introduction to dynamics - purdue university

Introduction to Dynamics (N. Zabaras)

Mechanical VibrationsProf. Nicholas Zabaras

Warwick Centre for Predictive Modelling

University of Warwick

Coventry CV4 7AL

United Kingdom

Email: [email protected]

URL: http://www.zabaras.com/

April 10, 2016

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mailto:[email protected]

http://www.zabaras.com/


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 (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.

2

Closely following the Vector Mechanics for Engineers, Beer and Johnston (Chapter 19), and Engineering Mechanics:

Dynamics Hibbeler (Chapter 21)

http://www.amazon.com/Vector-Mechanics-Engineers-Ferdinand-Beer/dp/0077295498

http://www.amazon.com/Engineering-Mechanics-Dynamics-13th-Edition/dp/0132911272


Since all forces

are

conservative:

Differentiate (with respect to time):

The mass-spring system

3


Divide through by :

(The Governing

Equation)

Consider new coordinate system:

Where, (from equilibrium)

Substitution into the governing equation, gives:

steady


4


Assume a solution of

the form:

A)

B)

C)

D) Substitution of the solution gives:

The natural frequency is then:

steady


5


Since all forces are

conservative:

Treat as fixed-axis rotation, so:

If is small ℎ = 𝑟𝜃𝑠𝑖𝑛𝜃

2≈ 𝑟𝜃2

1

2, so:

And differentiate:

⇒

Now divide by to get:

(The Governing Equation)

The Pendulum

6

𝐾. 𝐸. =1

2𝑚𝑣2 =

1

2𝑚 𝑟 𝜃

2=1

2𝑚𝑟2 𝜃2


mass-spring system

Governing equation:

Natural frequency:

The Pendulum

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

nn

nm

22

sin

0

:tt maF

• Consider tangential components of acceleration

and force for a simple pendulum,

0sin

sin

l

g

mlW

8


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,

0

kxxm

kxxkWFma st

• General solution is the sum of two particular solutions,

tCtC

tm

kCt

m

kCx

nn cossin

cossin

21

21

• x is a periodic function and n is the natural circular

frequency of the motion.

• C1 and C2 are determined by the initial conditions:

tCtCx nn cossin 21 02 xC

nvC 01 tCtCxv nnnn sincos 21 9



txx nm sin

n

n

2period

2

1 n

nnf natural frequency

20

20 xvx nm amplitude

1

0 0tan nx v phase angle

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

which rotate with constant angular velocity 21 CC

.n

02

01

xC

vC

n

10



txx nm 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.

2sin

cos

tx

tx

xv

nnm

nnm

tx

tx

xa

nnm

nnm

sin

sin

2

2

11


Sample Problem

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.

12


Sample Problem

mkN6mkN4 21 kk SOLUTION:

• Springs in parallel:

- determine the spring constant for equivalent spring

mN10mkN10 4

21

21

kkP

k

kkP

- apply the approximate relations for the harmonic

motion of a spring-mass system

410 N/m14.14 rad s

50kg

2

n

n

n

k

m

s 444.0n

srad 4.141m 040.0

nmm xv

sm566.0mv

2sm00.8ma

2

20.040 m 14.14 rad s

m m na x

13


Sample Problem

mkN6mkN4 21 kk• Springs in series:

- determine the spring constant for equivalent spring

- apply the approximate relations for the harmonic

motion of a spring-mass system

2400N/m6.93rad s

50 kg

2

n

n

n

k

m

s 907.0n

srad .936m 040.0

nmm xv

sm277.0mv

2sm920.1ma

2

20.040 m 6.93 rad s

m m na x

1 2

1 2 1 2

1 2

1 1

1 1 1 1 12.4 /

4 6

P PP

k k k k

k kN mk P k k

14


Free Vibrations of Rigid Bodies

• If an equation of motion takes the form

0or0 22 nn xx

the corresponding motion may be

considered as simple harmonic motion.

