ME 351 Machine Dynamics and Vibrations Core Course for VI Sem B.Tech.

Lecture Notes on Whirling of Rotating Shafts

Prepared by Dr. S. M. Murigendrappa

Dept. of Mech. Engg. NITK, Surathkal

CONTENTS

1) Introduction to whirling of rotating shafts

2) Critical speed of a light shaft having a single disc without damping

3) Example 1

4) Critical speed of a light shaft having a single disc with damping

5) Example 2

6) Critical speeds of a shaft having two discs without damping

Introduction to Whirling of Rotating Shafts

• In the rotor system, the shaft as well as the rotating disc is assumed to be rigid.

• In many practical applications such as turbines, compressors, electric motors and pumps, a heavy rotor is mounted on a lightweight, flexible shaft that is supported in bearings

• When a shaft is rotating about its longitudinal axis bends above that axis (line AB), the bent shaft will whirl about it original axis of rotation as well as continuing to rotate about its longitudinal axis.

A

B

A

B

• Whirling is defined as the rotation of the plane made by the line of centers of the bearings (AB) and the bent shaft.

• Whirling of shafts results from – unbalance in all rotating parts due to manufacturing errors, – Hysteresis damping in the shaft, – Gyroscopic forces, – Fluid friction in bearings, – Loose or worn bearings, – Eccentric mounting of the disc, – Bending of shaft, etc.

A

B

• When the natural frequency of the rotating system coincides with the external forcing frequency, it is called resonance, the speed at which resonance occurs are known as the critical speeds or whirling speeds or whipping speeds

• At these speeds the amplitude of vibration or transverse displacement of rotors is excessively large and the large amount of forces or stresses will be developed.

• In this region of critical speeds the system may fail.

• To avoid the occurrence of critical speeds, it is very important to find the natural frequency of the rotating system.

A

B

Case (i): Critical Speed of a Light Shaft having a Single Disc without Damping

• Let us consider the steady-state motion of the light shaft, which is whirling because of the disc attached at the centre to it and not balanced.

• Let, O- intersection of the bearing centre line with the disc S – geometric centre of the disc through which the centre line of the shaft passes G –centre of gravity of the disc which is displaced from the geometric centre e – eccentricity of the disc = SG r – lateral deflection of the shaft centre from O k - stiffness of shaft - uniform angular velocity of shaft n- lateral natural frequency of rotating system m – mass of the disc

A

B

O

S

+

G

e m2(r+e) kr

+ G

r + O S

• Considering the equilibrium of the disc, there are two forces acting on it – Centrifugal force at G acts radially outwards

– Restoring force at S acts radially inwards

• For the equilibrium of the disc, these two forces must act along the same line and therefore the points O, S and G must lie on the same straight line

• Thus,

• Which gives, amplitude ratio as,

• where, lateral natural frequency of rotating system,

)(2 erm

kr

krerm )(2 )1(

1 2

2

2

2

n

n

mkm

er

mkn

+ + O S G r e

m2(r+e) kr

• There are four possible combinations of speeds a) when = n Amplitude ratio becomes infinite which causes severe vibration

and excessive stresses on shaft and bearings will try to fly out. Thus the critical speed of the shaft is equal to the natural frequency of rotating system.

b) when < n amplitude ratio will be positive and S will lie between O and G this means that the disc rotates with heavy side (centre of gravity) it is corresponds to 0 phase difference

+ + O S G r e

c) when > n amplitude ratio will be negative and G will lie between O and S this means that the disc rotates with light side it is corresponds to 180 phase difference

d) when > > n r = - e this means that the point G approaches to O and the disc rotates

about its centre of gravity and thus vibrations are minimum.

+ + O S G

r e

+ + O S G

r e

Example 1

• The rotor of a turbo super charger weighing 9N is keyed to the centre of a 25mm diameter steel shaft 40cm between bearings. Determine:

(a) the critical speed of shaft (b) the amplitude of vibration of the rotor at a speed of 3200rpm, if the eccentricity is 0.015mm and (c) the vibratory force transmitted and that the shaft material has a density of 8gm/cm3. Take E=2.1 106 N/cm2.

• When damping is present in the form of air resistance, etc. in the system, three forces act on the system: – Centrifugal force at G along OG produced = m2a – Restoring force at S along SO = kr and – Damping force at S in direction a opposite to velocity of the point S = cr.

• Due to these forces, the points O, S and G don’t lie in a straight line as we found in case (i).

• Resultant of restoring force and damping force at point “S” must be equal to centrifugal force and parallel to it to give a clockwise moment which has to be overcome by the driving torque in the anti-clockwise direction.

