# Rotating Discs and Cylinders

**Contents**

## Introduction

The rotation of a fast moving disc or cylindrical member can set up axial, circumferential and radial stresses. For instance, in the turbine section of an aircraft gas turbine engine, the rotor disc frequently turns at speeds in the region of 10000 r.p.m. This generates large amounts of stresses and strains on the material of the rotor disc. Similar loads are bourne by the cylindrical turbine rotor shafts that transmit the rotational movement from the turbine to the compressor section. To design a disc or shaft that can withstand these stresses without catastrophic failure in flight, it is important to understand the loads acting on them. This section analyses the stress and strains acting on rotating discs and cylinders.##### MISSING IMAGE!

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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 (Hoop) and radial Stresses will occur. Both these Stresses are Tensile. If the radial shift at a radius is , then the Strain Equations are: Differentiating equation (2) and equating it to equation (1) gives:##### MISSING IMAGE!

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## A Solid Disc

Since the Stresses can not 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): From Which: At the centre r = 0 and so: This is the Maximum Stress. At the outside :##### MISSING IMAGE!

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## A Disc With A Central Hole

The Radial Stress is Zero at both the inner and outer radii. If the value of these is , then using equation (14) Solving the above two equations for A and B: Substituting from equation(15) p is a maximum when Note that if is very small which is tweiew the value which would be found in a solid disc.**At the outside**

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##### Example - Stress in a hollow uniform disc

^{-3}

^{-2}The maximum Shear Stress = 8250 lb.in.

^{-2}

## Long Cylinders

Assume that the Longitudinal Stress is 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 to that used for a rotating disc and the Strain Equations are: Using Equations (40) and (41) eliminate From Equation (41) From Equation (40) Substituting 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## A Disc Of Uniform Strength

Consider the condition of equal stress at all radii, i.e. .##### MISSING IMAGE!

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##### Example - A Disc Of Uniform Strength

^{2}. aqt 10,000 r.p.m.? Density of material = 0.28 lb./in

^{3}.