I have forgotten
my Password

# Circular Plates

Stresses and Strains in loaded Circular Plates and Rings.
View other versions (6)

## Circular Plates Symmetrically Loaded.

Consider a Diametral Section through a plate of thickness $\inline&space;t$. $\inline&space;O$ is the centre of the plate and $\inline&space;OX$ and $\inline&space;OY$ are the principal axes in the plane of the diagram. The axis $\inline&space;OZ$ is perpendicular to the screen.

##### MISSING IMAGE!

23287/Circular-Plates-0001.png cannot be found in /users/23287/Circular-Plates-0001.png. Please contact the submission author.

Let $\inline&space;C$ be the Centre of Curvature of a section $\inline&space;ab$ at a distance $\inline&space;x$ from $\inline&space;O$. Then if the deflection $\inline&space;y$ is small:
$\frac{dy}{dx}=\theta$

The radius of curvature in the plane $\inline&space;XOY$ is given by: $\inline&space;\displaystyle\frac{1}{R_{xx}}=\displaystyle\frac{d^2y}{dx^2}$ (Approximately)

Thus from equation (2)
$\frac{1}{R_{xx}}=\frac{d\theta&space;}{dx}$

Note that, on a circle of radius $\inline&space;x$ and centre $\inline&space;O$, lines such as $\inline&space;ab$ form part of a cone with $\inline&space;C$ as the apex. Hence $\inline&space;C$ is the Centre of Curvature in the plane $\inline&space;YOZ$ and:
$\frac{1}{R_{yx}}&space;=&space;\frac{\theta&space;}{x}$
(Approximately)

If $\inline&space;u$ is the distance of any "fibre" from the neutral axis (which is assumed to be central) then proceeding as for "Pure Bending" in the planes $\inline&space;XOY$ and $\inline&space;YOZ$ the linear Strains are:

$e_x=\frac{u}{R_{xy}}&space;=&space;\left(\frac{1}{E}&space;\right)\left(f_x&space;-&space;\frac{f_z}{m}&space;\right)$
And,
$e_z=\frac{u}{R_{xy}}&space;=&space;\left(\frac{1}{E}&space;\right)\left(f_z&space;-&space;\frac{f_x}{m}&space;\right)$

Where $\inline&space;f_x$ and $\inline&space;f_y$ are the Stresses in the directions $\inline&space;OX$ and $\inline&space;OZ$, $\inline&space;f_y$ is zero

Solving equations (5) and (6) for the Stresses and incorporating equations (3) and (4) gives:
$f_x=\frac{E\;u}{1-\displaystyle\frac{1}{m^2}}\left(\frac{1}{R_{xy}}&space;+&space;\frac{1}{m\;R_{yz}}&space;\right)=\frac{E\;u}{1&space;-&space;\displaystyle\frac{1}{m^2}}\left(\frac{d\theta&space;}{dx}&space;+&space;\frac{\theta&space;}{m\;x}&space;\right)$

$f_z&space;=&space;\frac{E\;u}{1&space;-&space;\displaystyle\frac{1}{m^2}}\left(\frac{1}{m\;R_{xy}}&space;+&space;\frac{1}{R_{yz}}&space;\right)&space;=&space;\frac{E\;u}{1&space;-&space;\displaystyle\frac{1}{m^2}}\left(\frac{1}{m}\times\frac{d\theta&space;}{dx}&space;+&space;\frac{\theta&space;}{x}&space;\right)$

The Bending Moment per unit length along $\inline&space;OZ$ is $\inline&space;\displaystyle&space;M_{xy}$ which is given by:

$M_{xy}\times&space;dz&space;=&space;\int_{-\frac{t}{2}}^{\frac{t}{2}}{f_x\times&space;u\;dz\times&space;du}$

$\therefore\;\;\;\;\;M_{xy}=D\left(\frac{d\theta&space;}{dx}&space;+&space;\frac{\theta&space;}{m\,x}&space;\right)$

By substitution from equation (7)
$D&space;=&space;\frac{E\;t^3}{12\left(1&space;-&space;\frac{1}{m^2}&space;\right)}$

Similarly if $\inline&space;M_{yz}$ is the Bending Moment per unit length about $\inline&space;OX$ then:
$M_{yz}\times&space;dx&space;=&space;\int_{-\frac{t}{2}}^{\frac{t}{2}}{f_z\times&space;u\,dx\times&space;du}$

