Register to reply 
(Need Verifications) Wear Rate for adhesive rubbing of 2 materials 
Share this thread: 
#1
Jul1813, 04:53 AM

P: 19

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). Rearrange 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 


Register to reply 
Related Discussions  
(Need Verifications) Wear Rate for adhesive rubbing of 2 materials  General Engineering  1  
How do you find specific wear rate?  Mechanical Engineering  1  
Experimental Verifications of the UPs?  Quantum Physics  11  
Materials for good wear?  Mechanical Engineering  3  
The rate of cooling of a material in air, or with no barrier between materials  Classical Physics  1 