Sınır eleman yönteminin çok bağımlı bir bölgeye uygulanması

Abstract

Düzlem Elastisite problemlerinin çözümlerinde, sınır eleman yöntemi bilindiği gibi Betti Karşıtlık teoremine dayanır. Çözümü belli bir problemle, çözülmek istenen problem arasında bu karşı tbk teoremi yazılarak, çözülecek problemin yer- değiştirme alam için bir integral denklem elde edilir. Bu integral denklemde bilin- miyenler problemin geçerli olduğu bölge sınırındaki yerdeğiştirme bileşenleridir. Integral denklem sayısal olarak çözülür ve lineer bir denklem takımına in dirgenir. Ancak çözümü bilinen problemin yerdeğiştirme ve yüzey gerilme vektörü bileşenlerinden ortaya çıkan çekirdekler sınırda Inr tekilliğine sahiptir. Gerilme-Şekil değiştirme bağıntıları kullanılarak hesaplanan gerilme bileşenleri için ortaya çıkan çekirdekler ise sınırda 1 / r tekilliğini içerirler. Bu çabşmada gerek yerdeğiştirme gerek gerilme bileşenleri hesabında or taya çıkan bu tekillikler ortadan kaldırılmış ve sınır eleman yöntemi çok bağımlı bir bölgeye hem düzlem gerilme hem düzlem şekil değiştirme hali için uygulanmıştır.

