Plane Thermoelastic Waves in Infinite Half-Space Caused FACTA UNIVERSITATIS Series: Mechanical Engineering Vol. 14, N o 2, 2016, pp. 159 - 168 Original scientific paper 1REDUCTION OF RESIDUAL SHEAR STRESS IN THE LOADED CONTACT USING FRICTION HYSTERESIS UDC 531.4 Adrian Kraft, Roman Pohrt Berlin Institute of Technology, Germany Abstract. We investigate the tangential contact problem of a spherical indenter at constant normal force. When the indenter is subjected to tangential movement, frictional shear stresses arise at the interface and do not vanish when it is moved backwards. We study the evolution of shear stress when the indenter is moved back and forth at falling amplitude. The method of dimensionality reduction (MDR) is employed for obtaining the distribution of stick and slip zones as well as external forces and the final stress distribution. We find that the shear stress decreases. For the special case of linearly falling amplitude of the movement, we observe uniform peaks in the shear stress. The absolute value of the shear stress peaks is reduced best for a high number of back-and- forth-movements with slowly decreasing amplitude. Key Words: Coulomb Friction, Residual Stress, Contact Mechanics, Stick, Slip 1. INTRODUCTION Loaded contacts exist in a broad variety of technical and natural situations. A basic frictional couple consists of two deformable bodies, which are pressed together and subjected to an additional tangential loading. When the tangential load is below the threshold of gross sliding, some micro-slip occurs nevertheless. In technical contact that experience cyclic loading, the relative movement of the bodies is often invisible to the naked eye. However, the friction involved can cause effects such as damping [3, 4] and fretting wear [8]. The same applies in a slightly more complicated fashion for biaxial loading [5]. Consider a contact where the normal load is held constant, so that the bodies cannot separate. Initially, the pure normal loading can take place without inducing shear stress in Received February 24, 2016 / Accepted July 10, 2016 Corresponding author: Roman Pohrt Technische Universität Berlin, Straße des 17. Juni 135, 10623 Berlin, Germany E-mail: roman.pohrt@tu-berlin.de 160 A. KRAFT, R. POHRT the contact interface. However, when some tangential load is applied, shear stress arises. Upon unloading, the stress does not simply vanish due to the dual nature of the contact (stick zones vs. slip zones). Tangential force and resulting tangential displacement together form a system with hysteresis, which can be described with a Preisach formalism [10]. In 1928, Prandtl developed a model for microtribology out of many micro-sliders, which also included hysteresis [12] (see [11] for English translation of the original paper). In his model, the exact state of the system at any moment of time depends on its prehistory and can be very complicated. Prandtl poses the question of whether it is possible to restore the virgin state. With very simple arguments, he shows that if the contact partners start to oscillate with large amplitude and the amplitude then decreases slowly, then each slider finally comes to a neutral, non-stressed position and the system returns to the virgin state. He compares this result with demagnetization through slowly decreasing oscillating magnetic fields, first studied by E. Madelung [6]. In this paper, we will apply this idea to the loaded contact with partial sliding. Assume a typical Hertzian contact. A spherical indenter with radius R is pressed into a plane. The sphere is approximated by a parabola and we assume that the plane is rigid. We keep normal force FN that we first applied constant throughout the following. This configuration is known as the Hertzian contact with contact radius a and has been studied in great detail. Normal stress distribution p as a function of radial coordinate r reads 1 2 2 max 2 2 / 3 1/ 3 * N max r p(r) p 1 a (6F ) E p R                . (1) Here we used the reduced modulus of elasticity: 2 2 1 2 * 1 2 1 11 E E E      , (2) for two elastic bodies with moduli of elasticity E1 and E2 , shear moduli G1 and G2 and Poisson's ratios 1 and 2, respectively. The equivalent reduced shear modulus is: 2 2 1 2 * 1 2 1 11 G G G      . (3) We now move the indenter horizontally along one spatial direction and assume that Coulombs law of friction is valid in its simplest form τ(r) ≤ μp(r). Here τ is the shear stress and μ is the coefficient of friction. This condition leads to the partitioning of the contact area into a stick area in the center and a slip area at the edges of it. For example, at the very edge of the contact surface, normal stress disappears and the slightest displacement of the indenter would lead to shear stress violating our friction condition if we assumed it would stick to the plane. The solution of this is that some parts of the contact area are being slipped over the plane while the others stick to it. At a certain point, the