JPD_2M Vibrations Notes

7/31/2019 JPD_2M Vibrations Notes

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1. Free Motion of Un-damped Single Degree of Freedom (SDOF) Systems

1.1 Objectives

Having followed these notes, you should be able:-

1. To represent any system having one degree of freedom as an equivalentspring-mass model;

2. To calculate the un-damped natural frequency of the system and obtain thefree response of the system for any given initial conditions.

1.2 Natural Frequency

The simplest vibrating systems can be represented by a model comprising arigid mass attached by a massless spring to a fixed abutment. Consider such asystem constrained to move in the horizontal plane with the mass resting on africtionless surface (Fig 1).

Fig 1.1 Horizontal spring-mass system Fig 1.2 FBD of mass

The spring has a stiffness k, meaning that if a force, Fis applied to the spring itwill change length by an amount x with F = kx. Consider the mass displaced (inthe positivex direction) by a distancex as shown in Fig 2.

For a positive displacement,x, the force exerted by the spring on the mass is inthe negativex direction. From Newton's second law, the force is equal to themass times the acceleration:

k x = mx or m x + k x = 0 (1.1)

Where x denotes the second derivative of the displacement with respect to

time (2

2

d

d

t

xx , the acceleration).This has a (harmonic) solution (see Appendix

1A for details),

x(t)= A cos nt + B sinnt ; where n = mk (1.2)

The displacement x varies periodically with time, t, a phenomenon known as

vibration!

n is called the natural frequencyof vibration (units rad s-1)

xx

kxk

m


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The periodof the vibration, t0, defined so that x(t) = x(t+N t0), where N is a

whole number) is 2/n so the frequency in Hertz (complete cycles of

vibration per second is)f= 1/t0 = n/2

A and B are arbitrary constants which may be found from the initial (t = 0)

conditions. For example, if the displacement at t = 0 is 0x and the initial

velocity is 0x then from Eq. 1.2 with t = 0, A= 0x

tx

txtxn

n

n

sincos)( 00

(1.3)

Thus if the mass is given an initial displacement, 0x , and is released from rest

(i.e, 0x =0 ) the motion is given by,

x(t) = x0cos

nt (1.4)

The vibration continues undiminished for an indefinite period. This is known assimple harmonic motion.

The natural frequency is not changed if the motion is in the vertical rather thanthe horizontal plane. In both cases, the mass oscillates about the staticequilibrium position, i.e. the rest position. If the system is horizontal, the springwill have zero extension in the equilibrium position, while if the system isvertical it will have extension mg/k(see Appendix 1B).

1.3 Energy

Suppose a spring-mass system is vibrating harmonically with amplitude, x 0.The strain energy, V, stored in the spring reaches a maximum at the instantwhen the displacement is a maximum and the velocity is zero, so :

Vmax=2

02

1kx (1.5)

The kinetic energy, T, of the mass reaches a maximum when the velocity is amaximum and the displacement is zero. The motion is simple harmonic and isof the form:

x(t) = 0x sinnt so that x (t) = 0x n cosnt (1.6)

and the maximumvelocity is 0x

n . Therefore, the maximum kinetic energy isgiven by:

In all vibration problems, the displacement origin (x =0)should be taken as the static equilibrium position.


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2

0max )(2

1nxmT (1.7)

But n2

= k/m, so

max

2

0

2

0max2

1

2

1Vkx

m

kmxT (1.8)

The maximum kinetic energy in the vibrating system equals the maximumstrain energy stored. Vibration may be thought of as an oscillation betweenstrain energy stored .in the spring and kinetic energy stored in the mass. Thetotal energy remains constant.

Equating the maximum kinetic and strain-energies is frequently a veryconvenient method for calculating the natural frequencies. See example 1.4.5.

1.4 Worked ExamplesThe spring-mass system may appear trivial, but in fact spring-mass modelscan be used to represent a very wide range of engineering systems.

1.4.1 Motor mounted on beam.

Calculate the natural frequency of a large electric motor, mass, m, mounted atthe centre of a simply supported beam of length L, second moment of area I,and Youngs modulus,E.

If the mass of the beam is small compared with that of the motor, the mass ofthe spring itself may be neglected. The stiffness, k, is the force per unitdeflection. For a simply supported beam loaded at midspan, the midspan static

deflection,, is given by,

WL3

/ 48EI (1.9)

so the stiffness, k, is,

kW / = 48E / L3

(1.10)

Hence, the natural frequency is,

n = m

k

= 3

48

mL

EI

(rad/s)

f =3

48

2

1

mL

EI

(Hz) (1.11)

1.4.2 Float

A circular ball of mass, m, and externalradius, r, floats exactly half submergedin a large cistern. What is the naturalfrequency of its free oscillations? (You

may ignore the effect of the linkageattached to the ball.) x

U


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Consider the equilibrium of the ball when it is displaced a small distance xdownwards from its static position.

The extra upthrust force U produced by the displacement x is given by theweight of extra water displaced. For small x, the volume of water displaced is

r

2

x so the upthrust is,U = wgr

2x (1.12)

where w is the density of water. This corresponds to the spring restoring forcekx considered in 1.2 The upthrust is in the negative x direction so the equationof motion is :

mbx = w gr2x (1.13)

where mb is the mass of the ball.

However, the ball floats exactly half-immersed, so the weight of the ball must

equal the weight of half its volume of water (the condition ofstatic equilibrium).

mbg = 2/3 w g r3 (1.14)

Substituting in (1.13) gives:

2/3 w r3x = w gr

2x

Or: 03

2 gxxr (1.15)

Now compare this with the equation of motion for a spring-mass system, Eq.(1.1),

mx + k x =0 (1.1)

The equation has the same form; comparing the coefficients suggests that the

effective stiffness is g and the effective mass is r3

2. The natural frequency is

therefore given by,

n = m

k

= r

g

2

3

(rad/s); f = r

g

2

3

2

1

(Hz) (1.16)

1.4.3 Torsion of a rotor system.

A propeller has mass moment of inertia, J, and is driven via a shaft of torsionalstiffness, kT, and negligible inertia. The other end of the shaft is connected to agearbox whose inertia is much larger than that of the propeller, so this end ofthe shaft can be considered as built-in. What is the natural frequency oftorsional vibration of the rotor system?

Consider the propeller rotated through an angle, . The resulting torque (or

moment) on the propeller is kT (The torque is negative because it acts inthe negative , sense.)

From Newtons second law, this applied moment is equal to the moment of


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inertiatimes the angular acceleration,

kT = J or, J + kT= 0

The natural frequency of the torsional oscillation is thereforeJ

kT rad/s.

Note that the units ofkT

are N m/rad and ofJare kg m2.

1.4.4 Pendulum

A pendulum comprises a uniform rod of length, L,pivoted-about one end with a mass, M, attached atits free end. The rod is slender and uniform withmass, m. What is the natural frequency of thependulum?

Consider the pendulum displaced through a small

angle, , as shown. The resulting moment, Q, aboutthe pivot due to the small angular displacement is

Q =(mg L/2 +Mg L ). (17)

The mass moment of inertia,J, of the rod about thepivot is mL2/3 and that of the end mass about thepivot isML

2.

From Newtons second law, the applied moment is equal to the moment of

inertiatimes the angular acceleration, .

Q = J (1.18)

Hence, substituting in (18) forCandJ, the equation of motion of the pendulumis :

L2

(M+ m/3) + (mgL/2 +MgL) = 0 (1.19)

Hence, the natural frequency is,

n = LmM

gmM

3

2

(rad/s) (1.20)

Question: Why does gravity control the natural frequency of the pendulumwhereas it does not affect the natural frequencies of the other systems wehave analysed?

L/2

L/2

Mg

mg


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1.4.5 Combined translation and rotation

A uniform wheel of radius R can roll without slipping on an inclined plane.Concentric with the wheel and fixed to it is a drum of radius raround which iswrapped one end of a string. The other end of the string is fastened to ananchored spring, of stiffness k, as shown in the diagram. Both spring and stringare parallel to the plane. The total mass of the wheel and drum assembly is m

and its moment of inertia about the axis through the centre of the wheel, 0, is I.Calculate the natural frequency of oscillation.

This problem is perhaps most easily solved by using the energy method.Consider the centre of the wheel vibrating with a small amplitude 0x up and

down the plane. If there is no slip, the rotation of the wheel associated with the

maximum displacement is 0 = 0x /R, and the total extension of the spring is

0x + r0= 0x + 0x r/R = 0x (R + r)/R (1.21)

If the amplitude of oscillation is 0x the maximum strain energy stored in the

spring is :

Vmax =2

0

2

2x

R

rRk

(1.22)

The maximum kinetic energy has rotational and translational components. Themaximum translational KE is (see section 1.3),

Tmax (trans)= 202

nxm (1.23)

Similarly, the maximum rotational KE is,

k

r

R


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Appendix 1A. Solution of Equation of Motion.

The equation of motion for free vibration is,

0 kxxm (1A1)

dividing by m gives,

02

xx n (1A2)

where mkn .

Let us first try a solution of the form:

ttDeCetx 21)(

(1A3)

where the values of C and D are arbitrary constants depending the initialconditions:

0x and 0x .

Differentiating (1A3):

ttDeCetx 21 21)(

; tt DeCetx 21 222

1)( (1A4)

Substituting in (1A2) gives:

02121 22222

1 t

n

t

n

ttDeCeDeCe

(1A5)

To make the left hand side zero for all t, we have to set 2

= n2, so = i

n

(similarly for where i =1

. Thus, selecting the positive root for andthe negative for:

titi nn DeCetx )( is a solution of (1A2) (1A6)

Rearranging gives,

)(2

)()(

2

)()(

titititi nnnn eeDC

eeDC

tx

(1A7)

so that,

x(t)= A cos nt + B sinnt (1A8)

When t= 0, x(t) =x0 , and substituting in (1A8) gives,x0 =A.

Also, when t= 0,0

xx and substituting in (1A8) gives Bxn

0

So finally we have:

tx

txtx nn

n

sincos)( 00

(1A8)

It is possible to show that this solution is unique.


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Appendix IB. Vertical System.

Consider the mass in Fig Bl deflected-downwards a distancex from the static

equilibrium position so the overall spring deflection, y, is given by,

y = x + mg/k (1B1)

Fig Bl. Vertical spring-mass system Fig B2 FBD of mass

From Fig B2, the net downward force acting on the mass, is

mgky = mg k(x + mg/k) = kx (1B2 )

so the final equation of motion is:

m x + k x = 0 (1B3)

which is the same as for the horizontal system.

k

m

ky

x = y mg/kmg

x


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1.5 Tutorial Questions

1. A Frahm's tachometer for measuring the rotational speed of machineryconsists of many small cantilever spring steel strips, each being rigidlyfastened to the frame of the instrument at one end and having a mass at theother (free) end. A strip will vibrate (and so give a visual indication) when the

speed of the machine is equal to the strip's natural frequency. One such strip is1 mm thick, 7 mm wide and 50 mm long. Calculate the mass required at thefree end so that the natural frequency will correspond to 1800 rev/min.

(0.079 kg)

Hint: Obtain the cantilever stiffness from data book.

2. A circular diaphragm, fixed all round its edge, has a natural frequency of5000 Hz. When an additional mass of 0.5 kg is attached to the centre of the

diaphragm, the natural frequency is reduced to 4900 Hz. What are the effectivemass and effective stiffness of the diaphragm, both referred to the central point

(12.14 kg,12.0 GN/m)

Hint: Write the natural frequency formula twice, considering the cases of withand without added mass.

3. Which of the following systems, when undergoing small oscillations, have anatural frequency which depends on the local value of the gravitationalacceleration ?

(a) A particle suspended on a string;

(b) A disc clamped normal to the end of a shaft, the other end of which is

fixed (torsional oscillations);

(c) A particle attached to the free end of a horizontal cantilever beam;

(d) A liquid in a vertical U tube;

(e) A cylinder with its axis vertical, floating in a liquid.

Hint: Try to establish the restoring force in each case.

4. What is the natural frequency of the systems depicted below? Levers are

massless and pivots frictionless.

(a)

(b)

[2

2

2

21

2

1

22212 ,,

l

g

ml

kl

ml

kl

m

kkn

]

mk2

k

x

l2

l1

m

m

l2

l1

k

(c)k1


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More difficult questions:

5. Part of a fluid-level measurementindicator consists of a sealed cylinder(outer diameterd, lengthL, massM) whichfloats half immersed inside a concentric

cylindrical container of fluid(density ).

(i) If the inner diameter of the containeris much larger than the diameter of thefloat, and if it floats freely exactly halfimmersed in the fluid, determine thenatural frequency at which oscillations ofthe float will take place it is disturbed fromits equilibrium position.

Hint: Use Upthrust force=stiffness displacement to obtain equivalent stiffness.(ii) In the complete assembly, the top of the float is connected to one end of auniform bar (mass m and length l) which is freely pivoted at its midpoint, asshown below. Derive and expression for the natural frequency of the completeindicator system.

Hint: Determine the additional effective mass (force/acceleration) due to thebar.

(i)L

g2 rad/s (ii)

MmLg

3

6

6. The mass in the diagram isconstrained by frictionless guides tomove along a vertical, straight line. Bothbars rigid, massless, and free to rotate inthe plane of the diagram. The mass is atthe centre of one bar, and the spring k3isattached at the centre of the other bar.Find the natural frequency of smallvertical oscillations of the mass fork1 = k2= 100 N/m; k3 = 200 N/m, k4 = 300 N/m;

m = 1 kg.

(2.76 Hz)

Hint: Start with k1 &k2. Add k3. Note that there will be rotation of the bar whenadding k4.

LM

l

m

m

k4k1 k2

k3


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7. The mechanism shown in the diagram consists of one central disc A, andof three smaller discs B. The system may be assumed to be undamped and to

be subject to torsional oscillation in the plane of the discs.

