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Dynamic Systems and Mechanical Systems tutorial
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5/22/2018 Dynamic Systems Mechanical Systems
1/64
1 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Mechanical Systems
Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell
j
UNSTABLESTABLE
TIME
FRF
FRF
TIME
FRF
TIME
FRF
TIME
TIME
TIME
> 1.0
= 1.0
= 0.7
= 0.3
= 0.1 = 0
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2 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Mechanical Systems-Translational Mass Element
Translation of a particle moving in space due to anapplied force is given by:
dt
dpf=
Where:
mvmomentump
forcef
==
=
Considering the mass to be constant:
madt
dvmfmdvfdt
dt
)mv(df ====
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3 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Mechanical Systems Translational Mass Element
Displacement, velocity, and acceleration are allrelated by time derivatives as:
2
2
dt
xddtda ==
xva &&&==
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4 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Mechanical Systems Rotational Mass Element
=
= &
dt
d
dt
d
dt
d 2=
=
Centroidal mass moment of inertia Ic (not to beconfused with I area moment of inertia used instrength of materials)
Angular acceleration
where: = angular velocity
= angular displacement
Then:
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5 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Mechanical Systems - Translational Spring Element
( )21relk xxkkxf ==
A linear spring is considered to have no massdescribed by:
(Torsional spring follows the same relationship)
fk fk
x1 x2
k
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6 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Translational Spring Element
Hardening Spring
Bi-Linear
Linear
Softening Spring
Gap Cubic
f f f
x x x
k
k = lb/in
= N/m
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7 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Damper Element
relc cf =
Viscous (fluid), Coulomb (dry friction), and structuraldamping (hysteretic)
Viscous Dashpot
fcfa
vrel
m
Coulomb DamperIn order to have motion, the applied force must
overcome the static friction. As soon as sliding occurs,the dynamic friction becomes appropriate.
fc
c
x&
Linear ( )21c cf =
fc fc
v1 v2
c
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8 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Equivalence - Springs in Parallel
Both springs see the same displacement
k2
x
f
k1
f
x
keq
f2=k2x
f
f2
f1=k1x
f1
f=f1 + f2
keqx=k
1x + k
2x
=(k1 + k2)x
21eq kkk +=
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9 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Equivalence - Springs in Series
Both springs see the same force but different displacements
But
k2
x1
f
k1
x2
f=keq
x
keq
f=k1(x1-x2)=k1
k1
k2
f2=k2x2=k2
1
2f
21 +=
22
11
eq k
f
k
f
k
f
+= 21eq k
1
k
1
k
1+=
21
21eq
kk
kkk
+=
21 fff ==
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10 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Translational Systems
Newtons Second Law- THE RIGHT WAY
=maF OR
= xx maF +
= yy maF +Note that this applies to the center of mass which is not
necessarily the center of gravity.
Free-Body Diagram & Sign Convention
kx
xc&m )t(f
xxx &&&+
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11 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Translational Systems Newtons 2ndLaw
Assume spring and dashpot are stretched
= xx maF
or in standard input-output differential form
xmFF)t(f kc &&=
xmkxxc)t(f &&& =
OR
)t(fkxxcxm =++ &&&m)t(fx
mkx
mcx =++ &&&
)t(fxx2x 2
nn =++ &&&
m
k
m2c
c
c
2n
nc
c
=
=
= - damping ratio
critical damping
n natural frequency
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12 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
DAlemberts Principle The Fictitious Force
The mass times acceleration is sometimes describedas a fictitious force, reverse effective force or apparent force
( ) =+ 0maFInitially developed since it looks like a classical
force balance but often confuses many students.
DO NOT USE DALEMBERT!!!!!USE NEWTONS SECOND LAW
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13 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Example Pendulum Problem
Mass at end of massless string
l
mg
2mJ l=
+x
cosmg
mg
sinmg
T
FBD
sinmgJ l=&& 0sinmgm 2 =+ ll &&ORThen
0sing
=+
l
&& 0
g =+
l&&for small .
