Circular Plates

Bending moment is the algebraic sum of all moments located between a cross section and one end of a structural member; a bending moment that bends the beam convex downward is positive, and one that bends it convex upward is negative.

A shear stress is defined as the component of stress coplanar with a material cross section. Shear stress arises from a force vector perpendicular to the surface normal vector of the cross section.

Deflection is a term that is used to describe the degree to which a structural element is displaced under a load.

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

Stresses and Strains in loaded Circular Plates and Rings

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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:

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

(1)

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)

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

(2)

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:

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

(3)

(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:

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

(4)

And

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

(5)

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:

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

(6)

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

(7)

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

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

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

(8)

By substitution from equation (#6)

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

(9)

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)

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

(10)

Note that:

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

(11)

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

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:

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

(12)

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.

Worked example #807 not found.

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 circumferentially (Except at $x = 0$ )

Substituting in Equation (#12) and Integrating:

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

(13)

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

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

(14)

Example 1

Problem

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

Workings

\f[2π \,x\times F\;=\;π x^2\times w\;+\;P\f] \f[\therefore\;\;\;\;\;\;\;F\;=\;\frac{w\,x}{2}\;+\;\frac{P}{2π \;x}\f]

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

Substituting in Equation (#19) and Integrating:

\f[θ \;=\;-\;\frac{w\;x^3}{16\;D}\;-\;\left(\frac{P\;x}{8π \;D} \right)\left(2\ln x\;-\;1 \right)\;+\;\frac{C_1\,x}{2}\;+\;\frac{C_2}{x}\f]#22 But From Equation (#1) \f[y=\int θ \;dx + C_3\f]#23

\f[\therefore\;\;\;\;\;y\;=\;-\;\frac{w\;x^4}{64\;D}\;-\;\left(\frac{P\;x}{8π \;D} \right)\left(2\ln x\;-\;1 \right)\;+\;\frac{C_1\;x}{4}\;+\;C_2\;\ln x\;+\;C_3\f]#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 (#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 $x=R$ , $M_{xy}=0$ do 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:

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

(15)

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

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:

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

(16)

From Equations (#15) to (#16) 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 $y$ 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 $\hat{f_z}=\left(\frac{6}{t^2} \right)\;M_{yz}=\frac{3\;P}{\pi \;t^2}\times \frac{\left(1 + \displaystyle\frac{1}{m} \right)}{R^2 - r^2}\times \ln\frac{R}{r}$