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1
THE TIME-DOMAIN RESPONSE OF A SINGLE-DEGREE-OF-FREEDOM
SYSTEM SUBJECTED TO AN IMPULSE FORCE
Revision C
By Tom Irvine
Email: [email protected]
December 19, 2013
Consider a single-degree-of-freedom system.
Where
m = mass
c = viscous damping coefficient
k = stiffness
y = displacement of the mass
f(t) = applied force
Note that the double-dot denotes acceleration.
The free-body diagram is
m
k y yc
y
m
k c
y
f(t)
f(t)
2
Summation of forces in the vertical direction
ymF (1)
)t(fkyycym (2)
)t(fkyycym (3)
Divide through by m,
)t(fm
1y
m
ky
m
cy
(4)
By convention,
n2)m/c( (5)
2
n)m/k( (6)
where
n is the natural frequency in (radians/sec)
is the damping ratio
By substitution,
)t(fm
1y
2nyn2y (7)
Now apply an impulse force at t=0 via a pair of step functions.
)t(u)t(uF)t(f (8)
where is a very small time step.
3
The governing equation becomes
)t(u)t(uFm
1yy2y
2nn (9)
Now consider that the system undergoes oscillation with < 1.
Take the Laplace transform of each side.
)t(u)t(uFm
1Lyy2yL
2nn (10)
)sexp(s
1
s
1
m
F)s(Y
)0(y2)s(sY2
)0(y)0(sy)s(Ys
2n
nn
2
(11)
)sexp(
s
1
s
1
m
F)0(y)0(y2s)s(Ys2s n
2nn
2 (12)
)0(y)0(y2s)sexp(s
1
s
1
m
F)s(Ys2s n
2nn
2
(13)
2n
2n
2n
2nn
2 ss2s (14)
22n
2n
2nn
2 1ss2s (15)
4
Let
2nd 1 (16)
Substitute equation (16) into (15).
2d
2n
2nn
2 ss2s (17)
)0(y)0(y2s)sexp(s
1
s
1
m
F)s(Ys n
2d
2n
(18)
)0(y
s
1)0(y
s
2s
s
)sexp(
sm
F
s
1
sm
F
)s(Y
2d
2n
2d
2n
n
2d
2n
2d
2n
(19)
Divide the right-hand-side of equation (19) into three parts.
)s(Y)s(Y)s(Y)s(Y 210 (20)
where
2d
2n
0s
1
sm
F)s(Y (21)
5
2d
2n
1s
)sexp(
sm
F)s(Y (22)
)0(y
2d
2ns
1)0(y
2d
2ns
n2s)s(2Y
(23)
Consider Y0(s) from equation (21).
2d
2n
0s
1
sm
F)s(Y (24)
Expand into partial fractions using Reference 1.
2d
2n
d
d
n2n
2d
2n
n2n
2n
2nn
2
s m
F
s
s
m
F
s
1
m
F
s2s
1
sm
F
(25)
6
The inverse Laplace transform per Reference 1 is
0t,tsintcostexpm
F)t(u
m
F)t(y d
d
ndn2
n2n
0
(26)
0t,tsintcostexp)t(um
F)t(y d
d
ndn2
n
0
(27)
Now consider the term
2d
2n
1s
)sexp(
sm
F)s(Y (28)
Expand into partial fractions using Reference 1.
)sexp(
s m
F
)sexp(s
s
m
F
)sexp(s
1
m
F
s2s
)sexp(
sm
F
2d
2n
d
d
n2n
2d
2n
n2n
2n
2nn
2
(29)
7
Take the inverse Laplace transform from Appendices A, B & C.
tfor
tsintexpm
F
tcostexpm
Ftu
m
F)t(y
dnd
n2n
dn2n
2n
1
(30)
tfor
tsintcostexptum
F)t(y d
d
ndn2
n
1
(31)
The inverse Laplace transform from Reference 2 for the natural response is
tsintexp)0(y)0(y
tcostexp)0(y)t(y dnd
ndn2
(32)
8
tdsintnexpd
1)0(y
tdsind
ntdcostnexp)0(y)t(2y
(33)
The final displacement solution is
)t(y)t(y)t(y)t(y 210 (34)
t0
,tsintcostexp)t(um
F
tsintexp1
)0(y
tsintcostexp)0(yty
dd
ndn2
n
dnd
dd
ndn
(35)
9
t
,tsintcostexptum
F
tsintcostexp)t(um
F
tsintexp1
)0(y
tsintcostexp)0(yty
dd
ndn2
n
dd
ndn2
n
dnd
dd
ndn
(36)
The response for a Dirac Delta impulse is derived in Appendix D.
References
1. T. Irvine, Partial Fractions in Shock and Vibration Analysis, Revision G,
Vibrationdata, 2011.
2. T. Irvine, Table of Laplace Transforms, Revision I, Vibrationdata, 2011.
3. T. Irvine, Free Vibration of a Single-Degree-of-Freedom System, Revision B,
Vibrationdata, 2005.
10
APPENDIX A
tum
F)sexp(
s
1
m
FL
2n
2n
1 (A-1)
APPENDIX B
Consider the generic term
)sexp(
s
s)s(F
22
(B-1)
Shift in the s-plane.
ss (B-2)
sexp
s
s)s(F
22 (B-3)
expsexp
s
s)s(F
22 (B-4)
expsexp
s expsexp
s
s)s(F
2222 (B-5)
11
t
,tsintexpexptcostexpexp)t(f
(B-6)
t
tsintcostexpexp)t(f
(B-7)
f(t) = 0 t < ρ (B-8)
Note that the texp term accounts for the phase shift.
