general forced response -...
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General forced response
Impulse response •Dirac delta function •Unit impulse response
arbitrary response •Convolution integral
F(t)
t
cam
Material being compacted
Follower
Cam motion
Platform x(t)
m
F(t)
t
The principle of Superposition
02 =+ xx nω
Linear equations
21, xx are both solutions
2211 xaxax += is also a solution
12 fxx n =+ω 1xsolution
22 fxx n =+ω 2xsolution
212 ffxx n +=+ω 21 xx +solution
21, aa are constants
Impact excitation
Impact is a force applied for a very short period of time
Impulse due to F(t): ∫= dttFI )(
= area under F-t curve
Dirac delta function
Dirac delta function is a unit impulse function
Properties of dirac delta function
0)( =τ−δ t
1)( =τ−δ∫∞
∞−
dtt
)()()( τ=τ−δ⋅∫∞
∞−
gdtttg
F(t)
t τ
τ≠tfor
)()( τ−δ⋅= tItF
Impulse
Linear impulse-momentum equation
maF = dtmadtF ∫∫ = )(
vmvvmI ∆=−= )( 12
A mass m initially at rest subject to an impact of size I N.s
impact
0mvIdtF ==∫mIv /0 =
An impact causes initial velocity change of v0 = I/m while keeping initial displacement unchanged
Unit impulse response (1)
EOM )(tkxxcxm δ=++
Initial conditions 0)0()0( == xx
Just after subject to a unit impact (t = 0+)
Time response of a system initially at rest but subject to a unit impact (I = 1)
EOM 0=++ kxxcxm
Initial conditions mxx /1)0(;0)0( ==
m
c k
x
Impact response is just free vibration of a system with initial velocity 1/m
Unit impulse response (2)
EOM 0=++ kxxcxm
Initial conditions mxx /1)0(;0)0( ==
Unit impulse response is just free vibration of a system with initial velocity 1 / m
tem
txth dt
d
n ωω
== ζω− sin1)()(
For impact applied at time t = τ
)(sin1 )( τ−ωω
τ−ζω− tem d
t
d
n
0
h(t) =
for 0 < t < τ
for t > τ
Before impact
After impact
underdamped
Unit impulse response (3)
EOM 0=++ kxxcxm
Initial conditions mxx /1)0(;0)0( ==
tem
txth dt
d
n ωω
== ζω− sin1)()(underdamped
Impulse response
EOM δ=++ Ikxxcxm
Initial cond. 0)0(;0)0( == xx
Unit impulse response is just free vibration of a system with initial velocity I / m
)(sin)( thItem
Itx dt
d
n ⋅=ωω
= ζω−
For impact applied at time t = τ
)(sin)( τ−ωω
τ−ζω− tem
Id
t
d
n
0
h(t) =
for 0 < t < τ
for t > τ
Before impact
After impact
underdamped
0=++ kxxcxm
mIxx /)0(;0)0( ==
Example 1-1
Given: m = 1 kg, c = 0.5 kg/s, k = 4 N/m
)(1.0)(2.0)( τ−δ+δ= tttF
24 ==ωn 125.0)2/( =ω=ζ nmc
)(2.0)( ttF δ=For
tetx t ⋅−−
= − 2)2)(125.0(
21 125.012sin)125.012)(1(
2.0)(
;
)(1.0)( τ−δ= ttF
)984.1sin(1008.0)( 25.01 tetx t−=
)(984.1sin0504.0)( )(25.02 τ−= τ−− tetx t
For τ>t
Example 1-2
)984.1sin(1008.0 25.0 te t−
)(984.1sin0504.0)984.1sin(1008.0 )(25.025.0 τ−+ τ−−− tete tt τ>t
)()()( 21 txtxtx +=
= τ<< t0
[ ] )()(984.1sin0504.0)984.1sin(1008.0)( )(25.025.0 τττ −Φ⋅−+= −−− ttetetx tt
Heaviside step function
=−Φ )( τtτ<t,0
τ≥t,1)( τ−tHor
Example 1-3
)984.1sin(1008.0 25.0 te t−
)(984.1sin0504.0)984.1sin(1008.0 )(25.025.0 τ−+ τ−−− tete tt τ>t
)()()( 21 txtxtx +=
= τ<< t0
Impact forces 5.0=τ Response
)()()( 21 txtxtx +=
0.5 )()( 1 txtx =
Example 2-1
)4()()(4)(2)( −δ−δ=++ tttxtxtx
mm/s 1;mm 1 00 −== vx
Compute and plot the response
Arbitrary input
Vibration of systems subject to arbitrary nonperiodic inputs.
Analysis tools: • Impulse response function • Super position • Convolution integral
Convolution integral (1)
Excitation Response
Convolution integral (2)
+++= 321)( xxxtx
+−+−+−=
)()()()(
33
2211
tthItthItthItx
+−⋅∆+−⋅∆+−⋅∆=
)()()()(
33
2211
tthtFtthtFtthtFtx
0→∆t
ττ−τ= ∫ dthFtxt
0
)()()(
Excitation Response
Convolution integral (3)
=++ kxxcxm 11 0),( tttF <≤
12 ),( tttF ≥
For the non-continuous function,
ττ−τ∫ dthFt1
01 )()(
ττ−τ+ττ−τ ∫∫ dthFdthFt
t
t
1
1
)()()()( 20
1
=)(tx10 tt <≤
1tt ≥
when
when
Example 3-1
=++ kxxcxm 00,0 tt <<
00 , ttF ≥
000 == vx
10 <ζ< (underdamped)
ττ−τ= ∫ dthFtxt
0
)()()(
tem
txth dt
d
n ωω
== ζω− sin1)()(ττ−ω
ω= ∫ τ−ζω− dte
mFtx
t
td
t
d
n
0
)(sin1)( )(0
ττ−ωω
= ∫ τζωζω− dteemFtx
t
td
t
d
nn
0
)(sin)( 0
Example 3-2
00)(
200 ];)(cos[
1)( 0 tttte
kF
kFtx d
ttn ≥θ−−ωζ−
−= −ζω−
2
1
1tan
ζ−
ζ=θ −
Static displacement
1.0=ζ
16.3=ωn
300 =F
1000=k00 =t
rad/s
N
N/m
Example
Example 4-1
=++ kxxcxm 10 0, ttF ≤<
1,0 tt >
000 == vx
10 <ζ< (underdamped)
Homework
Calculate and plot the response of an undamped system to a step function with a finite rise time of t1 for the case m = 1 kg, k = 1 N/m, t1 = 4 s, and F0 = 20 N. This function is described by
=)(tF1
1
0 0, ttt
tF≤≤
10 , ttF >0F
1t