How much deceleration is reasonable for a free fall impact?

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assafwei
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Hi,

Hope this is the correct forum for this kind of question.

I am doing an explicit finite elements simulation of a tank, weighs around 7.5 Kg with internal parts, falls from 1 meter. Since I never had a validation of such an experiment I am trying to understand if the numbers are right.

I get accelerations (after filtering the data and getting rid of the peaks) of internal parts in the tank of around 200G, there is a lot of deformation in there (plastic tank) and some whiplash movements of the internal parts,and the deformation data, velocity and dynamics of the graphical results look good, but I can't wrap my head around the 200G number... Is this number sane? is it possible to get these high Gs with a 1 meter free fall impact?

Thanks for your input.
 
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You can cross check your FEA results with some simple hand calculations:
1) Assume zero air drag, calculate the velocity at contact.
2) Assume 200 G constant acceleration, calculate the distance to decelerate to zero velocity.
3) Compare that distance to the deflection seen in the FEA results.
4) Do not expect exact agreement, but the two methods should agree within 50% or so.
 
assafwei said:
I can't wrap my head around the 200G number... Is this number sane? is it possible to get these high Gs with a 1 meter free fall impact?
Well, is it landing on mud? Sand? Concrete? Steel? You think that might make a difference? (This is the distance to decelerate mentioned by @jrmichler )

My point is that your question is incompletely formulated and a badly formulated question can make a solid answer impossible.
 
G, or the force of gravity is a constant accelleration of ~10 meters per second per second. When testing more fragile components like computer hard drives, the force you are looking for is the de-acceleration of the object (force needed in the opposite direction to stop the object and time necessary to stop it) . If I drop from one meter, the object picks up little speed but stops in a microsecond on hard surface, if I drop it on a rug or grass or foam, the surface "gives" the distance and time to stop are higher hence the G force over that distance is lower. The total force to stop the object is the same. The maximum force the object sees is the total kinetic energy of the object divided by distance.
 
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assafwei said:
but I can't wrap my head around the 200G number... Is this number sane?
Compare the time it takes to fall, accelerated by 1G, with the time it takes to decelerate to a stop.
That ratio will give you a good estimate for the G of a collision.
Different parts will take a different time to stop depending on where they are relative to the crumple zone.
 
assafwei said:
... but I can't wrap my head around the 200G number... Is this number sane? is it possible to get these high Gs with a 1 meter free fall impact?
Copied from
https://en.m.wikipedia.org/wiki/G-force

“Impact and mechanical shock are usually used to describe a high-kinetic-energy, short-term excitation. A shock pulse is often measured by its peak acceleration in ɡ0·s and the pulse duration. Vibration is a periodic oscillation which can also be measured in ɡ0·s as well as frequency. The dynamics of these phenomena are what distinguish them from the g-forces caused by a relatively longer-term accelerations.

After a free fall from a height
h
followed by deceleration over a distance
d
during an impact, the shock on an object is
{\displaystyle (h/d)}
· ɡ0. For example, a stiff and compact object dropped from 1 m that impacts over a distance of 1 mm is subjected to a 1000 ɡ0 deceleration.”
 
jrmichler said:
You can cross check your FEA results with some simple hand calculations:
1) Assume zero air drag, calculate the velocity at contact.
2) Assume 200 G constant acceleration, calculate the distance to decelerate to zero velocity.
3) Compare that distance to the deflection seen in the FEA results.
4) Do not expect exact agreement, but the two methods should agree within 50% or so.
1. The velocity at contact is the boundary condition for the simulation (to reduce simulation time).
2. Calculating deceleration to zero is problematic since there is a lot of warping and the body isn't rigid, so hand calculations will be very rough, but I will try this and see if I get something logical

Thanks.
 
phinds said:
Well, is it landing on mud? Sand? Concrete? Steel? You think that might make a difference? (This is the distance to decelerate mentioned by @jrmichler )

My point is that your question is incompletely formulated and a badly formulated question can make a solid answer impossible.
Sure, forgot to state - ground is infinitely rigid. Any other information missing?
 
Lnewqban said:
Copied from
https://en.m.wikipedia.org/wiki/G-force

“Impact and mechanical shock are usually used to describe a high-kinetic-energy, short-term excitation. A shock pulse is often measured by its peak acceleration in ɡ0·s and the pulse duration. Vibration is a periodic oscillation which can also be measured in ɡ0·s as well as frequency. The dynamics of these phenomena are what distinguish them from the g-forces caused by a relatively longer-term accelerations.

After a free fall from a height
h
followed by deceleration over a distance
d
during an impact, the shock on an object is
{\displaystyle (h/d)}
· ɡ0. For example, a stiff and compact object dropped from 1 m that impacts over a distance of 1 mm is subjected to a 1000 ɡ0 deceleration.”
Thanks, if this is true for rigid objects, than 200G seems logical in my case.
 
assafwei said:
Sure, forgot to state - ground is infinitely rigid. Any other information missing?
In that case, any capability of the falling object to deform itself, absorbing energy while increasing the deceleration time, will be the only thing avoiding a huge G number.

You are welcome :smile: