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Lecture 2
Read textbook CHAPTER 1.4, Apps B&D
Today:
Derive EOMs & Linearization
Fundamental equation of motion for mass-springdamper system (1DOF). Linear and nonlinear
system. Examples of derivation of EOMs
Appendix A Equivalence of principles of conservation
of mechanical energy and conservation of linear
momentum.
Appendix B: Linearization
Work problems:
Chapter 1: 5, 8, 13, 14, 15, 20, 43, 44, 56
1
Lecture 2
ME617
Kinetics of 1-DOF
mechanical systems
The fundamental elements in a
mechanical system and the
process to set a coordinate
system and derive an equation
of motion.
LINEARIZATION included.
Luis San Andres
Texas A&M University
M
K
C
2014 Luis San Andres©
2
A system with an elastic element
(linear spring)
Linear elastic element (spring-like)
F
Free length
l
F
F
ds
static deflection
Applied force on spring
K=dF/dd
F
deflection
d
ds
Notes:
a) Spring element is regarded as massless
b) Applied force is STATIC
c) Spring reacts with a force proportional to
deflection, Fs = -Kds, and stores potential
energy
Balance of static forces
Fs =-K ds
F
Fs
d) K = stiffness coefficient [N/m or lbf/in] is constant
4
Statics of system with elastic &
mass elements
Linear spring + added weight
W
Free length
l
F
W
ds
static
deflection
Reaction force from spring
K=dF/dd
W
static deflection
ds
Notes:
a) Block has weight W= Mg
b) Block is regarded as a point mass
Balance of static forces
W=Fs = K ds
d
W
Fs
6
Dynamics of system with elastic
& mass elements
Derive the equation of motion (EOM) for the system
Linear spring + weight (mass x g ) + force F(t)
F
F(t)
W
Free
length
l
Reaction force from spring
W+F(t)
W+F
ds
K=dF/dd
SEP
dynamic deflection
ds+X(t)
Notes:
a) Coordinate X describing motion has
origin at Static Equilibrium Position
(SEP)
b) For free body diagram, assume state
of motion, for example X(t)>0
c) Then, state Newton’s equation of
motion
d) Assume no lateral (side motions)
d
X+ds
Free Body diagram
W+F
M
acceleration
M Ax = W+F - Fspring
Fspring
8
Derive the equation of motion
W+F
M Ax = W+F - Fspring
ds
Free
length
SEP
l
ds+X(t)
FBD: X>0
W+F
X(t)
Fspring = K(X+ds )
M Ax = W+F – K(X+ds)
M Ax = ( W-Kds) +F – KX
Cancel terms from force balance at SEP to get
M Ax = +F – KX
M
Fspring
M Ax + K X= F(t)
d2 X
AX 
X
2
dt
Note: Motions from SEP
9
Another choice of coordinate system
Coordinate y has origin at free length of spring element.
W+F
M Ay = W+F - Fspring
Free
length
l
y(t)
Fspring = Ky
M Ay = W+F – K y
FBD: y>0
Weight (static force) remains in equation
W+F
M
Fspring
M Ay + K y= W + F(t)
d2 y
Ay  2  y
dt
10
EOM in two coordinate systems
W+F
Free
length
l
M Ay + K y = W + F(t)
ds
SEP
y(t) X(t)
y = X+ ds
yX
(1)
d2 y
Ay  2  y
dt
M Ax + K X= F(t)
(2)
d2 X
AX 
X
2
dt
Fspring = Ky = K(X+ ds )
Coordinate y has origin at free
length of spring element. X
has origin at SEP.
(?) Are Eqs. (1) and (2) the same?
(?) Do Eqs. (1) and (2) represent
the same system?
(?) Does the weight disappear?
(?) Can I just cancel the weight?
11
K-M system with dissipative
element (a viscous dashpot)
Derive the equation of motion (EOM) for the system
Recall: a linear elastic element
Fs
Free length
l
F
Fs
ds
static deflection
Applied force on spring
K=dF/dd
Fs
deflection
ds
d
Notes:
a) Spring element is regarded as massless
b) K = stiffness coefficient is constant
c) Spring reacts with a force proportional to deflection
and stores potential energy
13
(Linear) viscous damper element
F
F
Applied force on damper
F
V
velocity
F
D=dF/dV
velocity
V
Notes:
a) Dashpot element (damper) is regarded as massless
b) D = damping coefficient [N/(m/s) or lbf/(in/s) ] is constant
c) A damper needs velocity to work; otherwise it can not dissipate mechanical
energy
d) Under static conditions (no motion), a damper does NOT react with a force
14
Spring + Damper + Added weight
W
Free length
l
W
static
deflection
F
Reaction force from spring
ds
SEP
K=dF/dd
W
static deflection
ds
Notes:
a) Block has weight W= Mg and is regarded
as a point mass.
b) Weight applied very slowly – static
condition.