Disc A is free to rotate about its axis while the three discs, B, are attached to

clamped shafts each of which provides a torsional stiffness ofkt = 0.12 N m/rad.Attached to Disc A is a bar of mass M= 0.01 kg and length L = 0.02 m. Thenatural frequency of the system is controlled by adjusting the value k0of thespring which is attached to the bar, as shown in the diagram.

Disc A has moment of inertia about its axis IA = 10 10-6 kg m2 and radius,

RA = 0.03 m. Each disc B has a moment of inertia IB= 0.8 10-6 kg m2 and

radiusRB= 0.01 m.

(i) Calculate the natural frequency of the system for ko= 0, i.e. no attachedspring.

Hint: Use Tmax = Vmax. Note that the shafts will provide the strain energy terms,while the discs and rod will provide the kinetic energy terms.

(ii) Determine the value of ko that will be necessary to double the naturalfrequency of the system?

(41.4Hz, 6.075kN/m)

Hint: Use Tmax = Vmax, noting that the spring ko will give rise to just one additionalterm for strain energy.

k0

L/2 L/2

RARB

ktkt


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2. Free Motion of Damped Single Degree of Freedom Systems

2.1 Objectives


1. to represent any system having one degree of freedom and a source ofenergy dissipation (ie damping) as an equivalent spring-mass-dashpotmodel;

2. to state and explain the difference between the damped and undampednatural frequencies of the system, and calculate their values;

3. to obtain and sketch the response of the system for any given initialconditions;

4. given the response of the system, to calculate the logarithmic decrementand

hence obtain the damping ratio.

2.2 Free Motion

Real vibrating systems often have a source ofenergy dissipation and it is frequentlyconvenient to represent this by a masslessviscous damper as shown in Fig 1. Thisproduces a drag force opposing the motion andwhich is proportional to the velocityof the mass.Thus the damping results in an additionalforce on the mass of:

c x (t)

Fig 1. Spring-mass-damper systemwhere c is a constant, called the dampingcoefficientordamper rate.

Thus the equation of motion becomes :

m x (t)= k x(t) cx (t) (2.1)

or

mx + cx + k x = 0 (2.2)

It is useful to divide equation (2) by m and to rearrange so that we obtain,

x +2n x + n2x = 0 (2.3)

where n = mk is the un-damped natural frequency, as before and

km

c

2 is called the "viscous critical damping ratio".

When c = km2 then = 1 and the damping is critical (see Appendix 2A).

The solution of equation (2.3) has different forms depending on the value of

c

k

m

x


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For < 1,

)sincos()( tBtAetx ddtn

(2.4)

This is a decaying oscillation of frequency,2

1 nd which is called

the damped natural frequency. The derivation of this expression is given in the

Appendix 2A together with the results for = 1 and > 1. These two cases arevery unusual in structural vibration problems, but they do occur in the analysisof control systems.

2.3 Logarithmic Decrement

The solution for free motion given in equation (2.4) may be written as,

)sin()( tCetx dtn (2.5)

where tan =A/B. The ratio of theamplitudes of successive peaks (Fig 2)is given by :

)(

1

1

)(

)(

mmn tt

m

m etx

tx (2.6)

Fig 2.2. Decaying Oscillation

But, (tmtm+1)=(tm+1 tm) is the periodof the oscillation, t0 = 2/d so:

dnetx

tx

m

m /2

1)(

)(

(2.7)

Also,2

1 nd , so:

2

1 1

2exp

)(

)(

m

m

tx

tx

We now define the logarithmic decrement, , given by:

21 1

2

)(

)(ln

m

m

tx

txor 2 if is small (< 0.1 ) (2.8)

can be directly measured from vibration signals such as that shown in fig 2.2.

For example, if the amplitudes of the mth and (m + N)th peaks are measured, thelog of their ratio is :

N

N

tx

tx

Nm

m

21

2

)(

)(ln (2.9)

and so if is small, it is given by:

Nm

m

tx

tx

Nln

2

1

(2.10)

x(tm)x(tm+1)

tm+1tm


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2.4 Worked Example

A radio transmitter and its powersupplies are to be dropped byparachute to a polar expedition. Inorder to protect the electronics from

excessive dynamic loading, thepackage, which is of mass 20 kg, ismounted on a spring-damper systemas shown in the diagram.

The spring has. a stiffness of 10 kN/m,the damper has a rate c= 540 N s/m,and the base plate has negligiblemass. If the terminal velocity of theparachute is 8 m/s what is the

maximum distance which the spring iscompressed, assuming that theground on which the package lands isrigid?

mkn = 20104

= 22.36 rad/s.

km

c

2 = 540/2 20.104 = 0.604.

21 nd = 22.36 2604.01 = 17.83 rad/s.

At the instant of impact, there is no compression in the spring. However, in thestatic equilibrium position which we take as our datum, the spring iscompressed by mg /k Hence, taking x as positive downwards, the initialconditions are

0x = mg/k=20 9.81/104 = 0.0196 m; and we are given 0x = 8 m/s.

Now )sincos()( tBtAetx ddtn

so 0x = A = 0.0196m

)cossin()sincos()( tBtAetBtAetx ddt

ddd

t

n

nn

tBAtABe dnddndtn sin)(cos)(

Hence, BAx dd 0 and substituting for ndandA gives:

B = 0.434 m.

Therefore, the displacement response is given by:

ttetx ddtn

sin)m434.0(cos)m0196.0()(

The maximum displacement occurs when the velocity is zero, i.e. when:

m

ck


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nd

ndd

dnddnd

dnddnd

t

BA

ABt

tBAtAB

tBAtABe n

tan

sin)(cos)(

0sin)(cos)(

Substituting forA, B, ndand gives: tandt = 1.452 so dt= 0.97 rad

Hence, the maximum displacement is at t= 0.054 s.

Substituting t= 0.054 s in the expression forx(t) givesx = 0.167 m. This is thedisplacement of the mass down from the equilibrium position, so the overallcompression of the spring is this value plus the static deflection, i.e.

The maximum compression is 0.167 + 0.0196 0.186 m


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Appendix 2A. Solution of Equation of Motion

The equation of motion of a free, damped single degree of freedom system canbe written as,

022

xxx nn (2A1)

This has a solution of the form,

ttDeCetx 21)(

(2A2)

where the values ofCand D are given by the initial conditions, and 1 and 2

are found as before by substituting tAetx )( in Eq. (2A1) giving:

0)2( 22 nnt

Ae Hence, 0)2( 22 nn (2A3)

so nnnn

12

4)2( 222

2,1

There are now three possibilities for the solution of this equation depending on

the value of the damping ratio,

(a) Oscillatory motion.

When < 1 the roots of Eq. (2A3) are complex:

n

i 22,1

1

and the general solution becomes,titi nn DeCetx

)1()1( 22

)( (2A4)

titit nnn DeCeetx )1()1(

22

)( (2A5)

This equation can be re-written in terms of sin and cos, as shown in lecturesappendix 1A:

)1sin1cos()( 22 tBtAetx nntn (2A6)

It can be seen that the exponential term tne determines the decay rate and

the rest of the equation determines the oscillation. n2

1 is the damped

natural frequency, d. Eq. (2A6) can then be written as:

)sincos()( tBtAetx ddtn (2.4)


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(b) Non-oscillatory motion.

