Spring Damper

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1

THE VIBRATION RESPONSE OF SOME SPRING-DAMPER SYSTEMS

WITH AND WITHOUT MASSES

By Tom IrvineEmail: [email protected]

April 26, 2012

Introduction

Consider the single-degree-of-freedom system subjected to an applied force in Figure 1.

Figure 1.

where

Summation of forces in the vertical direction yields the equation of motion.

)t(Fkxxc (1)

Derive the transfer function by taking the Laplace transform.

)t(FLkxxcL (2)

F(t) = Applied force

c = viscous damping coefficient

k = stiffness

x = displacement

k c

)t(F

x



2

)s(F)s(Xk )0(xc)s(Xsc (3)

)0(xc)s(F)s(Xk sc (4)

k sc

)0(xc)s(F)s(X

(5)

Now assume that the initial displacement is zero.

k sc

)s(F

)s(X (6)

k sc

1

)s(F

)s(X

(7)

The dynamic stiffness in the Laplace domain is

k sc)s(X

)s(F

(7)

The dynamic stiffness in the frequency domain is

c jk

)(X

)(F (8)

The initial value problem for the free response is

k sc

)0(xc)s(X

(9)



3

c/k s

)0(x)s(X

(10)

The inverse Laplace transform yields the time domain response.

tc/k exp)0(x)t(x (11)

APPENDIX A

Consider the single-degree-of-freedom system subjected to an applied force.

Figure A-1.

The equation of motion is

)t(Fkxxcxm (A-1)

Derive the transfer function by taking the Laplace transform.

)t(FLkxxcxmL (A-2)

)s(F)s(Xk )0(xc)s(Xsc)0(msx)0(xm)s(Xms2 (A-3)

k c

)t(F

x m



4

)0(msx)0(xm)0(xc)s(F)s(Xk scms2 (A-4)

k scms

)0(msx)0(xm)0(xc)s(F)s(X

2 (A-5)

Now assume that the initial displacement is zero.

k scms

)s(F)s(X

2 (A-6)

k scms

1

)s(F

)s(X

2 (A-7)


k scms)s(X

)s(F 2 (A-8)


c jmk

)(X

)(F 2 (A-9)

The mechanical impedance in the frequency domain is

j

c jmk

)(V

)(F 2

(A-10)



5

m

k jc

)(V

)(F (A-11)

The apparent mass in the frequency domain is

j

mk

jc

)(A

)(F (A-12)

c

j

k

m)(A

)(F

2

(A-13)

m

c j

m

k 1m

)(A

)(F

2 (A-14)

Note that

m

k n (A-15)

m2c n (A-16)

2 (A-17)



6

By substitution,

n

2

2n 2

j1m)(A

)(F (A-18)

n2

2n j1m

)(A

)(F (A-19)



7

APPENDIX B

Consider the following two-degree-of-freedom system subjected to an applied force.

Figure B-1.

The free-body diagrams are

Figure B-2.

k

c

)t(F

x1

x2

k(x2-x1)

k(x1-x2)

)t(F

x2

1xc



8

The equations of motion are

0)t(F)x-k(x 21 (B-1)

)t(F)x-k(x 12 (B-2)

0)x-k(xxc 211 (B-3)

Solve for 2x using Equation (B-2).

12 xk

)t(Fx (B-4)

By substitution,

0)x-k(xxc 211 (B-5)

0xk

)t(F

-xk xc 111

(B-6)

0k

)t(F-k xc 1

(B-7)

)t(Fxc 1 (B-8)

Take the Laplace transform.

)t(FLxcL 1 (B-9)



9

)s(F)0(xc)s(Xsc 11 (B-10)

)0(xc)s(F)s(Xsc 11 (B-11)

sc

)0(xc)s(F)s(X 1

1

(B-12)

Recall

)t(F)x-k(x 12 (B-13)

Take the Laplace transform.

