Mechanical System Modeling
Dr. Kevin Craig
Professor of Mechanical Engineering
Rensselaer Polytechnic Institute
Mechanical System Modeling
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References for Mechanical Systems
• System Dynamics, E. Doebelin, Marcel Dekker,
1998. (This is the finest reference on system
dynamics available; many figures in these notes
are taken from this reference.)
• Modeling, Analysis, and Control of Dynamic
Systems, W. Palm, 2nd Edition, Wiley, 1999.
• Vector Mechanics for Engineers: Dynamics, 7th
Edition, F. Beer, E.R. Johnston, and W. Clausen,
McGraw Hill, 2004.
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Mechanical System Elements
• Three basic mechanical elements:
– Spring (elastic) element
– Damper (frictional) element
– Mass (inertia) element
• Translational and Rotational versions
• These are passive (non-energy producing) devices
• Driving Inputs
– force and motion sources which cause elements
to respond
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• Each of the elements has one of two possible
energy behaviors:
– stores all the energy supplied to it
– dissipates all energy into heat by some kind of
“frictional” effect
• Spring stores energy as potential energy
• Mass stores energy as kinetic energy
• Damper dissipates energy into heat
• Dynamic Response of each element is important
– step response
– frequency response
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Spring Element
•
•
•
•
Real-world design situations
Real-world spring is neither pure nor ideal
Real-world spring has inertia and friction
Pure spring has only elasticity - it is a
mathematical model, not a real device
• Some dynamic operation requires that spring
inertia and/or damping not be neglected
• Ideal spring: linear
• Nonlinear behavior may often be preferable and
give significant performance advantages
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• Device can be pure without being ideal (e.g.,
nonlinear spring with no inertia or damping)
• Device can be ideal without being pure (e.g., device
which exhibits both linear springiness and linear
damping)
f = K s ( x1 − x 2 ) = K s x
• Pure and ideal spring element:
T = K s ( θ1 − θ2 ) = K s θ
• Ks = spring stiffness (N/m or N-m/rad)
• 1/Ks = Cs = compliance (softness parameter)
x
Ks
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f
f
Cs
x = Cs f
θ = Cs T
x
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• Energy stored in a spring
Cs f 2 K s x 2
Es =
=
2
2
• Dynamic Response: Zero-Order Dynamic System
Model
– Step Response
– Frequency Response
• Real springs will not behave exactly like the
pure/ideal element. One of the best ways to
measure this deviation is through frequency
response.
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Spring Element
Differential Work Done
= ( f ) dx = ( K s x ) dx
Total Work Done
=∫
x0
0
K s x 02 Cs f 02
=
( K s x )dx =
2
2
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Frequency Response
Of
Spring Elements
f = f 0 sin ( ωt )
x = Cs f 0 sin ( ωt )
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Zero-Order Dynamic System Model
Step Response
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Frequency Response
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More Realistic Lumped-Parameter Model for a Spring
Ks
f, x
Ks
M
B
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B
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Linearization
for a
Nonlinear Spring
df
y = f (x 0 ) +
dx
y ≈ y0 +
df
dx
d 2f
( x − x0 ) + 2
dx
x =x0
( x − x0 )
2!
x =x0
2
+"
( x − x0 )
x =x0
y − y0 ≈ +
df
dx
( x − x0 )
x = x0
yˆ = Kxˆ
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• Real Springs
– nonlinearity of the
force/deflection curve
– noncoincidence of the
loading and unloading
curves (The 2nd Law of
Thermodynamics
guarantees that the area
under the loading f vs. x
curve must be greater
than that under the
unloading f vs. x curve.
It is impossible to recover
100% of the energy put
into any system.)
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• Several Types of Practical
Springs:
–
–
–
–
–
–
–
–
–
coil spring
hydraulic (oil) spring
cantilever beam spring
pneumatic (air) spring
clamped-end beam spring
ring spring
rubber spring (shock mount)
tension rod spring
torsion bar spring
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• Spring-like Effects in
Unfamiliar Forms
– aerodynamic spring
– gravity spring (pendulum)
– gravity spring (liquid
column)
– buoyancy spring
– magnetic spring
– electrostatic spring
– centrifugal spring
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Damper Element
• A pure damper dissipates all the energy supplied
to it, i.e., converts the mechanical energy to
thermal energy.
