Course Work: MSE 480 Materials Dissection
Selection Criteria and Materials Index
In order to minimize wear, while still having a part that will withstand the shear forces that are applied, we must take both constraints into account while coming up with a materials index…
Omega = specific wear rate
C = Constant
Pp = Pressure (perpendicular)
PpMax = Maximum Pressure (perpendicular)
Ka = wear rate constant
H = Hardness
Tau = Shear stress
Fs = Shear force
A = Cross sectional Area = normal area
Fn = Normal force
Omega = C*(Pp/PpMax)*ka*H
Tau = Fs/A
We are looking to minimize the wear rate, thus we are seeking to minimize Omega…
Omega = C*(Pp/PpMax)*ka*H
Since Fs/A cannot exceed Tau (failure), we can write Tau (failure) > Fs/A
We also know Pp = Fn/A
Substituting these into our equation for omega…
Omega = C*((Fn/A)/PpMax)*ka*H
Omega = C*((Fn/(Fs/Tau(failure)))/PpMax)*ka*H
Rearranging…
Omega = ((C*ka)/PpMax)*(Fp/Fs)*(H*Tau(failure))
Thus we see that our materials index is H*Tau(failure). So, we must maximize the hardness, while maximizing the shear strength at failure.
We must limit the hardness however so that the pad is no harder then the rim. If it were, it would cause the rim to wear instead of the pad, which is undesireable, as it is much more difficult to replace a rim then a brake pad. So, we set the upper limit of the hardness to an order of magnitude less than that of a typical rim material, 6000 series aluminum.
Perhaps the most critical criteria to consider is the coefficient of friction between the pad material and aluminum. This value must be high, so that there is sufficient friction between the pad and rim to slow down the rider.
The final constraint is price, this will be important in deciding between which of the final materials is used for what application.
Next: The selection process
this site last updated 10-21-2002 by pmiska@deepthought.org