Everything about contact stresses along with solved problems

1. Contact Stresses
Tayyab Khan
Advanced Strength of Materials
Department of Mechanical Engineering
Ataturk University
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2. Contents:
• Introduction
• Examples
• Geometry of Contact Surfaces
• Problem of Determining Contact Surfaces
• Expressions for Principal Stresses
• Notation and Meaning of Terms
• Method of Computing Contact Stresses
• Deflection of Bodies in Point Contact
• Related Example
• Significance of Stresses
• Orthogonal Shear Stress
• Related Example
• Stress for two bodies in line contact (Loads normal to contact area) 2

3. Introduction:
Contact Stresses are caused by pressure
of one solid on another over a limited
area of contact.
Most failures (by excessive elastic deformation, yielding and fracture) of
bodies are associated with stresses and strains in portions of the body far
away from the points of applications of the load (point of contact).
However, in certain cases the stresses on the surface of contact are the
major cause of failure for one or both of the bodies.
Examples:
• Between the teeth of a pair of gears in a mesh
• Between a ball and its race in a bearing
• Between the locomotive wheel and the railroad rail
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4. Examples:
a) b) c)
a) Mechanical Gears
b) Ball Bearing
c) Locomotive Wheel and rail
Importance:
Contact Stresses are often cyclic in nature and are repeated a very
large number of times, resulting in a fatigue failure that starts a
localized fracture (Crack).
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5. Geometryof the Contact Surfaces
Fundamental Assumptions:
Solution of the problem of contact stresses of two bodies in contact is based on
following assumptions:
a) Properties of Material:
Material of each body is homogenous, isotropic and elastic in accordance
with Hook’s Law
b) Shape of surfaces near Point of Contact, before loading:
Two bodies in contact have a common tangent plane to the surfaces at the
point contact. For solution, a relation for distance between corresponding
points on the surfaces near the point of contact is needed. Corresponding
points are points on the surfaces of the bodies and on a line perpendicular
to the common tangent line. Equation that determines the approximate
distance between corresponding points of any two surfaces is given by:
𝑑 = 𝐴𝑥2 + 𝐵𝑦²
Where, x and y are coordinates with respect to x and y axes with origin at
point of contact, and A and B are positive constant (Hertz,1895).
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6. Geometryof the Contact Surfaces(Continued)
C) Contact Surface Shape After Loading:
When the loads P are applied to the bodies, their surfaces deform elastically near
the point of contact so that a small area of contact is formed.
It is assumed that, as this small area of contact forms, points that come into
contact are the points on two surfaces that were originally at equal distance form
the tangent plane.
According to equation 𝑑 = 𝐴𝑥2
+ 𝐵𝑦² such equidistant points on the two
surfaces lie on an ellipse.
Hence the boundary line of area of contact is assumed to be an ellipse whose
equation is:
𝑥²
𝑎²
+
𝑦²
𝑏²
= 1
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8. Problemof DeterminingContact Stresses:
• To solve the problem relating contact stresses, it is important to first
determine the maximum principal (compressive) stress and shear
“contact stresses” on and beneath the contact area between two ideal
elastic bodies having curved surfaces that are pressed together by
external loads.
• The principal stresses on the contact area between two curved
surfaces that are pressed together are greater than at a point beneath
the contact area, whereas the maximum shear stress is usually greater
at a point a small distance beneath the contact surface.
• Principal stresses on a small cube taken from the
contact area are shown in the figure.
• Whereas, maximum shear stress at the point is:
𝜏max =
1
2
𝜎zz − 𝜎yy
where 𝜎zz and 𝜎yy are the maximum and minimum principal stresses. 8

