## Forcing function for rotating equipment foundations

## Forcing function for rotating equipment foundations

(OP)

Hi everyone,

I studied design of foundations for vibrating machines using the textbook "Design of structures and foundations for vibrating machines" by Suresh Arya.

In Example 1 of chapter 6 on page 95 while designing foundation for Reciprocating compressor, they used the following equation for VERTICAL forcing function:

Fo x cos (angularfrequency x time)

And for the horizontal they used:

Fo x sin (angularfrequency x time)

I thought it should be sine for vertical and cosine for horizontal. Please can someone explain to me why they used cos for vertical instead of sine?

Any support will be appreciated.

I studied design of foundations for vibrating machines using the textbook "Design of structures and foundations for vibrating machines" by Suresh Arya.

In Example 1 of chapter 6 on page 95 while designing foundation for Reciprocating compressor, they used the following equation for VERTICAL forcing function:

Fo x cos (angularfrequency x time)

And for the horizontal they used:

Fo x sin (angularfrequency x time)

I thought it should be sine for vertical and cosine for horizontal. Please can someone explain to me why they used cos for vertical instead of sine?

Any support will be appreciated.

## RE: Forcing function for rotating equipment foundations

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

## RE: Forcing function for rotating equipment foundations

But considering that the cross sectional area normal to the direction of the force will affect the stiffness in that direction which will in turn affect the damping ratio and the amplitude check, I have concern that using the horizontal force on the vertical direction may cause a error in the amplitude limit check for a particular direction

## RE: Forcing function for rotating equipment foundations

Force affects the displacement of a foundation with any given stiffness.

Its the same as a beam. Siffess does not change according to the force applied. Any beam of moment of inertia I and span L has a stiffness S. Force applied to that value of stiffness creates a corresponding deflection. Deflection = F/S. I (inertia) does not change with force. L (length) does not change with force. Therefore Stiffness of the beam does not change with force, only its deflection.

When you multiply force x the sin or the cosin, you separate the force's actions into its horizontal and vertical components. Thereafter the horizontal force acts in the horizontal direction only and affects the horizontal deflection only. The vertical force acts in the vertical direction and affects the vertical deflection only. The horizontal force is always in the horizontal direction and the vertical force is always in the vertical direction.

Due to the Sin an cos being a function of time, Fv and Fh vary with time and their values are always out of phase by π/2 If you reverse sin and cos, you still get the same values over the cycle, just with π/2 cycles difference.

OR (maybe you are just asking why sin Ω, not Cos on the picture)

http://mechengdesign.co.uk/PlannedWeb/mech226/rotu...

Notice that they start the displacement calculation from 0° (typical x axis to the right) and then begin rotating counter-clockwise, measuring X displacement in the

verticaldirection. In that frame of reference, the F x sinΩ/S describes that displacement, which they label as "X" displacement, which we often think of as being along the "Y" axis, but there they call it "X" displacement. Just tip the machine axis 90° Right, so Ω =0 is down, then x will be in X and it will make sense to you. But instead of starting at 0° being to the right, we will start with 0° being in the down direction. Arya just used a different frame of reference than you are used to seeing. You would prefer to measure "Y" displacement in the vertical direction, because you are more comfortable with that. You can do it, if you want to. The vibration displacements do not change either way. The problem is essentially the same. Just the machine is tipped over.--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

## RE: Forcing function for rotating equipment foundations

http://hyperphysics.phy-astr.gsu.edu/hbase/tord.ht...

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."