bending of plates

Image:PlateForcesMoments upd.png

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

The flexural rigidity is given by

:$$

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

The bending moments per unit length are given by

:$$

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

:$$

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

:$$

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

The shear forces per unit length are given by

:$$

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

:$$

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

:$$

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

:$$

\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

:$$

\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

:$$

\epsilon_{x} = \frac{\partial u}{\partial x} = -z\frac{\partial^2 w}{\partial x^2}

:$$

\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

:$$

\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

:$$

\epsilon_{x} = \frac{\partial u}{\partial x} + \frac{1}{2}\left(\frac{\partial w}{\partial x}\right)^2

:$$

\epsilon_{y} = \frac{\partial v}{\partial y} + \frac{1}{2}\left(\frac{\partial w}{\partial y}\right)^2

:$$

\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

:$$

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

:$$

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

In the Kirchhoff–Love plate theory for plates the governing equations areReddy, J. N., 2007, Theory and analysis of elastic plates and shells, CRC Press, Taylor and Francis.

:$$

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

and

:$$

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

In expanded form,

:$$

\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

:$$

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

:$$

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 $N$ has units of force per unit length. The quantity $M$ has units of moment per unit length.

For isotropic, homogeneous, plates with Young's modulus $E$ and Poisson's ratio $\backslash nu$ these equations reduce toTimoshenko, S. and Woinowsky-Krieger, S., (1959), Theory of plates and shells, McGraw-Hill New York.

:$$

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

where $w(x\_1,x\_2)$ is the deflection of the mid-surface of the plate.

This is governed by the Germain-Lagrange plate equation

:$$

\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

:$$

\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]

:$$

\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 $F$ 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 $z$ is the distance of a point from the midplane of the plate.

The governing equation in coordinate-free form is

:$$

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

In cylindrical coordinates $(r,\; \backslash theta,\; z)$,

:$$

\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, $w\; =\; w(r)$, and we have

:$$

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

Therefore, the governing equation is

:$$

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

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

:$$

w(r) = -\frac{qr^4}{64 D} + C_1\ln r + \cfrac{C_2 r^2}{2} + \cfrac{C_3r^2}{4}(2\ln r - 1) + C_4

where $C\_i$ are constants. The slope of the deflection surface is

:$$

\phi(r) = \cfrac{d w}{d r} = -\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 $r\; =\; 0$ implies that $C\_1\; =\; 0$. However, $C\_3$ need not equal 0, as the limit

of $r\; \backslash ln\; r\backslash ,$ exists as you approach $r\; =\; 0$ from the right.

For a circular plate with clamped edges, we have $w(a)\; =\; 0$ and $\backslash phi(a)\; =\; 0$ at the edge of

the plate (radius $a$). Using these boundary conditions we get

:$$

w(r) = -\frac{q}{64 D} (a^2 -r^2)^2 \quad \text{and} \quad

\phi(r) = \frac{qr}{16 D}(a^2-r^2) \,.

The in-plane displacements in the plate are

:$$

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

The in-plane strains in the plate are

:$$

\varepsilon_{rr} = \cfrac{d u_r}{d r} = -\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

:$$

\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 $2h$, the bending stiffness is $D\; =\; 2Eh^3/[3(1-\backslash nu^2)]$ and we

have

:$$

\begin{align}

\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{align}

The moment resultants (bending moments) are

:$$

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 $z\; =\; h$ and $r\; =\; a$:

:$$

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

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

:$$

\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} \,.

Image:RectangularPlateBending.svg

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

:$$

q(x,y) = q_0 \sin\frac{\pi x}{a}\sin\frac{\pi y}{b} \,.

Here $q\_0$ is the amplitude, $a$ is the width of the plate in the $x$-direction, and

$b$ is the width of the plate in the $y$-direction.

Since the plate is simply supported, the displacement $w(x,y)$ along the edges of

the plate is zero, the bending moment $M\_\{xx\}$ is zero at $x=0$ and $x=a$, and

$M\_\{yy\}$ is zero at $y=0$ and $y=b$.

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

solution

:$$

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

:$$

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

Analogous to flexural stiffness EI.Cook, R. D. et al., 2002, Concepts and applications of finite element analysis, John Wiley & Sons We can calculate the stresses and strains in the plate once we know the displacement.

For a more general load of the form

:$$

q(x,y) = q_0 \sin\frac{m \pi x}{a}\sin\frac{n \pi y}{b}

where $m$ and $n$ are integers, we get the solution

:$\backslash text\{(1)\}\; \backslash qquad$

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 $q(x,y)$ of the following form

:$$

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 $a\_\{mn\}$ is a Fourier coefficient given by

:$$

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:

:$$

\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 $w(x,y)$ of the following form

:$$

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

:$$

\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}

:$$

\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}

:$$

\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

:$$

\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

:$$

\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

:$$

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

:$$

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}

:{{multiple image

| align = right

| direction = vertical

| width = 200

| footer = Displacement and stresses along $x=a/2$ for a rectangular plate with $a=20$ mm, $b=40$ mm, $H=2h=0.4$ mm, $E=70$ GPa, and $\backslash nu=0.35$ under a load $q\_0\; =\; -10$ kPa. The red line represents the bottom of the plate, the green line the middle, and the blue line the top of the plate.

| image1 = wx rectangularPlate.svg

| caption1 = Displacement ($w$)

| image2 = sxx rectangularPlate.svg

| caption2 = Stress ($\backslash sigma\_\{xx\}$)

| image3 = syy rectangularPlate.svg

| caption3 = Stress ($\backslash sigma\_\{yy\}$)

}}

For a uniformly-distributed load, we have

:$$

q(x,y) = q_0

The corresponding Fourier coefficient is thus given by

:$$

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

:$$

a_{mn} = \frac{4q_0}{\pi^2 mn}(1 - \cos m\pi)(1 - \cos n\pi)

,

or alternatively in a piecewise format, we have

:$$

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

:$$

w(x,y) = \frac{16 q_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

:$$

M_{x} = \frac{16 q_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}

:$$

M_{y} = \frac{16 q_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 LévyLévy, M., 1899, Comptes rendues, vol. 129, pp. 535-539 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 $Y\_m(y)$ such that it satisfies the boundary conditions at $y\; =\; 0$ and $y\; =\; b$ and, of course, the governing equation $\backslash nabla^2\; \backslash nabla^2\; w\; =\; q/D$.

Let us assume that

:$$

w(x,y) = \sum_{m=1}^\infty Y_m(y) \sin \frac{m\pi x}{a} \,.

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

Consider the case of pure moment loading. In that case $q\; =\; 0$ and

$w(x,y)$ has to satisfy $\backslash nabla^2\; \backslash nabla^2\; w\; =\; 0$. Since we are working in rectangular

Cartesian coordinates, the governing equation can be expanded as

:$$

\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 $w(x,y)$ in the governing equation gives us

:$$

\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}{d y^2} \sin\frac{m\pi x}{a}

+ \frac{d^4Y_m}{dy^4} \sin\frac{m\pi x}{a}\right] = 0

or

:$$

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

This is an ordinary differential equation which has the general solution

:$$

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 $A\_m,\; B\_m,\; C\_m,\; D\_m$ are constants that can be determined from the boundary

conditions. Therefore, the displacement solution has the form

:$$

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 $x\; =\; 0$ and $x\; =\; a$ (same as before) and at $y\; =\; \backslash pm\; b/2$ (and not $y=0$ and

$y=b$). Then the moment boundary conditions at the $y\; =\; \backslash pm\; b/2$ boundaries are

:$$

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 $f\_1(x),\; f\_2(x)$ are known functions. The solution can be found by

applying these boundary conditions. We can show that for the symmetrical case

where

:$$

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

and

:$$

f_1(x) = f_2(x) = \sum_{m=1}^\infty E_m\sin\frac{m\pi x}{a}

we have

:$$

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

:$$

\alpha_m = \frac{m\pi b}{2a} \,.

Similarly, for the antisymmetrical case where

:$$

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

we have

:$$

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

:$$

q(x,y) = q_0

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

:$$

\begin{align}

&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{align}

\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{align}

\end{align}

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

:$$

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}

:$$

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 $y=\backslash pm\; b/2$,

:$$

M_{yy} = f_1(x) = \frac{4M_0}{\pi}\sum_{m=1}^\infty \frac{1}{2m-1}\,\sin\frac{(2m-1)\pi x}{a} \,.

:{{multiple image

| align = right

| direction = vertical

| width = 200

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

| image1 = surfRecBMIso_w.png

| caption1 = Displacement ($w$)

| image2 = surfRecBMIso_sy.png

| caption2 = Bending stress ($\backslash sigma\_\{yy\}$)

| image3 = surfRecBMIso_syz.png

| caption3 = Transverse shear stress ($\backslash sigma\_\{yz\}$)

}}

The resulting displacement is

:$$

\begin{align}

& 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{align}

where

:$$

\alpha_m = \frac{\pi (2m-1)b}{2a} \,.

The bending moments and shear forces corresponding to the displacement $w$ are

:$$

\begin{align}

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{align}

The stresses are

:$$

\sigma_{xx} = \frac{12z}{h^3}\,M_{xx} \quad \text{and} \quad

\sigma_{zx} = \frac{1}{\kappa h}\,Q_{zx}\left(1 - \frac{4z^2}{h^2}\right)\,.

Cylindrical bending occurs when a rectangular plate that has dimensions $a\; \backslash times\; b\; \backslash times\; h$, where $a\; \backslash ll\; b$ and the thickness $h$ is small, is subjected to a uniform distributed load perpendicular to the plane of the plate. Such a plate takes the shape of the surface of a cylinder.

For a simply supported plate under cylindrical bending with edges that are free to rotate but have a fixed $x\_1$. Cylindrical bending solutions can be found using the Navier and Levy techniques.

For thick plates, we have to consider the effect of through-the-thickness shears on the orientation of the normal to the mid-surface after deformation. Raymond D. Mindlin's theory provides one approach for find the deformation and stresses in such plates. Solutions to Mindlin's theory can be derived from the equivalent Kirchhoff-Love solutions using canonical relations.Lim, G. T. and Reddy, J. N., 2003, On canonical bending relationships for plates, International Journal of Solids and Structures, vol. 40,

pp. 3039-3067.

The canonical governing equation for isotropic thick plates can be expressed as

:$$

\begin{align}

& \nabla^2 \left(\mathcal{M} - \frac{\mathcal{B}}{1+\nu}\,q\right) = -q \\

& \kappa G h\left(\nabla^2 w + \frac{\mathcal{M}}{D}\right) =

-\left(1 - \cfrac{\mathcal{B} c^2}{1+\nu}\right)q \\

& \nabla^2 \left(\frac{\partial \varphi_1}{\partial x_2} - \frac{\partial \varphi_2}{\partial x_1}\right)

= c^2\left(\frac{\partial \varphi_1}{\partial x_2} - \frac{\partial \varphi_2}{\partial x_1}\right)

\end{align}

where $q$ is the applied transverse load, $G$ is the shear modulus, $D\; =\; Eh^3/[12(1-\backslash nu^2)]$

is the bending rigidity, $h$ is the plate thickness, $c^2\; =\; 2\backslash kappa\; G\; h/[D(1-\backslash nu)]$,

$\backslash kappa$ is the shear correction factor, $E$ is the Young's modulus, $\backslash nu$ is the Poisson's

ratio, and

:$$

\mathcal{M} = D\left[\mathcal{A}\left(\frac{\partial \varphi_1}{\partial x_1} + \frac{\partial \varphi_2}{\partial x_2}\right)

- (1-\mathcal{A})\nabla^2 w\right] + \frac{2q}{1-\nu^2}\mathcal{B} \,.

In Mindlin's theory, $w$ is the transverse displacement of the mid-surface of the plate

and the quantities $\backslash varphi\_1$ and $\backslash varphi\_2$ are the rotations of the mid-surface normal

about the $x\_2$ and $x\_1$-axes, respectively. The canonical parameters for this theory

are $\backslash mathcal\{A\}\; =\; 1$ and $\backslash mathcal\{B\}\; =\; 0$. The shear correction factor $\backslash kappa$ usually has the

value $5/6$.

The solutions to the governing equations can be found if one knows the corresponding

Kirchhoff-Love solutions by using the relations

:$$

\begin{align}

w & = w^K + \frac{\mathcal{M}^K}{\kappa G h}\left(1 - \frac{\mathcal{B} c^2}{2}\right)

- \Phi + \Psi \\

\varphi_1 & = - \frac{\partial w^K}{\partial x_1}

- \frac{1}{\kappa G h}\left(1 - \frac{1}{\mathcal{A}} - \frac{\mathcal{B} c^2}{2}\right)Q_1^K

+ \frac{\partial }{\partial x_1}\left(\frac{D}{\kappa G h \mathcal{A}}\nabla^2 \Phi + \Phi - \Psi\right)

+ \frac{1}{c^2}\frac{\partial \Omega}{\partial x_2} \\

\varphi_2 & = - \frac{\partial w^K}{\partial x_2}

- \frac{1}{\kappa G h}\left(1 - \frac{1}{\mathcal{A}} - \frac{\mathcal{B} c^2}{2}\right)Q_2^K

+ \frac{\partial }{\partial x_2}\left(\frac{D}{\kappa G h \mathcal{A}}\nabla^2 \Phi + \Phi - \Psi\right)

+ \frac{1}{c^2}\frac{\partial \Omega}{\partial x_1}

\end{align}

where $w^K$ is the displacement predicted for a Kirchhoff-Love plate, $\backslash Phi$ is a

biharmonic function such that $\backslash nabla^2\; \backslash nabla^2\; \backslash Phi\; =\; 0$, $\backslash Psi$ is a function that satisfies the

Laplace equation, $\backslash nabla^2\; \backslash Psi\; =\; 0$, and

:$$

\begin{align}

\mathcal{M} & = \mathcal{M}^K + \frac{\mathcal{B}}{1+\nu}\,q + D \nabla^2 \Phi ~;~~ \mathcal{M}^K := -D\nabla^2 w^K \\

Q_1^K & = -D\frac{\partial }{\partial x_1}\left(\nabla^2 w^K\right) ~,~~

Q_2^K = -D\frac{\partial }{\partial x_2}\left(\nabla^2 w^K\right) \\

\Omega & = \frac{\partial \varphi_1}{\partial x_2} - \frac{\partial \varphi_2}{\partial x_1} ~,~~ \nabla^2 \Omega = c^2\Omega \,.

\end{align}

For simply supported plates, the Marcus moment sum vanishes, i.e.,

:$$

\mathcal{M} = \frac{1}{1+\nu}(M_{11}+M_{22}) = D\left(\frac{\partial \varphi_1}{\partial x_1}+\frac{\partial \varphi_2}{\partial x_2}\right) = 0 \,.

Which is almost Laplace`s equation for w[ref 6]. In that case the functions $\backslash Phi$, $\backslash Psi$, $\backslash Omega$ vanish, and the Mindlin solution is

related to the corresponding Kirchhoff solution by

:$$

w = w^K + \frac{\mathcal{M}^K}{\kappa G h} \,.

Reissner-Stein theory for cantilever platesE. Reissner and M. Stein. Torsion and transverse bending of cantilever plates. Technical Note 2369, National Advisory Committee for Aeronautics,Washington, 1951. leads to the following coupled ordinary differential equations for a cantilever plate with concentrated end load $q\_x(y)$ at $x=a$.