stick area completely disappears and the whole contact is slipping. Tangential displacement ux (0) at this last point reads [2]: Reduction of Residual Shear Stress in the Loaded Contact Using Friction Hysteresis 161 (0) *x m x * a u E G   . (4) Let us now drag the indenter back and forth along the same spatial direction on the rigid plane. We observe different shear stress distributions depending on the motion history. The shear stress created from the first movement will not disappear if we simply return to our starting point. In the following, we will analyze the shear stress under a back and forth movement with falling amplitude with the aim of reducing this stress. For the analysis of the system, the Aleshins method of memory diagrams is well suited. Because the contact stress distribution accumulates more and information of past reversal points, the HH-mode described in sec. 31 of [9] should be applied iteratively. Using careful algebra, an explicit analytical solution might be achievable. However, we decided to use the method of dimensionality reduction for our analysis, which we describe in the next section. Alternatively, one could employ a full Boundary Elements solution [7]. 2. MDR AND DISCRETIZED 1D-MODEL We consider the three-dimensional contact of two elastic bodies. In the following, we presuppose axially symmetric profiles. Let z=f (r) be the difference between the profiles of the bodies. According to the theorems of the method of dimensionality reduction (MDR), this contact can be exactly replaced by a contact with a one-dimensional linearly elastic foundation with independent springs. To reduce the initial three-dimensional contact to a one-dimensional one, two steps are required. First, we replace the elastic bodies by the one-dimensional linearly elastic foundation. Normal stiffness Δkz and tangential stiffness Δkx of the springs are chosen according to: * z Ek x   (5) and * x k xG   , (6) where x denotes the distance between two springs. Second, we replace the three-dimensional profile z=f(r) with a one-dimensional profile in accordance to [2]: x 2 2 0 f (r) g(x) x dr x r     . (7) The reverse transformation is: r 2 2 0 2 g(x) f (r) dx r x   . (8) If now transformed profile g(x) is pressed with normal force FN and resulting indentation depth d into the elastic foundation, we obtain the displacement in the contact area: z u (x) d g(x)  . (9) 162 A. KRAFT, R. POHRT For non-adhesive contacts we can set: z a) 0 d (au g( : )    , (10) where a is the contact radius. We again denote the length a the ‘contact radius’ because following the theorems of MDR, all the lengths in the model are equal to the respective ones in a three-dimensional problem. The same applies to the contact area. The force of a spring at point x inside the contact area equals: z z z z (x) u (x) E * u (xF k x.)   (11) If we choose the spring separation distance to be infinitesimal, we get for the normal force: * a a a 0 * N z F : E u (x)dx 2E (d g(x))dx.      (12) After bringing the elastic bodies in normal contact with force FN, we apply tangential movement ux (0) to the indenter and assume that Coulomb's law of friction is valid following: (r) p(r) for stick,   (13) (r) p(r) for slip.  (14) Now every spring sticks to the indenter as long as ΔFx=Δkx ux (0) < μ ΔFz. This results in the following conditions: (0) x x x x z z z z x (x) , if k u (x) u (x) (x) u u F k F u , k (x) in a state of slip.         (15) where ux(x) denotes here the horizontal displacement of a spring at point x. The sign of the second equation depends on the direction of the indenter motion. We denote the radius of the stick-area by c and find: * *) x (0 E (d g(G )).u c  , (16) for which the whole contact area is in slip state. Similarly to the normal contact, the force of a spring at point x inside the contact area equals: * xx xx (x) u (x) G u (x)F k x    . (17) If we choose the spring separation distance to be infinitesimal, we get for the tangential force: * a x x a F G u (x)dx.    (18) Reduction of Residual Shear Stress in the Loaded Contact Using Friction Hysteresis 163 The distribution of tangential stress in the original three-dimensional contact can be calculated according to [2]:       r 22 x * zr dx rx )x(uG )r( . (19) For our problem, we choose the classical contact between a sphere with radius R and a plane. The modified profile of z=f(r)=r 2 /2R is g(x)=x 2 /R. Normal and tangential stiffness follow Eqs. (5) and (6). With a given indentation depth d, we calculate contact radius a according to Eq. (10). The length of the elastic foundation is set to 2a and consists of NS springs. Distance Δx between two adjacent springs is 2a/NS. The normal displacement is uz(x)=d-g(x)=d-x 2 /R. For the normal force, we get sN N k 1 z F 2 u (k)E x*    . With a given indenter displacement ux (0) , we first assume that every spring sticks. Next, we check if tangential spring force is exceeding the maximum tangential force according to the first part in Eq. (15) and where required, we adjust the displacement. For the tangential force, we get    SN 1k * xx xG)k(u2F . 