Accurate bending analysis of rectangular thin plates with corner supports by a unified finite integral transform method

Abstract

In this paper, for the first time, the double finite integral transform method is introduced to explore the analytical bending solutions of rectangular thin plates with corner supports. Introducing the concept of generalized simply supported edge, the general form analytical solutions for bending of the plates under consideration are obtained by applying the finite integral transform to the governing equation and some of the boundary conditions. The analytical bending solutions of plates under specific boundary conditions are then obtained elegantly by imposing the remaining boundary conditions. Compared with the conventional inverse/semi-inverse methods, the present method is straightforward without any predetermination of solution forms, which makes it very attractive for exploring new analytical bending solutions of plates with complex boundary conditions. Another advantage of the method is that the analytical solutions obtained converge rapidly due to utilization of the sum functions. Comprehensive analytical results obtained in this paper illuminate the validity and accuracy of the present method by comparison with those from the literature and the finite element method.

Appendix: Sum functions used in the main text

$$\begin{aligned} \Phi _{1n} \left( x \right)= & {} \frac{2}{a}\sum _{m=1}^\infty {A_{mn} \sin \left( {\alpha _m x} \right) },\;\;\;\Phi _{2n} \left( x \right) =\frac{2}{a}\sum _{m=1}^\infty {\left( {-1} \right) ^{m}A_{mn} \sin \left( {\alpha _m x} \right) }, \nonumber \\ \Phi _{3n} \left( x \right)= & {} \frac{2}{a}\sum _{m=1}^\infty {B_{mn} \sin \left( {\alpha _m x} \right) },\;\;\;\Phi _{4n} \left( x \right) =\frac{2}{a}\sum _{m=1}^\infty {\left( {-1} \right) ^{m}B_{mn} \sin \left( {\alpha _m x} \right) }, \nonumber \\ \Psi _{1m} \left( y \right)= & {} \frac{2}{b}\sum _{n=1}^\infty {C_{mn} \sin \left( {\beta _n y} \right) },\;\;\;\Psi _{2m} \left( y \right) =\frac{2}{b}\sum _{n=1}^\infty {\left( {-1} \right) ^{n}C_{mn} \sin \left( {\beta _n y} \right) }, \nonumber \\ \Psi _{3m} \left( y \right)= & {} \frac{2}{b}\sum _{n=1}^\infty {D_{mn} \sin \left( {\beta _n y} \right) },\;\;\;\Psi _{4m} \left( y \right) =\frac{2}{b}\sum _{n=1}^\infty {\left( {-1} \right) ^{n}D_{mn} \sin \left( {\beta _n y} \right) }, \nonumber \\ R_n \left( x \right)= & {} \frac{2}{a}\sum _{m=1}^\infty {E_{mn} \sin \left( {\alpha _m x} \right) },\;\;\;\;S_m \left( y \right) =\frac{2}{b}\sum _{n=1}^\infty {E_{mn} \sin \left( {\beta _n y} \right) }, \end{aligned}$$

(A1)

where $A_{mn} =\frac{\alpha _m }{\left( {\alpha _m^2 +\beta _n^2 } \right) ^{2}}$, $B_{mn} =\frac{\alpha _m \left[ {\alpha _m^2 +\left( {2-\mu } \right) \beta _n^2 } \right] }{\left( {\alpha _m^2 +\beta _n^2 } \right) ^{2}}$, $C_{mn} =\frac{\beta _n }{\left( {\alpha _m^2 +\beta _n^2 } \right) ^{2}}$, $D_{mn} =\frac{\beta _n \left[ {\left( {2-\mu } \right) \alpha _m^2 +\beta _n^2 } \right] }{\left( {\alpha _m^2 +\beta _n^2 } \right) ^{2}}$ and $E_{mn} =\frac{q_{mn} }{D\left( {\alpha _m^2 +\beta _n^2 } \right) ^{2}}$. When $m=1,3,5,\ldots , \quad \Phi _{2n} \left( x \right) =-\Phi _{1n} \left( x \right)$, $\Phi _{4n} \left( x \right) =-\Phi _{3n} \left( x \right)$, and the sum functions of the sine series give

$$\begin{aligned} \Phi _{1n} \left( x \right)&=\frac{1}{4\beta _n^2 \cosh \frac{\delta _n }{2}}\left\{ {\frac{\delta _n }{2}\tanh \frac{\delta _n }{2}\cosh \left[ {\beta _n \left( {x-\frac{a}{2}} \right) } \right] -\beta _n \left( {x-\frac{a}{2}} \right) \sinh \left[ {\beta _n \left( {x-\frac{a}{2}} \right) } \right] } \right\} , \end{aligned}$$