)t(FL)x-k(xL 12 (B-14)

)s(F)s(Xk )s(Xk 12 (B-15)

)s(Xk )s(F)s(Xk 12 (B-16)

sc

)0(xc)s(Fk )s(F)s(Xk 1

2 (B-17)

)0(xs

k

sc

k 1)s(F)s(Xk 12

(B-18)

)0(xs

1

sc

1

k

1)s(F)s(X 12

(B-19)



10

Now assume the initial displacement is zero.

sc

1

k

1)s(F)s(X2 (B-20)

sk c

k cs)s(F)s(X2 (B-21)

The receptance in the Laplace domain is

sk c

k cs

)s(F

)s(X2

(B-22)


k cs

sk c

)s(X

)s(F

2 (B-22)


c jk

k c j

)(X

)(F

2

(B-22)


)0(xs

1)s(X 12 (B-23)



11

The inverse Laplace transform yields the time domain response in terms of the unit step

function u(t).

)t(u)0(x)t(x 12 (B-24)

This can be simplified as

)0(x)t(x 12 (B-25)

Likewise

)0(x)t(x 11 (B-26)



12

APPENDIX C


Figure C-1.

Figure C-2.

k 1 (x2-x1)

k 1(x1-x2)

)t(F

x2

)t(F

k 1

c

x1

x2

k 2

-k 2 x2

1xc



13


0)t(Fxk )x-(xk 22211 (C-1)

)t(Fxk xk k 11221 (C-2)

0)x-(xk xc 2111 (C-3)

Solve for 2x using Equation (C-2).

121

1

212 x

k k k

k k )t(Fx

(C-4)

By substitution,

0)x-(xk xc 2111 (C-5)

0xk k

k k k )t(F-xk xc 1

21

1

21111

(C-6)

0)t(Fk k

k -x

k k

k 1k xc

21

11

21

111

(C-7)

0)t(Fk k

k -xk k

k k k k xc21

11

21

12111

(C-8)







16

21

21

1

21

1

21

2121

1

212

k k

k k sc

)0(xc

k k

k )s(F

k k

k k sc

1

k k

k

k k

1)s(X

(C-21)

21

21

1

21

1

21

21

1

212

k k

k k

sc

)0(xc

k k

k )s(F

k k

k k

sc

k 1

k k

1)s(X

(C-22)

Now assume the initial displacement is zero.

)s(F

k k

k k sc

k 1

k k

1)s(X

21

21

1

212

(C-23)


21

21

1

21

2

k k

k k sc

k 1

k k

1

)s(F

)s(X (C-24)



17

2121

1

21

2

k k sk k c

k

k k

1

)s(F

)s(X (C-25)

21212

21

21121212

k k k k sk k c

k k k k k sk k c

)s(F

)s(X

(C-26)

2121

2

21

211212

k k k k sk k c

k 2k k sk k c

)s(F

)s(X

(C-27)


21121

21212

21

2 k 2k k sk k c

k k k k sk k c

)s(X

)s(F

(C-28)


21

21

1

21

12

k k

k k sc

)0(xc

k k

k )s(X (C-29)

21

21

1

21

12

k k

k k

c

1s

)0(x

k k

k )s(X (C-30)



18

The inverse Laplace transform yields the time domain response.

t

k k

k k

c

1exp

k k

k )0(x)t(x

21

21

21

112 (C-31)

0xk xk k 11221 (C-32)

22111 xk k xk (C-33)

21

211 x

k

k k x

(C-34)

t

k k

k k

c

1exp)0(x)t(x

21

2111 (C-35)



19

APPENDIX D


Figure D-1.

Figure D-2.

k 1 (x2-x1)

1xc

k 1

c

x1

x2

k 2

F t

m

k 1(x1-x2)

x2

-k 2 x2

m

F(t)



20


222112 xk )x-(xk )t(Fxm (D-1)

)t(Fxk xk k xm 112212 (D-2)

0)x-(xk xc 2111 (D-3)

Take the Laplace transform of equation (D-2).

)t(FLxk xk k xmL 112212 (D-4)

)s(F)s(Xk )s(Xk k )0(xsm)0(xm)s(Xsm 112212222 (D-5)

)0(xsm)0(xm)s(F)s(Xk )s(Xk k sm 22112212 (D-6)

Take the Laplace transform of equation (D-3).

0)x-(xk xcL 2111 (D-7)

0)s(Xk )s(Xk )0(xc)s(Xsc 221111 (D-8)

)0(xc)s(Xk )s(Xk sc 12211 (D-9)

Consider equations (D-6) and (D-9) for the case of zero initial conditions.

