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Coiflets solutions for Föppl-von Kármán equations governing large deflection of a thin flat plate by a novel wavelet-homotopy approach

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Abstract

In this paper, a novel technique incorporated the homotopy analysis method (HAM) with Coiflets is developed to obtain highly accurate solutions of the Föppl-von Kármán equations for large bending deflection. The characteristic scale transformation is introduced to nondimensionalize the governing equations. The results are obtained for the transformed nondimensional equations, which are in very excellent agreement with analytical ones or numerical benchmarks performing good efficiency and validity. Besides, we notice the nonlinearity of the Föppl-von Kármán equations is closely connected with the load and length-width ratio of the plate. For the case of the plate suffering tremendous loads, the traditional linear theory does not work, while our Coiflets solutions are still very accurate. It is expected that our proposed approach not only keeps the outstanding merits of the HAM technique for handling strong nonlinearity, but also improves on the computational efficiency to a great extent.

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Funding

This work is partially supported by the National Natural Science Foundation of China (Approval No. 11272209 and 11432009).

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Correspondence to Hang Xu.

Appendices

Appendix 1: derivative process of governing equations

In classic elastic mechanics, U, V, and W denote displacement of mass point on plate. Thus, the relationship between strain and displacement by neglecting all other higher order terms can be written as

$$\begin{array}{@{}rcl@{}} {\varepsilon_{X}^{m}}&=&\frac{\partial U}{\partial X}+\frac{1}{2}\left( \frac{\partial W}{\partial X}\right)^{2}\quad {\varepsilon_{X}^{b}}=-Z\frac{\partial^{2} W}{\partial X^{2}}, \end{array} $$
(54a)
$$\begin{array}{@{}rcl@{}} {\varepsilon_{Y}^{m}}&=&\frac{\partial V}{\partial Y}+\frac{1}{2}\left( \frac{\partial W}{\partial Y}\right)^{2}\quad {\varepsilon_{Y}^{b}}=-Z\frac{\partial^{2} W}{\partial Y^{2}}, \end{array} $$
(54b)
$$\begin{array}{@{}rcl@{}} \gamma_{XY}^{m}&=&\frac{\partial U}{\partial Y}+\frac{\partial V}{\partial X}+\frac{\partial W}{\partial X}\frac{\partial W}{\partial Y}\quad \gamma_{XY}^{b}=-2Z\frac{\partial^{2} W}{\partial X \partial Y}. \end{array} $$
(54c)

where ε is the normal strain, γ is the shearing strain, the superscript m and b means that the strains are the film strain and the bending strain respectively while total strain is the sum

$$\begin{array}{@{}rcl@{}} \varepsilon_{X}& =&{\varepsilon_{X}^{m}}+{\varepsilon_{X}^{b}}, \end{array} $$
(55a)
$$\begin{array}{@{}rcl@{}} \varepsilon_{Y}& =&{\varepsilon_{Y}^{m}}+{\varepsilon_{Y}^{b}}, \end{array} $$
(55b)
$$\begin{array}{@{}rcl@{}} \gamma_{XY}& =&\gamma_{XY}^{m}+\gamma_{XY}^{b}. \end{array} $$
(55c)

Eliminating the bending strain from above equations, we obtain

$$ \frac{\partial^{2}{\varepsilon_{X}^{m}}}{\partial X^{2}}+\frac{\partial^{2}{\varepsilon_{Y}^{m}}}{\partial Y^{2}}-\frac{\partial^{2}\gamma_{XY}^{m}}{\partial X \partial Y}=\left( \frac{\partial^{2}W}{\partial X \partial Y}\right)^{2}-\frac{\partial^{2}W}{\partial X^{2}}\frac{\partial^{2}W}{\partial Y^{2}}. $$
(56)

Based on the Hooke law, the bending and film stress are written as

$$\begin{array}{@{}rcl@{}} \sigma_{X}& =&{\sigma_{X}^{m}}+{\sigma_{X}^{b}}=\frac{E}{1-\mu^{2}}(\varepsilon_{X}+\mu\varepsilon_{Y}), \end{array} $$
(57a)
$$\begin{array}{@{}rcl@{}} \sigma_{Y}& =&{\sigma_{Y}^{m}}+{\sigma_{Y}^{b}}=\frac{E}{1-\mu^{2}}(\varepsilon_{Y}+\mu\varepsilon_{X}), \end{array} $$
(57b)
$$\begin{array}{@{}rcl@{}} \tau_{XY}=\tau_{YX}& =&\tau_{XY}^{m}+\tau_{XY}^{b}=\frac{E}{2(1+\mu)}\gamma_{XY}. \end{array} $$
(57c)

