Bending of plates explained

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.

Bending of Kirchhoff-Love plates

Definitions

For a thin rectangular plate of thickness

H

, Young's modulus

E

, and Poisson's ratio

\nu

, we can define parameters in terms of the plate deflection,

w

.

The flexural rigidity is given by

D=

EH3
12\left(1-\nu2\right)

Moments

The bending moments per unit length are given by

Mx=-D\left(

\partial2w
\partialx2

+\nu

\partial2w
\partialy2

\right)

My=-D\left(\nu

\partial2w
\partialx2

+

\partial2w
\partialy2

\right)

The twisting moment per unit length is given by

Mxy=-D\left(1-\nu\right)

\partial2w
\partialx\partialy

Forces

The shear forces per unit length are given by

Qx=-D

\partial
\partialx

\left(

\partial2w
\partialx2

+

\partial2w
\partialy2

\right)

Qy=-D

\partial
\partialy

\left(

\partial2w
\partialx2

+

\partial2w
\partialy2

\right)

Stresses

The bending stresses are given by

\sigmax=-

12Dz
H3

\left(

\partial2w
\partialx2

+\nu

\partial2w
\partialy2

\right)

\sigmay=-

12Dz
H3

\left(\nu

\partial2w
\partialx2

+

\partial2w
\partialy2

\right)

The shear stress is given by

\tauxy=-

12Dz
H3

\left(1-\nu\right)

\partial2w
\partialx\partialy

Strains

The bending strains for small-deflection theory are given by

\epsilonx=

\partialu
\partialx

=-z

\partial2w
\partialx2

\epsilony=

\partialv
\partialy

=-z

\partial2w
\partialy2

The shear strain for small-deflection theory is given by

\gammaxy=

\partialu
\partialy

+

\partialv
\partialx

=-2z

\partial2w
\partialx\partialy

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

\epsilonx=

\partialu
\partialx

+

1\left(
2
\partialw
\partialx

\right)2

\epsilony=

\partialv
\partialy

+

1\left(
2
\partialw
\partialy

\right)2

\gammaxy=

\partialu
\partialy

+

\partialv
\partialx

+

\partialw
\partialx
\partialw
\partialy

Deflections

The deflections are given by

u=-z

\partialw
\partialx

v=-z

\partialw
\partialy

Derivation

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

N\alpha\beta,\alpha=0

and

M\alpha\beta,\alpha\beta-q=0

In expanded form,

\cfrac{\partialN11

} + \cfrac = 0 ~;~~ \cfrac + \cfrac = 0 and

\cfrac{\partial2M11

} + 2\cfrac + \cfrac = qwhere

q(x)

is an applied transverse load per unit area, the thickness of the plate is

H=2h

, the stresses are

\sigmaij

, and

N\alpha\beta:=

h
\int
-h

\sigma\alpha\beta~dx3~;~~ M\alpha\beta:=

h
\int
-h

x3~\sigma\alpha\beta~dx3~.

The quantity

N

has units of force per unit length. The quantity

M

has units of moment per unit length.

E

and Poisson's ratio

\nu

these equations reduce to[2]

\nabla2\nabla2w=-\cfrac{q}{D}~;~~D:=\cfrac{2h3E}{3(1-\nu2)}=\cfrac{H3E}{12(1-\nu2)}

where

w(x1,x2)

is the deflection of the mid-surface of the plate.

Small deflection of thin rectangular plates

This is governed by the Germain-Lagrange plate equation

\cfrac{\partial4w}{\partialx4} +2\cfrac{\partial4w}{\partialx2\partialy2} +\cfrac{\partial4w}{\partialy4} =\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.

