powerpoint presentationems.guc.edu.eg/download.ashx?id=175&file=control_engineering_tut… ·...
TRANSCRIPT
Control Engineering
Tutorial #2
Dynamics of Mechanical
Systems
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
Tutorial #2 Dynamics of Mechanical Systems
As far of now, we know how to model mechanical systems and get
the equations of motion.
System equation
of motion
Modeling
Mathematical
ModelPhysical Dynamic
system
In this course, we are interested in
• Linear time invariant systems.
• The equations of motion are ODE’s.
• If not linear, we linearize.
Free response of undamped systems
Undamped system “spring mass system”
System equation of motion
For this system, we define
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
1Problem
Plot the solution of a linear spring-mass system with frequency ωn = 2 rad/s,
x0 = 1 mm, and v0= 2.34 mm/s, for at least two periods.
Solution:
• The equation of motion of this system can be written as
0
0
0
2
xx
xm
kx
kxxm
n
Remember
• Substituting with the given natural frequency, we get the
following DE.
04 xx
The given initial conditions are
smmx
mmx
/34.2)0(
1)0(
Remember that the equation of
motion, together with the initial values,
is called “the initial value problem”.
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
04 xx smmx
mmx
/34.2)0(
1)0(
• To solve this initial value problem, we assume the solution
taetx )(
• Plugging this solution into the DE, we get
2
04
04
2,1
2
2
j
aeea tt
DE Characteristic equation
tjtj eaeatx 2
2
2
1)( DE General Solution
• From the lecture, this reduces to
)2sin()2cos()(21
tAtAtx
• Constants A1 and A2 are obtained from the initial conditions
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
smmx
mmx
/34.2)0(
1)0(
1
21
21
1
)0sin()0cos(1
)2sin()2cos()(
A
AA
tAtAtx
2
2
21
17.1
)0cos(2)0sin(234.2
)2cos(2)2sin(2)(
A
A
tAtAtx
Final Solution
)2sin(17.1)2cos()( tttx
Nice solution, but we still can simplify it more…
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
)2sin(17.1)2cos()( tttx
• From the lecture, we have
• So, the final answer can be written as;
)71.02sin(54.1)( ttx
04 xx
smmx
mmx
/34.2)0(
1)0(
)71.02sin(54.1)( ttx
The problem The solution
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
Free response of undamped systems (The General case)
• For any undamped system, we have
Free response solution
Where
Free response plot
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
• For the given problem, the final answer can be obtained directly from the
general solution
04 xxsmmx
mmx
/34.2)0(
1)0(
2n
)71.02sin(54.1)( ttx
Substituting by the given values, we get
A = 1.54 mm and Φ= 40 deg = 0.71 rad.
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
)7.02sin(54.1)( ttx
Finally, the solution is given by
The time period of this solution is T = 2π/ωn = 3.14 sec. i.e. the solution will be
repeated each 3.14 sec.
Plotting this function using MATLAB, we use the following code.
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
The resulting plot:
By examining the
resulting plot, we
find that it starts
from x = 1 mm
which is the initial
displacement
given, and the
slope at this point
is equal to the
initial velocity.
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
Free response of damped systems
Damped system “spring-mass-damper system”
System equation of motion
For this system we define
Undamped Natural Frequency
Damping Ratio
nm
c
2
02
0
2
xxx
xm
kx
m
cx
nn
Damping Ratio
The free response of the system is determined depending on the value of
the damping ratio
nm
c
2
1
1
1
Underdamped system, which means that the system will
oscillate until it reaches the equilibrium position
Critically damped system, which means that the system
will return to the equilibrium position without oscillation
Overdamped system, which means that the system will
return to the equilibrium position without oscillation
Note that the critically damped system is faster than overdamped system when
it returns back to the equilibrium position.
For this system, we define wd
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
2Problem
Solve the following initial value problems.
0)0(,1)0(,035 xxxxx
1)0(,3)0(,025.1 yyyyy
4)0(,1)0(,025.23 wwwww
Solution:
Before we solve, we find the natural frequency and the damping ratio for
each system just to relate the behavior of the response to the damping ratio
1,25.23
447.0,25.12
44.1,31
n
n
n
system
system
system
Underdamped system
Critically damped
Overdamped system
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
0)0(,1)0(,035 xxxxx
• Assume the solution
taetx )(
• The characteristic equation
0352
• The roots of the Characteristic
equation are
3.4,7.0 21
tt eaeatx 3.4
2
7.0
1)(
• The General solution is
• Invoking the initial conditions
gives the final solution as
tt eetx 3.47.0 2.02.1)(
)(tx
t
tt eetx 3.47.0 2.02.1)(
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
1)0(,3)0(,025.1 yyyyy
• Assume the solution
taety )(
• The characteristic equation
025.12
• The roots of the Characteristic
equation are
jj 5.0,5.0 21
)()( 21
5.0 tjjtt eaeaety
• The General solution is
• As in problem 1, this solution
can be recast in the form
))sin()cos(()( 21
5.0 tAtAety t
• Invoking the initial conditions
gives the final solution as
))sin(5.2)cos(3()( 5.0 ttety t
)(ty
t
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
4)0(,1)0(,025.23 wwwww
• Assume the solution
taetw )(• The characteristic equation
025.232
• The roots of the Characteristic
equation are
5.1,5.1 21
tt teaeatw 5.1
2
5.1
1)(
• Since the roots are repeated
here, the General solution is
given by
• Invoking the initial conditions
gives the final solution as
tt teetw 5.15.1 5.2)(
)(tw
t
tt teetw 5.15.1 5.2)(
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
A spring-mass-damper system has mass of 150 kg, stiffness of 1500 N/m and
damping coefficient of 200 kg/s. Calculate the undamped natural frequency, the
damping ratio and the damped natural frequency. Is the system overdamped,
underdamped, or critically damped? Does the solution oscillate?
