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Torsion problem for elastic cylinder with inserts and holes
COMPUTER METHODS IN APPLIED MECHANICS @ NORM-HOLLAND PUBLISHING COMPANY
AND ENGINEERING
23 (1980) 281-291
TORSION PROBLEM FOR ELASTIC CYLINDER WITH INSERTS AND HOLES
Department
R.D. BHARGAVA and S. PURANIK of Mathematics, Indian Institute of Technology, Bombay 4OW76, India Received 31 May 1980
An integral equation method to solve the classical torsion problem for an elastic cylinder with inserts and hoies is treated. The bounded region outside the inserts and the holes will be termed a matrix. As is well-known the solution depends on finding plane harmonic functions in the matrix and inserts such that (a) on the outer boundary of the matrix and the boundaries of the holes the harmonic function in the matrix takes the values $(x2+ y’) + ci, and (b) on the interfaces of the matrix and the inserts relations exist between the harmonic functions and between their normal derivatives. Here (x, y) are the coordinates of the point on the boundary and cj are unknown constants. The usual methods are cumbersome and lengthy. In this paper a straightforward method is presented which is easily programmable. The numerical solution is obtained by evaluating a few integrals either analytically or numerically and solving a system of linear simultaneous equations. An example of a cylinder with an eccentric insert is given which substantiates the theory developed in this paper and is found to agree with known resufts. However, the method is genera1 and may be applied to a variety of problems.
Introduction Recently the authors [l] have solved the torsion problem of an elastic cylinder with holes by the integral equation method. This paper is an extension of earlier work where one considers the case of a cylinder with inserts and holes.
1. The problem Let S be the cross section of a prism with its axis parallel to the z-axis and consisting of the regions $1, ~2, . . . , s,,, and a region so surrounding all the Sj (fig. 1). Let the boundaries of the nonintersecting regions Sj be & (j = 1, . . . . , m), and let so have outer boundary LO and inner boundaries L1, . . . . ,L,. Note that LO contains all the closed contours L1, . . . . , 12,. Let the regions si, s2, . . . . , Sk (k I m) consist of the regions corresponding to inserts, and let the remaining regions consist of cavities. If all the regions are cavities, one may solve the torsion problem by the method suggested in [l]. The conditions for the torsion problem are satisfied by assuming that the displacement components are u = --7zy,
V =
TXZ,
w = T@fX,Y,k
where T is the twist per unit length of the prism and @(x, y) is the torsion function. The strains
282
R.D. Bhargava, S. Puranik, Torsion problem for eiasric cylinder with inserts and hates
Fig. 1. Cross section of cylinder with k inserts and m -k
holes.
are obtained by strain-displacement relations, and stresses are obtained by Hooke’s law for the regions SO, sl,. . . , Sk. The nonzero stresses are
x =pj $7
(
)
(1)
Y*=&($+x),
)
where JQ is the shear modulus of materials in sj (i = 0, 1, . . . , k). Substitution of the stresses in the equilibrium equations show that function @(x, y) is harmonic in so, sl,. . . , Sk. The boundary conditions in these regions are (a) the external boundaries of the bar are free from tractions, (b) the stresses on the interfaces are equal and opposite, and (c) the displacements remain continuous across the boundaries L1, Lz, . . . , Lk. Condition (a) implies that x, cos(n,x)+ ~ndition
Y, cos(la,y)=O
on Lo and I&
(2)
(b) implies that [X* cos(n, x) + Y; cos(n, y)]o = [X, cos (12,x) + Y, cos(n, y)L on
Ljtj
=
1,2,. . . , k),
(3)
and condition (c) implies that [@lo = [@lj
on Lj(j=
1,2, ._. , k).
(4
Here the subscripts 0 and j refer to regions so and si(j = 1,2, . . . , k). Substituting the values of stresses from eq. (1) into eqs. (2) and (3), one obtains dQ, -& = y cos(lt, X) - X cos(n, y)
on&and&
(k
(5)
R.D. ~~arga~a, S. ~ranik,
283
Torsion problem fur elastic cy~i~~~r with inserts and holes
on L&=1,2
,...,
k), 6)
and of course eq. (4) must be satisfied. Note that the normal is taken positive when it is outwards from so. The problem may be solved by finding the function @ in the regions. It is possible to fo~ulate the solution in terms of the function (P]j = Pi which are conjugate of (@>)j3 @j in $j (i = 0, 1, +. . ) k). The stresses in eq. (1) may be written in terms of the conjugate functions as
x, = b%($--Y),
for31 u=O,
Y~=~j(-~+~)
1,. . . 1 k).
(7)
Using the Cauchy-Riemann equations &Pj/dn = dPj/ds in eqs. (5), (6) and integrating them over the respective boundaries, one obtains eqs. (8), (9) below. Also differentiating (4) with respect to s and using the equation aQ3i/as= -Z?$/dn, one gets eq. (10): on&
(i=O,
k+l,..
on&
~0~~~~~y)~~j~j(~~y)~~(y;0~~j)(~*+y2)+bj
a!P,
an
-22
aw, an
.,m),
(8) c=l,2
,...,
k),
on Lj (j = 1,2, . . . , k).
(9)
(10)
Here ci and bj are constants of integration-there are m -k + 1 C:! and kbj. Some comments now follow about these constants. Let go and gj be the harmonic functions in So and Sj 0’ = 1,2,. . . , k) such that
~c&, Y) = g&
Y I+
co
in Sn
and
(11)
in Si (i = 1,2, . . . , k). Substituting
these values in eqs. (8)-(lo), leads to
g&, y) = i(x”f y”) go(x,
y) =
on LO,
&x2-ty”) -t- c;
On
yogo(x,Y)-~jgj(x,Y)=~Olo-CLi)(n2+Y2)+Di
ago_itk an ai2
onLj
d.
I
pod0- bj Pi
on Lj tj = 1,2,
G=lt2,...,k),
where Dj =: bj + pjdj - podo. Let dj be
SO
(12)
Lj 0’ = k + 1, a + e , EE),
, k),
(13) (14)
chosen that Dj = 0, thus
(i = 1,2,. . . , k).
