unit_01 OF MECHANISMS

8/11/2019 unit_01 OF MECHANISMS

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Unit 01

Mechanical Vibrations

Vibration.A motion which repeats itself after an interval of time is called "vibration oroscillation". It is just a simple to-and-fromotion of a body.

Basic Components of a Vibrating System.

Spring (or gravity) for storing potential energy.

Mass for storing kinetic energy.

Damper to absorb energy without releasing it i.e. energy is gradually lost.

Spring. A device, such as, beam, coiled metal, etc., which, because of its elasticity, can

store and release energy when bent or twisted.

Mass. A property of matter that (with length and time) constitutes one of the fundamental

undefined quantities upon which all physical measurements are based and which is

intuitively associated with the amount of matter a body contains. Its role in this context is tostore and release kinetic energy.

Damping. A property of a mechanical system that decreases: the amplitude of vibration, or

oscillation of a body, or a wave motion. It is a complex parameter to physically realize andcaused by the internal friction within the body and also by the surrounding in which the

motion occurs.

Types of Vibration. In mechanical vibrations, some of the major types of vibration are as

follows.

Free and forced vibration, Damped and un-damped vibration,

Linear and nonlinear vibration, and

Deterministic and random vibration.

Why Vibration Analysis ?

Answer to this question can be summed up in a few of the following sentences. As

mechanical engineers, we design machines for the transportation and manufacturingindustries. These machines are expected to perform safely and efficiently under dynamic

loadings. Civil engineers design large-scale structures, which are expected to take safely

dynamic loads caused by natural events like earthquake, wind loading, ocean waves, etc.The designers need to know how such machines (or structures) perform under extreme

dynamic conditions. The dynamic response of a structure depends upon some of its very

basic characteristics such as: mass, material, shape (or geometry, which defines thedistribution of weight and strength). Also, how these machines/structures are constrained


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has a lot to do with their dynamics. In the analysis, such constraints are known as the

boundary conditions. We shall see in this study that the geometry, materials and strength

affect the vibration characteristics, which are natural frequency, mode shape, period ofoscillations, resonant behaviour, and others. In general, if there is sufficient number of

common dynamic characteristics between the machine and the loading, the end result is not

good and machine will fail. For example, amplitude and frequency of vibration of a system(or structure) under a time dependent force, are very much dependent upon the amplitudeand frequency of the loading function. If the frequency of the loading function is the same

or close to the natural frequency of the system, the system will fail due to very large

amplitude of the response.

For the vibration analysis of a complex mechanical system, the machine is idealized in terms

of several linear springs and concentrated masses. As mentioned above, positioning (or

distribution) of the stiffness and mass greatly affect the dynamic behaviour of the system.The idea here is to take a component and determine its stiffness and mass. It should be done

for all the components in a complicated mechanical system. Finally, the entire system is

reduced to a very simple spring-mass system which is easy to analyse and very wellunderstood by mechanical engineers. Many times, in design, a simple but good analysis is

all that is needed.

Vibration Analysis Procedure:

Step 1. Creation of a Mathematical Model. Engineering systems are complex. Vibrationanalysis of such systems is not possible, if we consider every detail of a mechanical system.

Therefore, it is essential to identify various elements one-by-one and put them together to

make one whole, while keeping in mind the purpose of analysis. For a quick and good

solution, engineers reduce the complex mechanical system to a simple one involving one ofeach of spring, mass and dashpot. The equation is a second order ordinary differential

equation in time.

Step 2. Derivation of Governing Equations. After defining the problem, (i.e. mathematical

model), differential equation using the principles of dynamics is derived. Methods to do this

are:

Newton's Law of motion, d'Alembert's principle, and the conservation of energy.

Step 3. Determination of constraints . There are two types of constraints. The first is timeconstraint, which is condition at the starting time. Such a constraint is known as the "Initial

Condition". The second constraint involves the physical restrictions on the motion of the

body. This is known as the boundary condition and applied to the geometry.

Step 4. Solution of the Differential Equation.

Step 5. Interpretation of the Results. An attempt to making sense of the results, i.e. responseacceptable or not.


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Galileo Galilei (15641642): Galileo, an Italian astronomer, philosopher and professor

of mathematics at the Universities of Pisa and Padua, was the first man to point atelescope to the sky in 1609. He also swung a simple a pendulum from the leaning tower

of Pisa to study its motion. His works on the oscillations of simple pendulum and the

vibrations of strings laid the foundation upon which the theory of vibrations still stands.

