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546 Appendix F Solving for pressure in CAPRI (the solver developed for similar materials in Purdue) involves finding the correct (a, e) pair corresponding to the applied force and moment. This is typically an iterative or search-type process. Guess values of “a” smaller than the actual value, for example, will lead to a pressure traction whose resultant is smaller than the applied force P. Fourier decompositions like the one in Equation (F.15) can be done very rapidly using the discrete sine-transform. Shear solution The case of full-sliding is trivially solved qx =px. However, different possibilities occur while computing shear tractions in partial sliding. The assumption is that there is a central stick zone with two slip zones, one at either end; the size and eccentricity of the stick zone is unknown a priori. Depending on the value of applied bulk-stress, the signs of shear in the slip-zones may be the same or opposing, positive or negative, which makes four possibilities in all. Positive shear stresses in both slip zones As described in [7], the shear traction is expressed as a superposition of two tractions qX =q X −q X =pX −q X (F.23) Clearly, q X must be zero in the slip zones. The term “dg/dx,” as described earlier is a history dependent term. In the very first shear loading step, it will be zero. In [2], a detailed analysis is carried out, considering shear-loading, unloading and re-loading, while keeping normal load constant. It is pointed out that after loading, relative slip is measured from the displacement state at Q and 0 for unloading. Here only the simple case when there is no prior history is considered. Substituting Equation (F.23) into Equation (F.3) and setting dg/dx =0, the stick-zone equation becomes 0 = 41− 2 E +a −a ps x −s ds− 41− 2 E +a −a q s x −s ds− 0 1− 2 E (F.24) Substituting for the pressure term and re-arranging, dh dx −C 1 − 0 1− 2 E = 41− 2 E +a −a q s x −s ds (F.25) which is similar in form to the original pressure equation, but with C 1 and dh/dx modified. Additionally, the equilibrium boundary condition for this equation is +a −a q x dx =P −Q (F.26) Appendix F 547 Also, q x has to vanish at both stick-slip boundaries; global coordinates are to be assumed in the following expression q −c +e c =q c +e c =0 (F.27) The solution procedure consists of finding c and e c such that all the prescribed conditions are satisfied. Different signs of shear in the two-slip zones This occurs when the bulk-stress exceeds a certain critical value in either direction; in this case, again qx is split into two parts as follows [2, 8]: q X =pX X ∈ −a +e−c +e c (F.28a) q X =p−c +e c −X +c −e c × pc +e c +p−c +e c 2c X ∈stick zone (F.28b) q X =−pX X ∈c +e c a+e (F.28c) The term in the middle, Equation (F.28b), simply represents a linear interpolation of the shear stress in the stick zone. Of course, the other possibility of different signs of shear is negative in the left slip-zone and positive in the right slip-zone. The corrective shear traction can be found again by using the principle that in the stick zone, no relative sliding can occur. The assumed shear function q x, however, produces a slope of displacement dg x/dx; for the case with no history, dg x dx − 0 1− 2 E = 41− 2 E +a −a q s x −s ds (F.29) subject to the equilibrium boundary condition +a −a q x dx =Q −Q (F.30) Again, the conditions in Equation (F.27) apply as boundary conditions for the corrective shear. This equation can also be solved using the sine-series technique described for pressure. For further details see [2]. History effects and general loading cases In most general loading case, it is possible there may be slip on only one end or no slip at all. An example loading case is when P and Q are applied together [9]. The “history” 548 Appendix F or path-dependence comes about due to growth of the stick zone into areas that were previously slipping (“slip-lock”). In such cases, a direct solution as outlined above cannot be used; the load step has to be broken into increments and the slope of slip dg/dx has to be updated at the end of every increment. CONTACT STRESSES IN A FINITE THICKNESS BODY, THE CARTEL CODE The analytical solution of the surface tractions produced by an indenter on a half space can be solved using SIEs [10] as outlined in the first section of this Appendix. Bodies that are dissimilar but still isotropic can still be solved analytically even with the coupling effects between P and Q [9]. The corresponding subsurface stress field can be obtained from the elasticity solution of any arbitrary traction on a half-space. The complete stress field due to any traction applied to the surface of a half-space can be solved using Fast Fourier Transforms (FFTs). SIE cannot be used when the elastic half-space is replaced by an infinite length sheet with finite thickness [11, 12]. SIEs have been derived to account for this finite thickness by [16] assuming a rigid pad indenting a finite thickness infinite layer. The derivation of the elasticity solution for the stress field of a half space due to any traction will be used in order