c) Damper does NOT react to a static
action, i.e. FD=0
SEP: static equilibrium position
Balance of static forces
W=Fs = K ds
d
W
Fs
15
Spring + Damper + Weight + force F(t)
F
F(t)
W
L-ds
W+F
ds
Reaction force from spring
Fspring
K=dF/dd
SEP
X(t)
dynamic deflection
X+ds
Notes:
a) Coordinate X describes motion from
Static Equilibrium Position (SEP)
b) For free body diagram, assume a
state of motion X(t)>0
c) State Newton’s equation of motion
d) Assumed no lateral (side motions)
d
W+F
Free Body diagram
M
acceleration
Fspring
M Ax = W+F - Fdamper - Fspring
Fdamper
16
Equation of motion: spring-damper-mass
W+F
M Ax = W+F(t) - Fdamper - Fspring
SEP
L-ds
X(t)
K
D
d2 X
AX 
X
2
dt
dX
VX 
X
dt
Fdamper = D VX
Fspring = K(X+ ds )
M Ax = W+F – K(X+ds) – D VX
FBD: X>0
W+F
M
Fspring
Fdamper
M Ax = ( W-Kds) +F – KX - DVX
Cancel terms from force balance at SEP to get
M Ax = +F – KX – D VX
M Ax + DVX + K X = F(t)
EOM:
Notes: Motions from SEP
M X  D X  K X  F(t )
17
Simple nonlinear mechanical
system
Derive the equation of motion (EOM) for the system
and linearize EOM about SEP
Recall a linear spring element
Fs
Free length
l
F
Fs
ds
static deflection
Applied force on spring
K=dF/dd
Fs
deflection
ds
d
Notes:
a) Spring element is regarded as massless
b) K = stiffness coefficient is constant
c) Spring reacts with a force proportional to
deflection and stores potential energy
19
A nonlinear spring element
F
Free length
l
F
F
ds
static deflection
Applied force on spring
F
d
ds
static deflection
F  K1 d  K 3 d 3
Notes:
a) Spring element is massless
b) Force vs. deflection curve is NON linear
c) K1 and K3 are material parameters [N/m, N/m3]
Linear spring + added weight
W
Free length
l
F
W
ds
static
deflection
Reaction force from spring
W
d
ds
static deflection
Notes:
a) Block has weight W= Mg
b) Block is regarded as a point
mass
c) Static deflection ds found from
solving nonlinear equation:
Balance of static forces
W  Fspring  K1 d s  K3 d s3
Fspring
W
Nonlinear spring + weight + force F(t)
F
F(t)
W
Free length
l
Reaction force from spring
W+F
W+Fs
SEP
ds+X(t)
deflection
d
X+ds
Notes:
a) Coordinate X describing motion has
origin at Static Equilibrium Position
(SEP)
b) For free body diagram, assume state
of motion, for example X(t)>0
c) Then, state Newton’s equation of
motion
d) Assume no lateral (side motions)
Free Body diagram
W+F
M
acceleration
M Ax = W+F - Fspring
Fspring
Nonlinear Equation of motion
W+F
Free
length
M X  F(t )  W  Fspring
SEP
ds+X(t)
l
d2 X
AX 
X
d t2
Fspring  K1 d s  X   K3 d s  X 
M X  F(t )  W  K1 d s  X   K3 d s  X 
3
3
Expand RHS:
FBD: X>0
W+F
M
Fspring
M X  F(t )  W  K1 d s  X   K 3 d s3  3d s2 X  3d s X 2  X 3 
M X  F(t )  W  K1 d s  K3d s3    K 1 3K3d s2  X  K3  3d s X 2  X 3 
Cancel terms from force balance at SEP to get
M X   K 1 3K3d s2  X  K3  3d s X 2  X 3   F(t )
Notes: Motions are from SEP
Linearize equation of motion
W+F
SEP
Free
length
EOM is nonlinear:
M X   K 1 3K3d s2  X  K3  3d s X 2  X 3   F(t )
ds+X(t)
l
Assume motions X(t) << ds
which means |F(t)| << W
and set X2 ~0, X3~0
FBD: X>0
W+F
M
Fspring
to obtain the linear EOM:
M X  Ke X  F(t )
where the linearized stiffness
K e   K 1 3K 3d s2 
is a function of the static condition
3
 dFspring  d  K1 d  K3 d 
Ke  
 K1  3K3 d s2

dd
 dd 
d
s
Linear equation of motion
W+F
M X  Ke X  F(t )
F
K e   K 1 3K 3d s2 
SEP
Free
length
ds+X(t)
l
W
deflection
ds
FBD: X>0
W+F
M
Fspring
d
X
3
 dFspring  d  K1 d  K3 d 
Ke  
 K1  3K3 d s2

dd
 dd 
d
Stiffness
(linear)
s
Ke
deflection
ds
d
Read & rework
Examples of derivation of EOMS for physical
systems available on class URL site(s)
2014 Luis San Andres©