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,

Question

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 occurs according to the maximum shear stress criterion of elastic failure. For steel, [latex]E = 200 GN/m^{2}[/latex], [latex]ν = 0.3[/latex], [latex]ρ = 7470 kg/m^{3}[/latex] and the yield stress [latex] σ_{y} [/latex] in simple tension [latex] = 500 MN/m^{2}[/latex]

Accepted Answer

Total mass of blades = 200 × 0.153 = 30.6 kg

Effective radius = 320 mm

Therefore centrifugal force on the blades[latex] = mω^{2}r = 30.6 × ω^{2} × 0.32[/latex]

Now the area of the disc rim [latex]= πdt = π × 0.6 × 0.004 = 0.024πm^{2}[/latex]

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

[latex]=\frac{30.6 imes \omega ^{2} imes 0.32 }{0.024 \pi }[/latex] [latex]=130\omega^{2} N/m^{2}[/latex] (tensile)

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. [latex] \sigma _{r}=A-\frac{B}{r^{2} } -\frac{(3+ u )}{8} ho \omega ^{2} r^{2}[/latex] (1) (4.7)
[latex]\sigma _{H} =A+\frac{B}{r^{2} } -\frac{(1+3 u )}{8} ho \omega ^{2} r^{2}[/latex] (2) (4.8)

When r = 0.05, [latex] σ_{r} = 0[/latex]

[latex]0=A-400B-\frac{3.3}{8} ho \omega ^{2} (0.05)^{2}[/latex] (3)

When r = 0.3, [latex] σ_{r} = +130ω^{2}[/latex]

[latex]130 \omega ^{2} =A-11.1B-\frac{3.3}{8} ho \omega ^{2} (0.3)^{2}[/latex] (4)

(4)-(3)

[latex]130 \omega ^{2} =3.88.9B-\frac{3.3}{8} ho \omega ^{2}(9-0.25)10^{-2}[/latex]

[latex]130 \omega ^{2}=388.9B-270\omega ^{2}[/latex]
[latex]B=\frac{(130+270)}{388.9} \omega ^{2}=1.03 \omega ^{2}[/latex]

Substituting in (3),

[latex]A=412\omega^{2}+\frac{3.3}{8} imes 7470(0.05)^{2}\omega^{2}[/latex]
[latex]=419.7\omega^{2}=420\omega^{2}[/latex]

Therefore substituting in (2) and (l), the stress conditions at the inside surface are

[latex]\sigma _{H} =420\omega^{2}+412\omega^{2}-4.43\omega^{2}=827\omega^{2}[/latex]

with [latex] σ_{r} = 0[/latex]

and at the outside [latex] \sigma _{H} =420\omega^{2}+11.42\omega^{2}-159\omega^{2}=272\omega^{2}[/latex]

with [latex] σ_{r} = 130ω^{2}[/latex]

The most severe stress conditions therefore occur at the inside radius where the maximum shear stress is greatest

i.e. [latex] au _{max} =\frac{\sigma _{1} -\sigma _{3} }{2}=\frac{827\omega^{2}-0 }{2}[/latex]

Now the maximum shear stress theory of elastic failure states that failure is assumed to occur when this stress equals the value of [latex]τ_{max}[/latex]at the yield point in simple tension,

i.e. [latex] au _{max} =\frac{\sigma _{1} -\sigma _{3} }{2}=\frac{\sigma _{y}-0 }{2}=\frac{\sigma _{y} }{2}[/latex]

Thus, for failure according to this theory,

[latex] \frac{\sigma _{y} }{2} =\frac{827 \omega^{2}}{2} [/latex]

i.e. [latex]827 \omega^{2}=\sigma _{y}=500 imes 10^{6}[/latex]
[latex]\omega^{2}=\frac{500}{827} imes 10^{6} =0.604 imes 10^{6}[/latex]
[latex]\omega =780 rad/s =7450 rev/min[/latex]

Question 4.2: A steel rotor disc which is part of a turbine assembly has a...

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