sdof-1211798306003307-8

Prof. A. Meher PrasadProf. A. Meher Prasad

Department of Civil EngineeringDepartment of Civil EngineeringIndian Institute of Technology MadrasIndian Institute of Technology Madras

email: [email protected]

Outline

Degrees of Freedom

Idealisation of SDOF System

Formulation of Equation of motion

Free vibration of undamped/damped systems

Forced vibration of systems

Steady state response to harmonic forces

Determination of natural frequency

Duhamel’s Integral and other methods of solution

Damping in structures

What is Dynamics ?

Basic difference between static and dynamic loading

P P(t)

Resistance due to internal elastic forces of structure

Accelerations producing inertia forces (inertia forces form a significant portion of load equilibrated by the internal

elastic forces of the structure)

Static Dynamic

Characteristics and sources of Typical Dynamic Loadings

(a)

(b)

Periodic Loading:

Non Periodic Loading:

Unbalanced rotating machine in building

Rotating propeller at stem of ship

(c)

(d)

Bomb blast pressure on building

Earthquake on water tank

(a) Simple harmonic (b) Complex (c) Impulsive (d) Long duration

Dynamic Degrees of Freedom

The number of independent displacement

components that must be considered to

represent the effects of all significant

inertia forces of a structure.

Examples

Massless spring

Spring with mass

(a) (b) (c)

Inextensible Spring

θ

1. 2.


3.Rigid bar with distributed mass

Massless spring

Flexible and massless

Flexible and massless

Point mass

Finite mass

(a) (b) (c)

4. Flexible and massless

Point mass

Flexible beam with distributed mass

Flexible beam with distributed mass

(a) (b) (c)


y(x) = c11(x)+ c22(x)+……

5.

Rigid deck

Massless columns


Idealisation of Structure as SDOF

Mathematical model - SDOF System

Mass element ,m - representing the mass and inertial characteristic of the

structure

Spring element ,k - representing the elastic restoring force and potential energy capacity of the

structure.

Dashpot, c - representing the frictional characteristics and energy losses of the structure

Excitation force, P(t) - represents the external force acting on structure.

P(t)

x

m

k

c

Newton’s second law of motion

Force = P(t) = Rate of change of momentum of any mass

=

When mass is not varying with time,

P(t) = m x(t) = mass x acceleration..

Inertia force

P(t)

x, x

m x..

D’Alembert’s Principle: This Principle states that “mass develops an inertia force proportional to its acceleration and opposing it”.

..

mg

N

kx mx..

The force P(t) includes ,

1) Elastic constraints which opposes displacement

2) Viscous forces which resist velocities

3) External forces which are independently defined

4) Inertia forces which resist accelerations

Equations of motion:

Spring force - fs x

Viscous damping force - fd x

Inertia Force - fI x

External Forces - P(t)

.

..

k

1

fs

x

c

1

fD

x.

Examples

P(t)

x

m

k

c

FBD for mass

δst = w/k

x(t) = displacement measured from position of static equilibrium

P(t)

1.

2.

fs = kx

fd = cxP(t).

(1)

(2)

f I = m x..

2

2;

dx d xx x

dt dt

( )mx cx kx P t

( )mx cx kx P t

P(t) w

Kx + w0+ cx

mx..

.

Rigid ,massless P(t) m

a

b

d

L

(3a) P(t)

m x..

ak x

L

x – vertical displacement of the mass measured from the position of static equilibrium

bc x

L

k c

2 2

( )b a d

mx c x k x P tL L L

(3)

x

ak x

L

P(t)

m x..

bc x

L

Rigid massless

m

a

b

d

L

k

c

2 2

( )b a W d

mx c x k x P tL L L L

P(t)

Stiffness term

(3b)

W=mg

(4)

Note: The stiffness is larger in this case

x

2 2

( )b a W d

mx c x k x P tL L L L

Rigid massless

m

a

b

d

L

k

cP(t)

ak x

L

P(t)

m x

..

Stiffness term

(3c)

bc x

L

.

(5)

Note: The stiffness is decreased in this case. The stiffness term goes to zero - Effective stiffness is zero – unstable - Buckling load

x

2 21

( )3

b a dm L x c x k x P t

L L L

(4a) μ m

a

b

d

L

k c

P(t)(distributed mass) P(t)

m x..

s

af k x

L

d

bf c x

L

.