3Problem
sradm
kn /16.3
21.02
nm
c
sec/02.31 2 radnd
m = 150 kg, k = 1500 N/m, c = 200 kg/s
The undamped natural frequency
The damping ratio
The damped natural frequency
It is obvious from the damping ratio that the system is underdamped because
< 1, and consequently the system will oscillate.
Solution:
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
The system of problem 3 is given an initial velocity of 10 mm/s and an initial
displacement of -5 mm. find the form of the response and plot it for as long as it
takes to die out.
Solution:
4Problem
)154.102.3sin(47.5)( 6636.0 tetx t
Now, we evaluate A and Φ
A = 5.47 mm
and Φ = -66.13 deg = -1.154 rad.
Finally the solution has the form:
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
Now we plot the solution using MATLAB using the following code
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
System equation of motion
where
Here, the solution has the form
Forced Response of damped systems
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
5Problem
For the shown system, find the particular solution for the given inputs
)(25.1 tfyyy
Solution:
)Re(10)2cos(10)()
)Im(5)sin(5)()
2
2
1
ti
it
ettfb
ettfa
)sin(525.1
)(25.1 1
tyyy
tfyyy
Here, we define the operator D
2
22,
dt
dD
dt
dD
iteyDD
tyDD
5~)25.1(
)sin(5)25.1(
2
2
)(DP
a)
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
iteyDD 5~)25.1( 2
)(DP
25.1)( 2 iiP
25.1
5~2
ii
ey
it
p
)25.0()06.1(
5~
25.0
5~
ie
y
i
ey
it
p
it
p
)]sin())(cos(25.0[(72.4~ titiyp
)]sin()sin(25.0
)cos()cos(25.0[72.4~
tti
tityp
)Im(5)sin(5)(1
itettf
Since the input is given by
Then the Particular solution will be
)~Im( pp yy
)]sin(25.0)cos([72.4 ttyp
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
)2cos(1025.1
)(25.1 2
tyyy
tfyyy
tieyDD
tyDD
22
2
10~)25.1(
)2cos(10)25.1(
)(DP
b)
25.12)2()( 2 iiP
25.12)2(
10~2
2
ii
ey
ti
p
)275.2()56.11(
10~
275.2
10~
2
2
ie
y
i
ey
ti
p
ti
p
)]2sin()2)(cos(275.2[(87.0~ titiyp
)]2sin(2)2sin(75.2
)2cos(2)2cos(75.2[87.0~
tti
tityp
))Re(10)2cos(10)( 2
1
tiettf
Since the input is given by
Then the Particular solution will be
)~Re( pp yy
)]2sin(2)2cos(75.2[87.0 ttyp
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
6Problem
For the shown system, find its total solution
1)0(,3)0(),sin(525.1 yytyyy
Solution:
The total solution is simply the superposition of the homogeneous solution shown
in problem 2 and the particular solution shown in problem 5
))sin()cos(()( 21
5.0
shomogeneou tAtAety t
)]sin(25.0)cos([72.4)( ttty particular
)]sin(25.0)cos([72.4))sin()cos((
)()()(
21
5.0
shomogeneou
tttAtAe
tytyty
t
particular
Then, invoke the initial conditions to get the constants values A1 and A2.
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
A 100-kg mass is suspended by a spring of stiffness mN /1030 3
with a viscous-damping constant of 1000 N.s/m. The mass is initially
at rest and in equilibrium. Calculate the steady-state displacement
amplitude and phase if the mass is exited by a harmonic cosine force
of 80N at 3 Hz.
7Problem
mN /1030 3
sradfn /84.182
)84.18cos(80 tF
M =100 kg, k =
Mass is initially at rest and in equilibrium means that the initial
displacement and velocity are equal to zero.
Exciting force natural frequency
The exciting force is
To find the solution of the system we have first to know the type of damping
in the system.
, and c = 1000 N.s/m
Solution:
F
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
sradm
kn /32.17
288.02
nm
c
The undamped natural frequency
The damping ratio
The value of the damping ratio shows that the system is underdamped and
the solution is given by
Now we can calculate X, and Θ
X = 0.04 m Steady state displacement amplitude.
Θ = 1.856 rad Steady state phase.
GUC Faculty of Engineering and Material Science
Department of Mechatronics
Control Engineering ENME 503
Dr. Ayman Ali El-Badawy
Summary