Ball Drop Impact test - equations?

[J_chem]

Ball Drop Impact test - equations??

I'm hoping someone can point me in the right direction here. =)

I'm testing a bunch of materials by dropping a steel ball from increasing heights until the materials break. So I have a bunch of data that looks like this:

Material A - broke at 18cm
Material B - broke at 20cm
Material C - broke at 12cm

etc.


How can I turn this data into usful information (i.e. how can I say Material A can take x amount of impact, Material B can take y amount of impact etc..)

Its been awhile since I've taken physics.. at first I was thinking force, but F=ma does not take height into consideration... height or velocity would have to be included...

Can someone please point me in the right direction?

Thanks!

 

[nasu]

You could "convert" into energy (potential energy = mgh = kinetic energy before impact). But I don't think it will be more meaningful than the heights. If the same energy is applied in different conditions (heavier ball, less speed for example) it does not guarantees that the material will break.

The "breaking" force itself depends on the elastic properties of materials so it will be difficult to calculate. During collision the force varies with time and the average force depends on how long the bodies are in contact.

 

[timmay]

You are describing a rough method of determining the toughness of your materials.

Toughness is often defined as the kinetic energy (per unit volume) required to cause failure of a sample. It's often estimated using a simple pendulum test such as a Charpy test, whereby the difference between the kinetic energy of the pendulum striker before impact and after impact is measured. There are a variety of different techniques that also suffice.

So, assuming your samples are all the same dimensions and the ball mass is kept constant throughout testing, the sample that breaks at a greater drop height is the toughest within your set.

As the previous poster said, the relevant equations to consider are:

Potential energy = mass x gravitational acceleration x drop height
Kinetic energy = 1/2 x mass x (velocity at impact)²

You'll be safe assuming that the potential energy of the ball is equal to its kinetic energy at the point of impact.

HTH,

Tim

 

[J_chem]

thanks for the answers.

I was also wondering if impact area matters? Is there a way to find out the impact area of a steel ball?

If a sphere hits a surface and we assume zero elasticity, it would just be a point right?

 

[timmay]

As per my response to your other thread, it is generally determined using Hertzian contact mechanics. Find a good tribology or contact mechanics textbook to explain it.

But, qualitatively speaking:

Take two impacts, both involving objects of mass 5 kg dropped from the same height. One object is a very wide flat plate. The other is a very long thin spike. Bearing in mind that as they both have the same mass and are dropped from the same height, they will both have the same kinetic energy upon impact, which one do you think will be more likely to cause a sample to break at a lower height?

This is why you can use your test to compare the toughness of the materials within those experimental settings. If you were to compare the toughness of a material with a flat plate dropped on it to that of a material with a spike dropped on it, you wouldn't really be making a fair comparison.

 

[J_chem]

Yes, the spike would break the material at a lower height.. I guess I'm trying to find out if there is some constant, or concept that could relate the two. ie. something that would take surface area of impact into consideration.

Thanks for the answers btw. very helpful =)