2

Lx

(2/3)L

x

(4b)

2 21 1

( )3 2

b a W dm L x c x k g x P t

L L L L

m

a

b

d

L

k

cP(t)

μ

P(t)

m x..

s

af k x

L

d

bf c x

L

2

Lx

(Negative sign for the bar supported at bottom)

(2/3)L

x

Special cases:

x

LL

k

θJ

0J k 0g

x xL

3

02

gx x

L

(4c) (4d) (4e)

Pe(t)

x

me

ke

ce

( )e e e em x c x k x P t

me - equivalent or effective mass

Ce - equivalent or effective damping coefficient

Ke - equivalent or effective stiffness

Pe - equivalent or effective force

(6)

Rigid ,masslessRigid with uniform mass μL/2 = m/2

L/4 L/4 L/8 L/8 L/4

k c

m/2

N

x(t)

P(t)

2

xk

2

xc

P(t)2 2 4

L x mx

2 2

m x

N

RL

7 1 1 16 31 ( )

24 4 4 4

Nm x c x k x P t

kL

(5)

o

N

N

Internal hinge

7 1 1 161

24 4 4e e e

Nm m c c k k

kL

3( ) ( )

4eP t P t

For N = - (1/16) k L ke = 0

This value of N corresponds to critical buckling load

(7)

Free Vibration

Undamped SDOF System

Damped SDOF systems

Free Vibration of Undamped System

General solution is,

x(t) = A cos pt + B sin pt (or)

x(t) = C sin (pt + α)

where,

2 0x p x

2 2C A B

2 kp

m

22 natural period

mT

p k

1natural frequency

2

pf

T

p - circular natural frequency of undamped system in Hz.

(9)

(10)

(11)

(12)

(13)

(14)

(15)

00( ) cos sin

vx t x pt pt

p

Amplitude of motion

t

x

vo

2

2 00

vx

p

2T

p

or

2

2 00( ) sin ( )

vx t x pt

p

where, 0

0

tanx

v p

(16)

(17)

x 0

X =initial displacement

V =initial velocity

0

0

t

Natural frequencies of other SDF systems

p – square root of the coefficient of displacement term divided by coefficient of acceleration

For Simple Pendulum, p g L

6 161

7

k Np

m kL

2k a g

pm L L

For system considered in (3b) ,

For system considered in (5) ,

For N=0 ,

and for

6

7o

kp p

m

, 0p 1

16N kL

(18)

(19)

(20)

(21)

Condition of instability

1

16 crN kL N

1ocr

Np p

N

p2

N

Ncr

(22)

(23)

Natural frequencies of single mass systems

/p k m

Letting m = W/g

and noting that W/k = δst

δst is the static deflection of the mass due to a force equal to its weight (the force applied in the direction of motion).

st

gp

1

2 st

gf

δst is expressed in m, 2 stT

(10)

(24)

(25)

(26)

(27)

Relationship between Simple oscillator and Simple pendulum

L

st

gp

L

gp

Hence, δst = L = 0.025 m f ≈ 3.1 cps δst = L = 0.25 m f ≈ 1.0 cps

δst = L = 2.50 m f ≈ 0.3 cps

Effective stiffness ke and static deflection δst

e

st

k gp

m

ke - the static force which when applied to the mass will deflect the mass by a unit amount.

δst - the static deflection of the mass due to its own weight the force (weight) being applied in the direction of motion.

(28)

1. Apply the static force ,F on the mass in the direction of motion

2. Compute or measure the resulting deflection of the mass ,∆

Then , ke = F / ∆ δst = ∆ due to F = W

Determination of Force - Displacement relation, F-∆

Examples

(a)

Rigid ,massless m

a

L

k

Fa

L

F

L F

a k

F

2L F

a k

2L F

a k

Therefore,

2

e

F ak k

L

or

2

st

L W

a k

From Equilibrium,

From Compatibility,

(29)

Rigid bar m

aL

k1

F = F1 + F2

k2

k1 k2

21

2

21

2

kkL

aFk

kkL

aF

e

(b) (c)Rigid bar

m

aL

k1

F

k2

k1 k2

k3

k3

∆

∆ = ∆1 + ∆2 =

21

23 kk

La

F

k

F

321

2

111

kkk

LakF e

(30) (31)

k1 kn

ke = k1 +k2 + ……+ kn

(d)

(e)k1

k2

kn

...

∆e = ∆1 + ∆2 + ……+ ∆n

nk

F......

k

F

k

F

21

ne k......

kkkF

1111

21

(f) Flexible but mass less

3

3

L

EIke

3

12

L

EIke

3

3

L

EIke

(32)

(34)

(33)

(g) Rigid deck; columns mass less & axially inextensible

E1I1E2I2

L1L2

Lateral Stiffness :

1 1 2 23 3

12 3e

E I E Ik

L L

(h)

EI

k1

k2

L

k2

k1kb

F

k2

kb+k1

2 1 213

1 1 1 1 13

e bEIk k k k k kL

(i)EI

L/2

k

L/2

R

F

∆

3 35 1

48 24a

FL RL R

EI EI k

where,

3

5

2 48R F

EIL

3 31 5

3 48

FL RL

EI EI

Eliminating R,3

3

3

768 7 1

3768 32

kLFLEI

kL EIEI

where, 3

3 3

768 323

768 7e

kLF EIEIk

kL LEI

(j)

