Wanting to do some impact force experiments

[LT72884]

I have 3d printed some 25mm cubes all with the same parameters. what i want to do is take some different weight dumbells, and drop them all from half a meter onto a cube. My goal is to collect data and see at what impact force the cube fails. For me, failure will be when layers break apart, not a total destruction of the part.

since the weights will be falling, it has KE, and we know that at the top, before the fall, PE=KE, meaning, milliseconds before the weight hits the cube, all the PE is now KE. This allows me to find its velocity right before impact. Since i now have its velocity, i can find KE on impact. All the KE will be transferred into the cube. A transfer of energy is work done

ok, so we all know that work done = force(distance) so force =work done/distance.

Here is my issue, i don't know distance. Usually distance is some sort of deformation or if the weight falls on a nail, its the distance the nail is driven into the wood. So i am not sure what to use as my distance?? should i use a value close to 0, but not 0??

i do not have access to a izod or charpy tester, nor do i have the time to make one.

i want to try it out this way and see what happens. Ill plot the data in excel and see what i can find:)

thanks

 

[tech99]

The impact does not transfer all the energy into the cube because energy is dissipated in heat and sound. We should use the loss of momentum. The formua to use is: Force x time = velocity x mass. I have done this experiment using a pointed cone made of soft clay (Plasticine UK). The stopping distance can then be measured and the duration time of the impact can be found.

 

[jbriggs444]

I am not sure that I agree with the premise that yield stress (in units of pressure) can be accurately assessed by a measurement of average force.

 

[LT72884]

I am not sure that I agree with the premise that yield stress (in units of pressure) can be accurately assessed by a measurement of average force.

I would usually agree with you on your exact argument, however, today i must not haha. I am specifically looking for impact force. Same method as the crumple zone of a falling car, or a person falling of a 4ft high bed. Except, we use 0.75 inches as the deform distance for a human body. I have a reason for this experiment haha

 

[LT72884]

The impact does not transfer all the energy into the cube because energy is dissipated in heat and sound. We should use the loss of momentum. The formua to use is: Force x time = velocity x mass. I have done this experiment using a pointed cone made of soft clay (Plasticine UK). The stopping distance can then be measured and the duration time of the impact can be found.

isnt force*(time) impulse? So if i follow you, i should do F=vm/t and since i know the v based on my equations above, of the weight, the mass of the weight and since i can find time based on v, i will have all i need?

v= sqrt(2gh)

thanks

 

[malawi_glenn]

The impact force will also be spread onto an area. Preassure could be a better idea to investigate.

 

[LT72884]

The impact force will also be spread onto an area. Preassure could be a better idea to investigate.

what kind of idea are you thinking? Once i find the force, i can find the pressure? I like that idea. Main reason, i am 3d printing parts soon that need to take a lot of force and pressure.

thanks

 

[malawi_glenn]

Once i find the force, i can find the pressure?

Then you need to find the contact area, which is not that easy to do I think. Why don't you consider something more static? Find a small bolt which you can fix in position, then apply weight on the bolt until your object cracks

might be hard to balance it, though.

 

[LT72884]

Then you need to find the contact area, which is not that easy to do I think. Why don't you consider something more static? Find a small bolt which you can fix in position, then apply weight on the bolt until your object cracks
View attachment 303903
might be hard to balance it, though.

i thought of this, but its all pinpointed to one spot. Its like walking on snow with your tippy toes, rather than using the entire snowshoe. Thats how my brain is thinking of it. I may be incorrect with my thinking on that analogy though

 

[malawi_glenn]

The impact of the dumbell kan also be on "one spot" depending on how it drops. Dumbells can have many shapes, and if you want to controll the outcome of the result, you have to make sure the area of the impact is the same each time.

 

[LT72884]

The impact of the dumbell kan also be on "one spot" depending on how it drops. Dumbells can have many shapes, and if you want to controll the outcome of the result, you have to make sure the area of the impact is the same each time.

excellent information. I agree.

 

[jrmichler]

I am specifically looking for impact force.

There are several ways to measure impact force.

1) Mount a load cell under the cube, and measure force vs time for the impact. A sample rate of about 10,000 samples per second would be about right depending on the stiffness of the block. It's easy to make a simple load cell if you know how to work with strain gages. The load cell will need its natural frequency to be at least 10 times higher than the duration of impact.

2) Use high speed video to observe the impact. You will need a frame rate at least 1000 frames per second, and possibly faster depending on the hardness of the block. (A) Measure position vs time, and calculate peak acceleration, and calculate the peak force from the mass and peak acceleration. Or (B), assume a constant spring constant (not a good assumption with elastomers), measure the peak deflection, calculate the spring constant from $KE = 0.5kx^2$, and calculate the peak impact force from $F = kx$, where $x$ is the peak deflection.

 

[tech99]

I have done a very similar experiment simulating a vehicle crash using an air track. I use a cone made of soft clay to take the impact. I actually make two identical cones, then subject one to the impact and the second to a static force which I adjust to give the same deformation. The impact deceleration can be found from the stopping distance, then we apply Pt=mv to find the force. We then compare this with the deformation of a second cone with a static load.

 

[hutchphd]

Use high speed video to observe the impact.

I'm a little bit lost. If you have a high speed camera why use anything else? Measure the acceleration profile and your done, right?

cone made of soft clay

Does a cone of soft clay produce a particularly constant acceleration? I understand it will provide inelasticity but what is P in equation? Lost again!

 

[LT72884]

well, here is what i think. for a free falling object, such as a metal dumbbell, F=MA. We also know that V=GT and since we know V and T then G works out to be 9.8, so all i need to do is take the mass of my metal dumbbell and multiply it by G and i will have the force, or in other words the weight of the object because MA is the same as the weight of a free falling body and now I am lost hahaha

working in english units for a moment. If i drop a 10 Lb metal block onto a cube, and it doesn't break. I know it can handle 10Lbs of force, but that's incorrect thinking because having 10Lbs just rest on me is not the same as having 10Lbs fall on me

 

[jbriggs444]

well, here is what i think. for a free falling object, such as a metal dumbbell, F=MA. We also know that V=GT and since we know V and T then G works out to be 9.8, so all i need to do is take the mass of my metal dumbbell and multiply it by G and i will have the force, or in other words the weight of the object because MA is the same as the weight of a free falling body and now I am lost hahaha

working in english units for a moment. If i drop a 10 Lb metal block onto a cube, and it doesn't break. I know it can handle 10Lbs of force, but that's incorrect thinking because having 10Lbs just rest on me is not the same as having 10Lbs fall on me

Correctly observed. The force required to support an object at rest is not the same as the force required to stop it in an impact. Not even close.

Perhaps "force" is not the parameter of interest for an impact. Perhaps neither "peak force" nor "average force" are the right parameters either.

Maybe damage done during an impact is more complicated than that.