FORCED VIBRATIONS

INTRODUCTION

When a mechanical system undergoes free vibrations, an initial force (causing somedisplacement) is impressed upon the system, and the system is allowed to vibrate under theinfluence of inherent elastic properties. The system however, comes to rest, depending uponthe amount of damping in the system.

In engineering situations, there are instances where in an external energy sourcecauses vibrations continuously acting on the system. Then the system is said to undergoforced vibrations, as it vibrates due to the influence of external energy source. The externalenergy source may be an externally impressed force or displacement excitation impressedupon the system. The excitation may be periodic, impulsive or random in nature. Periodicexcitations may be harmonic or non harmonic but periodic. The amplitude of vibrationsremains almost constant. Machine tools, internal combustion engines, air compressors, etcare few examples that undergo forced vibration.

3.2 FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONICEXCITATION

Consider a spring mass damper system as shown in Figure 3.1 excited by a sinusoidal forcingfunction F=Fo Sint

Figure 3.1

Let the force acts vertically upwards as shown in FBD. Then the Governing DifferentialEquation (GDE) can be written as

Ck

m

F = F0 Sin ωt

kx Cx∙F x

Fo

F

m = - K - C + F

m + C + Kx = F --------------- (3.1)is a linear non homogeneous II order differential equation whose solution is in two

parts.

1. Complementary Function or Transient ResponseConsider the homogenous differential equation,

m + C + Kx = 0 which is incidentally the GDE of a single DOF spring mass damper-system. It has been shown in earlier discussions that for different conditions of damping,the response decays with time. Thus the response is transient in nature and thereforetermed as transient response.

For an under damped system the complementary function or transient response.

xc = X1 ēζ Sin (ωdt + )

xc = X1 ēζ (A Sin ωdt + B Cos ωdt) ------------ (3.2)

2. Particular Integral or Steady State ResponseThis response neither builds up nor decays with time. It is steady state harmonicoscillation having frequency equal to that of excitation. It can be determined asfollows.Consider non-homogenous differential equation

m + C + K = Fo Sin ωt ----------- (3.3)

The particular integral or steady state response is a steady state oscillation of the samefrequency ω as that of external excitation and the displacement vector lags the force vectorby some angle.

Let x = X Sin (ωt - ) be the trial solutionX: Amplitude of oscillation: Phase of the displacement with respect to the exciting force (angle by which thedisplacement vector lags the force vector).

Velocity == ωX. Cos (ωt - )= ωX Sin [90 + (ωt - )]

Acceleration= - ω2 X. Sin (ωt - ), substitute these values in GDE, (equation 3.1)

We get-m ω2 X Sin (ωt - ) + Cω Sin [90 + (ωt - )]

+ KX Sin (ωt - ) = Fo Sin ωtm ω2 X Sin (ωt - ) - Cω Sin [90 + (ωt - )]

- KX Sin (ωt - ) + Fo Sin ωt = 0 ---------- (3.4)

x x x

xx

xx

x x x

x

x

ωn

t

ωn

t

x

The four terms in the above equation represent both in magnitude and direction, thefour forces namely: inertia force, damping force, spring force and impressed force, taken inorder, acting on the system and their sum is equal to zero. Thus they satisfy the D’Alembertsprinciple. F = 0. Now, if vector representation as shown in Figure 3.2, is employed todenote these forces the force polygon shown in Figure 3.3 should close.

Represent the force vectors and draw the force polygon as given below.

Figure 3.2

Figure 3.3

Impressed force: Fo Sin ωt: acts at an angle ωt from the reference axis.

X

A

O

B

Reference axis

) (ωt - )

Displacement vector X: Lags the force vector by an angle and hence shown at (ωt - ) fromthe reference axis.

Spring force: - KX Sin (ωt - ): which means that the vector – KX acting at (ωt - ) or KXacting in opposite direction to (ωt - ) = at [90 + (ωt - )]

Damping force: - CωX [Sin (90 + (ωt - )]- CωX acting at 90 + (ωt - ) or CωX acting in opposite direction to [90 +

(ωt - )]Inertia force: mω2X Sin (ωt - )

Vector mω2X acting at (ωt - )

From the force polygon, in figure 3.3

Consider the triangle OAB.OA2 = OB2 + BA2

= (CωX)2 + (KX-mω2X)2

Fo2 = X2 (Cω)2 + x2 (K-mω2)2

Fo2 = X2 [(K-mω2)2 + (Cω)2]

X = --------- (3.5)

and tan = = =

tan =

= tan-1 -------- (3.6)

If X and are expressed in non-dimensional form it enables a concise graphical presentationof results. Therefore, divide the numerator and the denominator by K.

