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Circular Plates

Stresses and Strains in loaded Circular Plates and Rings.

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Contents

Circular Plates Symmetrically Loaded.
Solid Circular Plate
Annular Ring , Loaded Around The Inner Edge
Page Comments

Circular Plates Symmetrically Loaded.

Consider a Diametral Section through a plate of thickness

is the centre of the plate and

and

are the principal axes in the plane of the diagram. The axis

is perpendicular to the screen.

Let

be the Centre of Curvature of a section

at a distance

from

. Then if the deflection

is small:

$\frac{dy}{dx}=\theta$

(2)

The radius of curvature in the plane

is given by: $\displaystyle\frac{1}{R_{xx}}=\displaystyle\frac{d^2y}{dx^2}$ (Approximately)

Thus from equation (2)

$\frac{1}{R_{xx}}=\frac{d\theta }{dx}$

(3)

Note that, on a circle of radius

and centre

, lines such as

form part of a cone with

as the apex. Hence

is the Centre of Curvature in the plane

and:

$\frac{1}{R_{yx}} = \frac{\theta }{x}$

(4)

(Approximately)

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

and

the linear Strains are:

$e_x=\frac{u}{R_{xy}} = \left(\frac{1}{E} \right)\left(f_x - \frac{f_z}{m} \right)$

(5)

And,

$e_z=\frac{u}{R_{xy}} = \left(\frac{1}{E} \right)\left(f_z - \frac{f_x}{m} \right)$

(6)

Where

and

are the Stresses in the directions

and

is zero

Solving equations (5) and (6) for the Stresses and incorporating equations (3) and (4) gives:

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

(7)

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

(8)

The Bending Moment per unit length along

is $\displaystyle M_{xy}$ which is given by:

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

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

(9)

By substitution from equation (7)

$D = \frac{E\;t^3}{12\left(1 - \frac{1}{m^2} \right)}$

(10)

Similarly if $M_{yz}$ is the Bending Moment per unit length about

then:

$M_{yz}\times dx = \int_{-\frac{t}{2}}^{\frac{t}{2}}{f_z\times u\,dx\times du}$

Using equation (8)

$M_{yz} = D\left[\left(\frac{1}{m} \right)\left(\frac{d\theta }{dx} \right) + \frac{\theta }{x} \right]$

(11)

Note that:

$f_x=M_{xy}\times \frac{12u}{t^3}$

(12)

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

The diagram shows the Forces and Moments per unit length acting on an element which subtends an angle $\delta \phi$ at the centre.

is the Shearing Force per unit length in the direction of

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 (9) and (11) 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:

$\left(\frac{d}{dx} \right)\left[\left(\frac{1}{x} \right)\times \frac{d(x\theta )}{dx} \right]\;= - \frac{F}{D}$

(13)

is known as a function of

, this equation can be integrated to determine $\theta$ and hence

. Bending Moments and Stresses can then be calculated.

Particular Case

A Plate Loaded with Uniformly distributed load of

per unit Area and a Concentrated load at Centre of

$2\pi \,x\times F = \pi x^2\times w + P$

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

Per unit length of circumferentially (Except at

)

Substituting in Equation (13) and Integrating:

$\theta \;= - \frac{w\;x^3}{16\;D} - \left(\frac{P\;x}{8\pi \;D} \right)\left(2\ln x - 1 \right) + \frac{C_1\,x}{2} + \frac{C_2}{x}$

(14)

But From Equation (2)

$y=\int \theta \;dx + C_3$

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

(15)

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

$2\pi \,x\times F\;=\;\pi x^2\times w\;+\;P$

(1)

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

(2)

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

Substituting in Equation (#19) and Integrating:

$\theta \;=\;-\;\frac{w\;x^3}{16\;D}\;-\;\left(\frac{P\;x}{8\pi \;D} \right)\left(2\ln x\;-\;1 \right)\;+\;\frac{C_1\,x}{2}\;+\;\frac{C_2}{x}$

(3)

But From Equation (#1)

$y=\int \theta \;dx + C_3$

(4)

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

(5)

Solid Circular Plate

Let the radius of the plate be

and the thickness

Uniformly Loaded, Edge Freely Supported

and since $\theta$ and

can not be infinite at the centre, then

from Equation (14) at

, and therefore

from equation (15).

Using equations (9) and (14). At

, $M_{xy}=0$ 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 =

. 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

by substitution from Equation (10)

$=\left(\frac{3\,w\,R^4}{16\,E\,t^3} \right)\left(5 + \frac{1}{m} \right)\left(1 - \frac{1}{m} \right)$

From Equation (7)

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

And at

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

From Equation (8)

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

As above when

, $f_z=\hat{f_z}$ . 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,

and

. Therefore

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.

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

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

$\therefore\;\;\;\;\;\;C_1 = \frac{w\;r^2}{8\;D}$

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

Eliminating

by using equation (10)

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

From Equation (7)

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

This Stress has its greatest numerical value when

( 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 (8)

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

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)

, $\theta=0$ therefore from equation (14)

and

. From equation(15)

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

, $M_{xy}=0$ do from equation (9)

$-\left(\frac{P}{8\;\pi \;D } \right)\left(2\;\ln\,R - 1 \right)\;\left(\frac{P\;R}{8\pi D} \right)\left(\frac{2}{R} \right) + \frac{C_1}{2} -$

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

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 (7)

$f_x = \frac{E\;u}{1 - \frac{1}{m^2}}\times \frac{P}{4\pi \,D}\left(1 + \frac{1}{m} \right)\ln\frac{R}{x}$

$=\left(\frac{3P}{2\pi \,t^2} \right)\left(1 + \frac{1}{m} \right)\ln\frac{R}{x}$

Note: $u=\displaystyle\frac{t}{2}$

And from equation (8),

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

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

be distributed around a circle of radius

It is necessary to divide the plate into two regions, one for

and the other for

. At

the values of $\theta$ ,

and $M_{xy}$ must be the same for both regions.

If , and

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

Since $\theta$ and

are not infinite at

then

, and since

when

and

, then above equations reduce to: $\theta =\displaystyle\frac{C_1 x}{2}$ And, $y=\displaystyle\frac{C_1\;x^2}{4}$

If and

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

And from Equation (15),

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

Equating the values of $\theta$ and $M_{xy}$ at

gives the following equations:

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

(16)

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

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

gives:

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

(17)

From Equations (16) to (17) the constants are found to be:

$C_1'=\frac{P}{4\pi \;D}\left[2\;\lnR + \frac{R^2 - r^2}{R^2}\left(\frac{1 - \displaystyle\frac{1}{m}}{1 + \displaystyle\frac{1}{m}} \right) \right]$

$C_2'=-\frac{P\;r^2}{8\pi \;D}$

$C_3'=\frac{P\;r^2}{8\pi \;D}\left(\ln r - 1 \right)$

The Central Deflection is given by the value of

and by substitution equation (15) reduces to:

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

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

Which has a maximum value at

Hence from equation (12)

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

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

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

around the inner edge and is freely supported around the outer edge. $M_{xy}=0$ at

and at

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

$-\left(\frac{C_2}{R^2} \right)\left(1 - \frac{1}{m} \right) = 0$

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

, therefore

$\hat{f_z}=\left(\frac{6}{t^2} \right)\;M_{yz}=\frac{3\;P}{\pi \;t^2}\times \frac{\left(1 + \displaystyle\frac{1}{m} \right)}{R^2 - r^2}\times \ln\frac{R}{r}$