Using equation (8)
$M_{yz}&space;=&space;D\left[\left(\frac{1}{m}&space;\right)\left(\frac{d\theta&space;}{dx}&space;\right)&space;+&space;\frac{\theta&space;}{x}&space;\right]$

Note that:
$f_x=M_{xy}\times&space;\frac{12u}{t^3}$
$f_z=M_{yz}\times&space;\frac{12u}{t^3}$

The diagram shows the Forces and Moments per unit length acting on an element which subtends an angle $\inline&space;\delta&space;\phi$ at the centre. $\inline&space;F$ is the Shearing Force per unit length in the direction of $\inline&space;OZ$.

##### MISSING IMAGE!

23287/Circular-Plates-0002.png cannot be found in /users/23287/Circular-Plates-0002.png. Please contact the submission author.

Now consider the equilibrium of the Couples in the Central Radial Plane.

i.e., $\inline&space;\left(M_{xy}&space;+&space;\delta&space;M_{xy}&space;\right)\left(x&space;+&space;\delta&space;x&space;\right)\delta&space;\phi&space;-&space;M_{xy}\times&space;x\delta&space;\phi&space;-&space;2M_{yz}\times&space;\delta&space;x\times&space;\sin\displaystyle\frac{1}{2}\delta&space;\phi&space;+f\;x\;\delta&space;\phi&space;\times&space;\delta&space;x&space;=&space;0$

Which in the Limit reduces to:
$M_{xy}&space;+&space;x\times&space;\frac{\delta&space;M_{xy}}{dx}&space;-&space;M_{yz}&space;+&space;Fx&space;=&space;0$
Substituting from Equations (9) and (11) gives:
$\frac{d^2\theta&space;}{dx^2}&space;+&space;\left(\frac{1}{x}&space;\right)\left(\frac{d\theta&space;}{dx}&space;\right)&space;-&space;\frac{\theta&space;}{x^2}&space;=&space;-&space;\frac{F}{D}$

This can be written as:
$\left(\frac{d}{dx}&space;\right)\left[\left(\frac{1}{x}&space;\right)\times&space;\frac{d(x\theta&space;)}{dx}&space;\right]\;=&space;-&space;\frac{F}{D}$

If $\inline&space;F$ is known as a function of $\inline&space;x$, this equation can be integrated to determine $\inline&space;\theta$ and hence $\inline&space;y$. Bending Moments and Stresses can then be calculated.

### Particular Case

A Plate Loaded with Uniformly distributed load of $\inline&space;w$ per unit Area and a Concentrated load at Centre of $\inline&space;P$.

$2\pi&space;\,x\times&space;F&space;=&space;\pi&space;x^2\times&space;w&space;+&space;P$
$\therefore\;\;\;\;\;\;\;F&space;=&space;\frac{w\,x}{2}&space;+&space;\frac{P}{2\pi&space;\;x}$

Per unit length of circumferentially (Except at $\inline&space;x&space;=&space;0$)

Substituting in Equation (13) and Integrating:

$\theta&space;\;=&space;-&space;\frac{w\;x^3}{16\;D}&space;-&space;\left(\frac{P\;x}{8\pi&space;\;D}&space;\right)\left(2\ln&space;x&space;-&space;1&space;\right)&space;+&space;\frac{C_1\,x}{2}&space;+&space;\frac{C_2}{x}$
But From Equation (2)
$y=\int&space;\theta&space;\;dx&space;+&space;C_3$

$\therefore\;\;\;\;\;y\;=&space;-&space;\frac{w\;x^4}{64\;D}&space;-&space;\left(\frac{P\;x}{8\pi&space;\;D}&space;\right)\left(2\ln&space;x&space;-&space;1&space;\right)&space;+&space;\frac{C_1\;x}{4}&space;+&space;C_2\;\ln&space;x&space;+&space;C_3$

Example:
##### Example - A particular case
Problem
A Plate Loaded with Uniformly distributed load of w per unit Area and a Concentrated load at Center of P.
Workings
$2\pi&space;\,x\times&space;F\;=\;\pi&space;x^2\times&space;w\;+\;P$
$\therefore\;\;\;\;\;\;\;F\;=\;\frac{w\,x}{2}\;+\;\frac{P}{2\pi&space;\;x}$

Per unit length of circumferentially (Except at x = 0)

Substituting in Equation (#19) and Integrating:

