Seventh Edition
CHAPTER
19
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Mechanical Vibrations
Lecture Notes:
J. Walt Oler
Texas Tech University
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Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Contents
Introduction
Sample Problem 19.4
Free Vibrations of Particles. Simple
Harmonic Motion
Forced Vibrations
Simple Pendulum (Approximate Solution)
Simple Pendulum (Exact Solution)
Sample Problem 19.1
Sample Problem 19.5
Damped Free Vibrations
Damped Forced Vibrations
Electrical Analogues
Free Vibrations of Rigid Bodies
Sample Problem 19.2
Sample Problem 19.3
Principle of Conservation of Energy
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Vector Mechanics for Engineers: Dynamics
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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Vector Mechanics for Engineers: Dynamics
Free Vibrations of Particles. Simple Harmonic Motion
• 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
mx + kx = 0
• General solution is the sum of two particular solutions,
 k 
 k 
x = C1 sin 
t  + C 2 cos 
t 
 m 
 m 
= 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 = v 0 ω n
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Vector Mechanics for Engineers: Dynamics
Free Vibrations of Particles. Simple Harmonic Motion
C1 =
v0
ωn
C 2 = x0
• Displacement is equivalent to the x component of the sum of two vectors
which rotate with constant angular velocity
ωn.
x = xm sin (ω n t + φ )
xm =
C1 + C 2
(v0 ω n )2 + x02 = amplitude
φ = tan −1 (v 0 x 0ω n ) = phase angle
τn =
fn =
2π
ωn
1
τn
= period
=
ωn
= natural frequency
2π
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Vector Mechanics for Engineers: Dynamics
Free Vibrations of Particles. Simple Harmonic Motion
• 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 = x m 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 n2 sin (ω n t + φ )
= xmω n2 sin (ω n t + φ + π )
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Vector Mechanics for Engineers: Dynamics
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 = ma t : − W sin θ = ml θ
g
θ + sin θ = 0
l
for small angles,
g
θ + θ =0
l
θ = θ m sin (ω n t + φ )
τn =
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2π
ωn
= 2π
l
g
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Vector Mechanics for Engineers: Dynamics
Simple Pendulum (Exact Solution)
An exact solution for
g
θ + sin θ = 0
l
leads to
l π 2
dφ
τn = 4
∫
g 0 1 − sin 2 (θ 2 ) sin 2 φ
m
which requires numerical solution.
2K 
 2π
τn =
π 
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l 

g
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Vector Mechanics for Engineers: Dynamics
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.
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Vector Mechanics for Engineers: Dynamics
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
10 4 N/m
=
= 14 .14 rad s
m
20 kg
ωn =
τn =
P = k1δ + k2δ
k=
P
δ
2π
ωn
τ n = 0 .444 s
vm = x m ω n
= (0 .040 m )(1 4.14 rad s )
= k1 + k2
4
= 10 kN m = 10 N m
vm = 0.566 m s
am = x m a n2
= (0.040 m )(1 4.14 rad s )2
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am = 8.00 m s 2
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.1
k1 = 4 kN m k2 = 6 kN m
• Springs in series:
- determine the spring constant for equivalent spring
- apply the approximate relations for the harmonic motion of a
spring-mass system
k
=
m
ωn =
τn =
2 400N/m
= 6.93 rad s
20 kg
2π
ωn
τ n = 0 .907 s
vm = x m ω n
P = k1δ + k2δ
k=
P
δ
= k1 + k2
= 10 kN m = 104 N m
= (0 .040 m )(6.93 rad s )
vm = 0.277 m s
am = x m an2
= (0 .040 m )(6 .93 rad s )2
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am = 1.920 m s 2
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Vector Mechanics for Engineers: Dynamics
Free Vibrations of Rigid Bodies
• If an equation of motion takes the form
x + ω n2 x = 0 or θ + ω n2θ = 0
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 ) = (mb θ) + I θ
[
]
1 m (2b )2 + (2b )2 = 2 mb 2 , W = mg
but I = 12
3
3g
3g
θ+
sin θ ≅ θ +
θ =0
5b
5b
3g
2π
5b
then ω n =
, τn =
= 2π
5b
3g
ωn
• For an equivalent simple pendulum,
l = 5b 3
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.2
SOLUTION:
k
• From the kinematics of the system, relate
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 as
shown.
Determine the period and natural
frequency of vibrations of the cylinder.
• Substitute the kinematic relations to arrive at
an equation involving only the angular
displacement and acceleration.
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Vector Mechanics for Engineers: Dynamics
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 = 2 rθ
α =θ
a = rα = rθ
a = r θ
• 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 (2 r ) = m a r + I α
but
T2 = T0 + kδ = 12 W + k (2 rθ )
• Substitute the kinematic relations to arrive at an equation
involving only the angular displacement and acceleration.
Wr − 1 W + 2 kr θ (2 r ) = m (rθ)r + 1 mr 2θ
(2
θ +
)
2
8k
θ =0
3m
8k
ωn =
3m
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2π
3m
τn =
= 2π
ωn
8k
fn =
ω n 1 8k
=
2π 2π 3m
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Vector Mechanics for Engineers: Dynamics
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
τ n = 1.13 s
τ n = 1.93 s
• With the natural frequency and moment of
inertia for the disk known, calculate the
torsional spring constant.
The disk and gear undergo torsional
vibration with the periods shown. Assume • With natural frequency and spring constant
that the moment exerted by the wire is
known, calculate the moment of inertia for
proportional to the twist angle.
the gear.
Determine a) the wire torsional spring
constant, b) the centroidal moment of
inertia of the gear, and c) the maximum
angular velocity of the gear if rotated
through 90o and released.
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• Apply the relations for simple harmonic
motion to calculate the maximum gear
velocity.
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Vector Mechanics for Engineers: Dynamics
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.
+ K θ = − I θ
∑ M O = ∑ (M O )eff :
K
θ + θ = 0
I
W = 20 lb
τ n = 1.13 s
τ n = 1.93 s
ωn =
K
I
τn =
2π
ωn
= 2π
I
K
• With the natural frequency and moment of inertia for the disk
known, calculate the torsional spring constant.
2
1  20  8 
2
I = 12 mr 2 = 
  = 0 .138 lb ⋅ ft ⋅ s
2  32 .2  12 
1.13 = 2π
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0.138
K
K = 4.27 lb ⋅ ft rad
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.3
• With natural frequency and spring constant known,
calculate the moment of inertia for the gear.
1.93 = 2π
I = 0.403 lb ⋅ ft ⋅ s 2
• Apply the relations for simple harmonic motion to calculate
the maximum gear velocity.
W = 20 lb
τ n = 1.13 s
I
4.27
τ n = 1.93 s
θ = θ m sin ω n t
ω = θ mω n sin ω n t
ω m = θ mω n
θ m = 90 ° = 1 .571 rad
ωn =
K
I
τn =
2π
ωn
= 2π
I
K
 2π 
2π 
 = (1 .571 rad )