Natl freq. lg
n =
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14 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Example Differential Equation about Equilibrium
y
x
+ = yy maFymmgky &&=+
( ) xmmgxxk st &&=++but stkxmg=
0kxxm =+ &&
Therefore, the equationscan be written about the
equilibrium point and theeffect of gravity makesno difference.
Source: Dynamic Systems Vu & Esfandiari
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15 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Systems with Displacement Input
In terms of naturalfrequency and damping
ratio
= xx maF+
m
ck
x
y
m x
)xy(c &&)xy(k
m
ck
x
y
FBD
( ) ( ) xmxycxyk &&&& =+ kyyckxxcxm +=++ &&&&OR
The force exerted canbe found to be
yy2xx2x 2
nn
2
nn
+=++ &&&&
( ) ( )xycxyk)t(f &&+=
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16 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Transfer Function and State Space
m
c
k
x
f(t) )t(fkxxcxm =++ &&&
00
2 XmmsXmX(s)s)x(m &&& =L
0)()( cXsscXxc =&L
)()( skXkx =L
)())(( sftf =L
)s(f)s(kXcX)s(scXXmmsX)s(mXs 0002 =++ &
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17 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Transfer Function and State Space
( ) 002 XmXcms)s(f)s(Xkcsms &+++=++
Grouping and rearranging:
Assume initial conditions are zero and rearrangingterms to obtain OUT/IN formThen:
kcsms1
)s(F)s(X)s(H
2 ++==
Sometimes written with
kcsms)s(b 2 ++=
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18 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
The frequency response function is the systemtransfer function evaluated along
Frequency Response Function - SDOF
= js
kcsms
1)s(h
2 ++=
Source: Vibrant Technology
Recall:
The complex valuedfunction defines the
surface shown
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19 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Polynomial Form
Pole-Zero Form
Partial Fraction Form
Exponential Form
kcsms
1)s(h
2 ++=
)ps)(ps(
m/1)s(h
*
11 =
)ps(
a
)ps(
a)s(h
*
1
*
1
1
1
+
=
tsinem
1)t(h d
t
d
=
SDOF Transfer Function
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20 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Polynomial Form
Pole-Zero Form
Partial Fraction Form
kcjm
1)h(j
2 ++=
)pj)(pj(
m/1)j(h
*11
=
)pj(
a
)pj(
a)j(h
*1
*1
1
1
+
=
SDOF Frequency Response Function
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21 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Transfer Function approach is used extensively indesign but is limited to linear, time-invariant systems.
SDOF Transfer Function
1. T.F. method to express output relative to input2. T.F. system property independent of the nature
of excitation
3. T.F. contains necessary units but does not providephysical structure of system
4. If T.F. is known, then response can be evaluateddue to various inputs
5. If T.F. is unknown, it can be establishedexperimentally by measuring output response due toknown measured inputs
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22 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
S-plane Plots
j
UNSTABLESTABLE
TIME
FRF
FRF
TIME
FRF
TIME
FRF
TIME
TIME
TIME
> 1.0
= 1.0
= 0.7
= 0.3
= 0.1 = 0
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23 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Experimental Determination of Damping Ratio
Determine decay of amplitudex1at t1and again at n cycleslater xnat t1+ (n-1)T
Then
x1 x2 xn
t1 t2 tn
T - period
( )T
TTt
t
2
1 n
nn
n
ee
1
e
e
x
x +
===
OR
( )( ) T1n
T1n
n
1 n
n
e
e
1
x
x ==
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24 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Log Decrement
Tx
xln1n
1
x
xln n
n
1
2
1 =
=
For damping < 10%
2d
n122
==
2n
1
1
2xxln
1n1
=
2xxln2
1
Note: This dampingratio formulation isapplicable to any 2nd
order system of this
form
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25 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Estimate of Response Time
whose time constant T of the exponential is
( )
= tcose1
X)t(x
d
t
2
0 n
The response of a mechanical system due to aninitial displacement is given as:
The exponential response envelope is
t
2
0 ne1
X
=
11
n
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26 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Estimate of Response Time
The response of the second-order system in termsof the settling time is
===
44
T4t ns
which will cause 2%
of the initial value
Source: Dynamic Systems Vu & Esfandiari
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27 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
State Space Representation
The state of the system can be described interms of the displacement and velocity as