(15)
R.D. Bhargava, S. Puranik, Torsion problem for elastic cylinder with inserts and holes
284
Note that the stresses in eq. (7) are not affected by substituting in terms of the functions go and gi from eq. (11). The nonzero stresses thus are given by
(
>
x,=pjr g-y
Y,
)
=
j.qr
(-2-X)
inSj
G=1,2,...,k). (16)
Hence, one may work with the harmonic functions gi 0’ = 0,1,. . . , m) in place of the Fj m). Note that in practice one does not determine the constants in eq, (9) and as o’=O,l,..., is also demonstrated by the particular problems solved in [2,3]. Thus one may solve the following equations: go(x,y)=;(x2+Y2)
on La, on Li
g0(x,y)=;(xz+y2)+c; P&0(%
%2& an
(17) (j=k+l,...,m), on Lj
Y)-~jgi(x,Y)=~o*U-cLi)(X2+Y2)
on Lj
an
(18) 0’ = 1,2, * 1 . ) k),
0’ = 1,2, . . . , k).
(19)
(20)
Note that in eq. (18) there are still m -k constants c;+~, . . . , CA, which are yet to be determined. These are determined by imposing the condition that the displacement component w is single-valued on the boundaries Lk+,, . . . . , I;,. Note that if @,(x, y) is singlevalued on each of these boundaries, then
or, in terms of the conjugate function ly,(x, y) using the Cauchy-Riemann
II L,
alu,
-m-e
8~
ax aP(&
as
ax
as
}
equations,
ds = 0,
or noting that $ = - ZY an
and
or
I *&=O L, an
(j = k + 1,. . . , m).
(21)
Eqs. (21) give another set of m - k equations to determine the m - k constants. At this stage it seems appropriate to remark whether one would get the solution for the torsion problem and multiply-connected region as a special case of the problem mentioned above taking the limiting values /Lj = 0 0’ = 1,2, . . . , k). This is probably physically true.
285
R.D. Bhargava, S, Puranik, Torsion problem for elastic cylinder with inserts and holes
Mathematically, one has to take the limiting value for eqs. (14) (15) by taking the limits when pi -0. Using heuristic arguments, one is not entitled to put Dj in eq. (13) equal to zero when gj = 0 since th is violates eq. (12). In fact, eq. (12) is available only for the case when there is at least one insert. One cannot therefore work with eqs. (17) (19), (21) and arrive at the solution by taking the limits when /.Ljtends to zero.
2. Theoretical analysis of the solution As suggested in [l] the problem terms of single-layer potentials:
may be solved by expressing the harmonic
functions
in
(22) g(pj)=-
[ Wj(s)log]s-pj]
ds
JL,
(j=1,2
,...,
k),
(23)
where PO,Pjlj= 1, 2,. . . , k) are points in regions SO and sj (j = 1,2, . . . , k), and ao, uj, yjyi,wj are the distributions on ~undaries Lo, I+ (i = 1,2,. . . , k), Lj (j =: k + 1,. . . , m) Lj (i = 1,2, .*. , k): These satisfy Holder condition. It may be noted that there are two different distributions uj and Wj on Lj 0’ = 1,2, . . . , k). The distributions Cj contribute to the potential function go for region so, while wj give rise to the potential functions gj for the regions Sj.Note that two distributions are required for the boundaries Lj('j = 1,2,. . . , k) to account for continuity of the normal derivative. For one distribution it is well-known (as can be seen from eqs. (4)a, (4)b in [I]) that the normal derivative will be discontinuous. Also values of go and gi given in (22), (23) are substituted for boundary points into eqs. (17)-(21). These give the requisite number of equations to determine the distributions go, aj, Vi, Wj and the constants on the boundary of the holes,
3. Numericall form~ation of the problem To simplify the analysis, consider the case of one insert whose boundary is L, and one hole with boundary Lz,being surrounded by a region whose outer boundary is Lo.It may be remarked that the method is perfectly general and may be trivially generalised to the case of k inserts and m - k holes (k I m). Let the boundaries Lo,Ll,L2 be su~ivided into &&N and K intervals, respectively, which are not necessarily equal. Let the intervals be denoted by I; = (a;, Qi+l) and assume that the distributions are constant in each of these intervals and are given by UOi (i = 172, *a e , M) on Lo by wli and gli (i = 1,2, . . . , N) on L1 and by yli (i = 1,2,. . , , K) on Lz.It is seen that these are altogether A4 + 21V+ K unknowns in o’s, w’s and y’s, Note that there is one more unknown ci on the boundary La-Replacing POby p. (where p. is on Lo) and
286
R.D. Bhargava, S. Puranik, Torsion problem for elastic cylinder with inserts and holes
P, by p1 on L,, the eqs. (18x20)
may be rewritten as
on Lo and Lz, I.Log&l)- ELlgl(Pl)= -PO ,R goi I,
logIs-PII ds+ ~0,$ gli I, logIs-pll ds (25)
(26) where 0 denotes the angle which the line joining pl and s makes with the normal at point pl. Details of derivation of eq. (26) are given in [l]. Identify p. by (pal, . . . , pOM) successively, which are the centres of the intervals 4. Replace the left side of eq. (24) by i(x&,+ y&,) o’=l,... , M) and evaluate the integrals on the right side numerically or analytically. Thus, M equations are obtained. Next, identify p. by (p&?),p$?!, . . . , ~$2) successively which are centres of the intervals 4 on L2. Replace the left side of eq. (24) by f(.xz;) + y @) + cl and evaluate the integrals on the right side numerically or analytically, thus getting K equations. Lastly identify pl by (~11,. . . , plN) successively, which are the centres of the intervals Ii on Lie Replace the left side of eq. (25) by &LO--PI) (xi,,+ yg,,),and on the right side evaluate the integrals numerically or analytically. These give N equations. Finally, use the eq. (25) for points plj , N) on L, and -get N equations. Thus, altogether A4 + 2N + K equations are o’=l,... obtained for M + 2N + K + 1 unknowns. One more equation is needed for the evaluation of c:. It is obtained by imposing the condition of single-valuedness of the displacement equation (21). This gives the equation ds = 0,
(27)
as is shown in [l]. Now the number of equations and that of unknowns is the same. Thus the unknowns (+oi,cli, Wli, yli and constant cl are determined. Once the values of these unknowns are known, the harmonic functions go(Po) and gl(P1) are known at any points POin so and PI in s1 from eqs. (22) and (23). These may be substituted in eq. (16) to evaluate the stresses.