Assuming that the plate is homogeneous, the bending moments MX,MY and shear moment MXY are obtained, by integrating stress along the thickness direction, as

$$\begin{array}{@{}rcl@{}} M_{X}& =&{\int}_{-h/2}^{h/2}\sigma_{X}ZdZ=-D\left( \frac{\partial^{2} W}{\partial X^{2}}+\mu\frac{\partial^{2}W}{\partial Y^{2}}\right), \end{array} $$
(58a)
$$\begin{array}{@{}rcl@{}} M_{Y}& =&{\int}_{-h/2}^{h/2}\sigma_{Y}ZdZ=-D\left( \frac{\partial^{2} W}{\partial Y^{2}}+\mu\frac{\partial^{2}W}{\partial X^{2}}\right), \end{array} $$
(58b)
$$\begin{array}{@{}rcl@{}} M_{XY}=M_{YX}& =&{\int}_{-h/2}^{h/2}\tau_{XY}ZdZ=-D(1-\mu)\frac{\partial^{2} W}{\partial X\partial Y}. \end{array} $$
(58c)

The axial forces NX,NY and circumferential force S are given by

$$\begin{array}{@{}rcl@{}} N_{X}& =&{\int}_{-h/2}^{h/2}\sigma_{X}dZ=\frac{Eh}{1-\mu^{2}}({\varepsilon_{X}^{m}}+{\mu\varepsilon_{Y}^{m}})=h{\sigma_{X}^{m}}, \end{array} $$
(59a)
$$\begin{array}{@{}rcl@{}} N_{Y}& =&{\int}_{-h/2}^{h/2}\sigma_{Y}dZ=\frac{Eh}{1-\mu^{2}}({\varepsilon_{Y}^{m}}+{\mu\varepsilon_{X}^{m}})=h{\sigma_{Y}^{m}}, \end{array} $$
(59b)
$$\begin{array}{@{}rcl@{}} S& =&{\int}_{-h/2}^{h/2}\tau_{XY}dZ=\frac{Eh}{1+\mu}\left( \frac{\partial U}{\partial Y}+\frac{\partial V}{\partial X}+\frac{\partial W}{\partial X}\frac{\partial W}{\partial Y}\right)=h\tau_{XY}^{m}. \end{array} $$
(59c)

In order to keep the plate element in balance, the independent equilibrium equations due to force analysis are given, by neglecting small amount of high orders, in the following forms

$$\begin{array}{@{}rcl@{}} && \sum F_{X}= 0 \quad \frac{\partial N_{X}}{\partial X}+\frac{\partial S}{\partial Y}= 0, \end{array} $$
(60a)
$$\begin{array}{@{}rcl@{}} && \sum F_{Y}= 0 \quad \frac{\partial N_{Y}}{\partial Y}+\frac{\partial S}{\partial X}= 0, \end{array} $$
(60b)
$$\begin{array}{@{}rcl@{}} &&\sum F_{Z}= 0 \quad \frac{\partial Q_{X}}{\partial X}+\frac{\partial Q_{Y}}{\partial Y}+Q +N_{X}\frac{\partial^{2} W}{\partial X^{2}}+N_{Y}\frac{\partial^{2} W}{\partial Y^{2}}+ 2S\frac{\partial^{2} W}{\partial X\partial Y}= 0, \end{array} $$
(60c)
$$\begin{array}{@{}rcl@{}} && \sum M_{YOZ}= 0 \quad \frac{\partial M_{Y}}{\partial Y}+\frac{\partial M_{XY}}{\partial X}-Q_{X}= 0, \end{array} $$
(60d)
$$\begin{array}{@{}rcl@{}} && \sum M_{XOZ}= 0 \quad \frac{\partial M_{X}}{\partial X}+\frac{\partial M_{XY}}{\partial Y}-Q_{Y}= 0. \end{array} $$
(60e)