L/2 L/2

3

48e

EIk

L

L/2 L/2

3

192e

EIk

L

L/2 L/2

3

768

7e

EIk

L

a b

2 2

3e

EILk

a b

(k)

(l)

(m)

2R

d

4

364

Gdk

nR

(n)

n – number of turns

(0)

(p)

(q)

EIk

L

AEk

L

L

L

GJk

L

I - moment of inertia of cross sectional areaL - Total length

J – Torsional constant of cross section

A – Cross sectional area

A, E, I, L

a

m

Natural frequencies of simple MDF systems treated as SDF

(i) (ii) (iii) (iv)

Columns are massless and can move only in the plane of paper

• Vertical mode of vibration

2 2e v

AE AEk p

L mL (35)

• For pitching or rocking mode

(36)

• For Lateral mode2

3 3

122 24 24e

EI EI AE rk

L L L L

r is the radius of gyration of cross section of each column2

3 3

122 24 24e

EI EI AE rk

L L L L (37)

1 20

2 2 31

061

06

63p v

a a AEy ya

LAE

ay yL

AEmy y

L

AEp p

mL

..

..

..

AE/L AE/L

(AE/L)y (AE/L)y

lateral < axial < pitchingppp

Free Vibration of Damped SDOF

Free Vibration of damped SDOF systems

km

c

mp

cζ

m

kp

22

(Dimensionless parameter)

(A)

(38)

where,

2

0

0

2 0

mx cx kx

c kx x x

m m

x ζpx p x

x

m

k

c

Solution of Eq.(A) may be obtained by a function in the form x = ert where r is a constant to be determined. Substituting this into (A) we obtain,

2 22 0rte r ζpr p

In order for this equation to be valid for all values of t,

2 2

21,2

2 0

1

r ζpr p

r p

or

Thus and are solutions and, provided r1 and r2 are different from one another, the complete solution is

trtr 21 ee

1 21 2

r t r tx c e c e

The constants of integration c1 and c2 must be evaluated from the initial conditions of the motion.

Note that for >1, r1 and r2 are real and negative

for <1, r1 and r2 are imaginary and

for =1, r1= r2= -p

ζζ

ζ

ζSolution depends on whether is smaller than, greater than, or equal to one.

For (Light Damping) :1

0

0021d

A x

vB x

p

2

cos sin

1

ptd d

d

x t e A p t B p t

p p

‘A’ and ‘B’ are related to the initial conditions as follows

(39)

(40)

(41) 2

cos sin1

pt oo d o d

d

vx t e x p t x p t

p

In other words, Eqn. 39 can also be written as,

where,

2

2Damped natural period

1 Damped circular natural frequency

d

d

Tp

p p

Extremum point ( )( ) 0

cos( ) 1d

t

p t

x

Point of tangency ( ) Td = 2π / pd

xn Xn+1

t

x

2

2Damped natural period

1 Damped circular natural frequency

dd

d

Tp

p p

Motion known as Damped harmonic motion

A system behaving in this manner (i.e., a system for which ) is said to be Underdamped or Subcritically damped

The behaviour of structure is generally of this type, as the practical range of is normally < 0.2

The equation shows that damping lowers the natural frequency of the system, but for values of < 0.2 the reduction is for all practical purpose negligible.

Unless otherwise indicated the term natural frequency will refer to the frequency of the undamped system

1

Rate of Decay of Peaks

2

1

2exp 2

1d

ppn

n

xe

x

(42)

1n

n

x

x

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 0.1 0.2 0.3 0.4

It is an alternative measure of damping and is related to by the equation

Defined as1

ln n

n

x

x

22 2

1

1ln n

n N

x

N x

When damping is quite small,

Logarithmic decrement

(43)

(44)

(46)

For small values of damping,

2n

n

x

x

(45)

Such system is said to be over damped or super critically damped.

1

i.e., the response equation will be sum of two exponentially decaying curve

In this case r1 and r2 are real negative roots.

( ) ( )1 2( ) t tx t C e C e

For (Heavy Damping)

xo

x

o t

Such system is said to be critically damped.

1 2( ) pt ptx t C e C te

1

The value of ‘c’ for which is known as the critical coefficient of damping

With initial conditions,

0 0( ) 1 ptx t x pt v t e

1

2 2crC mp km

Therefore,cr

C

C

For

(47)

(48)

Response to Impulsive Forces

Response to simple Force Pulses

Response to a Step Pulse

Response to a Rectangular Pulse

Response to Half-Sine Pulse

Response to Half-cycle Force Pulses

Response to Step force

Response to Multi-Cycle Force Pulses

Let the duration of force,t1 be small compared to the natural period of the system

The effect of the force in this case is equivalent to an instantaneous velocity change without corresponding change in displacement

The velocity,V0 ,imparted to the system is obtained from the impulse-momentum relationship

mV0 = I = Area under forcing function = α P0 t1

1 for a rectangular pulse

where , α 2 / π for a half-sine wave 1 / 2 for a triangular pulse

Therefore, V0 = 0 1

Pt

m

Response to Impulsive Forces

t

P(t)

Po

t1 << T

(50)

(49)

For an undamped system, the maximum response is determined from as ,

Therefore,

or

0 0 1 0 10 1max

( )stx P t P ktV x ptp m p mpk

max

0

112 2

( )st

txft

x T (51)

1 0max2 ( )stx ft x

•Damping has much less importance in controlling maximum

response of a structure to impulsive load.

