Numerical solution of the problems for plates on some complex partial internal supports

Journal of Computer Science and Cybernetics, V.35, N.4 (2019), 305–318 DOI 10.15625/1813-9663/35/4/13648 NUMERICAL SOLUTION OF THE PROBLEMS FOR PLATES ON SOME COMPLEX PARTIAL INTERNAL SUPPORTS TRUONG HA HAI1,∗, VU VINH QUANG1, DANG QUANG LONG2 1Thai Nguyen University of Information and Communication Technology 2Institute of Information Technology, VAST ∗haininhtn@gmail.com
Abstract. In the recent works, Dang and Truong proposed an iterative method for solving some problems of plates on one, two and three line partial internal supports (LPISs), and a cross internal support. In nature they are problems with strongly mixed boundary conditions for biharmonic equa- tion. For this reason the method combines a domain decomposition technique with the reduction of the order of the equation from four to two. In this study, the method is developed for plates on internal supports of more complex configu- rations. Namely, we examine the cases of symmetric rectangular and H-shape supports, where the computational domain after reducing to the first quadrant of the plate is divided into three subdo- mains. Also, we consider the case of asymmetric rectangular support where the computational domain needs to be divided into 9 subdomains. The problems under consideration are reduced to sequences of weak mixed boundary value problems for the Poisson equation, which are solved by difference method. The performed numerical experiments show the effectiveness of the iterative method. Keywords. Rectangular Plate; Internal Line Supports; Biharmonic Equation, Iterative Method, Domain Decomposition Method. 1. INTRODUCTION The plates with line partial internal supports (LPIS) play very important role in engi- neering. Therefore, recently they have attracted attention from many researchers. In the essence, the problems of plates on internal supports are strongly mixed boundary value pro- blems for biharmonic equation. There are some methods for analysis of these plates. It is worthy to mention the Discrete Singular Convolution (DSC) algorithm developed by Xiang, Zhao and Wei in 2002 [15, 16]. Essentially, DSC based on the theory of distributions and the theory of wavelets is an algorithm for the approximation of functions and their derivatives. Its efficiency has been proven in solving many complex engineering problems. To the best of our knowledge a rigorous justification of DSC has not been established yet. Later, in 2007, 2008 Sompornjaroensuk and Kiattikomol [11, 12] transformed the problem with one LPIS to dual series equations, which then by the Hankel transformation are reduced to the form of a Fredholm integral equation. It should be noted that the kernel and the right-hand side of the equation are represented in a series containing Hankel functions of both first and second kinds; therefore, the numerical treatment for this integral equation is very difficult. So, this result is of pure significance. Motivated by these mentioned works, some years ago