ON QUADRUPLE INTEGRAL EQUATIONS INVOLVING TRIGONOMETRIC KERNELS

A general technique is developed for the solution of quadruple integral equations involving trigonometric kernels. Four such sets are solved explicitly. Application is made to the problem of three-collinear cracks in linear plane elasticity.

lems in the linear theory of Elasticity and the solution of dual integral equations goes back to Busbridge in 1938 [1].Because of their importance in applicatons, dual integral equations have been considered by a large number of investigators, and a thorough account is given by Sneddon [2].Sneddon [2] also considers a particular case of Triple integral equations and those equations have recently been considered by Lowndes and Srivastava  [3], Chakrabarti  [4], Singh  [5,6], and others.A set of quadruple integral equations has been considered by Jain and Singh [7], and so on.
In this paper we outline a general method for solving quadruple integral equations involving trigonometric kernels and apply this method to a number of such sets.We later on discuss their application to the problem of three collinear cracks, not all of same length, situated symmetrically about the origin on the x-axis and evaluate a number of stress concentration factors.
We notice that if we write jfotA(t)cosxt dt f2(x), a < x < b (2.2) and if we are able to determine f2(x), then the equations (2.1) reduce to a pair of dual integral equations whose solution is known.We proceed to determine f2(x).
We write F(x) A(z) A(x) f3(z), O < z < l, where (2.3) stands for In terms of F(x) and g4(x), the solution of equations (2.1) is given by [8 (2.5c) where stands for differentiation w.r.t.z.
3. We now consider the set of equations This time we write r.]atA(t)sinzt dt fl(z), LA(t)sinzt dt g(z), fotA(t)sinzt dt fz(z), and A(t) is now given by A(t) -r r 2 fo J(ut)h(u)du -r 2 f uJ(ut)h2(u)du where (3.3a) as before.This time the equation to determine f(z) is found to be ']()K{.,)d.

a<<b
The constant D must again be determined by substitution into (3.4).
and proceed to determine ]2(z).This time, A(t) is given by A(t) uJo(ut)Fz(u)du + uJo(ut)G2(u)du D must again be determined by substitution into (4.4) or this time, by the requirement that f2(b-) i.e. by requiring F(x) to be continuous at the points x a and at x b.
5. We consider this time the same equations as in Article 4 with sin function there replaced by COS.
However, this time, the proper question to ask is [8]; .Find the constant C and the function A(t) a<x<b g,(x)dx, and F(x) f (x) f(x) @ f3(x) and this equation may be solved as before.The constts of integration may be found by requiring that S(x) (which denotes temperatures in a temperature problem, s ( [8]) is continuous at a and at x b.
We apply the equations in section 2 to the problem of finding stress in an elastic body in ple elasticity, where the body has thr colinear cracks lying along (-1,-b), (-a, a) nd (b, 1).
While the problems of one crack, that of two syetfil cracks, and of an infinite row of cracks open by the sme (equal) pressure on the surface of each crk have bn solved by Sneddon and Lowengrub [10], the problem of thr syetrical cracks, not all of them equal, does not seem to have mlv.
f2(), which is equal to -a in a < < b, in this case, is found to be (a + b 22) D A() 22 a2x/b2 v + V2 a2V,b2 2v/1 2 a < < b (6.3)where D must be found from v/1 z + V/1 ; fo' 41 z + V/1 f It is ey to double-check (by differentiating (6.4), for example) that D as given by (6.4) is independent of x in a < x < b.
Since f() is -a(, 0) in a < < b, the stress intensity factors k and k at the points (a, 0) and (b, O) are given by k, -lim 2(x-a) f(z) (6.5a) and ka -lim 2(b-x) (6.5b)Also the stress intensity factor k3 at (1,0) is easily found to be ;F12 (1), where after a substitution, F1 (1)  Values of stress concentration factors k, k, and k3 for various values of a and b are given in Table 1.It is to be seen that for a given value of b, i.e. for a given position of the crack b < x < 1, as a increases, the concentration factor k which is less than k: in the beginning overtakes k2.This happens for b >_ -3.If b << .3,i.e. if the outlying crack in b < x < is quite large, then the stress concentration factor k2 in it can be quite large indeed.The stress concentration factor k3 at (1,0)  in the crack b < x < is relatively less affected by the presence of the crack in 0 < x < a.The stress concentration factor for two equal cracks (a 0) and for three equal cracks (2a b) are given in Tables 2 and 3 respectively.If 2l is the length of a crack, then in the case of a single crack, the stress concentration factor in it is proportional to v/.To account for this effect, the factors in Table 2 and Table 3 have been divided by Vq where 21 is the length of the crack.It is to be seen from Table 2 that, as r lb-'-which is the distance between the cracks divided by the length of the crack, gets larger, the effect on stress concentration factors, due to the presence of other cracks, decreases.At r 2, this effect is less than 2% while at r the effect on k2 is about 6%, and at r the effect on k2 is about 47%.The effect on k3 is relatively smaller.kl 0 in this case.These observations are in accordance with the remarks made by Sneddon and Lowengrub [10, p. 44].In the case of three equal cracks as r (3b-1)/(2 2b) distance between cracks/length of each crack, gets smaller, the effect on the stress concentration factors kl is about 45% higher than what it would be without the two is more pronounced.At r g neighbouring cracks, ks is about 41% higher and k3 is only about 14% higher.In the case of three equal cracks, ka > k2 in every case, so that the middle crack has a higher stress concentration factor than its neighbour on either side.In each case, the effect on k3, the stress concentration factor on the outer edge of the outlying crack, is relatively small.Table 4 gives the values of k/v/-, where k is the largest stress-concentration factor, for 2, 3 and an infinite number of equal length cracks for various values of r, the ratio of the distance between cracks to the length of each crack.It is to be noted that k as r oo and k oo as r 0, so that the relationship k k(r) is hyperbolic in nature.On a log-log graph, however, this relationship becomes extremely close to linear (see Fig. 1) and we get the simple relationship where a .6070,.3591for the case of 2 cracks; a .6515,.3776for the case of 3 cracks; and a .6939,.4885for an infinite number of cracks.
These formulae give results very close to the computed values, for 0 < r < .Values of kl, k2 and k3 at (a,O), (b,O) and (1,0) respectively for three cracks situated along (-1,-b), (-a,a) and (b, 1).kl, k2 and k3 are listed in the first, second and third row respectively.3/V/ 1.1841 1.1200 1.0863 1.0326 1.0125 1.0036 1.0003 1.0001 1.0060  Values of kl/V/', k21v/' and k3/v for three equal cracks.2t b is the length of each crack.s (3b-I)/2 is the distance between the cracks.The ratio of the length of each crack divided by the distance between the cracks goes up from .0101 to 666 as we go down the  4 Values of k//' for two, three and an infinite number of cracks for various values of r. -s -'z -6 -s -, -3 " 1. Linear regression of F in the equation ln(/Vff) F(lnr) for the case of two cracks

Figure(
Figure (top), 3 cracks (middle), and an infinite number of cracks (bottom), for r < .1.k is the largest stress concentration factor.

Table 2
Value of k2/v/ and k3/v for two equal cracks 2t b is the length of each crack and s 2b is the distance between the cracks, kl 0 in this case.