The maximum will be reached in a very short time, before the damping forces can absorb much energy from the structure.

For this reason only undamped response to Impulsive loading is considered.

• Important: in design of Vehicles such as trucks, automobiles

or traveling cranes

= Static displacement induced by exciting force at time, t

2

2 2

( )2

2 ( )

( )( )

st

st

P tx px p x

m

x px p x p x t

P tx t

k

or

where,

( )P t

k

Response to simple Force Pulses

(52)

(53)

(54)

t

P(t)

General Form of solution:

x(t) = xhomogeneous + xparticular

Response to a Step Pulse

For undamped system, x + p2 x = p2 (xst)o

where (xst)o =

x(t) = A cos pt + B sin pt + (xst)o

At t = 0 , x = 0 and v = 0

A = - (xst)o and B = 0

x(t) = (xst)o [1 – cos pt] = [ 1 – cos 2π ]

k

Po

Tt

t

Po

0

P(t)

(55)

For damped systems it can be shown that:

0 2

( ) ( ) 1 cos sin1

d d

ptstx t x e p t p t

21max

0

1( )st

xe

x

Response to a Step Pulse….

(56)

2

1

0

( )

( )st o

x t

x

=0

(t / T)

For t t1, solution is the same as before,

For t t1, we have a condition of free vibration, and the solution can be obtained by application of Eq.17a as follows:

2

21( ) sin ( )ix

Vix t p t t

p

/tan i

iV

xp

where,

P(t)

t

Po


(55)

t1 0

1 cos pti st

x x

0

1 cos( ) ptst

x t x

1siniV p pt

11 1

1 11

1

22sin1-cos 2tan tan2sin 2sin cos

2 2

2

ptpt pt

pt ptpt

pt

hence,

2 2 11 1 10

( ) (1 cos ) sin ( ) sin ( )2st

tx t pt pt x p t t p


1 1 10 1 0

( ) ( ) 2(1 cos ) sin 2 sin sin2 2 2st st

t pt tx t x pt p t x p t

(57(a))

(57(b))

(Amplitude of motion)

So,

t1/T=1.5

t1/T=1 1/6

t1/T=2

In the plots, we have implicitly assumed that T constant and t1 varies;

Results also applicable when t1 = fixed and T varies

0 0

0 0

1 2 1 2

1 12

Response to a Rectangular Pulse…

t1/T=1/

t/T

t/T

t/T

t/T

1.682

22

x(t)

/(x s

t) 0x(

t)/(

x st) 0

Dynamic response of undamped SDF system to rectangular pulse force. Static solution is shown by dotted lines

Forced response

Free response

Overall maximum

Response to rectangular pulse force: (a) maximum response during each of forced vibration and free vibration phases; (b) shock spectrum

(a)

(b)

This diagram Is known as the response spectrum of the system for the particular forcing function considered.

Note that with xmax determined, the maximum spring force

Fmax = k xmax

In fact, max max max

00 0st st

F kx xPF x

max

0st

xx

2

1

021 3

Impulsive solution, 2π f t1

f t1 = t 1/T

3

Response to a Rectangular Pulse…

(58)

P(t) = Po sin ωt, where ω = π / t1

x + p2 x = p2 (xst)o sin ωt for t t1

= 0 for t t1

for t t1,

or

for t t1

or

20

( )( ) [sin sin ]

1

stx

x t t ptpp

0

21 1

1

1sin sin2211

4

( )x t T tst

t t TTt

x t

1 1

02

21( ) cos ( ) sin

2 21

st

p pt tx t x ptT

p

1 1

12

10

cos 1( ) ( ) sin 220.25

t tttT Tx t xst T Tt

T

Response to Half-Sine Pulse

P(t)

t

POsin ωt

t1

(59)

(60)

• Note that in these solutions, t1 and T enter as a ratio and that similarly, t appears as the ratio t /T. In other words, f t1 = t1 / T may be interpreted either as a duration or as a frequency parameter

• In the following response histories, t1 will be presumed to be the same but the results in a given case are applicable to any combination of t1 and T for which t1/T has the indicated value

• In the derivation of response to a half-sine pulse and in the response histories, the system is presumed to be initially at rest

Dynamic response of undamped SDF system to half cycle sine pulse force; static solution is shown by dashed lines

Response to half cycle sine pulse force (a) response maxima during forced vibration phase; (b) maximum responses during each of forced vibration and free vibration phases; (c) shock spectrum

Shock spectra for three force pulses of equal magnitude

t1

t1

t1

2ft1 4ft1

ft1

ft1

• For low values of ft1 (say < 0.2), the maximum value of xmax or AF is dependent on the area under the force pulse i.e, Impulsive-sensitive. Limiting value is governed by Impulse Force Response.

