ski impact 2

[andran]

hi all

I'm having a bit of trouble calculating impact forces on a 'ski'that I'm designing. the front of the ski is sloped, the slope has a length of 0.4m and a height of 0.2m. i need to calculate the forces on the front of the ski when it impacts a solid object as speeds ranging from 1 - 20 km/h. the ski will be attached to an object with a mass of 2131.53kg.

what i have done is used the conservation of energy. i said that the impact will raise the object 0.2m therefore increasing the potential energy. the kinetic energy will decrease by the same amount. i then calculated the change in horizontal velocity at various initial velocities. with this i calculated the impulse. i then used the average of the initial and final velocities and the 0.4m to calculate the time that the force would be applies. using this time and the impulse, i calculated the force.

i got a constant force for all speeds between 5 - 20 km/h and the impulse decreases with increasing velocity. this doesn't seem correct and i'm not sure where i went wrong.

any advice would be appreciated.

Thanks and regards
Andran

 

[hydroman247]

Purely out of interest, what are you trying to ski that is 2 tons at 20mph?

 

[andran]

it will be fitted to the bottom of an acv.

 

[rb1957]

"force due to impact" ... all impact problems revolve around the time duration of the impact. there are multiple threads on this topic.

the movement of the ski is obviously tied to the suspension of the ski. the shape of the object struck, is there a ramp or a bluff face ? but then that's why the ski-tip is curved up, no?; so that in contact even with a bluff face, the tip of the ski tries to move vertically.

i'd design the ski for all of the ground contact loads, then see how much fwd load the structure (the loadpath from the mass into th eski) can take.

or do a research project, like car impacts.

 

[andran]

hi rb, thanks for the advice. the problem i'm having is that the impact will cause a deflection, not a deformation. all the other threads i've come across involve a deformation.

thanks

 

[rb1957]

"I'm having a bit of trouble calculating impact forces" ... force due to impact has been discussed often. it comes down to how long does the impact take to transfer the momentum, how quickly does the vehicle come to a rest ?

if your ski hits a step, then the tip of the ski contacting the step will see a force, balancing the change in momentum. the worst case (i'd expect) would ben when the ski tip does not deflect upwards, and has an off-set compression load applied.

 

[andran]

okay. i understand all of that.

what i'd like to know is if i went about calculating the impact force correctly. my final answers do not make sense to me.

 

[rb1957]

i think your energy method is more like the vehicle going over a bump, rather than impacting a step.
since the change in PE is independent of speed, then so too the change in KE, then so too the force calculated from the change in momentum.

so, no, i don't think your method is calculating impact forces.

 

[andran]

thank you! that makes a lot of sense.

any idea how i could make the change in potential energy dependent on the initial velocity?

 

[rb1957]

"any idea how i could make the change in potential energy dependent on the initial velocity?" ... no.

if you are considering going over a bump, wouldn't the suspension absorb most of the energy ?

if you're running into a step there will be an impact with the step (even if the vehicle doesn't stop), a change in fwd velocity resulting from an arresting ground force.

 

[IRstuff]

If you read ALL the threads on impact forces, you'll find that the prevailing approach involves assuming some level of spring-like behavior, which would then allow you to determine a timescale, which then allows you to determine an acceleration.

TTFN
faq731-376

 

[zekeman]

"m having a bit of trouble calculating impact forces on a 'ski'that I'm designing. the front of the ski is sloped, the slope has a length of 0.4m and a height of 0.2m. i need to calculate the forces on the front of the ski when it impacts a solid object as speeds ranging from 1 - 20 km/h. the ski will be attached to an object with a mass of 2131.53kg."

If the object you are hitting is a step>.4 you have a horizontal impact problem whose solution depends on the model of the ski to attached mass using energy and momentum conservation equations,
including material energy losses which is very difficult.

If step<.4 you have a different impact problem where the force has a vertical component and again you use the energy/momentum equations with an energy loss model.

But the problem is you must deal with different obstruction shapes which lead to multiple force configurations.

So, in summary, these problems do NOT lend themselves to analytical solutions. Your best bet is to do the empirical stuff using accelerometers to get a realistic answer.

 

[Walterke]

[ignore]

you can quote people like this

[/ignore]

you can quote people like this

just a small fyi

NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000

 

[zekeman]

On the other hand, if you are looking for a worst case problem, then simply model the ski and attached mass connected by a "spring" and maybe break up the ski into a series of springs and masses. and allow the system to hit a wall.
Then, the solution to momentum/ energy equations are could yield some worst case data.

 

[zekeman]

Thank you Walter, I never knew.

 

[prex]

At v[sub]0[/sub]=20 km/h=5.55 m/s the velocity of your mass, assuming it will be able to pass over a step h=0.2 m high, will become
v[sub]1[/sub]=[&radic;](v[sub]0[/sub]-2gh)=5.18 m/s
So, if this is a method to reduce the speed of the AGV, it isn't very effective (not considering shock acceleration effects).
Note that when the tip of the ski is 0.4 m past the step, this doesn't mean the mass has lifted 0.2 m, more likely half of that, depends on the length of the ski and/or the mass. The mass will be at 0.2 m height when the CoG of the mass is at the step start.
Apart from this, your results are not necessarily wrong. In formula you did this:
(L is 0.4 m for you, but possibly wrong as noted above)
impulse I=m(v[sub]0[/sub]-v[sub]1[/sub])
time [&tau;]=2L/(v[sub]0[/sub]+v[sub]1[/sub])
force F=I/[&tau;]
Now rearranging a bit we find:
I=2mgh/(v[sub]0[/sub]+v[sub]1[/sub]) (decreases with increasing speed)
F=mgh/L (independent of velocity)

prex
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[andran]

thanks zekeman. but if i did calculated the force with the ski hitting a wall, there would be no way to determine the time interval.

it would be ideal if i could make the change in potential energy dependent on the initial velocity. there must be another term that i am missing out somewhere.

 

[andran]

thank you prex.

that is exactly where i am. it does not seem realistic though.

i've attached a spreadsheet with some of my calculations. i also assumed half the mass would be raised 0.2m.

thanks

 

[prex]

To make the change in potential energy dependent on the initial velocity you must send the ski over a climb, not a step. The vehicle will stop at changing heights with varying initial velocities.

prex
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[andran]

okay. so i have modified my calculations.

i stopped assuming that the displaced height would remain constant at 0.2m. this assumption meant that there were no solutions for speeds less than 5km/h as there wasn't enough energy.

instead, i calculated the theoretical maximum height that would be reached at each speed. kind of like projectile motion. with this, the potential energy is dependent on the speed.

both the impulse and force increase with increasing speed.

any thoughts?