Equivalent force on impact 9

[prelmes]

I'm trying to calculate the equivalent force on impact. Using the conservation of energy approach - the Physics based text books I've consulted suggest Force (Fe) x displacement (d) = kinetic energy (KE). However, the engineering texts suggest the relationship is:
Fe x d = 2 x KE.

Am I missing something ?

 

[GregLocock]

Engineering text is assuming an elastic collision and is talking about the peak force, the physics book is just restating conservation of energy, assuming a constant force.

Both are useful, neither is accurate, most of the time.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[rb1957]

two critical issues ...

greg mentions one ... elastic collision (like a billiard balls) or inelastic (like bird strike) ... how much energy in absorbed in shredding one (or both) of the participants ?

2nd ... how long does the collision last ? the change in momentum (of either participant) is easy to calculate. this then shows you the impulse of the collision. a very short collision then means a very high contact force.

 

[Cockroach]

Wrong, change in momentum equals impulse.

You can use the energy equations to get the velocity and then you need to consider elastic/inelastic collisons to estimate time of contact. The resulting quotient, change in momentum divided by time, will give you force.

This is a standard dynamics computation, college textbook stuff following static courses.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[rb1957]

yeah 'roach, of course change of momentum = impulse = Force*t. but how long does the collision last ? 0.01 sec ? 0.00001 sec ?? and how much energy is absorbed by the participants ? sure you can make a boatload of assumptions, but the OP needs to clarify the collision nad what he wants to know.

 

[prelmes]

That's exactly it rb1957 ...

I'm trying to calculate the equivalent force on impact for a PCB in an electronic device dropped from a height onto concrete. I'm using the expression: Fe = SQRT(2 x U x k),
where U = kinetic energy and k = the spring rate of the PCB. I did not want to make assumptions wrt elastic and plastic collisions or estimate a time of contact. I've derived a quasi spring rate for the PCB using FEA - applying an arbitrary load and determining the deflection. I then use this value of k in the above expression and calculate an equivalent force - which I subsequently plug back into the FEA simulation to predict the regions of maximum displacement of the PCB during impact. I'm aware of the limitations of this approach.

My original post was simply an attempt to understand where the factor of 2 comes from in the above expression. Despite my attempts to avoid elastic/plastic estimations - it appears, from Greg's post, that I may have assumed an elastic collision anyway.

 

[IRstuff]

It comes, as Greg suggested, from the calculation of the peak force of a spring:

x.max = SQRT(2*U/k)
F.max = k*x.max = SQRT(2*U*k), so yes, it's an elastic collision against an ideal spring.

Even so, is the spring rate of an unpopulated PCB? What about other impact geometries?

TTFN

FAQ731-376

 

[hydtools]

prelmes,
Energy stored in compressing a spring by an applied force F is F*d/2. If you equate that stored energy to the kinetic energy, KE = F*d/2.
Rearranging the equation you get F*d = 2KE.

Ted

 

[CavemanJones]

When i consider a force of a blow i typically use momentum in a deceleration form.

m = W*a/g

If my object has a volocity of 12 in/s. I consider my object decelertating in 1/20th of a second (almost instant depending on ur inpat).. so 12 * 20 = 240in/s^2

So if my load was 2000 lbs.. my resulting load would be

(2000lbs * 240in/s^2)/32.16in/s^2 = 14925 lbs

 

[desertfox]

Hi prelmes

If your treating the PCB as a spring, which is absorbing energy as it hits the concrete, then you are assuming that the on impact that there is no bounce. Therefore kinetic energy is equal to strain energy absorbed in the spring and
therefore:-

1/2 * M *V^2 = k*x^2/(2)

If you transpose to find x, the deflection of the spring and then multiply this by the stiffness k you will get the peak force in the spring. I think if you use the

Force * distance/2 you only get the average force.

Personally I am not sure how accurate your answer will be as I don't think that this is the right approach and that change in momentum equals linear impulse is actually a better one albiet that you have to assume an impact time.

desertfox

 

[rb1957]

you are, IMHO, making alot of assumptions ...

you're calculating the average force, as pointed out above ... but this may be ok for you.

you are assuming an impact time interval, which you could calculate from change in momentum = F*t

FE is a static solution (yes ?) which may (or may not) be accurate for a dynamic load.

btw, U is the kinetic energy of the PCB, yees ?

