Chapter 1

Introduction

• The branch of scientific analysis which deals with

motions, time, and forces is called mechanics and is

made up of two parts; statics and dynamics. Statics

deals with the analysis of stationary systems, and

dynamics deals with systems which change with time.

• As shown in next figure, dynamics is also made up of

two major disciplines, first recognized as seperate

entities by Euler in 1775.

The Science of Mechanics

Mechanics

Dynamics

Kinematics Kinetics

Statics

• The initial problem in the design of a mechanical

system is therefore understanding its kinematics.

Kinematics is the study of motion, quite apart from

the forces which produce the motion. More

particularly, kinematics is the study of position,

displacement, rotation, speed, velocity, and

acceleration. After the kinematics stage, kinetics is

considered.

CHAPTER 2

KINEMATICS

Mobility

• One of the first concerns in either the design or the

analysis of a mechanism is the number of degrees of

freedom, also called the mobility of the device. The

mobility of a mechanism is the number of input

parameters (usually pair variables) which must be

independently controlled in order to bring the device

into a particular position.

• Kutzbach criterion for the mobility of a spatial

mechanism:

• 𝑚 = 6 𝑛 − 1 − 5𝑗1 − 4𝑗2 − 3𝑗3 − 2𝑗4 − 𝑗5

• Kutzbach criterion for the mobility of a planar

mechanism:

• 𝑚 = 3 𝑛 − 1 − 2𝑗1 − 𝑗2

• m=mobility of the mechanism

• n = number of links

• ji = number of joints having i degrees of freedom

Examples

• An earlier mobility criterion named after Grübler

applies to mechanisms with only single degree of

freedom joints where the overall mobility of the

mechanism is unity. Putting 𝑗2 =0 and m=1 into

Kutzbach criterion, we find Grübler’s criterion for

planar mechanisms with constrained motion:

• 3𝑛 − 2𝑗1 − 4 = 0

• From this we can see, for example, that a planar

mechanism with a mobility of 1 and only single

degree of freedom joints can not have an odd number

of link. Also, we can find the simplest possible

mechanism of this type; by assuming all binary links

we find n = 𝑗1= 4. This shows why the the four-bar

and the slider crank linkage are so common in

application.

Grashof’s Law

• A very important consideration when designing a

mechanism to be driven by a motor, obviously, is to

ensure that the input crank can make a complete

revolution. For the four-bar linkage, there is a very

simple test of whether this is the case.

• Grashof’s law states that for a planar four-bar linkage, the sum of the shortest an the longest link lengths cannot be greater than the sum of the remaining two link lengths if there is to be continious relative rotation between two members. This is illustrated in next, where the longest link has length l, the shortest link has length s, and the other two links have lengths p and q. In this notation, Grashof’s law states that one of the links, in particular the shortest link, will rotate continiously relative to the other three links if and only if:

• 𝑠 + 𝑙 ≤ 𝑝 + 𝑞

• If the shortest link s is adjacent to the fixed link we

obtain a crank-rocker linkage as below, where s is the

crank and p is the rocker.

• The double-crank mechanism also called drag-link is

obtained by fixing the shortest link s as the frame.

• By fixing the link opposite to s we obtain the fourth

inversion, the double-rocker mechanism. Note that

although link s is able to make a complete revolution,

neither link adjacent to frame can do so; both must

oscillate between limits and are therefore rockers.

Kinematic Inversion

• Until a frame link has been chosen, a connected set of links is

called a kinematic chain. When different links are chosen as

the frame for a given kinematic chain, the relative motions

between various links are not altered, but their absolute

motions (those measured wrt the frame link) may be changed

drastically. The process of choosing different links of a chain

for the frame is known as kinematic inversion.

• In an n-link kinematic chain, choosing each link in turn as the

frame yields n distinct inversions; n different mechanisms. As

an example four-link slider-crank chain has four different

inversions.

• Figure (a) shows the basic slider-crank, as found in most

internal combustion engines. Figure (b) shows the basis of the

rotary engine found in early aircraft. The mechanism in figure

(c) used to drive the wheels of early steam locomotives, link 2

being a wheel. The mechanism in figure (d) is not found in

engines, but this mechanism can be recognized as part of a

garden water pump (rotate 90 cw).

Instantaneous Center of Velocity

• One of the more interesting concepts in kinematics is that of an instantaneous velocity axis for rigid bodies which move relative to one another. In particular, we shall find that an axis exists which is common to both bodies and about which either body can be considered as rotating with respect to the other.

• Since our study of these axes will be restricted to planar motions, each axis is perpendicular to the plane of the motion. We shall refer to them as instant centers or poles. These instant centers are regarded as a pair of coincident points, one attached to each body, about which one body has an apparent rotation relative to each other.

• This property is true only instantaneously, and a new

pair of coincident points will become the instant

center at the next instant.

• The instantaneous center of velocity is defined as the

location of a pair of coincident points of two different

rigid bodies for which the apparent velocity of one of

the points is zero as seen by the observer on the other

body. It may also be defined as instantaneous location

of a pair of coincident points of two different rigid

bodies for which the absolute velocities of the two

points are equal.

• Since we have adopted the convention of numbering the links of a mechanism, it is convenient to designate an instant center by using the numbers of the two links associated with it. Thus P32 identifies the instant center between links 3 and 2. A mechanism has as many instant centers as there are ways of pairing the link numbers. Thus the number of instant centers in an n-link mechanism is:

• N= 𝑛(𝑛−1)

2

The Aronhold-Kennedy Theorem of Three

Centers

• According to the equation (N= 𝑛(𝑛−1)

2) the number of

instant centers in a four-bar linkage is six. As shown

in below figue we can identify four of them by

inspection; we see that the four pins can be identified

as instant centers P12, P23, P34, and P14 since each

satisfies the definition.

• A good method of keeping track of which instant

centers have been found is to space the link numbers

around the perimeter of a circle. Then, as each pole is

identified, a line is drawn connecting the

corresponding pair of link numbers. Previous figure

shows that P12, P23, P34, and P14 have been found, it

also shows that missing lines where P13 and P24, have

not been located.

