Bending of plates

Bending of plates, or plate bending, refers to the deflection of a plate perpendicular to the plane of the plate under the action of external forces and moments. The amount of deflection can be determined by solving the differential equations of an appropriate plate theory. The stresses in the plate can be calculated from these deflections. Once the stresses are known, failure theories can be used to determine whether a plate will fail under a given load.

For a thin rectangular plate of thickness ${\displaystyle H}$, Young's modulus ${\displaystyle E}$, and Poisson's ratio ${\displaystyle \nu }$, we can define parameters in terms of the plate deflection, ${\displaystyle w}$.

The flexural rigidity is given by

${\displaystyle D={\frac {EH^{3}}{12\left(1-\nu ^{2}\right)}}}$

The bending moments per unit length are given by

${\displaystyle M_{x}=-D\left({\frac {\partial ^{2}w}{\partial x^{2}}}+\nu {\frac {\partial ^{2}w}{\partial y^{2}}}\right)}$

${\displaystyle M_{y}=-D\left(\nu {\frac {\partial ^{2}w}{\partial x^{2}}}+{\frac {\partial ^{2}w}{\partial y^{2}}}\right)}$

The twisting moment per unit length is given by

${\displaystyle M_{xy}=-D\left(1-\nu \right){\frac {\partial ^{2}w}{\partial x\partial y}}}$

The shear forces per unit length are given by

${\displaystyle Q_{x}=-D{\frac {\partial }{\partial x}}\left({\frac {\partial ^{2}w}{\partial x^{2}}}+{\frac {\partial ^{2}w}{\partial y^{2}}}\right)}$

${\displaystyle Q_{y}=-D{\frac {\partial }{\partial y}}\left({\frac {\partial ^{2}w}{\partial x^{2}}}+{\frac {\partial ^{2}w}{\partial y^{2}}}\right)}$

The bending stresses are given by

${\displaystyle \sigma _{x}=-{\frac {12Dz}{H^{3}}}\left({\frac {\partial ^{2}w}{\partial x^{2}}}+\nu {\frac {\partial ^{2}w}{\partial y^{2}}}\right)}$

${\displaystyle \sigma _{y}=-{\frac {12Dz}{H^{3}}}\left(\nu {\frac {\partial ^{2}w}{\partial x^{2}}}+{\frac {\partial ^{2}w}{\partial y^{2}}}\right)}$

The shear stress is given by

${\displaystyle \tau _{xy}=-{\frac {12Dz}{H^{3}}}\left(1-\nu \right){\frac {\partial ^{2}w}{\partial x\partial y}}}$

The bending strains for small-deflection theory are given by

${\displaystyle \epsilon _{x}={\frac {\partial u}{\partial x}}=-z{\frac {\partial ^{2}w}{\partial x^{2}}}}$

${\displaystyle \epsilon _{y}={\frac {\partial v}{\partial y}}=-z{\frac {\partial ^{2}w}{\partial y^{2}}}}$

The shear strain for small-deflection theory is given by

${\displaystyle \gamma _{xy}={\frac {\partial u}{\partial y}}+{\frac {\partial v}{\partial x}}=-2z{\frac {\partial ^{2}w}{\partial x\partial y}}}$

For large-deflection plate theory, we consider the inclusion of membrane strains

${\displaystyle \epsilon _{x}={\frac {\partial u}{\partial x}}+{\frac {1}{2}}\left({\frac {\partial w}{\partial x}}\right)^{2}}$

${\displaystyle \epsilon _{y}={\frac {\partial v}{\partial y}}+{\frac {1}{2}}\left({\frac {\partial w}{\partial y}}\right)^{2}}$

${\displaystyle \gamma _{xy}={\frac {\partial u}{\partial y}}+{\frac {\partial v}{\partial x}}+{\frac {\partial w}{\partial x}}{\frac {\partial w}{\partial y}}}$

The deflections are given by

${\displaystyle u=-z{\frac {\partial w}{\partial x}}}$

${\displaystyle v=-z{\frac {\partial w}{\partial y}}}$

In the Kirchhoff–Love plate theory for plates the governing equations are^[1]