X = tan =

Further, the above equations can be expressed in terms of the following quantities

= Xst - Zero frequency deflection

F0

√(K-mω2)2 + (cω)2

OBBA

CωxKX-mω2x

CωK-mω2

CωK-mω2

CωK-mω2

F0 /K

√ m ω2

K)2 +(1- cω

Kk

( )2

Cω /K

(1-mω2

K)

Fo

K

Deflection of spring mass system under the steady force Fo should not be mistaken as Δst =

=

=

Thus X =

= r = frequency ratio

--------- (3.7)

where is called magnification factor, amplification factor, or amplitude ratio.

M= : It is the term by which Xst is to be multiplied to get the amplitude.

ωtan = =

1 -

... tan = -------- (3.8)

Thus the steady state response xp = X Sin (ωt - ), in whichX and are as given above.Total solution x = xc + xp

For under damped conditions:as t , xc 0 i.e., the transient response dies out. Complete solution consists onlysteady state response only.x = X Sin (ωt - ) --------- (3.9)As mentioned above, the transient vibrations die out very soon and hence the system vibrateswith steady response amplitudes. The behaviour of the system can be best understood byplotting frequency response curves as given below, in figure 3.4 and 3.5.Frequency Response Curves:Magnification Factor vs Frequency Ratio for Different amounts of Damping

mgKm

K1ωn

2

CK

2ζKωn

Xst

√(1 - )2 + (2 ζ. ω)2ω2

ωn2 ωn

ωωn

X =Xst

√ (1 - r2)2 +( 2 ζ. r)2

XXst

XXst

2ζωn

ω2

ωn2

2ζr1-r2

2ζr1-r2

( )

ζ = 0ζ = 0

ζ = 0.25ζ = 0.375

Magnification

Figure 3.4

Phase lag vs frequency ratio for different amounts of damping.

Figure 3.5

The following characteristics of the magnification factor (M) can be observed.1) For damped system (ζ =0); M as r 1.2) Any amount of damping (ζ >0) reduces the magnification factor (M) for all values of

forcing frequency.3) For any specified value of r, a higher value of damping reduces the value of M4) When the force is constant (r =0), M =1.

ζ = 0

0.5 1.0 1.5 2.0 2.5 3.0

ζ = 0.25ζ = 0.5

ζ = 0.707

ζ = 1.0ζ = 2.0

Frequency Ratio r = (ω/ωn)

Phase Angle, ,

5) The amplitude of the forced vibrations becomes smaller with increasing value of forcedfrequency. i.e M 0,as r .

6) For 0< ζ < 1/ √2 (0 < ζ <0.707), the maximum value of M occurs when r=√(1-2ζ 2) orω= ωn√(1-2ζ 2), which is lower than the Undamped natural frequency ωn and thedamped natural frequency ωd = ωn√(1-2 ζ 2),

7) The maximum value of X (when r=√ (1 - 2 ζ 2) is given by (X/Xst)= 1/[2 ζ√ (1-ζ2)] andthe value of X at ω= ωn is given by (X/Xst ) = 1/2 ζ

8) For ζ >1/√2, the graphs of M decreases with increasing values of r.

The following characteristics of the phase angle can be observed from the graph

1) For undamped system the phase angle is 00 for 0<r<1, and 1800 for r>1. This impliesthat the excitation and response are in phase for 0<r<1 and out of phase for r>1 whenζ =0.

2) For ζ >0 and 0<r<1 the phase angle is given by 00<<900, implying that the responselags excitation.

3) For ζ >0 and r>1, the phase angle is given by 900<< 1800, implying that the responseleads excitation.

4) For ζ >0 and r=1, the phase angle is =900 implying that the phase difference betweenthe excitation and response is 900.

5) For ζ >0 and large values of r, the phase angle ω approaches 1800 implying that theresponse and excitation are out of phase.

The damping factor ζ has a large influence on amplitude and phase angle in the region wherer = 1(resonance).The phenomenon represented be frequency response curve can be furtherbetter understood by means of vector diagram as follows. Consider three different cases as(1) ω/ ωn << 1 (2) ω/ ωn = 1 (3) ω/ ωn >> 1

Case (1): ω/ ωn << 1 for which ω should be very small

At very low frequencies, when ω is very small, the inertia for m ω2x and the damping forceCωx are very small.