4. Test problem These ideas were checked against a rather difficult problem of an eccentric insert in a circular cross section. The problem has been solved analytically previously by Muskhelishvili
R.D. Bhargava,
[2], using the conformal
S. Puranik, Torsion problem for elastic cylinder with inserts and holes
287
mapping
_-L--
Z=l-a[
which maps the eccentric circles Lo, L, with radii Ri and centres (hi, 0) (i = 0,l) into concentric circles rO, rI with radii po, p1 in the l-plane such that 0 < pa < pl < lla (here a is a constant). The solutions for the insert and outside regions are given by q,(l) = C(S)
bI, + $,
(a;
sin
kt3 + bl cos k9)pk,
= b;: + k& (p ka ; - ppka ?k) Sin k9 + (p kb I$ i- pekb !k) cos ke,
where l= peie, b6 and known quantities.
bg
remain arbitrary constants, and a L, b;, at,
bi
(28)
are found in terms of
5. Numerical application For numerical purposes, two eccentric circles LO and L1 with radii Ri and centres (hi, 0) (i = 0,l) were taken. The arc lengths dso of LO and dsl of L, are given by dso = R. dGo and dsI = RI dG+, where Go and Q1 are the angles in the anticlockwise direction made at the centres of the respective circles by their radii with the lines parallel to the x-axis through the centres, respectively (fig. 2). Details about evaluation of integrals are given in [l]. After this the values of the coi, clit yli were obt ained by solving a system of linear simultaneous equations. The values of the goi, oli, yli and Wli are given in tables l-3 for 8, 16, 32 equal subdivisions of the outer boundary and the same number of subdivisions for the inner
Fig. 2. Eccentric cylinder with circular insert.
288
RD.
Bhargava, S. Puranik, Torsion problem for elastic cylinder with inserts and holes
Table 1. Results for eight subdivisions (Uo#= U*.O* U11= Ui.1,WI, = Wi.1) N~mericaI value 0.3332733 0.3061667 0.2~? 0.1705217 0.1341971 0.1705217 0.2~#7 0.3061667
Constant (Tli
VI.1 cTI.1 u3.1 U4~I
us.1 Vh.1 ff7.1
ox.1
Numeri~l value 0.0430129 0.0307941 0.0006756 - 0.0305065 - 0.0450162 - 0.0305~5 0.~756 0.0307941
Constant WI, W1.1 WZ.1 W3,I
W4.1
W5i.I WA.1 w7.1
WX.1
Numerica value 0.4492795 0.4098708 0.3129255 0.2133035 0.1668559 0.2133035 0.3129255 0.4098708
boundaries. The convergence may be seen from the table. The circular boundaries were su~ivided as in El]. Values of %Pwere calculated for some points in the matrix and the insert, giving !POand 9, analytically for these regions. Values of go and gl are evaluated for same points numerically and are compared with PO and ?P,. Values of go and PO, gl and P, differ from each other by constants co and dI. Note that co = PO-$(x2 -t-y”) on Lo and dI = IPI - gl on I,,. Thus these constants may be found from analytical results 121.Values of go, 3Po,g,, ?P1,
Table 2. Results for sixteen subdivisions (COi =
(7i.t.h Cli
Constant Uoi c+l.fl
EL0 03.0 f74.0
us,n
U&O lr7.D VX.0 (rs.0 fTlV>O
w11.0 (712.0 w13.0 u14.0 UlS.i?
orM.0
=
CFi.1, Wli
=
Wi.1)
Numerical value
Constant
WI.1 WZ.1
u7,1
0.0424846 0.0393574 0.0304175 0.0169206 0.~~9 - 0.0158295 - 0.0303727
(T8,l
-0.o404409
W&1
UP.1
-
0.0440873
e10.1
-
0.~
cTII.1
- 0.0303727 - 0.0158295 0.0007489 0.0169206 0.0304175 0.0393574
Numerical value
Constant
0.3320916 0.3251380 0.3053254 0.2756164 0.2403456 0.2053109 0.1716184 0.1459211 0.1350420 0.1459211 0.1716184 0.2043109 0.2403456 0.2756164 0.3055254 0.3251580
cTI.1
Grli
cr2.1 53.1 us.1
us.1 ufi.1
u12.i u13.1
u14.1 u15.1 ul6.1
Wli
w3.1
W4.1
WS.1 W4.1 w7.1
w9.1
w 10.1 W11.i w12.1 w13.1 Wl4.1
W15.1 wl6.1
Numerical value 0.~~~ 0.4384516 0.4092594 0.3652196 0.3125249 0.2586298 0.2115018 0.1789961 0.1672476 0.1789961 0.2115018 0.2586298 0.3125249 0.36521% 0.4092594 0.4384516
RD. Bhargaua, S. Puranik, Torsion problem for elastic cylinder with inserts and holes
289
Table 3. Resuits for thirty-two su~ivisions (IFOi= Cri.0,
Numerical value 0.3317199 0.3299717 0.3247939 0.3163834 0.3050592 0.2912478 0.2754641 0.2582852 0.2403203 0.2221732 0.2044028 0.1874093 0.1718425 0.1578917 0.1463398 0.1383993 0.1355200 0.1383993 0.1463398 0.1578917 0.1718425 0.1874903 0.2044028 0.2221732 0.2403203 0.2582853 0.2754641 0.2912478 0.3050592 0.3163854 0.3247939 0.32~7~7
Constant Uli ff1.1 cT2.1 U3,l (24.1 (r5.1 (76.1 o-7.1 us.1 ff9.1 UI0.I UI1.1 (Tn.1 u13.1 u14.1 U15.1 cTl6.1 u17.1 Ul8.1 m9.1 m.O.1 uz1.1 U2Z.l u23.1 u24.1 cT2s.1 U2h.l ff27.1 m&l U2Y.l u30.1 u31.1 u32.1
Numerical value 0.0422592 0.0414744 0.0391481 0.0353634 0.0302546 0.02~~ 0.01~2~ 0.0089833 O.OOD7432 - 0.0075999 - 0.0157439 -
0.0233815 0.0302051 0.0359146 0.0402326 0.0429820 0.0438449 0.~29280 0.~326 0.0359146 0.0302051 0.0233815 0.0157439 0.0075999 0.0007432 0.0089835 0.0168284 0.0240028 0.0302546 0.0353634 0.0391~1 0.~147~
Constant Wli
WI.1
W2.1 w3.1 W4.1 Ws.1 Wh.1 WY.1 W8.1 WQ.1 w10.1 w11.1 w12.1 w13.1 w14.1 WlS.1 w lb.1 w17.1 W18.1 wlQ.1 W20.1 w21.1 w22.1 Wk.1 w24.1 W2S.l W26.1 w27.1 ma.1 w29.1 W30.1 w31.1 w32.1
Numerical value 0.4484299 0.4458559 0.4382267 0.4258160 0.4090682 0.3885806 0.3650811 0.3394018 0.3124527 0.2851957 0.2586234 0.2337407 0.2115467 0.1930070 0.1790067 0.1702776 0.1673101 0.1702776 0.1790067 0.1930070 0.2115467 0.2337407 0.2586234 0.2851957 0.3124527 0.3394018 0.3650811 0.3885806 0.4090682 0.4258160 0.4382267 0.4458559
co and dl are given in tables 4 and 5, respectively, for some points. The results show excellent agreement with analytical results. The error is of O(10W5).