$\theta&space;\;=\;-\;\frac{w\;x^3}{16\;D}\;-\;\left(\frac{P\;x}{8\pi&space;\;D}&space;\right)\left(2\ln&space;x\;-\;1&space;\right)\;+\;\frac{C_1\,x}{2}\;+\;\frac{C_2}{x}$
But From Equation (#1)
$y=\int&space;\theta&space;\;dx&space;+&space;C_3$

$\therefore\;\;\;\;\;y\;=\;-\;\frac{w\;x^4}{64\;D}\;-\;\left(\frac{P\;x}{8\pi&space;\;D}&space;\right)\left(2\ln&space;x\;-\;1&space;\right)\;+\;\frac{C_1\;x}{4}\;+\;C_2\;\ln&space;x\;+\;C_3$

## Solid Circular Plate

Let the radius of the plate be $\inline&space;R$ and the thickness $\inline&space;t$.

### Uniformly Loaded, Edge Freely Supported

$\inline&space;P=0$ and since $\inline&space;\theta$ and $\inline&space;y$ can not be infinite at the centre, then $\inline&space;C_2=0$ from Equation (14) at $\inline&space;x&space;=&space;0$, $\inline&space;y&space;=&space;0$, and therefore $\inline&space;C_3=0$ from equation (15).

Using equations (9) and (14). At $\inline&space;x=R$, $\inline&space;M_{xy}=0$ therefore
$-\frac{3\,w\,R^2}{16&space;D}&space;+&space;\frac{C_1}{2}&space;-&space;\frac{w\,R^2}{16&space;D\,m}&space;+&space;\frac{C_1}{2&space;m}&space;=&space;0$
Thus,
$C_1=\left(\frac{w\,R^2}{8\,D}&space;\right)\left(\frac{3+\displaystyle\frac{1}{m}}{1+\displaystyle\frac{1}{m}}&space;\right)$
Central Deflection = $\inline&space;y$ at $\inline&space;x=R$. Thus,
$y=\frac{w\;R^4}{64\,D}&space;+&space;\frac{w\,R^4}{32&space;D}\times&space;\frac{3&space;+&space;\displaystyle\frac{1}{m}}{1&space;+&space;\displaystyle\frac{1}{m}}&space;=&space;\frac{w\;R^4}{64\;D}\left(\frac{5&space;+&space;\displaystyle\frac{1}{m}}{1&space;+&space;\displaystyle\frac{1}{m}}&space;\right)$

Eliminating $\inline&space;D$ by substitution from Equation (10)
$=\left(\frac{3\,w\,R^4}{16\,E\,t^3}&space;\right)\left(5&space;+&space;\frac{1}{m}&space;\right)\left(1&space;-&space;\frac{1}{m}&space;\right)$

From Equation (7)
$f_x&space;=&space;\frac{E\;u}{1&space;-&space;\displaystyle\frac{1}{m^2}}\left(-&space;\frac{w\,x^2}{16\,D}\times&space;\left\{3&space;-&space;\frac{1}{m}&space;\right\}&space;+&space;\frac{w\;R^2}{16\,D}\left\{3&space;+&space;\frac{1}{m}&space;\right\}&space;\right)$

And at $\inline&space;x=0$
$\hat{f}=\frac{E\times&space;\displaystyle\frac{t}{2}}{1&space;-&space;\displaystyle\frac{1}{m^2}}\times&space;\frac{w\;R^2}{16&space;D}\left(3&space;+&space;\displaystyle\frac{1}{m}&space;\right)=\frac{3w\;R^2\left(3&space;+&space;\displaystyle\frac{1}{m}&space;\right)}{8&space;t^2}$

From Equation (8)
$f_z&space;=&space;\frac{E\;u}{1&space;-&space;\displaystyle\frac{1}{m^2}}\left(-&space;\frac{w\,x^2}{16\;D}\left\{\frac{3}{m}&space;+&space;1&space;\right\}&space;&space;+&space;\frac{w\;R^2}{16\;D}\left\{3&space;+&space;\frac{1}{m}&space;\right\}\right)$

As above when $\inline&space;x=0$, $\inline&space;f_z=\hat{f_z}$. Therefore $\inline&space;\hat{f_x}=\hat{f_z}$

Note: the Maximum Stresses occur at the centre.