Figure 3.6

x

(ωt-)

mω2X

Kx

Fo

CωX

This results in small values of as shown in fig .The impressed force F0 is almost equal andopposite to spring force KX. Thus for very low frequencies, the phase angle tends to zero andthe impressed force wholly balance the spring force

Case (2): when ω/ ωn = 1

Figure 3.7

With increased value of ω, the damping force Cωx and inertia force m ω2 x increase. Thephase angle also increases. If ω is increased to such an extent that phase angle =900, theforce polygon becomes a rectangle as shown. The spring force and inertia vectors becomeequal and opposite.KX = m ω2 xω = √(K/m) = ωn

ω = ωn

ω / ωn =1This is the response condition of the system during which the forcing frequency is equal tonatural frequency of the system. Also the impressed force is completely balanced by thedamping force.CωX= F0

X= F0 /Cω = F0/K/ Cω/KX=Xst / 2 ζ (ω/ ωn)X=Xst / 2 ζ (ω/ ωn) = 1Xr /Xst = 1/2 ζ Xr = Amplitude at resonance

x

mω2X

KxFo

CωX

Case (3): when ω / ωn >>1

Figure 3.8

At very large values of ω > approaches 1800, the inertia force becomes very large, where asthe spring force and damping force vectors becomes negligibly small. The improved force iswholly utilized in balancing the inertia force. 1800 i.e., Fo = m ω2x

X = Fo / m ω2

NUMERICAL EXAMPLES:3.1) A machine part of mass 2.5 Kgs vibrates in a viscous medium. A harmonic excitingforce of 30 N acts on the part and causes resonant amplitude of 14mm with a period of0.22sec. Find the damping coefficient. If the frequency of the exciting force is changed to4Hz, determine the increase in the amplitude of forced vibration upon removal of the damper.

Data: m = 2.5Kg, F0 = 30N, X = 14mm, τ = 0.225sec

Part 1: At Resonanceωn = forcing frequency = 2/ τ = 28.56 rad/sec

At resonance: ω = ωn = 28.56 rad/secωn = √(K/m) = 28.56 rad/secK = 2039 N/m

Amplitude at resonance

X =

As ω/ ωn = 1, X = (F0/K)/2ζ = 0.014... ζ = 0.526

Damping coefficient = C = Cc ζ = 2m ωn ζ= 2*2.5*28.56*0.526= 75.04 N/m/s

Fo/K

√ [1 - r2] 2 + [2ζr] 2

x

mω2XKX

Fo

CωX

C = 0.07504 Ns/m

Part (2): When f = 4 HzForcing ω = 2*fn = 25.13 rad/secFrequency ωn = 28.56 rad/sec, unchangedAmplitude of vibration with damper

Xa = Fo/K

√ [1 - r2] 2 + [2ζr] 2

= 0.01544m

Amplitude of vibration without damperXb = (30/2039)/(0.2258)

= 0.0652mIncrease in Amplitude = 0.0652 – 0.0155 = 0.0497m

Amplitude = 49.7mm

3.2) A body having a mass of 15 kgs, is suspended from a spring which deflects 12mm due tothe weight of the mass. Determine the frequency of free vibrations. What viscous dampingforce is needed to make the motion a periodic at a speed of 1mm/sec.

If when, damped to this extent, a disturbing force having a maximum value of 100Nand vibrating at 6Hz is made to act on the body. Determine the amplitude of ultimate motion.

Solution:Data: m = 15Kg; F0 = 100 N; f = 6Hz; Δst = 12mm;

(a) fn = (1/2)√(g/ Δst) = 4.55Hz

(b) The motion becomes aperiodic, when the damped frequency is zero or when it is criticallydamped (ζ = 1).

ω = ωn = √(g/ Δ) = 28.59 rad/secC = Cc = 2m ωn = 2*15*28.59 = 857 N/m/s

= 0.857 N/mm/sThus a force of 0.857 N is required at a rate of 1mm/s to make the motion a periodic.

(c) X = F0

√(K-mω2)2 + (cω)2

ω = 2f = 2*6 = 37.7 rad/sec, f0 = 100 Nfn = (1/2)(√(K/m) ... K = 12,260 N/mX = 0.00298m

= 2.98mm.

Condition for peak amplitude of vibration

(Expression for peak amplitude)

The frequency at which the maximum amplitude occurs can be obtained as follows.

M = X =

i.e., for a system acted upon by a known harmonic force, the amplitude depends only on (ω/ωn). Hence for X to be maximum √ [1 – r2] 2 + [2ζr] 2 should be minimum.

([1 - r2] 2 + [2ζr] 2) = 0

2(1 - r2) 2 (-2r) + 4ζ2r = 0

2(1 - r2) + 4ζ2r = 0

42ζ2r = 0 = 2(1 - r2)

2ζ2 = 1-r2

r2 = 1 -2ζ2

r = √1 -2 ζ2

(ω/ ωn) peak = √1 -2 ζ2

= √1 -2 ζ2

= √1 -2 ζ2 ---------- (3.10)

ωp = frequency at which peak amplitude occurs.

Where ωp refers to the forcing frequency corresponding to the peak amplitude. No maximumor peak will occur when the expression within the radical sign becomes negative i.e., for ζ >

or for ζ > 0.707.

