Question 16.4: Design of a Flywheel–Shaft Assembly A 400 mm diameter flywhe...
Design of a Flywheel–Shaft Assembly
A 400 mm diameter flywheel is to be shrunk onto a 50 mm diameter shaft. Determine
a. The required radial interference
b. The maximum tangential stress in the assembly
c. The speed at which the contact pressure becomes 0
Requirement: At a maximum speed of n = 5000 rpm, a contact pressure of p = 8 MPa is to be maintained.
Design Decisions: Both the flywheel and shaft are made of steel having \rho=7.8 kN s^2 / m ^4 , E = 200 GPa, and \nu = 0.3.
a. Applying Equation 16.35, we have
\delta=\frac{a p}{E d}\left\lgroup \frac{a^2+b^2}{b^2-a^2}+\nu \right\rgroup +\frac{a p}{E_{ s }}(1-\nu)+\frac{a \rho \omega^2}{4 E_d}\left[(1-\nu) a^2+(3+\nu) b^2\right] (16.35)
\begin{aligned} \delta &=\frac{25\left(10^{-3}\right) p}{200\left(10^9\right)}\left\lgroup \frac{25^2+200^2}{200^2-25^2}+1 \right\rgroup+\frac{25(7.8) \omega^2}{4\left(200 \times 10^9\right)}\left[0.7(0.025)^2+3.3(0.2)^2\right] \\ &=\left(0.254 p+32.282 \omega^2\right) 10^{-12} \end{aligned} (a)
For p = 8 MPa and \omega = 5000(2π/60) = 523.6 rad/s, Equation (a) leads to \delta = 0.011 mm.
b. Using Equation 16.32,
\sigma_{\theta, \max }=p \frac{a^2+b^2}{b^2-a^2}+\frac{\rho \omega^2}{4}\left[(1-\nu) a^2+(3+\nu) b^2\right] (16.32)
\begin{aligned} \sigma_{\theta, \max } &=8 \frac{25^2+200^2}{200^2-25^2}+\frac{7800(523.6)^2}{4}\left[0.7(0.025)^2+3.3(0.2)^2\right] \\ &=8.254+70.80=79.05 MPa \end{aligned}
c. Inserting \delta=0.011 \times 10^{-3} m \text { and } p=0 into Equation (a) results in
\omega=\left\lgroup \frac{0.011 \times 10^9}{32.282} \right\rgroup ^{1 / 2}=583.7 rad / s
Therefore,
n=583.7 \frac{60}{2 \pi}=5574 rpm
Comment: At this speed, the shrink fit becomes completely ineffective.