${\displaystyle N_{\alpha \beta ,\alpha }=0}$

and

${\displaystyle M_{\alpha \beta ,\alpha \beta }-q=0}$

In expanded form,

${\displaystyle {\cfrac {\partial N_{11}}{\partial x_{1}}}+{\cfrac {\partial N_{21}}{\partial x_{2}}}=0~;~~{\cfrac {\partial N_{12}}{\partial x_{1}}}+{\cfrac {\partial N_{22}}{\partial x_{2}}}=0}$

and

${\displaystyle {\cfrac {\partial ^{2}M_{11}}{\partial x_{1}^{2}}}+2{\cfrac {\partial ^{2}M_{12}}{\partial x_{1}\partial x_{2}}}+{\cfrac {\partial ^{2}M_{22}}{\partial x_{2}^{2}}}=q}$

where ${\displaystyle q(x)}$ is an applied transverse load per unit area, the thickness of the plate is ${\displaystyle H=2h}$, the stresses are ${\displaystyle \sigma _{ij}}$, and

${\displaystyle N_{\alpha \beta }:=\int _{-h}^{h}\sigma _{\alpha \beta }~dx_{3}~;~~M_{\alpha \beta }:=\int _{-h}^{h}x_{3}~\sigma _{\alpha \beta }~dx_{3}~.}$

The quantity ${\displaystyle N}$ has units of force per unit length. The quantity ${\displaystyle M}$ has units of moment per unit length.

For isotropic, homogeneous, plates with Young's modulus ${\displaystyle E}$ and Poisson's ratio ${\displaystyle \nu }$ these equations reduce to^[2]

${\displaystyle \nabla ^{2}\nabla ^{2}w=-{\cfrac {q}{D}}~;~~D:={\cfrac {2h^{3}E}{3(1-\nu ^{2})}}={\cfrac {H^{3}E}{12(1-\nu ^{2})}}}$

where ${\displaystyle w(x_{1},x_{2})}$ is the deflection of the mid-surface of the plate.

This is governed by the Germain-Lagrange plate equation

${\displaystyle {\cfrac {\partial ^{4}w}{\partial x^{4}}}+2{\cfrac {\partial ^{4}w}{\partial x^{2}\partial y^{2}}}+{\cfrac {\partial ^{4}w}{\partial y^{4}}}={\cfrac {q}{D}}}$

This equation was first derived by Lagrange in December 1811 in correcting the work of Germain who provided the basis of the theory.

This is governed by the Föppl–von Kármán plate equations

${\displaystyle {\cfrac {\partial ^{4}F}{\partial x^{4}}}+2{\cfrac {\partial ^{4}F}{\partial x^{2}\partial y^{2}}}+{\cfrac {\partial ^{4}F}{\partial y^{4}}}=E\left[\left({\cfrac {\partial ^{2}w}{\partial x\partial y}}\right)^{2}-{\cfrac {\partial ^{2}w}{\partial x^{2}}}{\cfrac {\partial ^{2}w}{\partial y^{2}}}\right]}$

${\displaystyle {\cfrac {\partial ^{4}w}{\partial x^{4}}}+2{\cfrac {\partial ^{4}w}{\partial x^{2}\partial y^{2}}}+{\cfrac {\partial ^{4}w}{\partial y^{4}}}={\cfrac {q}{D}}+{\cfrac {H}{D}}\left({\cfrac {\partial ^{2}F}{\partial y^{2}}}{\cfrac {\partial ^{2}w}{\partial x^{2}}}+{\cfrac {\partial ^{2}F}{\partial x^{2}}}{\cfrac {\partial ^{2}w}{\partial y^{2}}}-2{\cfrac {\partial ^{2}F}{\partial x\partial y}}{\cfrac {\partial ^{2}w}{\partial x\partial y}}\right)}$

where ${\displaystyle F}$ is the stress function.

The bending of circular plates can be examined by solving the governing equation with appropriate boundary conditions. These solutions were first found by Poisson in 1829. Cylindrical coordinates are convenient for such problems. Here ${\displaystyle z}$ is the distance of a point from the midplane of the plate.

The governing equation in coordinate-free form is

${\displaystyle \nabla ^{2}\nabla ^{2}w=-{\frac {q}{D}}\,.}$

In cylindrical coordinates ${\displaystyle (r,\theta ,z)}$,

${\displaystyle \nabla ^{2}w\equiv {\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial w}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}w}{\partial \theta ^{2}}}+{\frac {\partial ^{2}w}{\partial z^{2}}}\,.}$

For symmetrically loaded circular plates, ${\displaystyle w=w(r)}$, and we have

${\displaystyle \nabla ^{2}w\equiv {\frac {1}{r}}{\cfrac {d}{dr}}\left(r{\cfrac {dw}{dr}}\right)\,.}$

Therefore, the governing equation is

${\displaystyle {\frac {1}{r}}{\cfrac {d}{dr}}\left[r{\cfrac {d}{dr}}\left\{{\frac {1}{r}}{\cfrac {d}{dr}}\left(r{\cfrac {dw}{dr}}\right)\right\}\right]=-{\frac {q}{D}}\,.}$

If ${\displaystyle q}$ and ${\displaystyle D}$ are constant, direct integration of the governing equation gives us

${\displaystyle w(r)=-{\frac {qr^{4}}{64D}}+C_{1}\ln r+{\cfrac {C_{2}r^{2}}{2}}+{\cfrac {C_{3}r^{2}}{4}}(2\ln r-1)+C_{4}}$

where ${\displaystyle C_{i}}$ are constants. The slope of the deflection surface is

${\displaystyle \phi (r)={\cfrac {dw}{dr}}=-{\frac {qr^{3}}{16D}}+{\frac {C_{1}}{r}}+C_{2}r+C_{3}r\ln r\,.}$

For a circular plate, the requirement that the deflection and the slope of the deflection are finite at ${\displaystyle r=0}$ implies that ${\displaystyle C_{1}=0}$. However, ${\displaystyle C_{3}}$ need not equal 0, as the limit of ${\displaystyle r\ln r\,}$ exists as you approach ${\displaystyle r=0}$ from the right. However, if ${\displaystyle C_{3}\neq 0}$, the second derivative of ${\displaystyle w}$ will diverge at the center, corresponding to a point-like defect.

For a circular plate with clamped edges, we have ${\displaystyle w(a)=0}$ and ${\displaystyle \phi (a)=0}$ at the edge of the plate (radius ${\displaystyle a}$). Using these boundary conditions and setting ${\displaystyle C_{3}=0}$, we get

${\displaystyle w(r)=-{\frac {q}{64D}}(a^{2}-r^{2})^{2}\quad {\text{and}}\quad \phi (r)={\frac {qr}{16D}}(a^{2}-r^{2})\,.}$

The in-plane displacements in the plate are

${\displaystyle u_{r}(r)=-z\phi (r)\quad {\text{and}}\quad u_{\theta }(r)=0\,.}$

The in-plane strains in the plate are

${\displaystyle \varepsilon _{rr}={\cfrac {du_{r}}{dr}}=-{\frac {qz}{16D}}(a^{2}-3r^{2})~,~~\varepsilon _{\theta \theta }={\frac {u_{r}}{r}}=-{\frac {qz}{16D}}(a^{2}-r^{2})~,~~\varepsilon _{r\theta }=0\,.}$

The in-plane stresses in the plate are

${\displaystyle \sigma _{rr}={\frac {E}{1-\nu ^{2}}}\left[\varepsilon _{rr}+\nu \varepsilon _{\theta \theta }\right]~;~~\sigma _{\theta \theta }={\frac {E}{1-\nu ^{2}}}\left[\varepsilon _{\theta \theta }+\nu \varepsilon _{rr}\right]~;~~\sigma _{r\theta }=0\,.}$

For a plate of thickness ${\displaystyle 2h}$, the bending stiffness is ${\displaystyle D=2Eh^{3}/[3(1-\nu ^{2})]}$ and we have

${\displaystyle {\begin{aligned}\sigma _{rr}&=-{\frac {3qz}{32h^{3}}}\left[(1+\nu )a^{2}-(3+\nu )r^{2}\right]\\\sigma _{\theta \theta }&=-{\frac {3qz}{32h^{3}}}\left[(1+\nu )a^{2}-(1+3\nu )r^{2}\right]\\\sigma _{r\theta }&=0\,.\end{aligned}}}$

