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

  1. 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
     
  2. jcsd
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