Question 13.2.3: Rotating machines such as pumps and fans must have their sha......
Rotating machines such as pumps and fans must have their shafts supported by bearings, and often are enclosed in a housing. Suppose that the rotor (the rotating element) of a specific machine has mass of 300 kg and a measured unbalance of m_uϵ = 0.5 kg · m. The machine will be run at a speed of 3500 rpm, and there is a clearance of 2 mm between the shaft and the housing. The shaft length from the bearings is L = 0.05 m. Assuming the shaft is steel, compute the minimum required shaft diameter. Model the shaft as a cantilever beam supported by the bearings, and neglect any damping in the system.
First convert the speed into rad/s: 3500 rpm = 3500(2π)/60 = 367 rad/s. From (13.2.6) with ζ = 0,
X=|X(j \omega)|=\frac{m_u \epsilon}{m} \frac{r^2}{\sqrt{\left(1 – r^2\right)^2 + (2 \zeta r)^2}} (13.2.6)
X=\frac{m_u \epsilon}{m} \frac{r^2}{\left|1 – r^2\right|}We are given that m = 300 kg. With m_uϵ = 0.5 and X = 0.002 m, we have
0.002=\frac{0.5}{300} \frac{r^2}{\left|1 – r^2\right|}Solve this for r² assuming that r² > 1:
0.002=\frac{0.5}{300} \frac{r^2}{r^2 – 1}which gives r² = 6 and thus
\omega_n^2=\frac{\omega^2}{r^2}=\frac{(367)^2}{6}=2.2387 \times 10^4But ω^2_n = k/m = k/300. Thus k = 300(2.2387 × 10^4) = 6.716 × 10^6 N/m.
From the formula for a cantilever spring,
where the area moment of inertia for a cylinder is I_A = πd^4/64. Solving for the diameter d in terms of k, we obtain
d^4=\frac{64 k L^3}{3 \pi E}=\frac{64\left(6.716 \times 10^6\right)(0.05)^3}{3 \pi\left(2.07 \times 10^{11}\right)}=2.75 \times 10^{-8}which gives a minimum shaft diameter of d = 12.9 mm.