Large deflection of thin rectangular plates

This is governed by the Föpplvon Kármán plate equations

\cfrac{\partial4F}{\partialx4}+ 2\cfrac{\partial4F}{\partialx2\partialy2}+ \cfrac{\partial4F}{\partialy4}= E\left[\left(\cfrac{\partial2w}{\partialx\partialy}\right)2- \cfrac{\partial2w}{\partialx2} \cfrac{\partial2w}{\partialy2}\right]

\cfrac{\partial4w}{\partialx4}+ 2\cfrac{\partial4w}{\partialx2\partialy2}+ \cfrac{\partial4w}{\partialy4}= \cfrac{q}{D}+\cfrac{H}{D}\left(\cfrac{\partial2F}{\partialy2}\cfrac{\partial2w}{\partialx2}+ \cfrac{\partial2F}{\partialx2}\cfrac{\partial2w}{\partialy2}- 2\cfrac{\partial2F}{\partialx\partialy}\cfrac{\partial2w}{\partialx\partialy} \right)

where

F

is the stress function.

Circular Kirchhoff-Love plates

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

\nabla2\nabla2w=-

q
D

.

In cylindrical coordinates

(r,\theta,z)

,

\nabla2w\equiv

1
r
\partial
\partialr

\left(r

\partialw
\partialr

\right)+

1
r2
\partial2w
\partial\theta2

+

\partial2w
\partialz2

.

For symmetrically loaded circular plates,

w=w(r)

, and we have

\nabla2w\equiv

1
r

\cfrac{d}{dr}\left(r\cfrac{dw}{dr}\right).

Therefore, the governing equation is
1
r

\cfrac{d}{dr}\left[r\cfrac{d}{dr}\left\{

1
r

\cfrac{d}{dr}\left(r\cfrac{dw}{dr}\right)\right\}\right]=-

q
D

.

If

q

and

D

are constant, direct integration of the governing equation gives us

w(r)=-

qr4
64D

+C1lnr+\cfrac{C2r2}{2}+

2}{4}(2ln
\cfrac{C
3r

r-1)+C4

where

Ci

are constants. The slope of the deflection surface is

\phi(r)=\cfrac{dw}{dr}=-

qr3
16D

+

C1
r

+C2r+C3rlnr.

For a circular plate, the requirement that the deflection and the slope of the deflection are finite at

r=0

implies that

C1=0

. However,

C3

need not equal 0, as the limit of

rlnr

exists as you approach

r=0

from the right.

Clamped edges

For a circular plate with clamped edges, we have

w(a)=0

and

\phi(a)=0

at the edge ofthe plate (radius

a

). Using these boundary conditions we get

w(r)=-

q
64D

(a2-r2)2and\phi(r)=

qr
16D

(a2-r2).

The in-plane displacements in the plate are

ur(r)=-z\phi(r)andu\theta(r)=0.

The in-plane strains in the plate are

\varepsilonrr=\cfrac{dur}{dr}=-

qz
16D

(a2-3r2)~,~~ \varepsilon\theta\theta=

ur
r

=-

qz
16D

(a2-r2)~,~~ \varepsilonr\theta=0.

The in-plane stresses in the plate are

\sigmarr=

E
1-\nu2

\left[\varepsilonrr+\nu\varepsilon\theta\theta\right]~;~~ \sigma\theta\theta=

E
1-\nu2

\left[\varepsilon\theta\theta+\nu\varepsilonrr\right]~;~~ \sigmar\theta=0.

For a plate of thickness

2h

, the bending stiffness is

D=2Eh3/[3(1-\nu2)]

and wehave

\begin{align} \sigmarr&=-

3qz
32h3

\left[(1+\nu)a2-(3+\nu)r2\right]\\ \sigma\theta\theta&=-

3qz
32h3

\left[(1+\nu)a2-(1+3\nu)r2\right]\\ \sigmar\theta&=0. \end{align}

The moment resultants (bending moments) are

Mrr=-

q
16

\left[(1+\nu)a2-(3+\nu)r2\right]~;~~ M\theta\theta=-

q
16

\left[(1+\nu)a2-(1+3\nu)r2\right]~;~~ Mr\theta=0.

