mechanic of machine-Mechanical Vibrations

Ninth
Edition

CHAPTER VECTOR MECHANICS FOR ENGINEERS:

19 DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Mechanical Vibrations
Lecture Notes:
J. Walt Oler
Texas Tech University

© 2010 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Dynamics
Edition
Ninth

Contents

Introduction Sample Problem 19.4
Free Vibrations of Particles. Simple Harmonic Motion
Forced Vibrations
Simple Pendulum (Approximate Solution) Sample Problem 19.5
Simple Pendulum (Exact Solution) Damped Free Vibrations
Sample Problem 19.1 Damped Forced Vibrations
Free Vibrations of Rigid Bodies Electrical Analogues
Sample Problem 19.2
Sample Problem 19.3
Principle of Conservation of Energy

© 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 2

Edition
Ninth

Introduction
• Mechanical vibration is the motion of a particle or body which
oscillates about a position of equilibrium. Most vibrations in
machines and structures are undesirable due to increased stresses
and energy losses.
• Time interval required for a system to complete a full cycle of the
motion is the period of the vibration.
• Number of cycles per unit time defines the frequency of the vibrations.
• Maximum displacement of the system from the equilibrium position is
the amplitude of the vibration.
• When the motion is maintained by the restoring forces only, the
vibration is described as free vibration. When a periodic force is
applied to the system, the motion is described as forced vibration.
• When the frictional dissipation of energy is neglected, the motion
is said to be undamped. Actually, all vibrations are damped to
some degree.

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Ninth

• If a particle is displaced through a distance xm from its
equilibrium position and released with no velocity, the
particle will undergo simple harmonic motion,
ma = F = W − k ( δ st + x ) = − kx
m + kx = 0
x
• General solution is the sum of two particular solutions,
 k   k 
x = C1 sin  t  + C 2 cos
 m   m t

   
= C1 sin ( ω n t ) + C 2 cos( ω n t )
• x is a periodic function and ωn is the natural circular
frequency of the motion.
• C1 and C2 are determined by the initial conditions:
x = C1 sin ( ω n t ) + C 2 cos( ω n t ) C 2 = x0
v = x = C1ω n cos( ω n t ) − C 2ω n sin ( ω n t )
 C1 = v0 ω n

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Ninth


v
C1 = 0
ωn
C 2 = x0

 
• Displacement is equivalent to the x component of the sum of two vectors C1 + C 2
which rotate with constant angular velocity ω n .

x = xm sin ( ω n t + φ ) xm = ( v0 ω n ) 2 + x0 = amplitude
2

φ = tan −1 ( v0 x0ω n ) = phase angle
2π
τn = = period
ωn
1 ωn
fn = = = natural frequency
τ n 2π

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Ninth

• Velocity-time and acceleration-time curves can be
represented by sine curves of the same period as the
displacement-time curve but different phase angles.
x = xm sin ( ω n t + φ )
v=x
= xmω n cos( ω n t + φ )
= xmω n sin ( ω n t + φ + π 2 )
a = 
x
= − xmω n sin ( ω n t + φ )
2

= xmω n sin ( ω n t + φ + π )
2


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Ninth

Simple Pendulum (Approximate Solution)
• Results obtained for the spring-mass system can be
applied whenever the resultant force on a particle is
proportional to the displacement and directed towards
the equilibrium position.
• Consider tangential components of acceleration and
force for a simple pendulum,
∑ Ft = mat : − W sin θ = mlθ 

θ + g sin θ = 0
l

for small angles,
 g
θ + θ = 0
l
θ = θ m sin ( ω n t + φ )
2π l
τn = = 2π
ωn g


Edition
Ninth

Simple Pendulum (Exact Solution)
 g
An exact solution for θ + sin θ = 0
l
l π 2 dφ
leads to τn = 4 ∫
g 0 1 − sin 2 (θ 2 ) sin 2 φ
m

which requires numerical
solution.
2K  l
τn =  2π
 
π  g



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Ninth

Sample Problem 19.1
SOLUTION:
• For each spring arrangement, determine
the spring constant for a single
equivalent spring.
• Apply the approximate relations for the
harmonic motion of a spring-mass
system.

A 50-kg block moves between vertical
guides as shown. The block is pulled
40mm down from its equilibrium
position and released.
For each spring arrangement, determine
a) the period of the vibration, b) the
maximum velocity of the block, and c)
the maximum acceleration of the block.

