Hertzian contact stress calculator
Calculate the contact stress between two curved bodies using Hertz theory. From the contact case, the radii, the force and the material properties E and ν of both bodies you obtain the maximum pressure p_max, the mean pressure, the contact size (contact radius a for point contact, half contact width b for line contact), the approach of the bodies and the maximum subsurface shear stress with its depth. Four standard cases (sphere–sphere, sphere–plane, cylinder–cylinder, cylinder–plane) can be switched to concave per counter body (hollow sphere, bearing shell). An optional check compares p_max with a selectable allowable pressure.
Hertzian contact stress – point and line contact
Model: homogeneous, linearly elastic bodies, frictionless normal contact, half-space assumption (a, b « R). Friction, tangential force, lubricant film and edge effects are not covered; the approach δ for line contact is an approximation for steel/steel. Generally curved bodies with an elliptical contact area are not modelled.
Results
Calculating …
Formulas and fundamentals
Both bodies are elastic and deflect; their compliances add up to the reduced modulus E* via 1/E* = (1 − ν1²)/E1 + (1 − ν2²)/E2. For steel on steel (E = 210,000 N/mm², ν = 0.3) this gives E* = 115,385 N/mm². Three notations of the reduced modulus circulate in the literature and differ by a factor of two; the calculator consistently uses the E* convention after Johnson so the results are unambiguous.
The curvatures of the two bodies add up to an equivalent radius R via 1/R = 1/R1 + 1/R2. If a body touches a plane, the second term vanishes (R = R1). If the counter body is concave, i.e. hollow (socket, bearing shell), it conforms and the radius enters with a negative sign: 1/R = 1/R1 − 1/|R2|. The better the conformity, the larger R, the larger the contact area and the lower the pressure – the design lever in deep-groove ball bearings and joint sockets. For |R2| = R1 the radius R becomes infinite (flat seating) and Hertz theory no longer applies.
For point contact the contact area is a circle of radius a = (3·F·R/(4·E*))^(1/3). The pressure is distributed as a hemisphere with the maximum p_max = 3·F/(2·π·a²) at the centre; the mean pressure is two thirds of that. The approach is δ = a²/R. For line contact the line load F' = F/L is used: the half contact width is b = √(4·F'·R/(π·E*)), the maximum pressure p_max = 2·F'/(π·b) = √(F'·E*/(π·R)), the mean pressure π/4 of that. Rule of thumb: for point contact p_max grows only with the cube root of the force (double force, +26 %), for line contact with the square root (+41 %).
What matters for failure is not the surface pressure but the shear stress inside the material. From the depth profile of the principal stresses, for ν = 0.3, the maximum shear stress is τ_max ≈ 0.31·p_max at a depth z ≈ 0.48·a for point and τ_max ≈ 0.30·p_max at z ≈ 0.79·b for line contact. This depth is the reference for the required case-hardening depth. The static check (onset of yielding below the surface) requires p_max ≤ 1.67·Rp0.2 for line and ≤ 1.61·Rp0.2 for point contact; higher, context-dependent guide values apply for hardened, lubricated rolling pairs.
Worked example
Example sphere on plane: A hardened sphere of diameter 10 mm (R1 = 5 mm) is pressed with F = 100 N against a flat surface, both of steel (E = 210,000 N/mm², ν = 0.3). The reduced modulus is E* = 115,385 N/mm², the equivalent radius R = 5 mm.
The contact radius is a = (3·100·5/(4·115,385))^(1/3) = 0.148 mm. This gives the maximum pressure p_max = 3·100/(2·π·0.148²) = 2176 N/mm², the mean pressure 1451 N/mm² and the approach δ = a²/R = 4.4 µm. The steel/steel control formula p_max = 2176·(F/d²)^(1/3) confirms the value.
Below the surface the maximum shear stress is τ_max = 0.31·2176 = 675 N/mm² at a depth z = 0.48·a = 0.071 mm. Against an allowable pressure of 4200 N/mm² (hardened rolling pair, static) the utilisation is 52 % and the safety factor S = 1.93 – the contact is statically capable.
Frequently asked questions
What is the difference between point and line contact?
For point contact (sphere–sphere, sphere–plane) the bodies initially touch at a point; under load a circular contact area of radius a forms. For line contact (cylinder–cylinder, cylinder–plane) they touch along a line; under load a rectangle of width 2b forms over the effective length L. Line contact is therefore computed with the line load F' = F/L and the pressure rises more slowly with force than for point contact.
How do I enter a concave counter body (hollow sphere, bearing shell)?
Select the concave switch for the second body. The radius R2 is then applied with a negative sign: 1/R = 1/R1 − 1/R2. The concave body conforms, the contact area grows and the pressure drops. The prerequisite is |R2| > R1; for nearly equal radii the area becomes elliptical and the simple theory underestimates it – the calculator warns in this case.
Which modulus and Poisson's ratio should I use?
Separately for both bodies: steel E = 210,000 N/mm² and ν = 0.3, grey cast iron E ≈ 110,000 N/mm² and ν = 0.25, bronze E ≈ 95,000 N/mm² and ν = 0.35, aluminium E ≈ 70,000 N/mm² and ν = 0.33. You can pick a material from the database that fills E and ν, or enter both values freely. ISO 6336 uses 206,000 N/mm² for steel; the difference affects p_max by only about 1.3 percent.
Why is the maximum shear stress more important than the surface pressure?
At the surface there is triaxial compression that does not cause yielding. Failure starts with the shear stress inside the material, whose maximum is about 0.31·p_max (point) or 0.300 to 0.304·p_max (line, depending on source: own numerics 0.3003, Niemann 0.304) at a depth of 0.48·a or 0.78 to 0.79·b. In addition there is the alternating shear stress τ_yz (0.25·p_max at 0.5·b for line, 0.215·p_max at 0.35·a for point contact, after Niemann), which reverses sign as the contact rolls over, lies closer to the surface and is regarded as the actual cause of pitting fatigue. This is exactly where the first cracks (pitting) form under rolling load. The hardening depth must therefore lie well below this zone.
Which allowable pressure is correct?
That depends on the application. For ductile steel under static load the yield onset is at 1.67·Rp0.2 (line) or 1.61·Rp0.2 (point). Hardened rolling pairs withstand 4000 to 4600 N/mm² statically (the level of the static load rating C0 with small permanent deformation), but only about 1500 N/mm² in continuous operation. Worm-wheel bronzes or grey cast iron have their own values. The calculator offers these guide values with context and source; do not mix the contexts.
What are the limits of this calculator?
Hertz theory assumes homogeneous, linearly elastic materials and the half-space: the contact dimensions must be small compared with the radii of curvature (a, b « R). Friction, tangential forces, lubricant film and residual stresses are not covered, nor is the edge stress at roller ends (which is why rollers are crowned). Generally curved bodies with an elliptical contact area (ball in a groove, crossed cylinders) require the elliptical theory with coefficients. For very soft materials such as plastics the values are only indicative because of viscoelasticity.