The maximum radial stress is at

z=h

and

r=a

:

\left.\sigmarr\right|z=h,r=a=

3qa2
16h2

=

3qa2
4H2

where

H:=2h

. The bending moments at the boundary and the center of the plate are

\left.Mrr\right|r=a=

qa2
8

~,~~ \left.M\theta\theta\right|r=a=

\nuqa2
8

~,~~ \left.Mrr\right|r=0=\left.M\theta\theta\right|r=0=-

(1+\nu)qa2
16

.

Rectangular Kirchhoff-Love plates

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.

Sinusoidal load

Let us assume that the load is of the form

q(x,y)=q0\sin

\pix\sin
a
\piy
b

.

Here

q0

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 ofthe plate is zero, the bending moment

Mxx

is zero at

x=0

and

x=a

, and

Myy

is zero at

y=0

and

y=b

.

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

w(x,y)=

q0\left(
\pi4D
1+
a2
1
b2

\right)-2\sin

\pix\sin
a
\piy
b

.

Where D is the flexural rigidity
D=Et3
12(1-\nu2)

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

q(x,y)=q0\sin

m\pix\sin
a
n\piy
b

where

m

and

n

are integers, we get the solution

(1)    w(x,y)=

q0\left(
\pi4D
m2+
a2
n2
b2

\right)-2\sin

m\pix\sin
a
n\piy
b

.

Navier solution

Double trigonometric series equation

We define a general load

q(x,y)

of the following form

q(x,y)=

infty
\sum
m=1
infty
\sum
n=1

amn\sin

m\pix\sin
a
n\piy
b

where

amn

is a Fourier coefficient given by

amn=

4
ab
b
\int
0
a
\intq(x,y)\sin
0
m\pix\sin
a
n\piy
b

dxdy

.The classical rectangular plate equation for small deflections thus becomes:

\cfrac{\partial4w}{\partialx4}+2\cfrac{\partial4w}{\partialx2\partialy2}+\cfrac{\partial4w}{\partialy4}= \cfrac{1}{D}

infty
\sum
m=1
infty
\sum
n=1

amn\sin

m\pix\sin
a
n\piy
b

Simply-supported plate with general load

We assume a solution

w(x,y)

of the following form

w(x,y)=

infty
\sum
m=1
infty
\sum
n=1

wmn\sin

m\pix\sin
a
n\piy
b

The partial differentials of this function are given by

\cfrac{\partial4w}{\partialx4}=

infty
\sum
m=1
infty
\sum\left(
n=1
m\pi
a

\right)4wmn\sin

m\pix\sin
a
n\piy
b

\cfrac{\partial4w}{\partialx2\partialy2}=

infty
\sum
m=1
infty
\sum\left(
n=1
m\pi
a

\right)2\left(

n\pi
b

\right)2wmn\sin

m\pix\sin
a
n\piy
b

\cfrac{\partial4w}{\partialy4}=

infty
\sum
m=1
infty
\sum\left(
n=1
n\pi
b

\right)4wmn\sin

m\pix\sin
a
n\piy
b

Substituting these expressions in the plate equation, we have
infty
\sum
m=1
infty
\sum
n=1

\left(\left(

m\pi
a

\right)2+\left(

n\pi
b

\right)2\right)2 wmn\sin

m\pix\sin
a
n\piy
b

=

infty
\sum
m=1
infty
\sum
n=1

\cfrac{amn

} \sin\frac\sin\fracEquating the two expressions, we have

\left(\left(

m\pi
a

\right)2+\left(

n\pi
b

\right)2\right)2wmn=\cfrac{amn

}which can be rearranged to give

wmn=

1
\pi4D
amn
\left(m2+
n2
b2
\right)2
a2

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

w(x,y)=

1
\pi4D
infty
\sum
m=1
infty
\sum
n=1
amn\sin
\left(m2+
n2
b2
\right)2
a2
m\pix\sin
a
n\piy
b

