DYNAMICS Ferdinand P. Beer Mechanical Vibrations E. Russell - - PowerPoint PPT Presentation

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Seventh Edition Ferdinand P. Beer

• E. Russell Johnston, Jr.

Lecture Notes:

• J. Walt Oler

Texas Tech University CHAPTER

• Mechanical Vibrations

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 2

Contents

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

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 3

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.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 4

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,

( )

= + − = + − = = kx x m kx x k W F ma

st

• δ
• General solution is the sum of two particular solutions,

( ) ( )

t C t C t m k C t m k C x

n n

ω ω cos sin cos sin

2 1 2 1

+ =

• +
• =
• x is a periodic function and ωn is the natural circular

frequency of the motion.

• C1 and C2 are determined by the initial conditions:

( ) ( )

t C t C x

n n

ω ω cos sin

2 1

+ =

2

x C =

n

v C ω

1 =

( ) ( )

t C t C x v

n n n n

ω ω ω ω sin cos

2 1

− = =

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 5

Free Vibrations of Particles. Simple Harmonic Motion

( )

φ ω + = t x x

n m sin

= =

n n

ω π τ 2 period = = = π ω τ 2 1

n n n

f natural frequency

( )

= + =

2 2

x v x

n m

ω amplitude

( ) =

=

− n

x v ω φ

1

tan phase angle

• Displacement is equivalent to the x component of the sum of two vectors

which rotate with constant angular velocity

2 1

C C

• +

.

n

ω

2 1

x C v C

n

= = ω

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 6

Free Vibrations of Particles. Simple Harmonic Motion

( )

φ ω + = t x x

n m sin

• 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.

( ) ( )

2 sin cos π φ ω ω φ ω ω + + = + = = t x t x x v

n n m n n m

• (

) ( )

π φ ω ω φ ω ω + + = + − = = t x t x x a

n n m n n m

sin sin

2 2

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 7

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. for small angles,

( )

g l t l g

n n n m

π ω π τ φ ω θ θ θ θ 2 2 sin = = + = = +

• :

t t

ma F =

• Consider tangential components of acceleration and force

for a simple pendulum, sin sin = + = − θ θ θ θ l g ml W

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 8

Simple Pendulum (Exact Solution)

sin = + θ θ l g

• An exact solution for

leads to

( )

• −

=

2 2 2

sin 2 sin 1 4

π

φ θ φ τ

m n

d g l which requires numerical solution.

• =

g l K

n

π π τ 2 2

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 9

Sample Problem 19.1

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. 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.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 10

Sample Problem 19.1

m kN 6 m kN 4

2 1

= = k k SOLUTION:

• Springs in parallel:
• determine the spring constant for equivalent spring

m N 10 m kN 10

4 2 1 2 1

= = + = = + = k k P k k k P δ δ δ

• apply the approximate relations for the harmonic motion of a

spring-mass system

n n n

m k ω π τ ω 2 s rad 14 . 14 kg 20 N/m 104 = = = = s 444 . =

n

τ

( )( )

s rad 4.14 1 m 040 . = =

n m m

x v ω s m 566 . =

m

v

2

s m 00 . 8 =

m

a

( )( )2

2

s rad 4.14 1 m 040 . = =

n m m

a x a

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 11

Sample Problem 19.1

m kN 6 m kN 4

2 1

= = k k

• Springs in series:
• determine the spring constant for equivalent spring
• apply the approximate relations for the harmonic motion of a

spring-mass system

n n n

m k ω π τ ω 2 s rad 93 . 6 kg 20 400N/m 2 = = = = s 907 . =

n

τ

( )( )

s rad .93 6 m 040 . = =

n m m

x v ω s m 277 . =

m

v

2

s m 920 . 1 =

m

a

( )( )2

2

s rad .93 6 m 040 . = =

n m m

a x a m N 10 m kN 10

4 2 1 2 1

= = + = = + = k k P k k k P δ δ δ

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 12

Free Vibrations of Rigid Bodies

• If an equation of motion takes the form
• r

2 2

= + = + θ ω θ ω

n n x

x

• the corresponding motion may be considered as

simple harmonic motion.

• Analysis objective is to determine ωn.

( ) ( )

[ ]

mg W mb b b m I = = + = , 2 2 but

2 3 2 2 2 12 1

5 3 sin 5 3 = + ≅ + θ θ θ θ b g b g

• g

b b g

n n n

3 5 2 2 , 5 3 then π ω π τ ω = = =

• For an equivalent simple pendulum,

3 5b l =

• Consider the oscillations of a square plate

( )

( )

θ θ θ

• I

mb b W + = − sin

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 13

Sample Problem 19.2

k A cylinder of weight W is suspended as shown. Determine the period and natural frequency of vibrations of the cylinder. SOLUTION:

• 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.

• Substitute the kinematic relations to arrive at

an equation involving only the angular displacement and acceleration.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 14

Sample Problem 19.2

SOLUTION:

• From the kinematics of the system, relate the linear displacement

and acceleration to the rotation of the cylinder. θ r x = θ δ r x 2 2 = = θ α

• r

r a = = θ α

• =

θ

• r

a =

• Based on a free-body-diagram equation for the equivalence of the

external and effective forces, write the equation of motion.

