Difference between revisions of "Hertzian contact"

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{{Box-round|title= Related papers |
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[https://e6.ijs.si/ParallelAndDistributedSystems/publications/32230439.pdf J. Slak, G. Kosec; Adaptive radial basis function-generated finite differences method for contact problems, International journal for numerical methods in engineering, vol. 119, 2019 [DOI: 10.1002/nme.6067]]
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}}
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Click on [[Solid Mechanics]] to go back.
 
Click on [[Solid Mechanics]] to go back.
  
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The solutions were verified and graphs were generated using this mathematica notebook: [[:File:hertzian_analytical.nb]]
 
The solutions were verified and graphs were generated using this mathematica notebook: [[:File:hertzian_analytical.nb]]
Copy-pasteable Matlab code:
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 +
Copy-pasteable Matlab code of the analytical solution:
 
<syntaxhighlight lang="matlab" line>
 
<syntaxhighlight lang="matlab" line>
 
function [sxx, syy, sxy] = analytical(x, z, b, p0)
 
function [sxx, syy, sxy] = analytical(x, z, b, p0)
Line 100: Line 106:
 
sxy = p0 / b * n .* ((m2 - z.^2) ./ mpn);
 
sxy = p0 / b * n .* ((m2 - z.^2) ./ mpn);
 
end
 
end
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</syntaxhighlight>
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 +
Copy-pasteable C++ code of the analytical solution:
 +
<syntaxhighlight lang="c++" line>
 +
std::function<Vec3d(Vec2d)> analytical = [=] (const Vec2d& p) {
 +
    double x = p[0], z = p[1];
 +
    double bxz = b*b - x*x + z*z;
 +
    double xz = 4*x*x*z*z;
 +
    double koren = std::sqrt(bxz*bxz + xz);
 +
    double m2 = 0.5 * (koren + bxz);
 +
    double n2 = 0.5 * (koren - bxz);
 +
    double m = std::sqrt(m2);
 +
    double n = signum(x) * std::sqrt(n2);
 +
    double mpn = m2 + n2;
 +
    double zmn = (z*z + n2) / mpn;
 +
    double sxx = -p0 / b * (m * (1 + zmn) + 2*z);
 +
    double syy = -p0 / b * m * (1 - zmn);
 +
    double sxy = p0 / b * n * ((m2 - z*z) / mpn);
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    return Vec3d(sxx, syy, sxy);
 +
};
 
</syntaxhighlight>
 
</syntaxhighlight>
  
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$s_{zz} = \frac{b p_0 z^3}{x^4}+O(x^{-5})$,  
 
$s_{zz} = \frac{b p_0 z^3}{x^4}+O(x^{-5})$,  
 
$s_{xz} = \frac{b p_0 z^2}{x^3}+O(x^{-5})$.
 
$s_{xz} = \frac{b p_0 z^2}{x^3}+O(x^{-5})$.
 
= Contact of cylinders under partial slip =
 
 
The second case we study is the application of a tangential force $Q$ to the previous problem. When the tangential force is less than the limiting force of friction, i.e., \[|Q| < \mu P,\] where $\mu$ is the coefficent of friction, sliding motion will not occur but the contact will be divided into regions of slip and stick zones that are unknown ''a priori''. For the case of cylinders the analysis is given in Hills & Nowell (1994), p. 44.
 
 
Besides the normal traction $p(x)$ we know have an additional shear traction given by
 
\begin{equation}
 
q(x) = \begin{cases}
 
-\mu p_0 \sqrt{1 - \frac{x^2}{b^2}}, \quad c \leq |x| \leq b \\
 
-\mu p_0 \left(\sqrt{1 - \frac{x^2}{b^2}} - \frac{c}{b}\sqrt{1 - \frac{x^2}{c^2}}\right), \quad |x| < c
 
\end{cases}
 
\end{equation}
 
where $b$ is the half-width of the whole contact, and $c$ the half-width of the central sticking region. The width of the central zone, i.e. the value of dimension $c$ is dependent on the applied tangential force $Q$:
 
\begin{equation}
 
\frac{c}{b} = \sqrt{1 - \frac{Q}{\mu P}}
 
\end{equation}
 
 
The distributions $q(x)$ and $p(x)$ as well as the widths of the stick and slip zones can be seen in the image below.
 