Simply-supported plate with uniformly-distributed load

For a uniformly-distributed load, we have

q(x,y)=q0

The corresponding Fourier coefficient is thus given by

amn=

4
ab
a
\int
0
b
\int
0
q\sin
0\sinm\pix
a
n\piy
b

dxdy

.Evaluating the double integral, we have

amn=

4q0
\pi2mn

(1-\cosm\pi)(1-\cosn\pi)

,or alternatively in a piecewise format, we have

amn=\begin{cases}

2
\cfrac{16q
0}{\pi

mn}&m~and~n~odd\\ 0&m~or~n~even \end{cases}

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

w(x,y)=

16q0
\pi6D
infty
\sum
m=1,3,5,...
infty
\sum
n=1,3,5,...
1\sin
mn\left(m2+
n2
b2
\right)2
a2
m\pix\sin
a
n\piy
b

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

Mx=

16q0
\pi4
infty
\sum
m=1,3,5,...
infty
\sum
n=1,3,5,...
m2+
\nun2
b2
a2
\sin
mn\left(m2+
n2
b2
\right)2
a2
m\pix\sin
a
n\piy
b

My=

16q0
\pi4
infty
\sum
m=1,3,5,...
infty
\sum
n=1,3,5,...
n2+
\num2
a2
b2
\sin
mn\left(m2+
n2
b2
\right)2
a2
m\pix\sin
a
n\piy
b

Lévy solution

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

Ym(y)

such that it satisfies the boundary conditions at

y=0

and

y=b

and, of course, the governing equation

\nabla2\nabla2w=q/D

.

Let us assume that

w(x,y)=

infty
\sum
m=1

Ym(y)\sin

m\pix
a

.

For a plate that is simply-supported along

x=0

and

x=a

, the boundary conditions are

w=0

and

Mxx=0

. Note that there is no variation in displacement along these edges meaning that

\partialw/\partialy=0

and

\partial2w/\partialy2=0

, thus reducing the moment boundary condition to an equivalent expression

\partial2w/\partialx2=0

.

Moments along edges

Consider the case of pure moment loading. In that case

q=0

and

w(x,y)

has to satisfy

\nabla2\nabla2w=0

. Since we are working in rectangularCartesian coordinates, the governing equation can be expanded as
\partial4w
\partialx4

+2

\partial4w
\partialx2\partialy2

+

\partial4w
\partialy4

=0.

Plugging the expression for

w(x,y)

in the governing equation gives us
infty
\sum\left[\left(
m=1
m\pi
a

\right)4Ym\sin

m\pix
a

-2\left(

m\pi
a

\right)2\cfrac{d2Ym}{dy2}\sin

m\pix
a

+

4Y
d
m
\sin
dy4
m\pix
a

\right]=0

or
4Y
d
m
dy4

-2

m2\pi2
a2
2}
\cfrac{d
m}{dy

+

m4\pi4
a4

Ym=0.

This is an ordinary differential equation which has the general solution

Ym=Am\cosh

m\piy
a

+

B\cosh
mm\piy
a
m\piy
a

+Cm\sinh

m\piy
a

+

D\sinh
mm\piy
a
m\piy
a

where

Am,Bm,Cm,Dm

are constants that can be determined from the boundary conditions. Therefore, the displacement solution has the form

w(x,y)=

infty
\sum
m=1

\left[\left(Am+

B
mm\piy
a

\right)\cosh

m\piy
a

+\left(Cm+

D
mm\piy
a

\right)\sinh

m\piy
a

\right]\sin

m\pix
a

.