to develop a numerical solution of the tractions on any finite thickness body [6, 13]. Previous researchers have studied the stress field of an infinite half-space using many techniques such as complex analysis [14] and Fourier transform technique [4] among others. For the FFT technique, the solution has been derived by satisfying the two- dimensional biharmonic equation, also known as the Airy stress function by using the compatibility relation in the absence of body forces and Fourier techniques [15]. The two-dimensional biharmonic equation 4 1 =0, can be rewritten in the frequency domain by taking the Fourier transform. − 4 1 e −ix dx = d 2 dy 2 − 2 2 − e −ix dx =0 x y = 1 2 − Gy e −ix d (F.31) Using the substitution of variables, the biharmonic equation can be rewritten as a second- order partial differential equation with a general solution, G = − e ix dx d 2 dy 2 − 2 2 G =0 G =A +Bye −y +C +Dye y (F.32) Appendix F 549 where A, B, C, and D are constants to be determined from boundary conditions. From the Fourier inversion theorem, the Airy stress function can be written as x y = 1 2 − Gy e −ix d (F.33) The stresses in rectangular coordinates can be written in terms of the Airy stress function using the equations of equilibrium. The stresses can then be rewritten in terms of G from the above Fourier technique. x = d 2 dy 2 − x e ix dx = d 2 G dy 2 x = 1 2 − d 2 G dy 2 e −ix d y = d 2 dx 2 − y e ix dx =− 2 G y =− 1 2 − 2 Ge −ix d (F.34) xy = d 2 dxdy − xy e ix dx =i dG dy xy = 1 2 − i dG dy e −ix d In a similar manner, the displacement field can be derived from the stress–strain relations. E 1+ u x =1 − x − y E 21 + u y + v x = xy (F.35) The displacements can also be rewritten in terms of G by substituting the general solution of the stress field found: u = 1+ 2E − 1− d 2 G dy 2 + 2 G ie −ix d v = 1+ 2E − 1− d 3 G dy 3 +2 + 2 dG dy e −ix d 2 (F.36) The above general stress- and displacement-traction relationships can be used to solve for the subsurface stresses and the surface tractions of two bodies in contact. Half-space solution The particular solution of the subsurface stress field can be evaluated by obtaining the correct boundary conditions (Figure F.2). These boundary conditions can then be applied to the stress and displacement field equations to solve for A, B, C, and D. For example, if the semi-infinite plate has a normal traction −px applied to the surface, then the 550 Appendix F Half-space p(x),q(x) x y 2a Finite Thickness Infinite Sheet p(x),q(x) –p(x),– q(x) x y 2b 2a Figure F.2. The coordinate system and applied boundary conditions for the half-space and finite thickness problem. boundary conditions in Equation (F.38) can be used to solve for G. The results can then be used in order to solve for the corresponding stress field. y =−px at y =0 xy =0aty =0 (F.37) x = y = xy →0 when y → From the conditions at infinity, C =0 and D =0 is obtained. From the conditions at y =0 the other two constants can be found: A = ¯p 2 B = ¯p (F.38) The particular solution of a normal traction on a half-space can now be obtained from the particular solution of G. x = 1 2 − −¯p 1−y e −y e −ix d y =− 1 2 − ¯p 1−y e −y e −ix d (F.39) xy = 1 2 − −i¯pye −y e −ix d Appendix F 551 The stresses due to a shear traction qx can be obtained by applying the boundary conditions in a similar manner. x =0aty =0 xy =qx at y =0 (F.40) x = y = xy →0 when y → The constants C and D are again zero, however, A and B are different in this case. A =0 B = ¯q i (F.41) The particular solution of a shear traction on a half-space can now be obtained from the particular solution of G. x = 1 2 − ¯q i −2 + 2 y e −y e −ix d y =− 1 2 − 2 ¯q i ye −y e −ix d (F.42) xy = 1 2 − i ¯q i 1−y e −y e −ix d Similarly, the solution for the displacement field of the body can be obtained. v p x = 1+ 2E − ¯p 21 − e −ix d v q x = 1+ 2E − −i¯q 5−2e −ix d u p x = 1+ 2E − −i¯q 1−2e −ix d u q x = 1+ 2E − −¯q 21 −e −ix d (F.43) Finite thickness solution In order to apply a similar methodology for solving the analytical solution of a layered body subjected to a normal and shear traction, the correct boundary values to the general solution must be chosen (Figure F.3). The boundary conditions stay the same on the 552 Appendix F –p(x) 2a p(x) h(x ) δ v 1 (x ) v 2 (x) h(x) – δ = v 1 (x) + v 2 (x) –a < x < a Figure F.3. Relationship between gap function h(x) and vertical displacements, v i x, due to normal traction p(x). surface and symmetry is used at the half thickness. First, the subsurface stresses due to a normal traction give the boundary conditions y =−px at y =b xy =−px at y =b vx =0aty = 0 xy =0aty =0 (F.44) From the symmetry boundary conditions, B =−D and A =C can be obtained. Similarly, from the surface boundary conditions the other two constants are obtained: A = 1 +b coshb sinhb B B = ¯p 2 2 sinhb b +coshb sinhb (F.45) The constants for a shear traction on the surface can be obtained in a similar manner. y =0aty = b xy =qx at y = b vx =0aty =0 xy =0aty = 0 (F.46) Appendix F 553 From the symmetry boundary conditions, B =−D and A =C are obtained again. How- ever, from the surface boundary conditions, A and B are obtained