Question 4.3: The cross-section of a turbine rotor disc is designed for un...
The cross-section of a turbine rotor disc is designed for uniform strength under rotational conditions. The disc is keyed to a 60 mm diameter shaft at which point its thickness is a maximum. It then tapers to a minimum thickness of 10 mm at the outer radius of 250 mm where the blades are attached. If the design stress of the shaft is 250 MN/m^{2} at the design speed of 12000 rev/min, what is the required maximum thickness? For steel p = 7470 kg/m^{3}
From eqn. (4.22) the thickness of a uniform strength disc is given by
t=t_{0}e^{(-\rho \omega ^{2}r^{2} )/(2\sigma )} (1)(4.22)
where t_0 is the thickness at r =0
Now at r = 0.25
\frac{\rho \omega ^{2}r^{2}}{2\sigma } =\frac{7470}{2\times 250\times 10^{6} } (12000\times \frac{2\pi }{60} )^{2}\times 0.25^{2}=1.47and at r = 0.03
\frac{\rho \omega ^{2}r^{2}}{2\sigma } =\frac{7470}{2\times 250\times 10^{6} } (12000\times \frac{2\pi }{60} )^{2}\times 0.03^{2}
=1.47\times \frac{9\times 10^{-4} }{625\times 10^{-4} } =0.0212
But at r = 0.25, t = 10 mm
Therefore substituting in (1)
0.01=t_{0}e^{-1.47}=0.2299 t_{0}
t_{0}=\frac{0.01}{0.2299} =0.0435m=43.5 mm
Therefore at r = 0.03
t = 0.0435e^{-0.021}=2 0.0435 × 0.98
= 0.0426 m = 42.6 mm