When > 1 the roots of Eq. (2A3) become realand the solution to Eq. (2A1)can be written as,

tt nn DeCetx )1()1( 22

)( (3A8)

In this case the displacement will not oscillate, but will gradually decay to zerodue to the high damping.

(c) Critically damped motion.

When = 1 so that kmc 2 , the roots of Eq. (2A3) are equal,

1 = 2 = n

The general solution then becomes,

tneDtCtx )()( (3A8)

As for > 1, The displacement will gradually decay to zero, the rate of decaybeing a maximum in the case of critical damping.


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1. The stiffness of the spring mounting of a heavy machine is 2105 N/m. The

machine is deflected and released. A trace of its subsequent motion is shownbelow. Find the actual and undamped natural frequencies, the mass of the

system, the damping ratio, , and the damping coefficient c.

4 rad/s and 12.6 rad/s 1260 kg , =0.081, c=2580 Ns/m

Hint. Obtain the damped natural frequency from the damped period (fromgraph). Obtain the log decrement from successive cycles. Convert it todamping ratio.

2. Derive expressions for the critical damping ratio and the damped natural

frequency of the system shown below, for small vertical vibrations of the pointmass? You may assume the levers have no mass and the pivot is frictionless.

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4

2

41

2

km

c

pr

q

m

k

p

r

kmrp

cq

d

x

k r

p

mcq


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More Difficult Question

3. In a type approval test, a helicopter is dropped under gravity from a height, h,with its rotors stationary. The pilot's seat is mounted on springs of combinedstiffness k, and of dampers of combined rate c. The combined mass of the(dummy) pilot and seat is m.

(i) Show that the equation of motion of the pilot and seat after impact with theground, which occurs at t= 0, is

mx + cx + k x = 0

with initial conditions:x(0) = mg/k, and )0(x = gh2 , the positive coordinate

direction being as shown in the diagram. x(t) is the displacement fromequilibrium at time, t.

Hint. Use static equilibrium condition.

(ii) Consider the case where m = 100 kg, k= 2.5 x 105 N/m, c = 3000 N s/m,and

h =5 m. Calculate the maximum acceleration experienced by the pilot after

the helicopter strikes the ground. To simplify the calculation, you may assumethat )0(x = 0. You should justify this assumption in the light of your calculations.

Hint: Unlike the case of un-damped motion, the maximum acceleration doesNOT occur at the time when the velocity is zero. Differentiate the expressionfor the acceleration and set it to zero. (Answer 404 m/s2 )

(iii) Compare the acceleration calculated above with that which would beobtained without the dampers. (Answer: 495 m/s2)

(iv) What would be the maximum acceleration of the pilot if the spring stiffness

were reduced to 6 10

4

N/m with no dampers fitted? What would be thedrawback of using a softer spring? (Answer: 242 m/s2)


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3. Forced Vibration of Single Degree of Freedom Systems

Objectives. .


1. to calculate the amplitude and phase of the steady-state vibration responseof a single degree of freedom system for a given harmonic exciting force;

2. to represent the motion of the system by a phasor diagram;

3. to sketch the frequency response curves for the system with different levelsof damping;

4. to describe qualitatively the effect of damping on the response of thesystem in different frequency regions;

5. to recognise that the full solution to forced vibration problems is given by atransient component (complementary function) plus a steady state component,

(particular integral) and be able to derive the full solution for simple cases suchas sinusoidal and step inputs.

3.1 Solution for Harmonic Force Input

Consider a single degree of freedom system

subject to an applied force, F(t) as shown in Fig 1.

The equation of motion is :

)(tFkxxcxm (3.1)

This is a linear, second order, inhomogeneousdifferential equation with constant coefficients.

The general solution is the sum of thecomplementary function (solution of thehomogenous equation formed by 3.1 with the righthand side set to zero) and a particular integral (asolution of the inhomogenous equation).

The complementary function is the solution of Eq. (2A1), which we havealready shown is:


where A and B depend on the initial conditions. The most common forcingfunction of interest is sinusoidal(or harmonic) excitationof the form, F(t) = F

sint, so Eq. (3.1) becomes,

tFkxxcxm sin (3.3)

It is a property of linear systems that if they are subject to harmonic excitation,the steady-state response (the particular integral) is also harmonic, and is atthe same frequency as the input. Therefore the steady-state response is givenby :

)sin()( 0 txtx (3.4)

Figure 3.1 Forced systemwith damping.

c

k

m

x

F(t)


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It is straightforward to show that (3.4) is a solution of (3.3) by differentiation andsubstitution. It therefore qualifies as a particular integral of (3.3) and thegeneral solution will be:

)sin()sincos()( 0 txtBtAetx dd

tn (3.5)

In other words, the full solution is a damped vibration at the (damped) natural

frequency, d which dies away as the damping takes effect (thecomplementary function), together with a steady harmonic vibration at the

frequency of the applied force, . It is this harmonic solution we are mostconcerned with here.

xo is the amplitude of the response and defines its phase with respect to the

exciting force. The values ofxoand may be found by substituting equation(3.4) into (3.1) and rearranging the terms (see appendix, 3.8). Alternatively, wemay use the phasor diagram approach as shown in section 3.4 below. Thesolution is,

2220

cmk

Fx

and 2

tan

mkc

(3.6)

Rearranging and setting r= /n gives,

2220

21

1

rrF

kx

; 212

tanr

r

(3.7)

where is the critical damping ratio.

Note that 0 180 sofrom equation (4), theresponse lags the force asillustrated in Fig 2.

Fig 3.2 Phase lag between force and response

3.3 Frequency Response Function

-The response given by equations (5) and (6) is a strong function of frequencyand the relationship xo/F is called the frequency response function(FRF). Forpractical values of damping the amplitude reaches a maximum at a frequency

close to n. This is known as resonance.

The phase varies between 0 180 and is 90 at = n. The FRF is

plotted in Fig 3 for= 0.1.

Amplitude,x0

x(t)

F(t)/k


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24

Fig 3.3 Frequency response functions - magnitude and phase.

NB: The response is given byxosin(t and the function has beenplotted in order to emphasise that the response lags behind the force.

Fig 3.4 Frequency response functions for different damping ratios

F

kx0

r

r

F

kx0

, r= /n

, r= /n

0.05


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25

3.4 Phasor diagrams

Consider the force, F(t) = F sin t. Thismay be represented as a phasor of

magnitude F rotating at speed , thevalue at any instant being given by its

projection on the vertical axis (Fig 4).

