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Replacing dynamic load with equivalent static load (deceleration profile is known)

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jaras_v

Automotive
Dec 22, 2020
1
Hi All,

Lets say i'm designing an anchor structure, which needs to withstand an impact. Structure caries a point load, and deceleration profile is known.
Lets say deceleration profile is a symmetric trapezium shaped, acceleration starts from 0, increases to 100m/s^2, and reduces back to 0. Each step lasts 20 ms, and a known point mass is attached to structure, say 10kg. Structure can yield, but can't rupture.

I have limited experience in dynamic load case analysis, but my logic would suggest F=m*a. So i would take F = 10*100 (assume structure mass itself not significant), and solve static load case using load F applied in appropriate direction, and keep VonMisses somewhere around the structure material yield range (i know VonMisses can not be used to predict structure rupture point, but that's a different topic).

Is this a reasonable approximation or am i missing something completely? Reason im asking is i was participating in a discussion (on unrelated topic, but to do with measuring shaft torque and sensor robustness against torque impacts), where an engineer from a well established sensor company told us a story about an interesting application of their torque sensors - in summary: their torque sensor should get damaged by plastic deformation of the shaft it is installed on, but there was this particular application where they have seen 4-5x higher torque impacts than actual shaft yield strength, but sensors kept working. His explanation was, that "impacts were too short to cause a permanent elastic deformation"...

Now this is a completely new concept to me, and hence i have started to question my F=m*a equivalency when analysing dynamic loads. Is it possible to have 'not enough time' to yield or rupture a structure even if impact loads are higher than material limits? How short is 'too short' - is 20ms too short?

Thanks
Jaras
 
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If you have the deceleration time (or profile function)....coming up with a impact force is pretty straight forward:

Favg=mv/(deceleration time)

Fmax=2Favg

As far as your question on impact time goes.....20 ms seems short....but I've seen shorter.

Me personally, I've gotten more realistic numbers by using a stiffness estimate of the system. (Do a thread search for Blodgett's method.)


 
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