# Critical speed of a vertical shaft

1. Dec 28, 2016

### okinaw

Hi everyone,

I have to calculate the critical speed of a vertical shaft. I have been looking on internet how to calculate, and I have found that this critical speed is based on the type of the supports of the shaft. The type of supports determine the deflection of the shaft (δ) and for this reason a fixed-fixed shaft has a higher critical speed than a supported-supported shaft.

If it is correct, how can I determine the critical speed of a vertical shaft?

Knowing that the deflection of the vertical shaft is lower than the horizontal one, I suppose that the critical speed is going to be higher. Am I correct?

2. Dec 29, 2016

### jack action

Whether vertical or horizontal, the deflection will be the same. It only depends on the type of supports.

The deflection is actually the static deflection under gravitational loading only when the shaft is horizontal. It is used with $g$ to find the ratio $k/m (= g/y)$ which is the relationship between stiffness and mass of the shaft such that we can evaluate its response to vibrations.

3. Dec 29, 2016

### Baluncore

4. Dec 31, 2016

### okinaw

Thanks to both of you,

I read the article of wikipedia but I have some doubts. The article says that the critical speed is calculated approximately as:

ω≈√g/ymax

being:
g=gravity
ymax= static deflections (under gravitational loading only)

Depending on the orientation of the shaft, the deflection is going to be different.(In vertical cases, the deflection is going to be much lower), and this implies higher critical speed. So I can not undertand why the relationship is valid no matter what the orientation of the shaft is.

I did some tests, and the critical speed in vertical shafts is higher but i do not know how to quantify how much higher.

Thanks in advance and happy new year.

5. Dec 31, 2016

### jack action

The critical speed for any shaft is $\omega = \sqrt{\frac{k}{m}}$. It relates the stiffness of the shaft (elastic restoring force) vs the inertia of the shaft (centripetal force). It comes from $kr = m\omega^2r$ where the mass of the shaft is assumed to be concentrated at one point where there is a shaft deflection $r$.

Values for $k$ in a bending shaft can be difficult to measure. But when you take an horizontal beam, you can easily evaluate how it will deform under its own load, usually it's a function $y(x)$ where a $y_{max}$ can be identified. If we assume the mass of the beam is concentrated at one point, say $x_c$, the only forces involved for a none rotating shaft are $mg = ky(x_c)$ where necessarily $y(x_c) = y_{max}$. This is the same equation than for our rotating shaft, but with a different acceleration. So we can rearrange the equation to $\frac{k}{m} = \frac{g}{y_{max}}$.

Therefore, you don't need to know $k$, as $y_{max}$ is a lot easier to determined.

6. Jan 2, 2017

### Dr.D

There is a vast literature available regarding the critical speed of shafts. Look under "rotor dynamics" in a search engine.

There are many different approaches to finding shaft critical speeds. The simple equation proposed by Jack above does not help understanding very much because it relates a single discrete stiffness (k), a single discrete mass (m), to the simple system natural frequency. Actual beams are invariably continuous distributions of both mass and flexibility.

The most common approach to realistic shaft geometries (numerous diameter changes, disks, etc) is via Rayleigh's method. This is detailed in the text by Mischke (Elements of Mechanical Analysis, Addison-Wesley, 1963, p. 279).