振动学系列(Vibration)6._Base_Excitation.doc

6. Base ExcitationNow that we can find the FRF for a mass/spring/damper system, let us tackle something slightly more complicated: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 tire to move up and down. These vibrations are transmitted through the suspension into the chassis.First, let us draw a free-body diagram of the systemfk=kx-yfc=kx-ymx=-fk-fcmx+cx-y+kx-y=0mx+cx+kx=cy+kyNow, let us assume that the base excitation is harmonicy=Yejty=jYejtalso, letx=Xejtx=jXejtx=-2XejtThen-m2Xejt+jcXejt+kXejt=jcYejt+kYejtFactoring out the exponential gives-m2+jc+kX=jc+kYXY=jc+k-m2+jc+kThis function is similar in form to the FRF found earlier, and is in fact another type of FRF.XY=output displacementinput displacement=displacement transmissibilityX/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 method is very powerful and simple to use.Measurement DevicesWe turn now to a practical application of some of this theory. Assume that we have a device as shown below:The device produces a voltage proportional to the movement of the mass relative to the base.voutx-yDefinez=x-yThen the equation of motion derived earlier becomesmx+cx-y+kx-y=0mz+y+cz+kz=0mz+cz+kz=-myAssume as beforez=Zejtz=jZejtz=-2Zejtalso, let us assume thaty=AbaseejtThen-m2Zejt+jcZejt+kZejt=-mAbaseejt-m2+jc+kZ=-mAbaseZ=-Abase-2+jcm+n2If n, thenZ2Y-2=YThus, the output is proportional to the displacement we have created a seismometer!Example Base Excitation Problem (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 displacement ofyt=0.01msinbtWe 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 speed do the cars experience resonance?3. What is the displacement of each car at resonance?1. Effect of speed on amplitudeThe car encounters a bump every 6m. The frequency at which it encounters bumps can be calculated asfb=v6mif v is in m/s. To express v in km/hr we must do a little unit conversionfb=vkmhr6m1000m1km1hr3600s=0.046vHzTo convert this into rad/s, we multiply by 2b=0.291vradsEarlier in the notes we derived an expression for the displacement transmissibilityXY=jbc+k-mb2+jbc+kWe wish to solve for displacement amplitude, so we rearrangeX=jbc+k-mb2+jbc+kYNext, we must determine the amplitude of the base excitation. We are given the base excitation in the formyt=0.01msinbtBut we must convert it to the formyt=ReYejbtFrom an earlier example problem, this is true ifY=-0.01j mThus, we can writeX=jbc+k-mb2+jbc+k(-0.01j)2. Speed of resonanceFor car 1, the natural frequency isn1=km1=40000Nm1007kg=6.303radsand for car 2n2=km2=40000Nm1585kg=5.024radsWe must now find the speeds that generate the above natural frequencies. For car 1v1=n10.291=21.7kmhrv2=n20.291=17.3kmhr3. Amplitude of displacement at resonanceFirst, let us simplify the amplitude equation by dividing through by m.X=jbcm+n2-b2+jbcm+n2(-0.01j)At resonance b = n, so that the denominator simplifies toX=jncm+n2jncm-0.01jX=jcm+njcm-0.01jX=jc+mnc-0.01X=j+mnc-0.01For car 1, we haveX1=j+1007 kg6.303rads2000kgs-0.01X1=-0.032-0.01j mAnd for car 2,X2=j+1585 kg5.024rads2000kgs