Dang c© 2019 Vietnam Academy of Science & Technology 306 TRUONG HA HAI et al. and Truong [3, 4] proposed a simple iterative method that reduces the problems with one and two LPISs to sequences of boundary value problems for the Poisson equation with weak mixed boundary conditions which can be solved by using the available efficient methods and software for second-order equations. This is achieved due to the combination of a domain decomposition technique and a technique for reduction of the order of differential equations. These techniques were used separately or together in the works [1, 6, 7, 8]. In this study, we develop the method for the problems of rectangular plate with more complex internal supports, namely for a symmetric rectangular support (lying in the center of the plate), asymmetric rectangular support (not lying in the center of the plate) and a symmetric H-shape support. Suppose that the plates are subjected to a uniformly distributed load (q), their bottom and top edges are clamped, while the left and right edges are simply supported. Then the problems are reduced to the solution of the biharmonic equation ∆2u = f for the deflection u(x, y) inside the plates, where f = q/D, D is the flexural rigidity of the plates, with boundary conditions on the plate edges and the conditions on the internal supports. As seen later, in the cases of the symmetric internal supports the problems will be reduced to ones in the domain divided into 3 subdomains. But in the case of asymmetric rectangular plate the domain of the problem must be divided into 9 subdomains. As was shown in [4], the boundary conditions on the fictitious boundary inside the plate are ∂u ∂ν = ∂∆u ∂ν = 0 and the conditions on the internal support are the same as clamped boundary conditions u = ∂u ∂ν = 0. In result of the domain decomposition method the problem for plates on internal supports will be reduced to sequences of boundary value problems for Poisson equation in the rectangles with weakly mixed boundary conditions. The rigorous theoretical proof of the convergence of the iterative method can be done in a similar way as the proof for one LPIS in [4] but due to the complexity of the internal supports we omit it. The paper is organized as follows. In Section 2 we consider the plate with a symmetric rectangular internal support. An iterative method for the problem with general boundary conditions is described and the numerical results are reported. In Section 3, omitting the description of iterative method, we briefly present the results of computation for the plate with a symmetric H-shape internal support. In Section 3 we extend the results of Section 2 to the case of asymmetric rectangular internal support. Some concluding remarks are given in the last section. 