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Circular Plates

Stresses and Strains in loaded Circular Plates and Rings

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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 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 is the deflection y is small:-

(1)

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

The radius of curvature in the plane XOY is given by:

(2)

$\frac{1}{R_{xx}}=\frac{d^2y}{dx^2}\;\;\;\;\;\;\;(Approximately)$

Thus from equation (1)

(3)

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

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:

(4)

$\frac{1}{R_{yx}}\;=\;\frac{\theta }{x}\;\;\;\;\;\;\;\;(Approximately)$

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:

(5)

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

and

(6)

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

Where

and

are the Stresses in the directions OX and OZ.

is zero

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

(7)

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

(8)

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

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

(9)

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

(10)

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

By substitution from equation (7)

(11)

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

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

(12)

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

Using equation (8)

(13)

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

Note that:

(14)

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

(15)

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

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.

(16)

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

Which in the Limit reduces to:

(17)

$M_{xy}\;+\;x\times \frac{\delta M_{xy}}{dx}\;-\;M_{yz}\;+\;Fx\;=\;0$

substituting from Equations (10) and (13) gives:

(18)

$\frac{d^2\theta }{dx^2} + \left(\frac{1}{x} \right)\left(\frac{d\theta }{dx} \right) - \frac{\theta }{x^2} = -\;\frac{F}{D}$

This can be written as:

(19)

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

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.

Example 1:

A particular cases will now be considered: A Plate Loaded with Uniformly distributed load of w per unit Area and a Concentrated load at Centre of P.

(20)

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

(21)

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

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

Substituting in Equation (19) and Integrating:-

(22)

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

But From Equation (1)

(23)

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

(24)

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

Solid Circular Plate

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

Uniformly Loaded, Edge Freely Supported:

and since $\theta$ and y can not be infinite at the centre then

from Equation (22) at

and therefore

from equation(24).

Using equations (10) and (22). At

, $M_{xy}=0$ therefore

(25)

$-\frac{3\,w\,R^2}{16 D} + \frac{C_1}{2} - \frac{w\,R^2}{16 D\,m} + \frac{C_1}{2 m} = 0$

Thus

(26)

$C_1=\left(\frac{w\,R^2}{8\,D} \right)\left(\frac{3+\frac{1}{m}}{1+\frac{1}{m}} \right)$

(27)

$\text{Central Deflection} = y\;\;\;\;\;at\;\;\;\;\;x=R$

Thus

(28)

$y=\frac{w\;R^4}{64\,D} + \frac{w\,R^4}{32 D}\times \frac{3 + \frac{1}{m}}{1 + \frac{1}{m}} = \frac{w\;R^4}{64\;D}\left(\frac{5 + \frac{1}{m}}{1 + \frac{1}{m}} \right)$

Eliminating D by substitution from Equation (11)

(29)

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

From Equation (7)

(30)

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

And at

(31)

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

(32)

$=\frac{3w\;R^2\left(3 + \frac{1}{m} \right)}{8 t^2}$

From Equation (8)

(33)

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

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(24) At

(34)

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

i.e. from Equation (22)

(35)

$-\frac{w\;R^3}{16\;D} + C_1\times \frac{R}{2}\;=\;0$

(36)

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

Using Equation (24):

(37)

$\text{Central Deflection}=-\frac{w\;r^4}{64\;D} + \frac{w\;R^4}{32\;D} = \frac{w\;R^4}{64\;D}$

Eliminating D by using equation (11)

(38)

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

from Equation (7)

(39)

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

This Stress has its greatest numerical value when

( i.e. at the clamped edge), thus

(40)

$\hat{f_x}=\frac{E\times \frac{t}{2}}{1 - \frac{1}{m^2}}\times \frac{w\;R^2}{16\;D}\times 2$

(41)

$=\frac{3\;w\;R^2}{4\;t^2}$

From Equation (8)

(42)

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

From which

(43)

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

(44)

$=\;\frac{3\;w\;R^2(1\;+\;\frac{1}{m})}{8\;t^2}\;\;\;\;\;\;\;(At\;the\;Centre)$

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

, $\theta=0$ therefore from equation(22)

and

. From equation(24)

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

, $M_{xy}=0$ do from equation(10)

(45)

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

From which

(46)

$c_1 = \frac{P}{4\pi \,D}\left(2\,\ln R + \frac{1 - \frac{1}{m}}{1 + \frac{1}{m}} \right)$

Thus

(47)

$\text{Central deflection} = \frac{P\,R^2}{8\pi \,D}\left(\ln R - 1 \right) + \frac{P\;R^2}{16\pi \,D}\left(2\,\ln R + \frac{1\;-\;\frac{1}{m}}{1 + \frac{1}{m}} \right)$

(48)

$=\frac{P\;R^2}{16\pi \,D}\times \frac{(3 + \frac{1}{m})}{(1\;+\;\frac{1}{m})}$

(49)

$=\frac{3P\;R^2}{4\pi\;E\;t^3}\times \left( 3 +\frac{1}{m} \right)\left(1\;-\;\frac{1}{m} \right)$

From Equation (7)

(50)

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

(51)

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

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

And from equation (8)

(52)

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

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

and the other for

. At

the values of $\theta$ , y and $M_{xy}$ must be the same for both regions.

If , and

Hence from Equation (22)

(53)

$\theta = \frac{C_1 x}{2} + \frac{C_2}{x}$

And from Equation (24)

(54)

$y=\frac{C_1 x^2}{4} + C_2\;\ln x + C_3$

Since $\theta$ and

are not infinite at

then

and since

when

and

then above equations reduce to:

(55)

$\theta =\frac{C_1 x}{2}$

And

(56)

$y=\frac{C_1\;x^2}{4}$

If and From Equation (22)

(57)

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

And from Equation (24)

(58)

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

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

gives the following equations:

(59)

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

(60)

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

And

(61)

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

(62)

$=\left(\frac{C_1}{2} \right)\left(1\;+\;\frac{1}{m} \right)$

$M_{xy}=0$ at

gives:

(63)

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

From Equations (59) to (63) the constants are found to be:

(64)

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

(65)

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

(66)

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

The Central Deflection is given by the value of y at x = R and by substitution equation (24) reduces to:-

(67)

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

For

(68)

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

Which has a maximum value at

Hence from equation (14)

(69)

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

(70)

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

Similarly:

(71)

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

(72)

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

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

and at

(73)

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

And

(74)

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

Subtracting and solving:

(75)

$C_2=\frac{P}{4\pi D}\times \frac{1 + \frac{1}{m}}{1 - \frac{1}{m}}\times \frac{R^2 r^2}{R^2 - r^2}\times \ln \frac{R}{r}$

And then

(76)

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

Then,

(77)

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

(78)

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