### Uniformly Loaded With The Edge Clamped

As in the last case, $\inline&space;P=0$ and $\inline&space;C_2=0$ at $\inline&space;x=0$, $\inline&space;y=0$. Therefore $\inline&space;C_3=0$ from equation(15)

Structural loads or actions are forces, deformations or accelerations applied to a structure or its components.Loads cause stresses, deformations and displacements in structures.

At $\inline&space;x=R$, $\inline&space;\displaystyle\frac{dy}{dx}=\theta&space;=0$

i.e. from Equation (14), $\inline&space;-\displaystyle\frac{w\;R^3}{16\;D}&space;+&space;C_1\times&space;\displaystyle\frac{R}{2}&space;=&space;0$
$\therefore\;\;\;\;\;\;C_1&space;=&space;\frac{w\;r^2}{8\;D}$

Using Equation (15),Central Deflection = $\inline&space;-\displaystyle\frac{w\;r^4}{64\;D}&space;+&space;\displaystyle\frac{w\;R^4}{32\;D}&space;=&space;\displaystyle\frac{w\;R^4}{64\;D}$

Eliminating $\inline&space;D$ by using equation (10)
$=\left(\frac{3\,w\;R^4}{16\;E\;t^3}&space;\right)\left(1&space;-&space;\frac{1}{m^2}&space;\right)$

From Equation (7)
$f_x&space;=&space;\left(\frac{E\;u}{1&space;-&space;\displaystyle\frac{1}{m^2}}&space;\right)\left(-&space;\frac{w\;x^2}{16\;D}\left\{3&space;+&space;\frac{1}{m}&space;\right\}&space;+&space;\frac{w\;R^2}{16\;D}\left\{1&space;+&space;\frac{1}{m}&space;\right\}&space;\right)$

This Stress has its greatest numerical value when $\inline&space;x&space;=&space;R$ ( i.e. at the clamped edge), thus
$\hat{f_x}=\frac{E\times&space;\displaystyle\frac{t}{2}}{1&space;-&space;\displaystyle\frac{1}{m^2}}\times&space;\frac{w\;R^2}{16\;D}\times&space;2=\frac{3\;w\;R^2}{4\;t^2}$

From Equation (8)
$f_z=\frac{E\;u}{1&space;-&space;\displaystyle\frac{1}{m^2}}\left(\frac{-&space;w\;x^2}{16\,D}\left\{\frac{3}{m}&space;+&space;1&space;\right\}&space;+&space;\frac{w\;R^2}{16\;D}&space;\left\{1+\frac{1}{m}&space;\right\}\right)$

From which, $\inline&space;\hat{f_z}&space;=&space;\displaystyle\frac{E\times\displaystyle\frac{t}{2}}{1&space;-&space;\displaystyle\frac{1}{m^2}}\times&space;\displaystyle\frac{w\;R^2}{16\;D}\times&space;\left(1&space;+&space;\displaystyle\frac{1}{m}&space;\right)=&space;\displaystyle\frac{3\;w\;R^2(1&space;+&space;\displaystyle\frac{1}{m})}{8\;t^2}$ (At the Centre)

### Central Load P, Edge Freely Supported (w=0)

At $\inline&space;x=0$, $\inline&space;\theta=0$ therefore from equation (14) $\inline&space;C_2=0$ and $\inline&space;y=0$. From equation(15) $\inline&space;C_3=0$

Note: $\inline&space;(L\times&space;t\times&space;(x\;\ln&space;x)=0)$

At $\inline&space;x=R$, $\inline&space;M_{xy}=0$ do from equation (9)
$-\left(\frac{P}{8\;\pi&space;\;D&space;}&space;\right)\left(2\;\ln\,R&space;-&space;1&space;\right)\;\left(\frac{P\;R}{8\pi&space;D}&space;\right)\left(\frac{2}{R}&space;\right)&space;+&space;\frac{C_1}{2}&space;-$
$-&space;\left(\frac{P}{8\pi&space;\;D\;m}&space;\right)\left(2\;\ln&space;R&space;-&space;1&space;\right)&space;+&space;\frac{C_1}{2\,m}&space;=&space;0$

From which, $\inline&space;c_1&space;=&space;\displaystyle\frac{P}{4\pi&space;\,D}\left(2\,\ln&space;R&space;+&space;\displaystyle\frac{1&space;-&space;\displaystyle\frac{1}{m}}{1&space;+&space;\displaystyle\frac{1}{m}}&space;\right)$

