'Y/170 = T L Cn c o s n ~ , ~ n o = T E bn s i n n ~ . (15)71.=0 71.=1

Consequently, the boundary condition of notraction at the edges of holes (i.e., 'Y/17=O, ~ , , = O , when 'Y/ = a) leads to

- 2co = 2I(a) -K cosh2a,and for n)1

(16)-21/;,,' sinha+2K(oln cosha-02n),2bn = 1/;71.-1' - 21/;71.' cosha

+1/;71.+/+2Ko1n sinha,where Omn is Kronecker delta which is equal tounity when m=n, or to zero when m ~ n , an d

TABLE 1. The maximum stresses (1'= 1).(1) All-around (2) Longitudinal (3) Transversetension tension tension>. 8= 0 8=71" 8=,,!2 8= 0 0=11"

1 2.894 00 2.569 3.869 oc1.5 2.255 2.887 2.623 3.151 3.2642 2.158 2.411 2.703 3.066 3.0203 2.080 2.155 2.825 3.020 2.9925 2.033 2.049 2.927 3.004 2.9978 2.014 2.018 2.970 3.001 2.99900 2.000 2.000 3.000 3.000 3.000

solved in general for any even stress systemacting in the plate.

Three fundamental cases will be considered inthe sequel, namely, (1) the plate is under anall-around tension T, so that we have

xo=!T(X2+y2),or (20)

1/;n=(n-l)n(n+l)n(a), 1/;n'=nn'(a). (17) (2) the plate is under a longitudinal tension T,The systems of equations may be replaced by

21(a) =K cosh2a-2co,"J

1/;1' = 2Ke-a sinha-2 E bme-",a,m=1

and for n ) 2 by1/;71.' sinha =1/;1' sinhna

-2K sinh(n-l)a sinha71.-1+2 E bm sinh(n-m)a,m=l

1/;" sinha=1/;I'(n coshna-cotha sinhna)+K[(n+l) sinh(n-2)a (19)

- ( n - I ) sinhnal0.--\-2 E !(n-m)bm cosh(n-m)am= l

- (mcm+bm cotha)Xsinh(n-m)al

The coefficients A" and B" can thus be expressedexplicitly in terms of Cn , bu , and K. An additionalequation is supplied by the condition (13) forthe determination of K. The problem is thusVOLUME 19, JANUARY, 1948

or (21)

and (3) the plate is under a transverse tension T,

or (22)Xo/aTJ = t sinh211/(cosh'Y/ - c o s ~ ) .

The stresses at 'Y/ = a derived from Xo are thengiven by

rfi//T= (1-cosha c o s ~ ) 2 / ( c o s h a - c o s ~ ) 2 , 1H/T=sinh 2a s i n 2 V ( c o s h a - c o s ~ ) 2 , [r,/T= -sinha s i n ~ ( 1 - c o s h a c o s ~ ) / I

(cosha - ( ' O S ~ ) 2 , and

(23)

(24)

~ / T = (1-cosha c o s ~ ) 2 / ( c o s h a - c o s ~ ) 2 , (25)'Y/1J/1'=sinh2a s i n 2 U ( c o s h a - c o s ~ ) 2 , f~ l I / T = s i n h a s i n ~ ( 1 - c o s h a c o s ~ ) j

(cosha - c o s ~ ) 2 , respectively. Hence, we have in the first case

co=1, cn=bn=O. (n)1) (26)

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andn the other two cases, we expand the first andthe third expressions, respectively, into Fourierseries between = -71" and = 71" with the aid ofthe following integrals, for n 0,

Co = e-a sinha, }cn=bn= -2e-na sinha(n sinha-cosha), (29)

( n ~ 1)

J" c o s n ~ d ~ _ ~ ( c o s h a - c o s ~ ) 2 respectively. With these coefficients we obtain2tP1(a) =K cosh2a-2co,(27) ifi"en = - 2K (n sinha+cosha)

271"e-"(n sinha+cosha)sinh3a

n(n2-1) sinha, ( n ~ 2 ) (30)ifin'en =2K sinha=Fn(n sinha-cosha).

Then we readily have ( n ~ 1)

Co = e- cosha, }cn =bn =2e-na sinha(n sinha-cosha), (28)

( n ~ 1)

Wherever the sign is ambiguous, the upper onegoes with the longitudinal tension case and thelower with the transverse tension case. In theall-around tension case, the terms with ambiguous signs are all absent. Hence, we have for n 1,

2K(e-na sinhna+ ne -a sinha) n(n+1)(e-na coshna - ne- a sinha)An=-------------------------------------------------n(n+ 1)(sinh2na+n sinh2a)2K(e-na sinhna+nea sinha)n(n-1)(e-na coshna-nea s i n h ~ ) ~ = ,n(n-1)(sinh2na+n sinh2a) (31)

with th e exception thatB1=!CK tanha cosh2a - 2co=Fe--2a coth2a).