• At high values of ft1, rate of application of load controls the AF. The rise time for the rectangular pulse, tr, is zero, whereas for the half-sine pulse it is finite. For all continuous inputs, the high-frequency limit of AF is unity.

• The absolute maximum value of the spectrum is relatively insensitive to the detailed shape of the pulse(2 Vs 1.7), but it is generally larger for pulses with small rise times (i.e, when the peak value of the force is attained rapidly).

• The frequency value ft1 corresponding to the peak spectral ordinate is also relatively insensitive to the detailed shape of the pulse. For the particular inputs investigated, it may be considered to range between ft1 = 0.5 and 0.8. * AF=Amplification factor

Response to Half-cycle Force Pulses

On the basis of the spectrum for the ‘ramp pulse’ presented next, it is concluded that the AF may be taken as unity when:

ftr = 2 (61)

For the pulse of arbitrary shape, tr should be interpreted as the horizontal projection of a straight line extending from the beginning of the pulse to its peak ordinate with a slope approximately equal to the maximum slope of the pulse. This can normally be done by inspection.

For a discontinuous pulse, tr = 0 and the frequency value satisfying ftr = 2 is, as it should be, infinite. In other words, the high-frequency value of the AF is always greater than one in this case

Conditions under which response is static:

0

0

sin( ) ( )

1 2( ) sin

2

st

st

r r

r r

t ptx t x

t pt

t T tx

t t T

For t tr

For t tr

0

0

sin 1( ) ( ) 1 sin ( )

( )1 1( ) 1 sin 2 sin 2

2 2

st rr r

rst

r r

ptx t x p t t

pt pt

t tT t Tx

t T t T

1 costan

sinr

rr

ptpt

pt

Response to Step force

P(t)

Po

ttr

=

+

0

( )r

r

t tP

t

0r

tP

t

tr

Differentiating and equating to zero, the peak time is obtained as:

Substituting these quantities into x(t), the peak amplitude is found as:

max

0

1 21 2(1 cos ) 1 sin

( ) 2

1[1 sin

rr

st r r

r

r

ptxpt

x pt pt

tT

t T

f tr

max

0( )st

x

x

For Rectangular Pulse:

Half-Sine Pulse:

0

1 10

( ) ( ) [1 cos ]

( ) 2sin sin2 2

st

st

x t x pt

pt tx p t

02

1 1

1

( ) 1( ) sin sin 2

211

4

stx t T tx t

t t TTt

01 1 12

1

( )( ) cos sin 2 -

0.25 ( )stx

ft ft f ft ftft

for t t1

for t t1

for t t1

P(t)

t

Po

P(t)

t

POsinωt

t1

• Line OA defined by equation.51 (i.e = 2 π α f t1 = 2 π α )

• Ordinate of point B taken as 1.6 and abscissa as shown

• The frequency beyond which AF=1 is defined by equation. 61

• The transition curve BC is tangent at B and has a cusp at C

Spectrum applicable to undamped systems.

Design Spectrum for Half-Cycle Force Pulses

AB

C D

ft1

ft1=0.6 ft1= 2

max

xt 0

x

(x )1

2

o

max

0( )st

x

x T

t1

• The high-frequency, right hand limit is defined by the rules given before

• The peak value of the spectrum in this case is twice as large as for the half-sine pulse, indicating that this peak is controlled by the ‘periodicity’ of the forcing function. In this case, the peak values of the responses induced by the individual half-cycle pulses are additive

• The peak value of the spectrum occurs, as before, for a value ft1=0.6

• The characteristics of the spectrum in the left-handed, low-frequency limit cannot be determined in this case by application of the impulse-momentum relationship. However, the concepts may be used, which will be discussed later.