In this study, an improvement is introduced to solve the problems of linear elasticity by boundary element method. It is considered that the problem which will be solved is the first fundamental problem. In this kind of problems the external stress vector is given on the surface which represents the boundary of the region. It is chosen that region related to the problem is multiply connected. Now, the aim is to determine the displacement vector in the region and on the boundary. In addition to this, the components of the stress tensor must be determined in the region- and also on the boundary. Let V be a region filled by a linear elastic material and S be its boundary. In plane problems, S and V are transformed to a closed curve and a plane region respectively. In the standard formulation of boundary element method any i component of the displacement vector (i=l,2) at a point y in the region V can be expressed as follows: Ui(y)= I t(x)V/x,y)ds- [ f(x,y)v(*)d8 (1) Where, t{x) is the surface traction vector given on the boundary S. The kernels U'(x,y) and T'(x,y) are given below U&V = -8MI^)[(3 ` ^^ ` ^-^-^ (2), 2(»< - ».X«j -_»X»* - lfc)j {i,j,k=l,2) (3) r4 VIi(x,y) = T{{x,y)n(x) (4) where n and 1/ represent the shear modulus and Poisson ' s ratio, respectively. n(x) is the unit outward normal vector of the surface S at the point x. 6{j (i, j =1,2) represents Kronecker ' s delta and defined as follows: A.=1 fori=j /?>. °«J 1=0 Jot i?j /°) Summation convention has been used in all expressions. For a multiply connected region, the boundary S contains a finite number of different curves which are closed. After examining the equation (1) it is clear that to obtain the displacement vector on the boundary it is enough to determine the displacement components at any point y of V. But before dealing with this problem, we have to explain the meanings of the kernels in the expression (1). An infinite medium having the same constants p, and u with the given prob lem is considered, x and^/ represent two different points of this medium. U*(x,y) and T*(x,y) are the displacement vector and the stress tensor at point x due to'a singular body force of unit magnitude acting at point y in the i (i=l,2) direction, respectively. ~ For solving the integral equation (1), boundary S is idealized as the sum mation of line segments. In this new boundary if the number of these line segments are N, the number of the end points of them will also be N. These end points are named as nodal points. It is assumed that the variation of the displacement components on any of these linear segments is linear. Then the unknowns of the problem are reduced to the values of the displacement components on the nodal points. 2N integral equations can be written by assuming there is a singular loading at every nodal point on each direction. In these integral equations, the integrals over the boundary are reduced to the summation of the integrals over the line segments. In addition, we shall define a new artificial boundary which includes line segments but not the nodal points. Around each nodal point a small arbitrary vucurvilinear part which leaves the point outside is added to complete this new artificial boundary. It is assumed that displacement components are constant on this small curvilinear parts. After necessary calculations, these small curvilinear parts will be şiirinken to the nodal points. After calculating integrals over this artificial boundary we obtain a linear system of 2N equations with 2N unknowns which are the displacement components at nodal points. After solving this linear systeııı using the displacement field and the ar tificial boundary by the help of constituve equations, one can calculate the stress components at any arbitrary internal point y as follows: Tij(y) = I tk(x)UİJ{x, v) ds - / T#(*, y)nt(x)Uk(x) ds (6) The expressions of the kernels in eq.(6) are given below `?<«> = -söbjK' - 2`»^* - ^` - ^'1 2 `irC*»' ~ vMxi ~ Vifak - »*)] (7) T^V = Mîhö t`^1 - 4`>M« + 2(1 - 2,)[^ + ^] +4(1 - 2»)[{Xk-yk)JXl-yi)6ij + İÎİZ»fcjS)M +4^Xi-Vi/Xk-m)6ji + (*i-nX*'-vi)Sjk 1 fi `7s (x> ~ &')(:Bj - %)(** - Vk)(xt - yi)] (8) VU1If we want to calculate the stress components using the expression (6) on the boundary we encounter to the 1 / r singularity. To eliminate this singularity the stress components in a new n, s coordinate system will be used instead of the stress components in x/,x2{ox x and y) cartesian system. But this new coordinate system will be different for each linear segment. Let n be the unit outward normal of any specific line segment whose number is (J), n axes coincides with n. s is directed from nodal point (J) towards nodal point (J + 1). The origin of n, s system is nodal point (J). The stress component Taa(y) in n,s system at any internal point y can be calculated as follows: Tss(V) = Tu{y)nl{J) + T22{y)n/{J) - 2T12(2/)n1(J)n2(J) (9) JV+l, JV+1 - ` Taa{y)=Y, uza(x,y)tk(X)ds+Y, Ws(*,y)jr[uk(*,y)]ds (10) 1^2 -`O ~ ~ ~ f^ Ji(l) - os - At a point on the boundary, we do not want to evaluate other stress com ponents Tna(y) and Tnn(i/) which coincide with the n,s components of the surface traction vector t(x) which is known. The kernels of U%s(x,y) and W%s(x,y) have been given as follows: U°ks(x, y) = Ul/x, y)nl(J) + Vf (x, y)n/{J) - 2U//x, y)nx{J)n2{J) (11) Wak/x,y) = Wl/x,y)n/{J) + Wt/x,y)n/{J) - 2WP(x,y)ni(J)n2(J) (12) n/(J) and n2(J) denote the components of n(J) which is the unit outward normal of a line segment whose number is J. Now an interior point C is considered. The nearest point of the boundary to C be a point B on a line segment whose number is J and it lies between (J)th and (J + /)th nodal points. But here we are introducing a restriction that the point B is neither ( J)th nor {J + l)th nodal point. At this point C, the stress component Tsa(y) will be calculated and c which IXrepresents the shortest distance from C to the artificial boundary will be taken as zero in the limit. After these calculations, Tas(y) stress component is calculated without any singularity at a point of the (J)th line segment of the artificial boundary. There is a s0 distance between this specific point and the (J)th nodal point. s0 can be equal to neither zero nor l(J) which is the length of the (J)th line segment. In this study two examples are solved as plane strain problems. Other two will be considered as plane stress problems. All given formulas are valid for the plane strain. Replacing v by v f (1 - u) will be enough to obtain the same expressions for plane stress. The distributions of the stress components have been calculated on some chosen lines in examples which have been solved. The graphs of the stress components and the displacement components have been drawn on these chosen lines. One can obtain the accuracy of results from the balance of any specific section. The accuracy has been determined as an acceptable value. The basic reason of the error came out because of the integrals calculated numerically at a point very near to the boundary. But stress components calculated on the boundary do not contain this kind of error.