Download Understanding Hertzian Contact: Surface Deformation and Force Distribution and more Exams Chemistry in PDF only on Docsity! Hertzian Contact Theory Authored by: Zachary Dowson Opti 421 – Optomechanics Introduction to Theory: In mechanical design and analysis we can usually get away with assuming that when two round surfaces make a mechanical contact that they are at contact at a single point. This however is untrue because we know surfaces will deform when they are given a force against another surface. Upon trivial inspection we should know that the surfaces will deform and make a larger contact with each other than a single point, as easily seen and understood when pushing a beach ball against the floor for example. Before I will explain applications and go through an example I feel it is appropriate to go through a mathematic derivation to show where this theory credited to Heinrich Hertz of SI unit Hertz fame. Introduction to Mathematical Derivation: Note that in order to understand this derivation some amount of knowledge in differential equations is required. Figure 1 - Hertzian Contact Layout Note that in this model there are few assumptions that are made behind the theory. The first assumption is that R is very long so we can making the point around the contact nearly flat for the length ±a before any forces are applied. The other assumptions are that the contact area ±a is proportional to the amount of force applied. The final assumption is that there is no singularity on x for ±a. d2u0 d x2 = −1 R u0 is function set we are looking for and it comes in form shown below: u0=C0− x2 2 R = C0− a2 cos2φ 4 R − a2 4 R Phi is the angle between our contact area ±a and the radius R. du0 dφ = −a2sin 2φ 2 R Solutions therefore include u1=0,u2= a2 2 R and un=0 (n>2 ) Putting these into a force distribution noted as p p0= −F πaa , p1=0, p2= 2μaa R (κ+1 ) , pn=0 (n>2 ) Which gives us the general addition of homogenous solutions, p (θ )= ( −F πaa + 2μaa R (κ+1 ) cos2θ) sinθ Solving for θ=0 , πa which will give us the singularity ideal cases at ±a F πaa = 2 μaa R (κ+1 ) =¿a=√ FR (κ+1 ) 2πaμa However when we need to look at force distribution, we can plug a back into the p equation p (θ )= −2 F sin θ πaa =¿ p ( x )= −2F √a2−x2 πa a2 If we look back to the diagram approach to this problem we are considering a round object against a flat, however we frequently have two round surfaces making contact. The results from this type of problem can be expressed as below a=√ F R1 R2 2 πa (R1+R2 ) ( κ1+1 μa1 + κ2+1 μa2 ) p ( x )= −2F √a2−x2 πa a2 Uses and Examples: We will often use this type of mathematics and force distributions to describe the loading on bearings in order to predict failure of rolling and moving parts. When there is too much contact, and thus too much forces we will create subsurface fractures. Another facet of use will be when predicting friction between bearings and fixed parts, because we need to know the area of contact since friction is a function of the contact area. This is especially important in creating systems with bearings or for example car tires. Finally in optics this data can be important when looking certain problems when loading lenses into tubes or other types of mounting systems although because of the complicated geometry this might be better analyzed in finite element software.