• Various physical mechanisms, usually associated
with some form of friction, can provide this
dissipative action, e.g.,
– Coulomb (dry friction) damping
– Material (solid) damping
– Viscous damping
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• Pure / ideal damper element provides viscous
friction.
• All mechanical elements are defined in terms of
their force/motion relation. (Electrical elements
are defined in terms of their voltage/current
relations.)
• Pure / Ideal Damper
– Damper force or torque is directly proportional
to the relative velocity of its two ends.
dx
⎛ dx1 dx 2 ⎞
f = B⎜
−
⎟=B
dt ⎠
dt
⎝ dt
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dθ
⎛ dθ1 dθ2 ⎞
T = B⎜
−
⎟=B
dt ⎠
dt
⎝ dt
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– Forces or torques on the two ends of the
damper are exactly equal and opposite at all
times (just like a spring); pure springs and
dampers have no mass or inertia. This is NOT
true for real springs and dampers.
– Units for B to preserve physical meaning:
• N/(m/sec)
• (N-m)/(rad/sec)
– Transfer Function
Differential
Operator
Notation
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dx
Dx dt
x
∫ (x)dt
D
d2x
D x 2
dt
x
∫ ⎡⎣ ∫ ( x ) dt ⎤⎦dt
2
D
2
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• Operational Transfer Functions
f = BDx
T = BDθ
f
( D ) BD
x
x
1
( D) f
BD
T
( D ) BD
θ
θ
1
( D) T
BD
• We assume the initial conditions are zero.
– Damper element dissipates into heat all
mechanical energy supplied to it.
⎛ dx ⎞
⎛ dx ⎞
Power ( force )( velocity ) = f ⎜ ⎟ = B ⎜ ⎟
⎝ dt ⎠
⎝ dt ⎠
2
• Force applied to damper causes a velocity in same
direction.
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• Power input to the device is positive since the force
and velocity have the same sign.
• It is impossible for the applied force and resulting
velocity to have opposite signs.
• Thus, a damper can never supply power to another
device; Power is always positive.
• A spring absorbs power and stores energy as a force
is applied to it, but if the force is gradually relaxed
back to zero, the external force and the velocity now
have opposite signs, showing that the spring is
delivering power.
• Total Energy Dissipated
2
⎛ dx ⎞
⎛ dx ⎞
∫ ( P )dt = ∫ B ⎜⎝ dt ⎟⎠ dt = ∫ B ⎜⎝ dt ⎟⎠dx = ∫ ( f )dx
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Damper Element
Step Input Force
causes instantly
(a pure damper
has no inertia) a
Step of dx/dt
and a
Ramp of x
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Frequency
Response
of
Damper
Elements
f = f 0 sin ( ωt )
=B
dx
dt
t
1
x − x 0 = ∫ f 0 sin ( ωt ) dt
B0
f0
⎡⎣1 − cos ( ωt ) ⎤⎦
=
Bω
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f0
A x Bω
1
=
=
Af
f0
Bω
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• Sinusoidal Transfer Function
x
1
( D) =
f
BD
D ⇒ iω
x
1
= M∠φ
( iω ) =
f
i ωB
– M is the amplitude ratio of output over input
– φ is the phase shift of the output sine wave with
respect to the input sine wave (positive if the
output leads the input, negative if the output lags
the input)
x
1
1
i
M
= ∠φ =
∠ − 90°
( ω) =
f
i ωB
Bω
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• Real Dampers
– A damper element is used to model a device
designed into a system (e.g., automotive shock
absorbers) or for unavoidable parasitic effects
(e.g., air drag).
– To be an energy-dissipating effect, a device
must exert a force opposite to the velocity;
power is always negative when the force and
velocity have opposite directions.
– Let’s consider examples of real intentional
dampers.
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Viscous (Piston/Cylinder) Damper
A relative velocity between the
cylinder and piston forces the
viscous oil through the clearance
space h, shearing the fluid and
creating a damping force.
