Jtam.dvi JOURNAL OF THEORETICAL AND APPLIED MECHANICS 3, 39, 2001 PROBLEM OF ELASTIC INTERACTION BETWEEN AN ANNULAR THICK PLATE AND ELASTIC STRATUM Bogdan Rogowski Dariusz Zaręba Department of Mechanics of Materials, Technical University of Łódź e-mail: brogowsk@ck-sg.p.lodz.pl; darek@kmm-lx.p.lodz.pl A solution to the contact pressure problem, bending moments and di- splacements of an axisymmetrically loaded thick annular transversely isotropic plate restingwithout friction on a transversely isotropic or gra- nular half-space is presented in the paper. No singularity occurs in the contact pressure because the extensional deformation of the plate is ta- ken into consideration.An approximate solution to the resulting integral equation is obtained using an effective numerical procedure. To assess the effects of anisotropyof the plate response, numerical results are obta- ined for threematerials:magnesium,which is nearly isotropic; cadmium, which ismoderately anisotropic; andgraphite epoxy,which is highly ani- sotropic and the supporting half-space is modelled as granularmaterial, like soil. The results are presented graphically. Key word: anisotropy, contact problem, thick plate 1. Introduction The considered linear elasticity contact problem models the actual pro- blem of interaction between an annular flat foundation and the supporting soil. Variousmodels used in studying soil-foundation interaction problems are discussed in the book by Selvadurai (1979). Extensive bibliography and com- prehensive accounts of various contact problems can be also found in thework byPopov (1971),Poulos andDavis (1974),Hooper (1978) andGladwell (1980). In modelling a raft foundation, it is usually assumed that it behaves li- ke a thin isotropic elastic plate, governed by the Kirchoff-Love plate theory. When the thickness of the plate is small, compared with the other dimensions of the plate, and the loading does not present any abrupt changes, the thin 638 B.Rogowski, D.Zaręba plate theory gives satisfactory results. However, when the plate is thick and anisotropic and is subjected to localized loads the influence of shearing defor- mations and transverse normal stresses on the plate response has to be taken into consideration. Plate theories taking into account shearing deformations and extensional deformations have been developed by Reissner (1945, 1947), Mindlin (1951),Goodier (1946) for isotropicmaterials andbyRogowski (1975) for orthotropic ones. Regarding the behaviour of the supporting soil, various models have be- en proposed and applied within the framework of the linear elasticity. The most common of them are: (a)Winkler springs, (b) half-space continuum, and (c) layered continuum. The half-space continuum is modelled as transversely isotropic medium. Its particular case was proposed by Weiskopf (1945). The Weiskopf model takes into consideration slipping of the granules of a gra- nular material, like soil, which causes appreciable shearing deflections. The Weiskopf model has been used previously by Misra and Sen (1975, 1976), by Ejike (1977) and byMastrojannis (1989) in analytical studies. The transverse isotropy proposed byWeiskopf introduces two shearmoduli in the plane of the isotropy and in the direction normal to the planes of the isotropy. ButYoung’s moduli of some soil masses are also dissimilar in both directions (Dahan and Predeleanu, 1981), so the soilmasses exhibit, in