Rotating Discs and Cylinders
The Stresses and Strains generated in a rotating disc or cylinder.
IntroductionThe 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.
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:
A Solid DiscSince 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):
A Disc With A Central HoleThe Radial Stress is Zero at both the inner and outer radii. If the value of these is , then using equation (14)
Example - Stress in a hollow uniform disc
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 and Density = 0.28 lb.in.-3
The maximum Principal Stress will be at the inside and is given by equation (31)
The maximum Principal Stress = 16,500 lb.in-2 The maximum Shear Stress = 8250 lb.in.-2
Long CylindersAssume 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:
A Disc Of Uniform StrengthConsider the condition of equal stress at all radii, i.e. . Le be the thickness of the disc at a radius and assume that the thickness will be at radius The mass of the element will be approximately
Example - A Disc Of Uniform Strength
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./in2. aqt 10,000 r.p.m.? Density of material = 0.28 lb./in3.
The required thickness = 2.5 inches
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