 

[hydtools]

Desertfox,
k*x = F, force

In your equation k*x^2/2 = F*x/2 F is peak force.

Plot force and distance to compress a spring. The area under the curve, a straight line equation k*x, is the work done or the energy stored. The area of the triangle formed by the equation k*x, distance from 0 to x and force from 0 to F is (1/2)F*k*x = k*x^2/2 the work done or energy stored.

The area of a right triangle is 1/2 the product of the legs.

Ted

 

[desertfox]

Hi hydtools

Yes you are correct the forces are the same assuming that the force P is the force on a spring.
However if the only information you have at the begining is the spring rate of the PCB board and the height it falls through before hitting the concrete, then you have to use
0.5*k*x^2 to enable the calculation to be done, if you try to use F*d/2 you would have two unknowns to begin with.


Regards

desertfox

 

[IRstuff]

0.5*k*x^2 = U, so x.max = SQRT(2*U/k)

F.max = k*x.max = SQRT(2*U*k)

conversely

F.max = 2*U/x.max, but x.max = SQRT(2*U/k), so substitution for x.max results in F.max = SQRT(2*U*k), which avoids solving for x.max

TTFN

FAQ731-376

 

[zekeman]

If you are dropping an electronic enclosure containing the PCB, and you are interested in the impact on the components, you have a different problem depending on the drop attitude to the ground.
If it is a parallel drop (PCB parallel to the ground ) you properly and conservatively treat it as an energy exchange, using the "spring constant" at the component location on the board and the KE of the component; however, if the drop orientation is where the plane of the board is perpendicular to the ground, you have a transverse shock where the character of the impact on the enclosure determines the resultant forces on the component, a vastly different and more complicated problem.

 

[prelmes]

Thanks Zekeman

As you suggest, I'm using the energy exchange approach for the PCB falling parallel to the ground. The way the PCB is fixed to the enclosure it has a different spring rate depending on whether it falls face-up or face-down. Despite the limitations, I felt it would provide a more conservative estimate of impact force than the alternatives.

For the vertical case (PCB perpendicular to ground)I've calculated the acceleration of the board on impact - and used that in the simulation.

 

[desertfox]

Hi prelmes

Have a look at this article:-

http://www.sandv.com/downloads/0702metz.pdf

It also discusses to approaches for this calculation, one based on force and time, the other energy based on the distance travelled after impact, here's a quote based on the latter method:-

The net work done during an impact is equal
to the average force of impact multiplied by the distance traveled Wnet = 1/2 mvfinal^2 - 1/2 mvinitial^2
during impact.
In a drop test application, Wnet = 1/2 mvfinal^2, since the initial velocity (vinitial) is equal to zero. Assuming one could easily measure the impact distance, the average impact force F is calculated
as follows: F= Wnet/d
where d = distance traveled after impact.

 

[Cockroach]

RB1957

I would compute the time of impact simply by noting the deceleration of the body over the depth of penetration upon impact. Noting the material of bodies involved in the collision, depth of penetration are typically a function of density, elasticity, that sort of thing.

Or equally, the force would equal the energy transfer from one body to the other in the collision divided by the depth of penetration.

This is not a technical involved problem.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[IRstuff]

The only issue I have with the article is that that Table 1 and 2 arrive at different conclusions, but in classical mechanics, either approach ought to yield the same answer.

TTFN

FAQ731-376

 

[zekeman]

"The only issue I have with the article is that that Table 1 and 2 arrive at different conclusions, but in classical mechanics, either approach ought to yield the same answer."

The only issue I have with that quote is that it is the ONLY issue.

The paper is seriously flawed and I would hope no reputable journal in engineering would publish it.

As on example,eq 2 is NOT a statement of peak acceleration but only the average of the absolute.

Table 1 data on steel is obviously wrong on steel is obviously wrong.

This is one of the worst examples of printed misinformation that I have ever seen.

As a further note, one should be very careful about papers that do not come from reputable journals, where they undergo some level of scrutiny before publication.