Impact force from tipped cabinet

[junkforJG]

I'm trying to calculate an impact force from a tipped electronic cabinet.

I realize that are a lot of unknowns, but I'll replace those with some reasonable assumptions and see what my range of results will be.

My goal is to see the range of forces that PWB's inside the cabinet may have experienced.

Thanks,
JunkforJG

 

[IRstuff]

acceleration = change in velocity/change in time

I would assume something on the order of 50 to 100 ms

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[hydtools]

Impulse = change in momentum.
F*t = m*(v2 - v1).

Ted

 

[JStephen]

I've seen a similar problem come up several times through the years. Basically, I'm going to drop Item X on a hard surface, and what's the force generated. Well, you can do the F=ma and all of that, but you can't solve the problem without detailed analysis of deformations in the object and in the surface it's hitting, which is usually impractical, and in many cases, impossible. You can assume the distance that it decelerates in, or assume the time it takes to decelerate, but then again, you could assume the force and save some effort.

One approach- see the diagram below- I'm showing widgets attached to a shelf in the cabinet. Assuming the cabinet tips over, hopefully, you can calculate the velocity when it hits. If you have any way of calculating the spring coefficient for that shelf, and know the mass of the widget, you can work it out from there. If the shelf is "heavy" relative to the widget, you're left assuming some portion of the shelf weight in that mass as well. Still very approximate, all in all, though.

 

[hydtools]

The velocity at impact will be sqrt(2*g*h), where h is the vertical distance through which mass m falls.

Ted

 

[3DDave]

It won't do any good unless one has a good idea what forces the contents can survive.

If you do, you can back out the acceleration and from that the distance over which the acceleration will take place. Then decide if the contents have that much travel.

I've had occasion to make such a calculation and unless the boards can accept 100 Gs or more it may be a problem as the travel will be rather large.

 

[IRstuff]

"If you do, you can back out the acceleration and from that the distance over which the acceleration will take place. "

This requires you to guess at a distance or a time. Alternately, you could use MIL-STD-810's Method 516.6 Table 516.6-II default terminal peak sawtooth pulse of 11-ms duration.

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[3DDave]

You know what velocity it will hit with, so you can directly calculate the time based on the acceptable acceleration.

 

[chicopee]

JunkforJG, double check on the definition of impact before you get too far on F=ma. Normally, an impact is defined as a jerk which is a change in acceleration I= dF/dt=d(ma)/dt.

 

[drawoh]

junkforJG,

Your cabinet will fall to the floor, and it will decelerate to a stop in some distance. You need to make assumptions about that distance. Everything else can be calculated. If you lift something and drop it, your kinetic energy is the height times the weight (not the mass).

--
JHG

 

[hydtools]

Is not F = m(v2-v1)/t sufficient to calculate impact force F? As IRSTUFF referenced, use t=.011secs.

Ted

 

[drawoh]

hydtools,

I don't know what the velocity is. I can very easily work out what the energy is, as I noted above. If I (think I) know the deceleration distance, then I can use work energy to figure out the force. This may be an iterative process, but that is what spreadsheets are for.

--
JHG

 

[hydtools]

Drawoh, the velocity, v2, is sqrt(2*g*h), h being the vertical fall distance. V1 is zero. It is a free fall from rest.

Ted

 

[rb1957]

yeah, but the velocity at time of impact doesn't tell you anything about the force of the impact. That depends on the time interval of the impact, and how the box (and it's contents) responds to the impact (is it rigid ?).

Do you want an accurate number or is a ballpark (0.1sec, 0.01sec) sufficient ??

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

 

[IRstuff]

Finding the deceleration distance or the time are essentially equivalent in complexity, since that requires some knowledge of how the structure reacts to the impact. For an electronic cabinet, there may be some flex that would stretch the impact time.

The only sure way is to do the actual test. You could possibly use mass simulators to minimize the potential damage to the subcomponents. Instrument as much as possible and tip the cabinet and see what you get.

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[GregLocock]

Predicting acceleration due to shocks is possible if you know an amazing amount of detail about the system, or if there is one dominant spring and one dominant mass. Since an accelerometer costs about $5 there is no excuse for not doing the test.

http://www.analog.com/en/products/mems/accelerometers/adxl335.html#product-overview

In fact your phone may well have accelerometers built in and there may be an app you can use to interrogate the accelerometers. Mine does. If yours does, stick your phone to the cabinet shelves with hot glue or modelling clay and push the sucker over.

Cheers

Greg Locock


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[hydtools]

The op was looking for a range of forces. That could be 1 to 100, depending on the impulse time 100 to 1 msecs. The calculation is good enough for that.

Ted

 

[junkforJG]

Thanks guys for the suggestions!

I do realize that the actual impact force will depend on a lot of factors that I can't determine easily and at low cost, such as the response of the cabinet, how far it bounces back up (I don't want to repeat this on a test set that is very expensive). Some of the attachment sheets of metal have bent as a result of the fall, so that would have absorbed some of the energy too. What I'm trying to do is bound the problem to determine maximum impact force and maybe also make some assumptions about the elasticity of the system that would reduce the impact force (by how much? order of magnitude?).

One of the product engineers has provided me the Cg of the cabinet. Would I simply use the diagonal height of Cg from the tipping point as the height from the ground and calculate the final velocity as v2 = sqrt(2*g*h)?

 

[hydtools]

Sure. The diagonal would become the max vertical cg height as the rack rotates about the tipping edge.

Ted

 

[rb1957]

but, again, what does final velocity tell you ? other than saying "ok, now how fast does it decelerate?" 0.01sec, 0.0001sec ??

and is it falling on a flat floor ? (it's been our assumption, but ...) if it falls can contacts in a specific area, away from the CG, the results are different.

and are you trying to answer "why did these things break/bend ?" ... 'cause this wouldn't tell you that.

again, do you want a precise answer or near enough ??

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