BEM Formulation plane orthotropic bodies a modification for exterior regions and its proof

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1 Ŕ periodica polytechnica Civil Engineering 5/ 7 3 web: ci c Periodica Polytechnica 7 BEM Formulation plane orthotropic bodies a modification for exterior regions its proof RESEARCH ARTICLE Received 7--7 Abstract Assuming linear displacements constant strains stresses at infinity our aim is to reformulate the equations of the direct boundary element method for plane problems of elasticity. We shall consider a body made of orthotropic material. The equations we have reformulated make possible to attack plane problems on exterior regions without replacing the region in question by a bounded one. Keywords Boundary element method Somigliana relations direct method orthotropic body plane problems exterior regions Acknowledgement The support provided by the Hungarian National Research Foundation within the framework of the project OTKA T46834 is gratefully acknowledged. Introduction As is well known a large literature studies plane problems for orthotropic bodies including for instance as well as the books the references therein. However as explained by Schiavone the stard formulation for exterior regions has the disadvantage that it is impossible to prescribe a constant stress state at infinity. The reason is that an assumption about the far field pattern of the displacements is needed in order to establish an appropriate Betti s formula to prove uniqueness existence for the exterior Dirchlet Neuman problems. Unfortunately this assumption excludes those problems from the theory for which the displacements are linear while the strains stresses are constant at infinity. To make progress on plane problems with such displacements we note that if the direct formulation reproduces this displacement field then the resulting strain stress conditions must also be constant at infinity. Consequently plane problems for the exterior regions can be attacked without replacing the region by a bounded one. The work in presents such direct formulations by assuming constant strains stresses at infinity for an isotropic body. For exterior regions reformulates the classical approach to plane problems. For the same class of problems but in a dual formulation sets up the equations of the direct method in terms of stress functions of order one. The present paper is an attempt to clarify how the formulation changes if we apply the ideas presented in paper to orthotropic bodies. György Szeidl Department of Mechanics University of Miskolc H-355 Miskolc- Egyetemváros Hungary Gyorgy.SZEIDL@uni-miskolc.hu Judit Dudra Department of Mechanics University of Miskolc H-355 Miskolc- Egyetemváros Hungary dudrajudit@fre .hu Preliminaries Throughout this paper x x are rectangular Cartesian coordinates referred to an origin O. Greek subscripts are assumed to have the range summation over repeated subscripts is implied. The doubly connected exterior region under consideration is denoted by A e is bounded by the contour. We stipulate that the contour admits a nonsingular parametrization in terms of its arc length s. The outer normal is denoted by n π. In accordance with the notations introduced δ κλ is the Kronecker BEM Formulation plane orthotropic bodies 7 5