(A2)

$$\begin{aligned} \Phi _{3n} \left( x \right)&=\frac{1-\mu }{4\cosh \frac{\delta _n }{2}}\left\{ {\left( {\frac{\delta _n }{2}\tanh \frac{\delta _n }{2}+\frac{2}{1-\mu }} \right) \cosh \left[ {\beta _n \left( {x-\frac{a}{2}} \right) } \right] -\beta _n \left( {x-\frac{a}{2}} \right) \sinh \left[ {\beta _n \left( {x-\frac{a}{2}} \right) } \right] } \right\} , \end{aligned}$$

(A3)

where $\delta _n ={an \pi }/b$. When $m=2,4,6,\ldots , \quad \Phi _{2n} \left( x \right) =\Phi _{1n} \left( x \right) , \quad \Phi _{4n} \left( x \right) =\Phi _{3n} \left( x \right)$, and the sum functions give

$$\begin{aligned} \Phi _{1n} \left( x \right)= & {} -\frac{1}{4\beta _n^2 \sinh \frac{\delta _n }{2}}\left\{ {\frac{\delta _n }{2}\coth \frac{\delta _n }{2}\sinh \left[ {\beta _n \left( {x-\frac{a}{2}} \right) } \right] -\beta _n \left( {x-\frac{a}{2}} \right) \cosh \left[ {\beta _n \left( {x-\frac{a}{2}} \right) } \right] } \right\} , \end{aligned}$$

(A4)

$$\begin{aligned} \Phi _{3n} \left( x \right)= & {} -\frac{1-\mu }{4\sinh \frac{\delta _n }{2}}\left\{ {\left( {\frac{\delta _n }{2}\coth \frac{\delta _n }{2}+\frac{2}{1-\mu }} \right) \sinh \left[ {\beta _n \left( {x-\frac{a}{2}} \right) } \right] -\beta _n \left( {x-\frac{a}{2}} \right) \cosh \left[ {\beta _n \left( {x-\frac{a}{2}} \right) } \right] } \right\} . \end{aligned}$$

(A5)

For uniform external load $q\left( {x,y} \right) =q_0$, $q_{mn} =4\big /{\left( {\alpha _m \beta _n } \right) } \left( {m,n=1,3,5,\ldots } \right)$, and the sum function gives

$$\begin{aligned} R_n \left( x \right)= & {} \frac{2}{a}\sum _{m=1,3,5,\ldots }^\infty {E_{mn} } \sin \left( {\alpha _m x} \right) \nonumber \\= & {} -\frac{q_0 }{D\beta _n^5 }\left\{ {\cosh \frac{\delta _n }{2}\left[ {\begin{array}{l} \left( {\frac{\delta _n }{2}\tanh \frac{\delta _n }{2}+2} \right) \cosh \left[ {\beta _n \left( {x-\frac{a}{2}} \right) } \right] \\ -\beta _n \left( {x-\frac{a}{2}} \right) \sinh \left[ {\beta _n \left( {x-\frac{a}{2}} \right) } \right] \\ \end{array}} \right] -2} \right\} . \end{aligned}$$

(A6)

The expressions of $\Psi _{1m} \left( y \right) -\Psi _{4m} \left( y \right)$ and $S_m \left( y \right)$ can be obtained by replacing the items of $\Phi _{1n} \left( x \right) -\Phi _{4n} \left( x \right)$ and $R_n \left( x \right)$ according to the following rules: $x\rightarrow y$, $m\rightarrow n$, $a\rightarrow b$, $\beta _n \rightarrow \alpha _m$, $\delta _n ={an\pi }/b\rightarrow \gamma _m ={bm\pi }/a$.