general, transversely isotropic mechanical behaviour. The contact conditions also influence the soil-foundation interaction. The usual assumption is that no shearing stresses develop at the interface of thebo- dies in contact. This assumption is used in the presentwork, too.Thepresence of the shearing stresses, due to friction or adhesion, decreases displacements and bending moments of a thin elastic plate (Hooper, 1981; Mastrojannis et al., 1988), which probably holds also for the case of a thick plate. Accordingly, the presented solution to a frictionless contact problem gives an upper bound for the actual response and the safety aspects of the structure are quaranted. The extensional transverse deformation of the plate, which is taken into con- sideration, yields that no singularity occurs in the contact pressure at the end of the contact region. This is in contrastwith thewell-known results (Dundurs and Lee, 1971; Adams and Bogy, 1976; Gecit, 1986). The purpose of this paper is: (a) to present a solution to the problem of a thick platewith shearing and extensional deformations taken into account and to half-space elasticity problem (with solution derived by making use of the transversely isotropic potential function method and integral transforms), to reduce contact conditions to the problem of solving the integral equation for unknown normal contact pressure, (b) to give an approximate solution to the Problem of elastic interaction... 639 resulting integral equation using an effective numerical procedure, and (c) to give numerical results, for some practicalmaterials which indicate dependence ofmechanical quantities on elastic constants of the plate andhalf-space on the plate thickness and other parameters over a certain parameter range. 2. Formulation of the problem Fig. 1. Geometry of the system An annular plate with the inner and outer radii a and b, and of the uniform thickness h is in smooth contact with the horizontal surface z = 0 of an elastic medium occupying the half-space z ­ 0 (Fig.1). Due to the axisymmetrically distributed load p(r) acting on the upper surface of the plate and the reactive normal pressure q(r) acting on the lower surface the plate deforms and its lower surface assumes a shape described by a function w0(r) for a ¬ r ¬ b. The normal surface displacement of the half-space in the contact region is also described by the same function w0(r). Taking into consideration the theory of thick plates (Rogowski, 1975), it is required to determine the normal displacement w(r,z), the contact pressure q(r), the bending moments Mr(r) and Mθ(r) and the shearing force Qr(r) induced by the plate. Materials of the plate and the half-space exhibit transversely 640 B.Rogowski, D.Zaręba isotropic behaviour. The planes of the isotropy are assumed to be parallel to the boundary z =0. The stress-strain relations for the transverse isotropy are σrr = c11εrr + c12εθθ+ c13εzz σθθ = c12εrr+ c11εθθ+ c13εzz σzz = c13εrr+ c13εθθ + c33εzz (2.1) σrz =2c44εrz where cij are the material constants of the transverse isotropy. The engine- ering constants Er, νrθ in the isotropic plane and Ez, νrz,Grz in the principal direction of anisotropy have the following relations between themoduli of ela- sticity cij c11 = Er(1−νrzνzr) ∆(1+νrθ) c12 = Er(νrθ +νrzνzr) ∆(1+νrθ) c13 = Erνrz ∆ c33 = Ez(1−νrθ) ∆ c44 = Grz ∆ =1−νrθ−2νrzνzr (2.2) The solution to the stated axisymmetric interaction