Chapter 10
Plates

10.1 Stress

Kirchhoff plate theory is the straightforward generalization of Euler-Bernoulli beam theory to plates. We abandon the plane situation in which all derivatives in \(y\)-direction vanish. The weak boundary conditions then become

\begin{align} \label {eq:beamweakforcex} Q_x(x,y) &= \int _h \dif z\, \tau _{xz}(x, y, z) \\ \label {eq:beamweakforcey} Q_y(x,y) &= \int _h \dif z\, \tau _{yz}(x, y, z) \\ \label {eq:beamweakmomentxx} M_{xx}(x, y) &= \int _h \dif z\, z \sigma _{xx}(x, y, z) \\ \label {eq:beamweakmomentyy} M_{yy}(x, y) &= \int _h \dif z\, z \sigma _{yy}(x, y, z) \\ \label {eq:beamweakmomentxy} M_{xy}(x, y) &= \int _h \dif z\, z \tau _{xy}(x, y, z), \end{align}

where the integral is over the height \(h\) of the plate. \(Q_x\) and \(Q_y\) are called shear forces, \(M_{xx}\) and \(M_{yy}\) are bending moments and \(M_{xy}\) is the torsional moment.

Note that employing static equilibrium \(\sigma _{ij,j}=0\) we obtain

\begin{equation} Q_{x,x} + Q_{y,y} = \int _{-h/2}^{h/2} \dif z\, \left ( \tau _{zx,x} + \tau _{zy,y} \right ) = -\int _{-h/2}^{h/2} \dif z\, \tau _{zz,z} \end{equation}

but

\begin{equation} \int _{-h/2}^{h/2} \dif z\, \tau _{zz,z} = \tau _{zz}(x, y, h/2) - \tau _{zz}(x, y, -h/2) \equiv p(x,y) \end{equation}

where \(p(x,y)\) is the pressure on the plate (cf. also the corresponding equation Eq. \eqref{eq:beambczz} for the beam). Similarly

\begin{equation} M_{xx,x} + M_{xy,y} = \int _{-h/2}^{h/2} \dif z\, z\left ( \tau _{xx,x} + \tau _{xy,y} \right ) = -\int _{-h/2}^{h/2} \dif z\, z\tau _{xz,z} \end{equation}

and

\begin{equation} \int _{-h/2}^{h/2} \dif z\, z\tau _{xz,z} = \left [z\tau _{xz}\right ]_{-h/2}^{h/2} - \int _{-h/2}^{h/2} \dif z\, \tau _{xz} = -Q_x. \end{equation}

where the last equality holds because \(\tau _{xz}(x,y,h/2)=-\tau _{xz}(x,y,-h/2)\). The condition for static equilibrium \(\sigma _{ij,j}=0\) therefore becomes

\begin{align} \label {eq:plateeq1} Q_{x,x} + Q_{y,y} &=-p(x,y) \\ \label {eq:plateeq2} M_{xx,x} + M_{xy,y}&=Q_x(x,y) \\ \label {eq:plateeq3} M_{xy,x} + M_{yy,y}&=Q_y(x,y) \end{align}

in the weak form. Note that this can be written in the compact form \(Q_{i,i}=-p\) and \(M_{ij,j}=Q_i\).

As in the Euler-Bernoulli case, we assume that the components \(\sigma _{xx}\), \(\sigma _{yy}\) and \(\tau _{xy}\) vary linearly with \(z\). We can write

\begin{align} \sigma _{xx}(x, y, z) &= \frac {M_{xx}(x, y)}{I} z \\ \sigma _{yy}(x, y, z) &= \frac {M_{yy}(x, y)}{I} z \\ \tau _{xy}(x, y, z) &= \frac {M_{xy}(x, y)}{I} z \end{align}

with \(I=\int dz\, z^2=h^3/12\). The remaining components of the stress tensor are obtained from static equilibrium. Static equilibrium yields

\begin{equation} \tau _{xz,z} = -\frac {z}{I} Q_x \quad \text {and}\quad \tau _{yz,z} = -\frac {z}{I} Q_y \end{equation}

which can be integrated under the condition \(\tau _{xz}(x, y, h/2)=\tau _{xz}(x, y, -h/2)=0\) to

\begin{equation} \tau _{xz}(x, y, z) = \frac {Q_x}{2I} \left (\frac {h^2}{4}-z^2\right ) \quad \text {and}\quad \tau _{yz}(x, y, z) = \frac {Q_y}{2I} \left (\frac {h^2}{4}-z^2\right ). \end{equation}

This is analogous to Eq. \eqref{eq:beamstressxz} for the beam.

We are finally left with finding an expression for \(\sigma _{zz}\). Again we use static equilibrium to obtain

\begin{equation} \sigma _{zz,z} = -\tau _{xz,x}-\tau _{yz,y} = \frac {p(x,y)}{2I} \left (\frac {h^2}{4}-z^2\right ). \end{equation}

Integration under the condition that the loads on top and bottom surface of the plate balance, \(\sigma _{zz}(x,h/2)=-\sigma _{zz}(x,-h/2)\), gives

\begin{equation} \label {eq:platestresszz} \sigma _{zz}(x, y, z) = \frac {p(x, y)}{2I} \left (\frac {h^2}{4} - \frac {z^2}{3}\right ) z. \end{equation}

At the top and bottom of the plate we find \(\sigma _{zz}(x,h/2)=-\sigma _{zz}(x,-h/2)=p(x,y)/2\).

