Chapter 3

A mechanical system is said to undergo forced vibration whenever external energy is supplied to the system during vibration

External energy can be supplied to the system through either an applied force or an impose displacement excitation

The applied force or displacement may be harmonic, nonharmonic but periodic, nonperiodic, or random

Harmonic or transient responses Dynamic response of a single degree of

freedom under harmonic excitations Resonance Examples: unbalanced rotating response, the

oscillation of a tall chimney due to vortex shedding and the vertical motion of an automobile on a sinusoidal road surface

tFkxxcxm

0 kxxcxm Homogenous solution;

The solution;

2chaptertxh

This free vibration dies out with time under each of the three possible conditions of damping and under all possible initial conditions.

tFkxxcxm

•The general solution eventually reduces to a particular solution xp (t), which represents the steady-state vibration•The steady-state motion is present as long as the forcing function

.....txp

Particular solution;

tFkxxm cos0

•The maximum amplitude of xp (t);

tCtCtx nnh sincos 21

The homogeneous solution;

The particular solution;

tXtxp cos

22

0

1

n

st

mk

FX

tFkxxm cos0

Using the initial conditions x(t=0) = x0 and v(t =0)=v0

tmk

FtCtCtx nnh

cossincos

2

021

The total solution is;

2

001

mk

FxC

n

xC

0

2

tFkxxm cos0

2

1

1

n

st

X

The maximum amplitude can be expressed;

Magnification factor, amplitude ratio

Frequency ratio, r

tFkxxm cos0

The response of the system can be identified to be of three types;

Case 1:

positive isr denominato the,10 n

tXtxp cos

The harmonic response is,

tFkxxm cos0

Case 2:

negative isr denominato the,1n

tXtxp cos

The harmonic response is,

1

2

n

stX

tFkxxm cos0

Case 3:

infinite become amplitude the,1n

tt

tx nnst

p

sin2

The harmonic response is,

Resonance: the forcing frequency is equal to the natural frequency

tFkxxm cos0

tt

tx

txtx nnst

n

n

n

sin2

sincos 00

The response of the system at resonance becomes;

tFkxxm cos0

The total response of the system can also be expressed

1;cos

1

cos)(2

n

n

stn forttAtx

1;cos

1

cos)(2

n

n

stn forttAtx

A weight of 50 N is suspended from a spring of stiffness 4000 N/m and is subjected to a harmonic force of amplitude 60 N and frequency 6 Hz. Find (a) the extension of the spring due to the suspended weight, (b) the static displacement of the spring due to the maximum applied force, and the amplitude of forced motion of the weight

Consider a spring –mass system, with k=4000 N/m and mass, m=10 kg, subject to a harmonic force F(t) = 400 cos 10 t N. Find the total response of the system under the following initial condition;

x0 = 0.1m, v0 = 0

The spring actuator shown in the figure operates by using air pressure from a pneumatic controller (p) as input and providing an output displacement to a valve (x) proportional to the input air pressure. The diaphragm, made of a fabric-base rubber, has an area A and deflects under the input air pressure against a spring of stiffness k. Find the response of the valve under a harmonically fluctuating input air pressure p(t)=p0 sinωt for the following data: p0=10 psi, ω=8 rad/s, A=10 in2, k=400 lb/in, weight of

spring = 15 lb, and weight of valve and valve rod = 20 lb.

tFkxxcxm cos0

tFtF cos0

The particular solution;

tXtxp cos)(

2/12222

0

][ cmk

FX

2

1tan

mk

c

Dividing X by k and substituting;

m

kn

mk

c

m

c

c

c

nc 22

nm

c2

n

r

k

Fst

0

222

2/12

2

221

1

21

1

rr

X

n

st

2

1

2

1

1

2tan

1

2

tanr

r

n

n

Some characteristics of the magnification factor; Any amount of damping reduces the

magnification factor

For any specified value of r, a higher value of damping reduces the value of M

In the degenerate case of a constant force (r=0), the value of M=1

The reduction in M in the presence of damping is very significant at or near resonance

The amplitude of the forced vibration becomes smaller with increasing values of the forcing frequency ( that is M→0 as r → ∞)

For 0 ≤ ζ≤ 1/√2, the maximum value of M occurs when

The maximum value of X is

22 21or 21 nr

2max 12

1

st

X

2

1

nst

X

For ζ= 1/√2, dM/dr =0 when r=0.

