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Contact stresses and Hertz assumptions

Short answer

Contact stresses develop when curved elastic bodies are pressed together, with small contact area and high subsurface shear. Assumptions: homogeneous, isotropic, linear elastic, frictionless, small deformations. For spheres: a = (3FR/4E*)^(1/3), p_max = 3F/(2πa^2). For cylinders: b = √(4FR/(πlE*)), p_max = 2F/(πbl). Maximum shear stress is subsurface, at depth ~0.48a (spheres) or ~0.78b (cylinders).

Write short notes on contact stresses. State the assumptions of Hertz theory and give the expressions for the contact width and the maximum pressure for (a) point contact between two spheres and (b) line contact between two parallel cylinders. Where do the principal stresses attain their maximum values?

Contact stresses

When two curved elastic bodies are pressed together, they touch over a very small area, and high compressive stresses (contact or Hertz stresses) develop at and below the surface. They govern the life of ball and roller bearings, gears, cams, wheels on rails, and rollers. Failure is by pitting, spalling or surface cracking, usually starting below the surface.

Assumptions (Hertz)

  1. The materials are homogeneous, isotropic and linearly elastic.
  2. The contact area is very small compared with the dimensions of the bodies and their radii of curvature.
  3. Surfaces are smooth and frictionless, so only normal pressure acts.
  4. Loads are static and act normal to the contact area.
  5. Deformations are small. In general the contact area is elliptical; it is a circle for two spheres and a narrow strip for long parallel cylinders.

Point contact: two spheres

With radii R1,R2R_1, R_2, load FF, and 1E∗=1−ν12E1+1−ν22E2\dfrac{1}{E^*} = \dfrac{1-\nu_1^2}{E_1} + \dfrac{1-\nu_2^2}{E_2}, 1R=1R1+1R2\dfrac{1}{R} = \dfrac1{R_1} + \dfrac1{R_2}:

a=(3FR4E∗)1/3,pmax=3F2πa2a = \left(\frac{3FR}{4E^*}\right)^{1/3},\qquad p_{max} = \frac{3F}{2\pi a^2}

The contact area is a circle of radius aa, with a hemispherical pressure distribution.

Line contact: two parallel cylinders (length ll)

b=4FRπl E∗,pmax=2Fπb lb = \sqrt{\frac{4FR}{\pi l\,E^*}},\qquad p_{max} = \frac{2F}{\pi b\,l}

The contact is a strip of width 2b2b with a semi-elliptical pressure distribution.

For a cylinder on a plane, R=R1R = R_1.

Stresses

The principal stresses are compressive and are greatest on the load axis, at the surface. They fall off rapidly with depth.

  • Line contact: σz=σy=−pmax\sigma_z = \sigma_y = -p_{max} and, along the cylinder axis, σx=−2νpmax\sigma_x = -2\nu p_{max}.
  • Spheres: the stress normal to the surface is −pmax-p_{max}, and the in-plane radial and tangential stresses are −(1+2ν)2 pmax-\dfrac{(1+2\nu)}{2}\,p_{max} (about −0.8 pmax-0.8\,p_{max} for ν=0.3\nu = 0.3).

The maximum shear stress is not on the surface but at a depth below it: τmax≈0.30 pmax\tau_{max} \approx 0.30\,p_{max} at z≈0.78bz \approx 0.78b for line contact, and ≈0.31 pmax\approx 0.31\,p_{max} at z≈0.48az \approx 0.48a for spheres (ν=0.3\nu = 0.3). This subsurface shear explains why fatigue cracks start below the surface.

✓ Answer checked by Claude Sonnet 5.5

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