Overview

Topic 1 - Shaft modes

Rotating shafts and disks have vibrational modes, just as do all the other mechanical components that engineers design and deal with.  However, due to the fact that they are rotating, there are additional effects which make them more complicated to deal with than static (i.e. non-rotating) components.

First, let’s consider the modes that a simply supported shaft (i.e. one supported by two flexible bearing supports, shown here by springs) would be expected to have:

For a constant diameter shaft, these shapes would be symmetric.   And the frequency would be determined by its mass and how resistant it is to bending, essentially the same two qualities, mass (m) and stiffness (k), which traditional vibration theory uses to determine a component’s natural frequency from w = (k/m)^1/2.

If we add disks to this shaft (representing things like gears, turbine or compressor wheels, electric motor windings, etc.) then these change the distribution of both mass and stiffness along the shaft and have and impact on both the frequency and mode shape.

We need some method for calculating the mode shape and responsive frequency
of complicated rotor/shaft systems. Permission of use granted MITCalc

Topic 2 - Predicting critical speeds and mode shapes

While there are some simplistic formulas for predicting the bending modes of a simply supported homogenous shaft, we will usually be dealing with shafts that carry disks or that are not constant in diameter, or that are coupled to other shafts.   In these cases, it will be advisable to use a rotordynamic modelling routine to predict the frequencies and mode shapes.  An example follows for the rotor that we have just been examining.

You will note that the previous figures referred to a “critical speed” rather than a “natural frequency.”  The two are similar, but have a distinct difference, which we now need to discuss.  In previous vibration discussions, you will have learned that every mechanical system has natural frequencies at which it has an inherent desire to respond with the response in the shape of a particular “mode”.  If randomly excited, the system will tend to vibrate at its natural frequency.  Under the influence of a forced vibration, when the frequency of that forcing function is near the natural frequency the system response becomes large (in fact if damping were to be zero, the response would become unbounded). These natural frequencies are usually expressed as f_n in units of hertz (hz) which is cylces/second, or as w_n in units of radians per second.

When we are dealing with a rotating system, we usually express this condition of exciting the vibrational mode in terms of a speed in units of revolutions per minute (rpm).  This makes sense because, as a rotating system, we are used to discussing it in terms of speed in rpm.  However, there is yet another difference.  The natural frequency of a static system is constant (assuming constant material, temperature, etc. conditions).  The critical speed of a rotating system is not constant.  Let us examine why.  Consider a gyroscope either in the form of a simple toy, or as a complex navigational device such as used in a spacecraft.

Wikipedia / Public Domain

If you have ever played with a toy gyroscope, you will have experienced the unique effect that rotation plays.   As the rotor spins about one axis, if you try to turn it about a perpendicular axis, it has a response motion in the direction of the third orthogonal axis.  This is referred to as the gyroscopic effect.  Now consider our spinning shaft and rotor.  The spin of the rotor is about the X axis.  The system will naturally tend to bow, as shown below, due to the unbalance forces.  This will create a rotation of the rotor about the Z axis.  Just as with the toy gyroscope, this will create a response pushing the rotor in the orthogonal direction, in this case, in the negative Y direction, which is back towards the undeformed, or un-bowed shape of the shaft.

Returning to gyroscopic forces, they are greater the faster the speed of the rotations.  This means that the faster the speed of the shaft, the greater the gyroscopic stiffening effect.  Thus, the shaft stiffness, k, increases with speed.   Since the natural frequency of the shaft would be calculated from the square root of k/m, this means that the natural frequency would rise as the shaft gets effectively stiffer as the speed increases.  This means that the natural frequency of the rotating system is not constant, but increases with speed. 

A critical speed is the speed at which a rotating natural frequency
and a rotating excitation source are coincident.

This all makes it much more difficult to predict critical speeds (and their associated modes shapes) using simple formulas.  Therefore, rotordynamic analyses most often have to use finite element modeling to predict these critical speeds and mode shapes.    The next figure shows how the first four rotational modes of a particular shaft and rotor system might be predicted. 

We have just discussed how the mode of the shaft can be excited when the rotating forces associated with the shaft’s own speed (e.g. unbalance forces) corresponds to the frequency of the mode.  However, there are other sources within a rotating system which can cause this mode to be excited, such as:

• Shock
• Friction/Rub
• Unbalance forces transmitted from other shaft/rotor systems

Let us particularly consider the last one.  What if we have a two spool gas turbine with the two rotors turning at different speeds.   As one rotor’s unbalance transmits forces through the bearing to the bearing support structure, these forces can transmit back through another bearing supporting the second rotor.  This means that the second rotor’s modes could potentially be excited by the unbalance on the first rotor.  This presents a situation where a particular rotor’s modes could be excited not only by its own speed, but also by the speed of a different rotor in the engine. 