Figure 3.5 Phasor diagramrepresentation of harmonicallyvarying quantity

Now, the steady state solution to equation (3.3) is of the form:

xtxtx

txtxtx

txtx

22

0

00

0

)sin()(

)2/sin()cos()(

)sin()(

so,

Hence the velocity leadsthe displacement by /2 while the acceleration leads

the velocity by /2 and is therefore in antiphaseto the displacement. The LHSof equation (3.3) may therefore be re-written as:

)sin()2/sin()sin( 02

00 txmtxctkx (3.8)

This may be represented on a phasor diagram as shown in Fig 3.6:

Fig 3.6. Phasor diagram for LHS of Eq. 1. Fig 3.7. Full phasor diagram of Eq. 1.

From equation (3.3), the resultant vector of this sum equals the force vector sothe full phasor diagram is as shown in Fig 3.7. By Pythagoras theorem:

F2

= x02

(m2 k)

2 +x0

2 (c )2or

2220

cmk

Fx

And tan =

2

mk

c

F(t) = Fsin t

F

t

m2x0

cx0

t t

F

m2x0

cx0

kx0 kx0


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26

3.5 Excitation by Rotating Out of Balance

Excitation by rotating out-of balance is a particularly common case. Forexample, a turbo-machine with a slightly out of balance rotor mounted on aplatform will excite the platform at a frequency equal to the rotational speed ofthe rotor :

The out of balance is representedby a mass, m', at a radius r(Fig 3.7) The centripetal forceexerted on m' to maintain the

motion is m'r2.The vertical component of thisforce is

2

sin t

Similarly, the horizontal componentis:

2cost

Fig 3.8; Rotating out-of-balance

There are therefore equal and opposite forces on the machine mass, m, so the

machine is subject to harmonic forces of magnitude 2 in both the

horizontal and vertical directions.

We are frequently only concerned about motion in one direction, in which caseequation (3.1) becomes,

(m+m')x + c x + k x= m'r2 sint (3.9)

Note that m'is usually very small compared with m and so m+m' m.

3.6 Worked ExampleA machine of mass 1000 kg creates a sinusoidally alternating vertical force ofmaximum value 1500 N at a frequency of 50 Hz. The machine is supportedequally by four spring mountings each of stiffness 4 MN/m and which exert atotal viscous damping force such that the damping is 20 % of the critical.

Calculate the steady state amplitude of the motion and the phase angle of thedisplacement with respect to the force. Sketch the corresponding phasediagram.

= c/2 mk. = 0.2.

Hence c =2 mk. = 0.2 2 10001044 6

= 5.06 l04 N s/m.

From equation 5,

2220

cmk

Fx

r m

c

m

k


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27

Now, F= 1500 N; = (50 x 2) = 100 rad/s; m = 1000 kg; k= 16 x 106 N/m;c = 5.06 x 10

4 N s/m. Hence,x0 = 1.8 x 10-5 m = 0.018mm

The phase angle between the force and the displacement is given by:

2tan

mk

c

; substituting values gives:

= 169, taking principal values 0 180.

This phase angle implies that the system is operating well above resonance.

Check: mkn = 100010166 = 126 rad/s; f = 20 Hz which is well

below the running speed of 50 Hz.

Phasor diagram values :

kx0= 16 106 1.8 10-5= 288 N = spring force

cx0 = 5.06 104 100 1.8 10-5= 286 N = damper force

mx0 = 1000 (100 )

2 1.8 l0-5= 1777 N = inertia force

mx0= 1777 N

cx0= 286 N

kx0= 288 N

F= 1500 N


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28

3.7 Glossary:

Vibration Cyclic variation of displacement

Natural frequency n angular frequency of freely vibrating systemwithout damping (rad/s)

Equation of motion Differential equation relating displacement andtime; follows from equilibrium (fbd) of mass

Viscous damping Retarding force proportional to velocity xcF Damped natural frequency d angular frequency of damped system (rad/s)

Amplitude Maximum value of harmonic function (e.g. ofdisplacement, velocity or acceleration)

Critical damping ratio kmc 2/ ratio of damping coefficient to that

which causes critical damping (onset of non-vibrating behaviour in free damped system

Harmonic Varying with time in a sinusoidal manner; x isharmonic if tBtAtx cossin)(

Excitation External (e.g. harmonic) force or displacementapplied to a (e.g. spring-mass) system

Response Displacement resulting from an external force

Phase angle Angle (rad) between harmonic functions ofidentical frequency

3.8 Appendix: Derivation of Amplitude Eq. (3.6)

Equation of motion: tFkxxcxm sin (3.3)

Consider a harmonic solution of the form: )sin()( 0 txtx (3.4)

We can express this as tBtAtx sincos)( (3.10)

(The amplitude ofx is 220 BAx )

Differentiating (3.10):

tBtAtx cossin)(

, tBtAtx sincos)(

22

Substitute into (3.3) and collect sine and cosine terms:

tFtkABcAmtBkAcBm sincossin 22 (3.11)

Since this must be true at all times, the coefficients of the sine and cosineterms must be the same on both sides of the equation.

Sine coefficients:

FBkAcBm 2 (3.12)

Cosine coefficients:

02 kABcAm


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29

Or:km

c

B

A

2

(3.13)

Substitute (3.13) into (3.12) to eliminate A:

222

2

2

2

)()(

)(

ckm

FkmB

FBkckm

cBBm

(3.14)

Likewise, eliminating B:

222

22

2

)()(

ckm

FcA

Fkc

kmAAcm

c

kmA

(3.15)

We have found the unknown coefficients A, B, in the displacement solution,

(3.10). Now determine the amplitudeofx(t): 220 BAx :

Square (3.14) and (3.15) and add together:

2222

2222222

0

)()(

)()(

ckm

FkmcBAx

222

22

0

)()( ckm

FBAx

(3.16)

This is the steady state amplitude of displacement in a 1DoF system excited bya sinusoidal force of amplitudeF.


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30


1. The phasor diagrams of Figs 3.6 and 3.7 correspond to the case when theforcing frequency is above the undamped natural frequency. Sketch thecorresponding diagrams when the forcing frequency is below the undampednatural frequency and is equal to the undamped natural frequency. What are

the relative values of the spring, damper and inertia forces in these cases?Discuss the diagrams with your tutor.

2. A simply supported girder carries a rotating machine at its centreThe mass of the girder is small compared with the mass of the machine. Thetotal mass of the machine is 500 kg and its out-of-balance is equivalent to amass of 12 kg acting at a radius of 50 mm.

The damping of the whole system can be determined from an equivalent forceproportional to the velocity acting at the middle of the girder. Such a force isestimated to be 400 N at a velocity of 20 mm/s

The weight of the machine causes the girder to deflect 0.25 mm.

(i) Determine the steady-state amplitude of forced vibration for a rotorspeed of 80 rad/s.

(ii) Determine the amplitude of forced vibration at resonance.

(0.23mm, 5.9mm)

Hint . Calculate the equivalent stiffness from F= kx and the effective damping

from F = c x . Use equation (3.5).


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31

4.Vibration Isolation

4.1 Objectives


1. to define vibration isolation and state the difference between isolation anddamping;

2. to calculate the magnitude and phase of the force transmitted to thefoundations of a vibrating system, given the magnitude and frequency of theforce exciting the system;

3. to sketch transmission vs frequency curves for different levels of dampingand state the region in which it is desirable to operate an isolation system.