Thus, Central deflection
$=&space;\frac{P\,R^2}{8\pi&space;\,D}\left(\ln&space;R&space;-&space;1&space;\right)&space;+&space;\frac{P\;R^2}{16\pi&space;\,D}\left(2\,\ln&space;R&space;+&space;\frac{1&space;-&space;\displaystyle\frac{1}{m}}{1&space;+&space;\displaystyle\frac{1}{m}}&space;\right)$
$=\frac{P\;R^2}{16\pi&space;\,D}\times&space;\frac{(3&space;+&space;\displaystyle\frac{1}{m})}{(1&space;+&space;\displaystyle\frac{1}{m})}=\frac{3P\;R^2}{4\pi\;E\;t^3}\times&space;\left(&space;3&space;+\frac{1}{m}&space;\right)\left(1&space;-&space;\frac{1}{m}&space;\right)$

From Equation (7)
$f_x&space;=&space;\frac{E\;u}{1&space;-&space;\frac{1}{m^2}}\times&space;\frac{P}{4\pi&space;\,D}\left(1&space;+&space;\frac{1}{m}&space;\right)\ln\frac{R}{x}$
$=\left(\frac{3P}{2\pi&space;\,t^2}&space;\right)\left(1&space;+&space;\frac{1}{m}&space;\right)\ln\frac{R}{x}$
Note: $\inline&space;u=\displaystyle\frac{t}{2}$

And from equation (8),
$f_z&space;=\left(\frac{3P}{2\pi&space;\,t^2}&space;\right)\left[\left(1&space;+&space;\frac{1}{m}&space;\right)\ln\frac{R}{x}&space;+&space;1&space;-&space;\frac{1}{m}&space;\right]$

These Stresses appear to become infinite at the centre, but it must be realised that the load can not be applied at a point but must extend over a finite area. If this area can be estimated then the maximum Stresses can be obtained.

### Loaded Round A Circle, Edge Freely Supported

Let a total load $\inline&space;P$ be distributed around a circle of radius $\inline&space;r$.

##### MISSING IMAGE!

23287/Circular-Plates-0001-1.png cannot be found in /users/23287/Circular-Plates-0001-1.png. Please contact the submission author.

It is necessary to divide the plate into two regions, one for $\inline&space;x&space;<&space;r$ and the other for $\inline&space;x&space;>&space;r$. At $\inline&space;x&space;=&space;r$ the values of $\inline&space;\theta$, $\inline&space;y$ and $\inline&space;M_{xy}$ must be the same for both regions.

• If $\inline&space;x, $\inline&space;w=0$ and $\inline&space;P=0$

Hence, from Equation (14) $\inline&space;\theta&space;=&space;\displaystyle\frac{C_1&space;x}{2}&space;+&space;\displaystyle\frac{C_2}{x}$. And from Equation (15), $\inline&space;y=\displaystyle\frac{C_1&space;x^2}{4}&space;+&space;C_2\;\ln&space;x&space;+&space;C_3$

Since $\inline&space;\theta$ and $\inline&space;y$ are not infinite at $\inline&space;x=0$ then $\inline&space;C_2=0$, and since $\inline&space;y=0$ when $\inline&space;x=0$ and $\inline&space;C_3=0$, then above equations reduce to: $\inline&space;\theta&space;=\displaystyle\frac{C_1&space;x}{2}$ And, $\inline&space;y=\displaystyle\frac{C_1\;x^2}{4}$

• If $\inline&space;x>r$ and $\inline&space;w=0$

From Equation (14), $\inline&space;\theta&space;=-\left(\displaystyle\frac{P&space;x}{8\pi&space;D}&space;\right&space;)\left(2&space;\ln&space;x-1&space;\right)&space;+&space;\displaystyle\frac{C_1'x}{2}&space;+&space;\displaystyle\frac{C_2'}{x}$

And from Equation (15),
$y=-\left(\frac{P\;x^2}{8\pi&space;\;D}&space;\right)\left(\ln&space;x&space;-&space;1&space;\right)&space;+&space;\frac{C_1'x^2}{2}&space;+&space;C_2'\;\ln&space;{x}&space;+&space;C_3'$