• After finding as many of the instant centers as

possible by inspection, others are located by applying

the Aronhold-Kennedy theorem of three centers.

• Aronhold-Kennedy theorem states that the three

instant centers shared by three rigid bodies in relative

motion to one another (whether or not they are

connected) all lie on the same straight line.

• Examples

The Angular Velocity Ratio Theorem

• In the below figure, P24 is the instant center common

to links 2 and 4. Its absolute velocity VP24 is the same

whether P24 is considered as a point of link 2 or 4.

• We can write:

• 𝑉𝑃24 = 𝑉𝑃12 + 𝜔2/1 × 𝑅𝑃24𝑃12 = 𝑉𝑃14 + 𝜔4/1 × 𝑅𝑃24𝑃14

• 𝑉𝑃12=𝑉𝑃14 = 0

• 𝜔2/1=𝜔2

• 𝜔4/1=𝜔4

• Thus: 𝜔4/1

𝜔2/1=

𝑅𝑃24𝑃12𝑅𝑃24𝑃14

• General notation : 𝜔𝑘/𝑖

𝜔𝑗/𝑖=

𝑅𝑃𝑗𝑘𝑃𝑖𝑗

𝑅𝑃𝑗𝑘𝑃𝑖𝑘

• The theorem states that the angular velocity ratio of any two

bodies in planar motion relative to a third body is inversly

proportional to the segments into which the common instant

center cuts the line of centers.

Freudenstein’s Theorem

• Freudenstein’s theorem makes use of the line connecting instant centers P13 and P24 called the collineation axis. The theorem states that at an extreme of the output to input angular velocity ratio of a four-bar linkage, the collineation axis is perpendicular to the coupler link.

•𝜔4

𝜔2=

𝑅𝑃24𝑃12𝑅𝑃24𝑃12+𝑅𝑃12𝑃14

• Since 𝑅𝑃12𝑃14 is the fixed link, the extremes of the velocity ratio occur when 𝑅𝑃24𝑃12 is either a maximum or a minimum. Such positions may occur on either or both sides of P12. Thus, the problem reduces to finding the geometry of the linkage for which 𝑅𝑃24𝑃12 is an extremum.

• During motion of the linkage, P24 travels along the line

P12P14 , but at an extreme value of the velocity ratio P24 must be instantaneously be at rest(its direction must be reversing). This occurs when the velocity P24 , considered as a point of link 3, is directed along the coupler link. This will be true only when the coupler link is perpendicular to the collineation axis, since P13 is the instant center of link 3.

• An inversion of the theorem (treating link 2 as fixed)

states that an extreme value of the velocity ratio 𝜔3

𝜔2 of a four bar linkage occurs when the

collineation axis is perpendicular to the follower (link

4).

Indexes of Merit Mechanical Advantage

• In this section we will study some of the various

ratios, angles and other parameters of mechanisms

which tell us whether a mechanism is a good one or

poor one.

• The mechanical advantage of a linkage is the ratio of

the output torque exerted by the driven link to the

necessary input torque requried at the driver: 𝑇𝑜𝑢𝑡

𝑇𝑖𝑛

• When the sine of the angle becomes zero, the

mechanical advantage becomes infinite, thus at such a

position only a small input torque is necessary to

overcome a large output torque load. This is the case

when the driver AB is directly in line with the coupler

BC.When the four-bar linkage is in either of these

positions, the mechanical advantage is infinite and the

linkage is said to be in a toggle position (dead-

centers).

The angle between coupler and the follower is called

the transmission angle. As this angle becomes small, the

mechanical advantage decreases and even a small

amount of friction will cause the mechanism to lock or

jam. The extreme values of the transmission angle occur

when the crank AB lies along the line of the frame AD.

•

Note that, above is the equation for the output to input

velocity ratio. Also the extremes of this ratio occur

when the collineation axis is perpendicular to the

coupler.

𝜔4

𝜔2=𝑅𝑃𝐴𝑅𝑃𝐷

If we assume that the linkage in the previous figure has

no friction or inertia forces during its operation

(negligible) the input power applied to link 2 is the

negative of the power applied to link 4 by the load:

𝑇2𝜔2 = −𝑇4𝜔4

Or

𝑇4

𝑇2= −

𝜔2

𝜔4 =

𝑅𝑃𝐷

𝑅𝑃𝐴

The mechanical advantage of a mechanism is the instantaneous ratio of the output force (torque) to the input force (torque). The mechanism is redrawn at the position where links 2 and 3 are on the same straight line. At this position, 𝑅𝑃𝐴 and 𝜔4 are passing through zero; hence an extreme value of the mechanical advantage (infinity) is obtained.

• Such toggle positions are often used to produce a

high mechanical advantage. An example is the

clamping mechanism.

• Proceeding further, construct 𝐵′𝐴 and 𝐶′𝐷

perpendicular to the line PBC.

• Also let and be acute angles made by the coupler.

• By similar triangles:

•𝑅𝑃𝐷

𝑅𝑃𝐴= 𝑅𝐶′𝐷

𝑅𝐵′𝐴=𝑅𝐶𝐷 sin 𝛾

𝑅𝐵𝐴 sin 𝛽

• Then:

•𝑇4

𝑇2= −

𝜔2

𝜔4 = −

𝑅𝑃𝐷

𝑅𝑃𝐴 = −

𝑅𝐶𝐷 sin 𝛾

𝑅𝐵𝐴 sin 𝛽

• This equation shows that the mechanical advantage is

infinite whenever the angle is 0 or .

• The angle between coupler and the follower is

called the transmission angle. This angle is also often

used as an index of merit. This equation also show

that the mechanical advantage diminishes when is

much less than a right angle. A common rule of

thumb is that :45° ≤ 𝛾 ≤ 135°

Centrodes

• We noted that the location of instant center of

velocity was defined instantaneously and would

change as the mechanism moves. If the locations of

the instant centers are found for all possible phases of

the mechanism, they describe curves or loci, called

centrodes.

• As the linkage is moved through all possible

positions, P13 traces out the curve called the fixed

centrode on link 1.