=
x
x
x
x
2
1
&
=
= )velocity(x)displ(x
x
x
X 2
1
&
u = f (force) and y = x (measured by sensor)
Then )t(fm
1x
m
cx
m
kx += &&&
ORu
m
1x
m
cx
m
kx 212 +=&
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28 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
State Space Representation
So that the state space representation is
u0
x
x10
x
x
m1
2
1
m
c
m
k
2
1
+
=
&
&State
Equation
OutputEquation [ ] u0x
x01y2
1 +
=
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29 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Lagranges Equations
nciiii
QqV
qT
qT
dtd =
+
T Kinetic energyV Potential energyQnci non-conservative generalized forcesqi
independent generalized coordinatesn total # independent generalized coordinates
Kinetic Energy is a function of
Potential energy is the sum of elastic potential Veand gravitational potential VgPotential Energy is a function of
),( tqT i
),( tqV i
dtdq /
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30 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Lagranges Equations
nci
ii
Q
q
L
q
L
dt
d=
&
iq&
One standard form of Lagranges Equation
where L = T-V
We can then write
( ) ( )nci
ii
VT
q
VT
dt
d=
&
(Note V is not a function of )
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31 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Lagranges Equations
2c
2
2
I2
1mv2
1TD2
mv2
1TD1
+=
=
Kinetic energy for a particle
If the mass is not located at apoint (such as a particle), then
a more complicated form ofthese equations is necessary
Potential Energy of anelastic element is
Potential Energy ofa mass is
2e k
2
1V =
mghVg=
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32 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Lagranges Equations
Non-conservative Forces are those that cannot bederived from a potential function (i.e., external forces,frictional forces)Generalized Forces are given by Virtual Work.
To determine Qj, obtain , then let all except
L++== 2211ii qQqQqQWW 0qi= jq
Thus
j
j
q
WQ
=
0qj=
ij
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33 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Lagranges Equations
Non-conservative Forces are then:L++== 22nc11ncincinc qQqQqQW
Then
i
nc
nci q
WQ
=
0qi=
ij
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34 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Example using Lagrange Equation
Use Lagrange EQ to obtaindifferential equation forSDOF system
Only one independent generalized coordinate exists: q=x
22
c xm2
1
mv2
1
T &
==
m
c
k
x
f(t)
Kinetic Energy
Potential Energy
Non-conservativeForces
( )0Vkx2
1VV g
2e ===
applied)t(f
edissipativxc &
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35 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Example using Lagrange Equation
Non-Conservative Forces[ ] xxc)t(fWnc = &
xQW ncnc =
xc)t(fx
WQ ncnc &=
=
Lagrange Equation
ncQ
x
V
x
T
x
T
dt
d=
+
&
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36 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Example using Lagrange Equation
Lagrange Equation ncQxV
xT
xT
dtd =
+
&
Where
Then substitutingxmx
xm
xT &
&
&
&=
=
)2
1( 2
xmdt
)
x
T(d
&&&
=
0x
)xm2
1(
x
T2
=
=
&
kx
x
)kx2
1(
x
V2
=
=
OR
xc)t(fkxxm &&& =+
)t(fkxxcxm =++ &&&
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37 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Example using Lagrange Equation
Now lets repeat this with the Lagrange function
22 kx2
1xm
2
1VTL == &
xmt
)
x
L(
&&&
=
xmx
)kxxm21(
x
L22
&
&
&
&=
=
)t(fkxxcxm =++ &&&
ncQx
L
t
)x
L(
=
&
xm)xL( && =
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38 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Example Translational Mechanical System
m
k
xst
xy
g
Undeformed position
Dynamic position
Static equilibrium
Solution
Source: Dynamic Systems Vu & Esfandiari
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22.451 Dynamic Systems Chapter 4
Example Translational Mechanical System
Source: Dynamic Systems Vu & Esfandiari
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40 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Example Translational Mechanical System
Source: Dynamic Systems Vu & Esfandiari
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22.451 Dynamic Systems Chapter 4
Example Translational Mechanical System
Source: Dynamic Systems Vu & Esfandiari
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42 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Example - Two DOF Systems
Consider
m1
f1(t)
x2x1
f2(t)
m2
c
k
FBD (assume x1> x2)
1121211 xm)xx(k)xx(c)t(f)1(maF &&&& ==
2221212 xm)xx(k)xx(c)t(f)2(maF &&&& =++=
221
211 f
)xx(c
)xx(kf
&&m2m1
STATEASSUMPTIONS!!!