• Analysis objective is to determine n.

mgWmbbbmI ,22but 23222

121

05

3sin

5

3

b

g

b

g

g

b

b

g

nnn

3

52

2,

5

3then

• For an equivalent simple pendulum,

35bl

• Consider the oscillations of a square plate

ImbbW sin

15


Principle of Conservation of Energy

• Resultant force on a mass in simple harmonic motion

is conservative - total energy is conserved.

constantVT2 21 1

2 2

2 2 2

constant

constantn

mx kx

x x

• Consider simple harmonic motion of the square plate,

01 T 2

1

212

1 cos 2sin 2m m

m

V Wb Wb

Wb

2 21 12 2 2

22 21 1 2

2 2 3

2 2 2 25 51 12 3 2 3

m m

m m

m n

T mv I

m b mb

mb mb

02 V

1 1 2 2

2 2 251 12 2 3

0 0m n

T V T V

Wb mb

bgn 53

16

Note that the max

velocity is m n

12

sin 2m m For small angles:

1: Position of max displacement

2: Position of equilibrium (also maximum

Velocity)


Sample Problem

Determine the period of small

oscillations 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.

17


Sample Problem

01 T

2

cos1

2

1

mrRW

rRWWhV

2 21 12 2 2

22 2 2 21 1 1

2 2 2

2 234

m m

m m

I

m

T mv I

R rm R r mr

r

m R r

02 V

SOLUTION:

• Apply the principle of conservation of energy

between the positions of maximum and minimum

potential energy.

2211 VTVT

18

𝑐𝑜𝑠𝜃 = 1 −𝜃2

2!+𝜃4

4!− ⋯ ≈ 1 −

𝜃2

2𝑓𝑜𝑟 𝜃 ≪ 1

Motion of the centre of mass:

Rotation around O.

Rotation of the disk. To compute

the angular velocity of the disk

use

m m mv R r r

Instanteneous

centre of rotation

m mv r


Sample Problem

2211 VTVT

02

0 22

43

2

mm rRmrRW

22

43

2

2 mnmm rRmrRmg

rR

gn

3

22g

rR

nn

2

32

2

01 T 221 mrRWV

22

43

2 mrRmT 02 V

• Solve the energy equation for the natural frequency

of the oscillations.

19


Sample Problem

20

Consider a 3 kg mass, A, suspended from the centre of a uniform pulley, B. A light

inextensible flexible cord passes around the periphery of the pulley and is

attached to a fixed upper surface, on one side via a spring of stiffness k =50 N/m.

The pulley has a mass of 2 kg, a radius of 60 mm, and a moment of inertia I = 3.6

×10-4 kg m2.

Using the principle of work and energy, derive the second-order differential

equation describing the motion of A, and calculate the frequency of vibration.


Sample Problem

21

What is the kinetic energy of the system?

1

2mAv 2

1

2mBv 2

1

2I

v

r

2

2.55v 2

What is the potential energy of the system?

P.E.=

(mA mb )gx 1

2k(2x)2 49.1x 100x2

Consider conservation of mechanical energy and take the time derivative =0.

2.55v2 49.1x 100x2 const

2.55 2vdv

dt

49.1

dx

dt100 2x

dx

dt

0

5.1dx

dt

d2x

dt2 49.1

dx

dt 200x

dx

dt 0

5.1d2x

dt2 200x 49.1

d2x

dt2 39.22x 0

39.22 6.26

Principle of work and energy: P.E. + K.E. = const. Therefore,

K.E. =

𝑥 = 𝑟𝜃, v = rω = 𝑟 𝜃Displacement of spring =2x


The damping force is:

where c is the damping constant (N⋅s/m)

Damping is a forcenon-conservative

Damping – The Dashpot

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Non-Conservative Forces

Non-conservative forces are not a function of

position only (e.g. air resistance, dry friction)

If a system has non-conservative forces, then:

work done on

system/mass by non-

conservative forces

23


Since damping force is non-

conservative, we use:

U is work done by damping:

(at x=0, spring extension=0)

⇒ Differentiate (with respect to time):

The mass-spring damper system

24


Divide through by :

If we redefine x as distance from

equilibrium:

(a damped 2nd-order

system) What happens if we move mass and release?