an
6. Earlier methods This method may be compared with other numerical methods, e.g. finite difference method and finite element methods. As remarked in [1], finite difference methods are extremely cumbersome for the case of the hole and also when the normal derivatives are involved. The
290
R.D. Bhargava,
Table 4. Comparison No. of divisions on each circle
8 16 32 8 16 32 8 16 32 8 16 32 8 16 32
S. Puranik, Torsion problem for elastic cylinder with inserts and holes
of rY, and go (analytic value of go- !P” is LI, = 0.2402157)
Analytical value of 90
0.05448998
0.0932139
0.1280981
0.1611351 0.1931183
Numerical value of gn
Difference
Error
0.2946716 0.2946449 0.2946842 0.3332749 0.3333735 0.3334127 0.3679432 0.3682403 0.3682977 0.40@7608 0.4012555 0.4013323 0.4325702 0.4332198 0.4333166
0.2401816 0.2401549 0.2401945 0.2400610 0.2401596 0.2401988 0.239845 1 0.2401422 0.2401996 0.2396277 0.2401224 0.2401993 0.2394519 0.2401015 0.2401980
0.0000341 0.0000608 0.0000214 0.0001547 0.0000561 0.0000169 0.0003706 o.oooo735 0.0000161 O.ooO5880 0.0000933 0.0000164 0.0007638 0.0001142 0.0000177
problem has also been done by finite element methods. In practice the problem is solved by finding the functions @ and rY. Details for the same are given in [4]. An analytical solution has been attempted for only one insert by Payne [3] and Mitra [6] by using conformal mapping of the region into a ring bounded by concentric circles. Chobanyan [7] solves the same problem of one insert by using the usual boundary conditions and taking infinite series. Both of these methods require the appropriate conformal mapping or series expansion only after this can numerical results be obtained. It is obvious that these methods do not succeed for several inserts and holes. Table 5. Comparison No. of divisions on each circle
8 16 32 8 16 32 8 16 32
of PI and gl (analytic value of gl - VY,is Lt = 0.6593873)
Analytical value of !P,
0.0044088
0.0264240
0.0483933
Numerical value of gl
Difference
0.6659550 0.6641423 0.6639171 0.6874037 0.6860353 0.6858932 0.7087885 0.7078610 0.7078002
0.6615462 0.6597355 0.6595083 0.6609797 0.6596113 0.6594692 0.6603950 0.6594577 0.6594069
Error 0.0021589 0.0003462 0.0001210 0.0015924 0.0002240 O.OMIO819 0.0010077 O.OOCW704 0.0000196
R.D. Bhargava, S. Puranik, Torsion problem for elastic cylinder with inserts and holes
291
7. Conclusion It is shown that the classical torsion problem for a composite cylinder (and also with holes) may be solved by the integral equation method suggested in this paper. The conditions at the outer boundary and at the interface of the insert and the holes may be translated and in terms of the distribution suggested in the paper. One gets the requisite number of equations to determine the distribution and the constants. The harmonic function can then directly be determined at any point on the boundary in place of only at the grid points as in other methods. The method is quite suitable for programming. It can be directly applied to any problem in engineering and physics where the governing equation is (i) the two-dimensional Laplace equation, (ii) the values of functions are prescribed on the outer boundary and the holes (values of functions of the holes are known up to arbitrary constants which are determined by the single-valuedness or multiple-valuedness of the conjugate function), and (iii) on interfaces between matrix and inserts the conditions on the normal derivatives and the relations between the harmonic functions are given. The method also solves the Hilbert problem in essence.
References [l] R.D. Bhargava and S.S. Puranik, Torsion problem for elastic cylinder with holes, Comp. Meths. Appl. Mech. Eng. 21 (1980) 63-74. [2] N.I. Muskhelishvili, Some basic problems of the mathematical theory of elasticity, 4th ed. (Noordhoff, Groningen, 1963). [3] L. Milne-Thomson, Antiplane elastic systems (Springer, Berlin, 1962). [4] H.C. Martin and G.F. Carey, Introduction to finite element analysis (McGraw-Hill, New York, 1973). [5] L.E. Payne, Torsion of composite sections, Iowa State Coll. J. Sci. 23 (1949) 381. [6] D.N. Mitra, Torsion of composite sections of different isotropic materials, Bull. Calcutta Math. Sci. 47 (1955) 191. [7] KS. Chobanyan, Application of stress function to problem of torsion of prismatic rods made of different materials (Russian), Ref. Zh. Mekh., No. 9 (1957).