μ = fluid viscosity
⎡
⎤
2
2
⎤
⎢
⎥
6πμL ⎡⎛
h⎞
R
R
−
2
2
1
B=
− h⎥
⎢⎜ R 2 − ⎟ − R 1 ⎥ ⎢
3
h ⎢⎣⎝
2⎠
⎥⎦ ⎢ R 2 − h
⎥
2
⎣
⎦
2
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Simple Shear Damper
And
Viscosity Definition
2Aμ
F=
V
t
F 2Aμ
B= =
V
t
μ fluid viscosity
shearing stress
F/A
=
velocity gradient V / t
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Examples
of
Rotary Dampers
πD04μ
B=
16t
πD3Lμ
B=
4t
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Commercial Air Damper
(Data taken with valve shut)
laminar flow
linear damping
turbulent flow
nonlinear damping
Air Damper
• much lower viscosity
• less temperature dependent
• no leakage or sealing problem
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•
•
•
•
•
Motion of the conducting
cup in the magnetic field
generates a voltage in the
cup.
A current is generated in
the cup’s circular path.
A current-carrying
conductor in a magnetic
field experiences a force
proportional to the current.
The result is a force
proportional to and
opposing the velocity.
The dissipated energy
shows up as I2R heating of
the cup.
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Eddy-Current Damper
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Temperature Sensitivity
Of
Damping Methods
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Other Examples
of
Damper Forms
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• The damper element can also be used to represent
unavoidable parasitic energy dissipation effects in
mechanical systems.
– Frictional effects in moving parts of machines
– Fluid drag on vehicles (cars, ships, aircraft, etc.)
– Windage losses of rotors in machines
– Hysteresis losses associated with cyclic stresses in
materials
– Structural damping due to riveted joints, welds,
etc.
– Air damping of vibrating structural shapes
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Hydraulic Motor Friction
and its Components
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Coulomb Friction: Modeling and Simulation
• In most control systems, Coulomb friction is a
nuisance.
• Coulomb friction is difficult to model and
troublesome to deal with in control system design.
• It is a nonlinear phenomenon in which a force is
produced that tends to oppose the motion of
bodies in contact in a mechanical system.
• Undesirable effects: “hangoff” and limit cycling
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• Hangoff (or dc limit cycle) prevents the steadystate error from becoming zero with a step
command input.
• Limit Cycling is behavior in which the steady-state
error oscillates or hunts about zero.
• What Should the Control Engineer Do?
– Minimize friction as much as possible in the design
– Appraise the effect of friction in a proposed control
system design by simulation
– If simulation predicts that the effect of friction is
unacceptable, you must do something about it!
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– Remedies can include simply modifying the design
parameters (gains), using integral control action, or
using more complex measures such as estimating the
friction and canceling its effect.
– Modeling and simulation of friction should contribute
significantly to improving the performance of motion
control systems.
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Modeling Coulomb Friction
Ff
Fstick
Fslip
V
"Stiction" Coulomb
Friction Model
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Case Study to Evaluate Friction Model
V0
V
k
m
m = 0.1 kg
k = 100 N/m
Fstick = 0.25 N
Fslip = 0.20 N (assumed independent of velocity)
V0 = step of 0.002 m/sec at t = 0 sec
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Ff
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Friction Model in Simulink
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Simulink Block Diagram
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Example with Friction Model
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Position, Velocity, Friction Force vs. Time
2*position, velocity, 0.1*Friction Force
0.025
0.02
0.015
0.01
0.005
0
0
0.5
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1
1.5
2
2.5
3
time (sec)
3.5
4
4.5
5
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Inertia Element
• A designer rarely inserts a component for the
purpose of adding inertia; the mass or inertia
element often represents an undesirable effect
which is unavoidable since all materials have
mass.
• There are some applications in which mass itself
serves a useful function, e.g., accelerometers and
flywheels.
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Accelerometer
Useful Applications
of
Inertia
Flywheels are used as
energy-storage devices or as
a means of smoothing out
speed fluctuations in engines
or other machines.
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– Newton’s Law defines the behavior of mass
elements and refers basically to an idealized
“point mass”:
∑ forces = ( mass )( acceleration )
– The concept of rigid body is introduced to deal
with practical situations. For pure translatory
motion, every point in a rigid body has identical
motion.