 1.93 s 
 τn 
ω m = θ m 
ω m = 5.11rad s
K = 4.27 lb ⋅ ft rad
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Vector Mechanics for Engineers: Dynamics
Principle of Conservation of Energy
• Resultant force on a mass in simple harmonic motion is
conservative - total energy is conserved.
1 mx 2 + 1 kx 2 = constant
T + V = constant
2
2
x 2 + ω n2 x 2 =
• Consider simple harmonic motion of the square plate,
T1 = 0
V1 = Wb(1 − cosθ ) = Wb 2 sin 2 (θ m 2 )
[
]
≅ 12 Wbθ m2
2
T2 = 12 mvm2 + 12 I ω m
2
= 12 m(bθm ) + 12
= 12
(53 mb 2 )θm2
V2 = 0
(23 mb 2 )ω m2
T1 + V1 = T2 + V2
(
)
0 + 12 Wbθ m2 = 12 53 mb 2 θ m2 ω n2 + 0
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ω n = 3 g 5b
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Vector Mechanics for Engineers: Dynamics
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.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.4
SOLUTION:
• Apply the principle of conservation of energy between the
positions of maximum and minimum potential energy.
T1 + V1 = T2 + V 2
T1 = 0
V1 = Wh = W ( R − r )(1 − cosθ )
(
≅ W ( R − r ) θ m2 2
)
T2 = 12 m v m2 + 12 I ω m2
V2 = 0
2
 R − r  2
= 12 m ( R − r )θm2 + 12 12 mr 2 
 θm
 r 
= 34 m ( R − r )2 θm2
(
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.4
• Solve the energy equation for the natural frequency of the
oscillations.
(
T1 = 0
V1 ≅ W ( R − r ) θ m2 2
T2 = 34 m ( R − r )2 θm2
V2 = 0
)
T1 + V1 = T2 + V2
0 + W (R − r )
θ m2
(mg )(R − r )
θ m2
ω n2
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2
2 g
=
3 R−r
2
= 34 m( R − r )2 θm2 + 0
= 34 m( R − r )2 (θ mω n )2m
2π
3 R−r
= 2π
τn =
2 g
ωn
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Vector Mechanics for Engineers: Dynamics
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 ) = mx
W − k δ st + x − δ m sin ω f t = mx
mx + kx = Pm sin ω f t
mx + kx = kδ m sin ω f t
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Vector Mechanics for Engineers: Dynamics
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,
− m ω 2f x m sin ω f t + kx m sin ω f t = Pm sin ω f t
Pm
Pm k
δm
=
=
xm =
k − mω 2f 1 − (ω f ω n )2 1 − (ω f ω n )2
mx + kx = Pm sin ω f t
mx + kx = kδ m sin ω f t
At ωf = ωn, forcing input is in
resonance with the system.
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Vector Mechanics for Engineers: Dynamics
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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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.5
SOLUTION:
• The resonant frequency is equal to the natural frequency of the
system.
m=
W = 350 lb
k = 4(350 lb/in)
350
= 10.87 lb ⋅ s 2 ft
32.2
k = 4(750) = 3000 lb in
= 36,000 lb ft
k
36,000
=
m
10.87
= 57.5 rad/s = 549 rpm
ωn =
Resonance speed = 549 rpm
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Vector Mechanics for Engineers: Dynamics
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
 1 lb 
 = 0.001941 lb ⋅ s 2 ft
m = (1 oz )

 16 oz  32.2 ft s 2 
W = 350 lb
k = 4(350 lb/in)
ω n = 57.5 rad/s
Pm = man = mrω 2
( )
6 (125.7 )2 = 15.33 lb
= (0.001941) 12
xm =
(
Pm k
1 − ω f ωn
)2
=
15.33 3000
1 − (125.7 57.5)2
= −0.001352 in
xm = 0.001352 in. (out of phase)
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Vector Mechanics for Engineers: Dynamics
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 = mx
mx + cx + kx = 0
• Substituting x = eλt and dividing through by eλt yields
the characteristic equation,
m λ 2 + cλ + k = 0
2
c
k
 c 
λ=−
± 
−

2m
m
 2m 
• Define the critical damping coefficient such that
2
k
 cc 

 − =0
m
 2m 
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k
cc = 2 m
= 2mω n
m
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Vector Mechanics for Engineers: Dynamics
Damped Free Vibrations
• Characteristic equation,
2
c
k
 c 
λ=−
± 
−

2m
m
 2m 
m λ 2 + cλ + k = 0
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 2 t ) e −ω n t
- double roots
- nonvibratory motion
• Light damping: c < cc
x = e − (c 2 m )t (C1 sin ω d t + C 2 cos ω d t )
 c 
2
ω d = ω n 1 −   = damped frequency
 cc 
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Vector Mechanics for Engineers: Dynamics
Damped Forced Vibrations
mx + cx + kx = Pm sin ω f t
x = x complement ary + x particular
xm
x
= m =
δ
Pm k
1
tan φ =
[1 − (ω
(
(
1− ω f ωn
)
2
ωn )
f
2(c c c ) ω f ω n
] + [2(c c )(ω
2 2
)
c
f
ω n )]2
= magnification
factor
= phase difference between forcing and steady state
response
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Vector Mechanics for Engineers: Dynamics
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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Vector Mechanics for Engineers: Dynamics
Electrical Analogues
• 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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Vector Mechanics for Engineers: Dynamics
Electrical Analogues
• The electrical system analogy provides a means of
experimentally determining the characteristics of a given
mechanical system.
• For the mechanical system,
m1 x1 + c1 x1 + c 2 ( x1 − x 2 ) + k1 x1 + k 2 ( x1 − x 2 ) = 0
m2 x2 + c2 ( x 2 − x1 ) + k 2 ( x2 − x1 ) = Pm sin ω f t
• For the electrical system,
q1 q1 − q 2
L1q1 + R1 (q1 − q 2 ) +
+
=0
C1
C2
q −q
L2 q2 + R2 (q 2 − 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.
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