Equating the values of $\inline&space;\theta$ and $\inline&space;M_{xy}$ at $\inline&space;x=r$ gives the following equations:
$-\left(\frac{P\,r}{8\pi&space;\;D}&space;\right)\left(2\;\lnr&space;-&space;1&space;\right)&space;+&space;\frac{C_1'\;r}{2}&space;+&space;\frac{C_2'}{r}&space;=&space;\frac{C_1\;r}{2}$
$-\left(\frac{P\,r^2}{8\pi&space;\;D}&space;\right)\left(\ln&space;r&space;-&space;1&space;\right)&space;+&space;\frac{C_1'\;r^2}{4}&space;+&space;C_2'\;\ln&space;r&space;=&space;\frac{C_1\;r^2}{4}$

And,
$\left(\frac{P}{8\pi&space;\;D}&space;\right)\left[\left(1&space;+&space;\frac{1}{m}&space;\right)2\;\ln&space;r&space;+&space;1&space;-\frac{1}{m}&space;\right]\:+\;\left(\frac{C_1'}{2}&space;\right)\left(1&space;+&space;\frac{1}{m}&space;\right)&space;-$
$-\left(\frac{C_2'}{r^2}&space;\right)\left(1&space;-&space;\frac{1}{m}&space;\right)=\left(\frac{C_1}{2}&space;\right)\left(1&space;+&space;\frac{1}{m}&space;\right)$

$\inline&space;M_{xy}=0$ at $\inline&space;x=R$ gives:
$\left(\frac{P}{8\pi&space;D}&space;\right)\left[\left(1&space;+&space;\frac{1}{m}&space;\right)2\;\ln&space;R&space;+&space;1&space;-&space;\frac{1}{m}&space;\right]&space;+&space;\left(\frac{C_1'}{2}&space;\right)\left(1&space;+&space;\frac{1}{m}&space;\right)&space;-&space;\left(\frac{C_2'}{R^2}&space;\right)\left(1&space;-&space;\frac{1}{m}&space;\right)&space;=0$

From Equations (16) to (17) the constants are found to be:
$C_1'=\frac{P}{4\pi&space;\;D}\left[2\;\lnR&space;+&space;\frac{R^2&space;-&space;r^2}{R^2}\left(\frac{1&space;-&space;\displaystyle\frac{1}{m}}{1&space;+&space;\displaystyle\frac{1}{m}}&space;\right)&space;\right]$
$C_2'=-\frac{P\;r^2}{8\pi&space;\;D}$
$C_3'=\frac{P\;r^2}{8\pi&space;\;D}\left(\ln&space;r&space;-&space;1&space;\right)$

The Central Deflection is given by the value of $\inline&space;y$ at $\inline&space;x&space;=&space;R$ and by substitution equation (15) reduces to:

$y=\left(\frac{P}{8\pi&space;\;D}&space;\right)\left[\left(R^2&space;-&space;r^2&space;\right)\times&space;&space;\frac{\left(3&space;+&space;\displaystyle\frac{1}{m}&space;\right)}{2\left(1&space;+&space;\displaystyle\frac{1}{m}&space;\right)}&space;-r^2\;\ln\frac{R}{r}&space;\right]$

• For $\inline&space;x>r$
$M_{xy}&space;=&space;\left(\frac{P}{8\pi&space;\right)\left[\left(1&space;+&space;\frac{1}{m}&space;\right)2\;\ln&space;x&space;+&space;\left(1&space;+&space;\frac{1}{m}&space;\right)\times&space;r^2\left(\frac{1}{x^2}&space;-&space;\frac{1}{R^2}&space;\right)\right]}$

Which has a maximum value at $\inline&space;x&space;=&space;r$

Hence from equation (12)

$\hat{f}=\left(\frac{6}{t^2}&space;\right)\;M_{xy}$

$=\left(\frac{3\;P}{4\pi&space;\;t^2}&space;\right)\left[\left(1&space;+&space;\frac{1}{m}&space;\right)2\;\ln\frac{R}{r}&space;+&space;\left(1&space;-&space;\frac{1}{m}&space;\right)\left(\frac{R^2\;-r^2}{R^2}&space;\right)&space;\right]$