2. PROBLEM FOR PLATE ON A SYMMETRIC RECTANGULAR INTERNAL SUPPORT 2.1. The problem setting In this section we consider the problem for plate on a symmetric rectangular internal support, i.e., a rectangular support which lies in the center of the plate as in Figure 1(a). As in [4] and [3], due to the two-fold symmetry it suffices to consider the problem in a quadrant of the plate. Associated conditions are given on the actual and fictitious boundaries, and on the parts of the support inside the quadrant as depicted in Figure 1(b). Thus, we have to solve the biharmonic equation ∆2u = f in the first quadrant of plate which is denoted by Ω with the boundary conditions given in Figure 1(b). For this purpose NUMERICAL SOLUTION OF THE PROBLEMS 307 (a) Rectangular support u = ∂u/∂ν = 0 u = ∂u/∂ν = 0 ∂u/∂ν = 0 ∂∆u/∂ν = 0u = ∂u/∂ν = 0 ∂u/∂ν = ∂∆u/∂ν = 0 u = ∆u = 0 (b) First quadrant of rectangular support Figure 1. Symmetric rectangular support and its quadrant with boundary conditions we set the problem with the general boundary conditions, namely, consider the problem ∆2u = f in Ω \ (MN ∪MQ), u = g0 on SA ∪ SD ∪MN ∪MQ, ∂u ∂ν = g1 on SA ∪ SB ∪ SC ∪MN ∪MQ, ∆u = g2 on SD, ∂ ∂ν ∆u = g3 on SB ∪ SC , (1) where Ω is the rectangle (0, a)×(0, b), SA, SB = SB1∪SB2, SC = SC1∪SC2, SD = SD1∪SD2 are its sides. See Figure 2. S A S B1 S B2 S D1 y S D2 S C1 S C2 x O e 2 N Ω 1 Ω 2 Ω 3 K Q e 1 M Figure 2. Domain decomposition for the problem considered in a quadrant of plate In the case if all boundary functions gi = 0 (i = 0, 3), the problem models the bending of the first quadrant of a rectangular plate. 308 TRUONG HA HAI et al. 2.2. Description of the iterative method To solve the problem, the domain Ω is divided into three subdomains Ω1, Ω2 and Ω3 as shown in Figure 2. Next, we set v = ∆u and denote ui = u|Ωi , vi = v|Ωi and by νi denote the outward normal to the boundary of Ωi (i = 1, 2, 3). It should be noted that on four sides of the subrectangle Ω3 there are defined boundary conditions sufficient for solving the biharmonic equation in this subdomain. This is a problem with weakly mixed boundary conditions, which can be performed by iterative method in a similar way as in [1]. After that it remains to solve the biharmonic problem in Ω1∪Ω2 by an iterative process. Thus, we must perform two iterative processes in sequence. But we do not handle so. Instead, we suggest the following combined iterative method for the problem (2), which is based on the idea of simultaneous update of the boundary functions ϕ1 = v1 on SA, ϕ2 = v2 on MQ, ϕ3 = v3 on MN , ξ = ∂v2 ∂ν2 on KM, η = ∂u2 ∂ν2 on KM as follows: Combined iterative method: 1. Given ϕ (0) 1 = 0 on SA; ϕ (0) 2 = 0 on MQ, ϕ (0) 3 = 0 on MN, ξ(0) = 0, η(0) = 0 on KM. (2) 2. Knowing ϕ (k) 1 , ϕ (k) 2 , ϕ (k) 3 , ξ (k), η(k), (k = 0, 1, ...), solve sequentially problems for v (k) 3 and u (k) 3 in Ω3, problems for v (k) 2 and u (k) 2 in Ω2, and problems for v (k) 1 and u (k) 1 in Ω1: ∆v (k) 3 = f in Ω3, v (k) 3 = ϕ (k) 2 on MQ, v (k) 3 = ϕ (k) 3 on MN, ∂v (k) 3 ∂ν3 = g3 on SB2 ∪ SC2,  ∆u (k) 3 = v (k) 3 in Ω3, u (k) 3 = g0 on MQ ∪MN, ∂u (k) 3 ∂ν3 = g1 on SB2 ∪ SC2, (3)  ∆v (k) 2 = f in Ω2, v (k) 2 = g2 on SD2, ∂v (k) 2 ∂ν2 = ξ(k) on KM, v (k) 2 = ϕ (k) 2 on MQ, ∂v (k) 2 ∂ν2 = g3 on SC1,  ∆u (k) 2 = v (k) 2 in Ω2, u (k) 2 = g0 on SD2 ∪MQ, ∂u (k) 2 ∂ν2 = η(k) on KM, ∂u (k) 2 ∂ν2 = g1 on SC1, (4)  ∆v (k) 1 = f in Ω1, v (k) 1 = g2 on SD1, v (k) 1 = ϕ (k) 1 on SA, ∂v (k) 1 ∂ν1 = g2 on SB1, v (k) 1 = ϕ (k) 3 on MN, v (k) 1 = v (k) 2 on KM,  ∆u (k) 1 = v (k) 1 in Ω2, u (k) 1 = g0 on SD1 ∪ SA ∂u (k) 1 ∂ν1 = g1 on SB1, u (k) 1 = g0 on MN u (k) 1 = u (k) 2 on KM (5) NUMERICAL SOLUTION OF THE PROBLEMS 309 3. Calculate the new approximation ϕ (k+1) 1 = ϕ (k) 1 − τ (∂u(k)1 ∂ν1 − g1 ) on SA, ϕ (k+1) 2 = ϕ (k) 2 − τ (∂u(k)2 ∂ν2 − g1 ) on MQ, ϕ (k+1) 3 = ϕ (k) 3 − τ (∂u(k)3 ∂ν2 − g1 ) on MN, ξ(k+1) = (1− θ)ξ(k) − θ∂v (k) 2 ∂ν2 , η(k+1) = (1− θ)η(k) − θ∂u (k) 2 ∂ν2 on KM (6) where τ and θ are iterative parameters to be chosen for guaranteeing the convergence of the iterative process. The convergence of the above iterative method can be proved in the same way as for the case of one and of two LPIS in [4]. But this is very cumbersome work, therefore we omit it. 2.3. Numerical example In order to realize the above combined iterative method we use difference schemes of second order of accuracy for mixed boundary value problems (3)-(5) and compute the normal derivatives in (6) by difference derivatives of the same order of accuracy. All computations are performed for uniform grids on rectangles Ωi (i = 1, 2, 3). The convergence of the discrete analog of the iterative method (2)-(6) was verified on some exact solutions for some sizes of the rectangular support and for some grid sizes. Performed experiments show that the convergence rate depends on the sizes e1, e2 (see Figure 2) and the values of the iteration parameters τ and θ. From the results of the experiments we observe that the values τ = 0.9 and θ = 0.95 give good convergence. The number of iterations for achieving the accuracy ‖u(k) − u‖∞ ≤ 10−4, where u is the exact solution, changes from 30 to 45. Using the chosen above iteration parameters τ and θ we solve the problem for computing the deflection of the symmetric rectangular support. As said in the end of the previous subsection, for the problem of bending of the plate gi = 0 (i = 0, 3). We perform numerical experiments for plate of the sizes pi× pi with the flexural rigidity D = 0.0057 under the load q = 0.3. The surfaces of deflection of the whole plate for some sizes of the support under uniform load are depicted in Figure 3(a) and 3(b). 3. PROBLEMS FOR PLATES ON H-SHAPE INTERNAL SUPPORT As in the case of a symmetric rectangular internal support, the problem for plate on a H-shape support (see Figure 4(a)) is reduced to boundary value problems with strongly mixed boundary conditions in a quadrant of the plate. For the latter one, the computational domain can also be divided into three subdomains (rectangles) as shown in Figure 4(b). The iterative method combining decrease of the equation order and domain decomposi- tion for these problems is constructed in an analogous way. The results of computation of deflection surfaces for some sizes of H-shape support are given in Figure 5(a) and 5(b). 