The moment resultants (bending moments) are

${\displaystyle M_{rr}=-{\frac {q}{16}}\left[(1+\nu )a^{2}-(3+\nu )r^{2}\right]~;~~M_{\theta \theta }=-{\frac {q}{16}}\left[(1+\nu )a^{2}-(1+3\nu )r^{2}\right]~;~~M_{r\theta }=0\,.}$

The maximum radial stress is at ${\displaystyle z=h}$ and ${\displaystyle r=a}$:

${\displaystyle \left.\sigma _{rr}\right|_{z=h,r=a}={\frac {3qa^{2}}{16h^{2}}}={\frac {3qa^{2}}{4H^{2}}}}$

where ${\displaystyle H:=2h}$. The bending moments at the boundary and the center of the plate are

${\displaystyle \left.M_{rr}\right|_{r=a}={\frac {qa^{2}}{8}}~,~~\left.M_{\theta \theta }\right|_{r=a}={\frac {\nu qa^{2}}{8}}~,~~\left.M_{rr}\right|_{r=0}=\left.M_{\theta \theta }\right|_{r=0}=-{\frac {(1+\nu )qa^{2}}{16}}\,.}$

For rectangular plates, Navier in 1820 introduced a simple method for finding the displacement and stress when a plate is simply supported. The idea was to express the applied load in terms of Fourier components, find the solution for a sinusoidal load (a single Fourier component), and then superimpose the Fourier components to get the solution for an arbitrary load.

Let us assume that the load is of the form

${\displaystyle q(x,y)=q_{0}\sin {\frac {\pi x}{a}}\sin {\frac {\pi y}{b}}\,.}$

Here ${\displaystyle q_{0}}$ is the amplitude, ${\displaystyle a}$ is the width of the plate in the ${\displaystyle x}$-direction, and ${\displaystyle b}$ is the width of the plate in the ${\displaystyle y}$-direction.

Since the plate is simply supported, the displacement ${\displaystyle w(x,y)}$ along the edges of the plate is zero, the bending moment ${\displaystyle M_{xx}}$ is zero at ${\displaystyle x=0}$ and ${\displaystyle x=a}$, and ${\displaystyle M_{yy}}$ is zero at ${\displaystyle y=0}$ and ${\displaystyle y=b}$.

If we apply these boundary conditions and solve the plate equation, we get the solution

${\displaystyle w(x,y)={\frac {q_{0}}{\pi ^{4}D}}\,\left({\frac {1}{a^{2}}}+{\frac {1}{b^{2}}}\right)^{-2}\,\sin {\frac {\pi x}{a}}\sin {\frac {\pi y}{b}}\,.}$

Where D is the flexural rigidity

${\displaystyle D={\frac {Et^{3}}{12(1-\nu ^{2})}}}$

Analogous to flexural stiffness EI.^[3] We can calculate the stresses and strains in the plate once we know the displacement.

For a more general load of the form

${\displaystyle q(x,y)=q_{0}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}}$

where ${\displaystyle m}$ and ${\displaystyle n}$ are integers, we get the solution

${\displaystyle {\text{(1)}}\qquad w(x,y)={\frac {q_{0}}{\pi ^{4}D}}\,\left({\frac {m^{2}}{a^{2}}}+{\frac {n^{2}}{b^{2}}}\right)^{-2}\,\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}\,.}$

We define a general load ${\displaystyle q(x,y)}$ of the following form

${\displaystyle q(x,y)=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }a_{mn}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}}$

where ${\displaystyle a_{mn}}$ is a Fourier coefficient given by

${\displaystyle a_{mn}={\frac {4}{ab}}\int _{0}^{b}\int _{0}^{a}q(x,y)\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}\,{\text{d}}x{\text{d}}y}$.