2 symbol α sts for the derivatives taken with respect to x α ɛ 3κλ is the permutation symbol. Assuming plane problems let u κ e κλ t κλ be the displacement field the in plane components of strain stress respectively. For orthotropic bodies the material constants are denoted by s s = s s. For homogenous orthotropic material the plane problem of classical elasticity is governed by the kinematic equations Hook s law e ρλ = ρu λ + λ u ρ t = c e + c e t = c e + c e t = t = c 66 e c = s d c = c = s d c = s d the equilibrium equations c 66 = d = s s s ; 3 t ρλ λ + b ρ = 4 which should be complemented with appropriate boundary conditions not detailed here since they play no role in the present investigations. The basic equation for u λ takes the form D ρλ u λ + b ρ = the differential operator D ρλ has the form Dρλ = c +c 66 c + c 66 c + c 66 c +c 66 5a. 5b Let Qξ ξ M x x be two points in the plane the source point the field point. We shall assume temporarily that the point Q is fixed. The distance between Q M is R the position vector of M relative to Q is r κ. The small circle as a subscript for instance M or Q indicates that the corresponding points i.e. Q or M are taken on the contour. It is obvious that r α M Q = x α M ξ α Q = x α ξ α. 6 Let us introduce the following notations λ + λ = s + /s 7 λ λ = s /s 8 A α = s λ α s 9 ρ α = λ αr + r D = π λ λ s. For our later considerations we note that Eqs. 7 8 imply λ = s + s ± s + s s s. The well-known singular fundamental solutions for the basic Eq. 5a 7 9 are given by the formulas U M Q = D λ A ln ρ λ A ln ρ λ λ r r U M Q = D A A arctan λ λ r + r 3 U M Q = U M Q A U M Q = D ln ρ A ln ρ λ λ λ A λ A T M Q = D r n + r n ρ λ A λ A T M Q = D r n T M Q = D ρ λ A ρ λ λ A ρ ρ ρ ρ λ A ρ r n } λ λ A r n ρ } λ A λ A r n λ A λ A T M Q = D r n + r n 4 ρ u λ M = U λκ M Qe κ Q t λ M = T λκ M Qe κ Q are the displacement vector stress vector on a line element with a normal n λ = n λ M to it at the point M due to the force e κ = e κ Q at Q. 3 Basic formulas for exterior regions Fig. depicts a triple connected region A e bounded by the contours L ε the circle with radius center at O. Here L ε is the contour of the neighborhood A ε of Q with radius R ε while is sufficiently large so that the region bounded by covers both L L ε. If R ε then clearly A e A e. Let u κ M g κ M be sufficiently smooth continuously differentiable at least twice but otherwise arbitrary displacement fields on A e. The stresses obtained from these displacement fields are denoted by t λκ uρ M t λκ gρ M respectively. ρ ρ Per. Pol. Civil Eng.

3 O x x Lo ds ds Q e n L A R A e we obtain A e u λ M µ D M λσ U σ κ M Q µ D M λσ u σ M U λκ M Q da M e κ Q = = u λ M T λκ M Q t λ M U λκ M Q ds M e κ Q+ u λ M T λκ M Q t λ M U λκ M Q ds M e κ Q L ε u λ M T λκ M Q t λ M U λκ M Q ds M e κ Q 8 since t λκ uρ M n κ M = t λ M is the stress on the contour obviously ds t λκ gρ M n κ M = T λκ M Qe κ Q. Fig.. Equation A e u λ M µ D M λσ g σ M g λ M µ D M λσ u σ M da M = uλ M t λκ gρ M n κ M g λ M t λκ uρ M n κ M ds M + L ε uλ M t λκ gρ M n κ M g λ M t λκ uρ M n κ M ds M + uλ M t λκ gρ M n κ M g λ M t λκ uρ M n κ M ds M 5 in which M over a letter denotes derivatives taken with respect to the point coordinates M n κ M is the outward normal is the primal Somigliana identity applied to the triple connected region A e. Let g λ Q = U λκ M Qe κ Q which is a non singular elastic state of the plane in A e. We regard u λm as a different elastic state in the region A e. Further we assume that u λ M has the far field pattern asymptotic behaviour ũ κ M = c κ + ε 3ρκ x ρ ω + e κβ x β 6 as x β or equivalently M tends to infinity. Here c