# Magnitude of contact normal force

Hi,

I am working on a triaxial compression test using rigid walls under 100 kPa confining pressure, and I am interesting to see the contact normal force distribution. I use i.phys.

1. The maximum normal contact force I got is about 40 N (under 100 kPa confining pressure), while the data I found in other papers usually could be several kN (such as 8 kN). Why there is such a big difference?

2. I guess one possible reason might be the young's module (E) of the material. But I use E = 2e8 Pa which is at the same magnitude of the E used in the papers I refer to, it should not lead to such a difference.

Do you have any ideas of why the forceN I get is so low? Or the result is just like this?

Many thanks!

Leonard

## Question information

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- For:
- Yade Edit question

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- Solved by:
- Leonard

- Solved:
- 2020-07-04

- Last query:
- 2020-07-04

- Last reply:
- 2020-07-02

Jan Stránský (honzik) said : | #1 |

Hello,

you cannot directly just compare young, confinement and force, they have different units.

Size of particles plays important role.

E.g. if you have half particle size, there are 4 times more particles on the same area to transmit the pressure, so the inter-particle forces will be 4 times smaller.

Also the maximum force is IMO not a very good indicator, there could be one "crazy" particle or interaction with much higher force then the rest of simulation (it is dynamic simulation after all), some average or mean or median or some distribution characteristics would IMO make more sense than plain maximum value.

You can easily check confinement, number of interactions and (average?) forces if the values roughly makes sense (and that you e.g. did not make mistake in physical units interpretation).

cheers

Jan

Leonard (z2521899293) said : | #2 |

Hi Jan,

Thanks for your reply.

1. The maximum force is IMO not a very good indicator.

>> Yes, I use maximum force here to say that I really got a small magnitude of ForceN.

2. You can easily check confinement, number of interactions and average forces.

Could you please explain more about how to relate these 3 values?

For my case, the size of sample is 0.07*0.07*0.14 (m), the size of particle is around 6 mm (from 2 to 9 mm), the number of spheres is 7000, the number of contacts is 13219 (at this moment), the average ForceN is around 5 N, the deviatoric stress q at this moment is around 160 kPa, the confinement is 100 kPa.

Thanks!

Leonard

Jan Stránský (honzik) said : | #3 |

> ... while the data I found in other papers usually could be several kN (such as 8 kN)

> Yes, I use maximum force here to say that I really got a small magnitude of ForceN.

you really got small maximum force compared to someone else's maximum force (at least it sounds like that). It is possible that the other results are "crazy" and yours are OK.

Do you use the same particle sizes as the references?

> Could you please explain more about how to relate these 3 values?

(veeeeery) rough and quick estimation for transverse direction:

At = transverse area = 0.07*0.14 = 0.0098 m2

Ft = transverse force = 100kPa * At = 100e3*0.0098 = 980 N

13219 interactions, dimensions 2:1:1 => 3300 per each transverse direction, 6600 per axial direction

V = volume = 0.07*0.07*0.14 = a*a*b = 2*a*a*a

3300 interactions per volume 2*a*a*a

We need estimation of number of interactions per area 2*a*a

2*a*a = (2*a*a*a) ^ (2/3) * 2^(1/3)

3300^(2/

980 N / 280 interactions = 3.5 N

so your average value 5 N makes perfect sense in this light.

cheers

Jan

Leonard (z2521899293) said : | #4 |

Hi Jan,

Do you use the same particle sizes as the references?

>>No, the particle size of the paper I mentioned is larger than mine. (size of sample: 90*90*180 mm, number of particles 1000).

So I guess this is the reason that they have larger ForceN.

2*a*a = (2*a*a*a) ^ (2/3) * 2^(1/3)

>> Why use this formula to transfer from per volume to per area ? Sorry I am slow about this :(

Thanks again!

Leonard

Jan Stránský (honzik) said : | #5 |

> the particle size of the paper I mentioned is larger than mine

> So I guess this is the reason that they have larger ForceN.

how much larger?

>> 2*a*a = (2*a*a*a) ^ (2/3) * 2^(1/3)

> Why use this formula to transfer from per volume to per area ? Sorry I am slow about this :(

This is a rough trick. You know how many interactions there is per volume. But pressure is related to area. So you need to estimate how many interactions you have in some area.

For a cube, volume is a*a*a and face area is a*a, so the volume->area formula is

a*a = (a*a*a)^(2/3)

A = V^(2/3)

You can relate (number of interactions per volume) to (number of interactions per area) with the same formula.

You can use this cube estimation for even more rough estimation, as the results are not too much different.

But you can use actual dimensions, in your case a=0.07, b=0.14

V = a*a*b = 2*a*a*a

A = a*b = 2*a*a

A = someFunction(V)

2*a*a = someFunction(

Turns out that the relation is:

2*a*a = (2*a*a*a) ^ (2/3) * 2^(1/3)

cheers

Jan

Leonard (z2521899293) said : | #6 |

Hi Jan,

Thanks for your reply.

> How much larger?

For my model, the size of particle is around 6 mm (from 2 to 9 mm).

For the paper I mentioned, I didn't find the value of (mean) particle size. The paper gives the information of size in another way, i.e., size of sample: 90*90*180 mm, number of particles 1000. If you are interested, you may find it here[1].

Cheers

Leonard

[1]https:/

Jan Stránský (honzik) said : | #7 |

From the dimensions:

V = 90mm*90mm*180mm = 0.00146 m3

Vs = V*0.6 = 0.000874 m3 (0.6 some estimation of porosity)

V1 = Vs / 1000 = 8.74e-7 m3

r = (V1/(4/

d = 12 mm

So in the other experiment uses roughly 2x larger particles, which (as mentioned above) would result in 4x higher forces (preserving external stress).

Which is still much lower than you "need"..

But for the case of large-large particles, the force could be higher..

Do they use some force measure other than maximum?

cheers

Jan

Leonard (z2521899293) said : | #8 |

Hi Jan,

Thanks for your further explanation.

Do they use some force measure other than maximum?

>No, in the paper[1], it shows the value of forceN by plotting the contact force distribution (force chain), you may find it at Figure 14(a) in [1] if you are interested. I am still curious to see why they have a larger foceN (than we roughly estimated). Maybe some "crazy" particles?

cheers

Leonard

Jan Stránský (honzik) said : | #9 |

Thanks for the reference.

you:

> I use E = 2e8 Pa which is at the same magnitude of the E used in the papers I refer to

paper:

> Normal stiffness [kn (MPa)] 600

should have no effect on forces (assuming applied pressure is same in both cases), just a note..

The estimation is done assuming forces are roughly the same. Which is not the case on figure 14.

From 14a I am not able to read any value, just that they are significantly smaller compared to maximum scale (15 kN).

14d shows two "red" 15 kN forces, than a few "smaller" forces, so let's say the total vertical force could be 40 kN, divided by area 90x90mm gives stress 5 MPa..

But I just very quickly went through the article and don't know what the results actually shows..

Contacting authors is also an option.

cheers

Jan

Leonard (z2521899293) said : | #10 |

Thanks Jan for your comments.

>From 14a I am not able to read any value, just that they are significantly smaller compared to maximum scale (15 kN).

>> Yes, I only read from Fig.14a that there are some pipes of orange color, so I roughly estimated that the ForceN could reach up to 8 kN at the beginning of my quesion.

Thanks for your suggestion.

Cheers

Leonard