AND ENGINEERING
23 (1980) 281-291
TORSION PROBLEM FOR ELASTIC CYLINDER WITH INSERTS AND HOLES
Department
R.D. BHARGAVA and S. PURANIK of Mathematics, Indian Institute of Technology, Bombay 4OW76, India Received 31 May 1980
An integral equation method to solve the classical torsion problem for an elastic cylinder with inserts and hoies is treated. The bounded region outside the inserts and the holes will be termed a matrix. As is well-known the solution depends on finding plane harmonic functions in the matrix and inserts such that (a) on the outer boundary of the matrix and the boundaries of the holes the harmonic function in the matrix takes the values $(x2+ y’) + ci, and (b) on the interfaces of the matrix and the inserts relations exist between the harmonic functions and between their normal derivatives. Here (x, y) are the coordinates of the point on the boundary and cj are unknown constants. The usual methods are cumbersome and lengthy. In this paper a straightforward method is presented which is easily programmable. The numerical solution is obtained by evaluating a few integrals either analytically or numerically and solving a system of linear simultaneous equations. An example of a cylinder with an eccentric insert is given which substantiates the theory developed in this paper and is found to agree with known resufts. However, the method is genera1 and may be applied to a variety of problems.
Introduction Recently the authors [l] have solved the torsion problem of an elastic cylinder with holes by the integral equation method. This paper is an extension of earlier work where one considers the case of a cylinder with inserts and holes.
1. The problem Let S be the cross section of a prism with its axis parallel to the z-axis and consisting of the regions $1, ~2, . . . , s,,, and a region so surrounding all the Sj (fig. 1). Let the boundaries of the nonintersecting regions Sj be & (j = 1, . . . . , m), and let so have outer boundary LO and inner boundaries L1, . . . . ,L,. Note that LO contains all the closed contours L1, . . . . , 12,. Let the regions si, s2, . . . . , Sk (k I m) consist of the regions corresponding to inserts, and let the remaining regions consist of cavities. If all the regions are cavities, one may solve the torsion problem by the method suggested in [l]. The conditions for the torsion problem are satisfied by assuming that the displacement components are u = --7zy,
V =
TXZ,
w = T@fX,Y,k
where T is the twist per unit length of the prism and @(x, y) is the torsion function. The strains
282
R.D. Bhargava, S. Puranik, Torsion problem for eiasric cylinder with inserts and hates
Fig. 1. Cross section of cylinder with k inserts and m -k
holes.
are obtained by strain-displacement relations, and stresses are obtained by Hooke’s law for the regions SO, sl,. . . , Sk. The nonzero stresses are
x =pj $7
(
)
(1)
Y*=&($+x),
)
where JQ is the shear modulus of materials in sj (i = 0, 1, . . . , k). Substitution of the stresses in the equilibrium equations show that function @(x, y) is harmonic in so, sl,. . . , Sk. The boundary conditions in these regions are (a) the external boundaries of the bar are free from tractions, (b) the stresses on the interfaces are equal and opposite, and (c) the displacements remain continuous across the boundaries L1, Lz, . . . , Lk. Condition (a) implies that x, cos(n,x)+ ~ndition
Y, cos(la,y)=O
on Lo and I&
(2)
(b) implies that [X* cos(n, x) + Y; cos(n, y)]o = [X, cos (12,x) + Y, cos(n, y)L on
Ljtj
=
1,2,. . . , k),
(3)
and condition (c) implies that [@lo = [@lj
on Lj(j=
1,2, ._. , k).
(4
Here the subscripts 0 and j refer to regions so and si(j = 1,2, . . . , k). Substituting the values of stresses from eq. (1) into eqs. (2) and (3), one obtains dQ, -& = y cos(lt, X) - X cos(n, y)
on&and&
(k
(5)
R.D. ~~arga~a, S. ~ranik,
283
Torsion problem fur elastic cy~i~~~r with inserts and holes
on L&=1,2
,...,
k), 6)
and of course eq. (4) must be satisfied. Note that the normal is taken positive when it is outwards from so. The problem may be solved by finding the function @ in the regions. It is possible to fo~ulate the solution in terms of the function (P]j = Pi which are conjugate of (@>)j3 @j in $j (i = 0, 1, +. . ) k). The stresses in eq. (1) may be written in terms of the conjugate functions as
x, = b%($--Y),
for31 u=O,
Y~=~j(-~+~)
1,. . . 1 k).
(7)
Using the Cauchy-Riemann equations &Pj/dn = dPj/ds in eqs. (5), (6) and integrating them over the respective boundaries, one obtains eqs. (8), (9) below. Also differentiating (4) with respect to s and using the equation aQ3i/as= -Z?$/dn, one gets eq. (10): on&
(i=O,
k+l,..
on&
~0~~~~~y)~~j~j(~~y)~~(y;0~~j)(~*+y2)+bj
a!P,
an
-22
aw, an
.,m),
(8) c=l,2
,...,
k),
on Lj (j = 1,2, . . . , k).
(9)
(10)
Here ci and bj are constants of integration-there are m -k + 1 C:! and kbj. Some comments now follow about these constants. Let go and gj be the harmonic functions in So and Sj 0’ = 1,2,. . . , k) such that
~c&, Y) = g&
Y I+
co
in Sn
and
(11)
in Si (i = 1,2, . . . , k). Substituting
these values in eqs. (8)-(lo), leads to
g&, y) = i(x”f y”) go(x,
y) =
on LO,
&x2-ty”) -t- c;
On
yogo(x,Y)-~jgj(x,Y)=~Olo-CLi)(n2+Y2)+Di
ago_itk an ai2
onLj
d.
I
pod0- bj Pi
on Lj tj = 1,2,
G=lt2,...,k),
where Dj =: bj + pjdj - podo. Let dj be
SO
(12)
Lj 0’ = k + 1, a + e , EE),
, k),
(13) (14)
chosen that Dj = 0, thus
(i = 1,2,. . . , k).
