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
H
E
\nu
w
The flexural rigidity is given by
D=
EH3 | |
12\left(1-\nu2\right) |
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 |
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)
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)
\tauxy=-
12Dz | |
H3 |
\left(1-\nu\right)
\partial2w | |
\partialx\partialy |
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 |
The deflections are given by
u=-z
\partialw | |
\partialx |
v=-z
\partialw | |
\partialy |
In the Kirchhoff–Love plate theory for plates the governing equations are[1]
N\alpha\beta,\alpha=0
M\alpha\beta,\alpha\beta-q=0
\cfrac{\partialN11
\cfrac{\partial2M11
q(x)
H=2h
\sigmaij
N\alpha\beta:=
h | |
\int | |
-h |
\sigma\alpha\beta~dx3~;~~ M\alpha\beta:=
h | |
\int | |
-h |
x3~\sigma\alpha\beta~dx3~.
N
M
E
\nu
\nabla2\nabla2w=-\cfrac{q}{D}~;~~D:=\cfrac{2h3E}{3(1-\nu2)}=\cfrac{H3E}{12(1-\nu2)}
w(x1,x2)
This is governed by the Germain-Lagrange plate equation
\cfrac{\partial4w}{\partialx4} +2\cfrac{\partial4w}{\partialx2\partialy2} +\cfrac{\partial4w}{\partialy4} =\cfrac{q}{D}
This is governed by the Föppl–von 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)
F
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
The governing equation in coordinate-free form is
\nabla2\nabla2w=-
q | |
D |
.
(r,\theta,z)
\nabla2w\equiv
1 | |
r |
\partial | |
\partialr |
\left(r
\partialw | |
\partialr |
\right)+
1 | |
r2 |
\partial2w | |
\partial\theta2 |
+
\partial2w | |
\partialz2 |
.
w=w(r)
\nabla2w\equiv
1 | |
r |
\cfrac{d}{dr}\left(r\cfrac{dw}{dr}\right).
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 |
.
q
D
wherew(r)=-
qr4 64D +C1lnr+\cfrac{C2r2}{2}+
2}{4}(2ln \cfrac{C 3r r-1)+C4
Ci
\phi(r)=\cfrac{dw}{dr}=-
qr3 | |
16D |
+
C1 | |
r |
+C2r+C3rlnr.
r=0
C1=0
C3
rlnr
r=0
For a circular plate with clamped edges, we have
w(a)=0
\phi(a)=0
a
The in-plane displacements in the plate arew(r)=-
q 64D (a2-r2)2 and \phi(r)=
qr 16D (a2-r2).
ur(r)=-z\phi(r) and u\theta(r)=0.
\varepsilonrr=\cfrac{dur}{dr}=-
qz | |
16D |
(a2-3r2)~,~~ \varepsilon\theta\theta=
ur | |
r |
=-
qz | |
16D |
(a2-r2)~,~~ \varepsilonr\theta=0.
\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.
2h
D=2Eh3/[3(1-\nu2)]
The moment resultants (bending moments) are\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}
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.
z=h
r=a
\left.\sigmarr\right|z=h,r=a=
3qa2 | |
16h2 |
=
3qa2 | |
4H2 |
H:=2h
\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 |
.
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)=q0\sin
\pix | \sin | |
a |
\piy | |
b |
.
q0
a
x
b
y
Since the plate is simply supported, the displacement
w(x,y)
Mxx
x=0
x=a
Myy
y=0
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 |
.
D= | Et3 |
12(1-\nu2) |
For a more general load of the form
q(x,y)=q0\sin
m\pix | \sin | |
a |
n\piy | |
b |
m
n
(1) w(x,y)=
q0 \left( \pi4D
m2 + a2
n2 b2 \right)-2\sin
m\pix \sin a
n\piy b .
We define a general load
q(x,y)
q(x,y)=
infty | |
\sum | |
m=1 |
infty | |
\sum | |
n=1 |
amn\sin
m\pix | \sin | |
a |
n\piy | |
b |
amn
amn=
4 | |
ab |
b | |
\int | |
0 |
a | ||
\int | q(x,y)\sin | |
0 |
m\pix | \sin | |
a |
n\piy | |
b |
dxdy
\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 |
We assume a solution
w(x,y)
w(x,y)=
infty | |
\sum | |
m=1 |
infty | |
\sum | |
n=1 |
wmn\sin
m\pix | \sin | |
a |
n\piy | |
b |
\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 |
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
\left(\left(
m\pi | |
a |
\right)2+\left(
n\pi | |
b |
\right)2\right)2wmn=\cfrac{amn
wmn=
1 | |
\pi4D |
amn | |||||||
|
w(x,y)=
1 \pi4D
infty \sum m=1
infty \sum n=1
amn \sin
\left( m2 + \right)2
n2 b2 a2
m\pix \sin a
n\piy b
For a uniformly-distributed load, we have
q(x,y)=q0
amn=
4 | |
ab |
a | |
\int | |
0 |
b | |
\int | |
0 |
q | \sin | ||||
|
n\piy | |
b |
dxdy
amn=
4q0 | |
\pi2mn |
(1-\cosm\pi)(1-\cosn\pi)
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 + \right)2
n2 b2 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 +
\nu n2 b2 a2 \sin
mn\left( m2 + \right)2
n2 b2 a2
m\pix \sin a
n\piy b My=
16q0 \pi4
infty \sum m=1,3,5,...
infty \sum n=1,3,5,...
