Jtam.dvi JOURNAL OF THEORETICAL AND APPLIED MECHANICS 47, 3, pp. 629-643, Warsaw 2009 SOLUTION OF INFINITE SLAB STATIC PROBLEM USING THE FINITE STRIP METHOD BY DIFFERENCE EQUATIONS Zdzisław Pawlak Jerzy Rakowski Poznan University of Technology, Institute of Structural Engineering, Poznań, Poland e-mail: jerzy.rakowski@put.poznan.pl An analytical approach to the solution of an infinite slab static problem using the finite stripmethod is presented. The structure simply suppor- ted on its opposite edges is treated as a discrete one. A regular mesh of identical finite strips approximates the continuous structure. This re- gular slab discretization enables one to derive a fundamental solution for the two-dimensional discrete strip structure in an analytical, closed form.Equilibrium conditions are derived from the finite elementmethod formulation. The set of the infinite number of equilibrium conditions is replaced by one equivalent difference equation. The solution to this equ- ation is the fundamental function, i.e. Green’s function for considered slab strip. Key words: finite element method, finite strips, difference equations, plane slab 1. Introduction The static problem of an infinite plane slab simply supported on its oppo- site edges is considered (see Fig.1).We assume that the longitudinal displace- ments and transverse stresses are equal to zero at both edges. The slab strips with such boundary conditions are commonly used at bridge structures. In the finite stripmethod (Loo andCusens, 1978) the structure is divided into a set of finite strips of the length L (L is width of the slab) and an arbitrary width b (see Fig.2). Theunknownsaredisplacements uand v (in the xand y directions) along the nodal line r which connects two adjacent strips. It is assumed that two 630 Z. Pawlak, J. Rakowski Fig. 1. An infinite slab strip Fig. 2. Finite strip discretization point forces Px and Py concentrated at an arbitrary nodal line load the slab. According to the finite strip procedure, the fields of loading and displacement functions are expressed in the form of harmonic series (Loo andCusens, 1978) p(x)= ∞∑ n=1 Px sin nπyo L u(x,y) = ∞∑ n=1 [( 1− x b ) ui,n+ x b uj,n ] sin nπy L (1.1) v(x,y) = ∞∑ n=1 [( 1− x b ) vi,n+ x b vj,n ] cos nπy L The four-degree-of-freedom finite strips (see Fig.3) are used in the calcu- lations. Solution of infinite slab static problem... 631 Fig. 3. The finite strip element Themaking use of displacement functions (1.1) and theminimization pro- cedure for the potential energy formula UI = t 2 L∫ 0 b∫ 0 σ ⊤ ε dxdy− L∫ 0 b∫ 0 [u,v] [ p(x) p(y) ] dxdy (1.2) where σ=    σx σy σxy   =   Ex 1−νxνy νyEy 1−νxνy 0 νxEx 1−νxνy Ey 1−νxνy 0 0 0 Gxy   ε ε=   εx εy γxy   =    ∂u ∂x ∂v ∂y ∂u ∂y + ∂v ∂x    leads to a set of infinite number of linear equations ∞∑ I=−∞ K I nW I n =P I n (1.3) 632 Z. Pawlak, J. Rakowski where W I n = [ui,n,vi,n,uj,n,vj,n] ⊤ P I n = [P x i,n,P y i,n,P x j,n,P y j,n] ⊤ The finite strip stiffness matrix (I-th strip) takes the form Kn =α1DxKDx+α2nDyKDy +α3n(D1KD1+D2KD2)+ (1.4) +Dxy(α2nK1+α1K2−α3nK3) where α1 = L 2b α2n = bn2π2 12L α3n =− nπ 4 Dxy =Gxy Dx = Ex (1−νxνy) Dy = Ey (1−νxνy) D1 = νyEy (1−νxνy) D2 = νxEx (1−νxνy) Ex, Ey are Young’s moduli, νx, νy – Poisson’s ratios, Gxy – Kirchhoff’s mo- dulus, t – thickness of the slab, Ki – the following numbermatrices KDx =   1 0 −1 0 0 0 0 0 −1 0 1 0 0 0 0 0   KDy =   0 0 0 0 0 2 0 1 0 0 0 0 0 1 0 2   KD1 =   0 −1 0 −1 0 0 0 0 0 1 0 1 0 0 0 0   KD2 =   0 0 0 0 −1 0 1 0 0 0 0 0 −1 0 1 0   K1 =   2 0 1 0 0 0 0 0 1 0 2 0 0 0 0 0   K2 =   0 0 0 0 0 1 0 −1 0 0 0 0 0 −1 0 1   K3 =    0 −1 0 1 −1 0 −1 0 0 −1 0 1 1 0 1 0    Solution of infinite slab static problem... 