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 mu ε = 0.5 kg · m. The

Question

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 [latex]m_{u} ε = 0.5 kg · m[/latex]. 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.

Accepted Answer

First convert the speed into rad/s: 3500 rpm = 3500(2π)/60 = 367 rad/s. From (13.2.6) with ζ = 0,

[latex]X = |X(jω)| = \frac{m_{u} ε}{m} \frac{r^{2}}{\sqrt{(1 − r^{2})^{2} + (2ζ r)^{2}}} [/latex] (13.2.6)

[latex]X = \frac{m_{u} ε}{m} \frac{r^{2}}{|1 − r^{2}|} [/latex]
We are given that m = 300 kg. With [latex]m_{u} ε = 0.5 [/latex] and X = 0.002 m, we have
[latex]0.002 = \frac{0.5}{300} \frac{r^{2}}{|1 − r^{2}|} [/latex]
Solve this for r² assuming that r² > 1:
[latex]0.002 = \frac{0.5}{300} \frac{r^{2}}{r^{2} − 1} [/latex]
which gives r² = 6 and thus
[latex]ω^{2}_{n} = \frac{ω^{2}}{r^{2}} = \frac{(367)^{2}}{6} = 2.2387 × 10^{4} [/latex]
But [latex]ω^{2}_{n} = k/m = k/300[/latex]. Thus [latex]k = 300(2.2387 × 10^{4}) = 6.716 × 10^{6} N/m[/latex].
From the formula for a cantilever spring,
[latex]k = \frac{3 E I_{A}}{L^{3}} [/latex]
where the area moment of inertia for a cylinder is [latex]I_{A} = π d6^{4}/64[/latex] . Solving for the diameter d in terms of k, we obtain
[latex]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}[/latex]
which gives a minimum shaft diameter of d = 12.9 mm.

Question 13.2.3: Shaft Design with Rotating Unbalance Rotating machines such ...

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