9. Expressionsfor PrincipalStresses:
The expressions for principal stresses at a point which is at a distance z from the
origin, which lies on the surface of contact of two elastic bodies are:
Where,
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10. Notationand Meaningof Terms:
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11. NotationandMeaningofTerms:
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12. Methodof ComputingContact Stresses:
1. Principal Stresses:
First of the values of constants A and B are found from the equations
mentioned earlier. The ratio B/A gives the values of constants k and k’.
Also the value of Δ can be found by following equation. Once the
constants have been found, the values of principal stresses can be
calculated from their formulae.
2. Maximum Principal Stress:
3. Maximum Shear Stress:
4. Maximum Octahedral Shear Stress:
Where, b is the semi-minor axis of the area of contact, and is given by:
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13. Figure 18.10: Stress and deflection coefficients for two bodies in contact
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14. Deflectionof Bodies in PointContact:
An expression for distance 𝛿 = 𝛿1 + 𝛿2 through which two bodies in
contact move toward each other under the action of load P is needed to
be specified.
This deflection is also called as “approach” because it expresses the sum of
the deflections of two bodies as they approach each other.
The relation for approach is as following:
This equation gives approximates values since the elastic strains in the two bodies
away from the contact region are neglected.
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15. Example1:
Figure 18.1
Contact Stresses between two semicircular discs
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16. Solution:
As mentioned earlier, in problems relating contact stresses, first of all we need to
calculate the values of the constants A and B, after that we can use these values to
extract other constants from the table shown above.
Next step is to find out the constant "∆“ and the ratio B/A
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17. Now, the coefficients needed to calculate the stresses can be read from
the figure for the value of B/A = 1.5 as following:
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18. Figure 18.10: Stress and deflection coefficients for two bodies in contact
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21. Significanceof Stresses:
As observed from the examples, the magnitude of the maximum principal stress is
quite large compared to the values of other stresses.
In these problems, the principal stresses at point of maximum values are all
compressive.
As a result, the maximum shear and maximum octahedral shear stresses are always
less than one half of the maximum principal stress (We recall that for a state of
uniaxial stress (one principal stress), the maximum shear stress is one-half the
principal stress.
Utilizable values of maximum shear stress and maximum octahedral shear stress are
not easily determined because in many problems additional factors are involved like
sliding friction, effect of a lubricant and the effect of repeated loads etc. For this
reason it becomes easy to estimate the stress values using maximum principal shear
stress.
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22. Figure 18.10: Stress and deflection coefficients for two bodies in contact
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23. OrthogonalShear stress:
As we noticed, maximum shear stress and maximum octahedral shear stress occur in
the interior of the contacting bodies at points located equidistant form the tangent
plane on a line perpendicular to the tangent plane.
Other shear stress components that are considered to be significant in the fatigue
failure of bearings and other rolling elements (e.g. cylinders) in contact are shear
stresses that occur on planes parallel and perpendicular to the plane tangent to the
contact area.
Among these stresses, 𝜎 𝑥𝑧 and 𝜎 𝑦𝑧 are known as orthogonal stresses as they are
perpendicular (orthogonal) to the plane of contact as shown in the figure below.
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24. OrthogonalShear stress:
Maximum Orthogonal Shear Stress:
Maximum orthogonal shear stress and the distance at which it takes place can be
found from following figures.
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27. Example2:
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29. Solution:
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34. 34

35. Stress for two bodiesin line contact: Loads
normal to contact area
If two cylindrical surfaces are in contact, the contact region is approximately along a
straight line element before loads are applied. Figure below illustrates contact
between two circular cylinders, the line of contact being perpendicular to the paper.
In such a case, the radii 𝑅1’ and 𝑅2’ which lie in a plane perpendicular to the paper
are each infinitively large, hence the values 1/𝑅1’ and 1/𝑅2’ become zero. As a
result, the equation for B and A become:
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36. Evaluationof Stresses:
Principal Stresses: k = 0
Value of b for K = 0 is shown as following, where w = load per unit length of contact
area.
Maximum Principal Stresses:
It can be seen from the above equations that the maximum value of principal
stresses is at z/b = 0. Hence the expressions become:
Maximum Shear Stress:
𝝉max =
𝟏
𝟐
𝝈zz − 𝝈yy
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37. Reference:
ADVANCE MECHANICS OF MATERIALS BY BORESI AND SCHMID
FIFTH EDITION
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