• This figure shows the inversion of the same linkage

in which link 3 is fixed and link 1 is movable. As the

linkage is moved through all possible positions, P13

traces a different curve on link 3. For the original

linkage (link 1 fixed), this is the curve traced by P13

on the coordinate system of the moving link 3; it is

called the moving centrode.

• This figure shows the the moving centrode, attached

to link 3, and the fixed centrode attached to link 1. It

is imagined here that links 1 and 3 have been

machined to the actual shapes of the respective

centrodes and links 2 and 4 have been removed. If the

moving centrode is now permitted to roll on the fixed

centrode without slip, link 3 will have exactly the

same motion as it had in the original linkage.

• This remarkable property, which stems from the fact

that a point of rolling contact is an instant center,

turns out to be quite useful in the synthesis of

linkages.

• We can restate the this property as follows: The plane

motion of one rigid body relative to an other is

completely equivalent to the rolling motion of one

centrode on the other.

• An other set of centrodes, both moving, is generated on links 2

and 4 when instant center P24 is considered. Below figure

shows these as two ellipses for the case of a crossed double-

crank linkage with equal cranks. These two centrodes roll

upon each other and describe the identical motion between

links 2 and 4 which would result from the operation of the

original four-bar linkage. This construction can be used as the

basis for the development of a pair of elliptical gears.

Chapter 3

KINETICS

Equation Of Motion

• Relationship between the generalized external forces and

accelerations. • Equation of motion can be determined by four different ways.

• 1. Direct application of Newton’s Second Law.

• 2. Dynamic equilibrium method; D’Alembert Principle.

• 3. Method of Virtual Work.

• 4. Energy Methods.

Reaction Forces

Newton’s Second Law

• The general equations of motion for a rigid body in plane

motion are:

• 𝐅 = 𝑚𝐚

• 𝑀𝐺 = 𝐼 𝛼

• Alternatively

• 𝑀𝑃 = 𝐼 𝛼 + 𝑚𝑎 𝑑

• Note that, when =0, point P becomes the mass center G and

above equation reduces to scalar form 𝑀𝐺 = 𝐼 𝛼.

• When point P becomes the fixed point O, then 𝐚𝑃=0 and above

equation reduces to scalar form 𝑀𝑂 = 𝐼𝑂 𝛼.

Translation

• Every line in a translating body remains parallel to its original

position at all times. In rectlinear translation all points move in

straight lines, whereas in curvelinear translation all points

move on congruent curved paths. In either case, there is no

angular motion of the translating body, so that both and

are zero.

• 𝐅 = 𝑚𝐚

• 𝑀𝐺 = 𝐼 𝛼 = 0

Rectilinear Translation

𝐹𝑥 = 𝑚𝑎𝑥

𝐹𝑦 = 𝟎

𝑀𝑃 = 𝑚𝑎𝑑

𝑀𝐴 = 0

or

Curvilinear Translation

𝐹𝑛 = 𝑚𝑎𝑛

𝐹𝑡 = 𝑚𝑎𝑡

𝑀𝐵 = 𝑚𝑎𝑡 𝑑𝐵

𝑀𝐴 = 𝑚𝑎𝑛 𝑑𝐴

or

Fixed Axis Rotation

• All points in the body describe circles about the rotation axis, and all lines of the body in the plane of motion have the same angular velocity and angular acceleration .

• 𝐅 = 𝑚𝐚

• 𝑀𝐺 = 𝐼 𝛼

• 𝐹𝑛 = 𝑚𝑟 𝜔2

• 𝐹𝑡 = 𝑚𝑟 𝛼

• For fixed axis rotation, it is generally useful to apply a moment

equation directly about the rotation axis O. • 𝑀𝑂 = 𝐼 𝛼 + 𝑚𝑎𝑡 𝑟

• 𝐼𝑂 = 𝐼 + 𝑚𝑟2

• 𝑀𝑂 = (𝐼𝑂−𝑚𝑟2)𝛼 + 𝑚𝑟2𝛼

• 𝑀𝑂 = 𝐼𝑂 𝛼

General Plane Motion

• Translation and rotation.

• 𝐅 = 𝑚𝐚

• 𝑀𝐺 = 𝐼 𝛼

• Alternatively

• 𝑀𝑃 = 𝐼 𝛼 + 𝑚𝑎 𝑑

Mass Moment of Inertia • Mass moments of inertia have units of mass times distance

square, it seems natural to define a radius value for the body as

follows:

• 𝐼𝐺 = 𝑘2𝑚

• 𝑘 =𝐼𝐺

𝑚

• This distance k is called the radius of gyration of the part, and

it is always calculated or measured from the center of mass of

the part about one of the principal axis.

Parallel Axis Theorem

• The form of the transfer, or parallel-axis formula for mass

moment of inertia is written:

• 𝐼 = 𝐼𝐺 +𝑚𝑑2

Inertia Forces and D’alembert’s Principle

• Consider a moving rigid body of mass m acted upon by forces

𝐅1, 𝐅2, 𝐅3. Resultant force: 𝐅 = 𝐅1+𝐅2+𝐅3.

• In general case the line of action of the resultant will not be

through the mass center G, but will be displaced by some

distance h.

• The effect of this unbalanced force system is to produce an

acceleration of center of mass of the body.

• 𝐅𝒊𝒋 = 𝑚𝑗𝐀𝐺

• Taking moments about the center of mass of the body, the

unbalanced moment effect of this resultant force about the

mass center causes angular acceleration of the body.

• 𝐌𝐺𝑖𝑗 = 𝐼𝐺𝜶𝑗

• Since, in the dynamic analysis of machines, the acceleration

vectors are usually known, an alternative form of above

equations is often convenient in determining the forces

required to produce these known accelerations.

• Thus, we can write

• 𝐅𝒊𝒋 + −𝑚𝑗𝐀𝐺 = 𝟎

• 𝐌𝐺𝑖𝑗 + −𝐼𝐺𝜶𝑗 = 𝟎

• The first equation states that the vector sum of all external forces acting upon the body plus the fictitious force −𝑚𝑗𝐀𝐺 is

equal to zero. This new fictitious force −𝑚𝑗𝐀𝐺 is called an

inertia force. It has the same line of action as the absolute acceleration 𝐀𝐺, but is opposite in sense.