[[File:Screenshot_2016-11-17_10-43-18.png|400px]]
 
 
 
== The effect of bulk stress ==
 
 
Additionally we might be interested in the addition of bulk stress. This type of stress occurs in fretting fatique experiments like the one shown below.
 
 
[[File:Screenshot_2016-11-17_10-55-59.png|500px]]
 
 
The previous solution for contact of cylinders under partial slip can be adjusted for the presence of bulk stresses $\sigma_\mathrm{axial}$. These cause an eccentricity $e$ to the solution given above. The shear traction $q(x)$ can be written as:
 
\begin{equation}
 
q(x) = \begin{cases}
 
-\mu p_0 \sqrt{1 - \frac{x^2}{b^2}}, \quad c \leq | x - e | \text{ and } |x| \leq b \\
 
-\mu p_0 \left[\sqrt{1 - \frac{x^2}{b^2}} - \frac{c}{b}\sqrt{1 - \frac{(x-e)^2}{c^2}}\right], \quad |x-e| < c
 
\end{cases}
 
\end{equation}
 
where once again \[ \frac{c}{b} = \sqrt{1 - \frac{Q}{\mu P}}\] and
 
\begin{equation}
 
e = \frac{b \sigma_\mathrm{axial}}{4 \mu p_0}.
 
\end{equation}
 
If larger values of $\sigma_\mathrm{axial}$ are applied, one edge of the stick zone will approach the edge of the contact ($e$ becomes larger). The solution for the shear stress traction is therefore only valid if $e + c \leq b$, i. e.
 
\[\frac{\sigma_\mathrm{axial}}{\mu p_0} \leq 4\left(1 - \sqrt{1 - \frac{Q}{\mu P}}\right).\]
 
  
 
= FreeFem++ numerical solution =
 
= FreeFem++ numerical solution =
  
For the numerical solution in FreeFem++ we choose parameters similar to those in Pereira et al. (2016): <ref> Pereira et al. (2016). '''On the convergence of stresses in fretting fatique'''. Materials, 9, 639.</ref>
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For the numerical solution in FreeFem++ we choose parameters similar to those in Pereira et al. (2016): <ref> K. Pereira et al., On the convergence of stresses in fretting fatigue, Materials
 +
9(8) (2016), doi:10.3390/ma9080639. </ref>  
  
 
*Modulus of elasticity: $E = 72.1$ GPa
 
*Modulus of elasticity: $E = 72.1$ GPa
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*Specimen length: $L = 40$ mm
 
*Specimen length: $L = 40$ mm
 
*Specimen height: $H = 10$ mm
 
*Specimen height: $H = 10$ mm
 +
*Specimen thickness: $t = 4$ mm
  
 
We assume that both the pad and specimen are from the same material, therefore the combined modulus is \[E^* = \frac{E}{2(1-\nu^2)}.\] According to Pereira et al. (2016) contact width is now defined with the specimen height in the denominator:
 
We assume that both the pad and specimen are from the same material, therefore the combined modulus is \[E^* = \frac{E}{2(1-\nu^2)}.\] According to Pereira et al. (2016) contact width is now defined with the specimen height in the denominator:
\[b = 2\sqrt{\frac{PR}{H\pi E^*}}\]
+
\[b = 2\sqrt{\frac{PR}{t\pi E^*}}\]
\[p_0 = \sqrt{\frac{PE^*}{H\pi R}}\]
+
\[p_0 = \sqrt{\frac{PE^*}{t\pi R}}\]
  
(this might not be correct, because $p_0$ should be equal to $2P/(\pi b)$? Does $R/H$ perhaps represent a dimensionless radius?)
 
