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Transcript of MSD System

  • 1

    Georgios H. Vatistas August 2006

    Mass-Spring-Damper Mechanical system and its Electrical Analogue.

    There is large number of mechanical devices where MSD (Mass-Spring-Damper) system constitutes the fundamental building block for their mathematical modeling. The basic mechanical components of this system are the spring and the damper. There are complete courses in the engineering curriculum that deal systematically with this topic. Some of these courses are: Modeling Simulation and Analysis of Physical Systems (MECH 370), Fundamentals of Control Systems (MECH 373), Mechanical Vibrations (MECH 373), Fluid Power Control (MECH 448), Basic Circuit Analysis ELEC 273), Principles of Electrical Engineering (ELEC 275), and several others. Here we will only give a brief introduction to the subject, always in the context of the first engineering math course (ENGR 213). The Spring For a linear spring:

    F s = k y The spring constant k is the slope of the Fs - y curve. It has units:

    k ! F sy

    = Nm

    =

    kg m

    s 2

    m =

    kg

    s 2

    Square bracket [] means units of the parameter(s) inside the bracket.

    If the spring is stiff then the value of k is relatively large, if it soft k is small.

    0

    y

    F s

    unstrechedposition

  • 2

    Fs (

    N )

    y ( m )

    k is the slope

    The Damper

    A damper is a dissipative device that converts mechanical energy into heat. This device is mathematically modeled by:

    F d = c V = c dy

    dt, F d always opposes the motion

    The units of c: c !

    F dV

    = Nm / s

    = N sm

    = kg m

    s 2 s 1

    m =

    kgs

    0

    y

    c

  • 3

    A relatively large value of c indicates hefty opposition to motion (and high

    dissipation), while a small value indicates the opposite. The Spring-Mass System

    0

    W = m g

    yo

    m

    no load (unstreched possition)

    loaded(static equilibrium)

    y

    k

    Fs (

    N )

    y ( m )

    k is the slope

    Fs o

    y o 0

  • 4

    When the solid body with mass m is attached the spring will stretch until Fs = W. At this point the system is in static equilibrium, or,

    forces acting on the body ! = 0 or W - F s o = W - k y o = 0 ! W = k y o

    Starting from this static position let us begin to pull the mass down with a varying velocity (an acceleration dV dt).

    0

    W = m g

    yo

    m

    loaded(static equilibrium)

    y

    k

    Fso

    Fso

    Fs

    Fs

    no loadpossition

    Newtons 3rd law

    0

    y

    positionwithout

    the mass

    Equilibrium possition(with the

    mass attached

    ________________________________________________________________________

    When a solid material is cut for the sake of analysis then the internal forces that bind the solid together must be included in the free-body diagram, see for example the solid rod before and after the cut.

    before after

    F i F i

    + ve x - direction Note: that when we put the two parts (on the right) together to obtain the uncut rod, the total force is from Newtons 2nd law Fi - Fi = 0. ________________________________________________________________________

    From Newtons 2nd law: ddt

    m V = forces acting external to the body !

  • 5

    d

    dt m V = m d V

    dt + V d m

    dt = W - F s o -F s = W - k y o - k y

    Since we are dealing with a solid body, dm/dt = 0. Also, from above W = k yo. Then, Newtons second law applied to the mass (free-body-diagram)gives,

    m d Vdt

    = - k y or m d 2 y

    dt 2 = - k y remember that V =

    d y

    dt

    Rearranging:

    d 2 y

    dt 2 + k

    m y = 0 with characteristic equation r 2 + k

    m = 0

    Alike to all equations, the differential equation must possess dimensional homogeneity i.e. all terms must have exactly the same units:

    d 2 y

    dt 2 !

    y

    t 2 = m

    s 2 i.e. units of acceleration

    d indicates an action; take the infinitesimal difference and as such it has no units. Similarly d 2 has no units.

    km

    y !

    kg

    s 2

    kg = m

    s 2 i.e. again units of acceleration

    Hence, the equation it does possess dimensional homogeneity.