Case (ii): Critical Speed of a Light Shaft having a Single Disc with Damping

kr

S r

+ + O

G

cr

m2a

• This is because of the damping present in the system which requires the driving torque equal to cr2.

• Referring to the triangle OSG,

• For equilibrium condition at a given time, we have

• Inserting equation (2) in to equation (3) , we get

• Re-arranging above equations, we get amplitude ratio as

+ + O S

G

r

m2a

kr

cr

y

x )2(

coscossinsin

eraea

)4()sin(0

)cos(02

2

emrcermkr

222

2

)()(

cmkm

er

)4(

)2(])(1[)(

222

2

ann

n

)3(sin0

sin02

2

amrcY

amkrX

• and phase angle as

• where damping factor,

• From above relations, following cases have been observed:

a) when < < n, then = 0 (i.e., heavy side out)

b) when < n, then 0< < 90 (i.e., heavy side out)

c) when = n, then = 90 (i.e., heavy side out)

d) when > n, then 90< < 180 (i.e., light side out)

e) when >> n, then = 180 and r = -e (i.e., light side out and disc rotates at centre of gravity)

2tan

mk

c

)4()(1

22 b

n

n

kmc

2

+ + O S

G

r

• A disc of mass 4kg is mounted midway between bearings which may be assumed to be simple supports. The span is 48cm. The steel shaft which is horizontal, is 9mm in diameter. The centre of gravity of the disc is displaced 3mm from the geometric centre. The equivalent viscous damping at the centre of the disc-shaft may be taken as 49Ns/m. If the shaft rotates at 760rpm, find the maximum stress in the shaft and compare it with dead load stress in the shaft. Also find the power required to drive the shaft at this speed. Take Young’s modulus as 196GPa.

• Data: m=4kg, d =0.009m, L = 0.48m, E= 1.961011, I = d4/64, e = 0.003m,

N=760rpm, c = 49Ns/m. Stiffness of the shaft, Natural frequency,

Example 2

A

B mkN

LEIk /4.27483

mk

n

srad /8.82

Forcing frequency, Damping factor,

Frequency ratio,

Amplitude ratio, Thus, the deflection of the shaft due to centrifugal effect, r = 0.017m.

• Dynamic load on the bearings is equal to the resultant force due to restoring force and damping force or is equal to centrifugal force of the disc, i.e.,

• Due to dynamic force, the shaft is deflected under operating conditions, but shaft is subjected with both dynamic load and static load, thus, the total or maximum load is given by

sradN /5.7960

2

074.02

kmc

96.0n

222

2

)2(])(1[)(

nn

n

er

NrckrFd 470)()( 22

NmgFFF staticd 2.50981.94470470max

• Bending stress in the shaft due to the load, say F acting at the midway of the shaft is obtained from moment equation. i.e.,

where,

• Thus,

• Total maximum bending stress due to the maximum load Fmax is given by

• Bending stress due to the static or dead load Fstatic is given by

yIM

mdydIFFLM 0045.02

,102.364

,12.04

104

F410168

max4

max 10168 F284 /1055.82.50910168 mN

staticstatic F410168 274 /1059.62.3910168 mN

• Damping force,

• Damping torque,

• Power required to overcome the damping torque,

NrcFDamping 2.66

NmrFT dampingdamping 125.1

WattsNT

P dampingdamping 90

602

Critical Speeds of a Shaft having Two discs without damping

• For a shaft having more than one disc, there will be as many critical speeds as the number of discs. – M1 and M2 be the masses of discs – r1 and r2 be the deflections of the two discs from the centre line of the bearings. – e1 and e2 be the distances from the center of gravity to the geometric centre. – F1 and F2 be the centrifugal forces are given by

– Deflections may be written in terms of Influent coefficients as

– Substituting values of F1 and F2 in to above equations,

A

B

r1 r2

F1 F2

)5()(

)(

222

22

112

11

erMFerMF

)6(2221212

2121111

FaFarFaFar

)7()()()()(

222

222112

1212

222

212112

1111

erMaerMarerMaerMar

• Equation (7) may be rearranged as

• Equation (8) can be written in matrix form as

• For non-trivial solutions is possible only if

• Thus, the natural frequency of the system is given by

where,

• -ve gives lower or first natural frequency or critical speed • +ve gives higher or second critical speed

)8()1()(0)()1(0

22

22212

121

22

21212

111

rMarMarMarMa

)9()1()(

)()1(00

2

12

2222

121

2212

2111

rr

MaMaMaMa

)10()1()(

)()1(0 2

2222

121

2212

2111

MaMaMaMa

)11(2

42

QQPP

n

)( 212221121

222111

aaaMMQMaMaP

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