The classical rectangular plate equation for small deflections thus becomes:

${\displaystyle {\cfrac {\partial ^{4}w}{\partial x^{4}}}+2{\cfrac {\partial ^{4}w}{\partial x^{2}\partial y^{2}}}+{\cfrac {\partial ^{4}w}{\partial y^{4}}}={\cfrac {1}{D}}\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }a_{mn}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}}$

We assume a solution ${\displaystyle w(x,y)}$ of the following form

${\displaystyle w(x,y)=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }w_{mn}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}}$

The partial differentials of this function are given by

${\displaystyle {\cfrac {\partial ^{4}w}{\partial x^{4}}}=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\left({\frac {m\pi }{a}}\right)^{4}w_{mn}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}}$

${\displaystyle {\cfrac {\partial ^{4}w}{\partial x^{2}\partial y^{2}}}=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\left({\frac {m\pi }{a}}\right)^{2}\left({\frac {n\pi }{b}}\right)^{2}w_{mn}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}}$

${\displaystyle {\cfrac {\partial ^{4}w}{\partial y^{4}}}=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\left({\frac {n\pi }{b}}\right)^{4}w_{mn}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}}$

Substituting these expressions in the plate equation, we have

${\displaystyle \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\left(\left({\frac {m\pi }{a}}\right)^{2}+\left({\frac {n\pi }{b}}\right)^{2}\right)^{2}w_{mn}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{\cfrac {a_{mn}}{D}}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}}$

Equating the two expressions, we have

${\displaystyle \left(\left({\frac {m\pi }{a}}\right)^{2}+\left({\frac {n\pi }{b}}\right)^{2}\right)^{2}w_{mn}={\cfrac {a_{mn}}{D}}}$

which can be rearranged to give

${\displaystyle w_{mn}={\frac {1}{\pi ^{4}D}}{\frac {a_{mn}}{\left({\frac {m^{2}}{a^{2}}}+{\frac {n^{2}}{b^{2}}}\right)^{2}}}}$

The deflection of a simply-supported plate (of corner-origin) with general load is given by

${\displaystyle w(x,y)={\frac {1}{\pi ^{4}D}}\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{\frac {a_{mn}}{\left({\frac {m^{2}}{a^{2}}}+{\frac {n^{2}}{b^{2}}}\right)^{2}}}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}}$

Displacement (${\displaystyle w}$)

Stress (${\displaystyle \sigma _{xx}}$)

Stress (${\displaystyle \sigma _{yy}}$)

Displacement and stresses along ${\displaystyle x=a/2}$ for a rectangular plate with ${\displaystyle a=20}$ mm, ${\displaystyle b=40}$ mm, ${\displaystyle H=2h=0.4}$ mm, ${\displaystyle E=70}$ GPa, and ${\displaystyle \nu =0.35}$ under a load ${\displaystyle q_{0}=-10}$ kPa. The red line represents the bottom of the plate, the green line the middle, and the blue line the top of the plate.

For a uniformly-distributed load, we have

${\displaystyle q(x,y)=q_{0}}$

The corresponding Fourier coefficient is thus given by

${\displaystyle a_{mn}={\frac {4}{ab}}\int _{0}^{a}\int _{0}^{b}q_{0}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}\,{\text{d}}x{\text{d}}y}$.

Evaluating the double integral, we have

${\displaystyle a_{mn}={\frac {4q_{0}}{\pi ^{2}mn}}(1-\cos m\pi )(1-\cos n\pi )}$,

or alternatively in a piecewise format, we have

${\displaystyle a_{mn}={\begin{cases}{\cfrac {16q_{0}}{\pi ^{2}mn}}&m~{\text{and}}~n~{\text{odd}}\\0&m~{\text{or}}~n~{\text{even}}\end{cases}}}$

The deflection of a simply-supported plate (of corner-origin) with uniformly-distributed load is given by

${\displaystyle w(x,y)={\frac {16q_{0}}{\pi ^{6}D}}\sum _{m=1,3,5,...}^{\infty }\sum _{n=1,3,5,...}^{\infty }{\frac {1}{mn\left({\frac {m^{2}}{a^{2}}}+{\frac {n^{2}}{b^{2}}}\right)^{2}}}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}}$

The bending moments per unit length in the plate are given by

${\displaystyle M_{x}={\frac {16q_{0}}{\pi ^{4}}}\sum _{m=1,3,5,...}^{\infty }\sum _{n=1,3,5,...}^{\infty }{\frac {{\frac {m^{2}}{a^{2}}}+\nu {\frac {n^{2}}{b^{2}}}}{mn\left({\frac {m^{2}}{a^{2}}}+{\frac {n^{2}}{b^{2}}}\right)^{2}}}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}}$

