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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
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cossinsincossinsinsin([
)coscossincoscossincoscos
cossinsinsinsincossin(
0)coscos(
)sincossincos()sinsin(
)cossincossin()()(
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• 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.
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