310 TRUONG HA HAI et al. 0 0.2 0.4 0.6 0.8 1 0 0.5 1 1.5 2 −8 −6 −4 −2 0 2 x 10−3 x/pi Deflection surfaces y/(pi/2) u(x ,y) /(q a4 /1 03 D) (a) 0 0.2 0.4 0.6 0.8 1 0 0.5 1 1.5 2 −15 −10 −5 0 5 x 10−3 x/pi Deflection surfaces y/(pi/2) u(x ,y) /(q a4 /1 03 D) (b) Figure 3. The surfaces of deflection of the whole plate for e1/pi, e2/pi equal 0.30, 0.50 (a) and 0.25, 0.50 (b), respectively (a) H-shape support y O Ω1 Ω2 Ω3 x e2 e1 (b) First quadrant of H-shape support Figure 4. H-shape support and its first quadrant 4. PROBLEM FOR PLATE ON AN ASYMMETRIC RECTANGULAR INTERNAL SUPPORT 4.1. The problem setting Now we consider the plate on an asymmetric rectangular internal support when the support lies not in the middle of the plate. In this case the problem has no symmetry, so it cannot be reduced to one in a quadrant of the plate. In order to construct a solution method for the problem, as in the previous section, we consider it with general boundary conditions. Namely, consider the following BVP ∆2u = f in Ω, u = g0 on OF ∪HC ∪ FH ∪OC ∪MN ∪MP ∪ PQ ∪NQ, ∂u ∂ν = g1 on FH ∪OC ∪MN ∪MP ∪ PQ ∪NQ, ∆u = g2 on OF ∪HC. (7) NUMERICAL SOLUTION OF THE PROBLEMS 311 0 0.2 0.4 0.6 0.8 1 0 0.5 1 1.5 2 −15 −10 −5 0 5 x 10−3 x/pi Deflection surface y/(pi/2) u(x ,y) /(q a4 /1 03 D) (a) 0 0.2 0.4 0.6 0.8 1 0 0.5 1 1.5 2 −0.04 −0.03 −0.02 −0.01 0 0.01 x/pi Deflection surface y/(pi/2) u(x ,y) /(q a4 /1 03 D) (b) Figure 5. The surfaces of deflection of the plate on a H-shape support for e1/pi, e2/pi equal 0.15, 0.30 (a) and 0.30, 0.40 (b), respectively where Ω = ∪9i=1Ωi (the interior of the plate excluding the internal support). See Figure 6. O A B C D E F G I H K L Ω9 Γ72 Γ79 Ω7 Ω2 Γ42 Ω4 Γ41 Ω1 P Ω6Γ61 M Ω3 N Γ59 Ω5 Γ58 Q Γ68 Ω8 e11 e12 e21 e22 Figure 6. Domain decomposition for the problem for plate on asymmetric rectangular internal support 4.2. Solution method To solve the problem, the domain Ω is divided into 9 subdomains {Ωi, i = 1, 2, ...9} by the fictitious boundaries Γ41, Γ42, Γ58, Γ59, Γ61, Γ68, Γ72, Γ79 as described in Figure 6. 312 TRUONG HA HAI et al. As usual, we set: ui = u|Ωi , vi = ∆ui|Ωi , i = 1, 2, ..., 9. Further, set ξ41 = ∂v4 ∂ν |Γ41 , η41 = ∂u4 ∂ν |Γ41 , ξ42 = ∂v4 ∂ν |Γ42 , η42 = ∂u4 ∂ν |Γ42 , ξ58 = ∂v5 ∂ν |Γ58 , η58 = ∂u5 ∂ν |Γ58 , ξ59 = ∂v5 ∂ν |Γ59 , η59 = ∂u5 ∂ν |Γ59 , ξ72 = ∂v7 ∂ν |Γ72 , η72 = ∂u7 ∂ν |Γ72 , ξ79 = ∂v7 ∂ν |Γ79 , η79 = ∂u7 ∂ν |Γ79 , ξ61 = ∂v6 ∂ν |Γ61 , η61 = ∂u6 ∂ν |Γ61 , ξ68 = ∂v6 ∂ν |Γ68 , η68 = ∂u6 ∂ν |Γ68 , and set ϕ1 = v1|OA, ϕ2 = v2|FG, ϕ6 = v6|AB, ϕ7 = v7|GI , ϕ8 = v8|BC , ϕ9 = v9|IH , ϕ34 = v3|MP , ϕ43 = v4|MP , ϕ35 = v3|NQ, ϕ53 = v5|NQ, ϕ36 = v3|PQ, ϕ63 = v6|PQ, ϕ37 = v3|MN , ϕ73 = v7|MN , I = {1, 2, 6, 7, 8, 9, 34, 43, 35, 53, 36, 63, 37, 73} , J = {41, 42, 58, 59, 61, 68, 72, 79} . Consider the following parallel iterative method with the idea of simultaneous update of ξ, η, ϕ on boundaries: NUMERICAL SOLUTION OF THE PROBLEMS 313 1. Given starting approximations ϕ (0) i , i ∈ I; ξ(0)j , η(0)j , j ∈ J on respective boundaries, for example, ϕ (0) i = 0, ξ (0) j = 0, η (0) j = 0. 