( )

:

• =

eff A A

M M

( )

α I r a m r T Wr + = − 2

2

( )

θ δ r k W k T T 2 but

2 1 2

+ = + =

• Substitute the kinematic relations to arrive at an equation

involving only the angular displacement and acceleration.

( )(

)

( )

3 8 2 2

2 2 1 2 1

= + + = + − θ θ θ θ θ m k mr r r m r kr W Wr

• m

k

n

3 8 = ω k m

n n

8 3 2 2 π ω π τ = = m k f

n n

3 8 2 1 2 π π ω = =

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 15

Sample Problem 19.3

s 13 . 1 lb 20

n =

= τ W s 93 . 1 =

n

τ

The disk and gear undergo torsional vibration with the periods shown. Assume that the moment exerted by the wire is proportional to the twist angle. 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. 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.

• With the natural frequency and moment of

inertia for the disk known, calculate the torsional spring constant.

• With natural frequency and spring constant

known, calculate the moment of inertia for the gear.

• Apply the relations for simple harmonic

motion to calculate the maximum gear velocity.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 16

Sample Problem 19.3

s 13 . 1 lb 20

n =

= τ W s 93 . 1 =

n

τ 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.

( )

:

• =

eff O O

M M = + − = + θ θ θ θ I K I K

• K

I I K

n n n

π ω π τ ω 2 2 = = =

• With the natural frequency and moment of inertia for the disk

known, calculate the torsional spring constant.

2 2 2 2 1

s ft lb 138 . 12 8 2 . 32 20 2 1 ⋅ ⋅ =

• =

= mr I K 138 . 2 13 . 1 π = rad ft lb 27 . 4 ⋅ = K

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 17

Sample Problem 19.3

s 13 . 1 lb 20

n =

= τ W s 93 . 1 =

n

τ rad ft lb 27 . 4 ⋅ = K K I I K

n n n

π ω π τ ω 2 2 = = =

• With natural frequency and spring constant known,

calculate the moment of inertia for the gear. 27 . 4 2 93 . 1 I π =

2

s ft lb 403 . ⋅ ⋅ = I

• Apply the relations for simple harmonic motion to calculate

the maximum gear velocity.

n m m n n m n m

t t ω θ ω ω ω θ ω ω θ θ = = = sin sin rad 571 . 1 90 = ° =

m

θ

( )

• =
• =

s 93 . 1 2 rad 571 . 1 2 π τ π θ ω

n m m

s rad 11 . 5 =

m

ω

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 18

Principle of Conservation of Energy

• Resultant force on a mass in simple harmonic motion is

conservative - total energy is conserved. constant = +V T = + = +

2 2 2 2 2 1 2 2 1

constant x x kx x m

n

ω

• Consider simple harmonic motion of the square plate,

1 =

T

( ) ( )

[ ]

2 2 1 2 1

2 sin 2 cos 1

m m

Wb Wb Wb V θ θ θ ≅ = − =

( )

( ) ( ) 2

2 3 5 2 1 2 2 3 2 2 1 2 2 1 2 2 1 2 2 1 2 m m m m m

mb mb b m I v m T θ ω θ ω

• =

+ = + =

2 =

V

( )

2 2 2 3 5 2 1 2 2 1 2 2 1 1

+ = + + = +

n m m

mb Wb V T V T ω θ θ b g

n

5 3 = ω

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 19

Sample Problem 19.4

Determine the period of small

• scillations of a cylinder which rolls

without slipping inside a curved surface. 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.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 20

Sample Problem 19.4

1 =

T

( )( ) ( )(

)

2 cos 1

2 1 m

r R W r R W Wh V θ θ − ≅ − − = =

( )

( )

( )

2 2 4 3 2 2 2 2 1 2 1 2 2 1 2 2 1 2 2 1 2 m m m m m

r R m r r R mr r R m I v m T θ θ θ ω

• −

=

• −

+ − = + =

2 =

V

SOLUTION:

• Apply the principle of conservation of energy between the

positions of maximum and minimum potential energy.

2 2 1 1

V T V T + = +

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 21

Sample Problem 19.4

2 2 1 1

V T V T + = +

( ) ( )

2

2 2 4 3 2

+ − = − +

m m

r R m r R W θ θ

• (

)( ) ( ) ( )2

2 4 3 2

2

m n m m

r R m r R mg ω θ θ − = − r R g

n

− = 3 2

2

ω g r R

n n

− = = 2 3 2 2 π ω π τ

1 =

T

( )(

)

2

2 1 m

r R W V θ − ≅

( )

2 2 4 3 2 m

r R m T θ

• −

=

2 =

V

• Solve the energy equation for the natural frequency of the
• scillations.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 22

