Question 13.2.3: Shaft Design with Rotating Unbalance Rotating machines such ...
Shaft Design with Rotating Unbalance
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_{u} ε = 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(jω)| = \frac{m_{u} ε}{m} \frac{r^{2}}{\sqrt{(1 − r^{2})^{2} + (2ζ r)^{2}}} (13.2.6)
X = \frac{m_{u} ε}{m} \frac{r^{2}}{|1 − r^{2}|}
We are given that m = 300 kg. With m_{u} ε = 0.5 and X = 0.002 m, we have
0.002 = \frac{0.5}{300} \frac{r^{2}}{|1 − r^{2}|}
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
ω^{2}_{n} = \frac{ω^{2}}{r^{2}} = \frac{(367)^{2}}{6} = 2.2387 × 10^{4}
But 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,
k = \frac{3 E I_{A}}{L^{3}}
where the area moment of inertia for a cylinder is I_{A} = π d6^{4}/64 . Solving for the diameter d in terms of k, we obtain
d^{4} = \frac{64 k L^{3}}{3 π E} = \frac{64(6.716 × 10^6)(0.05)^{3}}{3 π(2.07 × 10^{11})} = 2.75 × 10^{−8}
which gives a minimum shaft diameter of d = 12.9 mm.