Module 5 (CE-409: Introduction to Structural Dynamics and Earthquake Engineering)

University of Engineering and Technology

Peshawar, Pakistan

CE-409: Introduction to Structural Dynamics and

Earthquake Engineering

MODULE 5:

UNDAMPED & DAMPED VIBRATIONS IN S.D.O.F

SYSTEMS SUBJECTED TO HARMONIC FORCES

Prof. Dr. Akhtar Naeem Khan & Prof. Dr. Mohammad Javed

[email protected] [email protected]

1

CE-409: MODULE 5 (Fall-2013) 2

Harmonic force A harmonic force is one whose variation which with time is

defined by any one of the following equations

t)Cos(p t)Sin(p p(t) oo or Where po is the amplitude or maximum value of force and ω is its frequency

also called as exciting frequency or forcing frequency; T=2π/ω is the exciting

period or forcing period.

The equations used in this module are strictly applicable to po sin (ωt)

Time variation of harmonic force


A common source of such a sinusoidal force is unbalance in a

rotating machines (such as turbines, electric motors and electric

generators, as well as fans, or rotating shafts).

Unbalance cloth in a rotating drum of a washing machine is also

an harmonic force.

When the wheels of a car are not balanced, harmonic forces are

developed in the rotating wheels. If the rotational speed of the wheels

is close to the natural frequency of the car’s suspension system in

vertical direction , amplitude of vertical displacement in the car’s

suspension system increases and violent shaking occur in car due to

match of frequency of the force (due to vertical component of

harmonic forces acting at unbalanced mass centre) with natural

frequency of car’s suspension system in vertical direction, ωn

Harmonic forces

3


Response of undamped systems subjected

to harmonic forces


Response of undamped systems to harmonic forces

The equation of motion for harmonic vibration of undamped

system is: t)Sin(pku um o The solution to the equation is made up of two parts.

The first part is the solution which correspond to forced

vibration and is known as the Particular Solution. The

corresponding vibration is known as Steady state vibration, for its

present because of the applied force no matter what the initial

conditions .

The second part is the solution to the free vibration, which does

not require any forcing function, this part is known as the

Complimentary solution. The corresponding vibration is known as

Transient Vibration, which depends on the initial conditions


Particular solution of undamped harmonic vibrations

n2

n

op wheret)Sin(

-1

1.

k

p (t)u

up(t) is the displacements corresponding to the Particular

solution (i.e due to forced vibration).

ratiofrequency as termedisωω

n

n2

op wheret)Sin(

r-1

1

k

p (t)u

It can be derived that the particular solution of undamped

vibration is as follows:

For the sack of simplicity, we will use rω in our lectures to represent ω/ωn


Complimentary solution of undamped harmonic vibrations

t)BSin( t)ACos( (t)uc nn ωω uc(t) is the displacements corresponding to the Complimentary

solution (i.e due to free vibration) and depends on initial

conditions.

Complementary solution of undamped vibration is given as

follows:


Complete solution of undamped harmonic vibrations

Complete solution is the sum of complementary solution, uc(t),

and particular solution, up(t)

t)Sin(-1

1

k

p t)BSin( t)ACos( u(t)

2

onn

rωω

The constants ‘A’ and ‘B’ are determined by imposing initial conditions i.e., (0)u u and u(0)u

stateSteady

2

ω

o

Transient

n2

ω

ωo

n

n

t)(Sin r-1

1

k

p

t)Sin(ωr-1

r

k

p

ω(0)u

t)u(0)Cos(ω u(t)


Complete solution of undamped harmonic vibrations

The transient vibration exist even if . In such case the

complementary part of solution given on previous slide specializes to:

0(0)u u(0)

t)Sin(ωr-1

r

k

p

ω0

t)Cos(ω*0 (t)u n2

ω

ωo

n

nc

t)Sin(ωrt)(Sin r-1

1

k

p u(t)or nω2

ω

o

t)Sin(ω1r

r

k

p (t)uor n2

ω

ωoc

The complete solution is then specialized to the following form

CE-409: MODULE 5 (Fall-2013) 10

t)Sin(pku um o If the force is applied slowly then 0u of motion under harmonic force

and the equation

becomes: t)Sin(k

p uor t)Sin(pku o

sto The subscript “st” (standing for static) indicate the elimination of acceleration’s effect The maximum value of static deformation, (ust )o can be interpreted

as the deformation corresponding to the amplitude of p of the force

po:

For brevity we will refer to (ust)o as the static deformation

k

p u o

ost

Amplitude of ‘Static’ deflection due to harmonic force

CE-409: MODULE 5 (Fall-2013) 11

:as written becan t)Sin(r-1

1

k

p (t)u

2

op

t)Sin(r-1

1u (t)u

2ostp

ost

o u k

p where

Effect of Frequency ratio, rω, on the direction of structural displacements

r-1

12

It can be observed from this equation that up(t) has negative sign

when frequency ratio, rω >1 (i.e. ω>ωn), and vice versa. A Graph

on next slide is plotted b/w frequency ratio, rω and

up is positive if this term is positive and vice versa

CE-409: MODULE 5 (Fall-2013) 12


r-1

12

ωωr

nω

CE-409: MODULE 5 (Fall-2013) 13

Following observation can be made from the plot given on

previous slide

When rω < 1 ( i.e ω < ωn ), the displacement is positive,

indicating that up(t) and p(t) has same directions. The displacement

is said to be in phase with the applied force.

When rω > 1 ( i.e ω > ωn ), the displacement is negative ,

indicating that the u(t) and p(t) has apposite direction directions.

The displacement is said to be out of phase with the applied force.


CE-409: MODULE 5 (Fall-2013) 14

Structure displaces in the direction of force if

ω/ωn <1

t)Sin(po

n

u

t)Sin(po

n

u


Structure displaces opposite to direction of force if

ω/ωn >1

4.9 spring-mass second order system frequency response.flv









CE-409: MODULE 5 (Fall-2013) 15

t)Sin(r-1

1u t)Sin(

r-1

1

k

p (t)u

2ost2

op

Where Rd = Dynamic Magnification factor or Deformation (or

displacement) response factor, uo=Amplitude of dynamic

displacement, and, φ= Phase angle

Deformation response factor, Rd

2ost

od

r-1

1

u

u R Where

Another mathematical form of the above mentioned equation is:

)-tSin(Ru )-tSin(u (t)u dostop ωω

1r i.e., 180

1r i.e., 0 and

no

no

CE-409: MODULE 5 (Fall-2013) 16

Influence of Frequency ratio, rω, on Deformation response factor, Rd

rn

dR

Nearly static response . Load may be defined as quasi-static when rω ≤ 0.2

2 rat 1R ωd 2 r when 1R ωd

CE-409: MODULE 5 (Fall-2013) 17

Following observation can be made from the plot

When rω is small ( i.e force is ‘slowly varying’), Rd is only slightly

greater than 1 or in the other words amplitude of dynamic deformation,

uo, is almost same as amplitude of static deformation, (ust)o.

and the dynamic

deformation amplitude is less than static deformation

When become smaller and

becomes zero as

When rω is close to 1.0, Rd is many times larger than 1

1),2 (i.e 2rWhen n dR

dR ,2beyond increases r r

Influence of Frequency ratio, rω, on Deformation response factor, Rd

CE-409: MODULE 5 (Fall-2013) 18

A video camera, of mass 2.0 kg, is mounted on the top of a bank building

for surveillance. The video camera is fixed at one end of a tubular

aluminum rod whose other end is fixed to the building as shown in Fig.

The wind-induced force acting on the video camera, is found to be

harmonic with p(t) = 25 sin 75t N. Determine the cross-sectional

dimensions of the aluminum tube if the maximum amplitude of vibration

of the video camera is to be limited to 0.005 m. E Aluminum = 71 GPa

Posin ωt

Problem M5.1

CE-409: MODULE 5 (Fall-2013) 19

Response of damped systems under

Harmonic forces

CE-409: MODULE 5 (Fall-2013) 20

Response of damped systems under Harmonic forces

The equation of motion for harmonic vibration of damped

system is: t)Sin(pku uc um o

This equation is to be solved subjected to initial conditions

(0)u(t)u and u(0)u(t) The particular solution of this differential equation is

t)DCos( t)CSin( (t)up ωω

CE-409: MODULE 5 (Fall-2013) 21

Where

222

2o

r2r-1

r-1

k

pC

222

o

r2r-1

r2

k

pD


CE-409: MODULE 5 (Fall-2013) 22

The complementary solution is:

t)BSin( t)ACos( e (t)utζω

cn

DD ωω The complete solution is:

stateSteady

Transient

tζω

t)DCos( t)CSin(

t)BSin( t)ACos(eu(t) n

ωω

ωω DD


CE-409: MODULE 5 (Fall-2013) 23

Transient response (difference b/w total response and steady-state

response diminishes after few cycles of forced vibration in

damped systems and after this stage

u

u(t)

st o

t/T

Steady state response of damped systems under Harmonic forces

0 (t)u as (t)u u(t) i.e., cp (t)u(t)u u(t), pc

(t)u, p

23

CE-409: MODULE 5 (Fall-2013) 24

2 2 2ost

od

r2r-1

1

u

uR

Substituting the values of C and D (given on slide 20) in above

given equation results in:

Effect of rω and ζ on the Deformation response factor, Rd

)-tSin(Ru )-tSin(u (t)u dostop ωω The Steady state deformation can be rewritten as:

Tan and DCu Where 1-22o

C

D

r-1

r2Tan and

2

1-

CE-409: MODULE 5 (Fall-2013) 25


, rω

CE-409: MODULE 5 (Fall-2013) 26

Following observation can be made from the plot

Damping reduce Rd for all values of frequency ratio, rω. However rate

of reduction highly depend on the magnitude of rω (around 0.5 to 1.5)

If the rω is around 0.2 and below, ( i.e force is ‘slowly varying’), Rd is

only slightly greater than 1 and thus unaffected by damping. Thus

1r provided/k p uu oosto If the rω very high, around 2 and above,( i.e force is ‘rapidly varying’),

Rd tends to zero. In other words Rd is unaffected by damping. uo can be

approximated as: 1r provided /mp 2

o uo


CE-409: MODULE 5 (Fall-2013) 27

If the rω≈ 1( i.e frequency of force is close to natural frequency), Rd is very sensitive to damping. If rω =1, Rd as given by:

2 2 2ost

od

r2r-1

1

u

uR

21

1.2ζ1-1

1

u

uR

2 2 2ost

od becomes

This means that for a system with 5% of critical damping the

maximum displacement of the dynamic response is 10 times the

equivalent static displacement.


CE-409: MODULE 5 (Fall-2013) 28

n

2

n

ooost

o

2mωc

2

1.

mωp

2ζ1

.k

p

2ζu

u

The last equation on previous slide can be alternatively written as

n

on

2

n

oo

cωp

c

mω.

mωp

u


CE-409: MODULE 5 (Fall-2013) 29

An air-conditioning unit weighing 1200 lb is bolted at the middle of two

parallel simply supported steel beams. The clear span of the beams is 8 ft.

The second moment of cross-sectional area of each beam is 10 in4. The

motor in the unit runs at 300 rpm and produces an unbalanced vertical

force of 60 lb at this speed. Neglect the weight of the beams and assume

1% viscous damping in the system; for steel E = 30,000 ksi. Determine the

amplitudes of steady-state deflection.

Problem M 5.2

CE-409: MODULE 5 (Fall-2013) 30

Dynamic Response Factors

Now we introduce deformation (or displacement), velocity, and

acceleration response factors that are dimensionless and define the

amplitude of these three response quantities.

The steady-state displacement from slide 23 is reproduced below )-tSin(Ru )-tSin(u (t)u dostop ωω By differentiating both sides, a relation can be developed between

Velocity response factor, Rv and Rd ,which is given below.

RrR dωv

Similarly, another relation can also be developed between

Acceleration response factor, Ra and Rd ,which is given below.

vRrRrR ωd

2

ωa

CE-409: MODULE 5 (Fall-2013) 31 Rd,R

v and Ra for a damped system excited by harmonic force.

Rd

Rv

Ra

CE-409: MODULE 5 (Fall-2013) 32

1. A spring-mass-damper system is subjected to a harmonic force. The

amplitude is found to be 20 mm at resonance and 10 mm at a frequency

0.75 times the resonant frequency. Find the damping ratio of the system.

2. An air compressor of mass 100 kg is mounted on an elastic foundation.

It has been observed that, when a harmonic force of amplitude 100 N is

applied to the compressor, the maximum steady-state displacement of 5

mm occurred at a frequency of 300 rpm. Determine the equivalent

stiffness and damping constant of the foundation.

3. A 50-kg machine tool is mounted on an elastic foundation. An

experiment is run to determine the stiffness and damping properties of

the foundation. When the tool is excited with a harmonic force of

magnitude 8000 N at a variety of frequencies, the maximum steady–state amplitude obtained is 2.5 mm, occurring at a frequency of 32 Hz.

Use this information to determine the stiffness and damping ratio of the

foundation.

Problems

32

CE-409: MODULE 5 (Fall-2013) 33

4. The steel frame shown in figure

supports a rotating machine which

exerts a horizontal force at the

girder level, p(t)=200 Sin 5.3t lb.

Assuming 5% of critical damping,

determine:

(a) The amplitude of the dynamic displacement and

(b) The amplitude of equivalent static force.

