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Simple Hertz contact without friction.

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A gold ball on a steel table - effect of its own weight.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

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

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

\n

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

\n

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

\n

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.

\n

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

\n\n

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

\n

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

\n

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:

\n

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

\n

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.)

\n

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

\n\n

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

\n

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

\n

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

\n

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

\n

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

\n

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

", "advice": "\n

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

\n

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

\n\n

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

\n

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

\n\n

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

\n\n

$a = \\left({4 P R \\over \\pi E^*}\\right)^{1 \\over 2} = \\left({4 \\times\\left( \\var{load} \\times 10^3 \\div \\var{width} \\times 10^{-3} \\right) \\times \\var{siground(R*1000,3)} \\times 10^{-3} \\over \\pi \\times \\var{siground(ECM,3)} \\times 10^9}\\right)^{1 \\over 2} = \\var{siground(scw,3)}$mm.

\n\n

area = $2 \\times \\var{siground(scw,3)}$mm $\\times \\var{width}$mm $= \\var{siground(area,3)}$mm$^2$.

", "rulesets": {}, "variables": {"scw": {"name": "scw", "group": "Ungrouped variables", "definition": "(4 * P * R / (pi * ECM * 10^9))^(1/2) * 1000", "description": "

Semi-contact width. [Units: mm]

", "templateType": "anything"}, "p0": {"name": "p0", "group": "Ungrouped variables", "definition": "(P * ECM*10^9 / (pi * R))^(1/2) / 10^6", "description": "

Peak contact pressure. [Units: MPa]

", "templateType": "anything"}, "load": {"name": "load", "group": "Ungrouped variables", "definition": "random(4..7#0.5)", "description": "

Applied load. [Units: kN]

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Rail disc diameter. [Units: mm]

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Elastic Contact Modulus. [Units: GPa]

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Track width. [Units: mm]

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

", "templateType": "anything"}, "P": {"name": "P", "group": "Ungrouped variables", "definition": "(load*1000)/(width/1000)", "description": "

Load per unit length. [Units: N/m]

", "templateType": "anything"}, "R": {"name": "R", "group": "Ungrouped variables", "definition": "1/(2000/diawheel+2000/diarail)", "description": "

Equivalent radius. [Units: m]

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Wheel disc diameter. [Units: mm]

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Reference values:

\n\n

In a twin-disc test, a 'wheel' disc (machined from a steel train wheel) with diameter $\\var{diawheel}$ mm runs against a 'rail' disc (machined from a steel railway rail) with diameter $\\var{diarail}$ mm.

\n

The disc surfaces are cylindrical with the axes aligned. The width of the surfaces is $\\var{width}$ mm.

\n

The applied load is $\\var{load}$ kN.

\n

Calculate:

\n\n

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A gold ball on a steel table - effect of its own weight.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

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

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

\n

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

\n

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

\n

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.

\n

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

\n\n

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

\n

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

\n

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:

\n

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

\n

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.)

\n

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

\n\n

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

\n

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

\n

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

\n

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

\n

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

\n

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

", "advice": "\n

weight $= \\rho g \\times {4 \\over 3} \\pi r^3 = 19320 \\times 9.81 \\times {4 \\over 3} \\times \\pi \\times \\left( {\\var{diameter} \\times 10^{-2} \\over 2} \\right)^3 = \\var{siground(weight,3)}$N.

\n\n

${1 \\over E^*} = {1 - 0.4^2 \\over 79 \\times 10^9} + {1 - 0.3^2 \\over 209 \\times 10^9}$

\n

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

\n\n

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

\n\n

$a = \\left({3 P R \\over 4 E^*}\\right)^{1 \\over 3} = \\left({3 \\times \\var{siground(weight,3)} \\times \\var{diameter} \\times 10^{-2} \\div 2 \\over 4 \\times \\var{siground(ECM,3)} \\times 10^9} \\right)^{1 \\over 3} = \\var{siground(scw,3)}$mm.

\n\n

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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Weight of gold ball. [Units: N]

", "templateType": "anything"}, "ECM": {"name": "ECM", "group": "Ungrouped variables", "definition": "1/((1-0.4^2)/79+(1-0.3^2)/209)", "description": "

Elastic Contact Modulus. [Units: GPa]

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

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Peak contact pressure. [Units: MPa]

", "templateType": "anything"}, "scw": {"name": "scw", "group": "Ungrouped variables", "definition": "(3 * weight * R / (4 * ECM * 10^9))^(1/3) * 1000", "description": "

Semi-contact width. [Units: mm]

", "templateType": "anything"}, "diameter": {"name": "diameter", "group": "Ungrouped variables", "definition": "random(6..40)", "description": "

Diameter of gold ball. [Units: cm]

", "templateType": "anything"}, "volume": {"name": "volume", "group": "Ungrouped variables", "definition": "pi*R^3*4/3", "description": "

Volume of gold ball. [Units: m3]

", "templateType": "anything"}, "R": {"name": "R", "group": "Ungrouped variables", "definition": "diameter/200", "description": "

Radius of gold ball. [Units: m]

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Reference values:

\n\n

A gold sphere of diameter $\\var{diameter}$ cm sits on a flat steel table.

\n

Calculate:

\n\n

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Hertz contact calculation for ball joint.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

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

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

\n

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

\n

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

\n

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.

\n

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

\n\n

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

\n

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

\n

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:

\n

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

\n

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.)

\n

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

\n\n

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

\n

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

\n

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

\n

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

\n

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

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

", "advice": "

Reference values:

\n\n

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.

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Calculate:

\n\n

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

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

\n\n

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

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

\n\n

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

\n\n

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

\n\n

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:

\n\n

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.

\n

Calculate:

\n\n

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