Moreover, in the all-around tension case, K is determined from

{00 e- na sinhna+n sinha(n sinha+cosha)}K !+tanha sinh2a -4 :E = L

n=2 n(n2 -1 ) (sinh2na+n sinh2a) (32)

But in the longitudinal and the transverse tension cases, unity on the right-hand side of Eq. (32) isto be replaced by the quantities .'"t!=F2 sinh2a :E n/(sinh2na+n sinh2a). (33)

n= l

Most important is the stress at the edges of holes. With the same distinction to ambiguous signsas before, this stress is found to be given byHa {( oc sinhna c o s n ~ )- = 2 ( c o s h a - c o s ~ ) K sinha 1+4:E----T n=1 sinh2na+n sinh2a

'" n(n sinhna sinha-coshna cosha) c o s n ~ } =F2:E .n=1 sinh2na+n sinh2a (34)

80 JOURNAL OF ApPLIED PHYSICS




For convenience, we shall define 't] and 8 asshown in Fig. 1, which are connected to a, a) by

cosha = >-, c o s ~ = (1 +>- cosO) /(>-+cosO). (35)It appears that there are two points of maxi

mum (tensile) stress at th e edge of each hole.In the all-around and the transverse tensioneases, they are at 0=0 (or ~ = O ) an d at 0=7r (or~ = 7 r ) . In the longitudinal tension case, thesepoints are very nearly at 0= 7r/2, or moreprecisely at 101 slightly less than 7r /2 . Theyshift towards (J = 7r /2 when A increases. Fo rsimplicity in calculation, the stress at 0= 7r 2in the latter case may be taken as the maximumstress without any appreciable error.

The limiting case a= OCJ (or >-= OCJ) correspondsto a single hole in the plate. The results in thiscase are well-known. On the other hand, in th elimiting case a = 0 (or >- = 1), the hvo holes inthe plate become tangen tial to each other. Itcan be shown without much difficulty that whena=O, th e series in Eq. (32) takes the followinglimiting value

00 e-na sinhna+n sinha(n sinha+cosha)2 2: - - - - - - - - - - - - - -, , ~ 2 n(n2 -1 ) (sinh2na+n sinh2a)

= 2: 1 2 a 2 f " ' ~ s i n h 2 c p - c p 2 ) d C P . , , ~ 2 n(n2-1) 0 cp3(sinh2cp+2cp)

Since the sum of the series on the right-handside of the expression is equal to t, Eq. (32) thusbecomes in the limit

(36)

Also, the quantities in (33) become in the limit(37)

The maximum stresses take the following limitingvalues when a=O: (1) in the all-around tensioncase at ~ = O ,

...-3.6i3.4Iii13.2:I

2.82.62.42.2

(31 01 0.\- (3101 " .\.-

......-l -X-/"- (21 at >r12.

\ \ (I ) 01 1.'0.V \- (11101 . , , , ,:---....

2 I 2 3 4 & 6 7 8 9 10NO. OF OIA. Co TO C. OF HOLES, J\I-0 0- -0 0 -I

(I ) All-around (2) Longitudinaltension case tension case

0 '0I(3) Transversetension case

FIG. 2. The maximum stress.

an. i"'sinhcp coscp dcp- -=8(a 2K) - - - - - -T (I sinh2cp+2cpf. ",cp(coshcp - cp sinhcp) coscp dcp+4 ,

(J sinh2cp+2cp (39)

and (3) in the transverse tension case at ~ = O . foro sinhcp dcp- = 4 ( a 2K)

T n sinh2cp+2cpi oocp(cp sinhcp-coshcp)dcp+2 .o sinh2cp+2cp (40)Values of the maximum stresses are calculated

and shown, for T = 1, in the accompanying table.Figure 2 shows the resultfi graphically.I t may be noted that the integralti in thelimiting case a = 0 can be cOllveniently evaluatedwith the aid of the Illlmerkal values of thefollowing in tegrals, for n?:- 1,

(41)

which have heen calculated by Howland.! For(38) example:;:

(2) ill the longitudinal tension case at = a (or0= 7rj2)VOLUME 19, JANUARY, 1948

4 R. C. J_Howland, "On the stresses in the neighborhoodot a circular hole in a strip under tension," Phil. Trans.A 229, 67 (1930), Table I.81




(42)

An Alignment Chart Giving the Polarization Correction of Equi-InclinationWeissenberg Photographic IntensitiesW. L. BOND

Bell Telephone Laboratories, Murray Hill, New Jersey(Received August 15, 1947)

A nomograph is presented which gives, in te rms of film coordinates iJ. and T, the polarizationfactors to be applied to intensities taken from equi-inclination Weissenberg photographs tocorrect them for polarization. The remaining correction, the Lorentz factor, merely involvesdivision by COS 2iJ. sinT.

IN determining the structure factors F or Pfrom observed intensities of spots on equiinclination Weissenberg photographs,* a correction for polarization is necessary. The computation of the polarization correction factor, 1/p, issomewhat tedious. Here, P",1/Ll/p lo, wherelo is the spot intensity

and

1+cos220p=---2L (the Lorentz factor)

FIG. 1.

COS2/l sin')'

* M. J. Buerger and G. E. Klein, "Correction of. x-raydiffraction intensities for Lorentz and polarization factors,"J. App. Phys. 16,408 (1945).

Knowing /l the inclination angle and l ' the offaxis angle of the reflection, one can compute 20,the ray deviation, and hence 1/p. This can bedone very conveniently and with sufficient accuracy with an alignment chart. This paper presents such a chart which gives the polarizationcorrection for equi-inclination Weissenberg photographs.

Po20304050

60

70

80V.

90

,0 0

110

'2 0

13014 015 0.6 0.7 0.80

I035P 3,0H'00FiG. 2.

-

1.01.11.21.31.41 . ~ 1.61.71.6\.91.95

Z.O

1.9!)1.91.61.71.61.51.41.31.21.1

1.0

JOURNAL OF ,APPLmD PHYSICS

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