• The second equation states that the sum of all external moments and moments of all external forces acting upon the body about G plus the fictitious torque −𝐼𝐺𝜶𝑗 is equal to zero.

This new fictitious torque−𝐼𝐺𝜶𝑗 is called an inertia torque.

The inertia torque is opposite in sense to the angular acceleration vector 𝜶𝑗 .

• In a sense we can picture the fictitious inertia force and inertia

torque vectors as resistance of the body to the change of

motion required by the net unbalanced forces and torques.

• The equations above are known as D’Alembert’s principle.

• These equations can also be written as:

• 𝐅 = 𝟎 and 𝐌 = 𝟎

• ℎ =𝐼𝐺𝛼3

𝑚3𝐴𝐺

D’alembert’s Principle Fixed Axis Rotation

𝐅−𝑚𝐀𝐺 = 𝟎

𝐌𝑂 − 𝐼𝑂𝛂 = 𝟎

• When a rigid body has a motion of translation only, the

resultant inertia force and the resultant external force share the

same line of action which passes through the mass center of

the body. When a rigid body has rotation and angular

acceleration, the resultant inertia force and the resultant

external force have the same line of action, but this line does

not pass through the mass center but is off set from it.

• Taking moments about O from each figure:

• −𝑚𝑟𝐺𝛼 𝑙 = −𝐼𝐺𝛼 + (−𝑚𝑟𝐺𝛼)𝑟𝐺

• 𝑙 =𝐼𝐺

𝑚𝑟𝐺+ 𝑟𝐺

• 𝑙 =𝑘2

𝑟𝐺+ 𝑟𝐺

• Or

• 𝑙 =𝑘𝑂

2

𝑟𝐺

• The point P is called the center of percussion.As shown, the

resultant inertia forces passes through P, and consequently the

inertia force has zero moment about the center of percussion.

If an external force is applied at P, perpendicular to OG, an

angular acceleration will result, but the bearing reaction at O

will be zero except for the component caused by the

centrifugal inertia force −𝑚𝑟𝐺𝜔2.

• If the axis of rotation is coincident with the center of mass,

𝑟𝐺=0 and 𝑙 = ∞. Under these conditions there is no resultant

inertia force but a resultant inertia couple −𝐼𝐺𝛼.

Method of Virtual Work

ir

External forces are driving forces, gravitational forces, inertial forces and spring forces

iQ Generalized forces (including moments) associated with generalized coordinates. We

can write and in terms of generalized coordinates ( ). j iq

m

j

j

e

j

n

i

i

e

i TrFU

11

0

0...2211 FF qQqQqQU

0

1

F

i

ii qQU

,0...0,0

0

21

F

i

QQQ

iallfor

Q

Example

1m

2m

T

2r

1r

B

A 1k

2k

1m

2m

T

2r

1r

θ

gm2

2sF

1sF

2sF

gm1

By

Ay

'Ay

y

Find the spring deflections at the static equilibrium. (assuming

springs are in tension)

2'

2

222

11

111

2

1

)(

2

lAA

s

lB

s

A

fyy

kF

fy

kF

yq

q

F

012'

221 BsBAsAsA yFygmTyFyFygmU

222

1'

11'

ryCry

ryCry

BB

AA

0)()( 21221221 rFgrmTrFyFgm ssAs

okQk

gmkgmFs 11

2

122212 0,

Virtual work done by linear Hydraulic/Pneumatic

Actuator

sPU p

Example

23s

12 12T

B

A C

15

54s

sF

F

1r

The mechanical system works in a horizontal plane. The spring is unstreched when

The horizontal force at point B , F is known. Find the torque , hydraulic actuator force P

as function of piston variables and which is required to hold the system

in static equilibrium.

bs 54

152312 ,, s 54s

DOF=2

2312 , sq

12T

0)]cos()(cos[)]sin()(sin[

))sin())(cos(()sin(cos

sincoscos

)sin()cos(

'

cossincossinsinsin:Im

sincossincos

coscos:Re

23121554121212152354122312

12121523231215541212232312231212

1212232312122323

1212152323121554

1515541554121223122315541223

15155415541212231223

15541122354123

54231212

12

1512

ssbkFPssbkFsT

sssbkssFsPTU

ssxsxesr

sss

rulescramerfrom

ssssss

ssss

srsesres

sFxFsPTU

BBi

B

ii

sB

01 Q 02 Q

Assume the spring in compression

𝐹𝑠 = 𝑘(𝑏 − 𝑠54)

Dynamic Force Analysis via Virtual Work

By treating inertia forces and torques as external forces, one can perform dynamic

force analysis via virtual work method (D’alembert’s Principle).

0

1111

k

k

in

kGk

k

in

kj

b

j

e

ji

m

i

e

i TrFTrFU

Floating links: If acceleration is not known assume it in positive coordinate directions.

y

x ya

xaG

yma

xma

GI

nma

tma

t, nr ,

ta

na

GI

G

Gr

UnmaContribution of to is zero

is perpendicular to n Gr

Links rotating about a fixed pivot:

Tangent to path

Example

14s

12

12T13

14F

G

2a

14s

12

12T

C 13

14F

G

2a

144sm

gm3

Gyam3

Gxam3

13GI

02 m

14FGiven: ,

12T

033313131414141441212 GGGyGGxG ygmyamxamIsFssmTU

Motion velocity acceleration analysis:

okTQQU

okas G

121121

1413

00

,,

okfyxs GG )(,,, 121314

3a

3243121212 ,,,,,,, aamm

Find:

Example

x

m F(t) F(t)

Equation of Motion By Virtual Work Relationship between the external forces and generalized accelerations:

0iQ

kxxmtF

motionofequation

xxmkxtF

xxmxkxxtFU

)(

0))((

0)(

xk

inF

kx

Brief of Kinetic Energy

Translation

• The translating rigid body has a mass m and all of its particles

have a common velocity v.

• 𝑇 =1

2𝑚𝑣2

Fixed Axis Rotation

• The rigid body rotates with an angular velocity about O.

• 𝑇 =1

2𝐼𝑂𝜔

2

General Plane Motion

• The rigid body executes plane motion where, at the instant

considered, the velocity of its mass center G is 𝑣 and its

angular velocity is .