Let us choose the coordinate system such that the boundaries of the plate areat

x=0

and

x=a

(same as before) and at

y=\pmb/2

(and not

y=0

and

y=b

). Then the moment boundary conditions at the

y=\pmb/2

boundaries are

w=0,-D

\partial2w
\partialy2

r|y=b/2=f1(x),-D

\partial2w
\partialy2

r|y=-b/2=f2(x)

where

f1(x),f2(x)

are known functions. The solution can be found by applying these boundary conditions. We can show that for the symmetrical casewhere

Myyr|y=-b/2=Myyr|y=b/2

and

f1(x)=f2(x)=

infty
\sum
m=1
E
m\sinm\pix
a

we have

w(x,y)=

a2
2\pi2D
infty
\sum
m=1
Em
2\cosh\alpha
m
m

\sin

m\pix
a

\left(\alpham\tanh\alpham\cosh

m\piy
a

-

m\piy\sinh
a
m\piy
a

\right)

where

\alpham=

m\pib
2a

.

Similarly, for the antisymmetrical case where

Myyr|y=-b/2=-Myyr|y=b/2

we have

w(x,y)=

a2
2\pi2D
infty
\sum
m=1
Em
2\sinh\alpha
m
m

\sin

m\pix
a

\left(\alpham\coth\alpham\sinh

m\piy
a

-

m\piy\cosh
a
m\piy
a

\right).

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

Simply-supported plate with uniformly-distributed load

For a uniformly-distributed load, we have

q(x,y)=q0

The deflection of a simply-supported plate with centre

\left(a
2

,0\right)

with uniformly-distributed load is given by

\begin{align} &w(x,y)=

q0a4
D
infty
\sum
m=1,3,5,...

\left(

A
m\coshm\piy
a

+

B\sinh
mm\piy
a
m\piy
a

+Gm\right) \sin

m\pix
a

\\\\ &\begin{align} where &Am=-

2\left(\alpham\tanh\alpham+2\right)
\pi5m5\cosh\alpham

\\ &Bm=

2
\pi5m5\cosh\alpham

\\ &Gm=

4
\pi5m5

\\\\ and &\alpham=

m\pib
2a

\end{align} \end{align}

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

Mx=

2
-q
0\pi
infty
a
m=1,3,5,...

m2\left(\left(\left(\nu-1\right)Am+2\nu

B
m\right)\coshm\piy
a

+ \left(\nu

-1\right)B\sinh
mm\piy
a
m\piy
a

-Gm\right) \sin

m\pix
a

My=

2
-q
0\pi
infty
a
m=1,3,5,...

m2\left(\left(\left(1-\nu\right)Am+

2B
m\right)\coshm\piy
a

+

\left(1-\nu\right)B\sinh
mm\piy
a
m\piy
a

-\nuGm\right) \sin

m\pix
a

Uniform and symmetric moment load

For the special case where the loading is symmetric and the moment is uniform, we have at

y=\pmb/2

,

Myy=f1(x)=

4M0
\pi
infty
\sum
m=1
1\sin
2m-1
(2m-1)\pix
a

.

The resulting displacement is

\begin{align} &w(x,y)=

2M0a2
\pi3D
infty
\sum
m=1
1\sin
3\cosh\alpha
(2m-1)
m
(2m-1)\pix
a

x \\ &~~\left[ \alpham\tanh\alpha

-
m\cosh(2m-1)\piy
a
(2m-1)\piy\sinh
a
(2m-1)\piy
a

\right]\end{align}

where

\alpham=

\pi(2m-1)b
2a

.