Simple Pendulum.

Assume that the bob of mass mis fixed to a support by a chord of length . Using the

Newtons Second Law of motion, it can be easily shown that the differential equation forthe free oscillation of the pendulum is

0sin2

2

g

dt

d (a)

From the power series expansion of sine and cosine functions, we have

6!614

!412

!211cos and 7

!715

!513

!31sin .

By substituting the series for the sine function into equation (a), we can write,

0)( 7!715

!513

!31

2

2

g

dt

d (b)

Equations (a) and (b) are the same, but their looks are quite different. Definitely,equation (b) does not look like an ordinary second order differential equation that we are

Assume that the bob of mass m is fixed to a

support by a chord of length . Using the

Newtons Second Law of motion, it can be

easily shown that the differential equation for

the free oscillation of the pendulum is

0sin2

2

gdtd (a)

Note: The above equation comes from taking

moment about the top point.

2

2

sindt

dmmg

Simple Pendulum

cosmg

sinmg mg


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very familiar with. But, we already know that sin for small values of in whichcase higher order terms from the sine series are dropped.

02

2

g

dt

d (c)

Equation (c) defines the simple harmonic motion of a system and is valid for small

amplitude oscillation of a pendulum. It can also be written as

022

2

ndt

d, where

gn . (d)

Trifilar Suspension.

Now, we apply the Newtons Second Law of motion to get:

Jmgr

2

.

The equation of motion is then, 02

J

mgr,

Mass of the disc = m

Moment of inertia of the mass of the

disc about the zaxis =J.

The restoring force acting on the disc

is given by the horizontal components

of the tensions in the wire.

Total restoring force

mgmg sin)(331 for small

amplitude oscillation )(sin .

The arc length = r , where r=

radial distance of the point on the disc

to which the string is connected to.

Thus, the moment of the restoring

force about the center

2mgr .

The inertia torque J .

Z

mg31


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or, 022

2

ndt

d, where sec)/(

2

radJ

mgrn

. (e)

Solution of the second order ordinary differential equation (d) or (e) is obtained in the

following manner.

Assume, .tpeA (f).

Substitute equation (f) into (d or e) to get: 0)( 22 tpn eAp .

Since tpeA cannot be zero, then 022 np giving njp and .1j

The final solution is, therefore, .sincos tBtA nn

Here, the arbitrary constants A and B are unknowns and determined using the prescribedinitial conditions, i.e. how does the motion of the system begin?

Further simplification of the above solution.

t

BA

Bt

BA

ABA nn sincos

2222

22

)(cossinsincoscos 00 ttt nnn (g)

where, 220 BA and )/(tan 1 AB . 0and are known as the amplitude

of vibration and the phase angle respectively. The values of the amplitudes 0and the

phase angle depend upon the initial conditions.

If n = time taken by the oscillating mass to complete one full round (or one full

cycle), 2nn , where n is known as the period of oscillation.

Noting that nn f 2 , the natural frequency in (Hz) or (cycles per second) is

)(2//1 Hzf nnn or (cycles per second) .

Degree of Freedom. The minimum number of independent coordinates required to for

the complete determination (and/or definition) of the positions of all parts of a system at

any instant of time.

The simple pendulum has one degree of freedom, i.e. .


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Single Degree-of-Freedom System. As mentioned earlier, the most basic model of a

mechanical or structural system is composed of a spring and mass as shown below in

Figure 4.3.

k x

Spring Mass System

The spring resists the movement of the mass in the horizontal xdirection and the mass is

not allowed to move inyandzdirections, because there is no restoring mechanism that is

added to the model in the those directions. The system is in equilibrium in its naturalstate and there is no motion. If the mass is disturbed, the vibratory motion will take

place. Consider that the mass while in motion to the right is at a distance x from the

natural position. The equation of motion can be derived from the following free bodydiagram and the Newtons Second Law of motion.

We notice that acceleration )(x in the above equation is proportional to the magnitude of

the displacement (x) and simultaneously is in the opposite sense (due to the minus sign).

Hence, the mass (m) undergoes simple harmonic motion. Solution of this second orderhomogeneous equation is obtained in terms of the harmonic functions sine and cosine.