– Real physical bodies never display ideal rigid
behavior when being accelerated.
– The pure / ideal inertia element is a model, not
a real object.
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Rigid and Flexible Bodies: Definitions and Behavior
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– Newton’s Law in rotational form for bodies
undergoing pure rotational motion about a single
fixed axis:
∑ torques = ( moment of inertia )( angular acceleration )
– The concept of moment of inertia J also considers
the rotating body to be perfectly rigid.
– Note that to completely describe the inertial
properties of any rigid body requires the
specification of:
• Its total mass
• Location of the center of mass
• 3 moments of inertia and 3 products of inertia
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Rotational Inertia
J (kg-m2)
tangential force
= ( mass )( acceleration )
= ⎡⎣( 2πrLρ )( dr ) ⎤⎦ ( rα )
R
2
2
R
MR
total torque = ∫ ( 2πρLαr 3 ) dr = πR 2 Lρ
=
α = Jα
2
2
0
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Moments of Inertia
For
Some Common Shapes
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– How do we determine J for complex shapes
with possibly different materials involved?
• In the design stage, where the actual part exists only
on paper, estimate as well as possible!
• Once a part has been constructed, use experimental
methods for measuring inertial properties. How?
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Experimental Measurement
Of
Moment of Inertia
d 2θ
∑ ( torques ) = Jα = J dt 2
d 2θ
−K s θ = J 2
dt
d 2θ K s
+
θ=0
2
dt
J
(θ 0 = 0)
θ = θ0 cos ωn t
Ks
rad/sec
ωn J
ωn
fn cycles/sec
2π
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Ks
J= 2 2
4π f n
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– Actually the oscillation will gradually die out
due to the bearing friction not being zero.
– If bearing friction were pure Coulomb friction,
it can be shown that the decay envelope of the
oscillations is a straight line and that friction
has no effect on the frequency.
– If the friction is purely viscous, then the decay
envelope is an exponential curve, and the
frequency of oscillation does depend on the
friction but the dependence is usually negligible
for the low values of friction in typical
apparatus.
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Inertia Element
Real inertias may be
impure (have some
springiness and friction)
but are very close to
ideal.
x
1
θ
1
( D) = 2
(D) = 2
f
MD
T
JD
Inertia Element stores
energy as kinetic energy:
Mv 2
2
Jω2
or
2
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– A step input force applied to a mass initially at
rest causes an instantaneous jump in
acceleration, a ramp change in velocity, and a
parabolic change in position.
– The frequency response of the inertia element is
obtained from the sinusoidal transfer function:
x
1
1
°
=
∠
−
180
( iω ) =
2
2
ω
f
M
M ( iω )
• At high frequency, the inertia element becomes very
difficult to move.
• The phase angle shows that the displacement is in a
direction opposite to the applied force.
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Useful Frequency Range
for
Rigid Model
of a
Real Flexible Body
A real flexible body
approaches the
behavior of a rigid body
if the forcing frequency
is small compared to
the body’s natural
frequency.
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– Analysis:
2AE
= ρALx o
( xi − xo )
L
ρL2
x o + x o = x i
2E
⎛ D2 ⎞
2E
ωn ⎜ 2 + 1⎟ x o = x i
2
L
ω
ρ
⎝ n
⎠
xi
1
(D) = 2
D
xo
+1
2
ωn
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xi
1
1
=
( iω ) =
2
2
xo
⎛ iω ⎞
⎛ ω⎞
⎜ ω ⎟ +1 1− ⎜ ω ⎟
⎝ n⎠
⎝ n⎠
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xo
( iω) = 1.05 =
xi
1
⎛ ωmax ⎞
1− ⎜
⎟
ω
⎝ n ⎠
ωmax = 0.218ωn =
2
0.308 E
L
ρ
96200 cycles/min
for a 6-inch
steel rod
– ωmax is the highest frequency for which the real
body behaves almost like an ideal rigid body.
• Frequency response is unmatched as a technique
for defining the useful range of application for all
kinds of dynamic systems.
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Motion Transformers
• Mechanical systems often include mechanisms
such as levers, gears, linkages, cams, chains, and
belts.