Take E= 29,000 ksi and I= 69.2 in4

p(t) 15 kips

15 ft

HA1M5

Problems

Solve problems 1 and 4

CE-409: MODULE 5 (Fall-2013) 34

Vibration Isolation

CE-409: MODULE 5 (Fall-2013) 35

High vibration levels can cause machinery failure, as well as

objectionable noise levels.

A common source of objectionable noise in buildings is the

vibration of machines that are mounted on floors or walls. A typical

problem is a rotating machine (such as a pump, AC compressor,

blower, engine, etc) mounted on a roof, or on a floor above the

ground floor.

The problem is usually most apparent in the immediate vicinity

of the vibration source. However, mechanical vibrations can

transmit for long distances, and by very circuitous routes through

the structure of a building, sometimes resurfacing hundreds of feet

from the source.

Vibration Isolation

CE-409: MODULE 5 (Fall-2013) 36

A related problem is the isolation of vibration-sensitive

machines from the normally occurring disturbances in a building

(car or bus traffic, slamming doors, foot traffic, elevators…).

Examples of sensitive machines include surgical microscopes,

electronic equipment, lasers, and computer disk drives.

A common example of a vibration source is shown in figure, a

large reciprocating air conditioning compressor weighing 20,000

pounds, mounted on a roof. Annoying noise levels at multiples of

the compressor rotational frequency, predominantly 60 and 120

Hz, were measured in the rooms directly below the compressor.

Also, this type of compressor (reciprocating) is notorious for

high vibration levels. Centrifugal or scroll type compressors are

much quieter, but more expensive.

Vibration Isolation

CE-409: MODULE 5 (Fall-2013) 37

Vibration Isolation

CE-409: MODULE 5 (Fall-2013) 38

A Consider a vibrating machine, bolted to a rigid floor ( see

figure a on next slide). The force transmitted to the floor is equal to

the force generated in the machine.

The transmitted force can be decreased by adding a suspension

and damping elements (often called vibration isolaters) Figure b , or

by adding what is called an inertia block, a large mass (usually a

block of cast concrete), directly attached to the machine (Figure c).

Another option is to add an additional level of mass (sometimes

called a seismic mass, again a block of cast concrete) and

suspension (Figure d).

Vibration Isolators

CE-409: MODULE 5 (Fall-2013) 39

Vibration Isolators

Vibration isolation systems: a) Machine bolted to a rigid foundation b) Supported on

isolation springs, rigid foundation c) machine attached to an inertial block d)

Supported on isolation springs, non-rigid foundation (such as a floor); or machine on

isolation springs, seismic mass and second level of isolator springs

Input force,

p(t) = po sin ωt

Transmitted

force, fT(t)

CE-409: MODULE 5 (Fall-2013) 40

Typically vibration isolators employ a helical spring to provide

stiffness, and an elastomeric layer (such as neoprene) to provide

some damping.

Other types use a solid elastomeric element for both the stiffness

and the damping.

Vibration Isolators

Application of elastic sleeper pads for vibration

isolation, and adjustment of track stiffness

CE-409: MODULE 5 (Fall-2013) 41

Consider the mass-spring-damper

system subjected to a harmonic force.

The force transmitted to the base, fT,

is:

uckufff DST

By substituting, solving and rearranging we get:

2 ω

2 2ω

2ω

o

o

r2r-1

r21

pTR

Tf

Transmission of harmonic forces to base

Where TR is used to represent Transmissibility

m

k c

tSinptp o )(

Tf

CE-409: MODULE 5 (Fall-2013) 42


CE-409: MODULE 5 (Fall-2013) 43

The magnitude of transmitted force reduces with increase

in . The force transmitted to base can be

decreased by decreasing the value of ωn in such a way so that

The force transmitted to the base can also be reduced by

decreasing damping ratio. Although damping reduces the

amplitude of mass for all frequencies, it reduces maximum

force transmitted to the foundation only if . Below

that value, the addition of damping increases the transmitted

force

2beyond rω

2 rω

2 rω


CE-409: MODULE 5 (Fall-2013) 44

If the speed of a machine (forcing frequency) varies, we must

compromise in choosing the amount of damping to minimize the

transmitted force.