Response to Multi-Cycle Force Pulses

Effect of Full-Cycle Sine Pulse

The absolute maximum value of the spectrum in this case occurs at a value of, ft1=0.5

Where t1 is the duration of each pulse and the value of the peak is

approximately equal to: xmax = n (/2) (xst)o

Effect of n Half-Sine Pulses

(63)

(62)

x(t)I/mp

x(t) I/mp

x(t) 2I/mp

I

It1 t

t

Suppose that t1 = T/2

Effect of first pulse

Effect of second pulse

Combined effect of two pulses

Effect of a sequence of Impulses

• For n equal impulses, of successively opposite signs, spaced at

intervals t1 = T / 2 and xmax = n I/(mp)

• For n equal impulses of the same sign, the above equation holds

when the pulses are spaced at interval t1 = T

• For n unequal impulses spaced at the critical spacings noted

above, xmax = Σ Ij /(mp)

(summation over j for 1 to n). Where Ij is the magnitude of the jth

impulse

• If spacing of impulses are different, the effects are combined

vectorially

Effect of a sequence of Impulses

(64)

(65)

Response of Damped systems to Sinusoidal Force

Response of Damped systems to Sinusoidal Force

P(t) = P0 sinωt

where ω = / t1= Circular frequency

of the exciting force

P(t)

tt1

Solution:

The Particular solution in this case may be taken as

x(t) = M sinωt + N cosωt (a)

Substituting Eq.(a) into Eq.52, and combining all terms involving sinωt

and cos ωt, we obtain

2 2 2 2 20[- - 2 ]sin [- 2 ]cos ( ) sin stM p N p M t N p M p N t p X t

This leads to

(p2-ω2)M - 2 pωN = p2(xst)0

2 pωM+ (p2-ω2)N = 0

Where

The solution in this case is

x(t) = e- pt(A cos pD t +B sin pD t) + sin (ωt – ) (66)

where

(c)

2

2

st 02 2

2

ω1-

pM= (x )

ω ω1- +4ζ

p p

ζ

ζ

st 022 2

2

ω2ζ

pN=- (x )

ω ω1- +4ζ

p p

st 0

2 2 2 2

(x )

(1- ) +4ζ

1

1

p 2ft

2

2tan

(1- )

(67)

(68)

Steady State Response

2 2 2 2st 0

x(t) 1sin( t - )

(x ) (1- ) 4

(69)

max

2 2 2 2st 0

x 1AF

(x ) (1- ) 4

(70)

P(t)

t

x(t)

11, AF

p 2

Note that at (71)

Effect of damping

• Reduces the response, and the greater the amount of damping, the greater the reduction.

• The effect is different in different regions of the spectrum.

• The greatest reduction is obtained where most needed (i.e., at and near resonance).

• Near resonance, response is very sensitive to variation in ζ (see Eq.71). Accordingly, the effect of damping must be considered and the value of ζ must be known accurately in this case.

Resonant Frequency and Amplitude

2res

max 2

p 1- 2

1(AF)

2 1- 2

(72)

(73)

These equations are valid only for 1

2

11

For values of < 2

ωres = 0

(A.F.)max = 1(74)

Transmissibility of system

The dynamic force transmitted to the base of the SDOF system is

Substituting x from Eq.(69), we obtain

0

2 2 2 2

P 1 cF k [sin( t - ) cos( t - )]

k k(1- ) 4

2

c c2 2

k mp p

Noting that and combining the sine and

cosine terms into a single sine term, we obtain

2 2

2 2 2 20

1 4F(t)sin( t - )

P (1- ) 4

where is the phase angle defined by tan = 2ζ

(76)

(77)

(75)cxF kx cx k x

k

The ratio of the amplitudes of the transmitted force and the applied force is defined as the transmissibility of the system, TR, and is given by

2 20

2 2 2 20

1 4FTR

P (1- ) 4

(78)

The variation of TR with and ζ is shown in the following figure. For

the special case of ζ =0, Eq.78 reduces to

2

1TR

(1- )

which is the same expression as for the amplification factor xmax/(xst)0


Transmissibility for harmonic excitation

ω

p

TR


• Irrespective of the amount of damping involved, TR<1 only for values of (ω/p)2 >2. In other words, in order for the transmitted force to be less than the applied force, the support system must be flexible.

e steff

2p f 4.98

Noting that

in cms

stThe static deflection of the system, must be st 2

50

fe

5 10 40fe

0.5

0.1

fe is the frequency of the exciting force, in cps

in

cms

st

(79)

• In the frequency range where TR<1, damping increases the transmissibility. In spite of this it is desirable to have some amount of damping to minimize the undesirable effect of the nearly resonant condition which will develop during starting and stopping operations as the exciting frequency passes through the natural frequency of the system.

• When ζ is negligibly small, the flexibility of the supports needed to ensure a prescribed value of TR may be determined from

2

1TR

-1p

(80a)

Proceeding as before, we find that the value of corresponding to Eq.(80a) is

st

2

st st2 2e e

1 1 251 (in) or 1 (cm)

TR f TR f

(80b)

Application

Consider a reciprocating or rotating machine which, due to unbalance of its moving parts, is acted upon by a force P0 sinωt.

If the machine were attached rigidly to a supporting structure as shown in Fig.(a), the amplitude of the force transmitted to the structure would be P0 (i.e., TR=1).

If P0 is large, it may induce undesirable vibrations in the structure, and it may be necessary to reduce the magnitude of the transmitted force.