κ is a translation ω is a rotation in finite c κ + ε 3ρκ x ρ ω is the corresponding rigid body motion e κβ is a constant strain tensor at infinity e κβ x β is the corresponding displacement field. The stresses induced by the strains e κβ can be obtained by the Hooke law: In the sequel we shall assume that there are no body forces. This assumption has no effect on the result we will obtain. It is clear that one can omit e κ Q. Regarding the equation obtained by omitting e κ Q our goal is to compute its limit as R ε. As is well known the left h side vanishes under the conditions detailed abovesee hence + lim = R ε L ε u κ Q + u λ MT o λκ M Q t λ M U λκ M Q ds M. Consequently u κ Q = lim t λ M U λκ M Q u λ M T λκ M Q ds M + L R t λ M U λκ M Q u λ M T λκ M Q ds M. 9 In order to establish the first Somigliana formula for the exterior region we need to find the limit of the first integral on the right h side. 4 Somigliana formulas modified for exterior regions In this section our main objective is to prove that I κ = lim t λ M U λκ M Q u λ M T λκ M Q ds M = c κ + ε 3ρκ ξ ρ ω + e κβ ξ β = ũ κ Q. t = c e + c e t = c e + c e t = t = c 66 e 7 Using x β o M = n β o M Substituting the above quantities into the Somigliana identity BEM Formulation plane orthotropic bodies 7 5 3

4 we substitute t λρ n ρ M for t λ M c λ + ε 3ρλ x ρ ω + e κβ x β for u λ M. This implies I κ = I κ + I κ + 3 I κ + 4 I κ = lim c λ T λκ M Q ds M lim ε 3ρλω n ρ MT o λκ M Q ds M + L R lim t λρ n ρ M U λκ M Q ds M lim e λβ n β MT o λκ M Q ds M. REMARK.: In accordance with Eq. 6 the stresses strains are taken as constant quantities which are therefore independent of the arc coordinate s. Since T λκ M Q ds M = δ κλ ε 3ρλ r ρ T λκ M Q ds M = which is the moment about the origin of the stresses due to a unit force applied at Q one can write I κ + I κ = c κ lim ε 3ρλω ξ ρ +r ρ T λκ M Q ds M = L R c κ lim ε 3ρλ ωξ ρ T λκ M Q ds M + L R ε 3ρλ ω r ρ T λκ M Q ds M = c κ + ε 3ρκ ωξ ρ which clearly shows that reproduces the rigid body motion. Determining the limits 3 I κ 4 I κ requires long formal transformations. For this reason we confine ourselves to basic argument the results of the most important steps. First we have to exp U λκ T λκ into series in terms of to the power etc. These transformations are based on the relations: n α M o = n α 3a r α = x α M o ξ α Q = x α ξ α = n α ξ α 3b ρ α = λ α r + r λ α n + n λ α n ξ + n ξ λ α n + + λ α ξ + ξ } n λ α n + 3c n ln ρ α ln + λ ln α n + n λ α n ξ + n ξ λ α n + n ρα λ α n + λ α n ξ + n ξ + n λ α n + n arctanx + εa = arctan x + + λ α ξ + ξ λ α n + 3d n λ α ξ ξ λ α n + n εa + x 3e a is constant ε is a very small quantity. Consequently λ λ r r λ λ n n arctan λ λ r + arctan r λ λ n + + n λ λ λ λ n n n 3 λ ξ λ n 3 n n ξ n 4 λ λ + n 4 + λ + λ n. 3f n Using relations 3a... 3f for a sufficiently large we have U M Q λ D A ln + λ ln n + n λ n ξ + n ξ λ n + n λ A ln + λ ln n + n λ n ξ + n ξ λ n + 4 n U M Q A D ln + λ λ ln n + n λ n ξ + n ξ λ n + n A ln + λ λ ln α n + n λ α n ξ + n ξ λ α n + 5 n λ U M Q D A A arctan λ n n λ λ λ λ n + + n λ λ n n n 3 λ ξ λ n 3 n n ξ n 4 λ λ + n 4 + λ + λ n 6 n T M Q = D λ A λ n + + λ n ξ + n ξ n λ n + n D λ A λ n + + λ n ξ + n ξ n λ n + n ξ n + ξ n ξ n + ξ n 7 T M Q = D λ A λ n + n λ n n + λ n ξ + n ξ λ n + n λ ξ n ξ n D λ A λ n + n λ n n + λ n ξ + n ξ λ n + n λ ξ n ξ n 8 T M Q = D A λ λ n + n n λ n + λ n ξ + n ξ λ n + n D A λ λ n + n n λ n + λ n ξ + n ξ λ n + n ξ n ξ λ n ξ n ξ λ n 9 4 Per. Pol. Civil Eng.

