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


by nomisme
Tags: adhesive, materials, rate, rubbing, verifications, wear
nomisme
nomisme is offline
#1
Jul18-13, 04:53 AM
P: 17
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
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