Mechanical Systems II
System Analysis Spring 2014
2
Maximum vehicle acceleration
Rear wheel drive
System Analysis Spring 2014
4
systems with Gears
2 1 1
1 2 2
r N r N
1 2 1
2 1 2
T N
T N
Torque relationship?
Represent the system with gears
without gears.
2 22 1
1
( ) ( ) ( ) N
Js Ds K s T s
N
System Analysis Spring 2014
6
Solution of Differential Equation by Laplace Transformation
2
2
2 4 1 (0) 0 , (0) 2
L.T: ( ) (0) (0) 2 ( ) (0) 4 ( ) 1
1 2 1
( 2 4) ( ) 2
y y y y y
s Y s sy y sY s y Y s
s s s Y s s
s s
22 2
2 1 1 1 1 1
( )
( 2 4) 4 4 ( 1) 3
1 1 1
( ) cos 3 sin 3
4 4 4 3
t t
s s
Y s s s s s s
y t e
t e
t
System Analysis Spring 2014
8
Mechanical work : W = F x [N m] = [Joule]
= Force displacement
Energy : capacity or ability to do work. Electrical, Chemical, Mechanical, etc.
Mechanical energy : Potential energy – position Kinetic energy – velocity
Work , Energy, and Power
1 1
2
0 0
1 2
x x
dw F dx kx dx dw kx dx kx F kx
M Mg
X=0
h
E
1 mgx (Potential Energy) ex1)
ex2)
Potential Energy
System Analysis Spring 2014
10
1
22 mv
1
22 J
m v
:J :
Kinetic Energy
System + External Work = E
2Energy E
1W
Work and Energy
1 2
E W E
2 2 2
1
2
1 2
1 mv Fl mv
2 2 2
1
2
1 2
1 mv Fl mv
F m v
1m v
2m v
1m v
21 2
E W E
l
System Analysis Spring 2014
12
sec dw Nm
P Watt
dt
Ex) 2000 kg
V = 72km/h = 20m/s (in 10sec)
2 2
0
1 1
2 mv W 2 mv
1
2(2000)(20) 400 1000 400 [ ] 2
400 1000 10 sec
40 1000 / 40 1000 40
1 745.7 54
W Nm kNm kJ
W Nm
P t
Nm s W W
hp W P hp
V
0= 0
Power dissipated in a damper Power : time rate & doing work
Power
= 2
P Fv bx x bx
2 2 2
1 1 1
(2 )
2 2
w2
wKE mv m v I
Energy Method for Deriving Equivalent Mass and Inertia
Kinetic Energy of the total System
Kinetic Energy represented with a single variable
2 2
1 2 2
2
w w
KE m m I v
R
Equation of motion using equivalent mass
2 2 I
wm m m
System Analysis Spring 2014
14
Conservative system : No energy dissipation
Kinetic Energy T
Potential Energy U Δ (T+U) = Δ W
(the change in the total energy)
= (the net work done on the system by the external force)
no external force ; Δ W = 0
Δ (T+U) = 0 T+U = constant
1 2
2 1
E W E
E E W
Energy Method for Deriving Equations of Motion
x(t) k
m
ex1)
2 22 2
1 1
, ,
2 2
1 1
2 2 0
T mx U kx T U C
mx kx
( ) 0, ( ) 0
d T U mxx kxx mx kx x dt
0, ( ) 0
x mx kx
x(t) k
m
ex2) x 1
21
22 , 2
T mx U kx
( )
2, 0
( ) 0
d T U bx mxx kxx bxx dt
mx kx bx x
0, ( ) 0
x mx kx bx
Examples of Energy Method
System Analysis Spring 2014
16
x x l
ms
x l
d
x(t) k
m l
0
kx x m
m k
n
2
2 1kx U
2
2 1
mx T
Potential E Kinetic E
)
22 (
1 x
x l x l m d
dT
s
l x s
x l
d x x
m l dT
0
2 2 3
0
( )
1 2
1
2 3
3
1 1 1
( )
2 m x
s( ) 3 l x l x
)
23 ( 1 2
1 m
sx
2 2
2
)
3 ( 1 2 ) 1
3 ( 1 2 1 2
1 m x m x m m x
T
s
s ) 0
3
( m 1 m
sx kx
s n
m m
k
3
1
ex3)
Examples of Energy Method
Spring with mass
End of Lecture 3
System Analysis Spring 2014
18
To be revisited: Transfer function for systems with Gears
Impedances are reflected from the output to the input, therby eliminating the gears.
2 1 2
1 1
2 1
( ) N ( ) ( ) N
Js Ds K s T s
N N
2 2 2
1 2 1 1
1 1
2 2 2
( N N N ) ( ) ( )
J s D s K s T s
N N N
Rotational mechanical impedances can be reflected through gear trains by multiplying the mechanical impedance by the ratio,
2 2 2
2 1
1 2
N N
Number of teeth of gear on destination shaft
Number of teeth of gear on source shaft
