(Need Verifications) Wear Rate for adhesive rubbing of 2 materials

[nomisme]

Dear all members,

I just need anyone of you to verify my system and some equations I have worked out for my problem below. Please feel free to point out any flaws in my reasoning and formula applications.

Problem Definition: Two flat blocks (different materials, Fiberglass and Rubber) rubbing against each other on 1 flat plane in 1 direction. Find out when the amount of surface thickness loss of rubber reaches our defined limit, ℝ.

Assumptions:
1) Only Rubber deforms and fiberglass's deformation is neglected.

Logics behind:
1) During the wearing process , strain rate/loading force decreases as the thickness of rubber decreases SO the equation has to be an integral instead. It should be defined in the range from 0 to X(meter). Presumably, when rubber has rubbed against fiberglass for distance x, the amount of thickness loss on rubber is equal to our designated limit, ℝ.

Formula:

Wear formula (I assume it is an adhesive wear?):
w=k*L/H where k is a wear constant of rubber; L is the loading force and H is the hardness of material.
w will be in terms of Volume removed due to wear Per distance traveled by rubber(Contact surface, A) or fiberglass(infinite large Area).
Re-arrange w a little bit, dividing it by Area, A, and it become wt for which the unit is surface thickness loss/ distance traveled)



Equations Work Flow:


Total thickness loss from total distance x traveled equal to ℝ :

(intergral defined on range 0 to x)

∫ wt dx= ∫ (k*L/(H*A) ) dx<= ℝ

where
a) L= σ/A= Eε/A
b) ε=dL / L .....[dL denotes for current compressed thickness which is equal to dL0- ∫ wt dx(amount of thickness loss) WHILE L denotes for the current thickness of the rubber which is equal to L0(original thickness)- ∫ wt dx(amount of thickness loss)]


Define ε as a function of x (strain rate after traveling distance x)
Turns out ε is a function of itself which is a function of x.

Strain rate at distance x can be given by:

ε= (dL0- ∫ wt dx)/ (L0-∫ wt dx)

Becomes

ε= [dL0- ∫(E*k*ε)/(A*H) dx]/ [L0-∫(E*k*ε)/(A*H) dx]

Then we solve ε:
ε becomes a quadratic equation


then find ε in terms of those constants and variable x.

then put ε into the original function below to find x,

∫ wt dx =∫ (k*L/(H*A) ) dx <= ℝ where L=σ/A=(E*ε)/A


um...is that workable...or just plain wrong?
Thanks Report

 

[Achy47]

Thank you for sharing your problem and equations with us. I have reviewed your equations and have a few comments and suggestions for you to consider.

Firstly, I would like to clarify that the wear formula you have used is for adhesive wear, which occurs when two surfaces are in contact and slide against each other. There are other types of wear as well, such as abrasive wear, which occurs when particles or other hard surfaces rub against a material, causing it to wear down.

Secondly, I suggest including the specific values for the constants you have used in your equations. This will help in verifying the accuracy of your calculations and identifying any potential errors.

Thirdly, I noticed a potential issue with your assumption that only rubber deforms and fiberglass's deformation is neglected. While this may be true in some cases, it is possible for both materials to undergo some deformation during the rubbing process. I suggest considering this possibility in your equations.

Additionally, I recommend validating your equations with experimental data or by comparing them with established equations for wear. This will help in verifying the accuracy of your results and identifying any discrepancies.

Overall, your equations seem to have a logical flow and are based on sound principles. However, I would suggest further refining and validating them before using them for any practical applications.

I hope this helps and wish you all the best in your research.



Scientist