Chapter 2 Response to Harmonic Excitation 2018. 1. 30.آ  2 2 2 2 22 2 ( ) cos( tan ) ( ) (2 ) n p...

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Transcript of Chapter 2 Response to Harmonic Excitation 2018. 1. 30.آ  2 2 2 2 22 2 ( ) cos( tan ) ( ) (2 ) n p...

  • Chapter 2 Response to Harmonic Excitation

    Introduces the important concept of resonance

  • 2.1 Harmonic Excitation of Undamped Systems

    • Consider the usual spring mass damper system with

    applied force F(t)=F0cosωt

    • ω is the driving frequency

    • F0 is the magnitude of the applied force

    • We take c = 0 to start with

    M

    F=F 0 cosωt

    Displacement

    x k

  • 0

    2

    0

    0 0

    ( ) ( ) cos( )

    ( ) ( ) cos( )

    where ,

    n

    n

    mx t kx t F t

    x t x t f t

    F kf m m

     

      

     

     

    Equations of motion

    ( ) cos( )px t X t

    • Solution is the sum of

    homogenous and

    particular solution

    • The particular solution

    assumes form of forcing

    function (physically the

    input wins):

  • Substitute particular solution into the equation of motion:

    ( ) cos( )px t X t 2

    2 2

    0

    0

    2 2

    cos cos cos

    solving yields:

    p n px x

    n

    n

    X t X t f t

    f X

        

     

      

     

    Thus the particular solution has the form:

    0

    2 2 ( ) cos( )p

    n

    f x t t

      

  • Add particular and homogeneous solutions to get general

    solution:

    0 1 2 2 2

    homogeneous

    1 2

    ( )

    sin cos cos

    and are constants of integration.

    particular

    n n

    n

    x t

    f A t A t t

    A A

        

      

    (2.8)

  • Apply the initial conditions to evaluate the constants

    0 0 1 2 2 02 2 2 2

    0 2 0 2 2

    0 1 2 1 02 2

    0 1

    0

    (0) sin 0 cos0 cos0

    (0) ( cos0 sin 0) sin 0

    ( )

    n n

    n

    n n

    n

    n

    f f x A A A x

    f A x

    f x A A A v

    v A

    v x t

       

     

       

           

       

         

      

     0 00 2 2 2 2sin cos cosn n n n n

    f f t x t t  

       

         

      

    (2.11)

  • Comparison of free and forced response

    • Sum of two harmonic terms of different frequency

    • Free response has amplitude and phase effected by forcing

    function

    • Our solution is not defined for ωn = ω because it produces

    division by 0.

    • If forcing frequency is close to natural frequency the

    amplitude of particular solution is very large

  • Response for m=100 kg, k=1000 N/m, F=100 N, ω = ωn +5

    v0=0.1m/s and x0= -0.02 m.

    Note the obvious presence of two harmonic signals

    0

    Displacement (x)

    Time (sec)

  • 0

    What happens when ω is near ωn?

    0

    2 2

    2 ( ) sin sin (2.13)

    2 2

    n n

    n

    f x t t t

       

     

             

        

    When the drive frequency and natural frequency

    are close a beating phenomena occurs

    0

    2 2

    2 sin

    2

    n

    n

    f t

     

     

       

      

    Larger

    amplitude

    Time (sec)

    Displacement (x)

  • What happens when ω is ωn?

    0

    ( ) sin( )

    substitute into eq. and solve for

    2

    px t tX t

    X

    f X

    When the drive frequency and natural

    frequency are the same the amplitude

    of the vibration grows without bounds.

    This is known as a resonance

    condition.

    The most important

    concept in Chapter 2!

    0

    2 2

    2 sin

    2

    n

    n

    f t

     

     

       

      

    0 5 1 0

    1 5

    20

    25

    x(t) = A1 sinwt + A2 coswt +

    f0

    2w t sin(wt)

    grows with out bound

  • Example 2.1.1: Compute and plot the response for

    m=10 kg, k=1000 N/m, x0=0,v0=0.2 m/s, F=23 N, ω =2ωn.