Therefore, we can begin with the solution in the following manner.

tBtAx nn sincos (b)

Here,AandBare unknown constants, mkn / , is known as the circular frequency of

the system with unit in (radian/second) and t is the time. Equation (b) can be furtherreduced to

xm

kx

Equilibrium Equation: xmkx .

Alternatively, 02 xx n , where sec)/()/( radianmkn . (a)

m


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t

BA

Bt

BA

ABAx nn sincos

2222

22

)(cossinsincoscos 00 tXttX nnn (c)

where,22

0 BAX and )/(tan 1 AB .

Note here that both (b) and (c) represent the solution of equation (a) and each has two

constants. But equation (c) is normally used because it defines two important parameters

0X and which are known as the amplitude of vibration and the phase angle

respectively. The value of the amplitudesX0and the phase angle depend upon the

initial conditions.

Details on how to obtain: tBtAx nn sincos .

Assume the following expression for solution, which is a standard way to solve second

order ordinary differential equation.

pteAx , ptpt eApxandeApx 2 (e)

By substituting equation (e) into (d), the following characteristic equation is obtained.

0][ 22 ptn eAp

Since

pt

eA cannot be zero for the solution, therefore; 0

22

np . The roots of thisquadratic equation are: njp 2,1 , where 1j . Therefore, the solution is:

tjtj nn ebeax .

In order that the response function )(tx be real, constants a and b have to be complex

conjugate. Hence, by writing )(21 jBAa and )(

21 jBAb , we have

tBtA

j

eeB

eeAx nn

tnjtnjtnjtnj

sincos

22

, where

2

costnjtnj

n

eet

andj

eet

tnjtnj

n2

sin

.


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Systems Response with Initial Conditions: It is an initial value problem with initial

conditions: 0xx and 0vx at 0t . From these conditions, we get: 0xA and

n

vB

0 . Therefore, the amplitude and phase angle are

2

202

0

n

vxX

and

0

01tanx

v

n .

If )(100 kgm and )/(40 mkNk , sec)/(20100

40000rad

m

kn and the

frequency in cycle per second )(1831.32

Hzf nn

.

Spring Arrangements.

Springs in parallel. Figure (i) shows a massMis supported on a system of nsprings having

spring values of nkkk .....,, 21 respectively. The spring value is the amount of force required

to produce one unit of the elongation.

k1 k2 kn k

(i) (ii)

The layout suggests that the elongation (deformation) shall be the same for all springs, if

a force, which is shared by all springs, is applied. Figure 4.1(ii) shows an equivalent to

the system. If k= the overall stiffness, then: F= k .

Individually, F = k1 + k2 + k3 + ...... + kn

Therefore, k = k1+ k2+ k3+ ...... + kn

Springs in Series: Shown in Figure 4.2(i) are nsprings attached in a series like a chain link.

Same force is taken by all springs. In the figure below is a single spring having the same

effect as those in series. The net elongation comes from stretching of individual springs.

M M


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Table: Beam treated as springs.

Beam Deflection and Stiffness

F

L

Cantilevered Beam

3max

3

max

3;

3 L

EI

y

Fk

EI

FLy

F

L

Simply Supported Beam

3max

3

max48;

48 LEI

yFk

EIFLy

L

Simply Supported Beam. w= uniformlydistributed load per unit length of the

beam.

3max

4

max5

384;

384

5

L

EI

y

wLk

EI

wLy

F

L

FixedFixed Beam

3max

3

max

192;

192 L

EI

y

Fk

EI

FLy

Using the deflection of the beam under a load, beam stiffness is calculated. In the

vibration models, we shall treat beams as linear springs of one degree of freedom. Thefollowing example illustrates the procedure entailing how a system with many springs

and one mass is reduced to a simple spring-mass system.

Uniformly distributed load


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Example A. Obtain the equivalent spring-mass system of the following.

Given are: the Yongs modulusE = 210 (Gpa), the moment of inertiaI= 9.78 e05(m4), spring stiffness parameter k = 800 (kN/m), mass (m) = 100 (kg).

2 m

(E,I) kb

10k k

10k

2k 2k 2k keq

Figure (a) Figure (b) Figure (c) Figure (d)

Figure (a) above shows the beam, two springs and a mass. In Figure (b), the beam has

been replaced by an equivalent linear spring. In Figure (c), two springs in series are

combined and finally the equivalent single-degree-of-freedom system is shown in Figure

(d). The various steps are presented in the following.