  
 
Out of simplicity we only use traction boundaries at the top surface. On the left, right and bottom sides of the specimen Dirichlet boundaries are used. The center of the coordinate system is placed at $(x,y) = (L/2,0)$. The axises therefore go from $-L/2$ to $L/2$ in $x$-direction and from $0$ to $H$ in $y$-direction.
 
Out of simplicity we only use traction boundaries at the top surface. On the left, right and bottom sides of the specimen Dirichlet boundaries are used. The center of the coordinate system is placed at $(x,y) = (L/2,0)$. The axises therefore go from $-L/2$ to $L/2$ in $x$-direction and from $0$ to $H$ in $y$-direction.
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[[File:mss_contours.png|900px]]
 
[[File:mss_contours.png|900px]]
  
 +
= Meshless numerical solution for halfplane contact =
 +
We are solving the equation $$(\lambda + \mu) \nabla (\nabla \cdot \b{u}) + \mu \nabla^2 \b{u} = 0$$
 +
 +
The boundary conditions are zero except for the top layer which has zero traction in $x$ direction and traction $t_z$ in $z$ direction.
  
= MLSM numerical solution for halfplane contact =
+
* top: $\vec{t}(x) = (0, -p(x))$ or componentwise $(2 \mu+\lambda) \frac{\partial u}{\partial x}(x, 0) + \lambda \frac{\partial v}{\partial y}(x, 0) = -p(x)$ and $\mu \frac{\partial u}{\partial y}(x, 0) + \mu \frac{\partial v}{\partial x}(x, 0) = 0$.
 +
* bottom, left, right: $\vec{u}=0$
 
Problem parameters used:
 
Problem parameters used:
 
*Modulus of elasticity: $E = 72.1$ GPa
 
*Modulus of elasticity: $E = 72.1$ GPa
 
*Poisson's ratio: $\nu = 0.33$
 
*Poisson's ratio: $\nu = 0.33$
 
*Normal load: $P = 543$ N
 
*Normal load: $P = 543$ N
*Coefficient of friction: $\mu = 0.85$
+
*Cylinder radius: $R = 1$ m
*Cylinder radius: $R = 50$ mm
+
*Specimen height: $H = 10$ mm
*Specimen length: $L = 40$ mm
+
*Specimen length: $L = 2H = 20$ mm
*Specimen height: $H = 20$ mm
+
 
 +
Convergence with respect to uniform mesh is very bumpy, due to very few nodes (max. 100) actually being under the contact surface.
 +
 
 +
[[File:hertzian_convergence.png|600px]]
 +
 
 +
Much more can be achieved with refinement. Taking $H = 1$ m $\approx 75\,000\,b$ and refining the domain around the contact region.
 +
Refined domain is presented below, along tiwh the refinement density.
 +
 
 +
[[File:hertzian_refined_domain.png|600px]][[File:hertzian_refined_domain_density.png|600px]]
 +
 
 +
Convergence with respect to different refine levels is shown below.
  
The boundary conditions are zero except for the top layer which has zero traction in $x$ direction and traction $t_z$ in $z$ direction.
+
[[File:hertzian_refine_levels_convergence.png|600px]]
The solution matches the analytical solution well.
 
[[File:hertzian_psd_nonrefined.png|800px]]
 
  
[[File:hertzian_subsurface_nonrefined.png|800px]]
+
The convergence is more regular than before and every additional level greatly helps regarding the total error.
 +
The solution with the smallest error is shown below.
  
[[File:hertzian_midplane_nonrefined.png|800px]]
+
[[File:hertzian_solution_deformed_vm.png|600px]]
  
 
Execution time is shown below. Program ran on a single core. Most of the time is spent on solving the system of the equations.
 
Execution time is shown below. Program ran on a single core. Most of the time is spent on solving the system of the equations.
  
 
[[File:hertzian_time.png|800px]]
 
[[File:hertzian_time.png|800px]]
 +
 +
<!--
 +
 +
[[File:hertzian_subsurface_nonrefined.png|800px]]
 +
 +
[[File:hertzian_midplane_nonrefined.png|800px]]
  
 
The error decreases as the density of nodes increases, up to some limit when the system becomes too unstable.
 
