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.

 

[dvd]

One approach would be to determine the stored energy (PE=1/2kx^2) in the rod prior to failure and design your restraint to absorb this amount of energy without failing.

 

[chicopee]

F=K*x

 

[zdas04]

Was it launched by pressure or by a mechanical force?

If it was launched by pressure then the force is the magnitude of the pressure (in absolute units) times the area.

If it was a mechanical force then F=ma. The acceleration is the change in velocity from "normal operations" to "broke" applied over the duration of the breaking.

For example the force on a bridge abutment caused by a 3 ton SUV traveling at 70 mph is dependent on how long it took to stop. If it was 5 mS then the acceleration is 34000 ft/s^2 and F=3,800,000 lbf.

David

 

[handleman]

Has anyone there taken a course in Dynamics or Newtonian physics?

-handleman, CSWP (The new, easy test)

 

[TheTick]

You know how much mass the part has, and you know how high it went. From there you can calculate the kinetic energy and initial speed.

This isn't a "force" problem. It is one of energy absorption.

 

[IRstuff]

You don't have enough information to determine the force, anyway. You need either time, or distance. Your initial velocity of 7.5 m/s must have resulted from something that is described by accel*time or 1/2*k*dist^2. Your catastrophic fail probably occurred in a small fraction of a second, so say, 100 ms, would result in 7.7 g's.

You basically need something like a chain or other restraint, similar to those in garage door springs.



TTFN

FAQ731-376

 

[rb1957]

you calculated initial velocity from initial kinetic energy = final potential energy ?

i think you can equate the inital strain energy = stress^2/(2*E) = initial kinetic energy. this'll give you the stess in the part, and so the force acting on it.

 

[littlebum2002]

Well, here's the problem. The first 2 equations involve a spring coefficient and a displacement amount. However, these rods were simply subject to a large amount of tension before they catastrophically failed, releasing all the stored energy into kinetic energy, sending them flying. We don't know how much they displaced before failing, or their spring coefficient, and there's really no way of finding out.

As for F=ma, even though I calculated the initial velocity after fracture using V^2=V0^2+2ay, we have no way of determing the "acceleration" of the initial velocity, as we cannot determine how long the fracture took. 10ms, 100ms? We can't find that out.


I calculated the Kinetic Energy, using .5mv^2 to be 5.3kJ immediately after fracture. However, cable is, to my knowledge, rated by strength, so how could I convert the energy into a measure of force? Since KE is work, i tried using W=Fd, with d being the height of launch=2.9m, but that just gives me 1.83kN, the weight of the rod. That can't be right.

I think IR stuff might have the right idea, we may just have to estimate the time of failure, and use that to determine the acceleration at launch, and therefore the force. Using 7.7g's would mean the force is 14.2kN.

 

[IRstuff]

You can't use W=Fd to determine the initial acceleration, since that's the same equation you used to determine the initial velocity --> m*g*h = 1/2*m*v^2.

My time number is a swag. I originally wrote the post with 10 ms, but the force was 10 times larger, and that looked scary.

However, you should have the original tension applied, yes? and you should have some idea of the way the rod failed, from which you should be able to infer the length of deformation or the propagation time of the failure from physical mechanism.

However, I would tend to agree with others that knowing the force is irrelevant to some degree, since you already know the kinetic energy, which is what the restraint must absorb, without any elastic behavior, otherwise, you'd have a 188 kg object on a rubber band.

TTFN

FAQ731-376

 

[TheTick]

All things are springs. Springs absorb energy as they stretch. If absorbed strain energy results in stress less than tensile stress, then you win.

A longer cable will absorb more energy before breaking than a shorter cable. A cable with some sort of damper will also absorb shock.

Ever been fishing? You can catch big fish on light line, as long as the rod is long enough.

 

[ornerynorsk]

Ballistics calculations should give you theoretical terminal impact force, which would translate into restraint needed.

 

[rb1957]

i think you can determine the initial strain energy (from FEA if you have to, from calcs if you're smart enough).

i am assuming the your rod is a reasonably simple rod, and not some "wacked out" geometry.

if it is a link, for example, constant area A, apply a load P, stress is P/A, strain energy is X ... increase P unitll you have the required strain energy, no?

 

[racookpe1978]

Work a boundary problem:

From above: rb1957
"you calculated initial velocity from initial kinetic energy = final potential energy ?

i think you can equate the inital strain energy = stress^2/(2*E) = initial kinetic energy. this'll give you the stess in the part, and so the force acting on it."