Edition
Ninth

Sample Problem 19.1
k1 = 4 kN m k2 = 6 kN m SOLUTION:
• Springs in parallel:
- determine the spring constant for equivalent spring

- apply the approximate relations for the harmonic motion
of a spring-mass system
k 104 N/m
ωn = = = 14.14 rad s
m 20 kg
2π
τn = τ n = 0.444 s
ωn

P = k1δ + k2δ vm = x m ω n
P = ( 0.040 m )(14.14 rad s ) vm = 0.566 m s
k= = k1 + k2
δ
2
= 10 kN m = 10 N m 4 am = x m an
= ( 0.040 m )(14.14 rad s ) 2 am = 8.00 m s 2


Edition
Ninth

Sample Problem 19.1
• Springs in series:
k1 = 4 kN m k2 = 6 kN m
- determine the spring constant for equivalent spring
- apply the approximate relations for the harmonic motion
of a spring-mass system
k 2400N/m
ωn = = = 6.93 rad s
m 20 kg
2π
τn = τ n = 0.907 s
ωn

vm = x m ω n
P = k1δ + k2δ = ( 0.040 m )( 6.93 rad s ) vm = 0.277 m s
P
k= = k1 + k2 2
am = x m an
δ
= 10 kN m = 104 N m = ( 0.040 m )( 6.93 rad s ) 2 am = 1.920 m s 2


Edition
Ninth

Free Vibrations of Rigid Bodies
• If an equation of motion takes the form
2 2
 + ω n x = 0 or θ + ω nθ = 0
x 
the corresponding motion may be considered
as simple harmonic motion.
• Analysis objective is to determine ωn.
• Consider the oscillations of a square plate
− W ( b sin θ ) = ( mbθ) + I θ
 

[ ]
but I = 12 m ( 2b ) 2 + ( 2b ) 2 = 2 mb 2 , W = mg
1
3

 3 g sin θ ≅ θ + 3 g θ = 0
θ + 
5b 5b
3g 2π 5b
then ω n = , τn = = 2π
5b ωn 3g
• For an equivalent simple pendulum,
l = 5b 3

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Ninth

Sample Problem 19.2
SOLUTION:
• From the kinematics of the system, relate
k the linear displacement and acceleration
to the rotation of the cylinder.
• Based on a free-body-diagram equation
for the equivalence of the external and
effective forces, write the equation of
motion.
A cylinder of weight W is suspended • Substitute the kinematic relations to arrive
as shown. at an equation involving only the angular
Determine the period and natural displacement and acceleration.
frequency of vibrations of the
cylinder.


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Sample Problem 19.2
SOLUTION:
• From the kinematics of the system, relate the linear
displacement and acceleration to the rotation of the cylinder.
x = rθ δ = 2 x = 2rθ
  
α = θ a = rα = rθ a = rθ

• Based on a free-body-diagram equation for the equivalence of
the external and effective forces, write the equation of motion.
∑ M A = ∑ ( M A ) eff : Wr − T2 ( 2r ) = ma r + I α
but T2 = T0 + kδ = 1 W + k ( 2rθ )
2
• Substitute the kinematic relations to arrive at an equation
involving only the angular displacement and acceleration.
(2 )
Wr − 1 W + 2krθ ( 2r ) = m( rθ) r + 1 mr 2θ

2

 8 kθ =0
θ +
3m
8k 2π 3m ωn 1 8k
ωn = τn = = 2π fn = =
3m ωn 8k 2π 2π 3m

Edition
Ninth

Sample Problem 19.3
SOLUTION:
• Using the free-body-diagram equation for
the equivalence of the external and
effective moments, write the equation of
motion for the disk/gear and wire.
W = 20 lb
• With the natural frequency and moment
τ n = 1.13 s τ n = 1.93 s
of inertia for the disk known, calculate
The disk and gear undergo torsional the torsional spring constant.
vibration with the periods shown. • With natural frequency and spring
Assume that the moment exerted by the constant known, calculate the moment of
wire is proportional to the twist angle. inertia for the gear.
Determine a) the wire torsional spring • Apply the relations for simple harmonic
constant, b) the centroidal moment of motion to calculate the maximum gear
inertia of the gear, and c) the maximum velocity.
angular velocity of the gear if rotated
through 90o and released.