Stiffness of the beam: )/(77023

3 mkN

L

EIkb .

The beam and the spring with 10k stiffness are in series. Therefore, the resultingstiffness of the top two springs is

)/(392410

111mkNk

kkk b

The overall stiffness is then, )/(5524160039242 mkNkkkeq

The natural frequency sec)./(235100

035524rad

e

m

keqn

m mm


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Example C. Oscillation of an Inverted Pendulum. Consider a rigid bar of lengthL,

which has a massMattached at the tip. The other end is pinned at the bottom. The bar is

an inverted pendulum and can stay in the position as shown with the support of twosprings of equal stiffness k. We consider the small amplitude vibration of the system and

also examine its stability. Assume mass of the bar to be negligible.

Take moments (positive in ccwsense and negative in cw) of the forces shown in the free

body diagram about the pin and equate their sum to zero for equilibrium.

)sin()cos(sin2)cos( LMgakaLML

For small amplitude, assume that 1cossin;0 and . Therefore, the equationof motion is

02

2

2

ML

MgLka

Case 1. If 02

2

2

ML

MgLka, the system is stable and the natural frequency is given by

2

22

ML

MgLkan

Case 2. If 02

2

2

ML

MgLka, the system is unstable. The solution is: BtAt )( .

Case 3. If 02

2

2

ML

MgLka, the system is unstable.

Mg

L

sinak sinak

M

k k

L

a


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Example D. Figure below shows a uniform rod of mass (M) and length (L) is pinned at

O and at the top end it is attached to two springs, each with a stiffness of k. (a)

Determine an expression for the natural frequency in radian/sec. (b) What is thenumerical value of the circular frequency, if k= 2200 (N/m),M= 12 kgand L = 5 m.

k k kL31

kL31

L3

1

O

x

L32

dx

x

dxLmg )/(

Solution. Take moment about the pin O to obtain the differential equation of motion.

02

2

3

09

2

69

09

2

23

32

22

23/2

3/

23/2

)3/(

3

2

3/23/

3/23/

m

k

L

g

kLmgLmL

kLx

L

mgx

L

m

LkxdxLmgxxdx

Lm

L

L

L

L

LL

LL

The circular frequency, sec)/(2

2

3rad

m

k

L

gn .

Substitute, k= 2200 (N/m),M= 12 kgand L = 5 min the above to get:

.)sec/(225.19 radiann


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Example E. A 5 kg uniform rigid bar AC is pinned at the left end and supported by an

elastic spring (k = 1500N/m) at the right end C. The bar supports an additional mass of

15 kgat the center point. Determine (a) the frequency inHzand (b) maximum velocity atpoint C, if end C is pushed down by 25 mm and then released.

0.6 m 0.6 m

A C

B

M = 15 kg , Mass of the bar (m) = 5 kg k= 1500N/mL = Length of the rod = 1.2 m

Mass per unit length for the rigid bar =

L

mg

Solution. The free body diagrams are drawn for two conditions. The first correspond tothe static equilibrium position, while the second is for the vibrating condition. This is

done so, because the vibration occurs about the static equilibrium position. Note: If

reader should be careful in examining the forces on these diagrams. If any force is notshown, correct the free-body diagram.

Free Body Diagram for the Static Equilibrium.

Moment about A should balance. Therefore:

0)(2

: 00 LkLdxxL

mgLMgM LA Equation (A)

15 kg

x dx

A

B C

AR Mg

dxL

mg 0Lk

M


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Free Body Diagram for the Vibration Case. Assume from the static equilibrium

position in the counter-clock-wise sense. Accelerations are shown by dashed arrows.

L21 x

Mg dxL

mg )( 0kL

A

0

AR x dx

Apply Newtons Second Law of Motion: We take moments of the forces about A.

xxL

dxmLML

LkLdxxL

mgLMgM

L

LA

0

00

22

)(2

:

Equation (B)

Simplify using equation (A) to get: 0

43

12

mM

k

Part (a): Frequency inHz. Use: k= 1500N/m; M = 15 kg; m = 5 kgand L= 1.2 m.

)(65.22

.)sec/(64.16)5(4)15(3

)1500)(12(

43

12

Hzf

radmM

k

nn

n

Part (b): Maximum Velocity at C, if the displacement is 25 mm.