The error decreases as the density of nodes increases, up to some limit when the system becomes too unstable.
Line 234: Line 241:
 
[[File:hertzian_error_nonrefined_convergence.png|800px]]
 
[[File:hertzian_error_nonrefined_convergence.png|800px]]
  
= MLSM numerical solution for bulk load =
+
More can be gained using refinement:
== Case definition ==
 
  
We are solving the equation $$(\lambda + \mu) \nabla (\nabla \cdot \b{u}) + \mu \nabla^2 \b{u} = 0$$
+
[[File:hertzian_midplane_refined.png|800px]]
on the domain $D = [-L/2, L/2] \times [-H/2, 0]$. Componentwise $\b{u} = [u; v]$.
+
 
The boundary conditions are:
+
[[File:hertzian_subsurface_refined.png|800px]]
* top: $\vec{t}(x) = (q(x), -p(x))$ or componentwise $(2 \mu+\lambda) \frac{\partial u}{\partial x}(x, 0) + \lambda \frac{\partial v}{\partial y}(x, 0) = q(x)$ and $\mu \frac{\partial u}{\partial y}(x, 0) + \mu \frac{\partial v}{\partial x}(x, 0) = -p(x)$.
 
* left: $\b{u} = 0$ or componentwise $u(-L/2, y) = v(-L/2, y) = 0$.
 
* bottom: up-down symmetry conditions: componentwise $\frac{\partial u}{\partial y}(x, -H/2) = 0, v(x, -H/2) = 0$.
 
* right: this part is traction free, ie. $\vec{t}(y) = 0$ or componentwise $\mu \frac{\partial u}{\partial y}(L/2, y) + \mu \frac{\partial v}{\partial x}(L/2, y) = 0$ and $(2 \mu+\lambda) \frac{\partial v}{\partial y}(L/2, y) + \lambda \frac{\partial u}{\partial x}(L/2, y) = 0$.
 
  
== Phsyical parameters ==
+
[[File:hertzian_domain_refined.png|800px]]
Basic parameters are:
 
* $E = \unit{72.1}{GPa}$, Youngs modulus
 
* $\nu = 0.33$, Poissons ratio
 
* $L = \unit{40}{mm}$, length, width of the pad
 
* $H = \unit{4}{mm}$, height, thickness of the pad
 
* $\sigma_{ax} = \unit{100}{MPa}$, axial pressure
 
* $F = \unit{543}{N}$, normal force
 
* $Q = \unit{155}{N}$, tangential force
 
* $R = \unit{10}{mm}$ or $\unit{50}{mm}$, raduis of curvature of the cylindrical pad
 
* $COF = \mu = 0.3$ or $0.85$ or $2$, coefficient of friction
 
  
Derived parameters are, for choice of $R = \unit{10}{mm}$ and $\mu = 0.3$:
+
[[File:hertzian_psd_contour_refined.png|800px]] -->
* $E^\ast = \frac{E}{2(1-\nu^2)} = \unit{40.4}{GPa}$, combined Young's modulus
 
* $p_{max} = p_0 = \sqrt{\frac{FE^\ast}{HR \pi}} = \unit{418.10407}{MPa}$, maximal normal pressure
 
* $a = 2 \sqrt{\frac{FR}{HE^\ast \pi}} = \unit{0.2067}{mm}$, semi contact width, contact region is $[-a, a] \times \{0\}$
 
* $c = a \sqrt{1 - \frac{Q}{\mu F}} = \unit{0.04504}{mm}$, stick zone semi width
 
* $e = \frac{a \sigma_{ax}}{4 \mu p_{max}} = \unit{0.041197}{mm}$, eccentricity due to axial load
 
  
 
=References=
 
=References=
 
<references/>
 
<references/>

Latest revision as of 21:57, 22 October 2022

edit 

Related papers

J. Slak, G. Kosec; Adaptive radial basis function-generated finite differences method for contact problems, International journal for numerical methods in engineering, vol. 119, 2019 [DOI: 10.1002/nme.6067]


Click on Solid Mechanics to go back.