${\displaystyle M_{y}={\frac {16q_{0}}{\pi ^{4}}}\sum _{m=1,3,5,...}^{\infty }\sum _{n=1,3,5,...}^{\infty }{\frac {{\frac {n^{2}}{b^{2}}}+\nu {\frac {m^{2}}{a^{2}}}}{mn\left({\frac {m^{2}}{a^{2}}}+{\frac {n^{2}}{b^{2}}}\right)^{2}}}\sin {\frac {m\pi x}{a}}\sin {\frac {n\pi y}{b}}}$

Another approach was proposed by Lévy^[4] in 1899. In this case we start with an assumed form of the displacement and try to fit the parameters so that the governing equation and the boundary conditions are satisfied. The goal is to find ${\displaystyle Y_{m}(y)}$ such that it satisfies the boundary conditions at ${\displaystyle y=0}$ and ${\displaystyle y=b}$ and, of course, the governing equation ${\displaystyle \nabla ^{2}\nabla ^{2}w=q/D}$.

Let us assume that

${\displaystyle w(x,y)=\sum _{m=1}^{\infty }Y_{m}(y)\sin {\frac {m\pi x}{a}}\,.}$

For a plate that is simply-supported along ${\displaystyle x=0}$ and ${\displaystyle x=a}$, the boundary conditions are ${\displaystyle w=0}$ and ${\displaystyle M_{xx}=0}$. Note that there is no variation in displacement along these edges meaning that ${\displaystyle \partial w/\partial y=0}$ and ${\displaystyle \partial ^{2}w/\partial y^{2}=0}$, thus reducing the moment boundary condition to an equivalent expression ${\displaystyle \partial ^{2}w/\partial x^{2}=0}$.

Consider the case of pure moment loading. In that case ${\displaystyle q=0}$ and ${\displaystyle w(x,y)}$ has to satisfy ${\displaystyle \nabla ^{2}\nabla ^{2}w=0}$. Since we are working in rectangular Cartesian coordinates, the governing equation can be expanded as

${\displaystyle {\frac {\partial ^{4}w}{\partial x^{4}}}+2{\frac {\partial ^{4}w}{\partial x^{2}\partial y^{2}}}+{\frac {\partial ^{4}w}{\partial y^{4}}}=0\,.}$

Plugging the expression for ${\displaystyle w(x,y)}$ in the governing equation gives us

${\displaystyle \sum _{m=1}^{\infty }\left[\left({\frac {m\pi }{a}}\right)^{4}Y_{m}\sin {\frac {m\pi x}{a}}-2\left({\frac {m\pi }{a}}\right)^{2}{\cfrac {d^{2}Y_{m}}{dy^{2}}}\sin {\frac {m\pi x}{a}}+{\frac {d^{4}Y_{m}}{dy^{4}}}\sin {\frac {m\pi x}{a}}\right]=0}$

or

${\displaystyle {\frac {d^{4}Y_{m}}{dy^{4}}}-2{\frac {m^{2}\pi ^{2}}{a^{2}}}{\cfrac {d^{2}Y_{m}}{dy^{2}}}+{\frac {m^{4}\pi ^{4}}{a^{4}}}Y_{m}=0\,.}$

This is an ordinary differential equation which has the general solution

${\displaystyle Y_{m}=A_{m}\cosh {\frac {m\pi y}{a}}+B_{m}{\frac {m\pi y}{a}}\cosh {\frac {m\pi y}{a}}+C_{m}\sinh {\frac {m\pi y}{a}}+D_{m}{\frac {m\pi y}{a}}\sinh {\frac {m\pi y}{a}}}$

where ${\displaystyle A_{m},B_{m},C_{m},D_{m}}$ are constants that can be determined from the boundary conditions. Therefore, the displacement solution has the form

${\displaystyle w(x,y)=\sum _{m=1}^{\infty }\left[\left(A_{m}+B_{m}{\frac {m\pi y}{a}}\right)\cosh {\frac {m\pi y}{a}}+\left(C_{m}+D_{m}{\frac {m\pi y}{a}}\right)\sinh {\frac {m\pi y}{a}}\right]\sin {\frac {m\pi x}{a}}\,.}$

Let us choose the coordinate system such that the boundaries of the plate are at ${\displaystyle x=0}$ and ${\displaystyle x=a}$ (same as before) and at ${\displaystyle y=\pm b/2}$ (and not ${\displaystyle y=0}$ and ${\displaystyle y=b}$). Then the moment boundary conditions at the ${\displaystyle y=\pm b/2}$ boundaries are

${\displaystyle w=0\,,-D{\frac {\partial ^{2}w}{\partial y^{2}}}{\Bigr |}_{y=b/2}=f_{1}(x)\,,-D{\frac {\partial ^{2}w}{\partial y^{2}}}{\Bigr |}_{y=-b/2}=f_{2}(x)}$

where ${\displaystyle f_{1}(x),f_{2}(x)}$ are known functions. The solution can be found by applying these boundary conditions. We can show that for the symmetrical case where

${\displaystyle M_{yy}{\Bigr |}_{y=-b/2}=M_{yy}{\Bigr |}_{y=b/2}}$

and

${\displaystyle f_{1}(x)=f_{2}(x)=\sum _{m=1}^{\infty }E_{m}\sin {\frac {m\pi x}{a}}}$

we have

${\displaystyle w(x,y)={\frac {a^{2}}{2\pi ^{2}D}}\sum _{m=1}^{\infty }{\frac {E_{m}}{m^{2}\cosh \alpha _{m}}}\,\sin {\frac {m\pi x}{a}}\,\left(\alpha _{m}\tanh \alpha _{m}\cosh {\frac {m\pi y}{a}}-{\frac {m\pi y}{a}}\sinh {\frac {m\pi y}{a}}\right)}$

where

${\displaystyle \alpha _{m}={\frac {m\pi b}{2a}}\,.}$

Similarly, for the antisymmetrical case where

${\displaystyle M_{yy}{\Bigr |}_{y=-b/2}=-M_{yy}{\Bigr |}_{y=b/2}}$

we have

${\displaystyle w(x,y)={\frac {a^{2}}{2\pi ^{2}D}}\sum _{m=1}^{\infty }{\frac {E_{m}}{m^{2}\sinh \alpha _{m}}}\,\sin {\frac {m\pi x}{a}}\,\left(\alpha _{m}\coth \alpha _{m}\sinh {\frac {m\pi y}{a}}-{\frac {m\pi y}{a}}\cosh {\frac {m\pi y}{a}}\right)\,.}$

We can superpose the symmetric and antisymmetric solutions to get more general solutions.

For a uniformly-distributed load, we have

${\displaystyle q(x,y)=q_{0}}$

The deflection of a simply-supported plate with centre ${\displaystyle \left({\frac {a}{2}},0\right)}$ with uniformly-distributed load is given by

${\displaystyle {\begin{aligned}&w(x,y)={\frac {q_{0}a^{4}}{D}}\sum _{m=1,3,5,...}^{\infty }\left(A_{m}\cosh {\frac {m\pi y}{a}}+B_{m}{\frac {m\pi y}{a}}\sinh {\frac {m\pi y}{a}}+G_{m}\right)\sin {\frac {m\pi x}{a}}\\\\&{\begin{aligned}{\text{where}}\quad &A_{m}=-{\frac {2\left(\alpha _{m}\tanh \alpha _{m}+2\right)}{\pi ^{5}m^{5}\cosh \alpha _{m}}}\\&B_{m}={\frac {2}{\pi ^{5}m^{5}\cosh \alpha _{m}}}\\&G_{m}={\frac {4}{\pi ^{5}m^{5}}}\\\\{\text{and}}\quad &\alpha _{m}={\frac {m\pi b}{2a}}\end{aligned}}\end{aligned}}}$

The bending moments per unit length in the plate are given by

${\displaystyle M_{x}=-q_{0}\pi ^{2}a^{2}\sum _{m=1,3,5,...}^{\infty }m^{2}\left(\left(\left(\nu -1\right)A_{m}+2\nu B_{m}\right)\cosh {\frac {m\pi y}{a}}+\left(\nu -1\right)B_{m}{\frac {m\pi y}{a}}\sinh {\frac {m\pi y}{a}}-G_{m}\right)\sin {\frac {m\pi x}{a}}}$

