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Hertz contact calculation for ball joint.

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The size of the contact and the pressure between the contacting surfaces depends on:

the materials, and here we assume them to be linear, elastic and isotropic;
the geometry, and here we assume them to be either flat (infinite radius) or having a constant radius;
the applied load, and here we assume the load is applied normal to the contact, i.e., there is no friction.

Materials: A single material property - the Elastic Contact Modulus ($E^*$) - combines the elastic properties of both contacting surfaces into an equivalent stiffness:

\\[{1 \\over E^*} = {1 - \\nu_1^2 \\over E_1} + {1 - \\nu_2^2 \\over E_2}\\]

where $E_1$ and $E_2$ are the Young's elastic moduli of the two materials, and $\\nu_1$ and $\\nu_2$ are the corresponding Poisson's ratios.

Geometry: The two simplest forms of contact between curved surfaces are:

2D: cylindrical surfaces aligned along their length, or a cylinder on a flat surface, creating a rectangular contact area;
3D: spherical surfaces, or a sphere on a flat surface, creating a circular contact area.

In either case, an equivalent radius $R$ can be defined. If both surfaces are convex, e.g., two balls touching, then:

\\[{1 \\over R} = {1 \\over R_1} + {1 \\over R_2}\\]

where $R_1$ is the radius of Surface 1 and $R_2$ is the radius of Surface 2. If, however, one surface is concave, e.g., a ball in a cup, then:

\\[{1 \\over R} = {1 \\over R_1} - {1 \\over R_2}\\]

where $R_1$ is the radius of (convex) Surface 1 and $R_2$ is the radius of (concave) Surface 2. (The 'cup' radius must be larger than the 'ball' radius.)

Applied Load: This is a little tricky because the same notation, $P$, is used to mean different things:

In 2D, $P$ is the applied load per unit length of the cylindrical contact. [Units: N/m]
In 3D, $P$ is simply the applied load. [Units: N]

2D Contact: The peak contact pressure, $p_0$, and semi-contact width, $a$, are given by:

\\[p_0 = \\left({P E^* \\over \\pi R}\\right)^{1 \\over 2}\\]

\\[a = \\left({4 P R \\over \\pi E^*}\\right)^{1 \\over 2}\\]

3D Contact: The peak contact pressure, $p_0$, and semi-contact width, $a$, are given by:

\\[p_0 = \\left({6 P {E^*}^2 \\over \\pi^3 R^2}\\right)^{1 \\over 3}\\]

\\[a = \\left({3 P R \\over 4 E^*}\\right)^{1 \\over 3}\\]

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Reference values:

HDPE: Young's modulus, $E = 1$ GPa; Poisson's ratio, $\\nu = 0.45$.
Steel: Young's modulus, $E = 209$ GPa; Poisson's ratio, $\\nu = 0.3$.

A ball joint has a steel ball of diameter $\\var{diaball}$ mm in a HDPE cup of diameter $\\var{diacup}$ mm. The maximum applied load is $\\var{load}$ N.

Calculate:

Because the cup is concave not convex, the radius is effectively negative. The equivalent radius is therefore given by:

${1 \\over R} = {1 \\over R_1} - {1 \\over R_2} = {1 \\over \\var{diaball} \\div 2} - {1 \\over \\var{diacup} \\div 2}$ [units: mm$^{-1}$]

which can be rearranged to give $R = \\var{siground(R*1000,3)}$mm.

The Elastic Contact Modulus, $E^*$, is given by:

${1 \\over E^*} = {1 - \\nu_1^2 \\over E_1} + {1 - \\nu_2^2 \\over E_2} = {1 - 0.45^2 \\over 1 \\times 10^9} + {1 - 0.3^2 \\over 209 \\times 10^9}$

which can be rearranged to give $E^* = \\var{siground(ECM,3)}$GPa.

This is spherical (3D) contact, with the equivalent radius equal to the radius of the sphere. The peak contact pressure, $p_0$, is therefore given by:

$p_0 = \\left({6 P {E^*}^2 \\over \\pi^3 R^2}\\right)^{1 \\over 3} = \\left({6 \\times \\var{load} \\times \\left( \\var{siground(ECM,3)} \\times 10^9 \\right)^2 \\over \\pi^3 \\left( \\var{siground(R*1000,3)} \\times 10^{-3} \\right)^2}\\right)^{1 \\over 3} = \\var{siground(p0,3)}$MPa.

Similarly, the semi-contact width, $a$, is given by:

$a = \\left({3 P R \\over 4 E^*}\\right)^{1 \\over 3} = \\left({3 \\times \\var{load} \\times \\var{siground(R*1000,3)} \\times 10^{-3} \\over 4 \\times \\var{siground(ECM,3)} \\times 10^9} \\right)^{1 \\over 3} = \\var{siground(scw,3)}$mm.

The area of the contact patch is circular with radius equal to $a$, i.e.:

area $= \\pi a^2 = \\pi \\left( \\var{siground(scw,3)} \\times 10^{-3} \\right)^2 = \\var{siground(area,3)}$mm$^2$.

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Applied load. [Units: N]

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Elastic Contact Modulus. [Units: GPa]

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Peak contact pressure. [Units: MPa]

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Diameter of ball. [Units: mm]

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Diameter of cup. [Units: mm]

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Contact area. [Units: mm$^2$]

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Equivalent radius. [Units: m]

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Semi-contact width. [Units: mm]

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Reference values:

HDPE: Young's modulus, $E = 1$ GPa; Poisson's ratio, $\\nu = 0.45$.
Steel: Young's modulus, $E = 209$ GPa; Poisson's ratio, $\\nu = 0.3$.

A ball joint has a steel ball of diameter $\\var{diaball}$ mm in a HDPE cup of diameter $\\var{diacup}$ mm. The maximum applied load is $\\var{load}$ N.

Calculate:

the equivalent radius: $R =$ [[0]] [Units: mm]
the Elastic Contact Modulus: $E^* =$ [[1]] [Units: GPa]
the peak contact pressure: $p_0 =$ [[2]] [Units: MPa]
the semi-contact width: $a =$ [[3]] [Units: mm]
the area of the contact patch: [[4]] [Units: mm$^2$]

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