Impact Force 1

[loilfan]

I am designing a sampling rod that has a chain attached to it as a fail-safe in case the sampling rod's packing fails. My assumptions are that the rod is fully extended when the internal pressure of the vessel/pipe accelerates the rod backwards until the chain reaches the end of the length and absorbs all of the impact of the projectile rod.

Step 1etermine final velocity of rod
v2=SQRT(v1^2+2ad)

Step 2: Determine displacement of chain under loading
I treated the chain as a spring and used k=EA/L as my spring constant.
Kinetic Energy=Spring Energy
δ=√((L/(E*A))*(m*v2^2))

Step 3: Determine Chain Reaction Force
F=EAδ/L

The problem is that the force ends up being independent of length. So a rod that travels 1/4" ends up exerting the same force on the chain that travels 5'. The math seems right, but it doesn't make logical sense. Can anyone explain why?

 

[IRstuff]

The only effects that should be different are gravity and friction, neither of which seem to be in your equations, although the acceleration in Step 1 might include that.

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[loilfan]

IRStuff, I am ignoring gravity because it can be installed horizontally. Friction is factored into the acceleration in Step 1. Just note that the acceleration is the internal pressure * applied area / mass minus friction/drag.

 

[handleman]

What is the "d" in the top equation? Is it not the length you're saying it's independent of?

If you want an alternate way to calculate the kinetic energy of the rod, how about calculating work done on the rod using F dot d? That makes the assumption that your internal pressure is constant (it probably is actually decreasing so you're being conservative), but it allows you to subtract friction easily (assuming it's known) from the force exerted by the pressure.

-handleman, CSWP (The new, easy test)

 

[gruntguru]

Without going through your calculations in detail, it kinda makes sense. The longer travel increases the kinetic enrgy, but the longer chain has a lower spring constant so it will absorb more energy (greater displacement) to reach the same max force.

je suis charlie

 

[loilfan]

Handleman, the d value is the distance that the rod travels. It is the same value as the length of the chain.

Gruntguru, the math shows me that it shouldnt matter but if i look at the extreme case where d=L=0.000001", it doesn't make sense to me.

 

[chicopee]

What I don't see in your analysis is the application of momenta principles involving a force during the arresting period which is the time element between the full length of the chain and the final stop of the projectile rod. In all likelihood the time element will be greater for the longer chain than for the shorter one consequently the arresting force for the longer chain should be less than that of the shorter chain, however, having said that the length of the chains will contribute thru friction to the arresting force and that is the reason for my statement "in all lkelyhood".

 

[3DDave]

The beauty of math is that if the equations are correct and the units work out, that it automatically makes sense.

One factor not accounted for is that the pressure on the rod will also cause the chain to extend, much less when it is a stubby chain and more for a longer one. You also dropped V1 somewhere.

 

[gruntguru]

"Gruntguru, the math shows me that it shouldnt matter but if i look at the extreme case where d=L=0.000001", it doesn't make sense to me."

Why not? With such a short travel there is much less kinetic energy to convert to spring potential energy, however the spring is very short and therefore very stiff so the peak force is still high.

chicopee. No need to consider momentum or time. Energy in (f x s) = energy spring (1/2 k x^2)

Whoops - just noticed an issue. Analysis assumes travel where pv work is done (s) is equal to free length of chain, ie doesn't account for the additional travel while spring is stretching. Won't be a problem for normal "chain" stiffness but for "springy" chain where chain stretch is similar magnitude to free travel, this will introduce some error.

je suis charlie

 

[RolMec]

some ideas based on applied mech.-s:
to step 1) you'd need to treat v1 and 2ad only as vectors, if their directions would differ. Do they?
to step 2) you derive k as if you'd have an uniform cylindrical rod of homogenous material and shape properties. You don't have that, a chain is a rather (applied physics / mathematically ;-) difficult object, a joint creature of straight and highly bent elements, the stew if possible worsened by a welding in one of the two straight elements. Not talking about contact (=elongational) behaviour of two bent beams...
Concluding: the effective "spring length" of the chain will differ from the "physical" length of the chain L and thus from the distance d the rod may travel until the chain cuts in.

Proposal:
If you (could) choose the chain sufficiently strong (high Sf against MBL) to consider it a stiff element, you could apply the laws of impact. Metaphorically, the rod would hit the wall (of the chain).
Regards

 

[loilfan]

v1=0 so my equation in step 1 is v2=SQRT(2ad). I just left v1 in so everyone could recognize the formula easier.

I originally looked at momentum, but since I have no way of determining the time of the impulse force, I have to use energy methods.

I added a "joint efficiency" to the cross-sectional area of my chain to account for the fact that it is a welded chain and not a rod.

gruntguru, the additional error will be 0.6% for the extra spring expansion length.