5 T M Q = D λ A λ n + + λ n ξ + n ξ n λ n + n ξ n ξ n D λ A λ n + + λ n ξ + n ξ n λ n + n ξ n ξ n. 3 Knowledge of the series of U λκ T λκ allows us to calculate the integrals 3 I κ 4 I κ. The following observations are used in the calculations: The outward unit normal on is given by equation ϕ is the polar angle. The arc element on admits the form n a = sin ϕ cos ϕ 3 ds o M = dϕ. 3 3 As the coefficients of is are always an integral integrals of zero value. The integrals we have used are all listed in the Appendix. 4 The structure of the coefficients of to the power zero is similar; but these terms involve ξ α the trigonometric integrals that constitute these coefficients are not necessarily equal to zero. After performing the integrations for 3 I κ we have 3 I κ = lim t λρ n ρ M U λκ M Q ds M = = t ξ I κ + ξ I κ + t ξ I κ + ξ I κ + + t ξ 33 I κ + ξ 33 I κ + t } 3 I = π D A λ + A λ λ + λ } 3 I = π D A λ + A λ λ + λ ξ I κ + ξ I κ 33 3 I = 3 I = 33 λ λ I = π D A A λ + λ + 33 I = 34 I = 33 I 3 I = 33 I 3 I = 34 I 33 I = π D A λ + A } λ + } 34 A I = π D λ λ + A λ λ + In the same way we obtain for 4 I κ that 34 I = 3 I = 3 I = 33 I = 34 I =. 4 I κ = lim e λβ n β MT o λκ M Q ds M = = e ξ I κ + ξ I κ + e ξ I κ + ξ I κ + + e ξ 43 I κ + ξ 43 I κ + e 34a 34b 34c 34d 34e 34f 34g 34h ξ I κ + ξ I κ 35 4 I = π D A λ λ + A 4 λ I = π D A λ A 43 λ I = π D A λ λ 44 λ I = π D A λ λ 4 λ I = π D A λ 4 λ I = π D A 43 I = π D A } λ λ + } λ λ } A λ λ λ λ A } λ λ λ λ A λ } λ λ λ } λ λ A λ } λ λ λ + A λ + 44 λ I = π D A λ A } λ λ 4 I = 36a 4 I = 36b 43 I = 36c 44 I = 36d 4 I = 4 I = 36e 36f 43 I = 36g 44 I =. 36h Substituting the integrals with zero value from Eqs. 34a...34h 36a...36h into Eqs then utilizing the Hooke law 7 3 we obtain 3 I + 4 I = s e s e s s s e ξ s e s e s s s e ξ 3 I + 3 I + 33 I ξ I e ξ I + ξ I 4 ξ I I + ξ I + 43 I 3 I + 4 I = s e s e s s s e ξ + s e s e s s s e e ξ I + ξ I ξ 3 I + 3 I + 33 I + ξ 34 I e 37a ξ 4 I + ξ 4 I 4 ξ I I + ξ I + 43 I e ξ I + ξ I. 37b Using Eqs. 7 to one can prove that the following equations hold the formal calculations are presented in Appendix B.: s s s s s s s s 3 I + s 34 I 4 I = s s s 3 I + 33 I 4 I 43 I = 3 I s s s s 34 I 44 I = 38a BEM Formulation plane orthotropic bodies 7 5 5

6 s s s s s s s s 3 I + s 34 I 4 I = s s s 3 I + 33 I 4 I 43 I = 3 I s s s s 34 I 44 I =. 38b If we substitute Eqs. 38ab into equtions 37ab we find that 3 I κ + 4 I κ = e κβ ξ β. 39 Neglecting the rigid body motion i.e. setting I κ + I κ to zero utilizing 39 we obtain I κ = I κ = I κ + I κ + 3 I κ + 4 I κ = e κβ ξ β. Then the first modified Somigliana formula immediately follow from Eqs. 9 : u κ Q = e κβ ξ β Q + t λ M U λκ M Q u λ M T λκ M Q ds M Q A e 4 If Q = Q is on nothing changes concerning the limit of the integral taken on. Consequently C κρ u ρ Q = e κβ ξ β Q + t λ M U λκ M Q u λ M T λκ M Q ds M Q = Q 4 C κρ = δ κρ / if the contour is smooth at Q. This integral equation is that of the direct method or the second Somigliana formula for exterior regions. If Q is inside the contour this region is referred to as A i then it is easy to show that = e κβ ξ β Q+ t λ M U λκ M Q u λ M T λκ M Q ds M Q = Q A i 4 which is the third Somigliana formula for exterior regions. Using the formulae set up for the strains in Appendix C the Hooke law 7 we can calculate the stresses: t αβ Q = t αβ + t λ M ˆD λαβ M Q ds M L o u λ M S λαβ M Q ds M Q A e 43 ˆD λ = c D λ + c D λ ˆD λ = c D λ + c D λ ˆD λ = c 66 D λ = ˆD λ ˆ S λ = c S λ + c S λ ˆ S λ = c S λ + c S λ Sˆ λ = c 66 S λ = Sˆ λ Behaviour at infinity Our goal in this section is to compute the limit of representation 4 as Q. This will lead to a characterization of the asymptotic behaviour of u κ Q. If this behaviour is the same what we have assumed i.e. if the limit coincides with 6 provided that in the latter the rigid body motion is neglected then we confirm that the results of the previous section are correct. It is clear from representation 4 that it is sufficient to show that the following relations hold lim Qξ ξ t λ M U λκ M Qds M = lim Qξ ξ u λ M T λκ M Qds M =. 45 In order to find the limit of the above integrals we have to set up asymptotic relations for the fundamental solutions U λκ M Q T λκ M Q if Q. Using the notations introduced in Fig. as well as Eqs. 6 we have r α M o Q = x α M o ξ α Q = x α ξ α = 46a ˆR ˆn α x α ˆR ˆn α ˆn α = 46b ˆR ρ α = λ α r + r ˆR λ α ˆn + ˆn 46c ln ρ α ln ˆR + λ ln α ˆn + ˆn. 46d Substituting equations 46bbc into 3 then performing Fig.. x n O x ds M o r QM o r OQ = R some manipulations we obtain the following asymptotic relations for the fundamental solution of order one: U M Q D λ A λ A ln ˆR 47a U M Q = U M Q λ λ ˆn ˆn D A A arctan λ λ ˆn + 47b ˆn U M Q D Q n A A ln ˆR. λ λ 47c It is obvious that asymptotically U λκ M Q U λκ Q i.e. the kernel in integral 45 is independent of M. 6 Per. Pol. Civil Eng.

7 Consequently lim t λ M U λκ M Qds M = Qξ ξ = lim U λκq t λ M ds M Qξ ξ L } o } resultant =. 48 By repeating the line of thought leading to the asymptotic relations 47abc for the fundamental solutions of order two we obtain T M Q Ḓ λ A λ A R λ ˆn + ˆn λ ˆn + = T Q 49a ˆn ˆR T M Q 4 Iintegral Equations of the First Kind in Plane Elasticity Quarterly Ḓ λ A R λ λ ˆn + λ A ˆn λ λ ˆn + ˆn ˆn = T of Applied Mathematics LIII 995 no Q 5 Gradstein IS Ryzhik IM Table of Integrals Series Products Academic New York 98. ˆn ˆR 49b 6 Table of Integrals Series Products Moscow Nauka Pub. in Ḓ λ λ A λ λ A R λ ˆn + ˆn λ ˆn + ˆn T M Q ˆn ˆn = T Q ˆR 49c T M Q Ḓ λ A λ A R λ ˆn + ˆn λ ˆn + = T Q 49d ˆn ˆR T λκ = T λκ Q is defined by the above equations. Since for large ˆR the kernel T λκ M Q is independent of M tends to zero it follows that lim u λ M T λκ M Qds M = Qξ ξ T λκ Q = lim u λ M ds M =. 5 Qξ ξ ˆR This verifies that the asymptotic behaviour of the displacement representation 4 is as expected. Under this condition the strain energy density is bounded although the strain energy is not there is no need to replace the exterior region by a finite one if a constant stress condition is prescribed at infinity. This can be an advantage if one considers an infinite plane with holes or cracks subjected to constant stresses at infinity an attempt is made to determine the