Now what happens, if the two rotors are rotating in opposite directions?  This essentially creates a scenario where a mode of one rotor could be excited by the forces from the other one, but there will not be an associated gyroscopic stiffening.  In fact, due to the opposite rotation, there is essentially a de-stiffening, or softening effect, and the backward (BW) mode (same mode shape, but whirling in the opposite direction) could be excited at a much lower critical speed, as indicated in the next figure.

Some rotor and shafting configurations are designed to place the operating speed of the system above the firsts critical speed.  These are referred to as “supercritical” shaft systems.  Although the response of a system at its critical speed would be theoretically infinite with no damping, or at least very large with low damping, if a system had low enough unbalance and high enough damping, it would be possible to pass through the critical speed in the transient range of the engine, and then operate smoothly above it. 

The additional damping required to make this happen could be provided by mechanical isolators installed to support the bearing.  These would deflect under bearing load and the material deflection would provide some extra damping.  Another means would be creating an oil filled cavity behind the rolling element bearing, and the hydrodynamic forces created by the squeezing of the oil film would provide the required resistance and damping.  These configurations, which bear a resemblance to journal bearings, are known as “squeeze film dampers.”

Topic 3 - Failure modes

There are many potential failure modes in a gas turbine.  The table below addresses a few of them:

Topic 3 - Application 1

A small, high-speed turbine has a blade which is 2 inches tall.  It has a constant cross sectional area of  0.7 in² and the cross section has an area moment of inertia (second moment of area) I of 12 in⁴.  Material elastic modulus is E = 30,000,000 psi, and density = r = 490 lb/ft³.

Using hand calculations for a simple beam, we will estimate the first three bending mode natural frequencies of the blade and draw a Campbell Diagram (frequency-speed diagram) showing the natural frequencies of the blade recognizing that idle for the engine is 75,000 rpm and maximum speed is 100,000 rpm. 

Then we will assume that there are 72 airfoils in the vane row preceding the blade in question.  We will draw the operating range and the engine order line for the vane excitation and assess whether there is a mode to be concerned about.

Ideally we would have a FEM model which we could use in a FEA analysis to predict the frequencies and mode shapes of the blade.   However, if we have not reached that point yet, we could use fundamental equations for the natural frequency of a cantilever beam, since an unshrouded blade resembles a cantilever. 

f = C(gEI/wL⁴)^1/2

f = frequency

g = gravitational acceleration (in/sec²)

I = Area Moment of Inertial of cross-section (in⁴)

E = Modulus of Elasticity (psi)

W = weight distribution (lb/in)

L = length (in)

C = constant as shown below

First mode: 0.56

Second mode: 3.51

Third mode: 9.82

Using the formula for the simple beam, we can estimate the first three mode frequencies:

${\text{f}}_{\text{1}}=\text{C}{\left(\text{g}\text{E}\text{I}/\text{w}{\text{L}}^4\right)}^{1/2}$

$=\left(0.56\right){\left[\left(386\text{i}\text{n}/\text{s}\text{e}{\text{c}}^2\right)\left(30\text{x}{10}^6\text{l}\text{b}/\text{i}{\text{n}}^2\right)\left(12\text{i}{\text{n}}^4\right)/\left(490\text{l}\text{b}/\text{f}{\text{t}}^3\right){\left(\text{f}\text{t}/12\text{i}\text{n}\right)}^3\left(0.7\text{i}{\text{n}}^2\right){\left(2\text{i}\text{n}\right)}^4\right]}^{1/2}$

$=\left(0.56\right)\left(209175\right)=117,138 \text{r}\text{a}\text{d}/\text{s}\text{e}\text{c}=18,643 \text{h}\text{z}$

${\text{f}}_2=\left(3.51\right)\left(209175\right)=734,204\text{r}\text{a}\text{d}/\text{s}\text{e}\text{c} =116,852 \text{h}\text{z}$

${\text{f}}_3=\left(9.82\right)\left(209175\right)=2,054,098\text{r}\text{a}\text{d}/\text{s}\text{e}\text{c} = 326,919 \text{h}\text{z}$

Now we can draw a frequency-speed (Campbell) diagram, showing the first two mode frequencies, the operating range of the engine, and the 72nd engine order line, corresponding to the excitation frequency associated with the 72 vanes in front of the blade we are examining. Note that this excitation comes from the fact that every time the blade comes out from behind a vane, it will be hit by a force of the oncoming air. And it will do this 72 times each revolution. Thus there will be an excitation at 72 times engine speed, or 72nd engine order (72EO). Also note that a blade, although a rotating component, is not rotating about two axes, so there is no gyroscopic effect involved.