    The above equation is a second order ordinary linear equation with constant coefficients, its general solution is:

    y t = A cos km

    t + B sin km

    t

    The argument of transcendental functions such as for example the trigonometric, exponential, logarithmic etc must be dimensionless (clear number),

    km

    t !

    kg

    s 2

    kg s = 1 which is a clear number

    Let us now consider the case where the mass is pulled down until y = yin, it is stopped for a while, and then at t = 0 is released. The initial conditions are then:

  • 6

    i. t = 0, y = yin = 0.2 m ii. t = 0, Vin = dy/dt = 0 Application of the second initial condition yields,

    dy

    dt

    t = 0 = - A k

    msin k

    m 0 + B k

    m cos k

    m 0 = 0 or B = 0

    The first initial condition gives that A = yin. = 0.2 m. Therefore, the position of the top of the mass is then given by:

    y t = 0.2 cos ! t where ! (circular frequency) = k/m in rads / s

    Since the cosine function appears in the solution, the mass will oscillate. Let us now plot amplitude y as a function of time for different values of k/m. The results are given in the following figure.

    -0.4

    -0.2

    0

    0.2

    0.4

    0 2 4 6 8 10 12

    1 4 9

    y (

    t )

    in

    m

    t in s

    k / m = 1 4 9

    The period of oscillations T is,

    T =

    2 !"

    = 2 ! m/k in s while the frequency is given by:

    f = 1

    T = 1

    2 ! k/m in cycles / s or hz

    Frequency f is known as the natural frequency of the system and will designated henceforth by fn.

  • 7

    It is clear that this system will undulate indefinitely. The reason behind this

    unrealistic behavior is that in the mathematical modeling of this system the energy dissipation due to friction inside the spring and air have been neglected. The last paradox (contrary to what is expected) will be rectified in a subsequent section. Is also apparent that as the k/m value goes up, the number of oscillations within the same time interval (or fn) of the system increases. On one hand, if m is kept constant, increasing the stiffness of the spring (k) will result into a higher frequency of oscillations. On the other hand, increasing the mass m while keeping k constant will result into a lower frequency of the system.

    A practical way to find fn of the system, by only one simple measurement, is as follows.

    The static deflection of the spring-mass system was given previously by:

    W = k y o ! m g = k y o ! g

    y o = k

    m Then

    f n = 1

    2 ! k/m = 1

    2 ! g/y o in hz

    Therefore, the natural frequency of the system cam be obtained by loading the

    mass to the sprig and measure the static deflection yo. The Mass-Spring-Damper System

    As in the previous section, starting from the static equilibrium position let us now begin to pull the mass down with a varying velocity (an acceleration dV/ dt). Remember that the damper opposes the motion. Application of Newtons second law yields,

    m d 2 y

    dt 2 = - F s - Fd = - k y - c

    d y

    dt ! m

    d 2 y

    dt 2 + c

    d y

    dt + k y = 0

    or

    d 2 y

    dt 2 + c

    m d y

    dt + k

    m y = 0

  • 8

    position

    without

    the mass0

    W = m g

    m

    y

    k

    Fs

    Fs

    c

    Fd

    Fd

    0

    y

    Equilibrium

    position

    (with the mass

    attached

    position

    without the

    mass

    The above equation is a second order, ordinary, homogeneous, linear equation with constant coefficients. Its characteristic equation is:

    r 2 + c

    m r +

    k

    m = 0 ! r1, 2 =

    - c

    m c

    m

    2 - 4 k

    m

    2 Depending on the values of k, m, and c the roots of the characteristic equation could be (A) two real and distinct, (B) two real double, and (C) two complex conjugates. The three cases will be examined next. CASE A

    Let k = 1 N/m, m = 1 kg, and c = 3 Ns / m. Then

    r1, 2 = - 3

    2 5

    2

    The general solution is therefore

    y t = A exp - 3

    2 + 5

    2 t + B exp - 3

    2 - 5

    2 t

    At first the arguments of the exponential function appears to have the units of s ([t]

    s). This off course it is not true. In order to simplify the algebra we have not included the units of r 1,2 which are:

    r1, 2 ! -

    c

    m c

    m

    2 - 4 k

    m

    2 !

    - c

    m c

    m

    2 - 4 k

    m

    2

  • 9

    = - 3

    2

    kg

    s kg 3

    2

    kg

    s kg

    2 - 4

    kg

    s 2 kg = - 3

    2 1s

    32

    1s

    2 - 4 1

    s 2 1s

    = - 3

    2 5

    2 1

    s Therefore, the arguments of the exponentials functions are dimensionless,

    exp - 3

    2 5

    2 1s

    t

    Like before, let the initial conditions be:

    i. t = 0, y = yin = 0.2 m ii. t = 0, Vin = dy/dt = 0 From the first an