Contact of Cylinders - the Hertz problem

Detailed discussions of this problem can be found in Hills and Nowells (1994) as well as Williams and Dwyer-Joyce (2001). [1] [2]

If two circular cylinders with radii $R_1$ and $R_2$ are pressed together by a force per unit length of magnitude $P$ with their axes parallel, then the contact patch will be of half-width $b$ such that \begin{equation} b = 2\sqrt{\frac{PR}{\pi E^*}} \end{equation} where $R$ and $E^*$ are the reduced radius of contact and the contact modulus defined by \begin{equation} \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}, \end{equation} \begin{equation} \frac{1}{E^*} = \frac{1-{\nu_1}^2}{E_1} + \frac{1-{\nu_2}^2}{E_2}. \end{equation}

The resulting pressure distribution $p(x)$ is semielliptical, i.e., of the form \begin{equation} p(x) = p_0 \sqrt{1-\frac{x^2}{b^2}} \end{equation} where the peak pressure \begin{equation} p_0 = \sqrt{\frac{PE^*}{\pi R}}. \end{equation}

The coordinate $x$ is measured perpendicular to that of the cylinder axes. The cylinder is pressing on the lower halfplane from above. Analytical solutions for stresses in the plane are presented below. The coordinate ranges are $x \in (-\infty, \infty), z\in (-\infty, 0]$.

The surface stresses are given in the following equations. At the contact interface \begin{equation} \sigma_{xx} = \sigma_{zz} = -p(x); \end{equation} outside the contact region all the stress components at the surface are zero. Along the line of symmetry the following equations hold \begin{equation} \sigma_{xx} = -p_0\left( \frac{1+(z/b)^2}{\sqrt{1+(z/b)^2}} + 2z/b\right) \end{equation} \begin{equation} \sigma_{zz} = -p_0(1 + z^2/b^2)^{-1/2} \end{equation} These are the principal stresses so that the principal shear stress $\tau_1$ is given by \begin{equation} \tau_1 = -p_0\left(z/b + \frac{(z/b)^2}{\sqrt{1+(z/b)^2}}\right) \end{equation} from which \[(\tau_1)_\mathrm{max} = 0.30p_0, \quad \text{at } z = 0.78b\]

Note that these stresses are all independent of Poisson's ratio although, for plane strain, the third principal stress \[\sigma_{yy} = \nu(\sigma_{xx} + \sigma_{zz}). \]


At a general point $(x,z)$ the stresses may be expressed in terms of $m$ and $n$, defined by \begin{equation} m^2 = \frac{1}{2} \left(\sqrt{\left(b^2-x^2+z^2\right)^2+4 x^2 z^2}+b^2-x^2+z^2\right) \end{equation} \begin{equation} n^2 = \frac{1}{2} \left(\sqrt{\left(b^2-x^2+z^2\right)^2+4 x^2 z^2}-(b^2-x^2+z^2)\right) \end{equation} with $m = +\sqrt{m^2}$ and $n = \operatorname{sgn}(x)\sqrt{n^2}$. Whereupon \begin{equation} \sigma_{xx} = -\frac{p_0}{b}\left[m\left(1 + \frac{z^2 + n^2}{m^2 + n^2}\right)+2z\right] \end{equation}

\begin{equation} \sigma_{zz} = -\frac{p_0}{b}m\left(1 - \frac{z^2 + n^2}{m^2 + n^2}\right) \end{equation}

\begin{equation} \sigma_{xz} = \sigma_{zx} = \frac{p_0}{b}n\left(\frac{m^2 - z^2}{m^2 + n^2}\right) \end{equation}

Plots of this analytical solution are presented below, using $p_0 = 8.36\cdot10^{7}, b = 4.13\cdot10^{-4}$

