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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:-

$E\times \frac{du}{dr}\;=\;p\;-\;\frac{f}{m}$

(1)

$And\;\;\;\;\;E\times \frac{u}{r}\;-\;\frac{f}{m}$

(2)

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

$\left(f\;-\;p \right)\left(1\;-\;\frac{1}{m} \right)\;+\;r\times \frac{df}{dr}\;-\;\left(\frac{r}{m} \right)\left(\frac{dp}{dr} \right)\;=\;0$

(3)

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:-

$\left(\frac{\rho r\,\delta \theta \times \delta r}{g} \right)\times r\;\omega ^2$

(4)

Which for unit thickness:-

$\left(\frac{\rho r\,\delta \theta \times \delta r}{g} \right)\times r\;\omega ^2$

(5)

$=\;\frac{\rho \,r^2\,\omega ^2\;\delta r\times \delta \theta }{g}$

(6)

The equilibrium equation in the radial direction is:-

$2f\,\delta r\,\sin\frac{1}{2}\,\delta \theta \;+\;\rho r\,\delta \theta \;-\;(p\;+\;\delta p)(r\;+\;\delta r)\delta \theta \;=\;\frac{\rho \,r^2\,\omega ^2\;\delta r\times \delta \theta }{g}$

(7)

In the limit this equation reduces to:-

$f\;-\;p\;-\;r\,\frac{dp}{dr}\;=\;\frac{\rho \,r^2\;\omega ^2}{g}$

(8)

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

$\left(r\,\frac{dp}{dr}\;+\;\frac{\rho \,r^2\,\omega ^2}{g} \right)\left(1\;-\;\frac{1}{m} \right)\;+\;r\times \frac{df}{dr}\;-\;\left(\frac{r}{m} \right)\frac{dp}{dr}\;=\;0$

(9)

Rearranging

$\frac{df}{dr}\;+\;\frac{dp}{dr}\;=\;-\;\left(\frac{\rho \.r\,\omega ^2}{g} \right)\left(1\;+\;\frac{1}{m} \right)$

(10)

Integrating:-

$f\;+\;p\;=\;-\;\left(\frac{\rho \,r^2\,\omega ^2}{2g} \right)\left(1\;+\;\frac{1}{m} \right)\;+\;2A$

(11)

Subtracting equation (8)

$2p\;+\;r\,\frac{dp}{dr}=\;-\;\left(\frac{\rho \,r^2\,\omega ^2}{2g} \right)\left(3\;+\;\frac{1}{m} \right)\;+\;2A$

(12)

$\therefore\;\;\;\;\;\frac{1}{r}\;\frac{d(p\,r)^2}{dr}\;=\;-\;\frac{\rho \,r^2\,\omega ^2(3\;+\;1/m)}{2g}\;+\;2A$

(13)

Integrating:-

$p\,r^2\;=\;-\;\frac{\rho \,r^4\,\omega ^2(3\;+\;1/m)}{8g}\;+\;2A\,r^2\;-\;B$

(14)

$Or\;\;\;\;\;\;\mathbf{p\;=\;A\;-\;\frac{B}{r^2}\;-\;\left(3\;+\;\frac{1}{m} \right)\left(\frac {\rho \,r^2\;\omega ^2}{8g} \right)}$

(15)

Substituting from equation (11)

$\mathbf{f\;=\;A\;+\;\frac{B}{r}\;-\;\left(1\;+\;\frac{3}{m} \right)\left(\frac{\rho \;r\;\omega ^2}{8g} \right)}$

(16)

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:-

$p\;=\;0\;=\;A\;-\;\left(3\;+\;\frac{1}{m} \right)\left(\frac{\rho \,R^2\,\omega ^2}{8g} \right)$

(17)

From Which:-

$p\;=\;\left(\frac{\rho \;\omega ^2}{8g} \right)\left(3\;+\;\frac{1}{m} \right)\left(R^2\;-\;r^2 \right)$

(18)

$And\;\;\;\;\;f\;=\;\left(\frac{\rho \;\omega ^2}{8g} \right)\left[ \left(3\;+\;\frac{1}{m} \right)R^2\:-\:\left(1\;+\;\frac{3}{m} \right)r^2 \right]$

(19)

At the centre r = 0 and so:-

$p\;=\;f\;=\;\left(3\;+\;\frac{1}{m} \right)\left(\frac{\rho \;\omega ^2\,R^2}{8g} \right)$

(20)

This is the Maximum Stress.