Recall the term
)sexp(
s
s
m
F)s(Y
2d
2n
n2n
12
(B-9)
Compare the term in equation (B-9) to the generic term from (B-1).
)sexp(
s
s)s(F
22
(B-10)
12
The inverse Laplace transform of equation (B-9) is
tcostexpexp
m
F)t(y
dnn2n
12 (B-11)
tcostexp
m
F)t(y
dn2n
12 (B-12)
13
APPENDIX C
Consider the generic term
)sexp(
s )s(F
22
(C-1)
Shift in the s-plane.
ss (C-2)
sexp
s )s(F
22 (C-3)
expsexp
s )s(F
22 (C-4)
t
tsintexpexp)t(f
(C-5)
t
tsintexp)t(f
(C-6)
14
f(t) = 0 t < ρ (C-7)
Note that the texp term accounts for the phase shift.
)sexp(
s m
F)s(Y
2d
2n
d
d
n2n
13
(C-8)
Compare with the generic term
)sexp(
s )s(F
22
(C-9)
expsexp
s )s(F
22 (C-10)
t
tsintexpexp)t(f
(C-11)
t
tsintexpm
F)t(y dn
d
n2n
13
(C-12)
15
APPENDIX D
Dirac Delta Impulse
)t(fm
1y
2nyn2y (D-1)
Let I = total impulse
)t(I)t(f (D-2)
.
)t(m
Iyy2y
2nn (D-3)
m
I)s(Y
)0(y2)s(sY2
)0(y)0(sy)s(Ys
2n
nn
2
(D-4)
m
I)0(y)0(y2s)s(Ys2s n
2nn
2 (D-5)
)0(y)0(y2sm
I)s(Ys n
2d
2n (D-6)
16
)0(y
s
1)0(y
s
2s
s
1
m
I
)s(Y
2d
2n
2d
2n
n
2d
2n
(D-7)
The displacement is
tsintexp1
m
I
tsintexp1
)0(y
tsintcostexp)0(yty
dnd
dnd
dd
ndn
(D-8)
The displacement for zero initial conditions is
tsintexp1
m
Ity dn
d
(D-9)
The impulse response function is
tsintnexpm
1=)t(h d
dd
(D-10)
17
The corresponding Laplace transform is
2d
22n
2d
ns
1
sn2s
1=)s(H
m
1
m
1
(D-11)
tcostexpm
I
tsintexpm
I
tcostexp)0(y
tsintexp)0(y
tcostsintexp)0(y
tsintcostexp)0(yty
dn
dnd
n
dn
dnd
n
dnddn
dd
ndnn
(D-12)
tsintcostexpm
I
tsintcostexp)0(y
tsintexp)0(yty
dd
ndn
dd
ndn
ddd
2n
2
n
(D-13)
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tsintcostexpm
I
tsintcostexp)0(y
tsintexp)0(yty
dd
ndn
dd
ndn
d2d
2d
2n
2
n
(D-14)
The velocity for zero initial conditions is
tsintcostexp
m
Ity d
d
ndn
(D-15)
The impulse response function is
tsintcos)tnexp(-=)t(h d
d
ndv
m
1
(D-16)
The corresponding Laplace transform is
2d
22n
2v
ns
s
sn2s
s=)s(H
m
1
m
1
(D-17)
19
The acceleration for zero initial conditions is
)t(y)t(y2)t(m
Ity 2
nn
(D-18)
tsintexp1
tsintcostexp2)t(m
I
ty
dnd
2nd
d
ndnn
(D-19)
tsin
2tcos2texp)t(
m
Ity d
d
2n
2n
dnn
(D-20)
tsin12tcos2texp)t(
m
Ity d
2
d
2n
dnn
(D-21)
The Laplace transform is
2n
2
22n
2n
2n
asn2s
122s21
m
IsY
(D-22)
2n
2
2nn
asn2s
s21
m
IsY
(D-23)
20
2n
2
2nn
2n
2
2n
2
asn2s
s2
sn2s
sn2s
m
IsY
(D-24)
2n
2
2
asn2s
s
m
IsY
(D-25)
The corresponding transfer function in terms of the Laplace transform is
2d
2
2
2n
2
2
a
ns
s
sn2s
s=)s(H
m
1
m
1 (D-26)
21
APPENDIX E
Dirac Delta Impulse, Transmitted Force
Again, the impulse response function for displacement is
tsintnexpm
1=)t(h d
dd
(E-1)
The impulse response function for velocity is
tsintcos)tnexp(-=)t(h d
d
ndv
m
1
(E-2)
The transmitted force f t (t) is
kyyc (t) f t (E-3)
y
m
ky
m
cm (t) f t (E-4)
yy2m (t) f
2nnt (E-5)
22
The impulse response function for transmitted force is
tsintnexp
tsintcos)tnexp(-2=)t(h
dd
2n
dd
ndnft
(E-6)
tsintnexp
tsin2
tcos2)tnexp(-=)t(h
dd
2n
dd
2n
2
dnft
(E-7)
tsin
2tcos2)tnexp(-=)t(h d
d
2n
2
d
2n
dnft
(E-8)
The final transmitted force impulse response function is
tsin
2tcos2)tnexp(-=)t(h d
d
22
ndnft1
(E-9)
Note that the transmitted force impulse response function is essentially the same as the
absolute acceleration impulse response function for the case of base excitation.