Excitation of elastic waves at the boundary of a prismatic orthotropic body with thin inextensible face coatings

Abstract

In three-dimensional formulation we construct a numerical-analytic solution of the problem of describing the wave field in a semi-infinite orthotropic prismatic body at whose end dynamic forces are applied. We apply the method of series in basic homogeneous normal waves. We study numerically the dynamic boundary effects near the excitation surface. Five figures. Bibliography: 4 titles.Butko et al. [1] have developed a method of solving problems of determining the spectrum of normal waves in elongated prismatic orthotropic bodies of rectangular cross section on whose planar faces there are superthin inextensible isotropic coatings [2]. In the present paper an analogous method is used to solve problems of the excitation of wave fields in a semi-infinite prismatic orthotropic bodies.For an orthotropic body that in Cartesian coordinates Oxlx2x 3 occupies the region v={lxll<a, [x3l<b, 0<x2<~176 }, we consider the stationary dynamic problem whose relations include the equations of motion of the medium relative to complex amplitude components of the displacement vector (C71~2 +C66~ 2. 2 +C55~ 3. 2+s 1 + (C12+C66)~I~2U2 +(C13+C55)01~3U 3 =0;(4 + c:6)~,~2u, +(c~6~ + c;2~ + c:4~ +~2) u2 +(4 + c~,)~2~3u~ =0; (~) (c,3 + c55) ~,O3u, + (c23 + c44) ~2~u2 +(c;5~ 9 2 .... + c4,~ + c33~ +n 2) u3 = 0 and the boundary conditions on the bounding planes (U2)xt=+ a =(U3)x,=+ a = ((~,,)xt___A:a =0; (2) (Ul)x2=• b = (U3)x2=.• = (0"22)x2=:1:: b =0; (3) (o33)x3=o = F(x,, x2); ((s13)x3=o=((sZ3)x3=O =0. (4) In this variant of the boundary conditions a stationary field of dynamic elastic displacements arises in the body under consideration, whose amplitude characteristics possess the following symmetry: u,(x,, x~, x~)=-u,(-x,, x2, x3)= u,(xl,-~:, x~)=-u,(-x,,-x2, ~); U2(X,, X2, X3)=U2(--Xl, X2, X3)=--U2(XI,--X2, X3)=--U2(--X,,--X2, .X3) ; (5) u~(x,, x2, ~3)= u3(-Xl, ~2, x~)= u~(x,,-x2, x3)= u~(-~,,-x~, ~). The amplitude components of the displacement vector in basis normal waves with the given symmetry determined by the method of [I] have the form =,(~ sin)~,,x lcos~,,x2 exp(ikm~x3); u2,~p 2,,~ ul,,~ ~'l,,~p = u(~ c~ sin ~nX2 exp(ik.~pX3);,(0) COS~,.Xl COS~nX 2 exp(ikm~x3 ) (m,n = I, oo; p= 1, 3).Here 3.,,--(2m-l)~/2a; 5~ =(2n-l)~/2b are the transverse wave numbers; k,n,g , are the values of the longitudinal wave numbers determined from the bicubic algebraic dispersion equations.