633 2. The difference equations formulation Assembling two adjacent elements R and R+1 (see Fig.4), one obtains the equilibrium equations of force amplitudes for each n-th harmonic element (Pawlak et al., 1997) Px(r,r−1),n+P x (r,r+1),n =P x r,n (2.1) P y (r,r−1),n +P y (r,r+1),n =Pyr,n Fig. 4. Forces at the nodal line r Expression (2.1) can be written for a regular system in the form of diffe- rence equations (Pawlak and Rakowski, 1995) equivalent to the FEM matrix formulation (β1∆ 2+β0)ur+β2(E −1 −E)vr =ApxP x r (2.2) β5(E −1 −E)ur +(β3∆ 2+β4)vr =ApyP y r where β0 =6Dxyα2 β1 =Dxyα2−Dxα1 β2 =α3(Dxy +D1) β3 =Dyα2−Dxyα1 β4 =6Dyα2 β5 =−α3(Dxy +D2) Apx =sin nπy0x L Apy =sin nπy0y L ∆2r =∆ 2 =(E+E−1−2) is the second-order difference operator, Enr =E n is the shifting operator En(fr)= fr+n (Rakowski, 1991), P x r and P y r are forces acting at the nodal line r (with the co-ordinates y0x and y0y, respectively). 634 Z. Pawlak, J. Rakowski After elimination of the unknown vr, the equilibrium conditions can be written in the form of one fourth-order difference equation with the unk- nown ur (r-th nodal line transverse displacement amplitude for the n-th harmonic) (B4∆ 4+B2∆ 2+B0)ur =−Apx(β3∆ 2+β4)P x r +Apyβ2(E −1 −E)Pyr (2.3) where B4 =β2β5−β1β3 B2 =4β2β5−β0β3−β1β4 B0 =−β0β4 Elimination of the unknown ur leads to the equation with the unknown lon- gitudinal displacement amplitude vr (B4∆ 4+B2∆ 2+B0)vr =−Apy(β1∆ 2+β0)P y r +Apxβ5(E −1 −E)Pxr (2.4) For a regular infinite plate strip, equations (2.3) and (2.4) are equivalent to the set of infinite number of equilibrium conditions derived using the finite strip methodology (FSM). 3. The fundamental solution To simplify the calculations, the superposition principle is applied. Static ana- lysis of the structure loaded by the force Px = Pxr δr,0 = Pδr,0 (P y r = 0, δr,0 – the Kronecker delta) leads to the fundamental solution for the infinite slab strip (see Fig.5). The equilibrium conditions are expressed bymeans of FEM in the form of a difference equation (B4∆ 4+B2∆ 2+B0)ur =−Apx(β3∆ 2+β4)P x (3.1) In order to solve difference equilibrium equation (3.1) we use the discrete Fourier transform (Babuška, 1959; Pawlak andRakowski, 1996) in x-direction (variable r) F[fr] = f̃(α)= ∞∑ r=−∞ fre irα (3.2) F−1[f̃(α)] = fr = 1 2π π∫ −π f̃(α)e−irα dα Solution of infinite slab static problem... 635 Fig. 5. The infinite strip with a point load After transformations, one may obtain the solution ur = Px π π∫ 0 (S1cosα+S2)cos(rα) cos2α+Bmcosα+Cm dα (3.3) where Bm = B2−4B4 2B4 Cm = 4B4−2B2+B0 4B4 S1 =− β1Apx 2B4 S2 = (2β1−β0)Apx 4B4 The function of nodal displacement amplitude (3.3) may be expressed in the form of the following recurrent relation ur = Px π [S1F1(r)+S2F2(r)] (3.4) where F1(r)= 2 r−1C(r+1)− ( r 1 ) 2r−3C(r−1)+ + r 2 ( r−3 1 ) 2r−5C(r−3)− r 3 ( r−4 2 ) 2r−7C(r−5)+ . . . F2(r)= 2 r−1C(r)− ( r 1 ) 2r−3C(r−2)+ + r 2 ( r−3 1 ) 2r−5C(r−4)− r 3 ( r−4 2 ) 2r−7C(r−6)+ . . . 636 Z. Pawlak, J. Rakowski The integrals C(n) are analytically calculated coefficients C(n)= π∫ 0 cosnα cos2α+Bmcosα+Cm dα Fundamental solution (3.4) provides the horizontal displacement amplitudes for nodal lines r­ 0.When the horizontal force acts, the discrete function ur has the symmetric form u−r =ur (3.5) In the considered strip, the vertical amplitudes at nodal line r=0 are equal to zero, v0 = 0. Using equilibrium conditions (2.2), one may determine the vertical displacement amplitude for r=1 v1 = β1 β2 (u1−u0)+ β0 2β2 u0− Apx 2β2 Px (3.6) and all the following ones for r > 1 vr = β5 β3 (ur −ur−2)+2vr−1−vr−2− β4 β3 vr−1 (3.7) The vertical displacement functions have the antysymmetric form v−r =−vr (3.8) The real displacements at the nodal line r (for the co-ordinate y) are in the form of sums u(r,y) = N∑ n=1 ur(n)sin nπy L v(r,y) = N∑ n=1 vr(n)sin nπy L (3.9) where: ur(n) and vr(n) are amplitudes obtained from equations (3.4)-(3.8), respectively, N is the number of harmonic elements. 