mechanical vibrations chapter 4 - faculty server...

1 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4

Mechanical VibrationsChapter 4

Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell

Impulse Excitation

Impulsive excitations are generallyconsidered to be a large magnitudeforce that acts over a very shortduration timeThe time integral of the force is

When the force is equal to unity and the timeapproaches zero then the unit impulse exists andthe delta function has the property of

(4.1.1)∫= dt)t(FF̂

( ) ξ≠=ξ−δ tfor0t

Impulse Excitation

Integrated over all time, the delta function is

If this function is multiplied times any forcingfunction then the product will result in only onevalue at t=ξ and zero elsewhere

(4.1.2)∫∞

∞<ξ<=ξ−δ0

01dt)t(

∫∞

∞<ξ<ξ=ξ−δ0

0)(fdt)t()t(f (4.1.3)

Impulse Excitation

Considering impact-momentum on the system, asudden change in velocity is equal to the actualapplied input divided by the force.Recall that the free response due to initialconditions is given by

tcos)0(xtsin)0(xx nnn

ω+ωω

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

1)t(h nn

Impulse Excitation

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

which can be written as

(4.16)

tsine)0(xt1sin1e)0(xx d

=ζ−ωζ−ω

=σ−ζω− &&

t1sine1m

F̂x 2n

n ζ−ωζ−ω

= ζω−

)t(hF̂tsinem

F̂x dt

ω= σ−

Arbitrary Excitation

Using the unitimpulse responsefunction, theresponse due toarbitrary loadingscan be determined.The arbitrary forceis considered to bea series of impulses

Arbitrary Excitation

Since the system is considered linear, then thesuperposition of the responses of each individualimpulse can be obtained through numericalintegration

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

Convolution Integralor

Duhammel’s Integral

(4.2.1)∫ ξξ−ξ=t

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

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)

1)t(h nn

( ) ξξ−ωω

= ∫ dtsinm

F)t(xt

( )tcos1kF)t(x n

0 ω−=

Step Excitation

This implies that the peak response is twice thestatical displacement

(4.2.2)( )tcos1kF)t(x n

0 ω−=

0 5 10 15 20 25 300

Dis placement vers us Time

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

Step Excitation

When damping is included in the equation, then

(4.2.2)t1sin

1me)t(h 2

ζ−ωζ−ω

=ζω−

ψ−ζ−ω

ζ−ω−=

ζω−t1cos

kF)t(x 2

(4.2.3)

Step Excitation

This can be simplified as

tsinme)t(h d

ψ−ω

ω−=

σ−tcos

kF)t(x d

M=1 ; K=2

C=0.5C=1.0

Base Excitation

For base excitation,

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

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

(3.5.1)

(4.2.4)yzz2z 2nn &&&&& −=ω+ζω+

)yx(c)yx(kxm &&&& −−−−=

yxz −= (3.5.2)

Base Excitation

For and undamped systeminitially at rest, thesolution for the relativedisplacement is

(4.2.5)( ) ξξ−ωξω

−= ∫ dtsin)(y1)t(zt

Ramp Excitation and Rise Time

This solution must always be considered in twoparts - the time less than and greater than t1

The ramp of the force is

and h(t) for the convolution integral is

(4.4.1)

tF)t(f

(4.4.1)tsink

1)t(h nn

The response for the first part of the ramp is

and the response of the step portion after t1 is

(4.4.2)

tsintt

d)t(sint

ξξ−ωξω

10 ttt

ttsint

ttkF)t(x >

−ω−

−−=

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

(4.4.3)( )

n0 ttt

ttsint

tsin1kF)t(x >

−ω+

Rectangular Pulse

The rectangular pulse is the sum of two differentstep functions - one positive and one negativeshifted in time

Step Up

Step down

Combined

(4.4.4)( ) 1n0

tttcos1F

)t(kx<ω−=

(4.4.5)

(4.4.6)

( )( ) 11n0

ttttcos1F

)t(kx<−ω−−=

( )( ) ( )( ) 11nn0

ttttcos1tcos1F

)t(kx<−ω−−ω−=

( )( ) ( )( ) 11nn0

ttttcostcosF

)t(kx<−ω+ω−=

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

MATLAB Examples - VTB3_1VIBRATION TOOLBOX EXAMPLE 3_1

>> m=1; c=.1; k=2; tf=100; F0=1;>> vtb3_1(m,c,k,F0,tf)>>

0 10 20 30 40 50 60 70 80 90 1000

MATLAB Examples - VTB3_2VIBRATION TOOLBOX EXAMPLE 3_2

>> m=1; c=.1; k=2; tf=100; F0=1;>> VTB3_2(m,c,k,F0,tf)>>>>

MATLAB Examples - VTB1_4VIBRATION 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 5 10 15 20 25 300

MATLAB Examples - VTB1_4VIBRATION 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);>>

mechanical vibrations chapter 4 - faculty server...

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