We can write, )cos(0 tn , or )()cos(0 mtYy n .

The velocity equation is, )()sin(0 mtYyv nn .

If mmYC 25 , the amplitude for the velocity is )/(416 smmYV CnC .

M


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Free Vibration with Viscous Damping.

Damper. A device which decreases the amplitude of vibration, or oscillation of a body,

or a wave motion.

Damping. It is a process that produces resistance to motion.

Damping is inherent to nearly all systems. When a system is in motion, it opposes themotion and as a result the amplitude decreases with time.

Very complex factor, exact behavior is difficult to model in vibration theory. Thedamping factor is obtain experimentally.

The damping force is proportional to the velocity, i.e. vcFc , cF force due to a

damper, acting in the opposite direction of the velocity, and c = damping constant.

Spring-Mass-Dashpot System. It is well known that a mechanical system has damping

characteristics, which dissipate energy. This characteristic brings the system in motion toa complete halt, if there is no external action. The figure shown below represents asystem with damping.

The equation of motion for a case like the one shown above is

0 kxxcxm (a)

Assume the following expression for solution, which is a standard way to solve second

order ordinary differential equation.

)exp(ptAx , )exp()exp( 2 ptApxandptApx

By substituting equation (e) into (d), the following characteristic equation is obtained.

x

c

k

Spring-mass-damper system

m


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0)( 2 pteAm

kp

m

cp (b)

Since )exp(ptA cannot be zero, then 02

m

kp

m

cp . Hence, the roots of this

quadratic equation are

m

kmc

m

c

m

k

m

c

m

cp

2

4

2)2(2

2

2

2 (c)

The purpose of modeling a damped single-degree-of-freedom system is to examine and

also define some of its very basic characteristics.

By examining the discriminant )4( 2 mkc , we are able to introduce some parameters

commonly used in the vibration analysis of mechanical systems. These terms are also

used in electrical systems where the behavior is dynamic. It is possible to attach adamper with damping coefficient (c = cc) such that

042 mkcc (d)

From equation (d), we find nc mmkmmkc 2/22 .

Similarly, we can write nc

c m

c

c

c

m

c

22

In the above, ccc damping ratio. If the damping ratio is unity, the condition is

called the critical damping. Equation (c) can now be written as

dnnn

cn

nn

c

jj

ccjm

c

m

m

c

c

m

c

m

k

m

c

m

cp

2

2

2

2

22

2

2

1

)/(12

)2(

)2(

2

)2(2

(e)

In the above, we introduced a term 21 nd , which is known as the circular

frequency for the damped vibration. The damped frequency df in (Hz) and the period


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d of damped vibration can be obtained in the same manner as before for the un-damped

vibration i.e. vibration without damping.

There are two roots and we denote them by1

p and2

p respectively. Finally the general

solution is

)exp()exp( 21 tpBtpAx (f)

Substituting ,, 21 dndn jpjp and writingjeXA )2/( and

jeXB )2/( , equation (f) is simplified to:

)cos( teXx d

tn (g)

Here,X= the amplitude of vibration and = phase angle. Also, the equation of motion

(a) can be reduced to the following form.

02 2 xxx nn (h)

Note: The above form (meaning look) of the equation of motion is quite convenient and

we shall use it frequently in unit 2 which deals with the forced vibration.

Calculation of the amplitude and the phase angle. Equation (g) is the solution of

equation (h) and has two unknown constants inXand . These can be obtained using the

initial conditions given by

At time 0t , the displacement and velocity are prescribed by ox and ov respectively.

The first derivative of equation (g) is

)]sin()cos([ tteXx dddn

tn

By applying the initial conditions, we have

oxX cos andod

ono

x

xvX

sin

From the above, we can find

2

0 1

od

ono

x

xvxX

and

od

ono

x

xv

tan (i)


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Example F. Spring mass and dashpot system.

Free-body diagrams of the various components.

xx ,

F1 kx

F1+F2

F2

xcFF 21 due to zero moment at the pin. Second law of motion is applied to the mass

to get the following.

xmFkx 1 or, 0)()()( txktxctxm

A k

a

c a

BVertical bar has negligible mass compared to the top mass m.

m

m


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Example G. Shown in Figure is a mass-spring-damper system.