Edition
Ninth

Sample Problem 19.3
SOLUTION:
• Using the free-body-diagram equation for the equivalence
of the external and effective moments, write the equation
of motion for the disk/gear and wire.
∑ M O = ∑ ( M O ) eff : + Kθ = − I θ


 K
θ + θ = 0
W = 20 lb I
τ n = 1.13 s τ n = 1.93 s
K 2π I
ωn = τn = = 2π
I ωn K

• With the natural frequency and moment of inertia for the
disk known, calculate the torsional spring constant.
2
1  20  8 
I = 1 mr 2 = 
2   = 0.138 lb ⋅ ft ⋅ s
2
2  32.2  12 

0.138
1.13 = 2π K = 4.27 lb ⋅ ft rad
K


Edition
Ninth

Sample Problem 19.3
• With natural frequency and spring constant known,
calculate the moment of inertia for the gear.
I
1.93 = 2π I = 0.403 lb ⋅ ft ⋅ s 2
4.27

• Apply the relations for simple harmonic motion to
W = 20 lb
calculate the maximum gear velocity.
τ n = 1.13 s τ n = 1.93 s
θ = θ m sin ω nt ω = θ mω n sin ω nt ω m = θ mω n

θ m = 90° = 1.571 rad

 2π  2π 
ω m = θ m   = (1.571 rad ) 
τ   
K 2π I  n  1.93 s 
ωn = τn = = 2π
I ωn K
ω m = 5.11rad s
K = 4.27 lb ⋅ ft rad


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Ninth

Principle of Conservation of Energy
• Resultant force on a mass in simple harmonic
motion is conservative - total energy is conserved.
T + V = constant 1 mx 2 + 1 kx 2 = constant

2 2
x2 + ωn x2 =
 2

• Consider simple harmonic motion of the square
plate, 0
T1 = [
V1 = Wb(1 − cosθ ) = Wb 2 sin 2 (θ m 2 ) ]
2
≅ 1 Wbθ m
2
2 2
T2 = 1 mvm + 1 I ω m V2 = 0
2 2

2
2
2 3
(
= 1 m( bθm ) + 1 2 mb 2 ω m
2
)
(
= 1 5 mb 2 θm
2 3
2
)
T1 + V1 = T2 + V2

2 2 3
( 2 2
)
0 + 1 Wbθ m = 1 5 mb 2 θ mω n + 0
2
ω n = 3 g 5b

Edition
Ninth

Sample Problem 19.4
SOLUTION:
• Apply the principle of conservation of
energy between the positions of maximum
and minimum potential energy.
• Solve the energy equation for the natural
frequency of the oscillations.

Determine the period of small
oscillations of a cylinder which rolls
without slipping inside a curved
surface.


Edition
Ninth

Sample Problem 19.4
SOLUTION:
• Apply the principle of conservation of energy between the
positions of maximum and minimum potential energy.
T1 + V1 = T2 + V2

T1 = 0 V1 = Wh = W ( R − r )(1 − cosθ )
(
≅ W ( R − r) θm 2
2
)
2 2
T2 = 1 mvm + 1 I ω m V2 = 0
2 2

( )
2
1 m( R − r )θ 2  R − r  2
2
= 2
m + 1 1 mr 
2 2  θm
 r 
= 3 m( R − r ) 2 θm
4
2


Edition
Ninth

Sample Problem 19.4
• Solve the energy equation for the natural frequency of
the oscillations.
T1 = 0 (
V1 ≅ W ( R − r ) θ m 2
2
)
T2 = 3 m( R − r ) 2θm
4
2 V2 = 0

T1 + V1 = T2 + V2

2
θm 3
0 +W ( R − r) = 4 m( R − r ) 2 θm + 0
2
2
2
θm 3
( mg )( R − r ) = 4 m( R − r ) 2 (θ mω n ) 2
m
2

2 2 g 2π 3 R−r
ωn = τn = = 2π
3 R−r ωn 2 g


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Ninth

Forced Vibrations

Forced vibrations - Occur
when a system is subjected to
a periodic force or a periodic
displacement of a support.
ω f = forced frequency

∑ F = ma :
Pm sin ω f t + W − k ( δ st + x ) = m
x ( )
W − k δ st + x − δ m sin ω f t = m
x

m + kx = Pm sin ω f t
x m + kx = kδ m sin ω f t
x


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Forced Vibrations
x = xcomplementary + x particular
= [ C1 sin ω n t + C 2 cos ω n t ] + xm sin ω f t
Substituting particular solution into governing equation,
− mω 2 xm sin ω f t + kxm sin ω f t = Pm sin ω f t
f
Pm Pm k δm
xm = = =
( )
k − mω 2 1 − ω f ω n 2 1 − ω f ω n 2
f ( )

m + kx = Pm sin ω f t
x

m + kx = kδ m sin ω f t
x
At ωf = ωn, forcing input is in
resonance with the system.

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Ninth

Sample Problem 19.5
SOLUTION:
• The resonant frequency is equal to the
natural frequency of the system.
• Evaluate the magnitude of the periodic
force due to the motor unbalance.
Determine the vibration amplitude
from the frequency ratio at 1200 rpm.

