Calculating drop test height from a g-shock value 1

[deangardner]

I have currently had some small parts fail during a vibration test and it's more than likely that the shock part of the test that made it fail.

I've come up with some new ways to protect this part, but testing these methods will be very costly. Therefore to mitigate this risk i want to devise a drop test that will mimic the shock loadings seen during the vibration test.

I know that the max shock during the test is 50g for 10ms and the weight of the object is 0.05 Kg. Now all i need is a height, but the more i look into it the more it starts to look like einsteins style of work!

Has anybody ever come accross this type of rough and ready approach before and derived a simple formula for working out a drop height in meters?

 

[rb1957]

nah, this is more newton's work ...

you want u = ?, given v = 0, a = 50g = 160ft/sec^2, t = 0.01s ... (u-v)/t = a

then h = ? given mgh = 1/2*m*u^2 ... h = u^2/(2*g)

u = 50g*0.01 = 0.5g

h = 0.25*g^2/(2*g) = 0.125*g = 4 ft

 

[desertfox]

Hi rb1957

I thought anything falling under gravity could only reach a maximum acceleration of 9.81m/s^2, I have missed something?


desertfox

 

[IRstuff]

The math may not reflect the actual test parameters. A typical shock test uses a half-sine or sawtooth waveform, which does not correspond to a constant acceleration model.

Additionally, such shocks are highly dependent on the material of the part and the surface being hit.

Without doing the full analysis, you may understress or overstress the part.

TTFN

FAQ731-376

 

[rb1957]

desertfox ... it's the deceleration that's being asked for.

IRstuff ... the OP asked a question and i gave him an answer (to the question) ... it's hard enough gettting the right answer to the question, but getting the question right is a whole different matter !

maybe a way to calibrate the height used is to drop an original part so that it breaks (in some set up), then drop the new part.

 

[desertfox]

Hi rb1957

Ahhhhhhhhhhh I see now, so the acceleration would be a -Ve, sorry just saw the positive figure's anyway I agree with your 4feet answer.

desertfox

 

[deangardner]

Thanks guys. This seems reasonable for what i'm trying to achieve.

Dean

 

[IRstuff]

Dean,

Just be aware that your question as posed may not accurately reflect the actual test conditions.

"50g for 10ms" implies a step function to a constant 50g acceleration for a duration of 10ms.

The equivalent MIL-STD-810 sawtooth has only a peak value of 50g that is ramped at 50g/10ms rate. The amount of equivalent crash energy is less than half that calculated using a square wave acceleration.

Note also that MIL-STD-810 has a 15% tolerance on the peak acceleration, so it could be as high as 75g in an actual test.

TTFN

FAQ731-376

 

[GregLocock]

I think rb1957 has solved a different problem to what the OP wants. The problem with calculating impact acceleratrions is that you need to know the stiffness of the system, even for a linear single degree of freedom system.



Cheers

Greg Locock

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

 

[rb1957]

ok, dean estimated the acceleration and the time interval ... sure, probably not much more than a guess. that's why i suggested he calibrate the test set-up with a known result ... the current component fails under some shock load ... so drop a current part from 4' onto concrete, does it break ? what about 3' ? etc

 

[Dan320]

If the deceleration is given, the drop height is dependent upon the distance of the deceleration. Which is the deformation of the object if the impact surface is very rigid.
Maybe you can devise a device to control deceleration distance, i.e. a pad that is soft compared to the object and calculate drop height to give 50g for a reasonable deceleration distance (compression).
I'm not sure I understand this myself...

Dan

 

[RPstress]

You might be able to make something like a plate simply supported as a beam which you could be pretty sure of the stiffness of (or maybe supported as a plate - you could measure its actual stiffness).

You could then use something based on the formula for dynamic impact factors in Roark to estimate when you'd get 50 g by dropping the component on it. You could make sure that the impact was equal to or greater than 50 g for the required time.

The Roark formula is based on linear behaviour, so the load (and acceleration) are assumed vary linearly with deflection (hence the suggestion of a simple support).

 

[KENAT]

People have mentioned that it's the deceleration at the bottom of the drop that gives you the shock. So fundamentally what you are dropping it onto matters a lot.

Another thing is what orientation it is in when it impacts. Just dropping in free air will not give repeatable results, it will land differently each time. Even how it is released can make a difference.

When doing this kind of test for fuzing systems we used a rig with a quick release mechanism to get repeatable results. If the rig adds mass to the falling body this can be compensated though ideally you want to make it 'light'.

For an object as light as yours, you may be able to manufacture an OK rig from fishing line or similar hung vertically, with pieces of drinking straws or similar to run down the line. I did this for a physics project at school and it worked pretty well.

KENAT,

Have you reminded yourself of faq731-376 recently, or taken a look at posting policies:

http://eng-tips.com/market.cfm?

What is Engineering anyway: faq1088-1484

 

[FeX32]

Not such a simple test. What the others have mentioned above I agree with. Both the stiffness's of the material and of the object being hit need to be taken into account. (you could imagine dropping you cell phone on your bed as opposed to dropping it on a concrete floor).
I suggest starting with basic Newtonian dynamics with known stiffness's. Probably the Roark reference above is more then sufficient.
With this, you could then setup a basic test in which you attach an accelerometer to the object. (or use a vibrometer)
With this you can tell more accurately if what you have done is way off or sufficiently close.....
TTFN

Fe

 

[Sparweb]

I had a similar problem to solve once. I approached it from an "energy" point of view. I first worked out a strain energy in the flexure of the landing gear. I can't remember clearly now, but there is helpful info either in an advisory material on this topic or it's in a textbook like Niu.

When the strain energy absorbed by the gear equals the kinetic energy of the falling object, the object is effectively stopped. You have to simultaneously solve for all of the deflections in the landing gear system. Depending on the landing gear, that can either be easy or extremely hard. I had a helicopter skid and cross-tube system to work with, so equations in Roark worked out well. Assuming that the fuselage is perfectly rigid gets you pretty close.

I had to use a computer program to solve all of the simultaneous equations - not an easy task for hand calculations!

If it is a helicopter, you may find hints in the Maintenance Manual or SRM under "hard landing inspections".


Steven Fahey, CET

 

[KENAT]

A 0.05 Kg chopper would be pretty cool;-)

KENAT,

Have you reminded yourself of faq731-376 recently, or taken a look at posting policies:

http://eng-tips.com/market.cfm?

What is Engineering anyway: faq1088-1484

 

[EkOnkaar]

rb1957 I am still confused about the value of 160 ft/s^2 for "50g". If we are using ft/s^2, g=32, so how does 50g become 160ft/s^2 ?

tia. Ekonkaar.

 

[IRstuff]

160 ft/s^2 would be 5 g's not 50.

TTFN

FAQ731-376

 

[EkOnkaar]

IRstuff : precisely. Just wanted rb1957 to confirm. Thanks.

 

[IRstuff]

Anyway, your best bet is to find different suitable surfaces, like different woods, different metals with different backings, and drop a an object with accelerometer mounted on it and see what level of shock you get as a function of height.

While the 4 ft drop, might give you an adequate shock with a "solid wooden bench top" ala MIL-STD-810, you might get an overshock with a metal surface.

TTFN

FAQ731-376