problem involves the solu- tion to two coupledboundaryvalueproblems: one forbendingandcompression by the normal loads p(r) and q(r)with free-edge conditions, and the other for the stress anddisplacement fields inside the half-space z ­ 0when the surface z =0 is free to shearing traction with the normal displacement prescribed in the contact region and the normal stresses vanishing outside of the contact region. The contact surfaces separate in the neighbourhood of the pointwhere the contact stress changes from negative to positive. 2.1. The thick plate problem Accordingly (Rogowski, 1975), the normal displacement w(r,z) of an axi- symmetrically loaded thick transversely isotropic plate under bending satisfies the differential equation ∇4rw(r,z) =∇ 2 r[∇ 2 rw(r,z)] = [1−β(z)h 2∇2r] F(r) D a ¬ r ¬ b −h ¬ z ¬ 0 (2.3) where D = h3 12 ( c11− c213 c33 ) = Erh 3 12(1−ν2 rθ ) Problem of elastic interaction... 641 β(z) = 1 4 [c11− c213/c33 2c44 − c13 c33 ( 1+2 z(z +h) h2 )] = (2.4) = 1 4(1−νrθ) [Grθ Grz −νzr ( 1+2 z(z +h) h2 )] denote the plate flexural rigidity and variable correction coefficient, respecti- vely, and F(r)= p(r)−q(r) ∇2r = d2 dr2 + 1 r d dr (2.5) The shearing force Qr(r) and bending moments Mr(r) and Mθ(r) are given by expressions Qr(r)=−D dH dr Mr(r)=−D ( d2 dr2 + νrθ r d dr ) [w(r)+βh2H] (2.6) Mθ(r)=−D ( νrθ d2 dr2 + 1 r d dr ) [w(r)+βh2H] where H =∇2rw(r)+ βh2 D F(r) (2.7) β = 1 4 (c11− c213/c33 2c44 − 2 3 c13 c33 ) = 1 4(1−νrθ) (Grθ Grz − 2 3 νzr ) In equations (2.6)2− (2.7)1 w(r)= 1 h 0 ∫ −h w(r,z) dz is theweighted average normal displacement which satisfies a differential equ- ation similar to (2.3), with the weighted average correction coefficient β. For the axisymmetric and free-edge conditions of the considered annular plate the following boundary conditions apply Qr =0 Mr =0 for r = a ∧ r = b (2.8) 642 B.Rogowski, D.Zaręba The solution to equation (2.3), satisfying boundary condition (2.8), is expres- sed by w(r,z) = w0+ r2−2a2 ln r b 8D(b2−a2) b ∫ a F(ρ)ρ [ ηρ2+2(b2−a2)ln ρ b ] dρ− (2.9) − 1 4D r ∫ a F(ρ)ρ [ r2−ρ2+(r2+ρ2)ln ρ b ] dρ+ β(z)h2 D r ∫ a F(ρ)ρ ln ρ b dρ where ρ is the variable of integration and the constant w0 is determined by considering the contact condition, namely the equilibrium of forces in the vertical direction, i.e. b ∫ a F(r)r dr =0 and η = 1−νrθ 1+νrθ (2.10) For a circular plate, i.e. when a =0, w0 is the central deflection of the plate. The extensional deformation yields also transverse displacement of the plate which is expressed by equation (Rogowski, 1975) w′(r,z) =− (2z +h)h3 48D c11 c33 [p(r)+q(r)] (2.11) The transverse displacement is obtained as the superposition of bending deflection (2.9) with displacement (2.11). For z =0 this yields w(r,0)= w0+ r2−2a2 ln r b 8D(b2−a2) b ∫ a F(ρ)ρ [ ηρ2+2(b2−a2)ln ρ b ] dρ− − 1 4D r ∫ a F(ρ)ρ [ r2−ρ2+(r2+ρ2)ln ρ b ] dρ+ (2.12) + β0h 2 4D r ∫ a F(ρ)ρ ln ρ b dρ− h4 48D α0[p(r)+q(r)] a ¬ r ¬ b where β0 =4β(0)= c11− c213/c33 2c44 − c13 c33 = 1 1−νrθ (Grθ Grz −νzr ) (2.13) α0 = c11 c33 = Er(1−νzrνrz) Ez(1−ν2rθ) Problem of elastic interaction... 