4. Numerical examples Displacements functions (3.9) yield the fundamental solution for the infini- te plane slab simply supported on its opposite edges. The calculations were carried out for the following dimensionless physical parameters: Ex = 1.0, Ey = 0.5, νx = 0.25. The continuous body was divided into a regular set of Solution of infinite slab static problem... 637 finite strips. The values of displacements were determined at ten points along the transverse strip dimension L (y =0,1, . . . ,10). The assumption of finite strip width b = 1.0 means the discretization ratio b = L/10. The results of numerical calculations for the strip subjected to a unit horizontal force acting at the point yp =L/2 are presented in Table 1 and Table 2. Table 1.Horizontal displacements u(r,y) r=0 r=1 r=2 y=0 0 0 0 y=1 0.1795 0.1776 0.1597 y=2 0.4590 0.4053 0.3151 y=3 0.9269 0.5868 0.3949 y=4 0.5255 0.4711 0.3795 y=5 0.3155 0.3136 0.2930 y=6 0.2247 0.2209 0.2119 y=7 0.1502 0.1503 0.1474 y=8 0.0960 0.0951 0.0933 y=9 0.0458 0.0458 0.0453 y=10 0 0 0 Table 2.Vertical displacements v(r,y) r=0 r=1 r=2 y=0 0 −0.1057 −0.1577 y=1 0 −0.0906 −0.1603 y=2 0 −0.1440 −0.1367 y=3 0 −0.0162 −0.0347 y=4 0 0.1016 0.0659 y=5 0 0.0531 0.0832 y=6 0 0.0484 0.0711 y=7 0 0.0301 0.0613 y=8 0 0.0325 0.0540 y=9 0 0.0235 0.0505 y=10 0 0.0288 0.0491 The plot of strip deformation is shown in Fig.6. The numerical result, i.e. the obtained field of displacements is treated as the fundamental solution for the considered system. It can be used for solving 638 Z. Pawlak, J. Rakowski Fig. 6. Deformation of the infinite strip the static problem of a finite slab in an analogous way as in the boundary element method for continuous systems. Let us consider an orthotropic, square slab simply supported on two edges and free on the other ones (see Fig.7). Both slab dimensions are equal to 10b, the load is distributed in the middle of the slab. Fig. 7. The finite square slab We assume that the considered finite slab is a part of the infinite strip. Following the finite strip methodology we approximate the continuous body by the set of identical finite strips of thewidth b. In order to fulfil theboundary conditions at the nodal lines r = 0 and r = 10, we introduce an additional loading Xi,n (see Fig.8). Solution of infinite slab static problem... 639 Fig. 8. The strip with additional forces beyond the free edges The forces Xi,n are determined from the boundary equations (indirect boundary element method (Banerjee and Butterfield, 1981)) for free edges of finite slab (see Fig.7) Px0,n(Pr,n,Xi,n)= 0 P y 0,n(Pr,n,Xi,n)= 0 Px10,n(Pr,n,Xi,n)= 0 P y 10,n(Pr,n,Xi,n)= 0 (4.1) where n is the number of harmonic elements of sinus series, r is the number of the nodal line with acting forces (r=5b), i is the number of the additional force (i=1,2,3,4). The values of forces Xi,n for various numbers n are presented in Table 3. Table 3.Additional forces Xi,n n X1,n X2,n X3,n X4,n 1 18.4906 −18.1107 −18.1107 18.4906 2 0 0 0 0 3 −2.6113 −1.0518 −1.0518 −2.6113 4 0 0 0 0 5 – – – – 6 0 0 0 0 7 0.0001 0.0077 0.0077 0.0001 8 0 0 0 0 9 −0.0007 0.0002 0.0002 −0.0007 10 – – – – 640 Z. Pawlak, J. Rakowski The amplitude of the nodal line r displacement for the n-th harmonic element is found from the expression ur(n)= ∑ j Xj,nU(n,r−xj)+ ∑ p Pp,nU(n,r−xp) (4.2) where U(n,r) is the fundamental solution, the amplitudes are obtained from equations from (3.4)-(3.8) respectively, xj is the position of the additional force Xj,n, xp is the position of the load Pp,n. By adding the solutions for all used harmonic elements, one may obtain the total displacement of the finite slab u(r,y) = N∑ n ur(n)sin nπy L (4.3) where N is the number of used harmonic elements, r, y are the co-ordinates for the point of investigated displacement. The final results of numerical calculations are symmetric. The horizontal and vertical displacements for one quadrant of the finite strip are presented in Table 4 and Table 5. Table 4.Horizontal displacements u(r,y) r=0 r=1 r=2 r=3 r=4 r=5 y=0 0.000 0.000 0.000 0.000 0.000 0.000 y=1 −0.591 0.802 0.257 0.476 0.469 1.291 y=2 −0.106 0.152 0.467 0.836 1.249 1.811 y=3 −0.136 0.207 0.605 1.042 1.494 2.065 y=4 −0.152 0.239 0.677 1.142 1.608 2.182 y=5 −0.156 0.249 0.699 1.171 1.641 2.217 Table 5.Vertical displacements v(r,y) r=0 r=1 r=2 r=3 r=4 r=5 y=0 −0.048 0.183 0.316 0.363 0.268 0.000 y=1 −0.044 0.169 0.287 0.317 0.219 0.000 y=2 −0.033 0.133 0.217 0.222 0.136 0.000 y=3 −0.020 0.086 0.137 0.134 0.078 0.000 y=4 −0.009 0.042 0.065 0.062 0.036 0.000 y=5 0.000 0.000 0.000 0.000 0.000 0.000 Solution of infinite slab static problem... 