Download steady state solution

Course Outline
1. MATLAB tutorial
2. Motion of systems that can be idealized as particles
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Description of motion; Newton’s laws;
Calculating forces required to induce prescribed motion;
Deriving and solving equations of motion
3. Conservation laws for systems of particles
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Work, power and energy;
Linear impulse and momentum
Angular momentum
Exam topics
4. Vibrations
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Characteristics of vibrations; vibration of free 1 DOF systems
Vibration of damped 1 DOF systems
Forced Vibrations
5. Motion of systems that can be idealized as rigid bodies
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Description of rotational motion; Euler’s laws; deriving and solving
equations of motion; motion of machines
Particle Dynamics: Concept Checklist
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Understand the concept of an ‘inertial frame’
Be able to idealize an engineering design as a set of particles, and know when this
idealization will give accurate results
Describe the motion of a system of particles (eg components in a fixed coordinate system;
components in a polar coordinate system, etc)
Be able to differentiate position vectors (with proper use of the chain rule!) to determine
velocity and acceleration; and be able to integrate acceleration or velocity to determine
position vector.
Be able to describe motion in normal-tangential and polar coordinates (eg be able to write
down vector components of velocity and acceleration in terms of speed, radius of curvature
of path, or coordinates in the cylindrical-polar system).
Be able to convert between Cartesian to normal-tangential or polar coordinate descriptions
of motion
Be able to draw a correct free body diagram showing forces acting on system idealized as
particles
Be able to write down Newton’s laws of motion in rectangular, normal-tangential, and polar
coordinate systems
Be able to obtain an additional moment balance equation for a rigid body moving without
rotation or rotating about a fixed axis at constant rate.
Be able to use Newton’s laws of motion to solve for unknown accelerations or forces in a
system of particles
Use Newton’s laws of motion to derive differential equations governing the motion of a
system of particles
Be able to re-write second order differential equations as a pair of first-order differential
equations in a form that MATLAB can solve
Particle Kinematics
Inertial frame – non accelerating, non rotating reference frame
Particle – point mass at some position in space
Position Vector
Velocity Vector
j
r(t )  x(t )i  y(t ) j  z(t )k
v(t) dt
r(t)
v(t )  vx (t )i  v y (t ) j  vz (t )k
d
dx
dy
dz
O
 xi  yj  zk   i  j  k
k
dt
dt
dt
dt
dx
dy
dz
 vx (t ) 
v y (t ) 
vz (t ) 
dt
dt
dt
path of particle
r(t+dt)

• Direction of velocity vector is parallel to path
• Magnitude of velocity vector is distance traveled / time
Acceleration Vector


dv y
dvx
dv
d
a(t )  ax (t )i  a y (t ) j  a z (t )k 
vx i  v y j  vz k 
i
j z k
dt
dt
dt
dt
dv y
dvx
dvz
d 2x
d2y
d 2z
 ax (t ) 

a y (t ) 

az (t ) 

2
2
dt
dt
dt
dt
dt
dt 2
i
Particle Kinematics
• Straight line motion with constant acceleration
1

r   X 0  V0t  at 2
2


i

v  V0  at  i
a  ai
• Time/velocity/position dependent acceleration – use
calculus
t


r   X 0   v(t )dt  i


0


v (t )

t


v   V0   a(t )dt  i


0


x (t )
vdv 
V0
dv g (t )
a


dt f (v)
dx g (t )
v


dt f (v)

a( x)dx
0
v

f (v)dv   g (t )dt
V0
x (t )

X0
t
0
t
f ( x)dv   v(t )dt
0
Particle Kinematics
• Circular Motion at const speed
sin 
t
cos
  t s  R V   R
r  R  cos  i  sin  j
R
j

v   R   sin  i  cos  j  Vt
i
V2
a   R(cos  i  sin  j)   Rn 
n
R
2
2
• General circular motion
  d / dt   d  / dt  d 2 / dt 2
s  R V  ds / dt  R
r  R  cos i  sin  j
v   R   sin  i  cos j  Vt
a  R ( sin  i  cos j)  R 2 (cos i  sin  j)
dV
V2
  Rt   Rn 
t
n
dt
R
2
Rcos
n
Rsin 
Particle Kinematics
• Arbitrary path
v  Vt
dV
V2
a
t
n
dt
R
t
R
r  x( )i  y( ) j
n
dx d 2 y dy d 2 x

