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Circular Plates

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:

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 (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: (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:


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:

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}

By substitution from equation (7)

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 (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 \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 (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 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.

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 (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 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 (14) at x = 0, y = 0, and therefore C_3=0 from equation (15).

Using equations (9) and (14). 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
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 (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 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 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(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 x=R, \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 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 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, \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 (14) C_2=0 and y=0. From equation(15) C_3=0

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

At x=R, 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 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 (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 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 (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 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)

M_{xy}=0 at 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 y at 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 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 (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

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


- \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 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}