G-forces

[ginsoakedboy]

A fluid bearing supplier wants us to specify impact loads in terms of g-forces. Does anyone have any idea on how to convert expected impact loads into g-forces?

 

[MiketheEngineer]

That topic has been reviewed to death. What you need is the de-acceleration force. BUT you need to know the distance and time it takes to stop. Therein lies the problem.

If those could be accuratley predicted - the car and plane manufacturers would have little to test...

 

[FeX32]

acceleration force?
what the...
force is force, acceleration is acceleration.
1g=1 gravitational unit of acceleration




Fe

 

[ginsoakedboy]

FeX32,

OF course!

I am supposed to an impact load of (say) 5,000 N into an equivalent g-force. How do I go about doing that?

Mike,

I suspect you are right. I think we may be in for a tug of war on the issue of determining "stopping time".

 

[KENAT]

If you actually have impact loads then as fex32 points out, it's simple.

Force = Mass * Acceleration

Newton worked that out quite some time back and most of us would have learnt it early in highschool I'd hope. I'd also hope you can rearange it to.

Force/Mass = Acceleration

You then need to divide your acceleration by g to get your equivalent g force. Again, pretty basic stuff.

If on the other hand you have impact velocity or drop height etc then as Mike says, it's a bit more tricky.

Posting guidelines faq731-376

http://eng-tips.com/market.cfm?

(probably not aimed specifically at you)
What is Engineering anyway: faq1088-1484​

 

[ginsoakedboy]

Ok.

How would you convert a force of 5,000 N into an equivalent g-force?

 

[zdas04]

Did you read KNEAT's post??????????????? For mass use the fluid density, for force use your 5kN. That leaves acceleration, divide that by gravity and you get a unitless number.

David

 

[Twoballcane]

hmmm are they asking for a response from impact? Like 20g at .005 sec?

Tobalcane
"If you avoid failure, you also avoid success."

 

[drawoh]

htlyst,

What everyone is trying to explain to you is that acceleration and force are not equivalent. You will understand all this better if you purge the term "G-force" from your vocabulary.

A lot of components are said to be able to withstand an acceleration of some number of G's. This reflects some known accident condition that you may subject the component to. Perhaps you will drop your component onto the floor. Perhaps a vehicle accident will generate a known maximum deceleration. Your fluid bearing wants to know this.

Another possibility is that this is not a good description of your accident. Perhaps you will drop a screwdriver on your component, or back a forklift into it. Now, you have an external component, which will have kinetic energy which will have to be dissipated by your fluid bearing. Working out the forces is semi-complicated engineering.

JHG

 

[BobM3]

Since it's a fluid bearing (hydrostatic I'd bet), I'm guessing the bearing manufacturer is interested in how much force is developed at the bearing supports due to radial acceleration of the supported member. If so, a first approximation would be to take the mass of the supported member and multiply it by the radial acceleration of the supported member. That would get you the radial force required from the bearings. That would be a worst case situation that ignores any squeezing out of the oil from the gap.

 

[Twoballcane]

When I think of G-Loads (or G-Forces) it is F=m*#g which is the new force due to acceleration. For example if a car that weighs 1000lbs is accelerated 5gs will hit the wall with a force of (1000lbm*5*32ft/s^2)/(32lbm-ft/lbf-s^2)= 5000lbf. However, this will lead to a static anlysis and will not take in account the dynamics of the wall.
That is were a SRS comes into play.

May be if you can tell us how the shock is created (drop, hammer hit,...etc)



Tobalcane
"If you avoid failure, you also avoid success."

 

[ginsoakedboy]

Wow, I feel so loved ... j/k

I realize now that a statement like "impact load is 2x the operating load" is meaningless without any mention of the "interaction time".

But, some suppliers are satisfied with such a specification while others require this number in terms of "g-force".

I suppose, now the mass required to compute the g-force will be a combination of the mass of fluid being pumped and the rotor. However, simply adding the 2 masses will lead to a very high load (over-conservative). Also, without a prototype I cannot obtain a realistic measure of the impact force expected at the bearings.

Anyone know a way out of this dilemma?

 

[GregLocock]

The casual use of gs when you talk about a loading is that if the static load on the bearing is say 500 lbf, and it sees a maximum dynamic force of 4000 lbf, then we'd often describe that as 8g.

Does that help?

Please stop saying g-force it is driving me mad.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[FeX32]

yea, rather g-spot.
It has more meaning to us


Fe

 

[EspElement]

When I take into consideration of impact loads, i take a stopping time of 1/20th of a second.. so it decels from whatever volocity to zero in 1/20th of a second.. like if u dropped a hammer on a concrete floor the floor would see a load of - if the hammer was to be 5lbs and drops at 16ft/s
(5lb/32.14ft/s^2)(16ft/s/.05s) = 49.75 lbs

 

[ShaggyPE]

When I take into consideration of impact loads, i take a stopping time of 1/20th of a second

That seems awfully generic. What about dropping a 5lb rubber block onto carpet?

Here is a link to another thread on impact... not exactly related to OP, but interesting nonetheless.

http://www.eng-tips.com/viewthread.cfm?qid=218312

-Dustin
Professional Engineer
Certified SolidWorks Professional
Certified COSMOSWorks Designer Specialist
Certified SolidWorks Advanced Sheet Metal Specialist