Consider the spring mass dashpot system shown below which is acted upon by a harmonic excitation force F(t). The reference is taken at the equilibrium position of the system when no F(t) was present. At time t=0, F(t) acts and displaces the system from its equilibrium.

enter image description here

I'm interested in knowing the force transmitted to the support. I proceeded in the following manner: At time t, the spring would be stretched by $x-x_0$ and the end of damper will have a velocity equal to x(dot on the top) enter image description here

The resulting expression I get for transmitted force has an mg in it. However all the sources I'm referring to state the expression with no mg. Where am I wrong in the analysis?

  • $\begingroup$ could you provide which expressions you are referring to? (It seems as though you are trying to obtain the transmissibility ratio.) $\endgroup$
    – NMech
    Sep 30 at 4:45
  • $\begingroup$ Yes, eventually the transmissibility ratio, but prior to that - the force transmitted to the support as a function of time. $\endgroup$ Sep 30 at 5:27
  • $\begingroup$ there can be many reasons for that ( I still don't know your other references). However, I think is that the vibration will be largely unaffected by the gravity. Gravitational forces are a constant term that will be added to the mean value. The transmissability ratio offers a relationship between the excitation force (not gravity) and the transmitted force. $\endgroup$
    – NMech
    Sep 30 at 5:34
  • $\begingroup$ Ohh. Here is a screenshot from the textbook I'm referring to (with highlighted equation I'm talking about) - drive.google.com/file/d/1-Cg3qAYTk4niUEnZldXzO8Hm-1FoLw4w/… $\endgroup$ Sep 30 at 5:42
  • $\begingroup$ That looks like rao's mechanical vibration's right? $\endgroup$
    – NMech
    Sep 30 at 5:46

In the analysis you are referring to the authors are only concerned about the amplification of the excitation force (they are not concerned about the total force). I.e. they only care about the additional force that the excitation is causing on the supports.

(keep in mind that the vibration would still occur even in a zero-g environment if you added a harmonic excitation).

So, what they are doing is that they separate the static part (which is the mg) and the dynamic part.

It is easy to find the total force at the end by just adding mg to the total transmitted load (just like your equation).

  • $\begingroup$ Oh, so that is the reason. Got it. Thank you @NMech $\endgroup$ Sep 30 at 5:59
  • $\begingroup$ In many real-world applications, the accelerations caused by the vibration are hundreds or thousands of g, so the weight of the component is negligible. $\endgroup$
    – alephzero
    Sep 30 at 15:25

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