IMPACT VIBRATION CONTROL USING SEMI-ACTIVE DAMPER
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 seminar class Active In SP Posts: 5,361 Joined: Feb 2011 23-02-2011, 09:44 AM   vibration.pptx (Size: 356.78 KB / Downloads: 248) IMPACT VIBRATION CONTROL USING SEMI-ACTIVE DAMPER Introduction  Dynamic vibration absorbers are used for suppressing the steady state vibration of a machine.  It consists of spring-mass-damper system.  For an impact force, it can damp only the residual part of impact vibration  But maximum amplitude of the vibration is almost remain same that of undamped vibration DYNAMIC VIBRATION ABSORBER RESPONSE Of An IMPULSE FREE BODY DIAGRAM EQUATION OF MOTION  Control force of dynamic damper can be written as  fd=Mdx”=-c(x-xd)-Kd(x-xd)  By substituting in (1) we will get  Mx”+Kx=fim-fd RESPONSE CALCULATION  We calculate the response numerically by Runge-Kutta method. For this we take  M=100Kg  K=9.87MN/m  Fo=63.66kNs  Md=10Kg  C=1.055KNs/m  Kd=0.816MN/m RESPONSE DUE TO IMPACT TOTAL TRANSMITTED FORCE  Mx”+Kx=fim-fid  If we take Laplace transform of equation Ft(s) =G(s)Fim(s)-G(s)Fd(s)  Where G(s)=K/(Ms2+K)  Ft=KX(s)-transmitted force to the ground. TOTAL TRANSMITTED FORCE  If we take the inverse Laplace transform  ft= ftm-ftd  ft= Total transmitted force  ftm-transmitted force due to impact force fim  ftd=transmitted force due to control force fd COMPONENTS OF FORCE TRANSMITTED FORCE  Transmitted force is sum of transmitted force due to impact and due to damping force.  Both these forces are just opposite and cancel each other.  But in the first few waves ftd lacks sufficient magnitude CONCLUSION FROM GRAPH  Two methods we can adopt to suppress the force vibration of impact 1. Increase the amplitude of first wave of ftd by increasing certain action for driving damper mass 2. Utilise the large portion such as the second or third wave.  As a solution to check we will give an initial displacement to damper say xd(0) INITITAL DISPLACEMENT  Take the Laplace transform of given equation(1) with initial condition xd(0)  Ft(s)= (s2+2ζ ωds+ωd) ωt2fo+Kd ωt2s xd(0) Gd(s) Where Gd(s)=s4+2(1+µ) ζωd S3+[(1+ µ) ωd 2+ ωt2 ]s2+2 ζ ωd ωt2 s+ ωt2ωd 2  2ζ ωd =C/Md Kd/ Md = ωd 2 K/M= ωt2
 abhibhosale07 Active In SP Posts: 1 Joined: Feb 2011 25-02-2011, 10:35 PM PLZ, SEND ME THIS SEMINAR FULL
 sureshkharde Active In SP Posts: 6 Joined: Jul 2011 21-07-2011, 10:13 AM please can u send me this seminar and presentation report