Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, ()

Example 2.1. Obtain the torsional natural frequency of the system shown in Figure 2.10 using the transfer matrix method. Check results with closed form solution available. Take G = 0.81011 N/m2.

0.6m

0.1 m

22.6 kgm2 5.66 kgm2

Figure 2.10 Example 2.1

Solution: We have following properties of the rotor

The torsional stiffness is given as

Analytical method: The natural frequencies in the closed form are given as

Mode shapes are given as

For

and

Transfer matrix method: State vectors can be related between stations 0 & 1 and 1 & 2, as

The overall transformation of state vectors between 2 & 0 is given as

On substituting values of various rotor parameters, it gives

(A)

Since ends of the rotor are free, the following boundary conditions will apply

On application of boundary conditions, we get the following condition

Since , we have

which gives the natural frequency as

which are exactly the same as obtained by the closed form solution. Mode shapes can be obtained by substituting these natural frequencies one at a time into equation (A), as

For rigid body mode

and anti-phase mode

which are also exactly the same as obtained by closed form solutions.

Example 2.2. Find torsional natural frequencies and mode shapes of the rotor system shown in Figure 1. B is a fixed end and D1 and D2 are rigid discs. The shaft is made of steel with modulus of rigidity G = 0.8 (10)11 N/m2 and uniform diameter d = 10 mm. The various shaft lengths are as follows: BD1 = 50 mm, and D1D2 = 75 mm. The polar mass moment of inertia of discs are: Ip1 = 0.08 kg-m2 and Ip2= 0.2 kg-m2. Consider the shaft as massless and use (i) the analytical method and (ii) the transfer matrix method.

Solution:

Analytical method: From free body diagrams of discs as shown in Figure 2.12, equations of motion can be written as

The above equations for free vibrations and they are homogeneous second order differential equations. In free vibrations discs will execute simple harmonic motions.

k11 k2(2-1) k2(2-1)

1 2

(a) D1 (b) D2

Figure 2.12 Free body diagram of discs

For the simple harmonic motion, hence equations of motion take the form

On taking determinant of the above matrix, it gives the frequency equation as

which can be solved for , as

For the present problem the following properties are gives

Natural frequencies are obtained as

The relative amplitude ratio can be obtained as (Figure 2.13)

-10.7

0.2336

(a) (b)

Figure 2.13 Mode shapes

Transfer matrix method

k1 k2

0 1 2

Figure 2.14 Two-discs rotor system with station numbers

For Figure 2.14 state vectors can be related as

The above state vector at various stations can be related as

which can be combined to give

(A)

with

Boundary conditions are given as

At station 0  = 0 and T = 1 (assumed)

and at right of station 2 T = 0

On application of boundary conditions the second equation of equation (A), we get

since and on substituting for p, we get

which can be solved to give

It should be noted that it is same as obtained by the analytical method.

Exercise 2.1. Obtain the torsional critical speed of a rotor system as shown in Figure E.2.1 Take the polar mass moment of inertia, Ip = 0.04 kg-m2. Take shaft length a = 0.3 m and b = 0.7 m; modulus of rigidity G = 0.8  1011 N/m2. The diameter of the shaft is 10 mm. Bearing A is flexible and provides a torsional spring of stiffness equal to 5 percent of the stiffness of the shaft segment having length a and bearing B is a fixed bearing. Use either the finite element method or the transfer matrix method.

Figure E2.1An overhang rotor system

Exercise 2.2. Find the torsional critical speeds and the mode shapes of the rotor system shown in Figure E2.2 by transfer matrix method. B1 and B2 are frictionless bearings and D1 and D2 are rigid discs. The shaft is made of steel with modulus of rigidity G = 0.8 (10)11 N/m2 and uniform diameter d = 10 mm. The various shaft lengths are as follows: B1D1 = 50 mm, D1D2 = 75 mm, and D2B2 = 50 mm. The polar mass moment of inertia of discs are: Jd1 = 0.0008 kg-m2 and Jd2= 0.002 kg-m2. Consider shaft as massless.