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43 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Example - Two DOF Systems
Rearranging Terms)t(fkxkxxcxcxm 1212111 =++ &&&&
=
+
+
2
1
2
1
2
1
2
1
2
1
ff
xx
kkkk
xx
cccc
xx
m00m
&
&
&&
&&
)t(fkxkxxcxcxm 2212122 =++ &&&&
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44 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Example - Two DOF Systems
Consider
FBD (assume x2> x1)1112212111 xm)xx(k)xx(cxk)t(f)1(maF &&&& =++=
m1
f1(t)
x2x1
f2(t)
m2
c
k2
k1
22122122 xm)xx(k)xx(c)t(f)2(maF &&&& ==
)t(f)xx(k
)xx(c
xk
)t(f2
122
12
11
1
&&m2m1
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45 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Example - Two DOF Systems
Rearranging terms
)t(fxkx)kk(xcxcxm 1221212111 =+++ &&&&
=
++
+
2
1
2
1
22
221
2
1
2
1
2
1
f
f
x
x
kk
k)kk(
x
x
cc
cc
x
x
m0
0m
&
&
&&
&&
)t(fxkxkxcxcxm 222122122 =++ &&&&
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46 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Rotational Systems
A rotational system follows the same equationsdeveloped for translation
= 00 IM
dmrIorJ 20 =
Newtons Second Law
.accelangularw
intofmomentmassI
appliedmomentsM
0
0
&
Mass moment of inertia of rigid body about axis
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47 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Rotational Systems
Torsion spring stiffness similar to translation
)(KT 12Tk =
Dashpot similar to translation
Right hand rule convention determines +/-
)(KT RELTk =
)(CT 12DD = &&
)(CT RELDD =
usedoftenBorCD
RADLBFT /
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22.451 Dynamic Systems Chapter 4
Rotational Systems
FBD
OUTJ
IN
J DT
ST
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49 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Rotational Systems - Equations
Most systems we will treat will be 2D or planarsystems. Modeling of general 3D bodies is more
complex and beyond the scope of this course.
=
=+= Idt
dJTTJM
20
2
DS
)(KT OUTINTS =
OUTD BT = &
OUTOUTOUTINT JB)(K = &&&
INTOUTTOUTOUT KKBJ =++ &&&
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22.451 Dynamic Systems Chapter 4
Example - SDOF Torsional System
A torsional system: (a) physical system, (b) FBD
&BK
)t(T
J
K
B
)t(T J
)a( )b(
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22.451 Dynamic Systems Chapter 4
Example - SDOF Torsional System
This sign convention is simpler and useful for the given angle .Thus,
Consider a single-degree-of-freedom (SDOF) torsionalsystem. The system consists of a shaft of torsionalstiffness K, a disk of mass-moment of inertia J, and atorsional damper B. Derive the differential equation.
Solution. Applying the moment equation about the mass centeralong the longitudinal axis.