Similarly:
$M_{yz}=\left(\frac{P}{8&space;\pi&space;}&space;\right)\left\{&space;\left(1&space;+&space;\frac{1}{m}&space;\right)2&space;\ln\frac{R}{x}&space;+&space;\left(1&space;-&space;\frac{1}{m}&space;\right)\left[\frac{2R^2&space;-r^2}{R^2}&space;-&space;\frac{r^2}{x^2}\right]&space;&space;\right\}$
$\hat{f_z}=\left(\frac{3&space;P}{4\pi&space;\;t^2}&space;\right)\left[\left(1&space;+&space;\frac{1}{m}&space;\right)2&space;\ln\frac{R}{r}&space;+&space;\left(1&space;-&space;\frac{1}{m}&space;\right)\left(\frac{R^2&space;-r^2}{R^2}&space;\right)&space;\right]=f_x$

## Annular Ring , Loaded Around The Inner Edge

##### MISSING IMAGE!

23287/Circular-Plates-0004.png cannot be found in /users/23287/Circular-Plates-0004.png. Please contact the submission author.

The ring is loaded with a total load $\inline&space;P$ around the inner edge and is freely supported around the outer edge. $\inline&space;M_{xy}=0$ at $\inline&space;x=R$ and at $\inline&space;x=r$.

$-\left(\frac{P}{8\pi&space;D}&space;\right)\left[\left(1&space;+&space;\frac{1}{m}&space;\right)2\;\ln&space;R&space;+&space;1&space;-&space;\frac{1}{m}&space;\right]&space;+&space;\left(\frac{C_1}{2}&space;\right)\left(1&space;+&space;\frac{1}{m}&space;\right)&space;-$
$-\left(\frac{C_2}{R^2}&space;\right)\left(1&space;-&space;\frac{1}{m}&space;\right)&space;=&space;0$

And,

$-&space;\left(\frac{P}{8\pi&space;D}&space;\right)\left[\left(1&space;+&space;\frac{1}{m}&space;\right)2&space;\ln&space;r&space;+&space;1&space;-&space;\frac{1}{m}&space;\right]&space;+&space;\left(\frac{C_1}{2}&space;\right)\left(1&space;+&space;\frac{1}{m}&space;\right)&space;-$
$-\left(\frac{C_2}{r^2}&space;\right)\left(1&space;-&space;\frac{1}{m}&space;\right)&space;=&space;0$

Subtracting and solving:
$C_2=\frac{P}{4\pi&space;D}\times&space;\frac{1&space;+&space;\displaystyle\frac{1}{m}}{1&space;-&space;\displaystyle\frac{1}{m}}\times&space;\frac{R^2&space;r^2}{R^2&space;-&space;r^2}\times&space;\ln&space;\frac{R}{r}$
And then
$C_1=\frac{P}{4\pi&space;D}\left[\frac{2(R^2&space;\ln&space;R&space;-&space;r^2&space;\ln&space;r)}{R^2&space;-&space;r^2}&space;+&space;\frac{1&space;-\displaystyle\frac{1}{m}}{1&space;+&space;\displaystyle\frac{1}{m}}&space;\right]$

Then,
$\frac{M_{xy}}{D}&space;=&space;-\left(\frac{P}{8\pi&space;D}&space;\right)\left[\left(1&space;+&space;\frac{1}{m}&space;\right)2&space;\ln&space;x&space;+&space;1-\frac{1}{m}&space;\right]&space;+&space;\left(\frac{C_1}{2}&space;\right)\left(1&space;+&space;\frac{1}{m}&space;\right)&space;-&space;\left(\frac{C_2}{x^2}&space;\right)\left(1&space;+&space;\frac{1}{m}&space;\right)$
$\frac{M_{yz}}{D}\;=&space;-&space;\left(\frac{P}{8\pi&space;D}&space;\right)\left[\left(1&space;+&space;\frac{1}{m}&space;\right)2\;\ln&space;x&space;-&space;\left(1&space;+\frac{1}{m}&space;\right)&space;\right]&space;+&space;\left(\frac{C_1}{2}&space;\right)\left(1&space;+&space;\frac{1}{m}&space;\right)&space;-&space;\left(\frac{C_2}{x^2}&space;\right)\left(1&space;+&space;\frac{1}{m}&space;\right)$

The maximum Bending Moment is $\inline&space;M_{yz}$ at $\inline&space;x=r$, therefore
$\hat{f_z}=\left(\frac{6}{t^2}&space;\right)\;M_{yz}=\frac{3\;P}{\pi&space;\;t^2}\times&space;\frac{\left(1&space;+&space;\displaystyle\frac{1}{m}&space;\right)}{R^2&space;-&space;r^2}\times&space;\ln\frac{R}{r}$