(15)
R.D. Bhargava, S. Puranik, Torsion problem for elastic cylinder with inserts and holes
284
Note that the stresses in eq. (7) are not affected by substituting in terms of the functions go and gi from eq. (11). The nonzero stresses thus are given by
(
>
x,=pjr g-y
Y,
)
=
j.qr
(-2-X)
inSj
G=1,2,...,k). (16)
Hence, one may work with the harmonic functions gi 0’ = 0,1,. . . , m) in place of the Fj m). Note that in practice one does not determine the constants in eq, (9) and as o’=O,l,..., is also demonstrated by the particular problems solved in [2,3]. Thus one may solve the following equations: go(x,y)=;(x2+Y2)
on La, on Li
g0(x,y)=;(xz+y2)+c; P&0(%
%2& an
(17) (j=k+l,...,m), on Lj
Y)-~jgi(x,Y)=~o*U-cLi)(X2+Y2)
on Lj
an
(18) 0’ = 1,2, * 1 . ) k),
0’ = 1,2, . . . , k).
(19)
(20)
Note that in eq. (18) there are still m -k constants c;+~, . . . , CA, which are yet to be determined. These are determined by imposing the condition that the displacement component w is single-valued on the boundaries Lk+,, . . . . , I;,. Note that if @,(x, y) is singlevalued on each of these boundaries, then
or, in terms of the conjugate function ly,(x, y) using the Cauchy-Riemann
II L,
alu,
-m-e
8~
ax aP(&
as
ax
as
}
equations,
ds = 0,
or noting that $ = - ZY an
and
or
I *&=O L, an
(j = k + 1,. . . , m).
(21)
Eqs. (21) give another set of m - k equations to determine the m - k constants. At this stage it seems appropriate to remark whether one would get the solution for the torsion problem and multiply-connected region as a special case of the problem mentioned above taking the limiting values /Lj = 0 0’ = 1,2, . . . , k). This is probably physically true.
285
R.D. Bhargava, S, Puranik, Torsion problem for elastic cylinder with inserts and holes
Mathematically, one has to take the limiting value for eqs. (14) (15) by taking the limits when pi -0. Using heuristic arguments, one is not entitled to put Dj in eq. (13) equal to zero when gj = 0 since th is violates eq. (12). In fact, eq. (12) is available only for the case when there is at least one insert. One cannot therefore work with eqs. (17) (19), (21) and arrive at the solution by taking the limits when /.Ljtends to zero.
2. Theoretical analysis of the solution As suggested in [l] the problem terms of single-layer potentials:
may be solved by expressing the harmonic
functions
in
(22) g(pj)=-
[ Wj(s)log]s-pj]
ds
JL,
(j=1,2
,...,
k),
(23)
where PO,Pjlj= 1, 2,. . . , k) are points in regions SO and sj (j = 1,2, . . . , k), and ao, uj, yjyi,wj are the distributions on ~undaries Lo, I+ (i = 1,2,. . . , k), Lj (j =: k + 1,. . . , m) Lj (i = 1,2, .*. , k): These satisfy Holder condition. It may be noted that there are two different distributions uj and Wj on Lj 0’ = 1,2, . . . , k). The distributions Cj contribute to the potential function go for region so, while wj give rise to the potential functions gj for the regions Sj.Note that two distributions are required for the boundaries Lj('j = 1,2,. . . , k) to account for continuity of the normal derivative. For one distribution it is well-known (as can be seen from eqs. (4)a, (4)b in [I]) that the normal derivative will be discontinuous. Also values of go and gi given in (22), (23) are substituted for boundary points into eqs. (17)-(21). These give the requisite number of equations to determine the distributions go, aj, Vi, Wj and the constants on the boundary of the holes,
3. Numericall form~ation of the problem To simplify the analysis, consider the case of one insert whose boundary is L, and one hole with boundary Lz,being surrounded by a region whose outer boundary is Lo.It may be remarked that the method is perfectly general and may be trivially generalised to the case of k inserts and m - k holes (k I m). Let the boundaries Lo,Ll,L2 be su~ivided into &&N and K intervals, respectively, which are not necessarily equal. Let the intervals be denoted by I; = (a;, Qi+l) and assume that the distributions are constant in each of these intervals and are given by UOi (i = 172, *a e , M) on Lo by wli and gli (i = 1,2, . . . , N) on L1 and by yli (i = 1,2,. . , , K) on Lz.It is seen that these are altogether A4 + 21V+ K unknowns in o’s, w’s and y’s, Note that there is one more unknown ci on the boundary La-Replacing POby p. (where p. is on Lo) and
286
R.D. Bhargava, S. Puranik, Torsion problem for elastic cylinder with inserts and holes
P, by p1 on L,, the eqs. (18x20)
may be rewritten as
on Lo and Lz, I.Log&l)- ELlgl(Pl)= -PO ,R goi I,
logIs-PII ds+ ~0,$ gli I, logIs-pll ds (25)
(26) where 0 denotes the angle which the line joining pl and s makes with the normal at point pl. Details of derivation of eq. (26) are given in [l]. Identify p. by (pal, . . . , pOM) successively, which are the centres of the intervals 4. Replace the left side of eq. (24) by i(x&,+ y&,) o’=l,... , M) and evaluate the integrals on the right side numerically or analytically. Thus, M equations are obtained. Next, identify p. by (p&?),p$?!, . . . , ~$2) successively which are centres of the intervals 4 on L2. Replace the left side of eq. (24) by f(.xz;) + y @) + cl and evaluate the integrals on the right side numerically or analytically, thus getting K equations. Lastly identify pl by (~11,. . . , plN) successively, which are the centres of the intervals Ii on Lie Replace the left side of eq. (25) by &LO--PI) (xi,,+ yg,,),and on the right side evaluate the integrals numerically or analytically. These give N equations. Finally, use the eq. (25) for points plj , N) on L, and -get N equations. Thus, altogether A4 + 2N + K equations are o’=l,... obtained for M + 2N + K + 1 unknowns. One more equation is needed for the evaluation of c:. It is obtained by imposing the condition of single-valuedness of the displacement equation (21). This gives the equation ds = 0,
(27)
as is shown in [l]. Now the number of equations and that of unknowns is the same. Thus the unknowns (+oi,cli, Wli, yli and constant cl are determined. Once the values of these unknowns are known, the harmonic functions go(Po) and gl(P1) are known at any points POin so and PI in s1 from eqs. (22) and (23). These may be substituted in eq. (16) to evaluate the stresses.