For ζ> 1/√2, the graph of M monotonically decrease with increasing r

Some characteristics of the phase angle; For an undamped system (ζ=0), the phase angle is

0 for 0< r < 1 and 180° for r > 1. This implies that the excitation and the response are in phase for 0 < r < 1 and out phase for r > 1

For ζ > 1 and 0 < r < 1, the phase angle is given by 0 < φ < 90°, implying that the response lags the excitation

For ζ > 0 and r > 1, the phase angle is given by 90° < φ < 180°, implying that response leads the excitation

For ζ > 0 and r = 1, the phase angle is given by φ = 90°, implying that the phase difference between the excitation and the response is 90°

For ζ > 1 and large values of r, the phase angle approaches 180° , implying that the response and excitation are out of phase

The complete solution, for an underdamped system,

tXteXtx d

tn coscos 00

sincoscos

coscos

000

000

XXXx

XXx

dn

For the initial condition, 00 0 and 0 xtxxtx

2/12

00

2

000 sincos XXX 00

000

cos

sintan

X

X

222 21 rr

X st

2

1

1

2tan

r

r

For a vibrating system, m = 10 kg, k = 2500 N/m, and c = 45 N-s/m. A harmonic force 180 N and frequency 3.5 Hz acts on the mass. If the initial displacement and velocity of the mass are 15mm and 5 m/s, find the complete solution representing the motion of the mass.

QXX

nstsmallst

2

1

The quality factor, Q, is the value of the of the amplitude ratio at resonance.

For small values of damping, (ζ<0.05),

The points R1 and R2, where the amplification factor falls to Q/√2, are called half power points because the power absorbed by the damper is proportional to the square of the amplitude: 2XcW

122

1

nQ

The difference between the frequencies associated with the half power points is called bandwidth

21

2

12

1

2

1

n

rR

21

2

22

2

2

2

n

rR

n 212

The harmonic forcing function can be represented in complex form as F(t)=F0e

iωt

tiFekxxcxm Assuming the particular solution;

ti

p Xetx

icmk

FX

2

0

2

1tan

mk

c

ie

cmk

FX

2/1

2222

0

The steady-state solution;

The Frequency Response,

icmk

FX

2

0

responsefrequency complex iH

ti

p e

cmk

Ftx

2/12222

0

iHrirF

kX

21

12

0

The Frequency Response,

2/1

2220 21

1

rrF

kXiH

The absolute value ,

ti

p eiHk

Ftx 0

)(velocity 0 txieiHk

Fitx p

ti

p

)(accel. 202txeiH

k

Fitx p

ti

p

The figure shows a simple model of a motor vehicle that can vibrate in the vertical direction while traveling over a rough road. The vehicle has a mass of 1200 kg. The suspension system has a spring constant of 400 kN/m and a damping ratio of 0.5. If the vehicle speed is 20 km/hr, determine the displacement amplitude of the vehicle. The road surface varies sinusoidallywith an amplitude of Y = 0.05 m and a wavelength of 6m.

0)()( yxkyxcxm

yckykxxcxm

22 ckYA

k

c 1tan

tYty sin)( if

tYctkY cossin

tAsin

1

2/1222

22

sin][

)( tcmk

ckYtxp

2

1

1 tan

mk

c

The steady state response of the mass,

k

c 1tan

tAtxp sin)(

The response can be rewritten as,

2/1

222

22

1

222

22

21

21

rr

r

cmk

ck

Y

X

22

31

22

31

141

2tantan

r

r

cmkk

mc

Displacement transmissibility,

ti

p Yerir

ritx

21

21Re)(

2

iHrTY

Xd

2/1221

The harmonic excitation of the base expressed in complex form as

)Re()( tiYety

The response of the system ,

Some characteristics of the displacement transmissibility; The value of Td is unity at r=0 and close to unity for

small value of r

For an undamped system (ζ=0), Td →∞ at resonance (r=1)

The value of Td is less than unity for values r > √2

The value of Td =1 for all values of ζ at r= √2

For r< √2, smaller damping ratio lead to larger values of Td .