:$$

\begin{align}

& bD \frac{\mathrm{d}^4w_x}{\mathrm{d}x^4} = 0 \\

& \frac{b^3D}{12}\,\frac{\mathrm{d}^4\theta_x}{\mathrm{d}x^4} - 2bD(1-\nu)\cfrac{d^2 \theta_x}{d x^2} = 0

\end{align}

and the boundary conditions at $x=a$ are

:$$

\begin{align}

& bD\cfrac{d^3 w_x}{d x^3} + q_{x1} = 0 \quad,\quad

\frac{b^3D}{12}\cfrac{d^3 \theta_x}{d x^3} -2bD(1-\nu)\cfrac{d \theta_x}{d x} + q_{x2} = 0 \\

& bD\cfrac{d^2 w_x}{d x^2} = 0 \quad,\quad \frac{b^3D}{12}\cfrac{d^2 \theta_x}{d x^2} = 0 \,.

\end{align}

Solution of this system of two ODEs gives

:$$

\begin{align}

w_x(x) & = \frac{q_{x1}}{6bD}\,(3ax^2 -x^3) \\

\theta_x(x) & = \frac{q_{x2}}{2bD(1-\nu)}\left[x - \frac{1}{\nu_b}\,

\left(\frac{\sinh(\nu_b a)}{\cosh[\nu_b (x-a)]} + \tanh[\nu_b(x-a)]\right)\right]

\end{align}

where $\backslash nu\_b\; =\; \backslash sqrt\{24(1-\backslash nu)\}/b$. The bending moments and shear forces corresponding to the displacement

$w\; =\; w\_x\; +\; y\backslash theta\_x$ are

:$$

\begin{align}

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

& = q_{x1}\left(\frac{x-a}{b}\right) - \left[\frac{3yq_{x2}}{b^3\nu_b\cosh^3[\nu_b(x-a)]}\right]

\times \\

& \quad \left[6\sinh(\nu_b a) - \sinh[\nu_b(2x-a)] +

\sinh[\nu_b(2x-3a)] + 8\sinh[\nu_b(x-a)]\right] \\

M_{xy} & = (1-\nu)D\frac{\partial^2 w}{\partial x \partial y} \\

& = \frac{q_{x2}}{2b}\left[1 -

\frac{2+\cosh[\nu_b(x-2a)] - \cosh[\nu_b x]}{2\cosh^2[\nu_b(x-a)]}\right] \\

Q_{zx} & = \frac{\partial M_{xx}}{\partial x}-\frac{\partial M_{xy}}{\partial y} \\

& = \frac{q_{x1}}{b} - \left(\frac{3yq_{x2}}{2b^3\cosh^4[\nu_b(x-a)]}\right)\times

\left[32 + \cosh[\nu_b(3x-2a)] - \cosh[\nu_b(3x-4a)]\right. \\

& \qquad \left. - 16\cosh[2\nu_b(x-a)] +

23\cosh[\nu_b(x-2a)] - 23\cosh(\nu_b x)\right]\,.

\end{align}

The stresses are

:$$

\sigma_{xx} = \frac{12z}{h^3}\,M_{xx} \quad \text{and} \quad

\sigma_{zx} = \frac{1}{\kappa h}\,Q_{zx}\left(1 - \frac{4z^2}{h^2}\right)\,.

If the applied load at the edge is constant, we recover the solutions for a beam under a

concentrated end load. If the applied load is a linear function of $y$, then

:$$

q_{x1} = \int_{-b/2}^{b/2}q_0\left(\frac{1}{2} - \frac{y}{b}\right)\,\text{d}y = \frac{bq_0}{2} ~;~~

q_{x2} = \int_{-b/2}^{b/2}yq_0\left(\frac{1}{2} - \frac{y}{b}\right)\,\text{d}y = -\frac{b^2q_0}{12} \,.

• Bending
• Infinitesimal strain theory
• Kirchhoff–Love plate theory
• Linear elasticity
• Mindlin–Reissner plate theory
• Plate theory
• Stress (mechanics)
• Stress resultants
• Structural acoustics
• Vibration of plates

{{reflist}}

{{DEFAULTSORT:Bending Of Plates}}

Category:Continuum mechanics