(a) Find equivalent spring constant ek for the combination of five springs shown in

Figure 4. (b) Derive differentiation equation for the small amplitude vibration of the

system. (c) What are the values of: undamped natural frequency n in radian/second;

damped natural frequency d in radian/second; critical damping cc in (N.s/m); and the

damping ratio )( ? Given are: k= 1.2 kN/m, M = 60Kg, and the damping coefficient c

= 150N.s/m.

Solution.

2k kk

= =k k 2k 3k k 3k ke

Equivalent Spring Constant.

)/(720)1200(6.06.0;3

5

3

11

3

11mNkk

kkkkk e

e

Equation of motion :

0 xkxcxM e

0125.2

072015060

xxx

xxx

From the above:

36084.0)4641.3(2

5.2

2

5.2

sec)/(4641.3

5.22122

n

n

nn

radian

sec)/(231.31

)/.(7.4152887.0

150

2 radian

msNc

c

nd

c

ek

c

M

x

2k kk

k 2k

c

M


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Example H. An oscillating bracket of negligible mass is attached to a spring, mass and

damper as shown below. (a) Draw the free body diagrams of the bracket, damper, massand the spring. (b) Deduce the differential equation governing the motion in terms of k,

c, m, and displacementx(t). (c) Find values of the damping ratio )( , critical damping

cc in (N.s/m), un-damped )( n and damped )( d natural frequencies in radian persecond, if )/(2500),(80 mNkkgm and )/.(320 msNc . (d) What is the

percentage increase of mass (m) to increase the critical damping by 10 percent?

c

1.1 (m)

1.1 (m) x(t)

k

Figure

Free Body Diagrams.

F1F1

a

x(t)

a

F2 F2 kx kx

Equation of motion: 0)()()( txktxctxm

0)()()( txtxtx m

k

m

c 0)()(2)(

2 txtxtx nn

sec)/(59.5 radmk

n , mc

n 2 3578.0

sec)/(22.5)1( 2

radnd . Critical damping, )/.(4.894 msNcc .

To increase critical damping by 10 percent, we need to raise the mass by 21 percent. The

critical damping is given by .42

mkcc

xcFF 21

xmkxF 2

0 kxxcxm

m


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Example I. Using as the generalized coordinate, derive the differential equation of

motion in terms of m, M, a, c, and k. Determine the numerical values of the damping

ratio, critical damping coefficient and the undamped and damped natural frequencies in(radian/sec). The given data are: mass of the bar, m= 50 kg; the length unit a = 1 m;mass M = 75 kg; the stiffness of the spring k = 50,000 (N/m); and the damping

coefficients c= 5000 (N - s/m). The bar is uniform, rigid and pinned at C.

Solution. Free body diagram of the system for the static equilibrium case.

Sum of the moments of the forces about the pin at C is zero. This yields

03

42

22

udua

mgaMgkaka

a

aoo (a)

The free body diagram for the vibrating condition is shown below. Note, for this ismeasured from the static equilibrium position. In this, accelerations are shown by dotted-line vector and arc. Before writing the dynamic equilibrium equation, we obtain first the

moment of the inertia force due to the accelerating mass of the bar.

22

22

3)(

3amduu

a

muudu

a

mM

a

a

a

aPIN

(b)

Pin Mg

No resistance from damper o in static case.

oka

Spring Resistances oka2

a a a

A C B

c

k k

M


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From the sum of moments due to all forces about the pin and using equation (b), we get

22222

)()(4)(3

aMmcakakaudua

mgMga oo

a

a

Further, we use equation (a) into the above to get

05

,

05)(,

04)( 2222

mM

k

mM

cor

kcmMor

akakacamM

02 2 nn

20005075

)50000(55

405075

50002

2

mM

k

mM

c

n

n

Finally, 4472.0)72.44(2

40.)sec/(72.44 radiann

The critical damping, )/(7.111804472.05000 msNc cc

The damped frequency, .)sec/(401( 2 radiannd

Mg a

ak2

ac ka


26/32


27/32

27

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.015

-0.01

-0.005

0

0.005

0.01

0.015

Time (t)

Response-y(t)

k = 479556 (N/m), m = 100 (kg) and zeta = 0.1481

% Program to calculate the free vibration% response of a spring-mass-damper system

% subject to a given initial condition.

%

k = 479556.0; % (N/m)

m = 100.0; % (kg)

zeta = 0.1481; % damping ratio.

%

% natural frequencies.