The bending moments and shear forces corresponding to the displacement

w

are

\begin{align} Mxx&=-D\left(

\partial2w+\nu
\partialx2
\partial2w
\partialy2

\right)\\ &=

2M0(1-\nu)
\pi
infty1
(2m-1)\cosh\alpham
\sum
m=1

x \\ &~\sin

(2m-1)\pix
a

x \\ &~\left[ -

(2m-1)\piy\sinh
a
(2m-1)\piy
a

+\right.\\ &             \left.\left\{

2\nu
1-\nu

+\alpham\tanh\alpha

m\right\}\cosh(2m-1)\piy
a

\right]\\ Mxy&=(1-\nu)D

\partial2w
\partialx\partialy

\\ &=-

2M0(1-\nu)
\pi
infty1
(2m-1) \cosh\alpham
\sum
m=1

x \\ &~\cos

(2m-1)\pix
a

x \\ &~\left[

(2m-1)\piy\cosh
a
(2m-1)\piy
a

+\right.\\ &             \left.(1-\alpham\tanh\alpha

m)\sinh(2m-1)\piy
a

\right]\\ Qzx&=

\partialMxx-
\partialx
\partialMxy
\partialy

\\ &=

4M0
a
infty
\sum
m=1
1
\cosh\alpham

x \\ &~\cos

(2m-1)\pix\cosh
a
(2m-1)\piy
a

. \end{align}

The stresses are

\sigmaxx=

12z
h3

Mxxand\sigmazx=

1
\kappah

Qzx\left(1-

4z2
h2

\right).

Cylindrical plate bending

Cylindrical bending occurs when a rectangular plate that has dimensions

a x b x h

, where

a\llb

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.

Simply supported plate with axially fixed ends

For a simply supported plate under cylindrical bending with edges that are free to rotate but have a fixed

x1

. Cylindrical bending solutions can be found using the Navier and Levy techniques.

Bending of thick Mindlin plates

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.[5]

Governing equations

The canonical governing equation for isotropic thick plates can be expressed as[5]

\begin{align} &\nabla2\left(l{M}-

l{B
}\,q\right) = -q \\ & \kappa G h\left(\nabla^2 w + \frac\right) = -\left(1 - \cfrac\right)q \\ & \nabla^2 \left(\frac - \frac\right) = c^2\left(\frac - \frac\right) \endwhere

q

is the applied transverse load,

G

is the shear modulus,

D=Eh3/[12(1-\nu2)]

is the bending rigidity,

h

is the plate thickness,

c2=2\kappaGh/[D(1-\nu)]

,

\kappa

is the shear correction factor,

E

is the Young's modulus,

\nu

is the Poisson's ratio, and

l{M}=

D\left[
l{A}\left(\partial\varphi1
\partialx1

+

\partial\varphi2
\partialx2

\right) -(1-l{A})\nabla2w\right]+

2q
1-\nu2

l{B}.

In Mindlin's theory,

w

is the transverse displacement of the mid-surface of the plateand the quantities

\varphi1

and

\varphi2

are the rotations of the mid-surface normalabout the

x2

and

x1

-axes, respectively. The canonical parameters for this theoryare

l{A}=1

and

l{B}=0

. The shear correction factor

\kappa

usually has thevalue

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&=wK+

l{M
K}{\kappa

Gh}\left(1-

l{B
c

2}{2}\right) -\Phi+\Psi\\ \varphi1&=-

\partialwK
\partialx1

-

1
\kappaGh

\left(1-

1
l{A
} - \frac\right)Q_1^K + \frac\left(\frac\nabla^2 \Phi + \Phi - \Psi\right) + \frac\frac \\ \varphi_2 & = - \frac - \frac\left(1 - \frac - \frac\right)Q_2^K + \frac\left(\frac\nabla^2 \Phi + \Phi - \Psi\right) + \frac\frac \endwhere

wK

is the displacement predicted for a Kirchhoff-Love plate,

\Phi

is a biharmonic function such that

\nabla2\nabla2\Phi=0

,

\Psi

is a function that satisfies the Laplace equation,

\nabla2\Psi=0

, and

\begin{align} l{M}&=l{M}K+

l{B
}\,q + D \nabla^2 \Phi ~;~~ \mathcal^K := -D\nabla^2 w^K \\ Q_1^K & = -D\frac\left(\nabla^2 w^K\right) ~,~~ Q_2^K = -D\frac\left(\nabla^2 w^K\right) \\ \Omega & = \frac - \frac ~,~~ \nabla^2 \Omega = c^2\Omega \,. \end

Simply supported rectangular plates

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

l{M}=

1
1+\nu

(M11+M22)=D\left(

\partial\varphi1+
\partialx1
\partial\varphi2
\partialx2

\right)=0.

