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Circular Plates

Stresses and Strains in loaded Circular Plates and Rings.
+ View other versions (6)

Circular Plates Symmetrically Loaded.

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


23287/Circular-Plates-0001.png cannot be found in /users/23287/Circular-Plates-0001.png. Please contact the submission author.

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

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

Thus from equation (2)

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

If \inline 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 \inline XOY and \inline YOZ the linear Strains are:


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

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

The Bending Moment per unit length along \inline OZ is \inline \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 (7)

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

Using equation (8)

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 \inline \delta \phi at the centre. \inline F is the Shearing Force per unit length in the direction of \inline OZ.


23287/Circular-Plates-0002.png cannot be found in /users/23287/Circular-Plates-0002.png. Please contact the submission author.

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

i.e., \inline \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:

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

Particular Case

A Plate Loaded with Uniformly distributed load of \inline w per unit Area and a Concentrated load at Centre of \inline 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 \inline x = 0)

Substituting in Equation (13) and Integrating:

But From Equation (2)
y=\int \theta \;dx + C_3

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

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

Substituting in Equation (#19) and Integrating:

But From Equation (#1)

Solid Circular Plate

Let the radius of the plate be \inline R and the thickness \inline t.

Uniformly Loaded, Edge Freely Supported

\inline P=0 and since \inline \theta and \inline y can not be infinite at the centre, then \inline C_2=0 from Equation (14) at \inline x = 0, \inline y = 0, and therefore \inline C_3=0 from equation (15).

Using equations (9) and (14). At \inline x=R, \inline 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
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 = \inline y at \inline 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 \inline D 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 \inline 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 (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 \inline x=0, \inline f_z=\hat{f_z}. Therefore \inline \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, \inline P=0 and \inline C_2=0 at \inline x=0, \inline y=0. Therefore \inline C_3=0 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.

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

i.e. from Equation (14), \inline -\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 = \inline -\displaystyle\frac{w\;r^4}{64\;D} + \displaystyle\frac{w\;R^4}{32\;D} = \displaystyle\frac{w\;R^4}{64\;D}

Eliminating \inline D 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 \inline 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 (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, \inline \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 \inline x=0, \inline \theta=0 therefore from equation (14) \inline C_2=0 and \inline y=0. From equation(15) \inline C_3=0

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

At \inline x=R, \inline 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, \inline 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: \inline 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 \inline P be distributed around a circle of radius \inline r.


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

  • If \inline x<r, \inline w=0 and \inline P=0

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

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

  • If \inline x>r and \inline w=0

From Equation (14), \inline \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 \inline \theta and \inline M_{xy} at \inline 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}

\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)

\inline M_{xy}=0 at \inline x=R gives:

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 \inline y at \inline x = R 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]

  • For \inline 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 \inline x = r

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]

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


23287/Circular-Plates-0004.png cannot be found in /users/23287/Circular-Plates-0004.png. Please contact the submission author.

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


- \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]

\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 \inline M_{yz} at \inline 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}