Forced Vibrations

: ma F =

• (

)

x m x k W t P

st f m

• =

+ − + δ ω sin t P kx x m

f m

ω sin = +

• (

)

x m t x k W

f m st

• =

− + − ω δ δ sin t k kx x m

f m

ω δ sin = +

• Forced vibrations - Occur when

a system is subjected to a periodic force or a periodic displacement of a support. =

f

ω forced frequency

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 23

Forced Vibrations

[ ]

t x t C t C x x x

f m n n particular ary complement

ω ω ω sin cos sin

2 1

+ + = + =

( ) ( )2

2 2

1 1

n f m n f m f m m

k P m k P x ω ω δ ω ω ω − = − = − = t k kx x m

f m

ω δ sin = +

• t

P kx x m

f m

ω sin = +

• At ωf = ωn, forcing input is in

resonance with the system. t P t kx t x m

f m f m f m f

ω ω ω ω sin sin sin

2

= + − Substituting particular solution into governing equation,

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 24

Sample Problem 19.5

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. 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.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 25

Sample Problem 19.5

SOLUTION:

• The resonant frequency is equal to the natural frequency of the

system. ft s lb 87 . 10 2 . 32 350

2

⋅ = = m

( )

ft lb 000 , 36 in lb 3000 750 4 = = = k W = 350 lb k = 4(350 lb/in) rpm 549 rad/s 5 . 57 87 . 10 000 , 36 = = = = m k

n

ω Resonance speed = 549 rpm

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 26

Sample Problem 19.5

W = 350 lb k = 4(350 lb/in) rad/s 5 . 57 =

n

ω

• Evaluate the magnitude of the periodic force due to the motor
• unbalance. Determine the vibration amplitude from the

frequency ratio at 1200 rpm.

( )

ft s lb 001941 . s ft 2 . 32 1

• z

16 lb 1

• z

1 rad/s 125.7 rpm 1200

2 2

⋅ =

• =

= = = m

f

ω ω

( )( )( )

lb 33 . 15 7 . 125 001941 .

2 12 6 2

= = = = ω mr ma P

n m

( )

( )

in 001352 . 5 . 57 7 . 125 1 3000 33 . 15 1

2 2

− = − = − =

n f m m

k P x ω ω xm = 0.001352 in. (out of phase)

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 27

Damped Free Vibrations

• With viscous damping due to fluid friction,

: ma F =

• (

)

= + + = − + − kx x c x m x m x c x k W

st

• δ
• Substituting x = eλt and dividing through by eλt yields

the characteristic equation, m k m c m c k c m −

• ±

− = = + +

2 2

2 2 λ λ λ

• Define the critical damping coefficient such that

n c c

m m k m c m k m c ω 2 2 2

2

= = = −

• All vibrations are damped to some degree by forces

due to dry friction, fluid friction, or internal friction.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 28

Damped Free Vibrations

• Characteristic equation,

m k m c m c k c m −

• ±

− = = + +

2 2

2 2 λ λ λ = =

n c

m c ω 2 critical damping coefficient

• Heavy damping: c > cc

t t

e C e C x

2 1

2 1 λ λ

+ =

• negative roots
• nonvibratory motion
• Critical damping: c = cc

( )

t

n

e t C C x

ω −

+ =

2 1

• double roots
• nonvibratory motion
• Light damping: c < cc

( ) (

)

t C t C e x

d d t m c

ω ω cos sin

2 1 2

+ =

−

=

• −

=

2

1

c n d

c c ω ω damped frequency

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 29

Damped Forced Vibrations

( )

[ ]

( )(

) [ ]

( )(

) ( )

= − = = + − = =

2 2 2 2

1 2 tan 2 1 1

n f n f c n f c n f m m m

c c c c x k P x ω ω ω ω φ ω ω ω ω δ magnification factor phase difference between forcing and steady state response t P kx x c x m

f m

ω sin = + +

• particular

ary complement

x x x + =

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 30

Electrical Analogues

• Consider an electrical circuit consisting of an inductor, resistor

and capacitor with a source of alternating voltage sin = − − − C q Ri dt di L t E

f m

ω

• Oscillations of the electrical system are analogous to damped

forced vibrations of a mechanical system. t E q C q R q L

f m

ω sin 1 = + +

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 31

Electrical Analogues

• The analogy between electrical and mechanical systems

also applies to transient as well as steady-state

• scillations.
• 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.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 19 - 32

Electrical Analogues

• The electrical system analogy provides a means of

experimentally determining the characteristics of a given mechanical system.

• For the mechanical system,

( ) ( )

2 1 2 1 1 2 1 2 1 1 1 1

= − + + − + + x x k x k x x c x c x m

• (

) ( )

t P x x k x x c x m

f m

ω sin

1 2 2 1 2 2 2 2

= − + − +

• For the electrical system,

( )

2 2 1 1 1 2 1 1 1 1

= − + + − + C q q C q q q R q L

• (

)

t E C q q q q R q L

f m

ω sin

2 1 2 1 2 2 2 2

= − + − +

• The governing equations are equivalent. The characteristics of

the vibrations of the mechanical system may be inferred from the oscillations of the electrical system.