641 Fig. 9. Deformation of the finite square slab The plot of finite square slab deformation is shown in Fig.9. The analytical closed form of the derived fundamental functions provi- des the possibility of parametric analysis. The influence of the number N of harmonics used in the calculations on the results accuracy is shown in Fig.10. Fig. 10. Accuracy of the results The relation between thevalue of the selected displacement and themoduli ratio is shown in Fig.11. The next numerical test was carried out for a slab presented in Fig.7, where the following data were chosen: slab dimensions L = R = 2.0m (the finite strip width b=0.2m), Young’s moduli Ex =Ey =205GPa, Poisson’s ratios νx = νy =0.3, thickness of the slab t=1.0cm, load distribution in the middle of the slab Px =1.0MN/m.The result, i.e. thehorizontal displacement at the point located in the middle of the slab (1.0; 1.0m) coincides with the 642 Z. Pawlak, J. Rakowski Fig. 11. The influence of moduli ratio on the chosen displacement one given by the finite element method. The ratio between the displacement obtained here uFSM =0.5846cm and the displacement found from the finite element method uFEM =0.5964cm is uFSM/uFEM =0.9802. References 1. Andrermann F., 1996,Rectangular Slabs. Structural Analysis, Arkady,War- saw 2. Babuška I., 1959, The Fourier transform in the theory of differences equation and its applications,Arch. Mech. Stos., 4, 11 3. Banerjee P.K., Butterfield R., 1981,Boundary Element Methods in En- gineering Science, McGraw-Hill, Maidenhead 4. Loo Y.C., Cusens A.R., 1978, The Finite Strip Method in Bridge Engine- ering, Viewpoint Publications, NewYork 5. PawlakZ.,Rakowski J., 1995,Singular solutions for two-dimensionaldiscre- te systems by Difference Equation Method, Proc.of the Second International Conference on Difference Equations and Applications, Veszprem, Hungary 6. Pawlak Z., Rakowski J., 1996, Fundamental solutions for regular discrete slabs, Proceedings of the GAAM Conference – Gesellschaft für Angewandte Mathematik und Mechanik’96, Prague, 114-115 7. Pawlak Z., Rakowski J., Wielentejczyk P., 1997, Static and dynamics of infinite discrete strips,Proc. of GAMM-Tagung, Regensburg, Germany 8. Rakowski J., 1991, A critical analysis of quadratic beam finite elements, In- ternational Journal for Numerical Methods in Engineering, 31, 949-966 Solution of infinite slab static problem... 643 Zastosowanie równań różnicowych w rozwiązaniu problemu statyki nieograniczonej tarczy metodą pasm skończonych Streszczenie W pracy zaprezentowano analityczne rozwiązanie problemu statyki nieograniczo- nej tarczymetodą pasm skończonych. Tarcza swobodnie podparta na przeciwległych krawędziach jest rozwiązywana jako układ dyskretny. Ciągła struktura jest aproksy- mowana regularną siatką składającą się z identycznych pasm skończonych.Regularny podział pozwala na wyprowadzenie funkcji fundamentalnych dla dwuwymiarowego układu dyskretnego w analitycznej, zamkniętej formie. Warunki równowagi zosta- ływyprowadzone zgodnie ze sformułowaniemmetody elementów skończonych.Układ równańrównowagi składający się z nieskończonej liczby równańzostał zastąpiony jed- nym, równoważnym równaniem różnicowym.Rozwiązanie tego równania jest funkcją fundamentalną dla rozpatrywanej tarczy. Manuscript received February 6, 2009; accepted for print March 3, 2009