• 𝑇 =1

2𝑚𝑣 2 +

1

2𝐼 𝜔2

• Or

• 𝑇 =1

2𝐼𝐶𝜔

2

• Where C is the instantaneous center

of zero velocity.

Energy Methods

• In this method conservation of energy principle is applied.

•𝑑

𝑑𝑡𝐸𝑇 = 𝑃𝑛𝑒𝑡

• 𝐸𝑇 = 𝑇 + 𝑉

• 𝑇= Total kinetic energy

• 𝑉= Total potantial energy

• 𝑃𝑛𝑒𝑡 = 𝑃𝑖 − 𝑃𝑜 − 𝑃𝑑

• 𝑃𝑖 = Total power input by the external forces

• 𝑃𝑜 = Total power output

• 𝑃𝑑 = Total power output by heat dissipation due to damping

• There are many problems where the effect of damping is small

and may be neglected. • If there is no external force and the effect of damping is

neglected, then mechanical energy is conserved:

•𝑑

𝑑𝑡𝐸𝑇 = 0

• Or

•𝑑

𝑑𝑡 𝑇 + 𝑉 = 0

• If there is no external force and the effect of damping is

neglected, conservation of energy may also be used to

determine the period or frequency of vibration without having

to derive and solve the equation of motion. For a system which

oscillates with simple harmonic motion about the equilibrium

position, from which the displacement x is measured we may

write:

• 𝑇𝑚𝑎𝑥 = 𝑉𝑚𝑎𝑥

• From this relationship 𝜔𝑛 can be directly determined as

displayed in the following illustration.

• For the mechanical system given in the figure:

• 𝑇 =1

2𝑚𝑥 2

• 𝑉 =1

2𝑘𝑥2

• Note that 𝑘𝛿𝑠𝑡 = 𝑚𝑔. (Thus, a general rule: if the mass causes

deflection of spring, no need to use gravitational potantial

energy, because it cancels out when x is measured from the

equilibrium position).

• 𝑉1 = 𝑉2

•1

2𝑘𝛿𝑠𝑡

2 = −𝑚𝑔𝑥 +1

2𝑘 𝛿𝑠𝑡 + 𝑥 2

• 𝑘𝛿𝑠𝑡 = 𝑚𝑔

• Thus

• 𝑉 =1

2𝑘𝑥2

• 𝑇𝑚𝑎𝑥 =1

2𝑚(𝑥 𝑚𝑎𝑥)

2

• 𝑉𝑚𝑎𝑥 =1

2𝑘(𝑥𝑚𝑎𝑥)

2

• For the harmonic oscillator given in the figure the

displacement may be written as:

• 𝑥 = 𝑥𝑚𝑎𝑥 sin(𝜔𝑛𝑡 + 𝜓)

• So that max velocity is:

• 𝑥 𝑚𝑎𝑥= 𝜔𝑛𝑥𝑚𝑎𝑥

• Thus,

•1

2𝑚(𝜔𝑛𝑥𝑚𝑎𝑥)

2=1

2𝑘(𝑥𝑚𝑎𝑥)

2

• 𝜔𝑛 =𝑘

𝑚

• Note that eqn. of motion of the system is :𝑚𝑥 + 𝑘𝑥 = 0

Vibration of Rigid Bodies

• As an illustrative example consider the rotational vibration of

the uniform slender bar in Figure (a).

• Figure (b) depicts the FBD associated with the horizontal

position of static equilibrium. Equating to zero moment about

O yields:

• −𝑃2𝑙

3+𝑚𝑔

𝑙

6= 0

• 𝑃 =𝑚𝑔

4

• Where P is the magnitude of static spring force.

• Figure (c) depicts the FBD associated with an arbitrary

positive angular displacement . Using 𝑀𝑂 = 𝐼𝑂𝜃 we can

write:

• 𝑚𝑔𝑙

6cos 𝜃) − (

𝑐𝑙

3𝜃 cos 𝜃

𝑙

3cos 𝜃 −

𝑃 + 𝑘2𝑙

3sin 𝜃)

2𝑙

3cos 𝜃 + (𝐹0cos𝜔𝑡

𝑙

3cos 𝜃 =

1

9𝑚𝑙2𝜃

• Where 𝐼𝑂 =𝑚𝑙2

12+𝑚(

𝒍

𝟔)𝟐=

1

9𝑚𝑙2

• For small angular deflections; sin 𝜃 ≅ 𝜃 𝑎𝑛𝑑 cos 𝜃 ≅ 1 .

• With 𝑃 =𝑚𝑔

4, the equation of motion, upon simplification

becomes:

• 𝜃 +𝑐

𝑚𝜃 + 4

𝑘

𝑚𝜃 =

(𝐹0𝑙

3) cos 𝜔𝑡

𝑚𝑙2/9

• Note that the two equal and opposite moments associated with

static equilibrium forces canceled on the left side of the

equation of motion. Hence, it is not necessary to include the

static-equilibrium forces and moments in the analysis. (No

need to use mg and thus P).

• Also note that 𝑀0 =𝐹0𝑙

3

• At this point we observe that the above equation is identical in

form with the linear case:

• 𝑥 + 2𝜁𝜔𝑛𝑥 + 𝜔𝑛2𝑥 =

𝐹0 cos 𝜔𝑡

𝑚

• So we may write:

• 𝜃 + 2𝜁𝜔𝑛𝜃 + 𝜔𝑛2𝜃 =

𝑀0 cos 𝜔𝑡

𝐼𝑂

• Thus we may use all relations developed for linear quantities

by replacing with their rotational counterparts.

• The following table shows the results of this procedure as

applied to the rotating bar of this example.

• In the above table, the variable 𝑘𝜃 represents the equivalent

torsional spring constant of the system, and is determined by

writing the restoring moment of the spring. For a small angle

this moment of about O is:

• 𝑀𝑘 = −[𝑘(2𝑙/3) sin 𝜃][(2𝑙

3cos 𝜃)] ≅ −(

4

9𝑘𝑙2)𝜃

• Thus 𝑘𝜃 =4

9𝑘𝑙2.

