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

by nomisme
Tags: adhesive wear, material science, wear rate
 P: 6 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?
P: 694
 Quote by nomisme Assumptions: 1) Only Rubber deforms and fiberglass's deformation is neglected.
It is counter-intuitive but you must then consider the situation where grit or glass becomes embedded in the deformable rubber. That will protect the rubber and wear the resin binding the glass. Your assumption magnifies this effect.

A wooden shaft running in an iron bearing will cut it's way through the hard iron by the accumulation of abrasive material in the surface of the softer wood.

Your equations may be applicable to a clean world where there is no dust or grit. If water is present the rubber will not wear, it will simply generate heat because the water film separates the different materials.

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