Christian Jäger, summer term 2019
In 1881 the German physicist Heinrich Hertz published a theory to compute the stresses and forces acting during the contact of two elastic bodies. Previously, it was only possible to calculate an idealized case in contact mechanics. The two interacting bodies were originally assumed to be curved and rigid. Therefore, in the idealized case one receives only a point-shaped area of contact if the interacting bodies are either both spheres or a sphere and a cylinder. If two cylinders are pressed together the area of contact is line-shaped.
Corresponding to real conditions of two bodies being compressed, the two interacting surfaces will be deformed, and a flattening will be visible. Hertz used his theory to describe the size and form of the contact zone as well as the distribution of stresses induced by the contact.
The Hertzian Contact theory plays a major role in the fields of non-destructive-testing, mainly in the Impact Echo and the Local Acoustic Resonance Spectroscopy (LARS).2,3
To compute the contact mechanics, Hertz made several assumptions:
Firstly, the two interacting bodies are assumed to be of an elastic, isotropic and homogenous material. Secondly, the surface is assumed to be perfectly smooth such that no shear stresses occur in the interacting surfaces. And thirdly, it is assumed that only a relatively small part of the total surfaces is in contact.
The two bodies can thus be regarded as half-spaces and the half space theory is therefore applicable.
Hertz used his new theory to compute the size and form of the compressed area, the normal distribution of pressure between the two surfaces and the maximum pressure between the surfaces, from which the deformation of the two bodies depends. By computing all these parameters, he was finally able to compute the final approach of both bodies under the given load.
The tensile moduli and the form and size of the surfaces nearby the point of contact are known.
Hertz stated that the two surfaces meet in such a way that the surface normal in the contact point are orientated in the direction of compression and parallel to each other. An orthogonal cartesian coordinate system is defined with its origin placed exactly at the contact point (see Fig. 1). The two surfaces nearby the contact point can now be described as two paraboloids of the following form:1
(1) | z_{1} = A_{1}x^2 + B_{1}y^2 + C_{1}xy \\ z_{2} = A_{2}x^2 + B_{2}y^2 + C_{2}xy |
By the correct orientation of the axis of coordinates the xy-Terms will vanish. The two positive constants A and B can be computed from the principle radii of the surface curvatures:1
(2) | B + A = \frac {1}{2} * (\frac{1}{R_{11}} + \frac{1}{R_{12}} + \frac{1}{R_{21}} + \frac{1}{R_{22}}) \\ (B - A) = \frac{1}{2} * [(\frac{1}{R_{11}} - \frac{1}{R_{12}})^2 + (\frac{1}{R_{21}} - \frac{1}{R_{22}})^2 + 2*(\frac{1}{R_{11}} - \frac{1}{R_{12}}) * (\frac{1}{R_{21}} - \frac{1}{R_{22}}) * \cos(2\beta)]^\frac{1}{2} |
Fig. 1: Two bodies in Contact |
The constant \beta denotes the angle between the two normal planes containing the principal curvatures R_{11}and R_{21}. If body 1 is a sphere with radius R, then R_{11} = R_{12} = R. If one of the bodies is a cylinder with radius R, then the first principal radius is set to R and the second one equals \infty. Those and further simplifications are shown in the following Table:1
Type of contact | A | B | A/B |
---|---|---|---|
Body 1: Sphere of radius R_{1} Body 2: Sphere of radius R_{2} | \frac{1}{2R_{1}} + \frac{1}{2R_{2}} | \frac{1}{2R_{1}} + \frac{1}{2R_{2}} | 1 |
Body 1: Sphere of radius R_{1} Body 2: plane surface | \frac{1}{2R_{1}} | \frac{1}{2R_{2}} | 1 |
Body 1: Cylinder of radius R_{1} Body 2: Cylinder of radius R_{2}, placed crosswise to body 1 | \frac{1}{2R_{1}} | \frac{1}{2R_{2}} | \frac{R_{2}}{R_{1}} |
Body 1: Sphere of radius R_{1} Body 2: Cylinder of radius R_{2} | \frac{1}{2R_{1}} | \frac{1}{2R_{1}} + \frac{1}{2R_{2}} | \frac{R_{2}}{R_{1} + R_{2}} |
Body 1: Sphere of radius R_{1} Body 2: Cylindrical cup of radius R_{2}; R_{2} > R_{1} | \frac{1}{2R_{1}} - \frac{1}{2R_{2}} | \frac{1}{2R_{1}} | \frac {R_{2} - R_{1}}{R_{2}} |
Body 1: Sphere of radius R_{1} Body 2: Spherical cup of radius R_{2}; R_{2} > R_{1} | \frac{R_{2} - R_{1}}{2R_{1}R_{2}} | \frac{R_{2} - R_{1}}{2R_{1}R_{2}} | 1 |
Body 1: Sphere of radius R_{1} Body 2: Surface of rotation with principal radii R_{2} and R_{3}; R_{2} > R_{1} | \frac{1}{2R_{1}} - \frac{1}{2R_{2}} | \frac{1}{2R_{1}} + \frac{1}{2R_{3}} | \frac{1-\frac{R_{1}}{R_{2}}}{1 - \frac{R_{1}}{R_{3}}} |
The displacements of the two bodies can therefore be calculated as follows:1
(3) | w_{1} + w_{2} = \alpha + Ax^2 + By^2 = \\ = [\frac{1-\mu_{1}^2}{E_{1}\pi} + \frac{1-\mu_{2}^2}{E_{2}\pi}] \iint \frac{p}{s} \mathrm {d}x \mathrm {d}y = [\delta_{1} + \delta_{2}] \iint \frac{p}{s} \mathrm {d}x \mathrm {d}y |
The variables w_{1} and w_{2} represent the displacements of the two bodies, \alpha is used to display the approach, s equals the distance between any surface element in its initial position and the coordinate origin after the application of the pressure p, which must be deduced.