• Note that 𝑀𝑂/𝑘𝜃 is the static angular deflection which would

be produced by a constant external moment 𝑀𝑂.

Example

• Derive the equation of motion and determine the natural

frequency. 𝑟 = 0.9m and 𝑘𝑂 = 0.95m. Friction is negligible.

• 𝑀𝑂 = 𝐼𝑂𝜃 −𝑚𝑔𝑟 sin 𝜃 = 𝑚𝑘𝑂2𝜃

• or 𝜃 +𝑔𝑟

𝑘𝑂2 sin 𝜃 = 0

• when 𝜃 is small sin 𝜃 = 𝜃

• 𝜃 +𝑔𝑟

𝑘𝑂2 𝜃 = 0

• 𝜔𝑛 =𝑔𝑟

𝑘𝑂2

Example

• The uniform bar of mass m and length l is pivoted at its center.

The spring of constant k at the left end is attached to a

stationary surface, but the right end spring also of constant k,

is attached to a support which undergoes a harmonic motion

given by 𝑦𝐵 = 𝑏 sin𝜔𝑡 . Determine the natural frequency.

• Draw FBD

• 𝑀𝑂 = 𝐼𝑂𝜃

• − 𝑘𝑙

2sin 𝜃

𝑙

2cos 𝜃 − 𝑘

𝑙

2sin 𝜃 − 𝑦𝐵

𝑙

2cos 𝜃 =

1

12𝑚𝑙2𝜃

• Assuming small deflections:

• 𝜃 +6𝑘

𝑚𝜃 =

6𝑘𝑏

𝑚𝑙sin𝜔𝑡

• 𝜔𝑛 =6𝑘

𝑚

Example • Derive the eqn. of motion for the homogeneous circular

cylinder which rolls without slipping. Find the natural

frequency, damping ratio, damped natural frequency, and

period.

• The angle is taken positive cw to be kinematically consistent

with x. The direction of the friction force is to the right for x0

(maybe assumed in either direction).

• Fbd

• 𝐹𝑥 = 𝑚𝑥

• −𝑐𝑥 − 𝑘𝑥 + 𝐹 = 𝑚𝑥

• 𝑀𝐺 = 𝐼 𝜃

• −𝐹𝑟 =1

2𝑚𝑟2𝜃

• The condition of rolling with no slip is 𝑥 = 𝑟𝜃 . • Substituting this condition in to the moment eqn. gives:

𝐹 = −1

2𝑚𝑥

• Substituting this in to the force eqn. yields:

• −𝑐𝑥 − 𝑘𝑥 −1

2𝑚𝑥 = 𝑚𝑥

• 𝑥 +2𝑐

3𝑚𝑥 +

2𝑘

3𝑚𝑥 = 0

• 𝜔𝑛 =2𝑘

3𝑚

• 2𝜔𝑛 =2𝑐

3𝑚

• =𝑐

3𝑚𝜔𝑛

• 𝜔𝑑 = 𝜔𝑛 1 − 2

• 𝑇𝑑 =2𝜋

𝜔𝑑

Example

• Derive the eqn. of motion of the system.

• Fbd of the system (D’alembert).

𝑘𝑡

𝐼

𝑇(𝑡)

𝜃

𝑇(𝑡) 𝑘𝑡𝜃 𝐼𝜃

• 𝑇 𝑡 − 𝑘𝑡𝜃 − 𝐼𝜃 = 0

• Or 𝑇 𝑡 = 𝑘𝑡𝜃 + 𝐼𝜃

• By energy method:

• Total energy of the system 𝐸𝑇 =1

2𝑘𝑡𝜃

2 +1

2𝐼𝜃 2

• 𝑃𝑛𝑒𝑡 = 𝑇 𝑡 𝜃

•𝑑

𝑑𝑡𝐸𝑇 = 𝑃𝑛𝑒𝑡

• 𝑘𝑡𝜃𝜃 + 𝐼𝜃 𝜃 = 𝑇 𝑡 𝜃

• 𝑘𝑡𝜃 + 𝐼𝜃 = 𝑇 𝑡

Example

• Determine the equation of motion. At free length 𝑥 = 0 and

𝜃 = 0, y is the equilibrium position.

m

oI

kx

y

Or g

𝑓(𝑡)

• FBD’s are drawn according to the equilibrium position (use y,

no need for mg)

• From these equations:

• 𝑚+𝐼𝑂

𝑟2𝑦 + 𝑘𝑦 = 𝑓 𝑡

𝐼𝑂𝜃

𝑅𝑥

𝑅𝑦

𝐹𝑡 𝑘𝑦

𝐹𝑡

𝑚𝑦

𝑓(𝑡) 𝐹𝑡𝑟 − 𝑘𝑦𝑟 − 𝐼𝑂𝜃 = 0

𝑓 𝑡 − 𝐹𝑡 −𝑚𝑦 = 0

• By using energy method:

• Total energy of the system 𝐸𝑇 =1

2𝑚𝑦 2 +

1

2𝐼𝑂(𝑦 /𝑟)

2+1

2𝑘𝑦2

• 𝑃𝑛𝑒𝑡 = 𝑓 𝑡 𝑦

•𝑑

𝑑𝑡𝐸𝑇 = 𝑃𝑛𝑒𝑡

• 𝑚+𝐼𝑂

𝑟2𝑦 + 𝑘𝑦 = 𝑓 𝑡

Example

3

2

2

3

A

C

B

g

Two link robot manipulator is driven by two motors. Motor A is

mounted on the ground and motor shaft is connected to link 2 at

A. Motor B is mounted to link 2 and motor shaft is connected

to link 3 at B. Derive the equations of motion.

3

2

2

3

A

C

B

g

BT

AT

BT

y

x

cc ym cc xm

gmc 3

cIDOF=2

32 ,q

033322 ccccccBBA IygmyymxxmTTTU

cc Im ,

3332223322

3332223322

32

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sinsincoscos

32

cc

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3322

32

c

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x

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gm

mITTTU

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CBBA

0

0

0

0)]coscossincossincoscoscos

cossinsincossinsinsin([

)coscossincoscossincoscos

cossinsinsinsincossin(

0)coscos(

)sincossincos()sinsin(

)cossincossin()()(

2

1

3221

3332

3332

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32

2233222332

2333

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223

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22222

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22222332

• The small mass m is mounted on the light rod pivoted at O and

suuported at end A by vertical spring of stiffness k. End A is

displaced a small distance below the horizontal equilibirium

position and released. Determine the eqn. of motion for small

oscillations of the rod.