d d2 d d2
t
1

R 
2 3/2
2
 dx 
 dy  

 
 
d



 d   

• Polar Coordinates
dr
d
v  er  r
e
dt
dt
2
 d 2r
 d 2
dr d 
 d  
a
 r
 e
  er   r 2  2

 dt 2
dt
dt
dt

 
 dt


e
r
j

i
er
Newton’s laws
• For a particle
F  ma
• For a rigid body in motion without rotation, or
a particle on a massless frame
Mc  0
j
You MUST take moments
about center of mass
i
NA
W
TB
NB
Calculating forces required to cause
prescribed motion of a particle
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Idealize system
Free body diagram
Kinematics
F=ma for each particle.
M c  0 (for rigid bodies or frames only)
Solve for unknown forces or accelerations
Deriving Equations of Motion for particles
1. Idealize system
2. Introduce variables to describe motion
(often x,y coords, but we will see other
examples)
3. Write down r, differentiate to get a
4. Draw FBD
5. F  ma
6. If necessary, eliminate reaction forces
7. Result will be differential equations for coords
defined in (2), e.g. m d 2 x   dx  kx  kY sin t
dt
2
dt
0
8. Identify initial conditions, and solve ODE
Motion of a projectile
r  X 0i  Y0 j  Z 0k


t  0
dr
 Vx 0i  V y 0 j  Vz 0k 
dt

V0
k
X0
j
i
1 2

r   X 0  Vx0t  i  Y0  V y 0t j   Z 0  Vz 0t  gt  k
2



 

v  Vx 0  i  V y 0 j  Vz 0  gt  k
a   gk
Rearranging differential equations for MATLAB
• Example
d2y
dt 2
 2n
dy
 n2 y  0
dt
• Introduce v  dy / dt
• Then
v

d  y 




v
dt    2n v  n2 y 
• This has form
dw
 f (t , w )
dt
 y
w 
v 
Conservation laws for particles: Concept Checklist
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Know the definitions of power (or rate of work) of a force, and work done by a force
Know the definition of kinetic energy of a particle
Understand power-work-kinetic energy relations for a particle
Be able to use work/power/kinetic energy to solve problems involving particle motion
Be able to distinguish between conservative and non-conservative forces
Be able to calculate the potential energy of a conservative force
Be able to calculate the force associated with a potential energy function
Know the work-energy relation for a system of particles; (energy conservation for a closed
system)
Use energy conservation to analyze motion of conservative systems of particles
Know the definition of the linear impulse of a force
Know the definition of linear momentum of a particle
Understand the impulse-momentum (and force-momentum) relations for a particle
Understand impulse-momentum relations for a system of particles (momentum
conservation for a closed system)
Be able to use impulse-momentum to analyze motion of particles and systems of particles
Know the definition of restitution coefficient for a collision
Predict changes in velocity of two colliding particles in 2D and 3D using momentum and the
restitution formula
Know the definition of angular impulse of a force
Know the definition of angular momentum of a particle
Understand the angular impulse-momentum relation
Be able to use angular momentum to solve central force problems
Work and Energy relations for particles
k
Rate of work done by a force
(power developed by force)
Total work done by a force
P  F v
W   F  vdt W 
0
Kinetic energy
 F  dr
r0

