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• To: <seaint(--nospam--at)seaint.org>
• From: "Jake Watson" <jake_watson(--nospam--at)comcast.net>
• Date: Wed, 8 Oct 2003 07:23:13 -0600

```What has been said is correct.  You need to know how your stop is going to
work.  If you can determine that somehow, I would use a kinetic energy
approach.

energy of moving log = energy to "compress" the stop
(1/2)*m*v^2 = 1/2*k*x^2

where: (use consistent units)
m = mass of log
v = velocity of log
k = spring stiffness of log (k/in or similar number. Put a 1 kip load on the
stop, divide by deflection = k.)
x = distance to compress spring (should be your only unknown)

max then Force = k*x

This makes a lot of assumptions. Like it is a linear spring used to stop the
log. The log does not compress, the log hits evenly on all stops, etc.
Also, you can estimate the stiffness of your stop with any reasonable
accuracy.

Now you've done all that, add a healthy factor of safety.  This is an
educated guess after all.....Lastly, don't forget about fatigue.

Jake Watson, P.E.
Salt Lake City, UT

-----Original Message-----
From: Joseph Grill [mailto:j.grill(--nospam--at)comcast.net]
Sent: Wednesday, October 08, 2003 6:25 AM
To: seaint(--nospam--at)seaint.org

Thanks Scott and Steve,

I've gotten that far with this and with other problems of a similar nature.
I was just hoping that from that last guard rail type problem someone has
joined the list with a great "snap your fingers" easy way to go about it.
Guess I'm back to armwave the movement to get an answer.

Thanks
Joe
----- Original Message -----
From: "Scott Maxwell" <smaxwell(--nospam--at)engin.umich.edu>
To: <seaint(--nospam--at)seaint.org>
Sent: Tuesday, October 07, 2003 10:35 PM

> Joseph,
>
> The problem is easy, but not really.  <grin>
>
> If you know how quickly the log stops from 200 ft/s to zero and over what
> distance, then the problem is relatively simple.  You would be able to
> determine the decleration and multiply that by the mass of the log to get
>
> But, the problem is that determining how quickly it stops and over what
> distance is not easy.  Thus, the problem becomes difficult.  As Michael
> pointed out, it will be come a function of the stiffness of the stop (i.e.
> is it really stiff such that it does not compress much and stops the log
> rather quickly or somewhat soft so that the log takes a little longer to
> stop over a longer distance).
>
> Beyond the above, I am not much help.  I could only suggest that you take
> a "reasonable" guess.  If you take a guess as to the distance that it
> takes to stop (i.e. how much the stop compresses), you should be able to
> get an approximate decleration value.  But, to really be accurate, you are
> talking about dealing with the mechanics and dynamics of non-rigid
> bodies...stuff that make my head hurt (and not because my non-rigid head
> would rather be bounced on the more rigid desk than think of such things).
>
> HTH,
>
> Scott
> Ypsilanti, MI
>
>
> On Tue, 7 Oct 2003, Joseph Grill wrote:
>
> > Gentlemen and ladies of this fine list,
> >
> >
> > I know I am taking you California engineers away from the election this
evening, but I need to ask a question that I am not too familiar with.  I
have a situation where a 6200# log is moving along a support driven by
chains.  The log will contact a stop mechanism which is supposed to stop the
log.  The log will be moving at 200 ft/min.  I need to calculate the force
on the stop.  Can you give me any help or send me to a reference that I hope
I have that may explain what I need to do.  I have been looking at a 25 year
old dynamics book from college, but to be honest it isn't helping me very
much.  It is a little late this evening, but I need to get going on it in
the morning.  I can scratch out some numbers which I think will be pretty
conservative, but would like some other input if I can get it.  The stop
mechanism needs to be designed after the force is determined.
> >
> > Thanks,
> > Joseph R. Grill
>
>
>
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