CodeCogs

Stresses and Strains in loaded Circular Plates and Rings

You're viewing an older version of this page (#3537). View the current version.

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:-

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

(1)

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

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

(2)

Thus from equation (1)

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

(3)

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:

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

(4)

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:

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

(5)

and

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

(6)

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

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

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

(7)

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

(8)

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

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

(9)

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

(10)

By substitution from equation (#7)

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

(11)

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

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

(12)

Using equation (#8)

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

(13)

Note that:

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

(14)

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

(15)

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.

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

(16)

Which in the Limit reduces to:

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

(17)

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

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

(18)

This can be written as:

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

(19)

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.

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

(20)

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

(21)

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

Substituting in Equation (#19) and Integrating:-

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

(22)

But From Equation (#1)

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

(23)

Blank

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

(24)

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 (#22) at $x = 0$ , $y = 0$ and therefore $C_3=0$ from equation(24).

Using equations (10) and (22). At $x=R$ , $M_{xy}=0$ therefore

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

(25)

Thus

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

(26)

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

(27)

Thus

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

(28)

Eliminating D by substitution from Equation (#11)

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

(29)

From Equation (#7)

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

(30)

And at $x=0$

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

(31)

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

(32)

From Equation (#8)

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

(33)

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(#24) At $x=R$ ,

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

(34)

i.e. from Equation (#22)

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

(35)

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

(36)

Using Equation (#24):

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

(37)

Eliminating D by using equation (11)

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

(38)

from Equation (#7)

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

(39)

This Stress has its greatest numerical value when $x = R$ ( i.e. at the clamped edge), thus

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

(40)

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

(41)

From Equation (#8)

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

(42)

From which

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

(43)

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

(44)

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

At $x=0$ , $\theta=0$ therefore from equation(#22) $C_2=0$ and $y=0$ . From equation(#24) $C_3=0$

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

At $x=R$ , $M_{xy}=0$ do from equation(#10)

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

(45)

From which

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

(46)

Thus

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

(47)

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

(48)

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

(49)

From Equation (#7)

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

(50)

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

(51)

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

And from equation (#8)

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

(52)

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 (#22)

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

(53)

And from Equation (#24)

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

(54)

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:

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

(55)

And

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

(56)

If $x>r$ and $w=0$ From Equation (#22)

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

(57)

And from Equation (#24)

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

(58)

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

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

(59)

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

(60)

And

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

(61)

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

(62)

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

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

(63)

From Equations (#64) to (#70) the constants are found to be:

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

(64)

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

(65)

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

(66)

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

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

(67)

For $x>r$

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

(68)

Which has a maximum value at $x = r$

Hence from equation (#14)

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

(69)

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

(70)

Similarly:

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

(71)

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

(72)

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 $x=r$ .

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

(73)

And

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

(74)

Subtracting and solving:

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

(75)

And then

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

(76)

Then,

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

(77)

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

(78)

The maximum Bending Moment is $M_{yz}$ at $x=r$ , therefore

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

(79)

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

(80)