# Impulse divided by area equals granule strength?

• Jared409
In summary: E8A3C3B&version=1&jumpto=%2Farticle%2F"In summary, the Weibull distribution is a statistical model used to describe the shape of the strength-time curve of brittle materials. It is named after the Finnish engineer Antti Weibull, who developed the model in the early 1900s. The Weibull distribution is used to predict the amount of strength that a brittle material will have after being subjected to a certain amount of stress.""The Weibull distribution is a probability distribution that describes the shape of the strength-time curve of brittle materials. It is named after the Finnish engineer Antti Weibull, who developed the model in
Jared409
TL;DR Summary
I want to crush a single layer of granules from a single material that are 400-600um in size using a flat, circular probe, which will generate a force v time or force v distance curve. The quantity I am trying to measure is the granule (tensile) strength of the material.
Summary: I want to crush a single layer of granules from a single material that are 400-600um in size using a flat, circular probe, which will generate a force v time or force v distance curve. The quantity I am trying to measure is the granule (tensile) strength of the material.

I believe that the key value I am looking for is the area under the curve, the impulse. However, I do not know how to turn this into the granular strength that I am looking for. Thoughts? Is it impulse/area? Impulse/(total N number of granules)? Impulse/average size of granule? Something else? Highest force/average granule size maybe?

Thanks.

I am not familiar with this topic, but, since you are not getting other responses, I thought I'd make a comment.

Impulse seems a strange concept to involve here. Your probe applies pressure to one side of the sample and the container applies approx equal pressure to the opposite side. No impulse because no net force on the sample.
Also the force-time relation depends on how fast you crush. For reasonable variations in this, I can't see that you should get different results.
If you applied a constant force less than the crush strength for an indefinite (long) time, nothing would happen, even though the "impulse" force x time grows indefinitely large.

So much more likely to me, is the work done on the sample. It takes energy to break the particles, so work, force x distance, is an obvious input.

That said, it depends what you want to know. The actual strength of a solid material is, I think, measured in terms of pressure, force per unit area, causing fracture. That is ok when you have a simple block, but I'm not sure it would work for granules. Then I would expect the shape and size of the granules to affect the required pressure.
From one protocol for testing road aggregate that I have seen, the measure is the percent reduction in grain size under a defined load. That would seem to imply that smaller particles (of unspecified, maybe random, shape) can withstand greater pressure than larger ones. (I think you get a mixture of sizes and the smaller particles help distribute the load on larger ones. As with many things it is the stress concentrations which are the points of breakdown.)

It seems to me that there is more to your question than you have stated. Perhaps someone familiar with this area would understand implicitly these other details and so you do not bother to give them here. I would have thought that they would also know why they were using this procedure and how they would use the results. Do you you have a specific formula in mind and just want to understand why it (roughly) works?
For example, you suggest you are using strength = impulse /area = ∫(force x time)dt /area
If there is a constant low feed rate, then distance ∝ time for the movement of the disc. So your formula could really be strength ∝ ∫(force x distance)ds /area = work / area

256bits
This might be worth some investigation for the material science aspect.
https://en.wikipedia.org/wiki/Weibull_modulusfor brittle granules being crushing. Weaker ones fail first, so there ends up a distribution of particles failing.
https://en.wikipedia.org/wiki/Weibull_distribution
The "impulse" would have something to do with ageing of the particles under a constant stress, but you would only know if that is true by determining the distribution of failure in the first place.

Here is what these people say,

## 1. What is impulse divided by area equals granule strength?

Impulse divided by area equals granule strength is a formula used to calculate the strength of a granule based on the amount of force (impulse) applied to a specific area. It is commonly used in materials science and engineering to determine the durability and stability of granular materials.

## 2. How is impulse divided by area related to granule strength?

Impulse divided by area is directly related to granule strength because it represents the amount of force per unit area that a granule can withstand before breaking or deforming. This formula takes into account the size and shape of the granule, as well as the type of force applied.

## 3. Can impulse divided by area be used to compare different types of granules?

Yes, impulse divided by area can be used to compare the strength of different types of granules. However, it is important to note that other factors such as composition, density, and porosity may also affect the overall strength of a granule.

## 4. How is impulse divided by area measured in experiments?

In experiments, impulse divided by area is typically measured using specialized equipment such as a penetrometer or a compression tester. These instruments apply a known amount of force to a specific area of the granule and measure the resulting strength or deformation.

## 5. Are there any limitations to using impulse divided by area to determine granule strength?

Yes, there are some limitations to using impulse divided by area to determine granule strength. This formula may not accurately reflect the behavior of granules under certain conditions, such as extreme temperatures or high pressures. Additionally, it may not account for other factors that can affect granule strength, such as moisture content or chemical interactions.

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