643 2.2. The half-space problem The boundary conditions for the axisymmetric indentation half-space pro- blem are uz(r,0)= w0(r) for r ∈< a,b > σzz(r,0)= 0 for r ∈< 0,a)∪ (b,∞) σrz(r,0)= 0 for r ∈< 0,∞) (2.14) with the stress and displacement vanishing at infinity. The last equation of (2.14) states that the contact is frictionless. The normal displacement uz(r,0) on the surface z = 0 of the half-space is determined by means of the contact pressure σzz(r,0) =−q(r), as follows (Rogowski, 1982) uz(r,0)= 1 GzC b ∫ a ρq(ρ) dρ ∞ ∫ 0 J0(ξρ)J0(ξr) dξ (2.15) where Gz is the shear modulus in the z-direction, and C is a material constant (Rogowski, 1982) GzC = Er (1−ν2 rθ )s1s2(s1+s2) The improper integral in equation (2.15) has analytical representation, namely ∞ ∫ 0 J0(ρξ)J0(rξ) dξ = 2 π [H(r−ρ) r K (ρ r ) + H(ρ−r) ρ K (r ρ )] = 2 π K(k) r+ρ (2.16) The functions K(k) are complete elliptic integrals of the first kind (see equ- ation (3.14)); themodulus k =2 √ ρr/(ρ+r), and H(·) isHeaviside’s function. Then, the normal displacement is given as uz(r,0)= 2 πGzC b ∫ a q(ρ)ρK(k) r+ρ dρ (2.17) Substitution of equations (2.12) and (2.17) into equation (2.14) yields the following integral equation in terms of the contact pressure q(ρ) for the sta- ted contact problem 644 B.Rogowski, D.Zaręba 2 πGzC b ∫ a q(ρ)ρK(k) r+ρ dρ =− h4 48D α0[p(r)+q(r)]+ +w0+ r2−2a2 ln r b 8D(b2−a2) b ∫ a F(ρ)ρ [ ηρ2+2(b2−a2)ln ρ b ] dρ− (2.18) − 1 4D r ∫ a F(ρ)ρ [ r2−ρ2+(r2+ρ2)ln ρ r ] dρ+ β0h 2 4D r ∫ a F(ρ)ρ ln ρ r dρ 3. The solution to the integral equation It is convenient to transformequation (2.18) intodimensionless form.Write ρ = xb r = yb a = λb h = tb q(xb)= P0q(x) p(xb)= P0p(x) Kr = D GzCb 3 (3.1) where P0 is a constant with dimensions of pressure and Kr is the plate- to-half-space stiffness ratio. Then, the integral equation for determining the function q(x) takes the form 1 ∫ λ q(x)xK(k) x+y dx+ π 16Kr {y2−2λ2 lny 1−λ2 1 ∫ λ q(x)x[ηx2+2(1−λ2)lnx] dx− −2 y ∫ λ q(x)x [ y2−x2+(y2+x2)ln x y ] dx+2β0t 2 y ∫ λ q(x)x ln x y dx+ (3.2) + t4 s α0q(y) } = w∗0 + π 16Kr T(y) where w∗0 = πGzCw0 2P0b Problem of elastic interaction... 645 T(y) = ϕ(y)+2β0t 2 y ∫ λ p(x)x ln x y dx− t4 6 α0p(y) (3.3) ϕ(y) = y2−2λ2 lny 1−λ2 1 ∫ λ p(x)x[ηx2+2(1−λ2)lnx] dx− − 2 y ∫ λ p(x)x [ y2−x2+(y2+x2)ln x y ] dx We assume then, that the unknown function q(x), corresponding to the normal interfacial pressure, has the form q(x)= w∗0g (1)(x)+g(2)(x) λ ¬ x ¬ 1 (3.4) where g(1)(x) and g(2)(x) are new unknown functions. Thus the problem is reduced to solving the following two integral equations for the unknown functions g(1)(x) and g(2)(x) ψ(1)(y)= 1 ψ(2)(y)= π 16Kr T(y) λ ¬ y ¬ 1 (3.5) where (for m =1,2). ψ(m)(y)= 1 ∫ λ g(m)(x)xK(k) x+y dx+ + π 16Kr {y2−2λ2 lny 1−λ2 1 ∫ λ g(m)(x)x[ηx2+2(1−λ2)lnx] dx− (3.6) −2 y ∫ λ g(m)(x)x [ y2−x2+(y2+x2)ln x y ] dx+ +2β0t 2 y ∫ λ g(m)(x)x ln x y dx+ t4 6 α0g (m)(y) } The unknown constant w∗0 corresponding to the plate deflection w0, is evaluated by considering the equilibrium of forces in the transverse direction, see equation (2.10)1, i.e. w∗0 1 ∫ λ g(1)(x) dx+ 1 ∫ λ g(2)(x) dx = 1 ∫ λ p(x)x dx (3.7) 646 B.Rogowski, D.Zaręba In general, it is not easy to obtain an analytical expression for the solution g(m)(x) to equations (3.5) and (3.6). An analytical solution exists for the spe- cial case of Kr →∞ and λ =0, which corresponds to the problem of a rigid circular punch on a half-space. It is g(1) ∞ (x)= 4 π2 √ 1−x2 g(2) ∞ (x)≡ 0 w∗0∞ = π2 8 q(x)= 1 2 √ 1−x2 w0∞ = πP0b 4GzC (3.8) For an annular rigid plate (Kr →∞) the analytical approximate solution is (Rogowski, 1982) g(1) ∞ (x)∼= 4 π2 [ 1 √ 1−x2 ( 1+ 4λ3 3π2x2 ) + 2λ π √ x2−λ2 − 2 π arcsin λ x − − 4λ3 3π2 arccosx x3 ] (3.9) g(2) ∞ (x)≡ 0 w∗0∞ ∼= π2 8 ( 1− 4λ3 3π2 )−1 (1−λ2) q ∞ (x)= w∗0∞g (1) ∞ (x) w0∞ = πP0b 