v
j
F(t)
r1
1
2 1
T  m v  m vx2  v2y  vz2
2
2
P
O
i
t1
F
P
r0 k
j
O
i
r1

v
Power-kinetic energy relation
r1
Work-kinetic energy relation
r0 k
dT
P
dt
W   F  dr  T  T0
r0
i
P
O
j
r1
Potential energy
Type of force
Potential energy
m
Potential energy of a
conservative force (pair)
Gravity acting on a
particle near earths
surface
F
j
V  mgy
y
i
r
V (r )    F  dr  constant
r0
F   grad(V )
P
r0 k
i
Gravitational force
exerted on mass m by
mass M at the origin
Force exerted by a
spring with stiffness k
and unstretched length
1
2
V  k  r  L0 
2
j
r
F
i
j
r1
m
r
GMm
V 
r
r
L0
F
O
F
2
Force acting between
two charged particles
V
Q1Q2
4 r
j
+Q1
i
r
1
Force exerted by one
molecule of a noble gas
(e.g. He, Ar, etc) on
another (Lennard Jones
potential). a is the
equilibrium spacing
between molecules, and
E is the energy of the
bond.
6
 a 12
a 
E    2   
 r  
 r 
F
+Q2
2
F
j
i
1
r
Energy relations for
conservative systems subjected to external forces
F1ext
Internal Forces: (forces exerted by
one part of the system on another) R ij
m4
F3ext
m1
R13 R31
R12
m
R32 3
R23
R21
m2
External Forces: (any other forces) Fiext
System is conservative if all internal forces are
conservative forces (or constraint forces)
F2ext
Energy relation for a conservative system t0  t  t0  t
Kinetic and potential energy at time
t0
T0  V0 
Kinetic and potential energy at time
t
T  V 
Work done by external forces
Wext 
t0 t
 
forces
Fiext (t )  v(t )dt
t0
Wext  T  V  T0  V0 
Impulse-momentum relations
Impulse-momentum for a single particle
t1
Linear Impulse of a force
F(t)
   F(t )dt
k
t0
Linear momentum of a particle
p  mv
Impulse-momentum relations
F
dp
dt
v
m
j
O
i
  p1  p 0
Impulse-momentum for a system of particles
Total external impulse
 
tot 
particles
Total linear momentum
F1ext
t0 t
ptot 
ext
i
F (t )dt
t0

mi vi

Fiext (t ) 
particles
Special case (closed system)
dptot
dt
F3ext
m1
R13 R31
R12
m
R32 3
R23
R21
m2
F2ext
particles
Impulse-momentum
m4
tot  ptot
tot  0  ptot  0 (momentum conserved)
Collisions
vB0
x
A0
vx
A
mAvxA1  mB vxB1  mAvxA0  mB vxB 0
Momentum
B
Restitution formula
*
vA1
x
vB1
x
B
A
A
v
A0
B
Momentum
B0
v
v
A1
v B1  v B 0 
mA
(1  e)  v B 0  v A0 
mA  mB
v A1  v A0 
mB
(1  e)  v B 0  v A0 
mA  mB
mB v B1  mA v A1  mB v B 0  mA v A0
B1
A1
B0
A0
B0
A0
Restitution formula  v  v    v  v   (1  e)  v  v   n  n
n
A
v B1  v A1  e  v B 0  v A0 
v A1  v A0 
B
B1
v
v B1  v B 0 
mB
(1  e)  v B 0  v A0   n  n
mB  mA
mA
(1  e)  v B 0  v A0   n  n
mB  mA
Angular Impulse-Momentum Equations for a Particle
k
Angular Impulse

t0 t

r (t )  F(t )dt
Angular Momentum
h  r  p  r  mv
Impulse-Momentum relations
Special Case
r(t)
O
t0
dh
M
dt
  0  h1  h0
x
i
F(t)
j
z
y
  h1  h0
Angular momentum conserved
Useful for central force problems
Free Vibrations – concept checklist
You should be able to:
1. Understand simple harmonic motion (amplitude, period, frequency,
phase)
2. Identify # DOF (and hence # vibration modes) for a system
3. Understand (qualitatively) meaning of ‘natural frequency’ and
‘Vibration mode’ of a system
4. Calculate natural frequency of a 1DOF system (linear and nonlinear)
5. Write the EOM for simple spring-mass-damper systems by inspection
6. Understand natural frequency, damped natural frequency, and
‘Damping factor’ for a dissipative 1DOF vibrating system
7. Know formulas for nat freq, damped nat freq and ‘damping factor’ for
spring-mass system in terms of k,m,c
8. Understand underdamped, critically damped, and overdamped motion
of a dissipative 1DOF vibrating system
9. Be able to determine damping factor and natural frequency from a
measured free vibration response
10. Be able to predict motion of a freely vibrating 1DOF system given its
initial velocity and position, and apply this to design-type problems
Vibrations and simple harmonic motion
Typical vibration response
• Period, frequency, angular frequency
amplitude
Displacement
or
Acceleration
Period, T
y(t)
Peak to Peak
Amplitude A
time
Simple Harmonic Motion
x(t )  X 0  X sin t   
v(t )  V cos(t   )
a (t )  A sin t   )
V  X
A  V
Vibration of 1DOF conservative systems
Harmonic Oscillator
Derive EOM (F=ma)
m d 2s
 s  L0
k dt 2
Compare with ‘standard’ differential equation
xs
1
n2
Solution
s(t )  L0  (s0  L0 )2  v02 / n2 sin(nt   )
Natural Frequency
k
n 
m