3. REDUCTION OF RESIDUAL SHEAR STRESS MAGNITUDE We push a spherical indenter into an elastic plane and drag it horizontally. If we drag it not too far to one side, our contact area will be divided in a stick-area in the center and a slip-area on the outside. The stick-area will have the shape of a circle and the slip-area will encircle it in the shape of an annulus. By moving it further in the chosen direction, the stick-area gets smaller and smaller before it disappears completely and we reach macroscopic slipping. If we move the indenter in one direction without reaching macroscopic slipping and subsequently perform a small displacement in the opposite direction, we will observe that: the former stick-area is still in stick-state. More inner parts of the former slip-area are now also in stick-state. The outermost border of the former slip-area is in slip-state again. Thus, we will observe different displacement areas: in the center, there will be a circle where the displacement equals the macroscopic displacement of the indenter, then there is an annulus in a stick-state, which was at some former time in a slip-state, and finally on the outermost border there will always be an annulus in slip-state. That is, by simply alternating the direction of the indenter several times, we can create a strongly fragmented displacement field. If now, for example, we drag the indenter until the whole contact is in slip-state, we could create a growing stick-area with a fragmented displacement field, by simply moving in a certain back-and-forth motion with falling amplitudes. As the displacement field determines the shear stress distribution, we can manipulate the stress by moving the indenter in a given way. In order to minimize the shear stress, the following procedure is discussed. 164 A. KRAFT, R. POHRT On a rigid plane base, we push a spherical indenter until an indentation depth d and related normal force FN are reached. We keep this configuration for the normal contact throughout the following procedure. The origin of coordinate ux (0) is placed on the rigid plane. Subsequently we displace the indenter horizontally by uxmax (0) in ux (0) - direction so that the stick area disappears completely according to Eq. (4) and the indenter reaches macroscopic slipping. Now we drag the indenter back and forth with linearly falling amplitude around the origin of ux (0) -coordinate. Let NR be the total preselected amount of reversal points and k the current reversal point, then the analytic expression describing the position of the indenter relative to the rigid plane is:       (0) ( k 1)0 x max x R R R u u (k) (N 1 k) 1 for k 1, 2,..., N 1 N        . (20) For example, if we choose NR=1, the indenter moves to uxmax (0) and then back to 0. If we choose NR=2, the indenter goes to ux (0) =uxmax (0) , then to - uxmax (0) /2 and back to 0. Hence, the whole procedure is path-controlled by sequence ux (0) (k) . Shear stress τ(r) and tangential force Fx are only evaluated at each reversal point. Note that Eq. (20) only describes these points of the movement. The inertial forces of the indenter are not considered and the problem is seen as quasi-static. Therefore, it does not matter in what exact fashion the indenter moves back and forth. Instead of moving linearly, it might as well follow a sinusoidal movement, as depicted in Fig. 1. Finally, it should be pointed out that the movement always ends (for k=NR+1) at starting point ux (0) =0. Fig. 1 Normalized position of the indenter on the rigid plane versus the reversal points. uxmax (0) is the minimum displacement to reach macroscopic slipping. Total amount of reversal points is 20. Reduction of Residual Shear Stress in the Loaded Contact Using Friction Hysteresis 165 Fig. 2 Normalized tangential force versus reversal points. Total amount NR of reversal points is 20. FN is the constant normal force As aforementioned, we have chosen to move the indenter path-controlled. From this, we obtained the tangential forces at each reversal point. Fig. 2 shows the absolute value of the alternating forces. We would get exactly the same motion if we picked our reversal points according to this force sequence. For example the first reversal point (k=1) was defined so that we reach macroscopic slipping, that is: Fx= μFN. Fig. 3 Shear stress in the contact area after 6 and 6.5 reversal points (A and B in [6]). a is the contact radius and pmax the maximum normal stress in the contact surface of a spherical indenter and a rigid plane. Total amount NR of reversal points is 20. 166 A. KRAFT, R. POHRT At every state of the movement, we can calculate the shear stress distribution according to Eq. (19). Fig. 3 shows the stress distribution after six back and forth movements (continuous line; point A in Fig. 1), thus exactly at a reversal point. We can see that inside the outer ring of the contact area (approximately 0.5a