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# Hertzian contact equations for elliptical, spherical and cylindrical contacts

A theoretical background to the Hertz contact theory can be found here.

## Line Contact (Cylindrical contact)

In case of two cylinders in contact (with radii ), as shown in Fig. 1, the Hertzian radius of contact under applied normal load is given by the following equation:

(1)

where is the length of the cylinders.

**It is important to note here that the reduced elastic modulus is defined as follows:** . This definition is historically used in the field of hydrodynamic lubrication and it is different from the typical contact mechanics definition, where the reduced elastic modulus is given as . Therefore, the given equations may slightly differ from the classical Hertzian equations given in contact mechanics textbooks (but they are equivalent). Equivalent radius is given by the following relation:

(2)

The mean and maximum pressures are given by:

(3)

Corresponding Matlab code for Hertz solution can be found here. The online Hertz contact calculator can be found here. Further details of the contact theory can be found in Contact Mechanics by James Barber.

## Point Contact (Spherical contact)

For the case of two spheres in contact as shown in Fig.3, the Hertzian contact radius is given by the following equation:

(4)

, with is given by the following relation:

(5)

The elastic approach (also know as rigid body approach) is given by the following expression:

(6)

The mean and maximum pressures are given by:

(7)

Corresponding Matlab code for Hertz solution can be found here. The online Hertz contact calculator can be found here. Further details of the contact theory can be found in Contact Mechanics by James Barber.

## Elliptical Point Contact

For the case of two spheres in contact as shown in Fig.4. In this case the Hertzian contact is an ellipse and is described by major () and minor axes of the contact ellipse:

(8)

where .

(9)

The functions in the previous equations are approximated as follows:

(10)

The mean and maximum pressures are given by:

(11)

Load as a function of rigid body approach can be calculated as follows:

(12)

Stiffness of the contact is defined as follows:

(13)

Hence, for an elliptical contact, stiffness can be found from the following expression:

(14)

Corresponding Matlab code for Hertz solution can be found here. The online Hertz contact calculator can be found here. Further details of the contact theory can be found in Contact Mechanics by James Barber.

Here is an online calculator for an elliptical point contact.

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I believe the formula for the b (width of contact) is incorrect in two ways, firstly if b is the whole width it contradicts the diagram of the two cylinders in contact which shows 2b. So if it is the full width it should be b=2*(sqrt((4*F*R)/(pi*L*E’))) with a 4 instead of a 2 in the formula, OR if b is the half width as implied by the diagram it should be b=(sqrt((4*F*R)/(pi*L*E’))

Hi Owen,

Thank you for a comment. So b is the half-width, as shown in Figure 2. The equation for b that you posted in your comment assumes a different definition of the reduced elastic modulus (1/E’=…, while the equations posted in the wiki, assume 2/E’=…, see its definition after equation 1). If you substitute this formula to the equation in the article, you will get your equation.

Hi,

if we assume Steel material ,its ultimate strength is ~500MPa. if we calculate the contact pressure for a two sphere(0.1m) under contact in both elastic & plastic regime , the hertz contact pressure is reaching around 11.66E3 MPa. I have analytical & numerically(using abaqus) validated it. my concern is if the obtained contact pressure is so huge and crossing the ultimate strength of the material ,can i consider it for my design ?

Should Contact pressure or hertz contact stress be LESS than Ultimate stress ?

mat-reference: http://www.matweb.com/search/datasheet.aspx?MatGUID=abc4415b0f8b490387e3c922237098da

I believe, if the Hertzian stress exceeds the ultimate strength of the material in your design, it cannot be a good sign. You will get a lot of plastic deformation and probably a failure in a short term. So it is good to rethink the design.

A bit late to the party but, Hertz’s Contact Stress can be significantly greater than the ultimate stress, and not fail. Imagine that you are at the bottom of the ocean. In that case, you will be under considerable Stress from the water pressure, but you will not likely fail. All principal stress compression. It is similar to Hertz Stress, all three principal stresses are compression. The three principal stresses are not necessarily equal to each other. This sets up some shear internal to the component. So, it is the Shear Stress that is likely to fail your component rather than the compressible stresses. Thoughts anyone?

Hi Kevin! Good point! I think you are right, its the difference in stresses in different directions that cause the failure. And in Hertz stress that’s also the case. You have pressure gradient along the cross section in any direction. Here is how the pressure profile looks like: https://www.tribonet.org/calculators/hertz-pressure-calculator/.

Hi,

Are there any places that the derivation for the elliptical point contact equations are published? I’ve looked through some of the references and couldn’t find the exact equations.

Many thanks.

in case we have a pin within the hole-cylinder in cylinder contact, convex, and concave, respectively, should I put a negative value for R2 since it is a concave shape?

Olga, yes, you are right, as long as the resultant radius is positive!

Hi could you provide me the references for the above used equations. Thanks.