Spring Damper

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    THE VIBRATION RESPONSE OF SOME SPRING-DAMPER SYSTEMS

    WITH AND WITHOUT MASSES

    By Tom IrvineEmail: tomirvine@aol.com

    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

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

    )0(xc)s(F)s(Xksc (4)

    ksc

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

    (5)

    Now assume that the initial displacement is zero.

    ksc

    )s(F

    )s(X (6)

    ksc

    1

    )s(F

    )s(X

    (7)

    The dynamic stiffness in the Laplace domain is

    ksc)s(X

    )s(F

    (7)

    The dynamic stiffness in the frequency domain is

    cjk

    )(X

    )(F (8)

    The initial value problem for the free response is

    ksc

    )0(xc)s(X

    (9)

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    c/ks)0(x

    )s(X

    (10)

    The inverse Laplace transform yields the time domain response.

    tc/kexp)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

    xm

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    )0(msx)0(xm)0(xc)s(F)s(Xkscms2 (A-4)

    kscms

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

    2 (A-5)

    Now assume that the initial displacement is zero.

    kscms

    )s(F)s(X

    2 (A-6)

    kscms

    1

    )s(F

    )s(X

    2 (A-7)

    The dynamic stiffness in the Laplace domain is

    kscms)s(X

    )s(F 2 (A-8)

    The dynamic stiffness in the frequency domain is

    cjmk

    )(X

    )(F 2 (A-9)

    The mechanical impedance in the frequency domain is

    j

    cjmk

    )(V

    )(F 2 (A-10)

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    m

    kjc

    )(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

    cj

    m

    k1m

    )(A

    )(F

    2 (A-14)

    Note that

    m

    kn (A-15)

    m2c n (A-16)

    2 (A-17)

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    By substitution,

    n2

    2n 2j1m

    )(A

    )(F (A-18)

    n2

    2n j1m

    )(A

    )(F (A-19)

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    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

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    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

    -xkxc 111

    (B-6)

    0k

    )t(F-kxc 1

    (B-7)

    )t(Fxc 1 (B-8)

    Take the Laplace transform.

    )t(FLxcL 1 (B-9)

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    )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 11

    (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 12 (B-17)

    )0(xs

    k

    sc

    k1)s(F)s(Xk 12

    (B-18)

    )0(xs

    1

    sc

    1

    k

    1)s(F)s(X 12

    (B-19)

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    Now assume the initial displacement is zero.

    sc

    1

    k

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

    skc

    kcs)s(F)s(X2 (B-21)

    The receptance in the Laplace domain is

    skc

    kcs

    )s(F

    )s(X2

    (B-22)

    The dynamic stiffness in the Laplace domain is

    kcs

    skc

    )s(X

    )s(F

    2 (B-22)

    The dynamic stiffness in the frequency domain is

    cjk

    kcj

    )(X

    )(F

    2

    (B-22)

    The initial value problem for the free response is

    )0(xs

    1)s(X 12 (B-23)

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    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)

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    APPENDIX C

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

    Figure C-1.

    Figure C-2.

    k1(x2-x1)

    k1(x1-x2)

    )t(F

    x2

    )t(F

    k1

    c

    x1

    x2

    k2

    -k2x2

    1xc

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    The equations of motion are

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

    )t(Fxkxkk 11221 (C-2)

    0)x-(xkxc 2111 (C-3)

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

    121

    1

    212 x

    kkk

    kk)t(Fx

    (C-4)

    By substitution,

    0)x-(xkxc 2111 (C-5)

    0xkk

    kkk)t(F-xkxc 1

    21

    1

    21111

    (C-6)

    0)t(Fkk

    k-x

    kk

    k1kxc

    21

    11

    21

    111

    (C-7)

    0)t(Fkk

    k-xkk

    kkkkxc21

    11

    21

    12111

    (C-8)

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    21

    21

    1

    21

    1

    21

    2121

    1

    212

    kk

    kksc

    )0(xc

    kk

    k)s(F

    kk

    kksc

    1

    kk

    k

    kk

    1)s(X

    (C-21)

    21

    21

    1

    21

    1

    21

    21

    1

    212

    kk

    kk

    sc

    )0(xc

    kk

    k)s(F

    kk

    kk

    sc

    k1

    kk

    1)s(X

    (C-22)

    Now assume the initial displacement is zero.

    )s(F

    kk

    kksc

    k1

    kk

    1)s(X

    21

    21

    1

    212

    (C-23)

    The receptance in the Laplace domain is

    21

    21

    1

    21

    2

    kk

    kksc

    k1

    kk

    1

    )s(F

    )s(X (C-24)

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    2121

    1

    21

    2

    kkskkc

    k

    kk

    1

    )s(F

    )s(X (C-25)

    21212

    21

    21121212

    kkkkskkc

    kkkkkskkc

    )s(F

    )s(X

    (C-26)

    21212

    21

    211212

    kkkkskkc

    k2kkskkc

    )s(F

    )s(X

    (C-27)

    The dynamic stiffness in the Laplace domain is

    21121

    21212

    21

    2 k2kkskkc

    kkkkskkc

    )s(X

    )s(F

    (C-28)

    The initial value problem for the free response is

    21

    21

    1

    21

    12

    kk

    kksc

    )0(xc

    kk

    k)s(X (C-29)

    21

    21

    1

    21

    12

    kk

    kk

    c

    1s

    )0(x

    kk

    k)s(X (C-30)

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    The inverse Laplace transform yields the time domain response.

    t

    kk

    kk

    c

    1exp

    kk

    k)0(x)t(x

    21

    21

    21

    112 (C-31)

    0xkxkk 11221 (C-32)

    22111 xkkxk (C-33)

    21

    211 x

    k

    kkx

    (C-34)

    t

    kk

    kk

    c

    1exp)0(x)t(x

    21

    2111 (C-35)

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    APPENDIX D

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

    Figure D-1.

    Figure D-2.

    k1(x2-x1)

    1xc

    k1

    c

    x1

    x2

    k2

    F t

    m

    k1(x1-x2)

    x2

    -k2x2

    m

    F(t)

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    The equations of motion are

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

    )t(Fxkxkkxm 112212 (D-2)

    0)x-(xkxc 2111 (D-3)

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

    )t(FLxkxkkxmL 112212 (D-4)

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

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

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

    0)x-(xkxcL 2111 (D-7)

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

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

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

    )s(F)s(Xk)s(Xkksm 112212 (D-10)

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

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

    )s(Xksc

    k)s(X 2

    1

    21

    (D-13)

    )s(F)s(Xksc

    kk)s(Xkksm 2

    1

    21221

    2

    (D-14)

    )s(F)s(Xksc kkkksm 2121

    212

    (D-15)

    1

    2121

    22

    ksc

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