This can be done by the use of an approximately designed spring-dashpot support system, as shown in Fig (b) and (c).

P0 sinωt

m

P0 sinωt

k c

m

k c

P0 sinωt

m

mb

(a)(c)(b)

• If the support flexibility is such that is less than the value defined by Eq.(79), the transmitted force will be greater than applied, and the insertion of the flexible support will have an adverse effect.

• The required flexibility is defined by Eq.(80b), where TR is the desired transmissibility.

• The value of may be increased either by decreasing the spring stiffness, k, or increasing the weight of the moving mass, as shown in Fig.(c).

Application to Ground-Excited systems

x(t)

m

k

c

y(t)

For systems subjected to a sinusoidal base displacement, y(t) = y0 sinωt it can be shown that the ratio of the steady state displacement amplitude, xmax, to the maximum displacement of the base motion, yo, is defined in Eq.(78).

Thus TR has a double meaning, and Eq.(78) can also be used to proportion the support systems of sensitive instruments or equipment items that may be mounted on a vibrating structure.

For systems for which ζ ,may be considered negligible, the value of required to limit the transmissibility TR = xmax/y0 to a specified value may be determined from Eq.(80b).

Rotating Unbalance

Total mass of machine = M unbalanced mass = m eccentricity = e angular velocity = ω

2

2( ) ( sin ) 0

dM m x m x e t cx kx

dt

2 sinMx cx kx me t

ωte

M m

k/2 k/2c

x

Reciprocating unbalance

2 sin sin 2e

F me t tL

e - radius of crank shaft

L - length of the connectivity rod

e/L - is small quantity second term can be neglected

ωte

m

M

L

Structure subjected to a sinusoidally varying force of fixed amplitude for a series of frequencies. The exciting force may be generated by two masses rotating about the same axis in opposite direction

For each frequency, determine the amplitude of the resulting steady-state displacement ( or a quantity which is proportional to x, such as strain in a member) and plot a frequency response curve (response spectrum)

For negligibly small damping, the natural frequency is the value of fe for which the response is maximum. When damping is not negligible, determine p =2πf from Eq.72. The damping factor , ζ may be determined as follows:

Determination of Natural frequency and Damping

Steady State Response Curves


Resonant Amplification Method

Half-Power or Bandwidth Method

Duhamel’s Integral

Determine maximum amplification (A.F)max=(x0)max/ (xst)0

Evaluate from Eq.73 or its simpler version, Eq.71, when is small

Limitations: It may not be possible to apply a sufficiently large P0 to measure (xst)0 reliably, and it may not be possible to evaluate (xst)0 reliably by analytical means.

(a) Resonant Amplification Method


(b) Half-Power or Bandwidth Method:

In this method ζ is determined from the part of the spectrum near the peak steps involved are as follows,

1. Determine Peak of curve, (x0)max

2. Draw a horizontal line at a response level of , and

determine the intersection points with the response spectrum.

These points are known as the half-power points of the spectrum

0 max1/ 2 x

3. Evaluate the bandwidth, defined as f

f

(xo)max

(xo)max2

1

(xo)st

f

∆f fe

xo

f

4.For small amounts of damping, it can be shown that ζ is related to the bandwidth by the equation

Limitations:

Unless the peaked portion of the spectrum is determine accurately, it would be impossible to evaluate reliably the damping factor.

As an indication of the frequency control capability required for the exciter , note that for f = 5cps, and ζ = 0.01, the frequency difference

= 2(0.01)5 = 0.1cps

with the Cal Tech vibrator it is possible to change the frequency to a value that differs by one tenth of a percent from its previous value.

1

2

f

f

(81)

Determination of Natural frequency and Damping…

12

2 2

1 1 1

2 (1 ) (2 )2

22 2 2 2

2 2 2

2 2 2 21 1

2 2 2 22 2

1 2 1 21 2

2 1

1 2

1 1

8 1 4

1 2 2 1

1 2 2 1 1

1 2 2 1 1

( ) ( )1 1( )

2 2 2

( )

( )

f f f ff

f

f f

f f

Derivation

(c) Duhamel’s Integral

In this approach the forcing function is conceived as being made up of a series of vertical strips, as shown in the figure, the effect of each strip is then computed by application of the solution for free vibration, and the total effect is determined by superposition of the component effects

Other Methods for Evaluating response of SDF Systems

t

P(t)

x(t)

t

d

P()

o

or

The displacement at time t induced by integration as

The strip of loading shown shaded represents and impulse,

I = P() d

For an undamped SDF system, this induces a displacement

I=P(τ)dτ

( ) sin ( - )

P dx p t

mp

0

1( ) ( )

t

x t P sin p t dmp

0

( ) ( )t

stx t p x sin p t d

(82)

(83)

(84)

1

Implicit in this derivation is the assumption that the system is initially (at t=0) at rest. For arbitrary initial conditions, Eqn 84 should be augmented by the free vibration terms as follows

For viscously damped system with ,becomes

Leading to the following counterpart of Eqn.84

00

0cos sin sin

t

st

Vx t x pt pt p x p t d

p

( ) sinp td

d

P dx t e p t

mp

2 0

sin1

tp t

st d

px t x e p t d

The effect of the initial motion in this case is defined by Eqn. 41

Eqns. 84 and 87 are referred to in the literature with different names. They are most commonly known as Duhamel’s Integrals, but are also identified as the superposition integrals, convolution integrals, or Dorel’s integrals.