A motor weighing 350 lb is supported
by four springs, each having a constant
750 lb/in. The unbalance of the motor
is equivalent to a weight of 1 oz located
6 in. from the axis of rotation.
Determine a) speed in rpm at which
resonance will occur, and b) amplitude
of the vibration at 1200 rpm.

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Sample Problem 19.5
SOLUTION:
• The resonant frequency is equal to the natural frequency of
the system.
350
m= = 10.87 lb ⋅ s 2 ft
32.2

k = 4( 750 ) = 3000 lb in
= 36,000 lb ft
W = 350 lb
k = 4(350 lb/in)
k 36,000
ωn = =
m 10.87
= 57.5 rad/s = 549 rpm

Resonance speed = 549 rpm


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Ninth

Sample Problem 19.5
• Evaluate the magnitude of the periodic force due to the
motor unbalance. Determine the vibration amplitude from
the frequency ratio at 1200 rpm.
ω f = ω = 1200 rpm = 125.7 rad/s

 1 lb  1 
m = (1 oz )  
  = 0.001941 lb ⋅ s 2 ft
 16 oz  32.2 ft s 2 


W = 350 lb Pm = man = mrω 2
k = 4(350 lb/in) = ( 0.001941) (12 )(125.7) 2 = 15.33 lb
6
ω n = 57.5 rad/s
Pm k 15.33 3000
xm = =
(
1 − ω f ωn )2 1 − (125.7 57.5) 2
= −0.001352 in

xm = 0.001352 in. (out of phase)


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Damped Free Vibrations
• All vibrations are damped to some degree by
forces due to dry friction, fluid friction, or
internal friction.

• With viscous damping due to fluid friction,
∑ F = ma : W − k ( δ st + x ) − cx = m
 x
m + cx + kx = 0
x 

• Substituting x = eλt and dividing through by eλt
yields the characteristic equation,
2
c  c  k
mλ2 + cλ + k = 0 λ=− ±   −
2m  2m  m

• Define the critical damping coefficient such that
2
 cc  k k
  − =0 cc = 2 m = 2mω n
 2m  m m

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Ninth

Damped Free Vibrations
• Characteristic equation,
2
c  c  k
mλ2 + cλ + k = 0 λ=− ±   −
2m  2m  m
cc = 2mω n = critical damping coefficient
• Heavy damping: c > cc
x = C1e λ1t + C 2 e λ2t - negative roots
- nonvibratory motion
• Critical damping: c = cc
x = ( C1 + C 2t ) e −ω nt - double roots
- nonvibratory motion
• Light damping: c < cc
x = e −( c 2m ) t ( C1 sin ω d t + C 2 cos ω d t )
2
c
ω d = ω n 1 −   = damped frequency
c 
 c

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Ninth

Damped Forced Vibrations

m + cx + kx = Pm sin ω f t
x  x = xcomplementary + x particular

xm xm 1
= = = magnification
Pm k δ
[1 − (ω f ωn ) ] [
2 2
( )]
+ 2( c cc ) ω f ω n 2 factor
(
2( c cc ) ω f ω n )
tan φ = = phase difference between forcing and steady
(
1− ω f ωn ) 2
state response

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Ninth

Electrical Analogues
• Consider an electrical circuit consisting of an inductor,
resistor and capacitor with a source of alternating voltage
di q
E m sin ω f t − L − Ri − = 0
dt C
1
Lq + Rq + q = Em sin ω f t
 
C
• Oscillations of the electrical system are analogous to
damped forced vibrations of a mechanical system.


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Ninth

• The analogy between electrical and mechanical
systems also applies to transient as well as steady-
state oscillations.

• With a charge q = q0 on the capacitor, closing the
switch is analogous to releasing the mass of the
mechanical system with no initial velocity at x = x0.
• If the circuit includes a battery with constant voltage
E, closing the switch is analogous to suddenly
applying a force of constant magnitude P to the
mass of the mechanical system.


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Ninth

• The electrical system analogy provides a means of
experimentally determining the characteristics of a given
mechanical system.
• For the mechanical system,
m11 + c1 x1 + c2 ( x1 − x2 ) + k1 x1 + k 2 ( x1 − x2 ) = 0
x   

m2 2 + c2 ( x2 − x1 ) + k 2 ( x2 − x1 ) = Pm sin ω f t
x  
• For the electrical system,
q q −q
L1q1 + R1 ( q1 − q2 ) + 1 + 1 2 = 0
  
C1 C2
q −q
L2 q2 + R2 ( q2 − q1 ) + 2 1 = Em sin ω f t
  
C2
• The governing equations are equivalent. The characteristics
of the vibrations of the mechanical system may be inferred
from the oscillations of the electrical system.


mechanic of machine-Mechanical Vibrations

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