stresses in finite. Existing codes can be modified easily to perform computations. References Banarjee PK Butterfield R Boundary Element Methods in Engineering Science Mir Moscow 984. Banarjee PK The Bondary Element Methods in Engineering McGraw Hill New York Consta C The Boundary Iintegral Equation Method in Plane Elasticity Proceedings of the American Mathematical Society no. Russian Huang L Sun X Liu Y Cen Z Parameter identification for twodimensional orthotropic material bodies by the boundary element method Engineering Analysis with Boundary Elements 8 4 no Vable M Sikarskie DL Stress analysis in plane orthotropic material by the boundary element method Int. J. Solids Structures no.. 9 Rizzo RJ Shippy DJ A method for stress determination in plane anisotropic elastic bodies J. Composite Materials Schiavone P Chong-Quing Ru On the exterior mixed problem in plane elasticity Mathematics Mechanics of Solids Szeidl Gy Boundary integral equations for plane problems remark to the formulation for exterior regions Publications of the University of Miskolc Series D Natural Sciences Mathematics no Boundary integral equations for plane problems in terms of stress functions of order one Journal of Computational Applied Mechanics no Concluding remarks For the sake of completeness note that Consta 3 gives an asymptotic expansion for the displacements at infinity which ensures the validity of the Betti formula for exterior regions isotropic bodies. Under this condition the total strain energy stored in the region is bounded. In addition uniqueness existence proofs are easy to give. We have modified the Somigliana formulas for exterior regions of orthotropic bodies by assuming that the strains are constants accordingly the displacements are linear at infinity. BEM Formulation plane orthotropic bodies 7 5 7

8 A Trigonometric integrals When determining the integrals 3 I κ + 4 I κ we have used the following trigonometric integrals: n ln ds = ln n ln n ln π λ α n + n ds = λ α n + n ds = n λ α n + ds = n n n λ α n + ds = n n λ α n + ds = n π π π n3 n π π π cos ψ dψ = π π λ n + n λ n + ds = n λ n + n λ n + ds = n n n3 λ n + n λ n + ds = n n4 n ln ds = ln π sin ψ dψ = 5 cos ψ ln λ α cos ψ + sin ψ dψ = 5 sin ψ ln λ α cos ψ + sin ψ dψ = 53 cos ψ λ α cos ψ + sin ψ dψ = π λ α + λ α 54 cos ψ sin ψ λ α cos ψ + sin dψ = 55 ψ sin ψ λ α cos ψ + sin ψ dψ = π π π λa + 56 cos 3 ψ sin ψ λ cos ψ + sin ψ λ cos ψ + sin dψ = 57 ψ cos ψ sin 3 ψ λ cos ψ + sin ψ λ cos ψ + sin dψ = 58 ψ cos 4 ψ λ cos ψ + sin ψ λ cos ψ + sin ψ dψ = π λ + λ + λ λ λ + λ λ + λ + 59 λ n + n λ n + ds = n n n cos ψ sin ψ λ cos ψ + sin ψ λ cos ψ + sin ψ dψ = π λ + λ λ + λ + 6 n4 n 4 λ λ + n 4 + λ + λ n n ds = sin 4 ψ λ cos ψ + sin ψ λ cos ψ + sin ψ dψ = π λ + λ + λ λ λ + λ λ + λ +. 6 The integrals detailed above can be checked either by using Maple 9.5 or the Table of Integrals 5 6 by Gradstein Ryzhik. B Proof of equation 38a B.. First consider equation 38a. Substituting the integrals 3 I 34 I 4 I see equations 34a 34d 36a we have π s s s D s A λ + A λ λ + λ } λ + s D A A λ λ + λ + π D A λ λ + + π D A λ λ + =. 6 8 Per. Pol. Civil Eng.

