Force required to launch object? 4

[littlebum2002]

This seems like an easy equation, and I'm pretty sure I've seen it before, but I am having trouble finding the force required to launch an object.

Here is the background: We had a rod in tension that failed, sending the rod high into the air. We want to try to restrain the rod in the future, so in case it ever fails again it will be contained, and we want to know the force that launched it so we know how much it neds to be restrained.

It was a 188kg rod, that was launched 2.9m completely vertical in the air. Like I said, it seems this SHOULD be an easy calculation, but the best I have been able to find is the initial velocity (calculated at 7.54 m/s), but not the force required to create it.

Thanks for your help! I am sure I will feel stupid when i see how easy this was.

 

[DesignerGuy16]

I'd worry more about how to solve your fixture so it doesn't launch the rod anymore than trying to "absorb" the failure.

Fix the root cause, don't patch the failure mode. It sounds like whatever was restraining the rod wasn't strong enough. Based on the mass, height, etc you should be able to calculate how much force was exerted past the max it could hold, then add some for a safety factor.

Otherwise you could be in for a "look ma, watch this!" moment with someone getting seriously hurt.

James Spisich
Design Engineer, CSWP

 

[ornerynorsk]

I believe the 400 lb bicycle rider is rather in need of some good physical activity. Sorry, off the original topic, I know. Perhaps fluid dampers?

 

[zekeman]

Swearingen, you are probably on the right track. But when the rod breaks it ideally sets in motion a wave that imparts forces in both directions at the restraining end, as the "spring" alternately compresses and extends eventually decaying to zero.However in this case that constraint either fails or is not present in the opposite direction of launch and thus you get the launch .
If you have a free summer, you can solve this wave equation assuming no damping and zero restraint at the restraint end at a worst case time after the rod break, possibly during the compression phase.
By the way, what happened to the OP? Has he lost interest?

 

[littlebum2002]

First of all, we do plan on fixing the fixture for the future, but for all previous instalations, we want to make sure these rods, if they do fail, don't shoot up and hurt someone or something.

I'll try to explain this situation the best I can without a free body diagram, but if it doesn;t come across clearly, i'll draw one up and scan it.

This is a press. The middle has a large hydraulic cylinder with a diaphram on the end that presses down to extract water from towels and other laundry goods. There is a large 6" thick steel plate on the top and bottom, for the hydraulic cylinder to press against.
In the 4 corners are tie rods. They are threaded at the top and bottom, and tightened with hydraulic nuts to develop enough pretension to overcome the pressing force of the press. Between the 2 plates, around the tie rods, are 4 "sleeves" which absorb the compressive stress, only to relieve it when the press is activated.

In other words, it's a very large bolt in pretension. We cut off the nut, and the tension from the bolt caused an opposing force against the sheared-off face, causing launch.

 

[zekeman]

littlebum2002,
Thanks for clearing up the mystery. Now I see how this rod could launch on any break as designed.
I think you can prevent the launch by capturing the top tiedown nut to the upper plate. Then it would be impossible for the broken rod to be set free, But that capture design would have to be analyzed to make sure it will withstand the large impact force caused by release of the stored energy in the rod.
One way of doing it would have spring capture device designed with a spring constant K such that
1/2F^2/K= maximum energy stored in rod
and the energy stored in the rod is
s^2AL/2E
s= ultimate stress
A coss sectional area
L lentgh of rod
E modulus elasticity

 

[rb1957]

a solution might be not to cut the nut off ! ... ok, i know what you mean, the nut failed in-service and changed the bolt into a projectile.

i think you can equate the final proential energy to the original strain energy, determining the tensile stress in the bolt, to the load in the bolt. You can also work from the torque applied to tighten the bolt (T = P*d/5).

 

[swearingen]

Star for you, zeke, you beat me to it...


If you "heard" it on the internet, it's guilty until proven innocent. - DCS

http://www.eng-tips.com/supportus.cfm

 

[ivymike]

If I correctly understand the layout, you have a concentric sleeve and bolt, with very substantial plates at each end, through which the bolt passes (with clearance). I suspect that compression of the sleeve represents an important fraction of the stored energy.

Why not give the tensile rods some "bulges" at each end, so that they won't fit all the way through the holes in the plate? You could even thread the rod farther, and put a pair of nuts between the plates and inside the tubes, if space allows. That way the rods are captured if the nut breaks off (although the nut will still fly due to lengthening of the decompressing tube).