• They all serve a common basic function, the
transformation of the motion of an input member
into the kinematically-related motion of an output
member.
• The actual system may be simplified in many
cases to a fictitious but dynamically equivalent
one.
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• This is accomplished by “referring” all the
elements (masses, springs, dampers) and driving
inputs to a single location, which could be the
input, the output, or some selected interior point of
the system.
• A single equation can then be written for this
equivalent system, rather than having to write
several equations for the actual system.
• This process is not necessary, but often speeds the
work and reduces errors.
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Motion Transformers
Gear Train Relations:
θm N 2
=
≡N
θ ′m N 1
Tm N 1 1
=
≡
Tm′ N 2 N
N1
Tm θ m
N2
Tm′ θ ′m
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Translational Equivalent
for
A Complex System
Refer all elements and
inputs to the x1 location
and define a fictitious
equivalent system
whose motion will be
the same as x1 but will
include all the effects
in the original system.
Mechanical System Modeling
x 1, x 2, θ
are
kinematically related
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61
– Define a single equivalent spring element
which will have the same effect as the three
actual springs.
– Mentally apply a static force f1 at location x1
and write a torque balance equation:
⎛ L2
⎞
x 1K s
f1L1 = ( K s1x1 ) L1 + ⎜ x1K s2 ⎟ L 2 +
L1
⎝ L1
⎠
f1 = K se x1
2
⎡
⎤
⎛ L2 ⎞
1
K se ⎢ K s1 + ⎜ ⎟ K s2 + 2 K s ⎥
L1 ⎥
⎝ L1 ⎠
⎣⎢
⎦
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– The equivalent spring constant Kse refers to a
fictitious spring which, if installed at location
x1, would have exactly the same effect as all the
springs together in the actual system.
– To find the equivalent damper, mentally
remove the inertias and springs and again apply
a force f1 at x1: f1L1 = ( x 1B1 ) L1 + ( x 2 B2 ) L 2 + Bθ
L22
x 1
= x1B1L1 +
x1B 2 + B
L1
L1
f1 = Be x 1
2
⎡
⎛ L2 ⎞
1 ⎤
Be ⎢ B1 + ⎜ ⎟ B2 + 2 B⎥
L1 ⎥
⎢⎣
⎝ L1 ⎠
⎦
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63
– Finally, consider only the inertias present.
1
x1
x1
2 x
f1L1 ≈ ( M L ) + ( M 2 L 2 ) + ( J )
L1
L1
L1
f1 ≈ M e x1
2
1 1
2
⎡
⎛ L2 ⎞
1 ⎤
M e ⎢ M1 + ⎜ ⎟ M 2 + 2 J ⎥
L1 ⎥
⎝ L1 ⎠
⎣⎢
⎦
– While the definitions of equivalent spring and
damping constants are approximate due to the
assumption of small motions, the equivalent
mass has an additional assumption which may
be less accurate; we have treated the masses as
point masses, i.e., J = ML2.
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– To refer the driving inputs to the x1 location we
note that a torque T is equivalent to a force T/L1
at the x1 location, and a force f2 is equivalent to
a force (L2/L1)f2.
– If we set up the differential equation of motion
for this system and solve for its unknown x1,
we are guaranteed that this solution will be
identical to that for x1 in the actual system.
– Once we have x1, we can get x2 and/or θ
immediately since they are related to x1 by
simple proportions.
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– Rules for calculating the equivalent elements
without deriving them from scratch:
• When referring a translational element (spring,
damper, mass) from location A to location B, where
A’s motion is N times B’s, multiply the element’s
value by N2. This is also true for rotational elements
coupled by motion transformers such as gears, belts,
and chains.
• When referring a rotational element to a
translational location, multiply the rotational
element by 1/R2, where the relation between
translation x and rotation θ (in radians) is x = R θ.
For the reverse procedure (referring a translational
element to a rotational location) multiply the
translational element by R2.
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• When referring a force at A to get an equivalent
force at B, multiply by N (holds for torques).
Multiply a torque at θ by 1/R to refer it to x as a
force. A force at x is multiplied by R to refer it as a
torque to θ.
– These rules apply to any mechanism, no matter
what its form, as long as the motions at the two
locations are linearly related.
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Mechanical Impedance
• When trying to predict the behavior of an
assemblage of subsystems from their calculated or
measured individual behavior, impedance methods
have advantages.
• Mechanical impedance is defined as the transfer
function (either operational or sinusoidal) in which
force is the numerator and velocity the
denominator. The inverse of impedance is called
mobility.
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Mechanical Impedance for the Basic Elements
Ks
f
ZS ( D ) ( D ) =
v
D
f
ZB ( D ) ( D ) = B
v
f
ZM ( D ) ( D ) = MD
v
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69
• Measurement of impedances of subsystems can be
used to analytically predict the behavior of the
complete system formed when the subsystems are
connected. We can thus discover and correct
potential design problems before the subsystems
are actually connected.
• Impedance methods also provide “shortcut”
analysis techniques.
– When two elements carry the same force they are said
to be connected in parallel and their combined
impedance is the product of the individual impedances
over their sum.
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– For impedances which have the same velocity, we say
they are connected in series and their combined
impedance is the sum of the individual ones.
– Consider the following systems:
B
K
B
x 1, v 1
K
f, v
Series Connection
f, v
Parallel Connection
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71
– Parallel Connection
K
B
f
KB
D
=
(D) = K
v
+ B BD + K
D
– Series Connection
f
K BD + K
(D) = B + =
v
D
D
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72
Force and Motion Sources
• The ultimate driving agency of any mechanical
system is always a force not a motion; force causes
acceleration, acceleration does not cause force.
• Motion does not occur without a force occurring
first.
• At the input of a system, what is known, force or
motion? If motion is known, then this motion was
caused by some (perhaps unknown) force and
postulating a problem with a motion input is
acceptable.
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• There are only two classes of forces:
– Forces associated with physical contact between two
bodies
– Action-at-a-distance forces, i.e., gravitational, magnetic,
and electrostatic forces.
• There are no other kinds of forces! (Inertia force is a
fictitious force.)
• The choice of an input form to be applied to a system
requires careful consideration, just as the choice of a
suitable model to represent a component or system.
• Here are some examples of force and motion sources.
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Force and Motion Inputs
acting on a
Multistory Building
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A Mechanical Vibration
Shaker:
Rotating Unbalance
as a
Force Input
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Electrodynamic Vibration Shaker as a Force Source
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Force Source
Constructed from a
Motion Source
and a
Soft Spring
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• Energy Considerations
– A system can be caused to respond only by the source
supplying some energy to it; an interchange of energy
must occur between source and system.
– If we postulate a force source, there will be an
associated motion occurring at the force input point.
– The instantaneous power being transmitted through this
energy port is the product of instantaneous force and
velocity.
– If the force applied by the source and the velocity
caused by it are in the same direction, power is supplied
by the source to the system. If force and velocity are
opposed, the system is returning power to the source.
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– The concept of mechanical impedance is of some help
here.
– The transfer function relating force and velocity at the
input port of a system is called the driving-point
impedance Zdp.
f
Zdp (D) = (D)
v
f
Zdp (iω) = (iω)
v
– We can write an expression for power:
f
f2
P = fv = f
=
Zdp Zdp
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– If we apply a force source to a system with a high value
of driving-point impedance, not much power will be
taken from the source, since the force produces only a
small velocity. The extreme case of this would the
application of a force to a perfectly rigid wall (drivingpoint impedance is infinite, since no motion is produced
no matter how large a force is applied). In this case the
source would not supply any energy.
– The higher the driving-point impedance, the more a real
force source behaves like an ideal force source.
– The lower the driving-point impedance, the more a real
motion source behaves like an ideal motion source.
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– Real sources may be described accurately as
combinations of ideal sources and an output impedance
characteristic of the physical device.
– A complete description of the situation thus requires
knowledge of two impedances:
• The output impedance of the real source
• The driving-point impedance of the driven system
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Mechanical System Examples
Rack-and-Pinion Gear System
Problem Statement
Develop the equivalent rotational
model of the rack-and-pinion gear
system shown. The applied torque T is
the input variable, and the angular
displacement θ is the output variable.