Which is almost Laplace`s equation for w[ref 6]. In that case the functions

\Phi

,

\Psi

,

\Omega

vanish, and the Mindlin solution isrelated to the corresponding Kirchhoff solution by

w=wK+

l{M
K}{\kappa

Gh}.

Bending of Reissner-Stein cantilever plates

Reissner-Stein theory for cantilever plates[6] leads to the following coupled ordinary differential equations for a cantilever plate with concentrated end load

qx(y)

at

x=a

.

\begin{align} &bD

4w
d
x
dx4

=0\\ &

b3D
12
4\theta
d
x
dx4

-2bD(1-\nu)\cfrac{d2\thetax}{dx2}=0 \end{align}

and the boundary conditions at

x=a

are

\begin{align} &bD\cfrac{d3wx}{dx3}+qx1=0,

b3D
12

\cfrac{d3\thetax}{dx3}-2bD(1-\nu)\cfrac{d\thetax}{dx}+qx2=0\\ &bD\cfrac{d2wx}{dx2}=0,

b3D
12

\cfrac{d2\thetax}{dx2}=0. \end{align}

Solution of this system of two ODEs gives

\begin{align} wx(x)&=

qx1
6bD

(3ax2-x3)\\ \thetax(x)&=

qx2
2bD(1-\nu)

\left[x-

1
\nub

\left(

\sinh(\nuba)
\cosh[\nub(x-a)]

+\tanh[\nub(x-a)]\right)\right] \end{align}

where

\nub=\sqrt{24(1-\nu)}/b

. The bending moments and shear forces corresponding to the displacement

w=wx+y\thetax

are

\begin{align} Mxx&=-D\left(

\partial2w+\nu
\partialx2
\partial2w
\partialy2

\right)\\ &=qx1\left(

x-a
b

\right)-\left[

3yqx2
3[\nu
b
b(x-a)]

\right] x \\ &\left[6\sinh(\nuba)-\sinh[\nub(2x-a)]+\sinh[\nub(2x-3a)]+8\sinh[\nub(x-a)]\right]\\ Mxy&=(1-\nu)D

\partial2w
\partialx\partialy

\\ &=

qx2
2b

\left[1-

2+\cosh[\nub(x-2a)]-\cosh[\nubx]
2[\nu
2\cosh
b(x-a)]

\right]\\ Qzx&=

\partialMxx-
\partialx
\partialMxy
\partialy

\\ &=

qx1
b

-\left(

3yqx2
3\cosh
2b
4[\nu
b(x-a)]

\right) x \left[32+\cosh[\nub(3x-2a)]-\cosh[\nub(3x-4a)]\right.\\ &    \left.-16\cosh[2\nub(x-a)]+ 23\cosh[\nub(x-2a)]-23\cosh(\nubx)\right]. \end{align}

The stresses are

\sigmaxx=

12z
h3

Mxxand\sigmazx=

1
\kappah

Qzx\left(1-

4z2
h2

\right).

If the applied load at the edge is constant, we recover the solutions for a beam under aconcentrated end load. If the applied load is a linear function of

y

, then

qx1=

b/2
\int
-b/2
q
0\left(1
2

-

y
b

\right)dy=

bq0
2

~;~~ qx2=

b/2
\int
-b/2
yq
0\left(1
2

-

y
b

\right)dy=-

2q
b
0
12

.

See also

Notes and References

  1. Reddy, J. N., 2007, Theory and analysis of elastic plates and shells, CRC Press, Taylor and Francis.
  2. Timoshenko, S. and Woinowsky-Krieger, S., (1959), Theory of plates and shells, McGraw-Hill New York.
  3. Cook, R. D. et al., 2002, Concepts and applications of finite element analysis, John Wiley & Sons
  4. Lévy, M., 1899, Comptes rendues, vol. 129, pp. 535-539
  5. 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.
  6. E. Reissner and M. Stein. Torsion and transverse bending of cantilever plates. Technical Note 2369, National Advisory Committee for Aeronautics,Washington, 1951.