9 After multiplying throughout by the common denominator s s s λ λ some further manipulations we obtain s A λ λ + A λ λ + + s A A λ λ A λ λ + } λ λ If we divide throughout by λ λ substitute 9 then we find + s s s + λ λ s s s s A λ λ + λ λ =. 63 s λ s λ λ + s λ s s λ + s s s + s λ s λ λ + s λ s s λ + s s s λ λ s s s s λ + λ + s s λ s s λ s λ λ =. 64 Here we can again divide throughout by λ λ. Hence λ 3 λ 3 s 3 + λ s s + λ s s λ λ s3 λ λ s 3 λ λ 3 s 3 λ 3 λ s 3 + λ s s + λ s s s s s + λ λ s s + λ λ s s λ λ s s = 65 Consequently λ s s + λ s s = s s λ + λ = s s s + = s + s s s λ λ s3 λ λ s 3 = λ λ s 3 λ + λ = s s 3 s + = s + s s. s s Here λ 3 λ 3 s 3 λ λ 3 s 3 λ 3 λ s 3 + λ s s + λ s s s s s + λ λ s s + λ λ s s =. λ λ 3 s 3 λ 3 λ s 3 = s3 Therefore λ λ 3 + λ 3 λ = s 3 λ λ λ + λ = s 3 s λ + λ = s s s λ + λ λ s s + λ s s = s s λ + λ. λ 3 λ 3 s 3 s s s + λ λ s s + λ λ s s = λ 3 λ 3 s 3 + s s λ λ s λ 3 λ 3 + λ λ s s = s s s λ λ s λ 3 λ 3 s 3 + λ λ s s = s s s λ λ s s s s λ λ s =. B.. Consider equation 38a. If we substitute the integrals 3 I 33 I 4 I 43 I see equations 34b 34c 36b 36c perform some manipulations we obtain 4π D A λ λ + A λ λ + A λ λ A λ + λ + + A λ λ + A After substituting D multiplying throughout by the common denominator we have s λ λ λ + λ + A λ λ + A λ λ + A A λ λ A λ + λ A λ + λ λ λ + =. 66 s λ λ λ + λ + =. 67 BEM Formulation plane orthotropic bodies 7 5 9

10 If we make use of equation 9 perform some transformations we find s λ s λ + s λ s λ λ s λ s λ + s λ s λ + λ + s λ s s λ s λ λ s λ λ λ + λ + =. If we factor the above expression we have λ λ s + s λ λ + λ + λ s λ + λ s =. The first factor is obviously different from zero. Hence we should investigate the other two. We shall start with the last one: s λ + λ s = s + s s + s = s + s + =. Since the above expression is zero equation 38a is truly satisfied. B.3. The third equation to consider is 38a 3. If we substitute the integrals 3 I 34 I 4 I - see equations 34a 34d 36d we obtain π D s s s s A λ λ + A λ λ λ + s A A λ + λ + λ + + π D λ A λ + λ A λ + =. 68 Let us multiply throughout by the common denominator s s s λ + λ + substitute D. We have s A λ λ + A λ λ + + s A A λ λ + If we make use of equations 9 we find s s s A λ λ + A λ λ + =. 69 s s λ s λ λ + s s λ s λ λ + + s s λ s s λ s λ λ + s s s s λ s λ λ + s s s s λ s λ λ + =. 7 After factoring the above expression we obtain the product: λ λ s + s λ + λ + λ + λ + λ λ λ λ s s =. The first factor is apparently not equal to zero. Therefore we have to consider the other two. The last one clearly vanishes: s + λ λ s = s + s s = s + s =. s Consequently equation 38a 3 is also satisfied. C Proof of equation 38b C.. Third consider equation 38b. Upon substitution of the integrals 3 I 34 I 4 I see equations 34e 34h 36e we have π D s A A λ λ A s s s λ + λ + + s λ λ + A λ λ + A π D λ λ + A λ λ + =. 7 Per. Pol. Civil Eng.