C  L0 x0  s0
m
k
Vibration modes and natural frequencies
•Vibration modes: special initial deflections that cause entire
system to vibrate harmonically
•Natural Frequencies are the corresponding vibration frequencies
x1
x2
k
k
m
k
m
Number of DOF (and vibration modes)
In 2D: # DOF = 2*# particles + 3*# rigid bodies - # constraints
In 3D: # DOF = 3*# particles + 6*# rigid bodies - # constraints
Expected # vibration modes = # DOF - # rigid body modes
A ‘rigid body mode’ is steady rotation or translation of the entire
system at constant speed. The maximum number of ‘rigid body’
modes (in 3D) is 6; in 2D it is 3. Usually only things like a vehicle
or a molecule, which can move around freely, have rigid body
modes.
x1
x2
k
k
m
k
m
Calculating nat freqs for 1DOF systems – the basics
m
y
k,L0
EOM for small vibration of any 1DOF
undamped system has form
1 d2y
 y C
2
2
n dt
n is the natural frequency
1. Get EOM (F=ma or energy)
2. Linearize (sometimes)
3. Arrange in standard form
4. Read off nat freq.
Tricks for calculating natural frequencies of 1DOF undamped systems
• Using energy conservation to find EOM
s
2
1  ds  1
KE  PE  m    k ( s  L0 ) 2  const
2  dt  2
k, d
m
2
d
ds
 ds  d s
 ( KE  PE )  m   2  k ( s  L0 )  0
dt
dt
 dt  dt
d 2s
 m 2  ks  kL0
dt
• Nat freq is related to static deflection
n 
g

k,L0
m
L0+ 
Linearizing EOM
d2y
 f ( y)  C
2
dt
Sometimes EOM has form
We cant solve this in general…
Instead, assume y is small
d2y
df
m 2  f (0) 
dt
dy
d 2 y df
m 2 
dt
dy
y  ...  C
y 0
y  C  f (0)
y 0
There are short-cuts to doing the Taylor expansion
Writing down EOM for spring-mass-damper systems
Commit this to memory! (or be able to derive it…)
s=L0+x
k, L0
F  ma 
m
c
d2x
dt 2
d2x
dt 2
 2n

c dx k
 x 0
m dt m
dx
 n2 x  0
dt
n 
k
m
 
c
2 km
x(t) is the ‘dynamic variable’ (deflection from static equilibrium)
k1
k1
k2
Parallel: stiffness k  k1  k2
c1
c2
Parallel: coefficient c  c1  c2
k2
1 1
1
Series: stiffness  
k k1 k2
c1
c2
1 1
1


Parallel: coefficient c c c
1
2
Canonical damped vibration problem
2
EOM
m
d s
dt
c
2
ds
 ks  kL0
dt
Standard Form
sx
n 
k
m
with
s  s0
ds
 v0
dt
2 dx

 xC
x  x0
2
2

dt
n dt
n
c
x0  s0

C  L0
2 km
1 d 2x
s=L0+x
k, L0
t 0
dx
 v0
dt
m
t 0
c
d  n 1   2
Overdamped
 1
Critically Damped   1
Underdamped   1
Overdamped
 1
 v  (n  d )( x0  C )

v  (n  d )( x0  C )
x(t )  C  exp(nt )  0
exp(d t )  0
exp(d t ) 
2d
2d


Critically Damped   1
Underdamped   1
x(t )  C  ( x0  C)  v0  n ( x0  C)t exp(nt )


v  n ( x0  C )
x(t )  C  exp(nt ) ( x0  C )cos d t  0
sin d t 
d


Calculating natural frequency and damping factor
from a measured vibration response
Displacement
x(t0)
x(t1)
x(t2)
x(t3)
time
t0
t1
t2
t3
t4
T
Measure log decrement:
Measure period:
Then
1
n
 x(t0 ) 