2. Knowing ϕ (k) i , ξ (k) j , η (k) j (k = 0, 1, 2, ...), solve in parallel five problems for v (k) 3 , u (k) 3 in Ω3, problems for v (k) 4 , u (k) 4 in Ω4, problems for v (k) 5 , u (k) 5 in Ω5, problems for v (k) 6 , u (k) 6 in Ω6, problems for v (k) 7 , u (k) 7 in Ω7: On domain Ω3 ∆v (k) 3 = f in Ω3, v (k) 3 = ϕ (k) 34 on MP, v (k) 3 = ϕ (k) 35 on NQ, v (k) 3 = ϕ (k) 36 on PQ, v (k) 3 = ϕ (k) 37 on MN, { ∆u (k) 3 = v (k) 3 in Ω3, u (k) 3 = g0 on ∂Ω3. (8) On domain Ω4 ∆v (k) 4 = f in Ω4, v (k) 4 = g2 on ED, v (k) 4 = ϕ (k) 43 on MP, ∂v (k) 4 ∂ν = ξ (k) 41 on Γ41, ∂v (k) 4 ∂ν = ξ (k) 42 on Γ42,  ∆u (k) 4 = v (k) 4 in Ω4, u (k) 4 = g0 on ED ∪MP, ∂u (k) 4 ∂ν = η (k) 41 on Γ41, ∂u (k) 4 ∂ν = η (k) 42 on Γ42. (9) On domain Ω5 ∆v (k) 5 = f in Ω5, v (k) 5 = g2 on KL, v (k) 5 = ϕ (k) 53 on NQ, ∂v (k) 5 ∂ν = ξ (k) 58 on Γ58, ∂v (k) 5 ∂ν = ξ (k) 59 on Γ59,  ∆u (k) 5 = v (k) 5 in Ω5, u (k) 5 = g0 on NQ ∪KL, ∂u (k) 5 ∂ν = η (k) 58 on Γ58, ∂u (k) 5 ∂ν = ξ (k) 59 on Γ59. (10) On domain Ω6 ∆v (k) 6 = f in Ω6, v (k) 6 = ϕ (k) 6 on AB, v (k) 6 = ϕ (k) 63 on PQ, ∂v (k) 6 ∂ν = ξ (k) 61 on Γ61, ∂v (k) 6 ∂ν = ξ (k) 68 on Γ68,  ∆u (k) 6 = v (k) 6 in Ω6, u (k) 6 = g0 on PQ ∪AB, ∂u (k) 6 ∂ν = η (k) 61 on Γ61, ∂u (k) 6 ∂ν = η (k) 68 on Γ68. (11) 314 TRUONG HA HAI et al. On domain Ω7 ∆v (k) 7 = f in Ω7, v (k) 7 = ϕ (k) 73 on MN, v (k) 7 = ϕ (k) 7 on GI, ∂v (k) 7 ∂ν = ξ (k) 72 on Γ72, ∂v (k) 7 ∂ν = ξ (k) 79 on Γ79,  ∆u (k) 7 = v (k) 7 in Ω7, u (k) 7 = g0 on GI ∪MN, ∂u (k) 7 ∂ν = η (k) 72 on Γ72, ∂u (k) 7 ∂ν = η (k) 79 on Γ79. (12) 3. Solve parallel problems for v (k) 1 , u (k) 1 in Ω1, problems for v (k) 2 , u (k) 2 in Ω2, problems for v (k) 8 , u (k) 8 in Ω8, problems for v (k) 9 , u (k) 9 in Ω9 On domain Ω1  ∆v (k) 1 = f in Ω1, v (k) 1 = g2 on OD ∪OA, v (k) 1 = v (k) 6 on Γ61, v (k) 1 = v (k) 4 on Γ41,  ∆u (k) 1 = v (k) 1 in Ω1, u (k) 1 = g0 on OD ∪OA, u (k) 1 = u (k) 4 on Γ41, u (k) 1 = u (k) 6 on Γ61. (13) On domain Ω2  ∆v (k) 2 = f in Ω2, v (k) 2 = g2 on EF ∪ FG, v (k) 2 = v (k) 4 on Γ42, v (k) 2 = v (k) 7 on Γ72,  ∆u (k) 2 = v (k) 2 in Ω2, u (k) 2 = g0 on EF ∪ FG, u (k) 2 = u (k) 4 on Γ41, u (k) 2 = u (k) 7 on Γ72. (14) On domain Ω8  ∆v (k) 8 = f in Ω8, v (k) 8 = g2 on BC ∪ LC, v (k) 8 = v (k) 6 on Γ68, v (k) 8 = v (k) 5 on Γ58,  ∆u (k) 8 = v (k) 8 in Ω8, u (k) 8 = g0 onBC ∪ CL, u (k) 8 = u (k) 6 on Γ68, u (k) 8 = u (k) 5 on Γ58. (15) On domain Ω9  ∆v (k) 9 = f in Ω9, v (k) 9 = g2 on IH ∪HK, v (k) 9 = v (k) 5 on Γ59, v (k) 9 = v (k) 7 on Γ79,  ∆u (k) 9 = v (k) 9 in Ω9, u (k) 9 = g0 on IH ∪HK, u (k) 9 = u (k) 5 on Γ59, u (k) 9 = u (k) 7 on Γ79. (16) NUMERICAL SOLUTION OF THE PROBLEMS 315 4. Calculate the new approximations ϕ (k+1) 1 = ϕ (k) 1 − τ (∂u(k)1 ∂ν − g1 ) on OA, ϕ (k+1) 2 = ϕ (k) 2 − τ (∂u(k)2 ∂ν − g1 ) on FG, ϕ (k+1) 6 = ϕ (k) 6 − τ (∂u(k)6 ∂ν − g1 ) on AB, ϕ (k+1) 7 = ϕ (k) 7 − τ (∂u(k)7 ∂ν − g1 ) on GI, ϕ (k+1) 8 = ϕ (k) 8 − τ (∂u(k)8 ∂ν − g1 ) on BC, ϕ (k+1) 9 = ϕ (k) 9 − τ (∂u(k)9 ∂ν − g1 ) on IH, ϕ (k+1) 34 = ϕ (k) 34 − τ (∂u(k)3 ∂ν − g1 ) on MP, ϕ (k+1) 43 = ϕ (k) 43 − τ (∂u(k)4 ∂ν − g1 ) on MP, ϕ (k+1) 35 = ϕ (k) 35 − τ (∂u(k)3 ∂ν − g1 ) on NQ, ϕ (k+1) 53 = ϕ (k) 53 − τ (∂u(k)5 ∂ν − g1 ) on NQ, ϕ (k+1) 37 = ϕ (k) 37 − τ (∂u(k