At the outside :-

$f\;=\;\left(1\;-\;\frac{1}{m} \right)\left(\frac{\rho \,\omega ^2\,r^2}{g} \right)$

(21)

For a value of $\displaystyle \frac{1}{m}\;=\;0.3$

$\hat{f}\;=\;\left(\frac{3.3}{8} \right)\left(\frac{\rho \,\omega ^2\,R^2}{g} \right)\;=\;0.41\times \frac{\rho \,\omega ^2\,R^2}{g}\;\;\;\;(At\;the\;centre)$

(22)

And at the outside:-

$f\;=\;\frac{0.7}{4}\times \frac{\rho \,\omega ^2\,R^2}{g}\;=\;0.425\;\hat{f}$

(23)

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)

$0\;=\;A\;-=\;\frac{B}{R_1^2}\;-\;\left(3\;+\;\frac{1}{m} \right)\left(\frac{\rho \;R_1^2\,\omega ^2}{ 8g }\right)$

(24)

$0\;=\;A\;-=\;\frac{B}{R_2^2}\;-\;\left(3\;+\;\frac{1}{m} \right)\left(\frac{\rho \;R_2^2\,\omega ^2}{ 8g }\right)$

(25)

Solving the above two equations for A and B:-

$B\;=\;\left(3\;+\;\frac{1}{m} \right)\left(\frac{\rho \,\omega ^2}{8g} \right)\left(R_1^2\,R_2^2 \right)$

(26)

$A\;=\;\left(3\;+\;\frac{1}{m} \right)\left(\frac{\rho \,\omega ^2}{8g} \right)\left(R_1^2\;+\;R_2^2 \right)$

(27)

$Thus\;\;\;\;\;\;p\;=\;\left(3\;+\;\frac{1}{m} \right)\left(\frac{\rho \,\omega ^2}{8g} \right)\left(R_1^2\;+\;R_2^2\;-\;\frac{R_1^2\,R_2^2}{r^2}\;-\;r^2\right)$

(28)

Substituting from equation(15)

$And\;\;\;\;\;f\;=\;\left(\frac{\rho \;\omega ^2}{8g} \right)\left[\left(3\;+\;\frac{1}{m} \right)\left(R_1^2\;+\;R_2^2\;+\;\frac{R_1^2\,R_2^2}{r^2} \right)\;-\;\left(1\;+\;\frac{3}{m} \right)r^2 \right]$

(29)

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

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

(30)

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

(31)

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

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

(32)

$If\;\;\;R_1\rightarrow R_2\;=\;R\;\;\;\;Then\;\;\;\;\hat{f}\rightarrow \frac{\rho \;\omega ^2\;R^2}{g}$

(33)

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)

$\hat{f}\;=\;\frac{0.28}{4\times 32.2\times 12}\left(\frac{10,000\times 2\pi }{60} \right)^2\;\left(0.7\times 1^2\;+\;3.3\times 5^2 \right)$

(34)

$=\;16,500\;lb.in^{-2}$

(35)

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

$\frac{1}{2}(f\;-\;p)\;=\;\left(\frac{\rho \;\omega ^2}{8g} \right)\left[\left(3\;+\;\frac{1}{m} \right)\frac{R_1^2\;R_2^2}{r^2} \;+\;\left(1\;-\;\frac{1}{m} \right)r^2\right]$

(36)

It can be seen from the diagram that the greatest Stress difference occurs at

$r\;=\;R_1$

(37)

and the Maximum Shearing Stress is:-

$\frac{0.28}{8\times 32.2\times 12}\left(\frac{10.000\times 2\pi }{60} \right)^2\left(3.3\times\frac{1^2\times5^2}{1^2}\;+\;0.7\times 1^2 \right)\;=\;8250\;lb.in.^{-2}$

(38)

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:-

$E\,e\;=\;f_1\;-\;\left(\frac{1}{m} \right)\left(f\;+\;p \right)$

(39)

$E\;\frac{du}{dr}\;=\;p\;-\;\left(\frac{1}{m}\right)\left(f\;+\;f_1$

(40)

$E\;\frac{u}{r}\;=\;f\;-\;\left(\frac{1}{m}\right)\left(p\;+\;f_1 \right)$

(41)

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

From Equation (41)

$E\,.\,\frac{du}{dr}\;=\;f\;-\;\left(\frac{1}{m} \right)\left(p\;+\;f_1 \right)\;+\;\left[\frac{df}{dr}\;-\;\left(\frac{1}{m} \right)\left(\frac{dp}{dr} \right)\;-\;\left(\frac{1}{m}\right)\left(\frac{df_1}{dr} \right) \right]$

(42)

Blank

From Equation (40)

$=\;p\;-\;\left(\frac{1}{m} \right)\left(f\;+\;f_1 \right)$

(43)

$\therefore\;\;\;\;\;\;(f\;-\;p)\left(1\;+\;\frac{1}{m} \right)\;+\;r\times \frac{df}{dr}\;-\;\left(\frac{r}{m} \right)\left(\frac{dp}{dr} \right)\;-\;\left(\frac{r}{m} \right)\left( \right)\;=\;0$