4. Test problem These ideas were checked against a rather difficult problem of an eccentric insert in a circular cross section. The problem has been solved analytically previously by Muskhelishvili
R.D. Bhargava,
[2], using the conformal
S. Puranik, Torsion problem for elastic cylinder with inserts and holes
287
mapping
_-L--
Z=l-a[
which maps the eccentric circles Lo, L, with radii Ri and centres (hi, 0) (i = 0,l) into concentric circles rO, rI with radii po, p1 in the l-plane such that 0 < pa < pl < lla (here a is a constant). The solutions for the insert and outside regions are given by q,(l) = C(S)
bI, + $,
(a;
sin
kt3 + bl cos k9)pk,
= b;: + k& (p ka ; - ppka ?k) Sin k9 + (p kb I$ i- pekb !k) cos ke,
where l= peie, b6 and known quantities.
bg
remain arbitrary constants, and a L, b;, at,
bi
(28)
are found in terms of
5. Numerical application For numerical purposes, two eccentric circles LO and L1 with radii Ri and centres (hi, 0) (i = 0,l) were taken. The arc lengths dso of LO and dsl of L, are given by dso = R. dGo and dsI = RI dG+, where Go and Q1 are the angles in the anticlockwise direction made at the centres of the respective circles by their radii with the lines parallel to the x-axis through the centres, respectively (fig. 2). Details about evaluation of integrals are given in [l]. After this the values of the coi, clit yli were obt ained by solving a system of linear simultaneous equations. The values of the goi, oli, yli and Wli are given in tables l-3 for 8, 16, 32 equal subdivisions of the outer boundary and the same number of subdivisions for the inner
Fig. 2. Eccentric cylinder with circular insert.
288
RD.
Bhargava, S. Puranik, Torsion problem for elastic cylinder with inserts and holes
Table 1. Results for eight subdivisions (Uo#= U*.O* U11= Ui.1,WI, = Wi.1) N~mericaI value 0.3332733 0.3061667 0.2~? 0.1705217 0.1341971 0.1705217 0.2~#7 0.3061667
Constant (Tli
VI.1 cTI.1 u3.1 U4~I
us.1 Vh.1 ff7.1
ox.1
Numeri~l value 0.0430129 0.0307941 0.0006756 - 0.0305065 - 0.0450162 - 0.0305~5 0.~756 0.0307941
Constant WI, W1.1 WZ.1 W3,I
W4.1
W5i.I WA.1 w7.1
WX.1
Numerica value 0.4492795 0.4098708 0.3129255 0.2133035 0.1668559 0.2133035 0.3129255 0.4098708
boundaries. The convergence may be seen from the table. The circular boundaries were su~ivided as in El]. Values of %Pwere calculated for some points in the matrix and the insert, giving !POand 9, analytically for these regions. Values of go and gl are evaluated for same points numerically and are compared with PO and ?P,. Values of go and PO, gl and P, differ from each other by constants co and dI. Note that co = PO-$(x2 -t-y”) on Lo and dI = IPI - gl on I,,. Thus these constants may be found from analytical results 121.Values of go, 3Po,g,, ?P1,
Table 2. Results for sixteen subdivisions (COi =
(7i.t.h Cli
Constant Uoi c+l.fl
EL0 03.0 f74.0
us,n
U&O lr7.D VX.0 (rs.0 fTlV>O
w11.0 (712.0 w13.0 u14.0 UlS.i?
orM.0
=
CFi.1, Wli
=
Wi.1)
Numerical value
Constant
WI.1 WZ.1
u7,1
0.0424846 0.0393574 0.0304175 0.0169206 0.~~9 - 0.0158295 - 0.0303727
(T8,l
-0.o404409
W&1
UP.1
-
0.0440873
e10.1
-
0.~
cTII.1
- 0.0303727 - 0.0158295 0.0007489 0.0169206 0.0304175 0.0393574
Numerical value
Constant
0.3320916 0.3251380 0.3053254 0.2756164 0.2403456 0.2053109 0.1716184 0.1459211 0.1350420 0.1459211 0.1716184 0.2043109 0.2403456 0.2756164 0.3055254 0.3251580
cTI.1
Grli
cr2.1 53.1 us.1
us.1 ufi.1
u12.i u13.1
u14.1 u15.1 ul6.1
Wli
w3.1
W4.1
WS.1 W4.1 w7.1
w9.1
w 10.1 W11.i w12.1 w13.1 Wl4.1
W15.1 wl6.1
Numerical value 0.~~~ 0.4384516 0.4092594 0.3652196 0.3125249 0.2586298 0.2115018 0.1789961 0.1672476 0.1789961 0.2115018 0.2586298 0.3125249 0.36521% 0.4092594 0.4384516
RD. Bhargaua, S. Puranik, Torsion problem for elastic cylinder with inserts and holes
289
Table 3. Resuits for thirty-two su~ivisions (IFOi= Cri.0,
Numerical value 0.3317199 0.3299717 0.3247939 0.3163834 0.3050592 0.2912478 0.2754641 0.2582852 0.2403203 0.2221732 0.2044028 0.1874093 0.1718425 0.1578917 0.1463398 0.1383993 0.1355200 0.1383993 0.1463398 0.1578917 0.1718425 0.1874903 0.2044028 0.2221732 0.2403203 0.2582853 0.2754641 0.2912478 0.3050592 0.3163854 0.3247939 0.32~7~7
Constant Uli ff1.1 cT2.1 U3,l (24.1 (r5.1 (76.1 o-7.1 us.1 ff9.1 UI0.I UI1.1 (Tn.1 u13.1 u14.1 U15.1 cTl6.1 u17.1 Ul8.1 m9.1 m.O.1 uz1.1 U2Z.l u23.1 u24.1 cT2s.1 U2h.l ff27.1 m&l U2Y.l u30.1 u31.1 u32.1
Numerical value 0.0422592 0.0414744 0.0391481 0.0353634 0.0302546 0.02~~ 0.01~2~ 0.0089833 O.OOD7432 - 0.0075999 - 0.0157439 -
0.0233815 0.0302051 0.0359146 0.0402326 0.0429820 0.0438449 0.~29280 0.~326 0.0359146 0.0302051 0.0233815 0.0157439 0.0075999 0.0007432 0.0089835 0.0168284 0.0240028 0.0302546 0.0353634 0.0391~1 0.~147~
Constant Wli
WI.1
W2.1 w3.1 W4.1 Ws.1 Wh.1 WY.1 W8.1 WQ.1 w10.1 w11.1 w12.1 w13.1 w14.1 WlS.1 w lb.1 w17.1 W18.1 wlQ.1 W20.1 w21.1 w22.1 Wk.1 w24.1 W2S.l W26.1 w27.1 ma.1 w29.1 W30.1 w31.1 w32.1
Numerical value 0.4484299 0.4458559 0.4382267 0.4258160 0.4090682 0.3885806 0.3650811 0.3394018 0.3124527 0.2851957 0.2586234 0.2337407 0.2115467 0.1930070 0.1790067 0.1702776 0.1673101 0.1702776 0.1790067 0.1930070 0.2115467 0.2337407 0.2586234 0.2851957 0.3124527 0.3394018 0.3650811 0.3885806 0.4090682 0.4258160 0.4382267 0.4458559
co and dl are given in tables 4 and 5, respectively, for some points. The results show excellent agreement with analytical results. The error is of O(10W5).
