Circular Plates

Circular Plates Symmetrically Loaded.

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

Let $C$ be the Centre of Curvature of a section $ab$ at a distance $x$ from $O$. Then if the deflection $y$ is small:

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

Thus from equation (1)

Note that, on a circle of radius $x$ and centre $O$, lines such as $ab$ form part of a cone with $C$ as the apex. Hence $C$ is the Centre of Curvature in the plane $YOZ$ and:

(Approximately)

If $u$ 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 $XOY$ and $YOZ$ the linear Strains are:

And,

Where $f_x$ and $f_y$ are the Stresses in the directions $OX$ and $OZ$, $f_y$ is zero

Solving equations (4) and (5) for the Stresses and incorporating equations (2) and (3) gives:

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

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

By substitution from equation (6)

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

Using equation (7)

Note that:

$f_z=M_{yz}\times \frac{12u}{t^3}$

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

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

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

Which in the Limit reduces to: $M_{xy} + x\times \frac{\delta M_{xy}}{dx} - M_{yz} + Fx = 0$ Substituting from Equations (8) and (10) gives: $\frac{d^2\theta }{dx^2} + \left(\frac{1}{x} \right)\left(\frac{d\theta }{dx} \right) - \frac{\theta }{x^2} = - \frac{F}{D}$

This can be written as:

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

Particular case

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

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

Per unit length of circumference (Except at $x = 0$)

Substituting in Equation (12) and Integrating:

But From Equation (1) $y=\int \theta \;dx + C_3$

Solid Circular Plate

Let the radius of the plate be $R$ and the thickness $t$.

Uniformly loaded, Edge freely supported

$P=0$ and since $\theta$ and $y$ can not be infinite at the centre, then $C_2=0$ from Equation (13) at $x = 0$, $y = 0$, and therefore $C_3=0$ from equation (14).

Using equations (8) and (13). At $x=R$, $M_{xy}=0$ therefore $-\frac{3\,w\,R^2}{16 D} + \frac{C_1}{2} - \frac{w\,R^2}{16 D\,m} + \frac{C_1}{2 m} = 0$ Thus, $C_1=\left(\frac{w\,R^2}{8\,D} \right)\left(\frac{3+\displaystyle\frac{1}{m}}{1+\displaystyle\frac{1}{m}} \right)$ Central Deflection = $y$ at $x=R$. Thus, $y=\frac{w\;R^4}{64\,D} + \frac{w\,R^4}{32 D}\times \frac{3 + \displaystyle\frac{1}{m}}{1 + \displaystyle\frac{1}{m}} = \frac{w\;R^4}{64\;D}\left(\frac{5 + \displaystyle\frac{1}{m}}{1 + \displaystyle\frac{1}{m}} \right)$

Eliminating $D$ by substitution from Equation (9) $=\left(\frac{3\,w\,R^4}{16\,E\,t^3} \right)\left(5 + \frac{1}{m} \right)\left(1 - \frac{1}{m} \right)$

From Equation (6) $f_x = \frac{E\;u}{1 - \displaystyle\frac{1}{m^2}}\left(- \frac{w\,x^2}{16\,D}\times \left\{3 - \frac{1}{m} \right\} + \frac{w\;R^2}{16\,D}\left\{3 + \frac{1}{m} \right\} \right)$

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

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

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

Note: the Maximum Stresses occur at the centre.

Uniformly Loaded with the Edge Clamped

As in the last case, $P=0$ and $C_2=0$ at $x=0$, $y=0$. Therefore $C_3=0$ from equation(14)

At $x=R$, $\displaystyle\frac{dy}{dx}=\theta =0$

i.e. from Equation (13), $-\displaystyle\frac{w\;R^3}{16\;D} + C_1\times \displaystyle\frac{R}{2} = 0$ $\therefore\;\;\;\;\;\;C_1 = \frac{w\;r^2}{8\;D}$

Using Equation (14),Central Deflection = $-\displaystyle\frac{w\;r^4}{64\;D} + \displaystyle\frac{w\;R^4}{32\;D} = \displaystyle\frac{w\;R^4}{64\;D}$

Eliminating $D$ by using equation (9) $=\left(\frac{3\,w\;R^4}{16\;E\;t^3} \right)\left(1 - \frac{1}{m^2} \right)$

From Equation (6) $f_x = \left(\frac{E\;u}{1 - \displaystyle\frac{1}{m^2}} \right)\left(- \frac{w\;x^2}{16\;D}\left\{3 + \frac{1}{m} \right\} + \frac{w\;R^2}{16\;D}\left\{1 + \frac{1}{m} \right\} \right)$

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

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

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

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

At $x=0$, $\theta=0$ therefore from equation (13) $C_2=0$ and $y=0$. From equation(14) $C_3=0$

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

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

From which, $C_1 = \displaystyle\frac{P}{4\pi \,D}\left(2\,\ln R + \displaystyle\frac{1 - \displaystyle\frac{1}{m}}{1 + \displaystyle\frac{1}{m}} \right)$

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

From Equation (6) $f_x = \frac{E\;u}{1 - \frac{1}{m^2}}\times \frac{P}{4\pi \,D}\left(1 + \frac{1}{m} \right)\ln\frac{R}{x}$ $=\left(\frac{3P}{2\pi \,t^2} \right)\left(1 + \frac{1}{m} \right)\ln\frac{R}{x}$ Note: $u=\displaystyle\frac{t}{2}$

And from equation (7), $f_z =\left(\frac{3P}{2\pi \,t^2} \right)\left[\left(1 + \frac{1}{m} \right)\ln\frac{R}{x} + 1 - \frac{1}{m} \right]$

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 $P$ be distributed around a circle of radius $r$.

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

• If $x<r$, $w=0$ and $P=0$

Hence, from Equation (13) $\theta = \displaystyle\frac{C_1 x}{2} + \displaystyle\frac{C_2}{x}$. And from Equation (14), $y=\displaystyle\frac{C_1 x^2}{4} + C_2\;\ln x + C_3$

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

• If $x>r$ and $w=0$

From Equation (13), $\theta =-\left(\displaystyle\frac{P x}{8\pi D} \right )\left(2 \ln x-1 \right) + \displaystyle\frac{C_1'x}{2} + \displaystyle\frac{C_2'}{x}$

And from Equation (14), $y=-\left(\frac{P\;x^2}{8\pi \;D} \right)\left(\ln x - 1 \right) + \frac{C_1'x^2}{2} + C_2'\;\ln {x} + C_3'$

Equating the values of $\theta$ and $M_{xy}$ at $x=r$ gives the following equations:

$-\left(\frac{P\,r^2}{8\pi \;D} \right)\left(\ln r - 1 \right) + \frac{C_1'\;r^2}{4} + C_2'\;\ln r = \frac{C_1\;r^2}{4}$

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

$M_{xy}=0$ at $x=R$ gives:

From Equations (15) to (16) the constants are found to be: $C_1'=\frac{P}{4\pi \;D}\left[2\;\ln R + \frac{R^2 - r^2}{R^2}\left(\frac{1 - \displaystyle\frac{1}{m}}{1 + \displaystyle\frac{1}{m}} \right) \right]$ $C_2'=-\frac{P\;r^2}{8\pi \;D}$ $C_3'=\frac{P\;r^2}{8\pi \;D}\left(\ln r - 1 \right)$

The Central Deflection is given by the value of $y$ at $x = R$ and by substitution equation (14) reduces to:

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

• For $x>r$

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

Which has a maximum value at $x = r$

Hence from equation (11)

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

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

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

Annular Ring , Loaded around the Inner edge

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

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

And,

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

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

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

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