Divs hertzian analytical.png

Sxx hertzian analytical.pngSyy hertzian analytical.png

Sxy hertzian analytical.pngPsd hertzian analytical.png

The solutions were verified and graphs were generated using this mathematica notebook: File:hertzian_analytical.nb

Copy-pasteable Matlab code of the analytical solution:

 1 function [sxx, syy, sxy] = analytical(x, z, b, p0)
 2 % Returns the analytical solution for stress for hertzian contact.
 3 % x in (-inf, inf), z in (-inf, 0].
 4 bxz = b^2 - x.^2 + z.^2;
 5 xz = 4*x.^2.*z.^2;
 6 koren = sqrt(bxz.^2 + xz);
 7 m2 = 1/2 * (koren + bxz);
 8 n2 = 1/2 * (koren - bxz);
 9 m = sqrt(m2);
10 n = sign(x) .* sqrt(n2);
11 mpn = m2 + n2;
12 zmn = (z.^2 + n2) ./ mpn;
13 sxx = -p0 / b * (m .* (1 + zmn) + 2*z);
14 syy = -p0 / b * m .* (1 - zmn);
15 sxy = p0 / b * n .* ((m2 - z.^2) ./ mpn);
16 end

Copy-pasteable C++ code of the analytical solution:

 1 std::function<Vec3d(Vec2d)> analytical = [=] (const Vec2d& p) {
 2     double x = p[0], z = p[1];
 3     double bxz = b*b - x*x + z*z;
 4     double xz = 4*x*x*z*z;
 5     double koren = std::sqrt(bxz*bxz + xz);
 6     double m2 = 0.5 * (koren + bxz);
 7     double n2 = 0.5 * (koren - bxz);
 8     double m = std::sqrt(m2);
 9     double n = signum(x) * std::sqrt(n2);
10     double mpn = m2 + n2;
11     double zmn = (z*z + n2) / mpn;
12     double sxx = -p0 / b * (m * (1 + zmn) + 2*z);
13     double syy = -p0 / b * m * (1 - zmn);
14     double sxy = p0 / b * n * ((m2 - z*z) / mpn);
15     return Vec3d(sxx, syy, sxy);
16 };

The extreme of $\tau_1$ is achieved at $x = 0$, $z = -\sqrt{\frac{1}{2} \left(\sqrt{5}-1\right)} b = -0.786b$.

Midplane stress hertzian analytical.png Taumax stress hertzian analytical.png

When observing for $z \to -\infty$ it holds that $s_{xx} = \frac{\frac{b^3 p_0}{4}+b p_0 x^2}{z^3}+O(z^{-5})$, $s_{zz} = \frac{b p_0}{z}-\frac{b p_0 \left(b^2+4 x^2\right)}{2 z^3}+O(z^{-5})$, $s_{xz} = \frac{b p_0 x}{z^2}+\frac{-\frac{3}{2} b^3 p_0 x-2 b p_0 x^3}{z^4}+O(z^{-5})$.

When observing for $x \to \infty$ (and similarly for $-\infty$) it holds that

$s_{xx} = \frac{b p_0 z}{x^2}+\frac{\frac{3}{4} b^3 p_0 z-2 b p_0 z^3}{x^4}+O(x^{-5})$, $s_{zz} = \frac{b p_0 z^3}{x^4}+O(x^{-5})$, $s_{xz} = \frac{b p_0 z^2}{x^3}+O(x^{-5})$.

FreeFem++ numerical solution

For the numerical solution in FreeFem++ we choose parameters similar to those in Pereira et al. (2016): [3]

  • Modulus of elasticity: $E = 72.1$ GPa
  • Poisson's ratio: $\nu = 0.33$
  • Normal load: $P = 543$ N
  • Coefficient of friction: $\mu = 0.85$
  • Cylinder radius: $R = 50$ mm
  • Specimen length: $L = 40$ mm
  • Specimen height: $H = 10$ mm
  • Specimen thickness: $t = 4$ mm

