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1、16. Base ExcitationNow that we can find the FRF for a mass/spring/damper system, let us tackle something slightly more complicated:cmkx(t)y(t)Here, the system is excited by movement in the base, y(t). This is analagous to a car driving down the road bumps in the road cause the contact patch of the t
2、ire to move up and down. These vibrations are transmitted through the suspension into the chassis.road excitationFirst, let us draw a free-body diagram of the systemmx(t)fkfc= ( )= ( ) = + ( )+ ( )= 0 + + = + Now, let us assume that the base excitation is harmonic2 = = also, let = = = 2Then 2+ + = +
3、 Factoring out the exponential gives( 2+ + ) =( + ) = + 2+ + This function is similar in form to the FRF found earlier, and is in fact another type of FRF. = = X/Y is a measure of how much motion we get for a given input motion at the base. You can see by this example that the complex exponential me
4、thod is very powerful and simple to use.3Measurement DevicesWe turn now to a practical application of some of this theory. Assume that we have a device as shown below:cmkV0x(t)y(t)The device produces a voltage proportional to the movement of the mass relative to the base. Define = Then the equation
5、of motion derived earlier becomes + ( )+ ( )= 0( + )+ + = 0 + + = Assume as before = = = 2also, let us assume that = Then 2+ + = ( 2+ + ) = 4 = 2+ + 2If n, then 2 2= Thus, the output is proportional to the displacement we have created a seismometer!5cmkx(t)y(t)0.2m0.2m6m6mExample Base Excitation Pro
6、blem (Problem 2.43 in text) A very common example of base motion is the single-degree of freedom model of an automobile driving over a road or an airplane taxiing over a runway, indicated in the figure above. The road surface is approximated as sinusoidal in cross-section providing a base motion dis
7、placement of()=(0.01)sinWe will examine two cars: one with a mass of 1007kg (sports car) and the other with a mass of 1585kg (sedan). Both cars have a suspension stiffness of 40kN/m and a damping coefficient of 2000kg/s1. Determine the effect of speed on the amplitude of displacement. . 2. At what s
8、peed do the cars experience resonance? 3. What is the displacement of each car at resonance?1. Effect of speed on amplitude The car encounters a bump every 6m. The frequency at which it encounters bumps can be calculated as= 6if v is in m/s. To express v in km/hr we must do a little unit conversion=
9、() 61000 11 3600= 0.046To convert this into rad/s, we multiply by 2= 0.291Earlier in the notes we derived an expression for the displacement transmissibility6 = + 2+ + We wish to solve for displacement amplitude, so we rearrange = + 2+ + Next, we must determine the amplitude of the base excitation.
10、We are given the base excitation in the form()=(0.01)sinBut we must convert it to the form()= From an earlier example problem, this is true if = 0.01 Thus, we can write = + 2+ + ( 0.01)2. Speed of resonanceFor car 1, the natural frequency is1= 1=40000 1007= 6.303and for car 22= 2=40000 1585= 5.024We
11、 must now find the speeds that generate the above natural frequencies. For car 171=10.291= 21.72=20.291= 17.33. Amplitude of displacement at resonanceFirst, let us simplify the amplitude equation by dividing through by m. =+ 2 2+ 2( 0.01)At resonance b = n, so that the denominator simplifies to =+ 2
12、( 0.01) = + ( 0.01) = + ( 0.01) =( + )( 0.01)For car 1, we have1=( +1007 6.3032000)( 0.01)81=( 0.032 0.01) And for car 2,2=( +1585 5.0242000)( 0.01)2=( 0.04 0.01) In both cases we are asked for the amplitude of the displacement|1|= 0.033|2|= 0.041In both cases, the amplitude at resonance is larger than the amplitude of the excitation!