Example

• 𝑉 =1

2𝑘 𝑦 + 𝛿𝑠𝑡

2 −1

2𝑘𝛿𝑠𝑡

2 −𝑚𝑔(𝑏

𝑙𝑦)

• Substiting 𝑚𝑔𝑏 = 𝑘𝛿𝑠𝑡𝑙 into above equation ( 𝑀𝑜 = 0 static)

• 𝑉 =1

2𝑘𝑦2

• 𝑇 =1

2𝑚(

𝑏

𝑙𝑦 )2

•𝑑

𝑑𝑡𝑇 + 𝑉 = 0

• 𝑦 +𝑙2

𝑏2𝑘

𝑚𝑦 = 0

𝑇𝑚𝑎𝑥=𝑉𝑚𝑎𝑥 1

2𝑚(

𝑏

𝑙𝑦 𝑚𝑎𝑥)

2= 1

2𝑘𝑦𝑚𝑎𝑥

2

𝑦 = 𝑦𝑚𝑎𝑥 sin(𝜔𝑛𝑡) 𝑦 𝑚𝑎𝑥 = 𝑦𝑚𝑎𝑥𝜔𝑛 1

2𝑚(

𝑏

𝑙𝑦𝑚𝑎𝑥𝜔𝑛)

2= 1

2𝑘𝑦𝑚𝑎𝑥

2

𝜔𝑛 =𝑙

𝑏

𝑘

𝑚

𝜔𝑛 =𝑙

𝑏

𝑘

𝑚

EQUIVALENT SYSTEMS

• Converting complex systems to simpler equivalent systems.

• Enables us to obtain equations of motions easily.

• Converting lumped physical elements to ideal elements.

Converting Lumped Physical Elements To

Ideal Linear Elements

• Pure Elastic Elements

• Converting pure elastic elements (no mass, no damping) to

ideal spring.

• Force deflection relationship.

• For example, relationship between force (F) and deflection (y)

for an helical spring is:

• 𝑦 =8𝐹𝐷3𝑁

𝑑4𝐺

• Thus, it can be modeled as an ideal spring of constant:

• 𝑘 =𝐹

𝑦=

𝑑4𝐺

8𝐷3𝑁

• For a torsion bar of circular cross section:

• 𝜃 =32𝑙𝑇

𝜋𝐷4𝐺

• Thus, it can be modeled as an ideal torsional spring of

constant:

• 𝑘𝑡 =𝑇

𝜃=

𝜋𝑑4𝐺

32𝑙

• Simply supported beam loaded by a force F from a distance a.

• The deflection at a is equal to:

• 𝛿 =𝐹𝑎2(𝑙−𝑎)2

3𝐸𝐼𝑙

• Equivalent spring constant at a (same point) is equal to:

• 𝑘𝑒 =𝐹

𝛿=

3𝐸𝐼𝑙

𝑎2(𝑙−𝑎)2

• Lumped Mass

• Modelling of physical elements of lumped mass in terms of

ideal mass or inertia.

• Based on kinetic energy equality of the orijinal and equivalent

system.

• Example: Lumped mass helical spring:

• Displacement of free end is y(t), thus displacement of a point

which is away from the free end is:

• 𝑥 𝜉, 𝑡 =𝜉

𝐿𝑦(𝑡)

• Kinetic energy of the differential element is:

• 𝑑𝑇 =1

2(𝑀

𝐿𝑑𝜉))(

𝜉

𝐿𝑦 𝑡 )

2

• Total kinetic energy:

• 𝑇 = 1

2(𝑀

𝐿𝑑𝜉))(

𝜉

𝐿𝑦 )

2𝐿

0=

1

2

𝑀𝑦 2

𝐿3 𝜉2𝑑𝜉𝐿

0=

1

2

𝑀𝑦 2

𝐿3𝐿3

3=

1

2(𝑀

3)𝑦 2

• This is equal to the kinetic energy of a mass of 𝑀

3 which moves

with a speed of 𝑦 . Thus, this sytem can be modeled as an ideal

mass and spring system as shown in the figure:

• Example: Lumped mass rod: (For small )

• 𝑑𝑇 =1

2(𝑀

𝐿𝑑𝜉))(𝜃 𝜉)

2

• 𝑇 =1

2

𝑀

𝐿𝜃 2 𝜉2𝑑𝜉

𝐿

0=

1

2(𝑀𝐿2

3)𝜃 2

• Thus, equivalent system is given below, where the equivalent

inertia is equal to 𝐼𝑒 =𝑀𝐿2

3.

• Since: 𝑥 𝐴 = 𝑎𝜃

• 𝑇 =1

2(𝑀𝐿2

3𝑎2)𝑥 𝐴

2

• Thus, equivalent system is given below, where the equivalent

mass is equal to: 𝑀𝑒 =𝑀𝐿2

3𝑎2

Massless rod

Converting Multi Elements To Single

Elements

• Serially connected springs:

•𝐹

𝑘𝑒= 𝑥1 − 𝑥2

•1

𝑘𝑒=

1

𝑘1+

1

𝑘2+⋯+

1

𝑘𝑛

a) Original system

b) Equivalent system

• Parellel connected springs:

• 𝐹 = 𝑘𝑒(𝑥1 − 𝑥2)

• 𝑘𝑒 = 𝑘1 + 𝑘2 +⋯+ 𝑘𝑛

a) Original system b) Equivalent system

• Serially connected dampers:

•1

𝑏𝑒=

1

𝑏1+

1

𝑏2+⋯+

1

𝑏𝑛

• Parellel connected dampers:

• 𝑏𝑒 = 𝑏1 + 𝑏2 +⋯+ 𝑏𝑛

• Equivalent mass and inertia:

• Determine the equivalent mass or inertia for bodies which are

rigidly connected (no elastic connection) to each other.

