a = ∛(3FN·D/(8E*)); p0 = 3FN/(2πa²)

Hertz contact pressure for sphere-on-flat contact

Under load, the geometric point contact of a sphere on a flat becomes a small circular contact area; even small forces create very high pressures there because area grows only sub-proportionally with force (F^(2/3)).

MINTSI
01

Inputs

Peak pressure at the center of the circular contact area.

Radius of the circular contact area formed under load.

Contact force acting normal to the contact surface.

Diameter of the sphere.

Effective elastic modulus of the material pair, see the linked calculator.

02

Result

Select a target and calculate.

Calculation

a = ∛(3FN·D/(8E*)); p0 = 3FN/(2πa²)

Under load, the geometric point contact of a sphere on a flat becomes a small circular contact area; even small forces create very high pressures there because area grows only sub-proportionally with force (F^(2/3)).

Understand the inputs
  • Maximum Hertz pressure p0Peak pressure at the center of the circular contact area.
  • Contact radius aRadius of the circular contact area formed under load.
  • Normal force FNContact force acting normal to the contact surface.
  • Sphere diameter DDiameter of the sphere.
  • Reduced elastic modulus E*Effective elastic modulus of the material pair, see the linked calculator.
Example

500 N on a 20 mm steel sphere against a flat steel surface (E* ≈ 115.4 GPa) give a contact radius of a ≈ 0.32 mm and a maximum pressure of p0 ≈ 2,344 MPa.

Assumptions and limits

Classical Hertz point-contact theory: purely elastic, smooth surfaces, contact radius small compared with sphere diameter; at very high pressures real plastic deformation sets in, which is not captured here.

Technical article

Understand Hertz contact pressure for sphere-on-flat contact

This calculator finds the contact radius and maximum Hertz contact pressure between a sphere and a flat surface from normal force, sphere diameter and the reduced elastic modulus.

What does this quantity describe?

A sphere touches a flat surface geometrically at a single point. Under load, both deform elastically into a small circular contact area of radius a = ∛(3·FN·D/(8·E*)), with maximum pressure p0 = 3·FN/(2·π·a²) at its center.

Picture a finger pressing on a hard surface: at the fingertip (the "point"), a small but finite flattened area forms. For a hard steel ball, the flattening is tiny but real — and because area grows only sub-proportionally with force (as F^(2/3) rather than F), pressure rises sharply as load increases.

Formula and variables

a = ∛(3FN·D/(8E*)); p0 = 3FN/(2πa²)

  • a = ∛(3·FN·D/(8·E*))
  • p0 = 3·FN/(2·π·a²)
Symbol / inputMeaning
Maximum Hertz pressure p0Peak pressure at the center of the circular contact area.
Contact radius aRadius of the circular contact area formed under load.
Normal force FNContact force acting normal to the contact surface.
Sphere diameter DDiameter of the sphere.
Reduced elastic modulus E*Effective elastic modulus of the material pair, see the linked calculator.

Choose the inputs correctly

Normal force FN, sphere diameter D and the reduced elastic modulus E* set contact radius and maximum pressure.

How to use the calculator

First compute E* with the reduced elastic modulus calculator if not already known. Then enter force and sphere diameter to find contact radius and maximum pressure.

Worked example

500 N on a 20 mm steel sphere against a flat steel surface (E* ≈ 115.4 GPa) give a contact radius of a ≈ 0.32 mm and a maximum pressure of p0 ≈ 2,344 MPa.

Understand the result and units

A pressure of over 2,300 MPa at just 500 N shows how concentrated load becomes at point contact: the same force spread over, say, 1 cm² would give only 5 MPa — the difference of more than two orders of magnitude explains why point contacts (ball bearings, ball joints) are so sensitive to material hardness and surface finish.

Force is given in N, diameter in mm, the reduced elastic modulus in GPa, contact radius in mm, and the resulting pressure in MPa.

Why pressure becomes so high at point contact

Contact radius a grows only with the cube root of force (a ∝ FN^(1/3)), while contact area grows with a², i.e. FN^(2/3). Pressure p0 = 3FN/(2πa²) therefore grows with FN/FN^(2/3) = FN^(1/3) — slower than linear, but starting from a geometric point contact, absolute pressure remains very high because the area itself stays extremely small. This fundamentally distinguishes point contact from a flat bearing surface, where doubling force on the same area exactly doubles pressure.

Typical applications

The calculation is used to design ball bearings, ball joints, ball contacts in precision mechanics and metrology, and to estimate contact loading in ball-on-flat test methods (for example ball indentation hardness testing).

Assumptions, limits and common mistakes

Classical Hertz point-contact theory: purely elastic behavior, ideally smooth surfaces, contact radius small compared with sphere diameter. At very high pressures (often above a few gigapascals), real plastic deformation sets in, which this purely elastic theory no longer correctly represents.

Common mistake: A common mistake is comparing the high Hertz pressure directly against the base material's yield strength without recognizing that Hertzian pressures are inherently very high and common materials are designed for it (for example through hardening); comparison must be made against material-specific limits valid for Hertzian contact loading, not the uniaxial yield strength.

Frequently asked questions

Why is pressure so much higher at point contact than line contact?

Because load at point contact concentrates onto an even smaller, circular rather than strip-shaped area, which additionally grows only with force to the two-thirds power.

What happens when pressure exceeds the material limit?

Local plastic deformation sets in, which this purely elastic model no longer correctly represents; real ball bearings are therefore made from hardened materials that tolerate markedly higher Hertzian pressures than common structural steels.

How does pressure change with a larger sphere diameter?

A larger sphere spreads the same force over a larger contact area, lowering pressure; the relationship is nonlinear because of the cube-root terms involved.

Is this model transferable to sphere-on-sphere contact?

The basic principle yes, but that requires combining the reduced diameter of both spheres analogous to line contact, which this compact sphere-on-flat calculator does not represent.

What role does contact radius a play besides pressure?

It shows how small the actual contact area is, and is needed, for example, to estimate wear area or for further subsurface stress analysis.