(step response)

The mass-spring damper system

25


A) B)

C) D)

What is the step response?

t t

x

x

it depends on the particular combination of m, c and k

2nd Order System: Step Response

26


A) B)

C) D)

What is the form of the step response?

t t

x

x

Solution has a component of the form: What is the form of the step response if ω is...

purely imaginary: D)C)

§

A) & B)purely real: complex (both):


27


Trial

solution:

So:

Now use quadratic formula:

What happens to the step response if:

over damped

under damped

critically damped


28


Natural frequency is that with no

damping, i.e.:

For convenience, we define the

damping factor :

This gives:

over damped

under damped

critically damped


29


Damped Free Vibrations

• With viscous damping due to fluid friction,

:maF

0

kxxcxm

xmxcxkW st

• Substituting x = elt and dividing through by elt

yields the characteristic equation,

m

k

m

c

m

ckcm

22

220 lll

• Define the critical damping coefficient s.t.

ncc m

m

kmc

m

k

m

c220

2

2

• All vibrations are damped to some degree

by forces due to dry friction, fluid friction, or

internal friction.

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Damped Free Vibrations• Characteristic equation,

m

k

m

c

m

ckcm

22

220 lll

nc mc 2 critical damping coefficient

• Heavy damping: c > cc

tteCeCx 21

21ll

- negative roots

- nonvibratory motion

• Critical damping: c = cc

tnetCCx

21 - double roots

- nonvibratory motion

• Light damping: c < cc

tCtCex ddtmc cossin 21

2

2

1c

ndc

c damped frequency

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Forced Vibrations

:maF

xmxkWtP stfm sin

tPkxxm fm sin

xmtxkW fmst sin

tkkxxm fm sin

Forced vibrations - Occur

when a system is subjected

to a periodic force or a

periodic displacement of a

support.

f forced frequency

32


Forced Vibrations

txtCtC

xxx

fmnn

particulararycomplement

sincossin 21

22211 nf

m

nf

m

f

mm

kP

mk

Px

tkkxxm fm sin

tPkxxm fm sin

At f = n, forcing input is in

resonance with the system.

tPtkxtxm fmfmfmf sinsinsin2

Substituting particular solution into governing equation,

33


Sample Problem

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.

34


Sample Problem

SOLUTION:

• The resonant frequency is equal to the natural

frequency of the system.

ftslb87.102.32

350 2m

ftlb000,36

inlb30007504

k

W = 350 lb

k = 4(350 lb/in)

rpm 549rad/s 5.57

87.10

000,36

m

kn

Resonance speed = 549 rpm

35


Sample Problem

W = 350 lb

k = 4(350 lb/in)

rad/s 5.57n

• Evaluate the magnitude of the periodic force due to the

motor unbalance. Determine the vibration amplitude

from the frequency ratio at 1200 rpm.

ftslb001941.0sft2.32

1

oz 16

lb 1oz 1

rad/s 125.7 rpm 1200

2

2

m

f

lb 33.157.125001941.02

126

2

mrmaP nm

in 001352.0

5.577.1251

300033.15

122

nf

mm

kPx

xm = 0.001352 in. (out of phase)

36


Unforced governing equation:

Often we are interested in an

oscillatory external force (e.g.

vibration)

Treat external force in the

same way as we did for the

non-conservative damping

force

Governing equation with external

forcing is then:

The response is known as frequency response

Forced 2nd Order System

37


Substitution gives:

Governing equation with oscillatory forcing:

After initial transient response,

solution of the form:

Rearrange for normalized response:

when frequency is very low

(quasi-steady), G=1

Frequency Response Amplitude

38


Natural frequency:

Damping factor:

The frequency response can now

be written:

The amplitude is: The phase lag:

(between force and response)

Frequency Response: Amplitude

39

Introduction to Dynamics (N. Zabaras) 40


Damped Forced Vibrations

2

222

1

2tan

21

1

nf

nfc

nfcnf

m

m

m

cc

cc

x

kP

x

magnification

factor

phase difference between forcing and steady

state response

tPkxxcxm fm sin particulararycomplement xxx

41


Helicopter Ground Resonance

42


Summary

The 2nd-order equation describing damped

oscillation is:

over damped

under damped

critically damped

The response to step inputs or external forces

is characterised by the degree of damping:

If system is under damped ( ) driving near to

the natural frequency causes resonance

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introduction to dynamics - purdue university

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