${\displaystyle M_{y}=-q_{0}\pi ^{2}a^{2}\sum _{m=1,3,5,...}^{\infty }m^{2}\left(\left(\left(1-\nu \right)A_{m}+2B_{m}\right)\cosh {\frac {m\pi y}{a}}+\left(1-\nu \right)B_{m}{\frac {m\pi y}{a}}\sinh {\frac {m\pi y}{a}}-\nu G_{m}\right)\sin {\frac {m\pi x}{a}}}$

For the special case where the loading is symmetric and the moment is uniform, we have at ${\displaystyle y=\pm b/2}$,

${\displaystyle M_{yy}=f_{1}(x)={\frac {4M_{0}}{\pi }}\sum _{m=1}^{\infty }{\frac {1}{2m-1}}\,\sin {\frac {(2m-1)\pi x}{a}}\,.}$

Displacement (${\displaystyle w}$)

Bending stress (${\displaystyle \sigma _{yy}}$)

Transverse shear stress (${\displaystyle \sigma _{yz}}$)

Displacement and stresses for a rectangular plate under uniform bending moment along the edges ${\displaystyle y=-b/2}$ and ${\displaystyle y=b/2}$. The bending stress ${\displaystyle \sigma _{yy}}$ is along the bottom surface of the plate. The transverse shear stress ${\displaystyle \sigma _{yz}}$ is along the mid-surface of the plate.

The resulting displacement is

${\displaystyle {\begin{aligned}&w(x,y)={\frac {2M_{0}a^{2}}{\pi ^{3}D}}\sum _{m=1}^{\infty }{\frac {1}{(2m-1)^{3}\cosh \alpha _{m}}}\sin {\frac {(2m-1)\pi x}{a}}\times \\&~~\left[\alpha _{m}\,\tanh \alpha _{m}\cosh {\frac {(2m-1)\pi y}{a}}-{\frac {(2m-1)\pi y}{a}}\sinh {\frac {(2m-1)\pi y}{a}}\right]\end{aligned}}}$

where

${\displaystyle \alpha _{m}={\frac {\pi (2m-1)b}{2a}}\,.}$

The bending moments and shear forces corresponding to the displacement ${\displaystyle w}$ are

${\displaystyle {\begin{aligned}M_{xx}&=-D\left({\frac {\partial ^{2}w}{\partial x^{2}}}+\nu \,{\frac {\partial ^{2}w}{\partial y^{2}}}\right)\\&={\frac {2M_{0}(1-\nu )}{\pi }}\sum _{m=1}^{\infty }{\frac {1}{(2m-1)\cosh \alpha _{m}}}\,\times \\&~\sin {\frac {(2m-1)\pi x}{a}}\,\times \\&~\left[-{\frac {(2m-1)\pi y}{a}}\sinh {\frac {(2m-1)\pi y}{a}}+\right.\\&\qquad \qquad \qquad \qquad \left.\left\{{\frac {2\nu }{1-\nu }}+\alpha _{m}\tanh \alpha _{m}\right\}\cosh {\frac {(2m-1)\pi y}{a}}\right]\\M_{xy}&=(1-\nu )D{\frac {\partial ^{2}w}{\partial x\partial y}}\\&=-{\frac {2M_{0}(1-\nu )}{\pi }}\sum _{m=1}^{\infty }{\frac {1}{(2m-1)\cosh \alpha _{m}}}\,\times \\&~\cos {\frac {(2m-1)\pi x}{a}}\,\times \\&~\left[{\frac {(2m-1)\pi y}{a}}\cosh {\frac {(2m-1)\pi y}{a}}+\right.\\&\qquad \qquad \qquad \qquad \left.(1-\alpha _{m}\tanh \alpha _{m})\sinh {\frac {(2m-1)\pi y}{a}}\right]\\Q_{zx}&={\frac {\partial M_{xx}}{\partial x}}-{\frac {\partial M_{xy}}{\partial y}}\\&={\frac {4M_{0}}{a}}\sum _{m=1}^{\infty }{\frac {1}{\cosh \alpha _{m}}}\,\times \\&~\cos {\frac {(2m-1)\pi x}{a}}\cosh {\frac {(2m-1)\pi y}{a}}\,.\end{aligned}}}$