Neglect any twist in the shaft.
Bearings are frictionless. The pinion
gear mass moment of inertia about its
CG (geometric center) is Ip.
(I
Mechanical System Modeling
2 2
2
+
I
+
I
+
m
R
θ
+
cR
θ
+
kR
θ=T
)
m
s
p
r
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Multi-Gear System
Problem Statement
A load inertia I5 is driven through a
double-gear pair by a motor with inertia
I4, as shown. The shaft inertias are
negligible. The gear inertias are I1, I2,
and I3. The speed ratios are ω1/ω2 = 2
and ω2/ω3 = 5. The motor torque is T1
and the viscous damping coefficient c =
4 lb-ft-sec/rad. Neglect elasticity in the
system, and use the following inertia
values (sec2-ft-lb/rad): I1 = 0.1, I2 = 0.2,
I3 = 0.4, I4 = 0.3, I5 = 0.7. Derive the
mathematical model for the motor shaft
speed ω1 with T1 as the input.
2
2
2
⎧⎪
⎡
⎤ ⎛ 1 ⎞ 2 ⎫⎪
1⎞ ⎛1⎞
⎛1⎞
⎛
⎨( I 4 + I1 ) + ⎢( I5 + I3 ) ⎜ ⎟ + I 2 ⎥ ⎜ ⎟ ⎬ ω 1 + c ⎜ ⎟ ⎜ ⎟ ω1 = T1
⎝5⎠
⎝5⎠ ⎝ 2⎠
⎪⎩
⎣
⎦ ⎝ 2 ⎠ ⎪⎭
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Dynamic Vibration Absorber
Problem Statement
A dynamic vibration absorber consists of
a mass and an elastic element that is
attached to another mass in order to
reduce its vibration. The figure is a
representation of a vibration absorber
attached to the cantilever support. For a
cantilever beam with a force at its end, k
= Ewh3/4L3 where L = beam length, w =
beam width, and h = beam thickness. (a)
Obtain the equation of motion for the
system. The force f is a specified force
acting on the mass m, and is due to the
rotating unbalance of the motor. The
displacements x and x2 are measured
from the static equilibrium positions
when f = 0. (b) Obtain the transfer
functions x/f and x2/f.
Mechanical System Modeling
Physical Model
Physical System
m2 D2 + k 2
x
=
F mm 2 D 4 + [ m 2 ( k + k 2 ) + mk 2 ] D 2 + kk 2
x2
k2
=
F mm 2 D 4 + [ m 2 ( k + k 2 ) + mk 2 ] D 2 + kk 2
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Rigid Body Dynamics: Kinematics
y1
R1
y
P
O
R
x1
A
R
R
z1
z
Reference Frames
R - Ground xyz
R1 - Body x1y1z1
x
R
G
G
G
v A + ( R ωR1 × r AP ) +
R1
G
vP
G
G
G
G
G
G
G
a P = R a A + ⎣⎡ R ωR1 × ( R ωR1 × r AP ) ⎦⎤ + ⎡⎣ R α R1 × r AP ⎤⎦
R1 G P
R1 G P
R G R1
+ a + 2 ⎣⎡ ω × v ⎦⎤
G
q
Note: For any vector
Mechanical System Modeling
G
vP =
G R1 G
dq
dq R G R1 G
=
+ ω × q
dt
dt
R
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86
Rigid-Body Kinematics Example
G R1
ˆ = constant
ω
=
5i