4GzC ( 1− 4λ3 3π2 )−1 (1−λ2) When the plate is rigid, then the square root of the singularity exists for the contact pressure. When the bodies in contact are both elastic and the extensional deformation of the plate is taken into consideration in the governing integral equation then no singularity occurs in the normal contact pressure. This conclusion is in contrast with the well-known result (Dundurs and Lee, 1972; Adams and Bogy, 1976; Gecit, 1986), where the authors show thatwhen the bodies in contact are elastic, the power of the singularity of the contact pressure depends upon their elastic constants. We introduce an approximate solution using a numerical procedure. The contact length is divided into N equal segments of the length ∆x. In this procedure, the pressure distribution is assumed to be piecewise constant, that is g(m)(x)= g (m) j xj − ∆x 2 ¬ 2¬ xj + ∆x 2 (3.10) where xj =λ+(j −1/2)∆, j =1,2, ...,N. On this assumption, equations (3.5) and (3.6) become ψ (1) i =1 ψ (2) i = π 16Kr Ti i =1,2, ...,N (3.11) Problem of elastic interaction... 647 where (for m =1,2) ψ (m) i = N ∑ j=1 g (m) j I(yi,xj)+ π 16Kr {y2i −2λ 2 lnyi 1−λ2 N ∑ j=1 g (m) j [ S(x′j)−S(x ′ j−1) ] − −2 i ∑ j=1 g (m) j [ Q(yi,x ′ j)−Q(yi,x ′ j−1)−β0t 2 ( R(yi,x ′ j)−R(yi,x ′ j−1) )] + (3.12) + t4 6 α0g (m) i } and Ti = T(yi), etc., λ ¬ yi ¬ 1, x′j = λ + j∆x, Ni is the number of segments in contact of the length λ ¬ x ¬ yi, yi = λ + (i − 1/2)∆y, ∆y = (1−λ)/N, and S(x), Q(yi,x), R(yi,x) are like influence coefficients defined by the integrals in equations (3.6), which are easily integrated S(x)= x2 [ η x2 4 +(1−λ)2 ( lnx− 1 2 )] j =1,2, ...,N Q(yi,x)= x2 2 [ y2i − 5 4 x2+(2y2i +x 2)ln x yi ] j =1,2, ..., i R(yi,x)= x2 2 ( ln x yi − 1 2 ) j =1,2, ..., i (3.13) The integrals I(yi,xj)= xj+ ∆x 2 ∫ xj− ∆x 2 x x+yi K(ki) dx k 2 i = 4xyi (x+yi)2 (3.14) K(ki)= π 2 ∫ 0 dφ √ 1−k2i sin 2φ can be evaluated numerically over the N segments usingChebyshev’s quadra- ture rule. If the solution to two equations (3.11) is obtained, from condition (3.7) and equation (3.10) the value of w∗0 becomes w∗0 = I0 ∆xI1 − I2 I1 (3.15) where I0 = 1 ∫ λ xp(x) dx Im = N ∑ j=1 g (m) j xj m =1,2 (3.16) 648 B.Rogowski, D.Zaręba The discretized form of the integral equation for the stated contact pro- blem, given by (3.11) and (3.12), yields an effective numerical procedure for evaluating g(m)(x) in each segment of the contact area. The normal contact pressure takes the following form q(yb) P0 ≡ q(y)= w∗0g (1)(y)+g(2)(y) λ ¬ y ¬ 1 (3.17) Then, for instance, the transverse plate deflection and one of themoments are (function and discretized form) w(yb,ζh)GzC P0b ≡w(y,ζ)= 2 π ω∗0 + 1 8Kr { T(y)− t4 6 α0(2ζ +1)q(y)− − t4 3 α0ζp(y)− y2−2λ2 lny 1−λ2 1 ∫ λ xq(x)[ηx2+2(1−λ2)lnx] dx+ +2 y ∫ λ xq(x) [ y2−x2+(y2+x2)ln x y ] dx− −8β(ζ)t2 y ∫ λ xq(x)ln x y dx λ ¬ y ¬ 1 −1¬ ζ ¬ 0 (3.18) w(yi,ζ)= 2 π ω∗0 + 1 8Kr { Ti− t4 6 α0(2ζ +1)qi− t4 3 α0ζpi− − y2i −2λ 2 lnyi 1−λ2 N ∑ j=1 qj[S(x ′ j)−S(x ′ j−1)]+2 Ni ∑ j=1 qj [ Q(yi,x ′ j)−Q(yi,x ′ j−1)− −4β(ζ)t2[R(yi,x′j)−R(yi,x ′ j−1)] ]} i =1,2, ...,N Mr(yb) P0b 2 ≡ Mr(y)=− 1 8 (d2ϕ dy2 + νrθ y dϕ dy ) + y2(1+νrθ)+λ 2(1−νrθ) 4y2(1−λ2) · · 1 ∫ λ xq(x)[ηx2+2(1−λ2)lnx] dx+ + 1−νrθ) 4y2 y ∫ λ xq(x) [ y2−x2+2y2 ln x y ] dx− y ∫ λ xq(x)ln x y dx (3.19) Mr(yi)=− 1 8 (d2ϕ dy2 + νrθ y dϕ dy ) + y2i (1+νrθ)+λ 2(1−νrθ) 4y2i (1−λ2) · Problem of elastic interaction... 