4.2 Analysis

We are often concerned with the force transmitted by a vibrating system to itssurroundings. For example, a car engine produces large forces over a widefrequency range due to the rotation of an out of balance crank-connecting-rod-piston system and due to the combustion process. If this force weretransmitted directly to the car body, severe vibration would be produced andthe noise in the passenger compartment would be very unpleasant. The engineis therefore mounted on rubber blocks to isolate it from the body.

Consider a one degree of freedom system subject to a harmonic force, Fsint.The force transmitted to the base, FT, is given by the sum of the forcesgenerated by the spring and the damper. Hence:

FT= k x + cx (4.1)

and from the phasor diagram Figure (4.1)

220 ckxFT (4.2)

But we know that the amplitude of motion,x0 is given by :

2220

cmk

Fx

,

Fig 4.1. Phasor diagram showing transmitted force

so the transmissibility, T, is given by:

t

F

m2x0

kx0

cx0


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32

22222

cmk

ck

F

FT T

(4.4)

Rearranging and setting r= /n gives:

2222

21

21

rr

rT

(4.5)

where is the damping ratio.

The transmissibility is plotted inFig 2 as a function of frequencyfor different damping ratios. Notethat a log scale has been used.

The force transmitted is higherthan the exciting force for

0 < r< 2 and damping reducesthe transmitted force in this region.

Abover= 2 the force transmittedis less than the exciting force andthe base is isolated from theexcitation. In this region, dampingincreases the force transmission

and so is not beneficial.

4.3 Example

A machine of mass1000 kg creates a sinusoidally alternating vertical force of'maximum value 1000 N at a frequency of 40 Hz. The machine is supported onfour rubber mounts which act as both springs and dampers. The mounts aresymmetrically placed with respect to the centre of mass and the effectivedamping ratio is 0.15, Calculate the required stiffness of each mount if theamplitude of the steady state alternating force transmitted through the mountsto the floor is to be under 300 N.

From equation (4.5)

3.0

1000

300

21

21222

2

rr

rT

Setting =0.15 and solving forrgives r=2.2. (Or use the curves in figure 4.2!)

Now, r=n, and =240=80 rad/s, so n is 80 / 2.2=114 rad/s orfn= 18.2 Hz.

Now,n = mk so k = mn2 = 1000 114

2= 13 x 106 N/m.

Hence, the required stiffness ofeachmount is 13/4 = 3.2 MN/m.

Fig 4.2. Transmissibility

curves

= 0

= 0.1

increasing

= 0.5

r = /n

T


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33


1. For frequency ratios above 2 , the best isolation is obtained with nodamping. Explain this physically.

A machine operates at high speeds and it is proposed to mount it on metal

springs with minimal damping. Comment on the proposal.

2. An instrument of mass 7 kg is supported from a fixed base by a spring anda viscous damper which work in parallel. The system is constrained so thatonly vertical motion is possible. The instrument generates a harmonic

disturbing force of the form Fsintin the vertical direction. If the damping ratiois 0.3 and the frequency of the disturbing force is 50 Hz, calculate the springstiffness required so that only 10% of the force generated by the instrument istransmitted to the base.

(17.2kN/m)

Hint. Use equation (4.5) forT= 0.1.

3. At its running speed of 500 rev/min, a spring-mounted machine of mass850 kg produces a vertical harmonic excitation force of 275 N, the excitationfrequency being equal to the running speed. The design specification is asfollows:

At this running speed, the force transmitted to the floor must not exceed60 N.

The static deflection of the machine on its springs must not exceed 60mm.

The damping associated with the springs must be as high as possible

(subject to the force transmissibility constraint) in order to limit theamplitude at resonance.

Find the spring stiffness and the viscous damping coefficient necessary toachieve the required design conditions

(1.39 105 N/m, 8944 N m/s)

Hint. Obtain equivalent stiffness from static deflection. Use equation (4) forT= 60 / 275.


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34

5. Base Excitation of Single Degree of Freedom Systems

5.1 Objectives


1. to calculate the amplitude and phase of the steady-state vibration of asingle degree of freedom system for a given harmonic base motion;

2. to represent the motion of the system by a phasor diagram;

3. to sketch the frequency response curves for the system with different levelsof damping;

4. to recognise that the full solution to base excitation problems is given by atransient component (complementary function) plus a steady state component(particular integral) and be able to derive the full solution for simple cases suchas sinusoidal and step inputs.

5. 2 Solution for Harmonic InputIn many cases, we are concerned with theexcitation of a system via its base. Examplesare the vibration of a car driving along a roughroad and the vibration of buildings by motionof the ground caused by earthquakes.

Consider a single degree of freedom system(Fig 1) subject to base motion, y(t) . The

spring is compressed by (y x) so the force on

the mass in the positive x direction is k(y x).

Similarly, the force on the mass due to thedamper is c( y -x ).

Fig 5.1. Base ExcitationHence, the equation of motion is :

)()( xycxykxm (5.1)

Or, rearranging : yckykxxcxm (5.2)

This again is a linear ODE with constant coefficients so the general solution isthe sum of the complementary function and the particular integral. Thecomplementary function is the solution of equation (5.2) with the RHS set tozero which we have already shown is :


whereA andB depend on the initial conditions.

The most common base motion of interest is sinusoidal motion of the form,

tyty sin)( 0 (5.3)

for which the particular integral is also a harmonic function with frequency .

This means that the steady vibratory solution (the P.I.) can be represented, as

x

ck

m

y


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35

before, by a phasor rotating at an angular speed . Now, we are interested inthe amplitude and phase of the motion of the mass with respect to that of thebase. Fig 5.2 shows the phasor diagram representation of the LHS of equation(2) when the excitation, and hence the response is harmonic (sinusoidal inform). We know that for the equation to be satisfied, the vector formed by theRHS must equal the resultant vector of the LHS. The complete phasor diagramis shown in Fig 3.

Fig 5.2. Phasor diagram of LHS ofequation (5.2)

Fig 5.3. Full phasor diagram forharmonic base excitation.

From the geometry of Fig. 5.3, using Pythagoras theorem:

22222

0

222

0 )()()( cxmkxcky o

or

222

22

0

0

cmk

ck

y

x

(5.4)

Setting r =/ngives :

2222

0

0

21

21

rr

r

y

x

(5.5)

where is the damping ratio.

The ratio x0/y0 is sometimesknown as the motiontransmissibility and you willnotice that it is given .by thesame expression as the forcetransmissibility in the case of afixed base and harmonic force.The Frequency ResponseFunction (FRF) is shown in Fig

5.4 for = 0.1.

Fig 5.4, FRF for base excitation; = 0.1.


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36

5.3 Worked Example

A car has a body of mass 1500 kg which is mounted on four equal springs inparallel which deflect 0.20 m under the weight of the body. Each of the fourshock absorbers has a damping coefficient of 1100 Ns/m.