--

OK. So, (at least part of) the question becomes what do you know?

How high is the absolute maximum it could go: Top of room? (Did it hit the ceiling?) Top of building, tank edge? (Did it scratch or penetrate an outdoors structure of some kind?) What was minimum height? How far "over" did it go, and what angle was required to get there? What does that trajectory limit place on the height?

From potential energy = initial kinetic energy (after fracture) = stored potential energy (in the compressed parts) + energy lost in the fracture itself + heat lost (probably very, very little in comparison to the others)

You mentioned tension (stored energy in the "pulling" parts.

There is a "lot" of energy "lost" in the compressed parts that created the "pull" in the single broken part, but were NOT themselves broken. These other parts might not (won't be) be themselves broken - though it is possible/probable that a lot of energy was lost in stripping threads and bending other parts.

Example: An arrow is found penetrating a 12" thick target after a 300 foot flight. Energy from your arm and hand and back muscles was stored in the string and bow, and then was lost in air friction, the rise "up" and then (partially) regained in the fall "down" over the flight, and in the heat and shear and physical displacement of the target material away from itself. But some was also lost in compressing the arrow as it was driving into the target. If you know how high the trajectory, you can eliminate some of the smaller losses, and then make assumptions (from horizontal/vertical velocities) how "long" the reaction might have take.

Or you could break the part again> (Not recommended, but it would be fun.)

 

[GregLocock]

You are talking about something like the mass of a bicyclist moving at bicycle type speeds. Use chickenwire, if the rod is large in diameter, or chickenwire + something else if it is too small.



Cheers

Greg Locock

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

 

[mauricestoker]

If you know what the rod is made of, you should be able to find the modulus of elasticity. From the modulus, you should have a good estimate of stress at failure. If you need a precise number, maybe propose it for an undergrad mechanics of materials or metallurgy project for running tensile tests.

 

[handleman]

Greg, seriously, a 188kg bicycle rider? Not impossible, but I don't think I've ever seen a guy in the 400lb range ride a bicycle. ;-)

Get a fall arrest lanyard. 100 bucks from McMaster-Carr. They are designed to stop the fall of a 400lb guy in 3.5 feet. Should work fine to restrain your rod from flying away in 3.5 ft. or less with no worry about rebound.

-handleman, CSWP (The new, easy test)

 

[zekeman]

I have one small problem with this.
Please explain how a rod in tension fails and is launched.
If the rod is in tension initially , I would think that it is already restrained at one end and a balancing tensile force exists at the other end. I am guessing like the rest of us.
Why don't you end the mystery and tell us exactly how this thing is initially constrained and then where it fails and how it happens to get launched.
If you can do this<, I suspect you will get a very good answer.
You just can't take a bunch of equations and hope for an answer. Only people playing the stock market do this, but that isn't an exact science like this is.
You haven't stated whether springs or an energy storage system is initially holding it or after failure why it gets launched. I can't visualize how a single fracture would set it free.
I'm always amazed at how many wrong answers can be given to a problem that is not understood from the description of the OP.

 

[mauricestoker]

Sounds exactly the same as a tensile testing rig, except the test coupon on the rig is fastened at each end. Highest strss point is normally before plastic deformation, which is how the modulus is experimentally determined. Multiply the modulus, E, times cross-sectional area and you will have a good estimate of force exerted. If going to replace the support, I'd recommend securing both ends against tensile failure. Was the bottom end secured or did it also shoot off-that should have the same tensile force, plus gravity.

 

[swearingen]

zekeman, I was wondering that myself as I went through the thread.

The only thing I could think of is that when it snaps in tension and the large lateral force is imparted (from the moment induced by the propagation of the crack across the separation surface), bending and shock forces snap it at a second location further down the shaft near a lateral hard point (a bearing or guide).

That said, once it is free and it has its initial velocity, TheTick has it right - it becomes an energy problem.

I would recommend that some fabric straps, similar to the arrest lanyards handleman talked about, should be used to restrain it. The straps could easily be tested - you know the energy required - 5348J. That's 188kg over a 2.9m fall. If you used 4 straps and required that one strap take 1/3 of the energy (for safety's sake), you test the strap by dropping a 414lb load tied to the strap a little over 3'. If the strap does not break, you're good to go. It has absorbed the energy...


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

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