Mechanical Vibration by Palm

CHAPTER 4: HARMONIC RESPONSE WITH A SINGLE DEGREE OF

FREEDOM

Instructor: Dr Simin Nasseri, SPSU

© Copyright, 2010 1Based on a lecture from Brown university (Division of engineering)

External Forcing Base Excitation

Types of Forcing:

Rotor Excitation

All of these situations are of practical interest. Some subtle but important distinctions to consider, so we will look at each.

But the strategy is simple: derive Equation of Motion and put into the “Standard Form”

3

Rotating unbalanceRotor excitation

Rotor Excitation

Rotating machinery not always perfectly balanced

Unbalanced. More mass

one side than other.

e.g. your car’s wheels; Imperfectly machined rotating disks; Turbine engine with cracked turbine blades

Leads to a sinusoidally-varying force

Engineering Model:

Effective unbalanced mass m

Effective eccentricity e

In a machine of total mass M

k c

=t

mM

e x(t)

Rotor Excitation

M is total mass INCLUDING eccentric mass

Non-rotating part of the machine has position x,

Rotating piece has position:sinx e t

Equations of motion for the two masses (vertical direction only):-kx -cv

-FMm

FmM

R1R2

FBDs for the two masses:

)sin()(2

2

texdt

dmkxxcxmMFx

0)sin()( 2 kxcvtexmxmM

Standard Form but with

2meFo

k c

=t

mM

e x(t)

)sin(2 tmekxxcxM

Steady State Solution (same as before)

)sin()( tXtx

222

2

21 rr

r

MmeX

Same form as relative base excitation…….

)sin(22

2 tM

mexxx n

21

1

2tan

n

n

222

2

2

21

1

rrM

meX

n

rn

Standard Form but with

2meFo

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

8

9

10

222

2

21 rr

r

MmeX

MmeX

rn

Low frequencies: not moving fast enough to generate a force large enough to drive the mass

High frequencies: imbalance force grows at same rate as magnification factor decreases, leading to amplitude ratio =1

(not usually a problem)

Force Transmission to BaseWith vibrating machinery, forces exerted on the supporting structures can become large near resonance. Equipment is thus constructed on isolating mounts (springs and dashpots – dashpots suppress resonance)

What is the force transmitted?

Position, velocity are always 90o out of phase, so transmitted force is

Draw FBD for base structure:

sin cosA B Since max of over all is we have2 2A B

Maximum transmitted force:

k c

=t

mm

e x(t)

k c

kx

cv

)cos()sin( tXctkXFT

2222 )/2(1)/(1)()( nT kXkckXXckXF

2222

2

21

1

rrM

meX

n

2)/2(1 nT kXF

And we found earlier

222

22

21

21

rr

rrke

M

mFT

Force required to stretch the spring by a distance me/M

2

2

2 22

1 2

1 2T

rF me

r r

Example:

“Total mass of the device” = 10 kg

12e mm

i.e. m = 10 kg

Determine the two possible values of the equivalent spring stiffness k for the mounting to permit the amplitude of the force transmitted to the fixed mounting due to the imbalance to be 1500 N at a speed of 1800 rpm

kgm 1

2

2

1

1TF me

r

2 ; 0 nm

k Here,

2

2

2 22

1 2

1 2T

rF me

r r

i.e. we can choose a stiff spring and run the machine BELOW the natural freq

or we can choose a soft spring and run the machine ABOVE the natural freq

2 2 11

T

r meF

2

2

1

1TF me

r

2

2222

/1500

012.0 /)60/21800( 211

smkg

mxsxkg

Femr

T

568.01 2 r

ssn /6.1501.252

,/9.286 657.0

/1027.2 , /1023.8 552 mNxmNxmk n

12

Summary:

=e

R=e (eccentricity)

2

2

1

1TF me

r

0 if

2mRF

Force transmitted to the base: