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Simple Hertz contact without friction.
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "The size of the contact and the pressure between the contacting surfaces depends on:
\n\n- the materials, and here we assume them to be linear, elastic and isotropic;
\n- the geometry, and here we assume them to be either flat (infinite radius) or having a constant radius;
\n- the applied load, and here we assume the load is applied normal to the contact, i.e., there is no friction.
\n
\n\nMaterials: 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}\\]
\nwhere $E_1$ 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.
\nGeometry: The two simplest forms of contact between curved surfaces are:
\n\n- 2D: cylindrical surfaces aligned along their length, or a cylinder on a flat surface, creating a rectangular contact area;
\n- 3D: spherical surfaces, or a sphere on a flat surface, creating a circular contact area.
\n
\nIn 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}\\]
\nwhere $R_1$ 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}\\]
\nwhere $R_1$ 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.)
\nApplied Load: This is a little tricky because the same notation, $P$, is used to mean different things:
\n\n- In 2D, $P$ is the applied load per unit length of the cylindrical contact. [Units: N/m]
\n- In 3D, $P$ is simply the applied load. [Units: N]
\n
\n2D 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}\\]
\n3D 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- The equivalent radius, $R$, is given by:
\n
\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$^{-1}$]
\nwhich can be rearranged to give $R = \\var{siground(R*1000,3)}$mm.
\n\n- The Elastic Contact Modulus, $E^*$, is given by:
\n
\n${1 \\over E^*} = {1 - \\nu_1^2 \\over E_1} + {1 - \\nu_2^2 \\over E_2} ={1 - 0.3^2 \\over 209 \\times 10^9} + {1 - 0.3^2 \\over 209 \\times 10^9}$
\nwhich can be rearranged to give $E^* = \\var{siground(ECM,3)}$GPa.
\n\n- This is line (2D) contact (between parallel cylinders), so $P$ is the load per unit length, i.e., $\\var{load}$kN $\\div \\var{width}$mm, and the peak contact pressure, $p_0$, is given by:
\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- The semi-contact width, $a$, is likewise given by:
\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- The area of the contact patch is rectangular, and is thus the width of the contact (along the cylindrical axis) times twice the semi-contact width, $a$, from above:
\n
\narea = $2 \\times \\var{siground(scw,3)}$mm $\\times \\var{width}$mm $= \\var{siground(area,3)}$mm$^2$.
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\n\n- Steel: Density, $\\rho = 8\\ 050$ kg/m$^3$; Young's modulus, $E = 209$ GPa; Poisson's ratio, $\\nu = 0.3$.
\n
\nIn 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.
\nThe disc surfaces are cylindrical with the axes aligned. The width of the surfaces is $\\var{width}$ mm.
\nThe applied load is $\\var{load}$ kN.
\nCalculate:
\n\n- the equivalent radius: [[0]] [Units: mm]
\n- the Elastic Contact Modulus: $E^* =$ [[1]] [Units: GPa]
\n- the peak contact pressure: $p_0 =$ [[2]] [Units: MPa]
\n- the semi-contact width: $a =$ [[3]] [Units: mm]
\n- the area of the contact patch: [[4]] [Units: mm$^2$]
\n
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "The size of the contact and the pressure between the contacting surfaces depends on:
\n\n- the materials, and here we assume them to be linear, elastic and isotropic;
\n- the geometry, and here we assume them to be either flat (infinite radius) or having a constant radius;
\n- the applied load, and here we assume the load is applied normal to the contact, i.e., there is no friction.
\n
\n\nMaterials: 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}\\]
\nwhere $E_1$ 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.
\nGeometry: The two simplest forms of contact between curved surfaces are:
\n\n- 2D: cylindrical surfaces aligned along their length, or a cylinder on a flat surface, creating a rectangular contact area;
\n- 3D: spherical surfaces, or a sphere on a flat surface, creating a circular contact area.
\n
\nIn 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}\\]
\nwhere $R_1$ 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}\\]
\nwhere $R_1$ 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.)
\nApplied Load: This is a little tricky because the same notation, $P$, is used to mean different things:
\n\n- In 2D, $P$ is the applied load per unit length of the cylindrical contact. [Units: N/m]
\n- In 3D, $P$ is simply the applied load. [Units: N]
\n
\n2D 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}\\]
\n3D 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- The weight of the sphere is the volume times the density times the acceleration from gravity, i.e.:
\n
\nweight $= \\rho g \\times {4 \\over 3} \\pi r^3 = 19320 \\times 9.81 \\times {4 \\over 3} \\times \\pi \\times \\left( {\\var{diameter} \\times 10^{-2} \\over 2} \\right)^3 = \\var{siground(weight,3)}$N.
\n\n- The Elastic Contact Modulus, $E^*$, is given by:
\n
\n${1 \\over E^*} = {1 - 0.4^2 \\over 79 \\times 10^9} + {1 - 0.3^2 \\over 209 \\times 10^9}$
\nwhich can be rearranged to give $E^* = \\var{siground(ECM,3)}$GPa.
\n\n- 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:
\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- Similarly, the semi-contact width, $a$, is given by:
\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- The area of the contact patch is circular with radius equal to $a$, i.e.:
\n
\narea $= \\pi a^2 = \\pi \\left( \\var{siground(scw,3)} \\times 10^{-3} \\right)^2 = \\var{siground(area,3)}$mm$^2$.
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\n\n- Gold: Density, $\\rho = 19\\ 320$ kg/m$^3$; Young's modulus, $E = 79$ GPa; Poisson's ratio, $\\nu = 0.4$.
\n- Steel: Density, $\\rho = 8\\ 050$ kg/m$^3$; Young's modulus, $E = 209$ GPa; Poisson's ratio, $\\nu = 0.3$.
\n- Gravity: $g = 9.81$ m/s$^2$.
\n
\nA gold sphere of diameter $\\var{diameter}$ cm sits on a flat steel table.
\nCalculate:
\n\n- the weight of the ball: [[0]] [Units: N]
\n- the Elastic Contact Modulus: $E^* =$ [[1]] [Units: GPa]
\n- the peak contact pressure: $p_0 =$ [[2]] [Units: MPa]
\n- the semi-contact width: $a =$ [[3]] [Units: mm]
\n- the area of the contact patch: [[4]] [Units: mm$^2$]
\n
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "The size of the contact and the pressure between the contacting surfaces depends on:
\n\n- the materials, and here we assume them to be linear, elastic and isotropic;
\n- the geometry, and here we assume them to be either flat (infinite radius) or having a constant radius;
\n- the applied load, and here we assume the load is applied normal to the contact, i.e., there is no friction.
\n
\n\nMaterials: 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}\\]
\nwhere $E_1$ 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.
\nGeometry: The two simplest forms of contact between curved surfaces are:
\n\n- 2D: cylindrical surfaces aligned along their length, or a cylinder on a flat surface, creating a rectangular contact area;
\n- 3D: spherical surfaces, or a sphere on a flat surface, creating a circular contact area.
\n
\nIn 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}\\]
\nwhere $R_1$ 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}\\]
\nwhere $R_1$ 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.)
\nApplied Load: This is a little tricky because the same notation, $P$, is used to mean different things:
\n\n- In 2D, $P$ is the applied load per unit length of the cylindrical contact. [Units: N/m]
\n- In 3D, $P$ is simply the applied load. [Units: N]
\n
\n2D 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}\\]
\n3D 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": "Reference values:
\n\n- HDPE: Young's modulus, $E = 1$ GPa; Poisson's ratio, $\\nu = 0.45$.
\n- Steel: Young's modulus, $E = 209$ GPa; Poisson's ratio, $\\nu = 0.3$.
\n
\nA 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.
\nCalculate:
\n\n- Because the cup is concave not convex, the radius is effectively negative. The equivalent radius is therefore given by:
\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$^{-1}$]
\nwhich can be rearranged to give $R = \\var{siground(R*1000,3)}$mm.
\n\n- The Elastic Contact Modulus, $E^*$, is given by:
\n
\n${1 \\over E^*} = {1 - \\nu_1^2 \\over E_1} + {1 - \\nu_2^2 \\over E_2} = {1 - 0.45^2 \\over 1 \\times 10^9} + {1 - 0.3^2 \\over 209 \\times 10^9}$
\nwhich can be rearranged to give $E^* = \\var{siground(ECM,3)}$GPa.
\n\n- 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:
\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- Similarly, the semi-contact width, $a$, is given by:
\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- The area of the contact patch is circular with radius equal to $a$, i.e.:
\n
\narea $= \\pi a^2 = \\pi \\left( \\var{siground(scw,3)} \\times 10^{-3} \\right)^2 = \\var{siground(area,3)}$mm$^2$.
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\n\n- HDPE: Young's modulus, $E = 1$ GPa; Poisson's ratio, $\\nu = 0.45$.
\n- Steel: Young's modulus, $E = 209$ GPa; Poisson's ratio, $\\nu = 0.3$.
\n
\nA 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.
\nCalculate:
\n\n- the equivalent radius: $R =$ [[0]] [Units: mm]
\n- the Elastic Contact Modulus: $E^* =$ [[1]] [Units: GPa]
\n- the peak contact pressure: $p_0 =$ [[2]] [Units: MPa]
\n- the semi-contact width: $a =$ [[3]] [Units: mm]
\n- the area of the contact patch: [[4]] [Units: mm$^2$]
\n
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