A flexible rotor-shaft system comprises of a 10 kg rotor disc placed in the middle of a massless shaft of diameter 30 mm and length 500 mm between bearings (shaft is being taken mass-less as the equivalent mass of the shaft is included in the rotor mass) mounted at the ends. The bearings are assumed to simulate simply supported boundary conditions. The shaft is made of steel for which the value of E is 2.1 x 1011 Pa. What is the critical speed of rotation of the shaft?
90 Hz
Understanding Rotor Shaft Critical Speed
The critical speed of a rotating shaft is the rotational speed at which the shaft becomes unstable and experiences large vibrations. This happens when the rotational frequency of the shaft matches its natural frequency of vibration. For a simple rotor-shaft system like the one described, the critical speed is numerically equal to the fundamental natural frequency of the system in Hz.
To find the critical speed, we need to calculate the natural frequency of the system. For a shaft with a concentrated mass at the center and simply supported bearings, the natural frequency can be related to the static deflection of the shaft under the weight of the rotor.
Given Parameters for the Rotor Shaft System
- Rotor mass (m): 10 kg
- Shaft diameter (d): 30 mm = 0.03 m
- Shaft length (L): 500 mm = 0.5 m
- Modulus of Elasticity (E) for steel: 2.1 x 1011 Pa
- Bearing type: Simply supported
- Assuming acceleration due to gravity (g): 9.81 m/s2
Calculating Area Moment of Inertia (I)
The shaft has a circular cross-section. The area moment of inertia for a solid circular shaft is given by the formula:
$$\text{I} = \frac{\pi \text{d}^4}{64}$$
Substitute the given diameter (d = 0.03 m):
$$\text{I} = \frac{\pi \times (0.03 \text{ m})^4}{64}$$
$$\text{I} = \frac{\pi \times 0.00000081 \text{ m}^4}{64}$$
$$\text{I} \approx \frac{2.5447 \times 10^{-6} \text{ m}^4}{64}$$
$$\text{I} \approx 3.976 \times 10^{-8} \text{ m}^4$$
Calculating Static Deflection ($\delta$)
For a simply supported beam with a concentrated load (the rotor's weight) at the center, the maximum static deflection at the center is given by:
$$\delta = \frac{\text{W}\text{L}^3}{48\text{E}\text{I}}$$
Where W is the weight of the rotor, W = mg.
Calculate the weight (W):
$$\text{W} = \text{m} \times \text{g} = 10 \text{ kg} \times 9.81 \text{ m/s}^2 = 98.1 \text{ N}$$
Now substitute the values of W, L, E, and I into the deflection formula:
$$\delta = \frac{(98.1 \text{ N}) \times (0.5 \text{ m})^3}{48 \times (2.1 \times 10^{11} \text{ Pa}) \times (3.976 \times 10^{-8} \text{ m}^4)}$$
$$\delta = \frac{98.1 \times 0.125}{48 \times 2.1 \times 10^{11} \times 3.976 \times 10^{-8}}$$
$$\delta = \frac{12.2625}{401587.2}$$
$$\delta \approx 3.0534 \times 10^{-5} \text{ m}$$
Calculating Natural Frequency (Critical Speed)
The natural frequency ($\omega_n$) in radians per second can be found from the static deflection using the relationship:
$$\omega_n = \sqrt{\frac{g}{\delta}}$$
Substitute the values of g and $\delta$:
$$\omega_n = \sqrt{\frac{9.81 \text{ m/s}^2}{3.0534 \times 10^{-5} \text{ m}}}$$
$$\omega_n = \sqrt{321263.05} \text{ rad/s}$$
$$\omega_n \approx 566.8 \text{ rad/s}$$
To convert the natural frequency from radians per second ($\omega_n$) to Hertz ($f_n$), use the formula:
$$f_n = \frac{\omega_n}{2\pi}$$
$$f_n = \frac{566.8 \text{ rad/s}}{2\pi \text{ rad/cycle}}$$
$$f_n \approx \frac{566.8}{6.283}$$
$$f_n \approx 90.19 \text{ Hz}$$
The critical speed of rotation is equal to the natural frequency $f_n$.
Conclusion
The calculated critical speed of the shaft is approximately 90.19 Hz. Comparing this value to the given options, 90 Hz is the closest value.
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