10.2 Displacements

Now that we know the stress inside the plate, we can again compute the displacements from Hooke’s law. In the full three-dimensional case, Hooke’s law,

\begin{align} \label {eq:platehooke1} \varepsilon _{xx} \equiv u_{x,x} &= (\sigma _{xx} - \nu \sigma _{yy} - \nu \sigma _{zz})/E \\ \label {eq:platehooke2} \varepsilon _{yy} \equiv u_{y,y} &= (\sigma _{yy} - \nu \sigma _{xx} - \nu \sigma _{zz})/E \\ \label {eq:platehooke3} 2\varepsilon _{xz} \equiv u_{x,z} + u_{z,x} &= 2(1+\nu )\tau _{xz}/E \\ \label {eq:platehooke4} 2\varepsilon _{yz} \equiv u_{y,z} + u_{z,y} &= 2(1+\nu )\tau _{yz}/E \\ \label {eq:platehooke5} 2\varepsilon _{xy} \equiv u_{x,y} + u_{y,x} &= 2(1+\nu )\tau _{xy}/E, \end{align}

and taking the derivative of Eq. \eqref{eq:platehooke3} with respect to \(x\) and of Eq. \eqref{eq:platehooke4} with respect to \(y\), we obtain

\begin{align} \label {eq:ux1} u_{x,xz} + u_{z,xx} &= 2(1+\nu )\tau _{xz,x}/E \\ \label {eq:uy1} u_{y,yz} + u_{z,yy} &= 2(1+\nu )\tau _{yz,y}/E \end{align}

and

\begin{align} \label {eq:ux2} u_{x,xz} + u_{z,xx}=\partial _z( u_{x,x}) + u_{z,xx}&=u_{z,xx} + (\sigma _{xx,z} - \nu \sigma _{yy,z} - \nu \sigma _{zz,z})/E \\ \label {eq:uy2} u_{y,yz} + u_{z,yy}=\partial _z( u_{y,y}) + u_{z,yy}&=u_{z,yy} + (\sigma _{yy,z} - \nu \sigma _{xx,z} - \nu \sigma _{zz,z})/E. \end{align}

Combining Eqs. \eqref{eq:ux1}, \eqref{eq:ux2} and Eqs. \eqref{eq:uy1}, \eqref{eq:uy2} and noting that \(\sigma _{zz,z}=-\tau _{xz,x}-\tau _{xz,y}\) yields

\begin{align} u_{z,xx} &= \left [(2+\nu )\tau _{xz,x} - \nu \tau _{yz,y} - \sigma _{xx,z} + \nu \sigma _{yy,z} \right ]/E \\ u_{z,yy} &= \left [(2+\nu )\tau _{yz,y} - \nu \tau _{xz,x} - \sigma _{yy,z} + \nu \sigma _{xx,z}\right ]/E. \end{align}

We now create linear combination of these expressions such that \(\sigma _{xx,z}=M_{xx}/I\) or \(\sigma _{yy,z}=M_{yy}/I\) drop out,

\begin{align} u_{z,xx} + \nu u_{z,yy} &= \left [(2+\nu -\nu ^2)\tau _{xz,x} - \nu (1+\nu )\tau _{yz,y} - (1-\nu ^2)M_{xx}/I \right ]/E \\ u_{z,yy} + \nu u_{z,xx} &= \left [(2+\nu -\nu ^2)\tau _{yz,y} - \nu (1+\nu )\tau _{xz,x} - (1-\nu ^2)M_{yy}/I \right ]/E. \end{align}

We now only consider the displacement at the surface, \(w(x,y)\equiv u_z(x,y,h/2)\). Since the surfaces are traction free, all terms involving \(\tau _{xz}\) and \(\tau _{yz}\) vanish. Hence

\begin{align} \label {eq:plateMxx} M_{xx} &= -K (w_{,xx} + \nu w_{,yy}) \\ \label {eq:plateMyy} M_{yy} &= -K (w_{,yy} + \nu w_{,xx}) \end{align}

with the flexural rigidity \(K=EI/(1-\nu ^2)=Eh^3/[12(1-\nu ^2)]\).

Finally, we are looking for an expression for \(M_{xy}=I\tau _{xy,z}\). We have from Eqs. \eqref{eq:platehooke3}-\eqref{eq:platehooke5}

\begin{equation} \frac {2(1+\nu )}{EI} M_{xy} = u_{x,yz} + u_{y,xz} = \frac {2(1+\nu )}{E} (\tau _{xz,y} + \tau _{yz,x}) - 2u_{z,xy}, \end{equation}

which yields

\begin{equation} \label {eq:plateMxy} M_{xy} = -K(1-\nu ) w_{,xy}, \end{equation}

the desired expression.

We now plug Eqs. \eqref{eq:plateMxx}, \eqref{eq:plateMyy} and \eqref{eq:plateMxy} into the equilibrium conditions Eqs. \eqref{eq:plateeq2} and \eqref{eq:plateeq3}. This yields

\begin{align} -K(w_{,xxx} + w_{,xyy}) &= Q_x(x, y) \\ -K(w_{,yyy} + w_{,xxy}) &= Q_y(x, y) \\ -K(w_{,xxxx} + 2w_{,xxyy} + w_{,yyyy}) &= -p(x, y). \end{align}

The last expression is Kirchhoff’s equation,

\begin{equation} w_{,xxxx} + 2w_{,xxyy} + w_{,yyyy} = \nabla ^2 \nabla ^2 w = \nabla ^4 w = \frac {p}{K}, \end{equation}

that governs the deformation of plates.

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