%

wn = sqrt(k/m); % Undamped (radian/sec.)

wd = wn*sqrt(1. - zeta*zeta); %damped (radian/sec)

%

% Initial Conditions

%

y0 = 0.010; % (meters)

v0 = 0.100; % (meter/sec.)%

A = y0;

B = (v0 + zeta*wn*y0)/wd;

Y0 = sqrt(A*A + B*B);

phi = atan(B/A);

p = - zeta*wn;

% open files to write response.

dt = 0.0005;

t = 0.0;

for j=1:1000

p1 = -zeta*wn*t;

q1 = wd*t - phi;

u(j) = t;

z(j) = Y0*exp(p1);

y(j) = Y0*exp(p1)*cos(q1);

t = t+ dt;

end

%

plot(u,y,u,z,':',u,-z,':')

grid on

xlabel('Time (t)')

ylabel('Response - y(t)')

gtext('k = 479556 (N/m), m = 100 (kg) and zeta = 0.1481')


28/32

28

Logarithmic Decrement. It has been shown that the amplitude in the case of a free

damped vibration decreases exponentially with time.

The envelope curve is represented by: )exp(0 tXX n .

The oscillating part of the response is given by: )cos( tX d . The oscillating parttouches the envelope at the peak of the cosine curve and the time taken to develop a peak

after the previous one is the period of oscillation )( d .

If the amplitude at time1t is

1x and

2x being the amplitude after time dt 1 , then

)exp())(exp()exp(

)exp(21

20

10

2

1dnn

n

n tttX

tX

x

x

The logarithmic decrement can be defined as dnx

x

2

1

ln

Using dd /2 and nd 21 , we get:

22

1

1

2ln

x

x (o)

If the damping ratio is small (i.e. )1 , the logarithmic decrement is

2ln2

1 x

x (p)

% Program to calculate and plot the

% logarithmic decrement for a

% spring-mass-damper system

%

x = linspace(0,1,100);

y = 2.*pi*x;

z = sqrt(1. - x.*x);

y1= y./z;

plot(x,y,'--',x,y1)

xlabel('Damping Ratio Zeta')

ylabel('Logarithmic Decrement - Delta')


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29

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

5

10

15

20

25

30

35

40

45

Damping Ratio Zeta

LogarithmicDecrement-Delta

Figure shown above has two plots of the logarithmic decrement .

22

1

1

2ln

x

xand 2ln

2

1 x

x


30/32

30

PROBLEMS

1. Mass of the bent rod ABCD is m. Derive the equation governing the vibration of the

system. Determine also the frequency of the system in terms of a, kand m.

2. The 15 kg uniform rod AB is attached to springs at A and B. The springs act in both

tension or compression. Determine: (a) the natural frequencies (rad/sec) andf(Hz) and(b) amplitude of the angular motion of the rod, when the maximum velocity of point B is

0.6 (m/s). Use: )/(1.1 mkNk .

A B

k k

1200 (mm) 800 (mm)

D

B

a

2a

A

k

k

C

a


31/32

31

3. Determine: (a) an equivalent system comprising one spring and the same mass and

(b) the natural frequencies (rad/sec),f(Hz) and the period (sec) of the system. Use:

)/(3.1 mkNk and )(30 kgm .

4k

2k

k k

4. For the system shown below, the values of kand care given as 2400(N/m) and 300

(N-s/m) respectively. (a) Derive the differential equation of motion. (b) Find thevalues of the damping ratio, critical damping (N-s/m), and undamped and damped natural

frequencies in radian per second.

k 2k c

3k 2k

m

75 kg


32/32

A

1k

0.9 m

0.7 m

2k c

B

c

2k

k k

5. The 15 kg uniform rod AB is attached

to springs at A and B. The springs act inboth tension or compression. (a) Derive

the differential equation of the system,(b) damped and undamped the naturalfrequencies in (rad/sec) and in (Hz), (c)

value of the critical damping cc and (d)

damping ratio . Use: )/(8001 mNk ,

)/(6002 mNk and damping factor

)/(250 msNc

6. (a) Derive differential equation for thefollowing spring-mass-dashpot system. (b)

Find the values of: the damping ratio )( ,

critical damping )( cc , the undamped and

damped frequencies sec)/(radn and

sec)/(radd respectively , and damped

period (sec)d . Use: )/(5.1 mkNk ,

)/(400 msNc and )(50 kgm .

m

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