We can see from this that the second mode and the 72EO line intersect inside the operating range.  This means, during the engine’s steady state operation, the second mode of the blade is likely to be excited by the 72^nd engine order pulses coming from the vane row in front of it.  This is an undesirable situation, and thus something should be done to alter the design so as to remove the possible excitation from the operating range.  Possibilities include:

• Raise the frequency of the blade so that the intersection occurs outside the operating range, by changing one of the following:
  • Raising the material modulus
  • Lowering the material density
  • Modified cross-section I to make the blade stiffer
  • Add shrouds
• Move the engine order line by reducing the number of vanes until the intersection is outside the operating range
• Lower maximum speed (not likely as this will impact engine performance)
• NOTE: For any change, the impact on other modes must be examined, because it is possible that the change might also move a mode that was not previously a concern into a situation where it is a concern

Topic 3 - Application 2

If we have built a FEM model of the rotor system, then we can apply to it the appropriate boundary and environmental conditions and have our FEA routine predict the stresses.  However, if we are not to that stage yet, and we wish a rough estimate, there are equations which we can use to estimate the wheel stress:

${\text{σ}}_{\text{tan}\text{g}\text{e}\text{n}\text{t}\text{i}\text{a}\text{l}}=\frac{\left(3+\text{ν}\right)}{8}\text{ρ}{\text{ω}}^2\left[{{\text{r}}_{\text{O}}}^2+{{\text{r}}_{\text{i}}}^2+\frac{{{\text{r}}_{\text{O}}}^2{{\text{r}}_{\text{i}}}^2}{{\text{r}}^2}-\frac{\left(1+3\text{ν}\right){\text{r}}^2}{\left(3+\text{ν}\right)}\right]$

${\text{σ}}_{\text{r}\text{a}\text{d}\text{i}\text{a}\text{l}}=\frac{\left(3+\text{ν}\right)}{8}\text{ρ}{\text{ω}}^2\left[{{\text{r}}_{\text{O}}}^2+{{\text{r}}_{\text{i}}}^2-\frac{{{\text{r}}_{\text{O}}}^2{{\text{r}}_{\text{i}}}^2}{{\text{r}}^2}-{\text{r}}^2\right]$

where

$\sigma$ = stress (Pa)

$\text{r}$ = radius of interest (m)

$r_o$ = outer radius (m)

$r_o$ = inner radius (m)

$\rho$ = material density (kg/m3)

$\omega$ = rotational speed (rad/sec)

$\nu$ = Poisson's Ratio of material

Using this formulae:

${\text{α}}_{\text{tan}\text{g}\text{e}\text{n}\text{t}\text{i}\text{a}\text{l}}=\frac{\left(3+0.3\right)}{8}\left(8200\right){\left[12017\left(2\right)\text{π}/60\right]}^2\left[0.4^2+0.1^2+\frac{0.4^{2}0.1^2}{0.1^2}-\frac{\left[1+3\left(0.3\right)\right]{\left(0.1\right)}^2}{\left(3+0.3\right)}\right]$

$=1736 \text{M}\text{P}\text{a}$

The centrifugal forces on the wheel material will be outward.  Thus the stress created in the material by the centrifugal force will be a tensile stress. 

We calculated the burst stress at the inner diameter of the wheel, as that is where they will be highest and thus where the failure will initiate if the wheel bursts during the test.  Note that we can assure ourselves of this by looking at the formula.  As r becomes larger, the term r_o²r_i²/r² becomes smaller, and since it is a positive number, that means the stress is lower.  Similarly, if r becomes larger, the term (1 +3n)r²/(3 + n) becomes larger, but since it is negative in the equation, it reduces the stress.

The value calculated exceeds the ultimate tensile strength of the material at 1335 MPa.  Thus we expect failure before we reach the 122% overspeed.

Summary

Fellah-Jahromi, A., Rama, B., & Wenfang, X. (2014). “Forward and Backward Whirling of a Rotor with Gyroscopic Effect.” Mechanisms and Machine Science. 23(10). Doi:1007/978-3-319-09918-7_78.

Fioeide, H., Katia, C., Cavalca, L. & Nordmann, R. (2008).  “Whirl and Whip Instabilities in Rotor-Bearing System Coinsidering a Nonlinear Force Model.”  Journal of Sound and Vibration 317(1-2).

Saravanamutto, H., Rogers, G., Cohen, H. & Strznicky, P. (2009).  Gas Turbine Theory.  Essex, UK: Pearson.

Sforza, P. (2011). Theory of Aerospace Propulsion. New York: Butterworth-Heinemann.

Swanson, E. & Powell, C. (2005)  “A Practical Review of Rotating Machinery Critical Speeds and Modes”. Sound and Vibration, May 2005.

Wilson, D & Korakianitis, T. (2014).  The Design of High-Efficiency Turbomachinery and Gas Turbines.  Cambridge, USA:  MIT Press.