To compute the final pressure distribution, which Hertz assumed to be in a potential form, the following equilibrium conditions need to be fulfilled:
The pressure potential was assumed to be a quadratic function of x and y and an ellipsoid with a certain mass density. By assuming F as the total mass of the ellipsoid the formula for the pressure distribution reduces to the following:7
(4) | p = p_{0}\sqrt{1 - \frac{x^2}{a} - \frac{y^2}{b}} = \frac{3F}{2\pi ab} \sqrt{1 - \frac{x^2}{a} - \frac{y^2}{b}} |
The two constants a and b constitue the length of the two semiaxes of the elliptic compression area. Including the evaluation of means of Legendre polynomials the length of those semiaxes can be computed as:1
(5) | a = q_{a} \sqrt[3]{\frac{3F(\delta_{1} + \delta_{2})}{4(A + B)}} \\ b = q_{b} \sqrt[3]{\frac{3F(\delta_{1} + \delta_{2})}{4(A + B)}} |
And the total Force F calculates as:1
(6) | F =\frac{4q_{k}}{3(\delta_{1} + \delta_{2}) \sqrt{(A+B)}} = k_{2}a^\frac{3}{2} |
The next table shows some values for the constants q_{a}, q_{b} and q_{k} depending on the ratio \frac{A}{B} or the auxiliary angle \theta = \arccos(\frac{B-A}{B+A}):1
\theta | \frac{A}{B} | q_{a} | q_{b} | q_{k} |
---|---|---|---|---|
90° | 1.000 | 1.4645 | 1.4645 | 0.3180 |
80° | 0.7041 | 1.6517 | 1.3074 | 0.3215 |
70° | 0.4903 | 1.8797 | 1.1741 | 0.3322 |
60° | 0.3333 | 2.1759 | 1.0502 | 0.3505 |
50° | 0.2174 | 2.5690 | 0.9383 | 0.3819 |
40° | 0.1325 | 3.1277 | 0.8308 | 0.4300 |
30° | 0.0718 | 3.9991 | 0.7220 | 0.5132 |
20° | 0.0311 | 5.5327 | 0.5974 | 0.6662 |
10° | 0.00765 | 9.6833 | 0.4666 | 1.1450 |
0° | 0 | \infty | 0 | - |
To compute the occurring forces for the dynamic contact of two bodies, a calculation of the contact time of both bodies is necessary. By integration of Newton’s second law of motion1
(7) | F = -m_{1}\ddot{w_{1}} = -m_{2}\ddot{w_{2}} = - \frac{m_{1}m_{2}}{m_{1}+m_{2}}\ddot{\alpha} |
where \alpha marks the compression of the two colliding elements and substituting Equation (6), the following correlation of the initial conditions \alpha_{0} = 0 and \dot{\alpha_{0}} = v_{0} is compiled1.
(8) | \frac{1}{2}(\dot{\alpha}^2-v_{0}^2) = -\frac{2}{5}k_{1}k_{2}\alpha^\frac{5}{2} |
For the closest approach of the colliding bodies \dot{\alpha}=0 and the maximum compression computes as:4
(9) | \alpha_{m} = (\frac{5v_{0}^2}{4k_{1}k_{2}})^\frac{2}{5} |
The total duration of contact can therefore be calculated as4
(10) | T = 2t = 2 \int \limits_0^{\alpha_{m}}\frac{1}{\sqrt{v_{0}^2-\frac{4}{5}k_{1}k_{2}\alpha^\frac{5}{2}}}\mathrm{d}\alpha = 2\eta\frac{\alpha_{m}}{v_{0}} |
where4
(11) | \eta = \int \limits_0^1 \frac{1}{\sqrt{1-\epsilon^\frac{5}{2}}}\mathrm{d}\epsilon = 1.4716 |
Compression as a function of time can therefore be determined as follows:5
(12) | \alpha = \alpha_{m}\sin{\frac{1.068v_{0}t}{\alpha_{m}}} |
The force-time-correlation can thus be approximated as:1
(13) | \[F =\begin{cases} \frac{1.140}{k_{1}\alpha_{m}}\sin{\frac{1.068v_{0}t}{\alpha_{m}}} \quad \text{for } 0\leq t \leq \frac{\pi\alpha_{m}}{1.068v_{0}}\\ 0 \quad \text{for } t > \frac{\pi\alpha_{m}}{1.068v_{0}} \end{cases} \] |
The shape and size of the compression-figure can now be computed analogue to the static case.