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Rotating Discs and Cylinders

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The Stresses and Strains generated in a rotating disc or cylinder

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Discs Of Uniform Thickness.

For a "thin" disc it can be assumed that the Stress in the Axial direction is zero. Due to the rotation of the disc, Circumferential \displaystyle f (Hoop) and radial \displaystyle p Stresses will occur. Both these Stresses are Tensile. If the radial shift at a radius r is u, then the Strain Equations are:-

Differentiating equation (2) and equating it to equation (1) gives:-

13108/img_disc_3_0001.jpg
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If the angular velocity of rotation is \displaystyle \omega and the density of the disc material is \displaystyle \rho then the element shown in the diagram is subjected to a centrifugal force which:-

Which for unit thickness:-

The equilibrium equation in the radial direction is:-

In the limit this equation reduces to:-

From equation (8) obtain \displaystyle f\;-\;p and substitute into equation (3)

Rearranging

Integrating:-

Subtracting equation (8)

Integrating:-

Substituting from equation (11)

A Solid Disc

Since the Stresses can no be infinite at the centre of a solid Disc B must be zero.

If R is the outside radius of the disc then rewriting equations (14) and (15) then:-

From Which:-

At the centre r = 0 and so:-

This is the Maximum Stress.

At the outside :-

13108/img_disc_3_0002.jpg
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For a value of \displaystyle \frac{1}{m}\;=\;0.3

And at the outside:-

The variations of the Hoop and Radial Stresses with Radius are shown on the diagram.

A Disc With A Central Hole

The Radial Stress is Zero at both the inner and outer radii. If the value of these is \displaystyle R_1\;\;and\;\;R_2 then using equation (14)

Solving the above two equations for A and B:-

Substituting from equation(15)

p is a maximum when \displaystyle r\;=\;\sqrt{R_1\,R_2}

Note that if \displaystyle R_1 is very small \displaystyle \hat{f}\rightarrow \left(3\;+\;\frac{1 } {m} \right)\left(\frac{\rho \;\omega ^2\;R_2^2}{4g} \right) which is tweiew the value which would be found in a solid disc.

At the outside

13108/img_disc_4_0003.jpg
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As in the case of a thin rotating cylinder. The variations in Stress are shown on the diagram.
Example 1:
A thin uniform disc of 10 in. diameter with a central hole of 2 in. , runs at 10,000 r.p.m. Calculate the maximum Principal Stress and the maximum Shear Stress in the disc. Take \displaystyle \frac{1}{m}\;\;\;=\;\;0.3\;\;\;\;\;and\;\;\;\;\;Density\;=\;0.28\;lb.in.^{-3}

The maximum Principal Stress will be at the inside and is given by equation (31)

The maximum Shearing Stress at any radius is given by:-

It can be seen from the diagram that the greatest Stress difference occurs at and the Maximum Shearing Stress is:-

Long Cylinders

Assume that the Longitudinal Stress is \displaystyle f_1 and that the Longitudinal Strain e is constant. i.e. the cross sections remain plane which must be true away from the ends. The analysis is similar that used for a rotating disc and the Strain Equations are:-

Using Equations (40) and (41) eliminate \displaystyle \frac{du}{dr}

From Equation (41)

From Equation (40)

Substituting \displaystyle \frac{df_1}{dr}\;=\;\left(\frac{1}{m} \right)\left(\frac{df}{dr}\;+\;\frac{dp}{dr} \right) which was obtained from equation (39)

The equilibrium equation is as equation (8) and subtracting equation (46) we get.

Integrating;-

This can be compared to equation (11) which is for a thin disc. It can be seen that the result for a long cylinder can be obtained from those for a thin disc by writing \displaystyle \frac{1}{1\;-\;\frac{1}{m}}\;instead\;of\;1\;+\;\frac{1}{m}

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{Solid Cylinder}

The maximum Stress occcurs at the centre where p and f are equal

( This compares with \displaystyle 0.41\times\frac{\rho \times \omega ^2\times R^2}{8g} for a solid disc)

Hollow Cylinder

\displaystyle \hat{p}\;=\;\left(\frac{3\;-\;\frac{2}{m}}{1\;-\;\frac{1}{m}} \right)\times \frac{\rho \times \omega ^2}{8g}\times\left(R_2\;-\;R_1 \right)^2

\displaystyle \hat{f}\;=\;\frac{\rho \times \omega ^2}{4g\left(1\;-\;\frac{1}{m} \right)}\left[\left(1\;-\;\frac{2}{m}\right)R_1^2\;+\;\left(3\;-\;\frac{2}{m} \right) R_2^2 \right]

These values do not differ greatly from those for a thin disc.

A Disc Of Uniform Strength

Consider the condition of equal stress at all radii. i.e. \displaystyle p\;=\;f\;=\;Constant

13108/img_5.jpg
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Le \displaystyle t be the thickness of the disc at a radius \displaystyle r and

assume that the thickness will be \displaystyle t\;+\;\delta t at radius \displaystyle r\;+\;\delta r

The mass of the element will be approximately

And the centrifugal Force will:-

Hence the equilibrium Equation is:-

Which in the limit is:-

Integrating:-

Example 2:
A Turbine rotor Disc is 24 in. Diameter at the blade ring and is keyed to a 2 in diameter shaft. If the minimum thickness is 3/8 in. what should be the thickness at the shaft for a uniform stress of 30,000 lb./sq.in. aqt 10,000 r.p.m.? Density of material = 0.28 lb./cu.in.


Last Modified: 2009-03-26 08:16:20     Page Rendered: 2010-03-10 21:42:41

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