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Equivalent force on impact(9)

prelmes (Materials) (OP) 
13 May 10 22:16 
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 ? 

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
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(2) rb1957 (Aerospace) 
14 May 10 7:53 
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. 

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 SturniHueston Engineering Inc Edmonton, Alberta Canada 

rb1957 (Aerospace) 
18 May 10 7:42 
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 (Materials) (OP) 
18 May 10 17:34 
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 (Aerospace) 
18 May 10 18:14 
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
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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 

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


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 (Aerospace) 
19 May 10 7:54 
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 ? 

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 

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 

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
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zekeman (Mechanical) 
20 May 10 9:01 
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 (Materials) (OP) 
20 May 10 18:29 
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 faceup or facedown. 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.


Hi prelmes Have a look at this article: http://www.sandv.com/downloads/0702metz.pdfIt 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. 

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 SturniHueston Engineering Inc Edmonton, Alberta Canada 

IRstuff (Aerospace) 
21 May 10 10:13 
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
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zekeman (Mechanical) 
21 May 10 14:13 
"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.


IRstuff (Aerospace) 
21 May 10 16:35 
The article is a salesoriented document, so everything past the tables is irrelevant to the discussion at hand. Given the level of detail that this thread was taking, there should be no expectation that the article is doing much more than an approximation, in any case. TTFN
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I had a similar issue in trying to determine impact force of a speeding train upon an object placed on the tracks. The assumptions I was forced to make regarding friction, duration of impulse, elasticity, etc. in this complicated system made the resulting answer correct only to ROM at best.
We ended up doing actual impact tests to proof the design, using the plastic deformation from the test on the initial design to allow us to judge relative strength increases in the improved versions, which allowed us to sidestep the need to estimate the actual impact force accurately.
Is it feasible to slap an accelerometer on a prototype and measure the acceleration upon impact directly? I've seen those "G force" stickers that you can just stick on a box to see if it was roughly handled during shipment, which you can purchase and do yourself, although they may not be ideal or practical for this instance. I would think this would give you much more realistic numbers than anything you could calculate. Contact your local testing labs; they may be able to do for you right for a few hundred bucks or so. Heck, you could build your own test setup for future use if you wanted.
Mike 



