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 r is u, then the Strain Equations are:
If the angular velocity of rotation is and the density of the disc material is then the element shown in the diagram is subjected to a centrifugal force which:
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:
As in the case of a thin rotating cylinder. The variations in Stress are shown on the diagram.
Example:
[imperial]
Example - Stress in a hollow uniform disc
Problem
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
Workings
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 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 that used for a rotating disc and the Strain Equations are:
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
Solid Cylinder
The maximum Stress occcurs at the centre where p and f are equal
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.