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’))

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 ?

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,
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.

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,

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.