 

[rb1957]

"a" is the deceleration, "d" is the distance it decelerates over.

the problem is "simply" the initial inertia force, ma, to reduced to zero over some time interval that is undefined by the math. If the thing stops in a 1/1000th of a second, or 1/1E6th, or in 1 second the resulting force changes (by three orders of magnitude).

another day in paradise, or is paradise one day closer ?

 

[gruntguru]

No, d is the acceleration distance - constant acceleration due to pressure in the vessel. Deceleration due to spring tension is increasing linearly (k.x) but you don't have to go there because the area under the curve (energy) is 1/2.k.x^2. Energy in (acceleration phase) = f.d so f.d = 1/2.k.x^2

je suis charlie

 

[rb1957]

ok, so "a" and "d" determine v2, the velocity prior to impact. you still need to determine, separately, the time of the impact event to determine the impact force from the impact energy.

another day in paradise, or is paradise one day closer ?

 

[loilfan]

I don't need to know the time, but it is another way that it can be calculated:
I know δ from step two and v2 from step 1. t=(δ)/(v2) to go from my impact velocity to 0
I can determine my deceleration using a=v2/t and use F=ma to determine my reaction force.

It comes up with the same answer as Step 3 in my original post.

Currently my calculations show that a chain trying to stop a 3/8" diameter projectile being launched from a 2000 psig source requires 41,000lbf to stop it. This is independent of the chain length or mass of the object. A normal 0.311" chain can withstand 1400 lbf. This is ~30x undersized!

I have tried to factor in friction from the packing (because this assumes that the packing fails and lets the rod be projected out), the stopper nut at the end of the 3/8" rod being stripped and drag from the projectile through the air. All of this combined knocks off only 1/4 of the 41,000 lbf.

There must be a better way to justify this design without needing to attach a ship anchor chain to it.

 

[handleman]

The great thing about hydraulics (as opposed to pneumatics) is that the fluid is incompressible. The rod will only accelerate until its velocity is limited by the rate at which the fluid can flow through the orifice of the fitting and into the cylinder. Find your max volumetric flow through the fitting at 2000psi via fluid dynamics, calculate rod velocity from there.

-handleman, CSWP (The new, easy test)

 

[rb1957]

I don't appreciate the details of the design, but from our math, F is not independent of L ...
F = EA/L*d
d = sqrt[L/EA*m*v^2] = sqrt[L/EA]*sqrt(m)*v
so F = sqrt(EA/L)*sqrt(m)*v

or are you also saying v = sqrt(2ad), ie is "d" and "δ" the same ? then ...
v = sqrt(2ad)
d = sqrt(L/EA)*sqrt(m)*v ... sqrt(d) = sqrt(L/EA)*sqrt(m) *sqrt(2a) ... d = (L/EA)*m*2a
F = m*2a*d

but as I understand your situation, a pressure is driving a mass against a spring. so initially the mass is at rest, being accelerated by the pressure and restained by the spring ? so the spring will stretch until the restraining force is equal to the applied fluid pressure. but then your dynamic equations are all wrong ... the final state will have zero velocity ... the mass will accelerate initially (as pressure > spring force) but as the spring force increases and the mass will slowly come to rest when the spring force opposes the fluid force.

another day in paradise, or is paradise one day closer ?

 

[loilfan]

handleman, I considered the hydraulic effects but held back since the rod will be in a variety of fluids so I won't have consistent results. I will try it with a compressible fluid and see what I can calculate.

rb1957, the mass is at rest and accelerated by pressure. Once it goes a distance 'd', it is restrained by a chain which I am treating as a spring. The chain will stretch a distance 'δ' as it absorbs all of the energy of the moving rod. So d does not equal δ. In fact, δ/d=strain. Note that the spring force won't act until the chain gets tight at a distance 'd'.

In F = sqrt(EA/L)*sqrt(m)*v, the L is independent because L = d and v=SQRT(2ad)=SQRT(2aL) which turns the equation into:
F=sqrt(E*A*m*2*a)

 

[3DDave]

I'd ask blah and blah-blah, but without an exact drawing of what is proposed it's a game of 20 questions.

 

[gruntguru]

rb1957. It is not a pressure driving a mass against a spring. It is a pressure accelerating a mass then brought to a sudden stop by a very stiff spring (chain). The work done by the pressure during the final, spring-extension phase is negligible so the analysis only considers the kinetic energy transfer to the spring.

handleman. If the rod extends into the pressure vessel there is no orifice and therefore no terminal velocity.

loilfan. A normal 0.311" chain can withstand 1400 lbf. This is ~30x undersized! But E=1/2kx^2. If the chain safety factor is 10, the energy absorbed in exting the chain to yeild will be 100 times the safe rating. Anything you can do to "soften" the impact (eg deformable or "springy" brackets or connecting members) will dramatically reduce the peak force.

je suis charlie