11 After multiplying throughout by the common denominator s s s λ λ λ + λ + making use of equations 9 we obtain λ λ s 3 λ + λ s s + s 3 + λ 3 λ 3 s 3 λ s s λ s s s s s λ λ s s + λ λ 3 s s λ λ 3 s s } } + λ 3 λ s s λ 3 λ s s } } +λ λ s s λ s s s λ s s s + +λ λ 3 s s + λ 3 λ s s + λ λ s s λ λ s s } } Let us divide throughout by λ λ cancel the terms braced. If in addition we take into account that we shall have s s + s 3 = s3 + s3 = =. λ s 3 + λ s 3 + λ 3 λ 3 s 3 λ s s in which λ s s s s s λ λ s s + λ λ s s λ λ 3 s s + + λ λ 3 s s λ 3 λ s s + λ 3 λ s s + λ λ s s λ s s s λ s s s + λ λ 3 s s + λ 3 λ s s λ λ s s s = λ 3 λ 3 s 3 s s s λ λ s s + λ λ s s λ s s s λ s s s + λ λ 3 s s + λ 3 λ s s = λ λ 3 s s + λ 3 λ s s λ s s s λ s s s = Therefore = s s λ λ λ + λ s s s λ + λ = s s λ + λ s s s λ + λ = s λ 3 λ 3 s 3 + λ λ s s s s s λ λ s s = s = s s s λ + λ s s s λ + λ =. = s λ λ s λ λ + s s s s + s λ λ = s s λ λ + s s s s + s λ λ = s s = s s λ λ + s s s s s s s s λ λ =. C.. Consider equation 38b. If we substitute the integrals 3 I 33 I 4 I 43 I see equations 34b 34c 36b 36c perform some manipulations we obtain π D A λ + A λ + A A λ λ λ + λ + π D A λ + A λ + π D A λ λ + A λ =. 7 λ + BEM Formulation plane orthotropic bodies 7 5

12 Let us multiply throughout by the common denominator λ + λ + make some manipulations. In this way we have A λ + A λ + If simplify?? then we have A A λ λ A λ + A λ + A λ λ + A λ λ + λ λ s λ + λ + =. 73 s λ s λ + s λ s λ + s λ s s λ s λ λ + s λ s λ + λ + + s λ s λ + λ λ λ s λ + λ + =. 74 If we factoring this expression we shall get λ λ s + λ s + λ s + λ λ s s + λ s + λ s =. The first factor is evidently different from zero. Hence we have to investigate the other two. We shall start with the last one s + λ s + λ s = s +s λ +λ = s +s s + s = s +s + =. C.3. Third consider equation 38b 3.Upon substitution of the integrals 3 I 34 I 4 I see equations 34e 34h 36e we have π D s A A λ λ s s s λ + λ + + A λ λ + A λ λ + A π D λ + A =. 75 λ + Then multiplying throughout by the common denominator s s s λ λ λ + λ + making some manipulations we obtain s A A λ λ λ λ s A λ λ + A λ λ + s s s λ λ A λ + A λ + λ λ s s s s λ λ λ + λ + =. 76 If we substitute equation 9 we have s s λ s s λ + s s λ λ s s λ λ + s s λ s s λ + + s s λ λ s s λ λ + s s λ λ 3 s s λ 3 λ s λ λ 3 + s λ 3 λ =. 77 After factoring this equations we shall find λ λ s + λ s + λ s + λ λ s s λ λ s =. One can check easily that the third factor is zero: Consequently 38b 3 is fulfilled. s λ λ s = s s s = s s =. s Per. Pol. Civil Eng.

13 D Formulae for strains Making use of equations from the kinematic equations we obtain e αβ = t λ M D λαβ M Q ds M u λ M S λαβ M Q ds M Q A e 78 D λ = U λ S λ = T λ If we introduce the notations D λ = U λ + U λ = D λ D λ = U λ S λ = T λ + T λ = D λ S λ = T λ. 79 b = λ ρ k = λ 3 A ρ b = λ ρ k = λ 3 A ρ c = λ λ r + r d = λ λ r A A f = f λ 3 = r + r λ 3 r + r for the derivatives in equation 79 we have U = D b λ A r b λ A r U = D A A + d r λ λ dr cd = U U = D b A r b A r U = D b A r b A r U = D A A dr + d r r c λ λ r = U A U = D r ρ A r λ ρ λ 8 8 T = D T = D T = D T = D b λ A r ρ b λ A r ρ b λ A r ρ b λ A r ρ b λ A r ρ r n + r n D b A b A n b λ A r ρ b λ A r ρ b λ A r ρ r n D b A b A n D f λ r f λ r r n r n D b λ A b λ A n D r n + r n D b A b A n b A r T = D b A r r n + r n D b A b A n T = D T = D T = D ρ b A r ρ b λ A r ρ b A r ρ ρ b A r ρ b λ A r r n D f r f r r n + D f f n b λ A r r n D ρ b A r ρ b A r ρ r n + r n D b A b A n. ρ b λ A r ρ r n b A r ρ r n + D b A b A n 8a 8b BEM Formulation plane orthotropic bodies 7 5 3