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What are the units of mechanical wear? 2

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buoy

Marine/Ocean
Feb 18, 2011
34
I need to calculate the mechanical wear of a machine. I have the velocity and applied force (load), and have been told to approximate Wear as ([Cubic mean load]^3)*Velocity*[Duration of wear], which would give units of (Newtons^3)*meters. Are these the right units for wear? Can these units be rearranged into a more familiar combination? What other physical quantities have units of N^3*m? If you can suggest a basic textbook or reference, that would be helpful.
 
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In my first Applied Mechanics examination, we studied one of the basic hypothesis employed to deal with wear. It's called Reye Hypothesis.
It states that the volume of removed material per unit time is proportional to the power dissipated by friction:

d = k * f * p * w

d = thickness of material removed per unit time
k = coefficient
f = friction coefficient
p = pressure exchanged by the surfaces in contact
w = relative velocity between the surfaces

I don't know the origin of your formula, however I would say that you should obtain something like an energy (N*m) or a volume (m^3).
Maybe the cubic mean load must have the exponent 1/3.

Hope it helps.

Stefano
 
d has dimensions m/sec
w has dimensions m/sec
p has dimensions N/m2
f has no dimensions
k has dimensions m2/N ?

i notice that d is defined as volume removed/sec, and figure you've divided both sides by A area of contact (since p would act over A to give you a normal force, to multiply by f to give you the friction force ...)
 
rb1957,

Yes, the area is eliminated since it is found on both sides of the equation. Obviously the coefficient k must take into account the difference in measuring units between d and fpw.
 
It's like corrosion, can be mass per unit time, perhaps length per unit time. Mass loss is obvious, it relates to the volume through density. The length would be something more along the lines of thickness change over time.

Regards,
Cockroach
 
"...(Newtons^3)*meters''"
Probably meant
M^3/NM

where

N^3 is the volume removed

and

NM is the friction energy that causes the removal
 
I am getting more information on what I have asked to calculate - any confusion is all mine! The wear I am calculating is based on bearing life with non-adhesive contact between ball bearings and the race. Apparently there is an empirical basis that doubling the load reduces the life to 1/8, and the wear is applied over the total distance the bearing moves, which leads to units of [Newtons^3]*[Meters]. I will try to get better oriented and repost once I have the big picture. But I still don't "like" these units since I can't relate them to a familiar physical quantity like friction, energy dissipation, etc.
 
Very interesting topic, it seems that a version of the Archard equation uses a dimensional wear coefficient of the units mm^3/Nm, any ideas on how you would go about obtain a coefficient like this for a material? Thanks.
 
Older Ball bearing L10 life equations carry that load^3 function. It is not really wear related, but a prediction of subsurface fatigue and spalling.

If the lube viscosity is high enough for the rpm and bearing geometry, Elastohydrodynamic (EHD) lubrication and full surface separation is achieved, and there is no wear of the ball and races.

If the equivalent load is low enough, ( FAG says CO/8, SKF publishes fatigue load limits) and lubrication conditions are good then the bearing life can be "infinite".

If your ball and roller bearings are actually wearing, there is likely something wrong with the lubricant or its cleanliness.
Surface distress from wear, or just non EHD conditions can induce surface fatigue at otherwise modest bearing loads, and cause failures when the life >could< be "infinite."

There are some who don't buy the infinite life theory,

but I don't think many deny that at lower loads and with good lube the l10 life calculations are an order of magnitude or more too conservative
 
Wear coefficients are friction based and typically obtained in laboratory settings. Friction is application dependent and nonlinear over lifetime of the machine. I find it the most elusive of theoretical calculations with any degree of certainty. I venture to say impossible, actually.

Regards,
Cockroach
 
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