Exercise 2.3. Obtain the torsional critical speed of an overhang rotor system as shown in Figure E2.3. The end B1 of the shaft is having fixed end conditions. The disc is thin and has 0.02 kg-m2 of polar mass moment of inertia. Neglect the mass of the shaft. Use (i) the finite element and (ii) the transfer matrix method.

Exercise 2.4 Find the torsional natural frequencies and the mode shapes of the rotor system a shown in Figure E2.4 by ONLY transfer matrix method. B1 and B2 are fixed supports and D1 and D2 are rigid discs. The shaft is made of steel with modulus of rigidity G = 0.8 (10)11 N/m2 and uniform diameter d = 10 mm. The various shaft lengths are as follows: B1D1 = 50 mm, D1D2 = 75 mm, and D2B2 = 50 mm. The polar mass moment of inertia of discs are: Jd1 = 0.08 kg-m2 and Jd2= 0.2 kg-m2. Consider shaft as massless.

Exercise 2.5 Find all the torsional natural frequencies and draw corresponding mode shapes of the rotor system shown in Figure E2.5. B and D represent bearing and disc respectively. B1 is fixed support (with zero angular displacement about shaft axis) and B2 and B3 are simply supported (with non-zero angular displacement about shaft axis). The shaft is made of steel with modulus of rigidity G = 0.8 (10)11 N/m2 and uniform diameter d = 10 mm. The various shaft lengths are as follows: B1D1 = 50 mm, D1B2 = 50 mm, B2D2 = 25 mm, D2B3 = 25 mm, and B3D3 = 30 mm. The polar mass moment of inertia of the discs are: Ip1 = 2 kg-m2, Ip2 = 1 kg-m2, and Ip3 = 0.8 kg-m2. Use both the transfer matrix method and the finite element method so as to verify your results. Give all the detailed steps in obtaining the final system equations and application of boundary conditions. Consider the shaft as massless and discs as lumped masses.

Exercise 2.6 Obtain the torsional critical speed of turbine-coupling-generator rotor as shown in Figure E2.6 by the transfer matrix and finite element methods. The rotor is assumed to be supported on frictionless bearings. The polar mass moment of inertias are IpT = 25 kg-m2, IpC = 5 kg-m2 and IpG = 50 kg-m2. Take modulus of rigidity G = 0.8  1011 N/m2. Assume the shaft diameter throughout is 0.2 m and lengths of shaft between bearing-turbine-coupling-generator-bearing are 1 m each so that the total span is 5 m. Consider shaft as massless.

Figure E2.6 A turbine-generator set

Exercise 2.7 In a laboratory experiment one small electric motor drives another through a long coil spring (n turns, wire diameter d, coil diameter D). The two motor rotors have inertias I1 and I2 and are distance l apart, (a) Calculate the lowest torsional natural frequency of the set-up (b) Assuming the ends of the spring to be “built-in” to the shafts, calculate rotational speed (assume excitation frequency will be at the rotational frequency of the shaft) of the assembly at which the coil spring bows out at its center, due to whirling.

Example 2.3. For geared system as shown in Figure 2.16 find the natural frequency and mode shapes. Find also the location of nodal point on the shaft (i.e. the location of the point where the angular twist due to torsional vibration is zero). The shaft ‘A’ has 5 cm diameter and 0.75 m length and the shaft ‘B’ has 4 cm diameter and 1.0 m length. Take modulus of rigidity of the shaft G equals to 0.8  1011 N/m2, polar mass moment of inertia of discs are IA = 24 Nm2 and IB = 10 Nm2. Neglect the inertia of gears.

Figure 2.17 Example problem 2.1

Solution: On taking shaft B has input shaft (or reference shaft) as shown in Figure 2.18 the gear ratio can be defined as

10 cm B

d = 4 cm

Input

lB = 1 m

OutputIPB=10 Nm2

d = 5 cm

lA = 0.75 m

IPA= 24 Nm2 20 cm

Figure 2.18 A geared system

The area moment of inertia and the torsional stiffness can be obtained as

On replacing shaft A with reference to the shaft B by an equivalent system, the system will look as shown in Figure 2.19. The equivalent system of the shaft system A has the following torsional stiffness and mass moment of inertia properties

which gives the equivalent length as:

Gear location

Figure 2.19 Equivalent single shaft system

The equivalent stiffness of the full shaft is given as

The equivalent shaft length is given as

The natural frequency of the equivalent two mass rotor system as shown in Figure 2.19 is given as

Gear location

=1.0

0.8358 m

node point

Figure 2.20 Mode shape and nodal point location in the equivalent system

The node location can be obtained from Figure 2.20 as

which can be written as noting equation (9), as

The negative sign indicates that both discs are either ends of node location. The absolute location of the node position is given as

Also from Figure 2.20 we have

which gives

The node is at 0.8356 m from end B. Alternatively, from similar triangle of the mode shape (Figure 2.20), we have

The mode shape and node location in the actual system is shown in Figure 2.21.

Gear pair

node location

B = 1

A = - 0.8333 0.8358 m

0.75 m 1m

Gear pair position

Figure 2.21 Mode shape and nodal point location in the actual system

Alternative way to obtain natural frequencyis to use the equivalent two mass rotor (Figure 2.19) can be considered as two single DOF systems (one such system is shown in Figure 2.22).

Figure 2.22 A single DOF system

The stiffness and mass moment of inertia properties of the system is given as

It gives the natural frequency as

which is same as obtained earlier.

The whole analysis can be done by replacing shaft B with reference to shaft A speed by an equivalent system. For more clarity some of the basic steps are given as follows.

IpB

IpA

lA = 0.75 = 0.61

le = 1.36 m.

IpA

Figure 2.23 Actual and equivalent geared systems

It is assumed here that we are choosing reference shaft as input shaft (i.e. for present case shaft A is reference shaft hence it is assumed to be input shaft and according the gear ratio will be obtained).

It is assumed that equivalent shaft (i.e. B) has same diameter as the reference shaft (i.e. A). The equivalent mass moment of inertia and stiffness can be written as

which gives the equivalent length as

since

The total equivalent length and the torsional stiffness would be

Alternatively the effective stiffness can be obtained as

The natural frequencies of two mass rotor system is given by

and

A factor 9.81 is used since IPA is in Nm2.

1.36 m

0.75m

Gear location

node location

Figure 2.24 Equivalent two mass rotor system

The node location can be obtained as

we have

which gives

The stiffness of will be (equivalent stiffness corresponding to shaft A speed)

The shaft stiffness corresponding to shaft B speed can be defined in two ways i.e.

and

On equating above equations the location of the node in the actual system can be obtained as

which is same as by previous method.

Exercise Problem 2.8. For a geared system as shown in Figure E2.8 find the natural frequencies and mode shapes. Find also the location of nodal point on the shaft (if any). The shaft ‘A’ has 5 cm diameter and 0.75 m length and the shaft ‘B’ has 4 cm diameter and 1.0 m length. Take modulus of rigidity of the shaft G equals to 0.8  1011 N/m2, polar mass moment of inertia of discs and gears are IA = 24 Nm2, IB = 10 Nm2, IgA = 5 Nm2, IgB = 3 Nm2.

Figure E2.8 A geared system

Example 2.4. Obtain the torsional critical speeds of the branched system as shown in Figure 2.26. Take polar mass moment of inertia of rotors as: IPA = 0.01 kg-m2, IPE = 0.005 kg-m2, IPF = 0.006 kg-m2, and IPB = IPC = IPD = 0. Take gear ratio as: nBC= 3 and nBD= 4. The shaft lengths are: lAB = lCE = lDF = 1 m and diameters are dAB = 0.4 m, dCE = 0.2 m and dDF = 0.1 m. Take shaft modulus of rigidity G = 0.8  1011 N/m2.

Solution: The branched system has the following mass moment of inertias

kg-m2; kg-m2; kg-m2

For branch A the state vector at stations are related as

with

For branch B the state vector at stations are related as

with

Similarly, for branch C, we have

with

From equation (53), the frequency equation can be written as

On substitution, we get

which can be simplified to

The roots of the polynomial are

and

Natural frequencies are given as

; rad/s and rad/s

It can be seen that the rigid body mode exist since ends of the gear train is free.