n2 +
\nu m2 a2 b2 \sin
mn\left( m2 + \right)2
n2 b2 a2
m\pix \sin a
n\piy 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
Ym(y)
y=0
y=b
\nabla2\nabla2w=q/D
Let us assume that
w(x,y)=
infty | |
\sum | |
m=1 |
Ym(y)\sin
m\pix | |
a |
.
x=0
x=a
w=0
Mxx=0
\partialw/\partialy=0
\partial2w/\partialy2=0
\partial2w/\partialx2=0
Consider the case of pure moment loading. In that case
q=0
w(x,y)
\nabla2\nabla2w=0
\partial4w | |
\partialx4 |
+2
\partial4w | |
\partialx2\partialy2 |
+
\partial4w | |
\partialy4 |
=0.
w(x,y)
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 |
+
| \sin | |||||||
dy4 |
m\pix | |
a |
\right]=0
| |||||||
dy4 |
-2
m2\pi2 | |
a2 |
2} | |
\cfrac{d | |
m}{dy |
+
m4\pi4 | |
a4 |
Ym=0.
Ym=Am\cosh
m\piy | |
a |
+
B | \cosh | ||||
|
m\piy | |
a |
+Cm\sinh
m\piy | |
a |
+
D | \sinh | ||||
|
m\piy | |
a |
Am,Bm,Cm,Dm
Let us choose the coordinate system such that the boundaries of the plate areatw(x,y)=
infty \sum m=1 \left[\left(Am+
B
m m\piy a \right)\cosh
m\piy a +\left(Cm+
D
m m\piy a \right)\sinh
m\piy a \right]\sin
m\pix a .
x=0
x=a
y=\pmb/2
y=0
y=b
y=\pmb/2
w=0,-D
\partial2w | |
\partialy2 |
r|y=b/2=f1(x),-D
\partial2w | |
\partialy2 |
r|y=-b/2=f2(x)
f1(x),f2(x)
Myyr|y=-b/2=Myyr|y=b/2
f1(x)=f2(x)=
infty | |
\sum | |
m=1 |
E | ||||
|
wherew(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)
\alpham=
m\pib | |
2a |
.
Myyr|y=-b/2=-Myyr|y=b/2
We can superpose the symmetric and antisymmetric solutions to get more generalsolutions.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).
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)
\begin{align} &w(x,y)=
q0a4 D
infty \sum m=1,3,5,... \left(
A
m\cosh m\piy a +
B \sinh
m m\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)\cosh m\piy a + \left(\nu
-1\right)B \sinh
m m\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)\cosh m\piy a +
\left(1-\nu\right)B \sinh
m m\piy a
m\piy a -\nuGm\right) \sin
m\pix a
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
where\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}
\alpham=
\pi(2m-1)b | |
2a |
.
w
\begin{align} Mxx&=-D\left(
\partial2w | +\nu | |
\partialx2 |
\partial2w | |
\partialy2 |
\right)\\ &=
2M0(1-\nu) | |
\pi |
| ||||
\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
|
\right]\\ Mxy&=(1-\nu)D
\partial2w | |
\partialx\partialy |
\\ &=-
2M0(1-\nu) | |
\pi |
| ||||
\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
|
\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}
\sigmaxx=
12z | |
h3 |
Mxx and \sigmazx=
1 | |
\kappah |
Qzx\left(1-
4z2 | |
h2 |
\right).
Cylindrical bending occurs when a rectangular plate that has dimensions
a x b x h
a\llb
h
For a simply supported plate under cylindrical bending with edges that are free to rotate but have a fixed
x1
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]
The canonical governing equation for isotropic thick plates can be expressed as[5]
\begin{align} &\nabla2\left(l{M}-
l{B | |
q
G
D=Eh3/[12(1-\nu2)]
h
c2=2\kappaGh/[D(1-\nu)]
\kappa
E
\nu
l{M}=
D\left[
|
+
\partial\varphi2 | |
\partialx2 |
\right) -(1-l{A})\nabla2w\right]+
2q | |
1-\nu2 |
l{B}.
w
\varphi1
\varphi2
x2
x1
l{A}=1
l{B}=0
\kappa
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 |
wK
\Phi
\nabla2\nabla2\Phi=0
\Psi
\nabla2\Psi=0
\begin{align} l{M}&=l{M}K+
l{B | |
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.
\Phi
\Psi
\Omega
w=wK+
l{M | |
K}{\kappa |
Gh}.
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)
x=a
\begin{align} &bD
| |||||||
dx4 |
=0\\ &
b3D | |
12 |
| |||||||
dx4 |
-2bD(1-\nu)\cfrac{d2\thetax}{dx2}=0 \end{align}
x=a
\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}
\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}
\nub=\sqrt{24(1-\nu)}/b
w=wx+y\thetax
\begin{align} Mxx&=-D\left(
\partial2w | +\nu | |
\partialx2 |
\partial2w | |
\partialy2 |
\right)\\ &=qx1\left(
x-a | |
b |
\right)-\left[
3yqx2 | ||||||
|
\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] | ||||||
|
\right]\\ Qzx&=
\partialMxx | - | |
\partialx |
\partialMxy | |
\partialy |
\\ &=
qx1 | |
b |
-\left(
3yqx2 | ||||||||||||
|
\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}
\sigmaxx=
12z | |
h3 |
Mxx and \sigmazx=
1 | |
\kappah |
Qzx\left(1-
4z2 | |
h2 |
\right).
y
qx1=
b/2 | |
\int | |
-b/2 |
q | ||||
|
-
y | |
b |
\right)dy=
bq0 | |
2 |
~;~~ qx2=
b/2 | |
\int | |
-b/2 |
yq | ||||
|
-
y | |
b |
\right)dy=-
| |||||||
12 |
.