Example1: Evaluate response to rectangular pulse, take and

For

For

0

0 0 0x v 1t t

1t t

1

10 00sin - 0 cos - - cos

t

st stx t p x p t d x p t t pt

0

0

sin cos

1 cos

tt

st sto o

st o

x t p x p t d x p t

x pt

P(t)

t

Po

t1

Generalised SDOF System

Generalised SDOF System

* * * *

*

*

*

*

( ) ( ) ( ) ( )

( )

m q t c q t k q t p t

q t

m

c

k

p

Single generalised coordinate expressing the motion of the system

generalised mass

generalised damping coefficient

generalised stiffness

generalised force

(a)

x1

m1,j1 m2, j2m(x)(b)

tqxt,xy

* 2 2 2

0

( ) ( )l

i i i im m x x dx m j

c1 c2

a1(x)

L

(c) c(x)

k(x)

k1 k2

(d)

* 2 " 2 2 ' 2

0 0 0

( ) ( ) ( ) ( ) ( ) ( )l l l

i ik k x x dx EI x x dx k N x x dx

N(e)

P(x,t)

Note: Force direction and displacement direction is same (+ve)

*

0

( , ) ( ) ( ) ( )l

i ip p x t x dx p t x

Pi(t)

Effect of damping

Viscous damping

Coulomb damping

Hysteretic damping

Effect of damping

Energy dissipated into heat or radiated away

• The loss of energy from the oscillatory system results in the decay of amplitude of free vibration.

• In steady-state forced vibration ,the loss of energy is balanced by the energy which is supplied by the excitation.

Energy dissipated mechanism may emanate from

(i) Friction at supports & joints

(ii) Hysteresis in material ,internal molecular friction, sliding friction

(iii) Propagation of elastic waves into foundation ,radiation effect

(iv) Air-resistance,fluid resistance

(v) Cracks in concrete-may dependent on past load –history etc..,

Exact mathematical description is quite complicated &not suitable for vibration analysis.

Effect of damping

Simplified damping models have been proposed .These models are found to be adequate in evaluating the system response.

Depending on the type of damping present ,the force displacement relationship when plotted may differ greatly.

Force - displacement curve enclose an area ,referred to as the hysteresis loop,that is proportional to the energy lost per cycle.

d dW F dxIn general Wd depends on temperature,frequency,amplitude.

For viscous type

2

2 2 2 2 2 2

0

cos ( )

d d

d

d

W F dx

F cx

W c X cxdx cx dt c X t dt c X

Fd(t) = c

c - coefficient of damping

Wviscous - work done for one full cycle =

x

2c X

cωX

Fd

-X X(t)

(a) Viscous damping

(b) Equivalent viscous damping:

2

2

2

22,

4

eq d

deq

sc s

d

c s

C X W

WC

XWk

C where k W strain energyX

WC

C W

x

ellipseFd+kx

x1 x2 ∆

4Fd/k

Linear decay

Frequency of oscillation

kp

m x-1

It results from sliding of two dry surfaces

The damping force=product of the normal force & the coefficient of friction (independent of the velocity once the motion starts.

(b) Coulomb damping:

-X X(t)

Fd

F

4coulombW F X

2 21 1 1 1

1 1

1( ) ( ) 0

21

( )2

d

d

k X X F X X

k X X F

The motion will cease ,however when the amplitude becomes less than ∆, at which the spring force is insufficient to overcome the static friction.

1 2

4 dFX X

k Decay in amplitude per cycle

(c) Hysteretic damping (material damping or structural damping):

2 22

D

D

xf k x

x

W kX X

Energy dissipated is frequency independent.

- Inelastic deformation of the material composing the device

14hysteretic yW F X

Fy is the yield force

Xy Displacement at which material first yields

Kh=elastic damper stiffness

xy

- xx

Fy

Fd

(d) Structural dampingy

x

x

a) Coulomb Ceq=

b) Hysteretic Ceq=

c) Structural Ceq=

4F

WX

2 sk

4 1yF

X

Equivalent viscous Coefficient

Reference

Dynamics of Structures: Theory and Application to Earthquake Engineering – Anil K. Chopra, Prentice Hall India

Reading Assignment

Course notes & Reading material

sdof-1211798306003307-8

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