+=+ cc IM
= &&&JBK)t(T
)t(TKBJ =++ &&&The differential equation in the input-output form is
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22.451 Dynamic Systems Chapter 4
Example - Two DOF Torsional System
Source: Dynamic Systems Vu & Esfandiari
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22.451 Dynamic Systems Chapter 4
Example - Two DOF Torsional System
Source: Dynamic Systems Vu & Esfandiari
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22.451 Dynamic Systems Chapter 4
Example - Rigid Body in Planar Motion
As before
= 00 IM+= &&)ml0(mgsinL 2
Point mass on string Moment method
0sin
2
=+ mgLmL &&
0sin =+ Lg
&&
L
mg
0
sinL
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22.451 Dynamic Systems Chapter 4
Example Pendulum Problem
Thin uniform rod of mass m and length l is a pendulum
Then
= 00 IM
&&
0sin2 Img
L
=
FBD
0sin2
lmgI0 =+&&
0
mg
Lg
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22.451 Dynamic Systems Chapter 4
Linearization: For small
where 2c0 mdII +=
or
sin
02
0 =+ mgL
I &&
02 0
=+ I
mgL&&
2
2
)L
m(Ic+=
2
03
1mLI =
2
12
1mLIc=
Example Pendulum Problem
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22.451 Dynamic Systems Chapter 4
Mixed Translation and Rotation
Pulley system
Rotation of pulley
Newtons second law for mass m
and
Txm =&&
Icof pulley, mass m of radius r,Tension in string
m x
r
k
cI
(everything measured from equilibrium so no mg term)
For small angle
or
The natural frequency is
kxrTrJ =&&
=rx kxrrxmJ = &&&&then
0kr)mrJ( 22 =++ &&
0mrJ
kr2
2
=+
+&&
2
2
n mrJ
kr
+=
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22.451 Dynamic Systems Chapter 4
Example Cart-Pendulum Problem
Consider the pendulum systemshown attached to a horizontalcart
This is a mixed problem. First solve the pendulum and
then the cart translation.
Cart moves horizontally onfrictionless surface. Mass oninextensible string
The general moment about point P (where the string isattached to the cart mass) is needed to sum the forces
for Newtons Second Law.
M
c L
k
m
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59 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
pp/cp axmrIMp +=
yP
xPP
L
mg
sinL
xap &&=
Lr p/c1
=
1C
Example Cart-Pendulum Problem
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60 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
The cross product term is
Using the parallel axis theorem, the mass of thependulum at a distance L gives
The moment about P due to the mass on the pendulum is== &&2
p
2
p mLImLI
= sinLmgM
sinararpc/ppc/p
=== cosxL)90sin(xL &&&&
The general moment equation becomes
+= cosxLmLmsinLmg 2 &&&&
Example Cart-Pendulum Problem
E l d l l
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61 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
Now the translational equation is evaluated. Only horizontalis considered.
M )t(fkx
L
msinL
xc& xapx &&=
2L& &&L
sinL 2& cosL&&
centripetaltangential
Example Cart-Pendulum Problem
E l P d l P bl
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62 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
For the cart, Newtons Second Law
xckx)t(fmaF &=
The acceleration of the pendulum mass
sinLcosLx 2&&&&& +(cart) (tangential) (centripetal)
xM)sinLcosLx(m 2 &&&&&
&& ++=
0sinmgLcosxmLmL2 =++ &&&&
( ) )t(fkxxcsinLcosLmxmMx 2 =++++ &&&&&&
Example Cart-Pendulum Problem
E l C P d l P bl
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63 Dr. Peter AvitabileModal Analysis & Controls Laboratory
22.451 Dynamic Systems Chapter 4
For small motion, theequations can be linearized
0mgLxmLmL2 =++ &&&&
or in matrix forms as
0;sin;1cos 2
&
( ) )t(fkxxcmLxmM =++++ &&&&&
=
+
+
+ )t(f
0
xk0
0mgL
xc0
00
x)mM(mL
mLmL2
&
&
&&
&&
Example Cart-Pendulum Problem
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Chapter 4 - Review Slide
k2
xf
k1
21eq kkk +=
k2
x1
f
k1
x2
21
21eq
kk
kkk
+=
2x
xln
2
1x1 x2 xn
t1 t2 tn
T