Some characteristics of the displacement transmissibility;

For r> √2, smaller damping ratio lead to smaller values of Td

Td attains a maximum for 0<ζ<1 at the frequency ratio r = rm < 1;

2/12 181

2

1

mr

xmyxcyxkF

A force is transmitted to the base or support due to the reaction from the spring and the dashpot,

2/1

222

2

2

21

21

rr

rr

kY

FT

Force transmissibility,

tFtXmF T sinsin2

2/1

222

2

2

21

21

rr

rr

kY

FT

Force transmissibility,

Relative Motion, z = x – y,

tYmymkzzcxm sin2

1

2/1222

2

sin

sin

tZ

cmk

tYmtz

Relative Motion, z = x – y,

222

2

222

2

21 rr

rY

cmk

YmZ

2

1

2

1

11

2tantan

r

r

mk

c

The figure shows a simple model of a motor vehicle that can vibrate in the vertical direction while traveling over a rough road. The vehicle has a mass of 1200 kg. The suspension system has a spring constant of 400 kN/m and a damping ratio of 0.5. If the vehicle speed is 20 km/hr, determine the displacement amplitude of the vehicle. The road surface varies sinusoidallywith an amplitude of Y = 0.05 m and a wavelength of 6m.

A heavy machine, weighing 3000 N supported on a resilient foundation. The static deflection of the foundation due to the weight of the machine is found to be 7.5 cm. It is observed that the machine vibrates with an amplitude of 1 cm when the base of the foundation is subjected to harmonic oscillation at the undamped natural frequency of the system with an amplitude of 0.25 cm. Find (a) the damping constant of the foundation, (b) the dynamic force amplitude on the base, and (c) the amplitude of the of the machine relative to the base.

A precision grinding machine is supported on an isolator that has a stiffness of 1 MN/m and a viscous damping constant of 1 kN-s/m. The floor on which the machine is mounted is subjected to a harmonic disturbance due to the operation of an unbalanced engine in the vicinity of the grinding machine. Find the maximum acceptable displacement amplitude of the floor if the resulting amplitude of vibration of the grinding wheel is to be restricted to 10-6 m. Assume that the grinding machine and the wheel are a rigid body of weight 5000 N

One of the tail rotor blades of a helicopter has an unbalanced mass of m=0.5 kg at a distance of e = 0.15 m from the axis of rotation , as shown in the figure. The tail section has a length of 4 m, a mass of 240 kg, a flexural stiffness (EI) of 2.5 MN – m2 , and a damping ratio of 0.15. The mass of the tail rotor blades, including their drive system, is 20 kg. Determine the forced response of the tail section when the blades rotate at 1500 rpm

tmekxxcxM sin2

tmetF sin2

The particular solution;

ti

n

p

eiHM

me

tXtx

2

Im

sin)(

iH

M

me

cMk

meX

n

2/12222

2

][

2

1tan

Mk

c

iHr

rr

r

me

MX 2

222

2

21

2

1

1

2tan

r

r

The following observation can be made from the equations;

All the curves begin at zero amplitude. The amplitude near resonance is markedly affected by damping. Thus if the machine is to be run near resonance, damping should be introduced purposefully to avoid dangerous amplitudes.

At very high speeds(ω large), MX/me is almost unity, and the effect of damping is negligible.

The following observation can be made from the equations;

For 0 < ζ < 1 /√2, the maximum of MX/me is

▪ The peaks occur to the right of the resonance value r=1

Forζ > 1 /√2, MX/me does not attain a maximum. Its value grows from 0 at r=0 to 1 at r→∞.

2max 12

1

me

MX