)s(F)s(Xk )s(Xk k sm 112212 (D-10)



21

0)s(Xk )s(Xk sc 2211 (D-11)

)s(Xk )s(Xk sc 2211 (D-12)

)s(Xk sc

k )s(X 2

1

21

(D-13)

)s(F)s(Xk sc

k k )s(Xk k sm 2

1

21221

2

(D-14)

)s(F)s(Xk sc

k k k k sm 21

2121

2

(D-15)

1

2121

22

k sc

k k k k sm

)s(F)s(X (D-16)


1

2121

2

2

k sc

k k k k sm

1

)s(F

)s(X (D-17)


0xk xk k xm 112212 (D-18)

0)x-(xk xc 2111 (D-19)





23

211212

1121212

k k k sck k sm

)0(xk c)0(xk scsm)0(xk scm)s(X

(D-27)

211212

212

11212

212

k k k k k smsck k sm

)0(xk c)0(xk smscm)0(xk mscm)s(X

(D-28)

211212

1213 11212

2

2122k k k k k sk msk k cscm

)0(xk c)0(xk sm)0(xscm)0(xk m)0(xscm)s(X

(D-29)

2121

21

311212212

2

2k sk k csk mscm

)0(xk c)0(xk m)0(xscm)0(xk sm)0(xscm)s(X

(D-30)

2121

21

311212212

2

2k sk k csk mscm

)0(xk c)0(xk msm)0(xc)0(xk )0(xscm)s(X

(D-31)

cm

k s

m

k k s

c

k s

)0(xm

1)0(x

c

1k s)0(x)0(x

c

k )0(xs

)s(X2121213

121221

22

2

(D-32)



24

APPENDIX E

Consider the following two-degree-of-freedom system subjected to base excitation.

Figure E-1.

Figure E-2.

k 1 (x2-x1)

yxc 1

k 1

c x1

x2

k 2

m

y

k 1(x1-x2)

x2

-k 2 (x2-y)

m



25

yxk )x-(xk xm 222112 (E-1)

0yxk )x-(xk xm 221212 (E-2)

0)x-(xk yxc 2111 (E-3)

Let

yxz 11 (E-4)

yxz 22 (E-5)

2121 x-xz-z (E-6)

By substitution,

ymzk )z-(zk zm 221212 (E-7)

ymzk -zk k zm 112212 (E-8)

0)z-(zk zc 2111 (E-9)

Take the Laplace transform of equation (E-2).

ymLzk zk k zmL 112212 (E-10)

)s(Ym)s(XZ)s(Zk k )0(zsm)0(zm)s(Zsm 112212222 (E-11)

)0(zsm)0(zm)s(Zk )s(Zk k sm 2211221

2

(E-12)



26

Take the Laplace transform of equation (E-3).

0)z-(zk zcL 2111 (E-13)

0)s(Zk )s(Zk )0(zc)s(Zsc 221111 (E-14)

)0(zc)s(Zk )s(Zk sc 12211 (E-15)

Consider equations (E-6) and (E-9) for the case of zero initial conditions.

)s(Ym)s(Zk )s(Zk k sm 112212 (E-16)

0)s(Zk )s(Zk sc 2211 (E-17)

)s(Zk )s(Zk sc 2211 (E-18)

)s(Zk sc

k )s(Z 2

121

(E-19)

)s(Ym)s(Zk sc

k k )s(Zk k sm 2

1

21221

2

(E-20)

)s(Ym)s(Zk sc

k k k k sm 2

1

2121

2

(E-21)



1

2121

22

k sc

k k k k sm

)s(Ym)s(Z (E-22)

The relative acceleration )s(Z2 is

22

2 Zs)s(Z (E-23)

1

2121

2

2

2

k sc

k k k k sm

)s(Yˆ

sm)s(Z (E-24)

)s(Y

k sc

k k k k sm

)s(Ysm)s(X

1

2121

2

2

2

(E-25)

)s(Y1

k sc

k k k k sm

sm)s(X

1

2121

2

2

2

(E-26)

1

k sc

k k k k sm

sm

)s(Y

)s(X

1

2121

2

22 (E-27)

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