Given:
R1 G R 2
ω = 4kˆ 1 = constant
Find: R aG P
R
R
r = 0.06 m
R1
O
⎡ ˆi1 ⎤ ⎡1
0
0 ⎤ ⎡ ˆi ⎤
⎢ ⎥ ⎢
⎢ ⎥
⎥
⎢ ˆj1 ⎥ = ⎢0 cos α sin α ⎥ ⎢ ˆj ⎥
⎢ kˆ ⎥ ⎢0 − sin α cos α ⎥ ⎢ kˆ ⎥
⎦⎣ ⎦
⎣ 1⎦ ⎣
y1
y1
y2
α
y
x2
φ
x1
R2
z
θ = 30º
Reference Frames:
R → ground: xyz
R1 → shaft: x1y1z1
R2 → disk: x2y2z2
O
O
Mechanical System Modeling
z1
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87
R
G
G
G
a P = R a O + ⎡⎣ R ωR 2 ×
(
R
G
G
G
G
ωR 2 × r OP ) ⎤⎦ + ⎡⎣ R α R 2 × r OP ⎤⎦
R2 G P
R2 G P
R G R2
+ a + 2 ⎡⎣ ω × v ⎤⎦
GO
a = 0 Point O at end of rotating shaft fixed in R
R2 G P
a =0
Point P fixed in R2 (disk)
R2 G P
v =0
G
G
R G R2
ω = R ωR1 + R1 ωR 2 = 5iˆ + 4kˆ 1
G OP
r = ( r cos θ ) ˆi1 + ( r sin θ ) ˆj1
R R G R2
R
d
d ˆ
ω
G
R R2
⎡⎣5i + 4kˆ 1 ⎤⎦
α =
=
dt
dt
G
R
dk1
R G R1
= 0+4
= 4 ω × kˆ 1
dt
= 4 5iˆ1 × kˆ 1 = −20ˆj1
R
(
)
(
(
)
)
(
= −20 ˆjcos α + kˆ sin α
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88
After Substitution and Simplification:
R
G
a P = ( −16r cos θ ) ˆi1 + ( −41r sin θ ) ˆj1 + ( 40r cos θ ) kˆ 1
Alternate Solution:
R
G
G
G
a P = R a O + ⎡⎣ R ωR1 ×
GO
a =0
R G R1
ω = 5iˆ = constant
R R G R1
d ω
R G R1
α =
=0
dt
R
Mechanical System Modeling
(
R
G
G
G
G
ωR1 × r OP ) ⎤⎦ + ⎡⎣ R α R1 × r OP ⎤⎦
R1 G P
R1 G P
R G R1
+ a + 2 ⎡⎣ ω × v ⎤⎦
G OP
r = ( r cos θ ) ˆi1 + ( r sin θ ) ˆj1
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89
R1
GP
a =
R1
GO
R1 G R 2
a + ⎣⎡ ω ×
GO
a =0
R1 G R 2
ω = 4kˆ 1
R1
(
R1
G R 2 G OP
G OP
R1 G R 2
ω × r ) ⎦⎤ + ⎡⎣ α × r ⎤⎦
(P is fixed in R2)
G R 2 R1
d ω
d ˆ
R1 G R 2
⎡⎣ 4k1 ⎤⎦ = 0
α =
=
dt
dt
G OP
R1 G P
R1 G O
R1 G R 2
v = v +( ω × r )
R1 G O
v =0
R1
R1
G OP
r = ( r cos θ ) ˆi1 + ( r sin θ ) ˆj1
After Substitution and Simplification:
R
G
a P = ( −16r cos θ ) ˆi1 + ( −41r sin θ ) ˆj1 + ( 40r cos θ ) kˆ 1
(same result)
Mechanical System Modeling
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90
Rigid Body Dynamics: Kinetics
Point C: mass center of a rigid body of mass m.
G
Linear Momentum L = m R vG C
Angular Momentum about point C
G
H = H x1 ˆi1 + H y1 ˆj1 + H z1 kˆ 1
y
⎡ H x1 ⎤ ⎡ Ix1x1
⎢
⎥ ⎢
⎢ H y1 ⎥ = ⎢ Iy1x1
⎢H ⎥ ⎢ I
⎣ z1 ⎦ ⎣ z1x1
Ix1y1
Iy1y1
Iz1y1
Ix1z1 ⎤ ⎡ ω ⎤
⎥ ⎢ R R1 ⎥
Iy1z1 ⎥ ⎢ ωy1 ⎥
Iz1z1 ⎥⎦ ⎢⎣ R ωzR11 ⎥⎦
Equations of Motion
Mechanical System Modeling
R
R1
x1
R R GC
G
d v
∑F = m
dt
G
R
G
dH
∑M =
dt
y1
y1 C
R1
A
x1
’
z1
z
O
R
x
Reference Frames
R - Ground xyz
R1 - Body x1y1z1
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