• Determine the kinetic energies of the original and equivalent

systems. Equalize them to obtain the equivalent mass or

inertia.

• Velocity of the equivalent system is equal to the velocity of

one of the original mass or inertia.

• Example: Determine the equivalent inertia of the gear pair,

which moves with an angular velocity of Ω1.

• For the original system:

• 𝑇 =1

2𝐽1𝛺1

2 +1

2𝐽2𝛺2

2

• 𝑇 =1

2𝐽1𝛺1

2 +1

2𝐽2(

𝑟1

𝑟2𝛺1)

2

• For the equivalent system:

• 𝑇𝑒 =1

2𝐽𝑒𝛺1

2

• Equating kinetic energies:

• 𝐽𝑒 = 𝐽1 + 𝐽2(𝑟1

𝑟2)2

• Example: For the mass and pulley system given in Fig.a, determine the equivalent mass of the system given in Fig.b, and determine the equivalent inertia of the system given in Fig.c.

• For the original system:

• 𝑇 =1

2𝑀𝑥 2 +

1

2𝐽𝜃 2

• 𝑇 =1

2𝑀𝑥 2 +

1

2𝐽(𝑥

𝑟) 2

• For the equivalent system at Fig.b:

• 𝑇𝑒 =1

2𝑀𝑒𝑥

2

• Thus:

• 𝑀𝑒 = 𝑀 +𝐽

𝑟2

• For the equivalent system at Fig.c:

• 𝑇𝑒 =1

2𝐽𝑒𝜃

2

• Kinetic enerygy of the original system:

• 𝑇 =1

2𝑀(𝑟𝜃 )2+

1

2𝐽𝜃 2

• Thus:

• 𝐽𝑒 = 𝑀𝑟2 + 𝐽

Obtaining Simple Equivalent Systems of 1-

DOF Complex Systems

• 1-DOF complex systems are converted to an equivalent simple mass, spring and damper systems.

• These equivalent systems must have the same kinetic and potential energy with the original system.

• Further, the power dissipation from dampers, and the power input from the external forces must be the same for the original and equivalent systems.

• Thus, natural frequency and damping ratio will be the same for the original and equivalent systems.

• Initial step for this method is, selecting the velocity of the equivalent mass as the velocity of a point on the original system.

• Example: Determine the equivalent of the system given in

Fig.a, as in Fig.b and as in Fig.c.

• Geometric relationships:

•𝑥𝐴

𝑎=

𝑥𝐵

𝑏= −

𝑥𝐶

𝑐= −

𝑥𝐷

𝑑 (𝑥 = 𝑟𝜃)

• Kinetic Energies:

• Original System: 𝑇𝑎 =1

2𝑀𝑥 2

𝐷

• Equivalent at b: 𝑇𝑏 =1

2𝑀𝑒𝑥

2𝐴

• Equivalent at c : 𝑇𝑐 =1

2𝑀𝑒 𝑥 2

𝐷

• Equating the kinetic energies and using geometric

relationships equivalent masses are determined:

• 𝑀𝑒 =𝑑

𝑎

2𝑀

• 𝑀𝑒 = 𝑀

• Potantial Energies:

• Original System: 𝑉𝑎 =1

2𝐾𝑥2

𝐴

• Equivalent at b: 𝑉𝑏 =1

2𝐾𝑒𝑥

2𝐴

• Equivalent at c : 𝑉𝑐 =1

2𝐾𝑒 𝑥

2𝐷

• Equating the potantial energies and using geometric

relationships equivalent stiffnesses are determined:

• 𝐾𝑒 = 𝐾

• 𝐾𝑒 =𝑎

𝑑

2𝐾

• Power dissipation from dampers:

• Original System: 𝑃𝑑𝑎 = 𝐵𝑥 2𝐵

• Equivalent at b: 𝑃𝑑𝑏 = 𝐵𝑒𝑥 2𝐴

• Equivalent at c : 𝑃𝑑𝑐 = 𝐵𝑒 𝑥 2𝐷

• Equating the power outputs and using geometric relationships

equivalent damping coefficients are determined:

• 𝐵𝑒 =𝑏

𝑎

2𝐵

• 𝐵𝑒 =𝑏

𝑑

2𝐵

• Power input from external forces:

• Original System: 𝑃𝑎 = −𝐹𝑥 𝐶

• Equivalent at b: 𝑃𝑏 = −𝐹𝑒𝑥 𝐴

• Equivalent at c : 𝑃𝑐 = −𝐹𝑒 𝑥 𝐷

• Equating the power inputs and using geometric relationships :

• 𝐹𝑒 = −𝑐

𝑎𝐹

• 𝐹𝑒 =𝑐

𝑑𝐹

• Equation of motion of the equivalent system at b:

• 𝑀𝑒𝑥 𝐴 + 𝐵𝑒𝑥 𝐴 + 𝐾𝑒𝑥𝐴 = 𝐹𝑒

• Equation of motion of the equivalent system at c:

• 𝑀𝑒 𝑥 𝐷 + 𝐵𝑒 𝑥 𝐷 +𝐾𝑒 𝑥𝐷 = 𝐹𝑒

• Natural frequency and damping ratio of equivalent system at b:

• 𝜔𝑛 =𝐾𝑒

𝑀𝑒=

𝐾

𝑑

𝑎

2𝑀

• 𝜁 =𝐵𝑒

2 𝐾𝑒𝑀𝑒=

𝑏

𝑎

2𝐵

2 𝐾𝑑

𝑎

2𝑀

(Remember that: 𝜁 =𝐵

2𝑀𝜔𝑛)

• Natural frequency and damping ratio of equivalent system at c:

• 𝜔𝑛 =𝐾𝑒

𝑀𝑒 =

𝑎

𝑑

2𝐾

𝑀 =

𝐾

𝑑

𝑎

2𝑀

• 𝜁 =𝐵𝑒

2 𝐾𝑒 𝑀𝑒 =

𝑏

𝑑

2𝐵

2𝑎

𝑑

2𝑀 = ⋯ =

𝑏

𝑎

2𝐵

2 𝐾𝑑

𝑎

2𝑀

• Thus, natural frequency and damping ratio is the same.