an
6. Earlier methods This method may be compared with other numerical methods, e.g. finite difference method and finite element methods. As remarked in [1], finite difference methods are extremely cumbersome for the case of the hole and also when the normal derivatives are involved. The
290
R.D. Bhargava,
Table 4. Comparison No. of divisions on each circle
8 16 32 8 16 32 8 16 32 8 16 32 8 16 32
S. Puranik, Torsion problem for elastic cylinder with inserts and holes
of rY, and go (analytic value of go- !P” is LI, = 0.2402157)
Analytical value of 90
0.05448998
0.0932139
0.1280981
0.1611351 0.1931183
Numerical value of gn
Difference
Error
0.2946716 0.2946449 0.2946842 0.3332749 0.3333735 0.3334127 0.3679432 0.3682403 0.3682977 0.40@7608 0.4012555 0.4013323 0.4325702 0.4332198 0.4333166
0.2401816 0.2401549 0.2401945 0.2400610 0.2401596 0.2401988 0.239845 1 0.2401422 0.2401996 0.2396277 0.2401224 0.2401993 0.2394519 0.2401015 0.2401980
0.0000341 0.0000608 0.0000214 0.0001547 0.0000561 0.0000169 0.0003706 o.oooo735 0.0000161 O.ooO5880 0.0000933 0.0000164 0.0007638 0.0001142 0.0000177
problem has also been done by finite element methods. In practice the problem is solved by finding the functions @ and rY. Details for the same are given in [4]. An analytical solution has been attempted for only one insert by Payne [3] and Mitra [6] by using conformal mapping of the region into a ring bounded by concentric circles. Chobanyan [7] solves the same problem of one insert by using the usual boundary conditions and taking infinite series. Both of these methods require the appropriate conformal mapping or series expansion only after this can numerical results be obtained. It is obvious that these methods do not succeed for several inserts and holes. Table 5. Comparison No. of divisions on each circle
8 16 32 8 16 32 8 16 32
of PI and gl (analytic value of gl - VY,is Lt = 0.6593873)
Analytical value of !P,
0.0044088
0.0264240
0.0483933
Numerical value of gl
Difference
0.6659550 0.6641423 0.6639171 0.6874037 0.6860353 0.6858932 0.7087885 0.7078610 0.7078002
0.6615462 0.6597355 0.6595083 0.6609797 0.6596113 0.6594692 0.6603950 0.6594577 0.6594069
Error 0.0021589 0.0003462 0.0001210 0.0015924 0.0002240 O.OMIO819 0.0010077 O.OOCW704 0.0000196
R.D. Bhargava, S. Puranik, Torsion problem for elastic cylinder with inserts and holes
291
7. Conclusion It is shown that the classical torsion problem for a composite cylinder (and also with holes) may be solved by the integral equation method suggested in this paper. The conditions at the outer boundary and at the interface of the insert and the holes may be translated and in terms of the distribution suggested in the paper. One gets the requisite number of equations to determine the distribution and the constants. The harmonic function can then directly be determined at any point on the boundary in place of only at the grid points as in other methods. The method is quite suitable for programming. It can be directly applied to any problem in engineering and physics where the governing equation is (i) the two-dimensional Laplace equation, (ii) the values of functions are prescribed on the outer boundary and the holes (values of functions of the holes are known up to arbitrary constants which are determined by the single-valuedness or multiple-valuedness of the conjugate function), and (iii) on interfaces between matrix and inserts the conditions on the normal derivatives and the relations between the harmonic functions are given. The method also solves the Hilbert problem in essence.
References [l] R.D. Bhargava and S.S. Puranik, Torsion problem for elastic cylinder with holes, Comp. Meths. Appl. Mech. Eng. 21 (1980) 63-74. [2] N.I. Muskhelishvili, Some basic problems of the mathematical theory of elasticity, 4th ed. (Noordhoff, Groningen, 1963). [3] L. Milne-Thomson, Antiplane elastic systems (Springer, Berlin, 1962). [4] H.C. Martin and G.F. Carey, Introduction to finite element analysis (McGraw-Hill, New York, 1973). [5] L.E. Payne, Torsion of composite sections, Iowa State Coll. J. Sci. 23 (1949) 381. [6] D.N. Mitra, Torsion of composite sections of different isotropic materials, Bull. Calcutta Math. Sci. 47 (1955) 191. [7] KS. Chobanyan, Application of stress function to problem of torsion of prismatic rods made of different materials (Russian), Ref. Zh. Mekh., No. 9 (1957).