    0 0

    30

    2 2 2 2 2 2

    1000 N/m 10 rad/s, 2 20 rad/s

    10 kg

    23 N 0.2 m/s 2.3 N/kg, 0.02 m

    10 kg 10 rad/s

    2.3 N/kg 7.9667 10 m

    (10 20 ) rad / s

    Equation (2.11) then yields:

    ( ) 0.02sin10 7.96

    n n

    n

    n

    k

    m

    vF f

    m

    f

    x t t

      

     

        

        

         

      367 10 (cos10 cos 20 )t t 

  • Example 2.1.2 Given zero initial conditions a

    harmonic input of 10 Hz with 20 N magnitude and k=

    2000 N/m, and measured response amplitude of 0.1m,

    compute the mass of the system.

     

     

    0

    2 2

    0

    2 2

    0.1 m

    0

    2 2 2

    ( ) cos 20 cos for zero initial conditions

    2 trig identity ( ) sin sin

    2 2

    2 2(20 / ) 0.1 0.1

    2000 (20 )

    0.405 kg

    n

    n

    n n

    n

    n

    f x t t t

    f x t t t

    f m

    m

    m

       

       

     

      

      

              

        

         

     

  • Example 2.1.3 Design a rectangular mount for a

    security camera.

    0.02 x 0.02 m

    in cross section

    Compute l > 0.5 m so that the mount keeps the camera from

    vibrating more then 0.01 m of maximum amplitude under a

    wind load of 15 N at 10 Hz. The mass of the camera is 3 kg.

  • Solution: Modeling the mount and camera as a beam

    with a tip mass, and the wind as harmonic, the equation

    of motion becomes:

    03

    3 ( ) cos

    EI mx x t F t 

    From strength of materials:

    3

    12

    bh I 

    Thus the frequency expression is: 3 3

    2

    3 3

    3

    12 14 n

    Ebh Ebh

    m m   

    Here we are interested computing l that will make the

    amplitude less then 0.01m:

    2 20

    2 2

    0

    2 2 2 20

    2 2

    2 ( ) 0.01 , for 0

    2 0.01

    2 ( ) 0.01, for 0

    n

    n

    n n

    n

    f a

    f

    f b

       

       

     

         

           

     

  • Case (a) (assume aluminum for the material):

    3 2 2 20

    0 02 2 3

    3 3

    2

    0

    2 0.01 2 0.01 0.01 0.01 2 0.01

    4

    0.01 0.02 0.272 m 4 (0.01 2 )

    n

    n

    f Ebh f f

    m

    Ebh

    m f

        

            

         

    Case (b):

    3 2 2 20

    0 02 2 3

    3 3

    2

    0

    2 0.01 2 0.01 0.01 2 0.01 0.01

    4

    0.01 0.012 0.229 m 4 (2 0.01 )

    n

    n

    f Ebh f f

    m

    Ebh

    m f

        

           

         

  • 3(2.7 10 )(0.22)(0.01)(0.01)

    0.149 kg

    m bh

     

    Remembering the constraint that the length must be at

    least 0.2 m, (a) and (b) yield

    0.2 0.229, or choose = 0.22 m  To check, note that

      3

    22 2

    3

    3 20 1609 0

    12 n

    Ebh

    m       

  • A harmonic force may also be represented by sine or a

    complex exponential. How does this

    change the solution?

    2

    0 0( ) ( ) sin or ( ) ( ) sin (2.18)nmx t kx t F t x t x t f t     

    The particular solution then becomes a sine:

    ( ) sin (2.19)px t X t

    Substitution of (2.19) into (2.18) yields:

    0

    2 2 ( ) sinp

    n

    f x t t

      

    Solving for the homogenous solution and evaluating the constants yields

    0 0 0 0 2 2 2 2

    ( ) cos sin sin (2.25)n n n n n n

    v f f x t x t t t

       

         

     