x
(
t
)
 n 
  log 
T


4 2   2
4 2   2
n 
T
Forced Vibrations – concept checklist
You should be able to:
1. Be able to derive equations of motion for spring-mass systems subjected
to external forcing (several types) and solve EOM using complex vars,
or by comparing to solution tables
2. Understand (qualitatively) meaning of ‘transient’ and ‘steady-state’
response of a forced vibration system (see Java simulation on web)
3. Understand the meaning of ‘Amplitude’ and ‘phase’ of steady-state
response of a forced vibration system
4. Understand amplitude-v-frequency formulas (or graphs), resonance, high
and low frequency response for 3 systems
5. Determine the amplitude of steady-state vibration of forced spring-mass
systems.
6. Deduce damping coefficient and natural frequency from measured forced
response of a vibrating system
7. Use forced vibration concepts to design engineering systems
EOM for forced vibrating systems
L0
k, L0
x(t)
1 d 2x
F(t)=F0 sin t
n2 dt 2
m



2 km
,
K
1
k
x(t)
L0
k, L0
1 d 2x
m
n2 dt 2
Base Excitation

n 
y(t)=Y0 sint
L0 x(t)
k, L0

k
,
m
n 
External forcing
2 dx
 x  KF0 sin  t
n dt



2 dx
2 dy 
 x Ky

n dt
n dt 

k
,
m


,
K 1
2 km
y(t)=Y0sint
m

m0
Y0 2
2 dx
K d2y
Rotor Excitation  2 dt 2   dt  x    2 dt 2  K  2 sin t
n
n
n
n
1 d 2x
n 
k
M


2 kM
m
K  0 M  m  m0
M
Steady-state and Transient solution to EOM
Equation
1 d 2x
n2
Full Solution
dt
2

2 dx
 x  C  KF0 sin(t )
n dt
Initial Conditions x  x0
dx
 v0
dt
x(t )  C  xh (t )  x p (t )
Steady state part (particular integral) x p (t )  X 0 sin  t   
X0 
KF0
1/2

2
2
2 2
 1   / n   2 / n  




  tan 1
Transient part (complementary integral)
Overdamped
 1
Critically Damped
 1
Underdamped
 1
2 / n
1   2 / n2
h
h


v h  (n  d ) x0h
 v  (n  d ) x0

xh (t )  C  exp(nt )  0
exp(d t )  0
exp(d t ) 
2d
2d






xh (t )  C  x0h  v0h  n x0h  t exp(nt )




v h  n x0h


xh (t )  C  exp(nt )  x0h cos d t  0
sin d t 
d




d  n 1   2
x0h  x0  C  x p (0)  x0  C  X 0 sin 
v0h  v0 
dx p
dt t 0
 v0  X 0 cos
t 0
Canonical externally forced system (steady state solution)
Steady state solution to
1 d 2x
n2 dt 2
M
2 dx
 x  C  KF0 sin(t )
n dt
1
1/2

2
2
2 2
 1   / n   2 / n  



  2 / T
x p (t )  X 0 sin  t   
k
c
1
n 

K
m
k
2 km
X 0  KF0 M ( / n ,  )


  tan
1
2 / n
1   2 / n2
s=L0+xF (t )  F0 sin t
k, L0
m
c
Canonical base excited system (steady state solution)
1 d 2x
Steady state solution to
n 
n2 dt 2
k
c
 
K 1
m
2 km
X 0  KY0 M ( ,  n ,  )
M


2 dx
2 dy 
 xCKy

n dt
n dt 

x p (t )  X 0 sin  t   
1   2 / n 






2
1   2 / n2

2 1/2
1/2
2
  2 / n  

1  (1  4 2 ) 2 / n2
 = 0.01
0
0
Magnification X /KY
  tan
2 3 / n3
1
10
 = 0.05
1
 = 0.1
 = 0.2
 = 0.4
10
0
 = 0.6
10
 = 1.0
-1
0
0.5
1
1.5
2
Frequency Ratio /n
2.5
3
Canonical rotor excited system (steady state solution)
c
2 dx
K d2y

xC
n2 dt 2 n dt
n2 dt 2
1 d 2x
Steady state solution to
n 
m0
k
c
 
K
m  m0
m  m0
2 k (m  m0 )
X 0  KY0 M ( ,  n ,  )
M
x p (t )  X 0 sin  t   
 2 / n2
1/2

2
2
2 2
 1   / n   2 / n  




  tan 1
2 / n
1   2 / n2
y  Y0 sin t