Jumping beam calculations?
Jumping beam calculations?
(OP)
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?
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





RE: Jumping beam calculations?
RE: Jumping beam calculations?
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
RE: Jumping beam calculations?
TTFN
RE: Jumping beam calculations?
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.
RE: Jumping beam calculations?
you can't have t=0 (instant failure)
The strain energy is released over the time
RE: Jumping beam calculations?
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.
RE: Jumping beam calculations?
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
RE: Jumping beam calculations?
RE: Jumping beam calculations?
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.