Substituting (58a-c) into (60d,e), we obtain the vertical forces QX,QY by integration in the following forms

$$\begin{array}{@{}rcl@{}} Q_{X}& =&{\int}_{-h/2}^{h/2}\tau_{XZ}dZ=-D\frac{\partial}{\partial X}\left( \frac{\partial^{2} W}{\partial X^{2}}+\frac{\partial^{2} W}{\partial Y^{2}}\right), \end{array} $$
(61a)
$$\begin{array}{@{}rcl@{}} Q_{Y}& =&{\int}_{-h/2}^{h/2}\tau_{YZ}dZ=-D\frac{\partial}{\partial Y}\left( \frac{\partial^{2} W}{\partial X^{2}}+\frac{\partial^{2} W}{\partial Y^{2}}\right). \end{array} $$
(61b)

To make simplification of the original Föppl-von Kármán equations, the Airy function Ψ is introduced satisfying

$$ N_{X}=h\frac{\partial^{2} {\Psi}}{\partial X^{2}},\quad N_{Y}=h\frac{\partial^{2} {\Psi}}{\partial Y^{2}},\quad S=-h\frac{\partial^{2}{\Psi}}{\partial X\partial Y}. $$
(62)

Substituting (61a, b) and (62) into (60c), (1a) can be obtained. On the other hand, from (59a-c) and (58a-c), we can find forces are only contributed by film stresses which are not related to bending stresses while moments are the opposite. we substitute film stresses expressed by Airy function Ψ into (57ac) to obtain film strains by combining (59a-c) and (62), then substitute the resulting equations into (56) with consideration of deformation compatibility condition, the full Föppl-von Kármán equations are finally given.

Appendix 2: symbolic definitions and test functions

Operator ∙ is given in order to emphasize multiplication of tensor matrix with straight vector and matrix A is a tensor product expressed \(\tilde {\mathbf {A}}\). \(\hat {\mathbf {G}}\) is straight vector of point value of G(x, y) and elements are coefficients of Coiflets series, which is used to estimate its derivatives vector \(\hat {\mathbf {G}}_{u,v}^{d,j}\) by (63) with tensors by resolution Coiflets.

$$ \hat{\mathbf{G}}_{u,v}^{d,j}=\left( \widetilde{\mathbf{\Phi}}_{u}^{j}\bigotimes\widetilde{\mathbf{\Phi}}_{v}^{j}\right)^{T}\bullet\hat{\mathbf{G}} $$
(63)

where straight vectors of point value and tensors are

$$\begin{array}{ll} &\widetilde{\mathbf{\Phi}}_{u}^{j}=\left\{ a_{k,s}^{j}=\phi_{j,k}^{(u)}\left( \frac{s}{2^{j}}\right)\right\}, \quad\hat{\mathbf{G}}=\left\{g_{o}=G\left( \frac{k}{2^{j}},\frac{l}{2^{j}}\right) \right\},\\ &\hat{\mathbf{G}}_{u,v}^{d}=\left\{{g_{p}^{d}}=\frac{\partial^{u+v} }{\partial x^{u}\partial y^{v}}G\left( \frac{s}{2^{j}},\frac{t}{2^{j}}\right)\right\},\\ &o = 2^{j}k+l + 1,p = 2^{j}s+t + 1, \quad k,l,s,t = 0\sim2^{j}. \end{array} $$

Definition 1 (Straight Vector)

If Matrix \(\mathbf {A}=\{a_{k,l}\}_{m\times n} \in \mathbb {R}^{m\times n} \), then its horizontal straight vector \(\mathbf {\hat {A}}\) and vertical straight vector \(\check {\mathbf {A}}\) are defined

$$\begin{array}{@{}rcl@{}} \mathbf{\hat{A}}&=&\{c_{r}\}_{mn\times1}\quad a_{k,l}=c_{l+(k-1)m}, \end{array} $$
(64)
$$\begin{array}{@{}rcl@{}} \mathbf{\check{A}}&=&\{c_{r}\}_{mn\times1}\quad a_{k,l}=c_{k+(l-1)m}. \end{array} $$
(65)