The car is placed with all four wheels on a test platform which is moved

sinusoidally up and down at approximately the resonance frequency (= n)with an amplitude of 25 mm Find the steady-state amplitude of the car bodyrelative to the fixed ground, assuming the centre of mass to be in the centre ofthe wheel base.

The centre of mass of the car is in the centre of the wheel base so whensubject to equal excitation at the four wheels, the motion will be purely verticaland a single degree of freedom model can be used.

The total spring stiffness, k = mg/= 1500 9.81 / 0.2 = 73575 N/m.The overall damping coefficient, c= 4 1100 = 4400 N s/m.

The critical damping ratio, = c /2 km = 4400 / 2 15007357 = 0.209.

From equation 5.5 :

2222

0

0

21

21

rr

r

y

x

and setting r= l and substituting forgives x0/y0 = 2.59 (>1 i.e. amplification).

Therefore, sincey0 = 25 mm,x0 = 64.7 mm.


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37


1. A simple way to study the vibration of a vehicle travelling over arough road is to model it as a SDOF system of mass m, stiffness k, anddamper rate c, and subject it to base excitation.

(i) Consider the case where the car is travelling at uniform speed V over asinusoidally undulating surface with wavelength L and peak-to-peak amplitude2h. Derive an expression for the amplitude of vibration of the vehicle body by

assuming steady state conditions.

Hint. Excitation frequency, in Hz, is = V/L(frequency of going over onewavelength). Use equation (5) on Page 26.

(ii) For the specific case where m =750 kg, k= 6104 N/m, c = 2650 Ns/m,L = 10 m, h =0.03 m, calculate the amplitude of vibration when the vehicletravels at a steady speed of 17.5 m/s.

(iii) Estimate the speed at which the vibration amplitude of the vehicle will be amaximum.

Hint. Max response when excitation frequency = natural frequency(0.047 m, 14.2 m/s)

2. A body of mass m lying on a smooth horizontal plane has a spring ofstiffness k fixed to it, the axis of the spring being horizontal and passingthrough the centre of mass of the body. The other end of the spring is suddenlygiven, in the line of the spring,

(i) a constant velocity v,

(ii) a constant acceleration .

Find the displacement of the body as a function of time in cases (i) and (ii).

For case (i), ifm = 1.0 kg, k = 400 N/m and v = 3 m/s, find the maximum forcein the spring and the time taken for this value to be reached first.

(60N, 0.0785 s)

Hint: Start from equation (5.2) and ignore damping. Note that vy and

y = vt + yo in case (i). Similarly, y in case (ii).

x

k c

V

h

L


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6. Two Degree of Freedom Systems

6.1 Objectives


1. to define the term 'degree of freedom' and recognise that a system with ndegrees of freedom has n natural frequencies and corresponding mode shapes;

2. to calculate the natural frequencies and mode shapes for any given twodegree of freedom system;

3. to recognise that the free motion of an n degree of freedom system may berepresented as a sum of oscillations at each of the n natural frequencies, theproportion of each mode depending on the initial conditions;

4. to recognise that the frequency response function for an n degree offreedom system will have n peaks corresponding to the n natural frequencies.

6.2 Degrees of Freedom

A system has n degrees of freedom if its position may be completely defined byn coordinates. The systems we have considered in parts 1-5 were completelydefined by one coordinate (a translation or a rotation) and so were termedsingle degree of freedom systems. An unconstrained rigid body has sixdegrees of freedom: translation along the x, y and z axes and rotation abouteach axis.

6.3 The Analysis of Two Degree of Freedom Systems

Consider the system shown in Fig 1 in which each of the masses is

constrained to move in the line of the springs. The position of the system maytherefore be completely described by the two displacements x1 and x2 so thesystem has two degrees of freedom.

Fig 6.1. Two degree of freedom System

As in the case of single degree of freedom systems, we take the datum of eachcoordinate as the static equilibrium position so if the system is vertical, gravityforces may be omitted from the analysis.

m1 m2

x1 x2

k1 k2


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39

Step 1. Write Down Spring Deflections

The deflections of the springs k1 and k2 in Fig 1 are governed by thedisplacementsx1 and x2. In general, for any displacements x1 and x2, we maywrite:

Compression of spring k1 = x1 (6.1)

Compression of spring k2

= x1

x2

(6.2)

Step 2. Obtain Forces (and moments if applicable) on Each Mass

Now, the force on m1 in the positivex1 direction is

{(compression ofk1)k1} {(compression ofk2) k2}

as shown in the free body diagram of Fig 6.2a. Hence,

F1 = (x1) k1 k2 (x1x2)

Likewise, the force on m2 in the positivex2 direction is: (compression ofk2) k2

as shown in Fig 6.2b.

Fig 6.2a FBD for mass m1 Fig 6.2b. FBD for mass m2

kx1 k2 (x1 x2)

x1

k2 (x1 x2)

x2

m1m

2


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40

Step 3. Obtain Equations of Motion

Applying Newton's second law to mass m1 gives:

m1 x 1 = F1 = k1x1 k2 (x1x2) or,

m1 x 1 + (k1 + k2 )x1 k2x2 = 0 (6.3)

Similarly for mass m2,

)( 212222 xxkFxm

0)( 21222 xxkxm (6.4)

Step 4. Write in Matrix Form

Assume harmonic solutions of the form :

)sin( 111 tXx and )sin( 222 tXx

Then, differentiating twice with respect to time gives :

1

2

11

2

1 )sin( xtXx and

2

2

12

2

2 )sin( xtXx

Substituting in equations (6.3) and (6.4) gives :

0)( 2112112

1 xkxkkxm (6.5)

0122222

2 xkxkxm (6.6)

Equations (6.5) and (6.6) may now be re-written in matrix form as

0

0

2

1

2

222

2

2

121

x

x

mkk

kmkk

(6.7)

or,

0)( 2 xMK (6.8)

where,

22

221

kk

kkkK ;

2

1

0

0

m

mM ;

2

1

x

xx

K is known as the stiffness matrix, M is the mass matrix and x is thedisplacement vector.


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41

Step 5. Check Symmetry

The stiffness and mass matrices should each be symmetricaland should havepositiveterms on the leading diagonal. It is very easy to get a sign wrong in theanalysis and this check frequently reveals the error. The mass matrix of asystem with rigid masses or inertias is always diagonal.

Step 6. Obtain Natural Frequencies

Equation (6.8) has a null right hand side (mathematically it is a homogeneous

equation) so non-trivial solutions forx can only exist if the matrix (KM2) issingular, i.e. its determinant is zero. Hence:

02

222

2

2

121

mkk

kmkk (6.9)

Expanding the determinant gives:

0))(( 222

22

2

121 kmkmkk (6.10)

or multiplying out brackets: ,

0)()( 222212

12221

4

21 kkkkmkmkkmm (6.11)

This is a quadratic equation in 2

which may readily be solved to give two

natural frequencies, n1 and n2. The usual convention is to call the lower

natural frequency n1 and the higher one n2.

