co = 1033 rad/s = 9860 rev/min Example 42

A steel rotor disc which is part of a turbine assembly has a uniform thickness of 40 mm. The disc has an outer diameter of 600 mm and a central hole of 100 mm diameter. If there are 200 blades each of mass 0.153 kg pitched evenly around the periphery of the disc at an effective radius of 320 mm, determine the rotational speed at which yielding of the disc first occurs according to the maximum shear stress criterion of elastic failure.

For steel, E — 200 GN/m2, v = 0.3, p = 7470 kg/m3 and the yield stress ay in simple tension = 500 MN/m2.


Total mass of blades = 200 x 0.153 = 30.6 kg Effective radius = 320 mm Therefore centrifugal force on the blades = mafr = 30.6 x co2 x 0.32 Now the area of the disc rim = ndt = rc x 0.6 x 0.004 = 0.0247rm2

The centrifugal force acting on this area thus produces an effective radial stress acting on the outside surface of the disc since the blades can be assumed to produce a uniform loading around the periphery.

Therefore radial stress at outside surface


Now eqns. (4.7) and (4.8) give the general form of the expressions for hoop and radial stresses set up owing to rotation, i.e.

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