The amount of damping should be sufficient to limit the

amplitude of displacement and the transmitted force, while passing

through the resonance, but not so much to increase unnecessarily

the force transmitted at the operating speed (see the effect of

damping on force transmission, from the graph, when )

Luckily, natural rubber is a very satisfactory material and is

often used for the isolation of vibration

2 rω


CE-409: MODULE 5 (Fall-2013) 45

A rotating machine with a 600 kg mass operating at a constant

speed produces harmonic force in vertical direction. The

harmonic force is expressed as p(t)= 5000 Sin 150t, where p(t)

is in N. If the damping ratio of isolators at the foundation of

machine is 7.5%, determine the stiffness of isolators so that the

Transmissibility at the operating speed does not exceed 0.15.

Also determine the amplitude of force transmitted to the

foundation

Problem M 5.3

CE-409: MODULE 5 (Fall-2013) 46

2 ω

2 2ω

2ω

go

ot

r2r-1

r21

u

uTR

Base Excitations (Transmission of harmonic displacements from base)

If the ground motion is defined as , it can be

shown that the amplitude of the total displacement

of the mass can be calculated from the same formula that is used

for transmission of force from a system to its foundation. i.e.,

t)sin(uu got

g u

to (t) u t

CE-409: MODULE 5 (Fall-2013) 47

m

k c

Base

tu

ug


CE-409: MODULE 5 (Fall-2013) 48

tu

ug

k

c

m Base


CE-409: MODULE 5 (Fall-2013)

An automobile is modeled as SDOF system vibrating in

Vertical Direction. It is driven along a road where the elevation

varies sinusoidally. The distance from peak to trough is 0.2m and

the distance along the road b/w the peaks is 35 m. If the natural

frequency of automobile is 2 Hz and the damping ratio of the shock

absorbers is 0.15, determine the maximum displacement by which

the automobile jump while moving at a speed of 60km/hr.

Do you think that the shock absorbers have appropriate damping

ratio.?

If the speed of the automobile is varied, find the most unfavorable

speed for passengers.

Problem M 5.4

49

CE-409: MODULE 5 (Fall-2013) 50

m

k c

Base

tu

ug

tu

ug

k

c

m Base

Base Excitations (Transmission of harmonic

accelerations from base)

gouotuTR

nω ωωr

CE-409: MODULE 5 (Fall-2013) 51

2 ω

2 2ω

2ω

go

ot

r2r-1

r21

u

uTR

Base Excitations (Transmission of harmonic

accelerations from base)

If accelerations acting at the base of a system varies sinusoidally,

it can be proved that:

CE-409: MODULE 5 (Fall-2013) 52

1. The steel frame shown in figure is subjected to a sinusoidal

ground motion ug(t)=0.2 Sin 5.3t inches. Determine the amplitudes

of displacement , the equivalent static force and acceleration at the

top end. Take E= 29,000 ksi and I= 69.2 in4 and ζ = 0.05

15 kips

15 ft

Problems

CE-409: MODULE 5 (Fall-2013) 53

2: A delicate instrument is to be spring mounted to the floor of a test

laboratory where it has been determined that the floor vibrates freely with

harmonic motion of amplitude 0.1in at 10 cycles per second, If the

instrument weighs 100lb, determine the stiffness of isolation springs

required to reduce the vertical motion amplitude of the instrument to 0.01

in. neglect damping.

3. When the person stands in the centre of the floor system shown, he

causes a Deflection of 0.2 in. of floor under his feet. He walks (or runs

quickly) in the same area , how many steps per second would cause the

floor to vibrate with the greatest vertical amplitude

53

Problems

CE-409: MODULE 5 (Fall-2013) 54

Problems

54

4. What is the required column stiffness of single one-story structure to limit

its acceleration amplitude to 2.1 m/s2 during an earthquake whose

acceleration amplitude is 150 mm/s2 at a frequency of 50 rad/s? The mass of

structure is 1800 kg. Assume a damping ratio of 0.05.

5. A 10-kg laser flow-measuring device is used on a table in a laboratory.

Because of operation of other equipment, the table is subject to vibration.

Accelerometer measurements show that the dominant component of the

table vibrations is at 300 Hz and has an amplitude of 4.3 m/s2. For effective

operation, the laser can be subject to an acceleration amplitude of 0.7 m/s2.

(a) Design an undamped isolator to reduce the transmitted acceleration, to

an acceptable amplitude.

(b) Design the isolator such that it has a damping ratio of 0.04.

CE-409: MODULE 5 (Fall-2013) 55

6. A 150-kg engine operates at speeds between 1000 and 2000 rpm. It is

desired to achieve at least 85 percent isolation at all speeds. The only

readily available isolator has a stiffness of 5x 105 N/m. How much mass

must be added to the engine to achieve the desired isolation?

Problems

HA2M5

Solve problems 3,4 and 6