ME22457_Chapter4_021703_MACL

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1 Dr. Peter AvitabileModal Analysis & Controls Laboratory

22.457 Mechanical Vibrations - Chapter 4

Mechanical Vibrations

Chapter 4

Peter Avitabile Mechanical Engineering Department University of Massachusetts Lowell

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Impulse Excitation

Impulsive excitations are generally considered to be a large magnitude

force that acts over a very short duration time

The time integral of the force is

When the force is equal to unity and the time approaches zero then the unit impulse exists and

the delta function has the property of

(4.1.1)

∫= dt)t(FF̂

( ) ξ≠=ξ−δ tfor 0t

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Impulse Excitation

Integrated over all time, the delta function is

If this function is multiplied times any forcing function then the product will result in only one value at t=

and zero elsewhere

(4.1.2)

∫

∞

∞

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Impulse Excitation

Considering impact-momentum on the system, a sudden change in velocity is equal to the actual

applied input divided by the force.Recall that the free response due to initial conditions is given by

tcos)0(xtsin)0(x

x nnn

ω+ωω

= &

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Impulse Excitation

Then the velocity initial condition yields

and it can be seen that the solution includes h(t)

(4.1.4)

(4.1.5)

)t(hF̂tsinm

F̂x

nn

=ωω

=

tsinm

1)t(h n

n

ωω

=

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Impulse Excitation

When damping is considered in the solution, the free response is given as (x(0)=0)

which can be written as

or as

(4.16)

tsine)0(x

t1sin1

e)0(xx d

d

t2n2

n

tn

ωω

=ζ−ωζ−ω

= σ−ζω− &&

t1sine1m

F̂x 2n

t

2n

n ζ−ωζ−ω

= ζω−

)t(hF̂tsinem

F̂x d

t

d

=ωω

= σ−

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Arbitrary Excitation

Using the unit impulse response

function, the response due to arbitrary loadings can be determined.

The arbitrary force is considered to be a series of impulses

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Arbitrary Excitation

Since the system is considered linear, then the superposition of the responses of each individual

impulse can be obtained through numerical integration

This is called the superposition integral. But it is also referred to as the

Convolution Integral or Duhammel’s Integral

(4.2.1)∫ ξξ−ξ=t

0

d)t(h)(f )t(x

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Step Excitation

Determine the indamped response due to a step.

For the undamped system,

which is substituted into (4.2.1) to give

(4.2.2)

tsinm

1)t(h n

n

ωω

=

( ) ξξ−ωω

= ∫ dtsinmF

)t(x

t

0

nn

0

( )tcos1k

F)t(x n

0 ω−=

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Step Excitation

This implies that the peak response is twice the statical displacement

(4.2.2)( )tcos1k F)t(x n

0 ω−=

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Time

D i s p l a c e m e n t

Displacem ent versus Time

Note: Force selected such that F/k ratio is 1.0

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Step Excitation

When damping is included in the equation, then

and

(4.2.2)t1sin1m

e

)t(h2

n2n

tn

ζ−ωζ−ω=

ζω−

ψ−ζ−ω

ζ−ω−=

ζω−

t1cos1m

e1k F)t(x 2n2

n

t

0n

(4.2.3)

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Step Excitation

This can be simplified as

tsinm

e)t(h d

d

t

ωω=

σ−

ψ−ωω−=

σ−

tcosm

e

1k

F

)t(x dd

t0

M=1 ; K=2

C=0

C=0.1 C=0.5

C=1.0

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Base Excitation

For base excitation,

the equation of motion is expressed as z=x-y and will result in

Notice that the F/m term is replaced by the negative of the base acceleration (ie, F=ma)

(3.5.1)

(4.2.4)yzz2z2

nn &&&&& −=ω+ζω+

)yx(c)yx(k xm &&&& −−−−=

yxz −= (3.5.2)

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Base Excitation

For and undamped system initially at rest, the

solution for the relative displacement is

(4.2.5)( ) ξξ−ωξω−= ∫ dtsin)(y1

)t(z

t

0n

n &&

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Ramp Excitation and Rise Time

This solution must always be considered in two parts - the time less than and greater than t 1

The ramp of the force is

and h(t) for the convolution integral is

(4.4.1)

=

10 t

tF)t(f

(4.4.1)tsink

tsinm

1)t(h

n

n

nn

ωω

=ωω

=

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The response for the first part of the ramp is

and the response of the step portion after t 1 is

(4.4.2)

1

1n

n

1

0

n

t

0 10

n

ttt

tsin

t

t

k

F

d)t(sintFk )t(x

ω

−ω

−

−

−=

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The superposition of the two pieces of the solution gives the total response due to the force as

(4.4.3)( )

11n

1n

1n

n0 ttt

ttsin

t

tsin1

k

F)t(x >

ω

−ω+

ωω

−=

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Rectangular Pulse

The rectangular pulse is the sum of two different step functions - one positive and one negative

shifted in time Step Up

Step down

Combined

(4.4.4)( ) 1n0

tttcos1F

)t(kx

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0 10 20 30 40 50 60 70 80 90 100-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Time

D i s p l a c e m e n t

Dis placeme nt versus Time

MATLAB Examples - VTB3_1

VIBRATION TOOLBOX EXAMPLE 3_1

>> m=1; c=.1; k=2; tf=100; F0=1;

>> vtb3_1(m,c,k,F0,tf)>>

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0 10 20 30 40 50 60 70 80 90 1000

0.5

1

Time

D i s p l a c e m e

n t

Dis placeme nt versus Time


VIBRATION TOOLBOX EXAMPLE 3_2

>> m=1; c=.1; k=2; tf=100; F0=1;

>> VTB3_2(m,c,k,F0,tf)>>

>>

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VIBRATION TOOLBOX EXAMPLE 1_4 – pulse

>> clear; p=1:1:1000; pp=p./p; ppp=[(p./p-p./p) pp (p./p-p./p) (p./p-p./p)];

>> x0=0; v0=0; m=1; d=.5; k=2; dt=.01; n=4000;

>> u=ppp; [x,xd]=VTB1_4(n,dt,x0,v0,m,d,k,u);

>> t=0:dt:n*dt; plot(t,x);plot(t,x);

>>

0 5 10 15 20 25 30 35 40-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

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0 5 10 15 20 25 300

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5


VIBRATION TOOLBOX EXAMPLE 1_4 – ramp up (basically a static problem)

>> clear; x0=0; v0=0; m=1; d=.5; k=2; dt=.01; n=3000;

>> t=0:dt*100:n; u=t./3000; [x,xd]=VTB1_4(n,dt,x0,v0,m,d,k,u);

>> t=0:dt:n*dt; plot(t,x);

>>

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