Jumping beam calculations?

[sprintcar]

No, this isn't homework.
Simply supported I-beam, 48" between supports.
One support is a hydraulic jack, the other end rests on the support.
==============
___^_____|______^___

Center of beam is chained to a test specimen.
Beam is 6x6 w20 A36 steel, weight 80lbs.
Applying a 10k lb force to the test part with the jack results in beam stress of 10ksi with total deflection in the center of 0.020"
Question: When the test part snaps, how high will the beam jump?

"If A equals success, then the formula is: A = X + Y + Z, X is work. Y is play. Z is keep your mouth shut."
-- by Albert Einstein

 

[boo1]

What force is on the beam at test part failure and what is the elapse time of failure?

 

[sprintcar]

Assume instantaneous failure - ie Snap.
Force on the beam (by the jack at one end) would be at 10,000lbf

Kindly include any equations - I'm at a loss for this one!
Thanks!


"If A equals success, then the formula is: A = X + Y + Z, X is work. Y is play. Z is keep your mouth shut."
-- by Albert Einstein

 

[IRstuff]

If the beam is modeled like a spring, couldn't you assume that all the stored energy is transferred to the part that broke?

TTFN

 

[JStephen]

Calculate the potential energy in the spring (the beam), figure at worst case, it is 100% converted to kinetic, and that gives you a velocity. From that, you can figure maximum height, assuming uniform motion (not spinning).

In reality, some of that energy might taken by the broken part, and some would go into beam vibration, but that should get you into the ballpark, which is probably about all you can do.

It would probably be quicker to throw a loop of chain over each end of the beam than it would be to calculate if you needed to.

 

[boo1]

KE=PE
you can't have t=0 (instant failure)
The strain energy is released over the time

 

[GregLocock]

True, but in energy methods that doesn't matter. That's why we use them.

I reckon IRstuff/JStephen have the answer.

The truth is that some of the energy would be retained in the beam as vibration, and some would be lost to damping, but as a first estimate that should work.

There's a far more exciting answer you can get if you start thinking about 10000 lbf forces and 80 lb mass beams, but you'll get an F for that.




Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[sprintcar]

I was in favor of just strapping it down, but our chief engineer suggested that the jumping beam might be good to know anyway - and we all know the effect of curiosity amongst engineering types!
Hence the instant release, weightless chain, etc. to simplify and create worst case. In reality it may be a slower, controlled failure.

Greg - I'll take the "F" - I seem to remember a couple of those back in college, mostly in courses NOT related to Mech Eng - and I probably got some exciting answers then too!
...now where did I leave my spring constant equations...

"If A equals success, then the formula is: A = X + Y + Z, X is work. Y is play. Z is keep your mouth shut."
-- by Albert Einstein

 

[yates]

When I was in College, they were called 'jumping beans' not 'jumping beams'

 

[GregLocock]

energy in a spring is 1/2FX

If I can still do imperial units that means it'll jump an inch into the air.




Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.