Step 7. Obtain Mode Shapes

Having found the two natural frequencies, these may then be substituted intoequation (6.8) to obtain the corresponding displacement vectors, x. Since the

matrix (KM2) is singular, equation (6.8) has no unique solution; the ratio of

the components of the displacement vector can be found, but the amplitude ofthe vector is arbitrary.

Physically, this means that at a natural frequency the ratio of thedisplacements of the two masses is defined, but that any absolute values ofthe displacements are possible. In other words, the amplitude of vibration at aresonance can build up to any value. Since we know that the amplitude of thedisplacement vector is arbitrary, we can solve for the ratio x2 /x1 (called themode shape). The first row of equations (6.7) gives

0)( 2212

121

xkxmkk (6.12)

so at the first natural frequency, n1:


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42

2

2

1121

1

2 )(

1

k

mkk

x

x n

n

(6.13)

This ratio x2/x1 defines the mode shape at the first natural frequency, n1.

Similarly, the mode shape at the second natural frequency n2is:

2

2

2121

1

2 )(

2

kmkk

xx n

n

(6.14)

6.4 Complete Solution for Free Motion

Let A1 and A2 be the amplitudes of motion of the mass m1 in the first and

second modes respectively. Then in the first mode:

)sin( 1111 tAx n ; )sin( 11112 tAx n

where 1 =x2 /x1 in the first mode. Similarly in the second mode:

)sin( 2221 tAx n ; )sin( 22222 nAx

where 2 =x2 /x1 in the second mode. The general solution of equation (8) is

obtained by adding (superposing) these two 'normal modes', i.e.:

)sin( 1111 tAx n )sin( 222 tA n (6.15a)

)sin( 11112 tAx n )sin( 2222 nA (6.15b)

The constants A1, A2,1and 2 are found from the initial conditions (the initial

displacements and velocities of the two masses). The free motion consists ofcomponents from both modes, the proportion of each mode depending on theinitial conditions.

6.5 Forced Motion

Consider a harmonic force, F1sintapplied to mass m1 in Fig 6.1. The matrix

equation (6.8) now becomes

FxMK )(2 (6.16)

where

0

1FF is termed the force vector. This equation may be solved forx

at any given excitation frequency, . However, if=n1 or=n2, the matrix

)( 2MK is singular so the amplitude of the response is indeterminate as in

the case for a single degree of freedom system discussed earlier.


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43

Fig 6.3 shows a typical form of the function x1/F1 for the case wherek1= k2 = 1 N/m; m1 = m2 = 1 kg. The response rises sharply to maxima at thenatural frequencies and is zero ata frequency between the twonatural frequencies. This null

response even though a force isbeing applied is exploited in somevibration absorbers. There arethree other frequency responsefunctions for a two degree offreedom system: x2/F1, x1/F2,x2/F2which can be obtained in a similarway.

Fig 6.3. Typical 2 Degree of Freedom FRF

6.6 Worked Example

A box containing delicate electronics is mounted on two springs as shown inthe diagram. Find the natural frequencies andcorresponding mode shapes of thesystem, when k1 = 1 kN/m; k1 = 2 kN/m; the massof the box , m = 1 kg and its moment of inertiaabout its centre of mass, I = 0.01 kg m

2. The

dimension a = 0.05 m. The centre of mass is

constrained to move vertically and only motion inthe plane of the diagram is to be considered.

Step 1. Write Down Spring Deflections

Compression of spring axk 1

Compression of spring axk 2

opposite signs for rotational terms.

Step 2. Obtain Forces and Moments

Now, the force on mass in the positivex direction is

{(compression ofk1)k1} {(compression ofk2) k2}

Hence,

)()( 21 axkaxkFx

The moment applied by the springs about the centre of mass in the positive direction is :

)()( 21 axakaxakM

x1/F1

a a

m, I

k1 k2

x

k(xa) k(xa)


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Step 3. Obtain Equations of Motion

Applying Newton's second law in translation gives :

0)()(

)()(

1221

21

kkaxkkxm

axkaxkFxmx

(6.17)

Similarly in rotation:

0)()(

)()(

21

2

12

21

kkaxkkaI

axakaxakMI

(6.18)

Step 4. Write in Matrix Form

Assume harmonic solutions of the form :

)sin( 10 txx and 20 sin t

Then differentiating twice gives xx 2 and 2

.

Substituting in equations (6.17) and (6.18) and writing in matrix form gives:

0

02

21

2

12

12

2

21

x

Ikkakka

kkamkk (6.19)

or,

0)( 2 xMK

(6.8)

Where, in this case:

21

2

12

1221

)(

)(

kkakka

kkakkK ;

I

m

0

01M ;

xx

Step 5. Check Symmetry

The stiffness matrix is symmetrical and has positive terms on the leadingdiagonal.

Note also that the leading diagonal term corresponding to the rotational

equation has dimensions (stiffness length2). It is a very common mistake toomit one of the length terms. The mass matrix is diagonal.

Step 6. Obtain Natural Frequencies

The frequency equation is

221212

12

2

21

()(

)(

Ikkakka

kkamkk

= 0

Expanding the determinant and substituting the numerical values gives:

n1 = 25.4 rad/s (4.04 Hz) and n2 = 55.7 rad/s (8.87 Hz)


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Step 7. Obtain Mode Shapes

The first row of equations (6.19) gives :

0)())( 122

21 kkaxmkk

so at the first natural frequency, n1 = 25.4 rad/s, the ratio /x is given by,

12

2

121

1kka

mkk

x

n

n

and substituting numerical values gives x = 47 rad/m. Similarly, at the

second natural frequency, n2= 55.7rad/s,/x = +2.05 rad/m. Note that had been defined in the opposite sense, the signs of these ratios would have beenreversed.


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1. How many degrees of freedom does the (free-free) system shown in figureQ1 have? What is its natural frequency?

[21

21 )(

mm

mmkn

]

2. A simplified model of an automobile suspension system is shown in Fig Q2.Determine the two natural frequencies and mode shapes for this system whenm2 = 250 kg, m2 = 20 kg, k2 = 40 kN/m and k1 = 250 kN/m.

[Mode 1:1.88 Hz,x2= 7.2x1 & Mode 2: 19.2 Hz,x1 = 90x2 ]

3. Fig Q3 shows a motor car mounted on springs. The springs are constrainedso that the centre of mass can move in the vertical direction only. The car hasa mass of 1200 kg, the spring pairs have stiffnesses: k1 = 24 kN / mk2 = 26 kN / m and the dimensions are L1 = 1.2 m, L2 =1.5 m. The radius ofgyration of the body about its centre of mass for in plane rotation is 1.1 m.

Find the natural frequencies and corresponding mode shapes, consideringmotion only in the plane of the diagram.

[Mode 1:1.0 Hz, /x = + 0.27 rad/m & Mode 2:1.3 Hz, /x = 2.9 rad/m.]

Fig Q2 Fig Q3

Fig. Q1m1 m2k

L1 L2

m2

m1

k2

k1k1 k2

JPD_2M Vibrations Notes

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