649 · N ∑ j=1 qj[S(x ′ j)−S(x ′ j−1)]+ 1−νrθ 4y2 Ni ∑ j=1 qj[Q ∗(yi,x ′ j)−Q ∗(yi,x ′ j−1)]+ + Ni ∑ j=1 qj[R(yi,x ′ j)−R(yi,x ′ j−1)] i =1,2, ...,N where ζ = z/h ∈ [−1,0]. 4. Numerical results and discussion Aparticular case of the transversely isotropic half-space continuum,which is used in numerical calculations, is constituted by themodel proposed byWe- iskopf (1945). In thismodel Γ = Gr/Gz > 1 and thematerial parameter C is C = Γ √ 2 (1−ν)(Γ +1−2ν) (4.1) Write the relative stiffness parameter Kr, equation (3.1), for a granular material as Kr = Kmt 3 √ Γ −ν 2(1−ν) + 1 2 (4.2) where Km = 1 6 Er E 1−ν2 1−ν2 rθ (4.3) The relative stiffness parameter Kr, the reduced relative stiffness Km and the plate parameters νrθ, β0 and α0 are shown in Table 1. In the case of a uniformly distributed load, the procedure described in the preceding sections gives the following result for the input functions from equations (3.11) p(x) = 1 λ ¬ x ¬ 1 ϕ(y) = 1 8 ( y2−2 1+3νrθ 1+νrθ ) y2+ λ2 8 (8y2−5λ2)− (4.4) − λ2 ( y2+λ2 2+νrθ 1+νrθ − 1+3νrθ 2(1+νrθ) ) lny+ 1 2 λ4(1+4lny)lnλ = ϕ0(y) T(y) = ϕ(y)− 1 2 β0t ( y2−λ2+2λ2 ln λ y ) − 1 6 t4α0 650 B.Rogowski, D.Zaręba Table 1 Parameter Magnesium Cadmium Graphite/epoxy Isotropy νrθ 0.3711 0.1163 0.0292 0.3000 β0 1.1312 1.6996 22.0696 1.0000 α0 0.9144 2.3729 8.1245 1.0000 Km 375 604 1120 220 Γ =1 t =0.1 0.375 0.604 1.12 0.22 t =0.4 24 38.7 71.7 14.1 Kr Γ =2.5 t =0.1 0.550 0.886 1.644 0.323 t =0.4 35.2 56.7 105.2 20.7 Γ =5 t =0.1 0.757 1.220 2.261 0.444 t =0.4 48.4 78.1 144.7 28.4 Below is shown (Fig.2) a numerical result of the problem of a uniformly distributed load when the evaluation is done for two different material para- meters Kr, the contact area iswithin a =1¬ r ¬ b =2, and it is divided into N = 10 segments. The presented curves are approximated for the values for the middle of each segment. The continuous line corresponds to Kr =0.604, the dashed one – to Kr =0.02. Fig. 2. Contact stresses q(yb)/P0, λ = a/b =0.5 Figures 2, 3, 4 show that the effect of the relative stiffness parameter Kr (anisotropy) on thephysical quantities is strong.When this parameter is large, for instance when the plate is very rigid then the contact stresses are greater Problem of elastic interaction... 651 Fig. 3. Transverse deflection of the plate, λ = a/b =0.5 Fig. 4. Bending moment M r (yb)/(P0b 2), λ = a/b =0.5 652 B.Rogowski, D.Zaręba at the outer diameter than at the inner one the transverse deflection is almost constant and the moments have greater values than it is observed in plates made of muchmore elastic materials. References 1. Selvadurai A.P.S., 1979, Elastic Analysis of Soil – Foundation Interaction, Developments in Geotechnical Engineering, Amsterdam, Elsevier 2. 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Mech., 34, 2, 119-126 Zagadnienie sprężystego oddziaływania między pierścieniową grubą płytą i sprężystym podłożem Streszczenie Otrzymano wzory dla naprężeń kontaktowych, momentów zginających i prze- mieszczeń osiowo symetrycznie obciążonej pierścieniowej płyty poprzecznie izotropo- wej, która kontaktuje się bez tarcia z poprzecznie izotropowym lub ziarnistympodło- żem. Naprężenia kontaktowe nie wykazują osobliwości, dzięki temu, że uwzględniono odkształcenia wywołane ściskaniem płyty. Równanie całkowe zagadnienia rozwiąza- no w sposób przybliżony za pomocą efektywnego numerycznego algorytmu. Wyniki przedstawiono na wykresach. Manuscript received December 5, 2000; accepted for print March 20, 2001