We assume that both the pad and specimen are from the same material, therefore the combined modulus is \[E^* = \frac{E}{2(1-\nu^2)}.\] According to Pereira et al. (2016) contact width is now defined with the specimen height in the denominator: \[b = 2\sqrt{\frac{PR}{t\pi E^*}}\] \[p_0 = \sqrt{\frac{PE^*}{t\pi R}}\]


Out of simplicity we only use traction boundaries at the top surface. On the left, right and bottom sides of the specimen Dirichlet boundaries are used. The center of the coordinate system is placed at $(x,y) = (L/2,0)$. The axises therefore go from $-L/2$ to $L/2$ in $x$-direction and from $0$ to $H$ in $y$-direction.

                top
      ________________________
     |                        |
     |                        |
left |                        | right
     |                        |
     |________________________|
                
               bottom

Boundary conditions: \[ t_z(x) = \begin{cases} -p_0\sqrt{1 - \frac{x^2}{b^2}}, \quad |x| \leq b \\ 0, \quad \text{otherwise} \end{cases} \quad \text{on } \Gamma_\mathrm{top} \] \[ (u_x,u_z) = (0,0) \quad \text{on } \Gamma_\mathrm{left},\Gamma_\mathrm{right},\Gamma_\mathrm{bottom} \]

The numerical solution is obtained for plane strain conditions using quadratic finite elements. The final mesh after adaptation had ~$40000$ dof. The images below show the raw values of the surface stresses, sub-surface stresses along the axis of symmetry and a close-up of the maximum shear stress contours close to the contact. The contact width $b \approx 0.29$ mm for the parameters described at the beginning of this section.

Surface stress.png

Subsurface stress.png

Mss contours.png

Meshless numerical solution for halfplane contact

We are solving the equation $$(\lambda + \mu) \nabla (\nabla \cdot \b{u}) + \mu \nabla^2 \b{u} = 0$$

The boundary conditions are zero except for the top layer which has zero traction in $x$ direction and traction $t_z$ in $z$ direction.

  • top: $\vec{t}(x) = (0, -p(x))$ or componentwise $(2 \mu+\lambda) \frac{\partial u}{\partial x}(x, 0) + \lambda \frac{\partial v}{\partial y}(x, 0) = -p(x)$ and $\mu \frac{\partial u}{\partial y}(x, 0) + \mu \frac{\partial v}{\partial x}(x, 0) = 0$.
  • bottom, left, right: $\vec{u}=0$

Problem parameters used:

  • Modulus of elasticity: $E = 72.1$ GPa
  • Poisson's ratio: $\nu = 0.33$
  • Normal load: $P = 543$ N
  • Cylinder radius: $R = 1$ m
  • Specimen height: $H = 10$ mm
  • Specimen length: $L = 2H = 20$ mm

Convergence with respect to uniform mesh is very bumpy, due to very few nodes (max. 100) actually being under the contact surface.

Hertzian convergence.png

Much more can be achieved with refinement. Taking $H = 1$ m $\approx 75\,000\,b$ and refining the domain around the contact region. Refined domain is presented below, along tiwh the refinement density.

Hertzian refined domain.pngHertzian refined domain density.png

Convergence with respect to different refine levels is shown below.

Hertzian refine levels convergence.png

The convergence is more regular than before and every additional level greatly helps regarding the total error. The solution with the smallest error is shown below.

Hertzian solution deformed vm.png

Execution time is shown below. Program ran on a single core. Most of the time is spent on solving the system of the equations.

Hertzian time.png


References

  1. Hills, D. A. and Nowell, D. (1994). Mechanics of Fretting Fatique, p. 20-25. Springer Science+Business Media, Dordrecht.
  2. Williams, John A. and Dwyer-Joyce, Rob S. (2001). Contact Between Solid Surfaces, p. 121 in Modern Tribology Handbook: Volume 1, Principles of Tribology, editor: Bushan, Bharat. CRC Press LLC, Boca Raton.
  3. K. Pereira et al., On the convergence of stresses in fretting fatigue, Materials 9(8) (2016), doi:10.3390/ma9080639.