= √1 -2 ζ2

and peak amplitude is given by

(X/Xst)max = 1/[2 ζ(√1- ζ)] --------- (3.11)

XXst

Xst

√ [1 - r2] 2 + [2ζr] 2

dxd(ω/ ωn)

dxd(r)

ωωn( )

peak

( )peak

ωp

ωn

1

√2

ωp

ωn( )

3.3) A machine of mass 25 kgs, is placed on an elastic foundation. A sinusoidal force ofmagnitude 25N is applied to the machine. A frequency sweep reveals that the maximumsteady state amplitude of 1.3mm occurs when the period of response is 0.22sec. Determinethe equivalent stiffness and damping ratio of the foundation.

Solution:

Data: F0 = 25N; m = 25 Kgs; Xmax = 1.3mm; τ = 0.22sec

For a linear system, the frequency of response is same as frequency of excitation.

... Excitation frequency = ω = 2f = 2/ τ = 28.6 rad/secthus Xmax = occurs, when ω = 28.6 rad/s

Condition for maximum amplitude to occur:

r = √1 -2 ζ2 = ω/ωn

... ωn = ω /(√1 -2 ζ2 ) = 28.6/(√1 -2 ζ2 ) --------------(1)

also we have,

X/Xst = 1 for Xmax = r =√1 -2 ζ2

√ [1 - r2] 2 + [2ζr] 2

Xmax/Xst = 1

√ [1 – (1 -2 ζ2 )] 2 + [4ζ2(1 -2 ζ2 )]

= 1

2 ζ √ (1 -ζ2 )

Xmax/(F0/K) = 1

2 ζ √ (1 -ζ2 )

Xmaxmωn2/F0 = 1

2 ζ √ (1 -ζ2 )

25*0.013* ωn2/25 = 1

2 ζ √ (1 -ζ2 )

Now substitute for ωn2 from eq.(1);

0.013*28.6/(√1 -2 ζ2) = 1

2 ζ √ (1 -ζ2)

1.0633/(√1 -2 ζ2) = 1

2 ζ √ (1 -ζ2)Squaring and rearranging,

ζ4 - ζ2 +0.117 = 0Z2 – Z + 0.117 = 0 where ζ2 = Z.

Solving the quadratic equation

ζ = 0.368, 0.93

The larger value of ζ is to be discarded because the amplitude would be maximum onlyfor ζ < 0.707 ... take ζ = 0.368

... natural frequency ωn = ω√ (1 – 2(0.368)2 )

ωn = 33.5 rad/secstiffness of the foundation,

K = mωn2 = 25(33.5)2

= 28.05*103 N/m

3.4) A weight attached to a spring of stiffness 525 N/m has a viscous damping device. Whenthe weight is displaced and released, without damper the period of vibration is found to be1.8secs, and the ratio of consecutive amplitudes is 4.2 to 1.0. Determine the amplitude andphase when the force F=2Cos3t acts on the system.

Solution:Data: K = 525 N/m; τ = 1.8secs: x1 = 4.2; x2 = 1.0; F = F0sinωt = 2cos3t... F0 = 2N, ω = 3 rad/sec

X = Fo/K

√ [1 - r2] 2 + [2ζr] 2

ωn = 2/ τ = 3.49rad/secδ = ln(4.2/1.0) = 1.435ζ = δ = 0.22

√ (42 + δ2)

r = ω/ωn = 2/3.49 = 0.573

r2 = 0.328

X = 2/525

√ [1 – 0.328] 2 + [4*0.484*0.328]X = 5.3mm

= tan-1(2ζr)(1-r2)

= tan-1(2*0.22*0.573)(1-0.328)

= tan-1(0.375)

= 20.560

3.5) The damped natural frequency of a system as obtained from a free vibration test is 9.8cps. During a forced vibration test with a harmonic excitation on the same system, thefrequency of vibration corresponding to peak amplitude was found to be 9.6 cps.Determine the damping factor for the system and natural frequency.

ωd = 9.8 cps, (ωp / ωn) = √1 -2ζ2

ωp = 9.6 cps.ωn = ωd/√1 -2ζ2

ωp√1 -2ζ2 /ωd = √1 -2ζ2

Solving for ζ: ζ = 0.196

ωn = ωd/ √1 -2ζ2 = 10 cps.

3.6) A reciprocating pump of mass 300 Kgs is mounted at the middle of a steel plate ofthickness 12 mm and width 500 mm and length 2.5 m damped along two edges as shown.During the operation of the pump, the plate is subjected to a harmonic excitation of F(t) = 50cos 60 t N. Determine the amplitude of vibration of the plate.

m = 300 KgsF0 = 50 Nω = 60K = 192EI/l3 = 176.94*103 N/m

ζ = 0 X = F0 /(K-m ω2)2

X = 6.13*10-8mm

2.5 m 500

12.

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