Definition 2 (Hadamard/Schur Product)

If Matrix A = {ak, l}m×n and \(\mathbf {B}=\{b_{k,l}\}_{m\times n} \in \mathbb {R}^{m\times n} \) (vectors when m = 1 and n = 1 ), then their Hadamard/Schur Product \(\bigodot \) is defined

$$ \mathbf{A}\bigodot\mathbf{B}=\{c_{k,l}=a_{k,l}b_{k,l}\}_{m\times n}. $$
(66)

Definition 3 (Kronecker Tensor Product)

If Matrix A = {ak, l}m×n and \(\mathbf {B}=\{b_{k,l}\}_{p\times q} \in \mathbb {R}^{p\times q} \) (vectors when m = 1 and n = 1 ), then their Kronecker Tensor Product \(\bigotimes \) is defined

$$ \mathbf{A}\bigotimes\mathbf{B}=\{a_{k,l}\mathbf{B}\}_{mp\times nq}. $$
(67)

Definition 4 (Dot Product)

If Matrix \(\mathbf {A}=\{a_{k,l}\}_{m\times n} \in \mathbb {R}^{m\times n} \) is a tensor product, as expressed \(\tilde {\mathbf {A}}\). \(\hat {\mathbf {B}}=\{b_{k}\}_{n\times 1}\in \mathbb {R}^{n\times 1}\) is straight vector of matrix \(\mathbf {B}=\{b_{k,l}\}_{p\times q}\in \mathbb {R}^{p\times q} \) while n = pq, then their matrix product ∙ is emphasized

$$ \tilde{\mathbf{A}}\bullet\hat{\mathbf{B}}=\{a_{k,l}b_{k}\}_{m\times1}. $$
(68)

In linear cases, test Function T(x, y) with exact solution U(x, y)

$$ \begin{array}{ll} T(x,y)&= 294912\left[x^{6} (5 y^{2}-5 y + 1)-3 x^{5} (5 y^{2}-5 y + 1)\right.\\ &+x^{4} (25 y^{4}-50 y^{3}+ 45 y^{2}-20 y + 3)\\ &+x^{3} (-50 y^{4}+ 100 y^{3}-65 y^{2}+ 15 y-1)\\ &+x^{2} y (5 y^{5}-15 y^{4}+ 45 y^{3}-65 y^{2}+ 36 y-6)\\ &\left.+x (-5 y^{6}+ 15 y^{5}-20 y^{4}+ 15 y^{3}-6 y^{2}+y)+(y-1)^{3} y^{3}\right].\\ U(x,y)&= 2^{12} (1-x)^{3} x^{3} (1-y)^{3} y^{3}. \end{array} $$
(69)

In nonlinear validation case, for simply supported boundary

$$ \begin{array}{ll} q_{1}(x,y)=&\frac{1}{8} \left[-9 \pi^{4} \cos (\pi x+\pi y)+ 25 \pi^{4} \cos (3 \pi x+\pi y)\right.\\ &-100 \pi^{4} \cos (\pi x + 2 \pi y)+ 25 \pi^{4} \cos (\pi x + 3 \pi y)\\ &-9 \pi^{4} \cos (3 \pi x + 3 \pi y)-9 \pi^{4} \cos (\pi x-\pi y)\\ &+ 25 \pi^{4} \cos (3 \pi x-\pi y)+ 100 \pi^{4} \cos (\pi x-2 \pi y)\\ &\left.+ 25 \pi^{4} \cos (\pi x-3 \pi y)-9 \pi^{4} \cos (3 \pi x-3 \pi y)\right]. \end{array} $$
(70)
$$ \begin{array}{ll} p_{1}(x,y)=&\frac{1}{50} \left[625\pi^{4} \cos(2\pi x+\pi y)+ 625\pi^{4}\cos(2\pi x-\pi y)\right.\\ &\left.-9\pi^{4}\cos(2\pi x)-9\pi^{4}\cos(4\pi y)\right] \end{array} $$
(71)

with the analytical solutions

$$ \begin{array}{l} w_{1}(x,y)=\sin (2\pi x) \sin (\pi y),\;\;\; \varphi_{1}(x,y)=\sin (\pi x) \sin (2\pi y). \end{array} $$
(72)

For circled clamped boundary

$$ \begin{array}{ll} q_{2}(x,y)&= 2\pi^{4} \cos(\pi x + 2 \pi y)-3\pi^{4} \cos (3\pi x + 2\pi y)\\ &+ 50\pi^{4} \cos (2\pi x + 4\pi y)-3 \pi^{4} \cos (\pi x + 6 \pi y)\\ &-2 \pi^{4} \cos (\pi x-2 \pi y)+ 3 \pi^{4} \cos (3\pi x-2\pi y)\\ &+ 50 \pi^{4} \cos (2\pi x-4\pi y)+ 3 \pi^{4} \cos (\pi x-6\pi y)\\ &-4\pi^{4} \cos(2\pi x)-64 \pi^{4} \cos (4\pi y), \end{array} $$
(73)
$$ \begin{array}{ll} p_{2}(x,y)&= \frac{1}{100} \left[-1250\pi^{4} \cos(\pi x + 2\pi y)+ 18\pi^{4} \cos(2\pi x + 4\pi y)\right.\\ &-9 \pi^{4}\cos (4 \pi x + 4\pi y)-9 \pi^{4} \cos (2\pi x + 8\pi y)\\ &+ 1250 \pi^{4} \cos (\pi x-2 \pi y)+ 18 \pi^{4} \cos (2\pi x-4\pi y)\\ &-9 \pi^{4} \cos (4\pi x-4\pi y)-9\pi^{4} \cos (2\pi x-8\pi y)\\ &-18 \pi^{4} \cos(2 \pi x)+ 18 \pi^{4} \cos (4\pi x)\\ &\left.-18\pi^{4} \cos (4\pi y)+ 18\pi^{4} \cos(8 \pi y)\right]. \end{array} $$
(74)

with the analytical solutions

$$ \begin{array}{l} w_{2}(x,y)=\frac{1}{4} (1-\cos 2 \pi x) (1-\cos 4 \pi y),\;\;\; \varphi_{2}(x,y)=\sin (\pi x) \sin (2 \pi y). \end{array} $$
(75)

and combined boundary corresponding to

$$ \begin{array}{ll} q_{3}(x,y)&=\frac{1}{16} \left[-11\pi^{4}\sin(\pi x+\pi y)-100 \pi^{4} \sin (2 \pi x+\pi y)\right.\\ &+ 25 \pi^{4} \sin (3 \pi x+\pi y)+ 27 \pi^{4} \sin (\pi x + 3 \pi y)\\ & -9 \pi^{4} \sin (3 \pi x + 3 \pi y)-11 \pi^{4} \sin (\pi x-\pi y)\\ &+ 100 \pi^{4} \sin (2 \pi x-\pi y)+ 25 \pi^{4} \sin (3 \pi x-\pi y)\\ &+ 27 \pi^{4} \sin (\pi x-3 \pi y)-9 \pi^{4} \sin (3 \pi x-3 \pi y)\\ &\left.+ 8 \pi^{4} \sin (\pi y)\right], \end{array} $$
(76)
$$ \begin{array}{ll} p_{3}(x,y)&= \frac{1}{400} \left[-5000 \pi^{4} \cos (\pi x + 2 \pi y)+ 9\pi^{4}\cos(2 \pi x + 2 \pi y)\right.\\ &+ 5000\pi^{4} \cos(\pi x-2 \pi y)+ 9\pi^{4} \cos (2\pi x-2\pi y)\\ &\left.-18 \pi^{4} \cos (2 \pi x)+ 18 \pi^{4} \cos (4 \pi x)-18 \pi^{4} \cos (2 \pi y)\right.] \end{array} $$
(77)

with the analytical solutions

$$ \begin{array}{l} w_{3}(x,y)=\frac{1}{2} (1-\cos 2 \pi x) \sin(\pi y),\;\;\; \varphi_{3}(x,y)=\sin (\pi x) \sin (2\pi y) \end{array} $$
(78)

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Yu, Q., Xu, H. & Liao, S. Coiflets solutions for Föppl-von Kármán equations governing large deflection of a thin flat plate by a novel wavelet-homotopy approach. Numer Algor 79, 993–1020 (2018). https://doi.org/10.1007/s11075-018-0470-x

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