(44)

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)

$(f\;-\;p)\left(1\;+\;\frac{1}{m} \right)\;+\;r\times \frac{df}{dr}\left(1\;-\;\frac{1}{m^2} \right)\;-\;\left(\frac{r}{m} \right)\left(1\;+\;\frac{1}{m}\right)\left(\frac{dp}{dr} \right)\;=\;0$

(45)

$Or\;\;\;\;\;f\;-\;p\;-\;r\times \left(1\;-\;\frac{1}{m} \right)\left(\frac{dp}{dr} \right)\;=\;0$

(46)

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

$-\;r\left(1\;-\;\frac{1}{m} \right)\left(\frac{df}{dr} \right)\;-\;r\left(1\;-\;\frac{1}{m} \right)\left(\frac{dp}{fdr} \right)\;=\;\frac{\rho \times r\times \omega ^2}{g}$

(47)

$Or\;\;\;\;\;\frac{df}{dr}\;+\;\frac{dp}{dr}\;=\;\frac{\rho \times r\times \omega ^2}{\left(1\;-\;\frac{1}{m} \right)g}$

(48)

Integrating;-

$Or\;\;\;\;\;f\;+\;p\;=\;\frac{\rho \times r^2\times \omega ^2}{2\left(1\;-\;\frac{1}{m} \right)g}\;+\;2A$

(49)

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 $\displaystyle \frac{1}{1\;-\;\frac{1}{m}}\;instead\;of\;1\;+\;\frac{1}{m}$

\B{Solid Cylinder}

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

$\hat{f}\;=\;\frac{3\;-\;\frac{2}{m}}{1\;-\;\frac{1}{m}}\times \frac{\rho \times\omega ^2\times R^2}{8g}$

(50)

$If\;\;\;\;\;\frac{1}{m}\;=\;0.3$

(51)

$Then\;\;\;\;\;\hat{f}\;=\;\frac{2.4}{5.6}\times \frac{\rho \times\omega ^2\times R^2}{8g}\;=\;0.43\;\frac{\rho \times \omega ^2\times R^2}{8g}$

(52)

( 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. $\displaystyle p\;=\;f\;=\;Constant$

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

$\rho \times r\times \delta \theta \times \delta r\times \frac{t}{g}$

(53)

And the centrifugal Force will:-

$\rho \times r^2\times\omega ^2\times \delta \theta \times \delta r\times \frac{t}{g}$

(54)

Hence the equilibrium Equation is:-

$2\;f\;\delta r\times \sin\frac{1}{2}\delta \theta \;+\;f\;r\;\delta \theta \times t\;=\;f(r\;+\;\delta r)\;\delta \theta \;(t\;+\;\delta t)\;+\;\frac{\rho \;r^2\;\psi ^2\;t\;dr}{g}$

(55)

Which in the limit is:-

$f\;t\times dr\;=\;f\;r\times dt\;+\;f\;t\times dr\;+\;\frac{\rho\;\omega^2\;t\times dr}{g}$

(56)

Integrating:-

$\lnt\;=\;-\;\frac{\rho \;r^2\;\omega ^2}{2\;f\;g}\;+\;Constant$

(57)

$Or\;\;\;\;\;\;t\;=\;A\;e^{-\;\frac{p\;r^2\;\omega ^2}{2\;f\;g}}$

(58)

$Or\;\;\;\;\;\;t\;=\;t_0\;e^{-\;\frac{p\;r^2\;\omega ^2}{2\;f\;g}}$

(59)

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.

$T\;=\;A\;e^{-\frac{\rho \,r^2\,\omega ^2}{2f\;g}}$

(60)

$At\;\;\;\;\;\;\;r\;=\;12\;in.$

(61)

$t\;=\;\frac{3}{8}\;=\;A\;e^{-\rho \,\omega ^2\times\frac{144}{2f\,g}}$

(62)

$At\;\;\;\;\;\;\;r\;=\;1\;in.$

(63)

$t\;=\;A\;e^{-\frac{\rho \,\omega ^2}{2f\,g}}\;=\;\frac{3}{8}\;e^{\rho \,\omega ^2\times \frac{143}{2f\,g}}$

(64)

$Where\;\;\;\;\;\rho \;\omega ^2\times \frac{143}{2f\;g}\;=\;0.28\left(\frac{10,000\pi }{30} \right)^2\times \frac{143}{2}\times30,000\times 32.2\times 12$

(65)

$=\;1.89$

(66)

$\therefore\;\;\;\;\;\;t\;=\;\frac{3}{8}\;e^{1.89}\;\;=\;2.5\;inches$

(67)

Rotating Discs and Cylinders

Discs of Uniform Thickness.

A Solid Disc

A Disc with a Central Hole

Example 1

Long Cylinders

A Disc of Uniform Strength

Example 2