in terms of the shear traction ¯q, and the thickness b. B = coshb b sinhb A A = i¯q 2 2 b sinhb b +coshb sinhb (F.47) The equations in this form can be used to solved for the subsurface stress field, Equation (F.34), of any finite thickness infinite sheet subjected to a given shear and normal traction without iterating. However, a different approach must be implemented from the above relationships in order to obtain the particular surface tractions. Surface tractions The surface tractions can be obtained by applying a different set of boundary conditions and by deriving a different set of relations between the surface displacements and the corresponding tractions. The equations in this form can be used to solved for the subsurface stress field of any finite thickness infinite sheet subjected to a given shear and normal traction [16]. The relationship between the displacements and the surface tractions can be used to obtain a solution for the surface tractions, Equation (F.36). At the surface where y = b, the total displacement u and v due to each traction can be written as a function of the two forms of G from the normal and shear tractions with the constants corresponding to Equations (F.45) and (F.47). vx = 1+ 2E − −¯p 21 −sinh 2 b b +coshb sinhb e −ix d + 1+ 2E − −i¯q 5−2 − 21 −b b +coshb sinhb e −ix d ux = 1+ 2E − −i¯p 1−2 − 21 −b b +coshb sinhb e −ix d + 1+ 2E − ¯q 21 −cosh 2 b b +coshb sinhb e −ix d (F.48) The above displacement-traction relationships may be used for deriving the solution for the tractions inside the contact zone. It must be noted that Equation (F.48) differs slightly from previous results [9, 17]. The vertical displacement vx due to qx has a 5−2 term instead of 1 −2. This result can be first compared to the half-space solution in 554 Appendix F Equation (F.43). The solution for vx in Equation (F.48) must be equal to the solution in Equation (F.43) for a large thickness b for any given shear stress qx. This is true since the hyperbolic sine and cosine terms for vx go to zero for b →in Equation (F.48). This result was also validated with the same procedure using the finite thickness constants from Equation (F.47). This difference is shown for completeness if the correct vertical displacement must be calculated correctly for a half-space or a finite thickness infinite length plate. The shear traction qx acts equal and opposite on each body in contact, therefore, the 5 −2 half-space term must disappear when the gap function h 0 x is calculated from the relative displacement v 1 −v 2 of the two bodies, otherwise, the shear traction qx would influence the normal traction px even for a half-space As previous work has shown [10], the displacement field can be used to derive the tractions. When two bodies come in contact with one another, there is a gap function, hx, which describes the distance the two bodies are away from each other at each point before deformation (Figure F.3). If one body is allowed to inter-penetrate the other by a rigid body displacement dv, then there is a normal traction, px, which can be applied equal and opposite to each body which would produce no inter-penetration inside the contact zone, and one which integrates to the total load P, Equation (F.49). x≤ahx =v 1 x +v 2 x + v1 + v2 =v p1 px +v q1 qx +v p2 px +v q2 qx + v (F.49) x≥a px =0 qx =0 P = +a −a pxdx The boundary condition to obtain the shear traction is slightly different (Figure F.4). The shear traction is defined to be qx =px in the slip zone. The relative horizontal slip can next be assumed to be ux = u in the stick-zone, where u is the amount of relative q(x) g (x) u 1(x) u 2(x) g(x ) – δu = u 1 (x ) + u 2 (x ) –c < x < c δu 2c 1 2 1 2 1 2 μ p μ p g(x) = 0 Figure F.4. Relationship between slip function gx and horizontal displacements u i x due to a shear traction qx. Appendix F 555 tangential displacement between the pad and specimen before the shear traction is applied. By specifying the stick-zone size 2c and knowing the total shear load Q applied, the shear traction can be iterated for simultaneously with the normal traction in such a way that an equal and opposite shear traction on each body causes there to be no inter-penetration, and the shear traction is continuous throughout the contact zone, Equation (F.50). x≤c gx =u 1 x +u 2 x + u1 + u2 =u p1 px +u q1 qx +u p2 px +u q2 qx + u a ≥x> c qx =px x≥c px = 0 qx =0 Q = +a −a qxdx (F.50) When there is a moment M present, then the pad profile h 0 x is rotated by an angle m such that the resulting normal traction satisfies the moment boundary condition M for a particular pad rotation angle, Equation (F.51). M = +a −a xpxdx (F.51) Note that the equilibrium conditions for contact on a finite thickness plate, given in Equations (F.49), (F.50) and (F.51), are identical to the equilibrium conditions for the half-space given earlier as Equations (F.5), (F.6), and (F.7). Bulk stress effect can be accounted for by assuming that there is a constant strain in the stick zone due to the remote bulk stress that causes each point to displace by an amount based on the property of the material, Equation (F.52). XX = bulk 1− 2 E u 1 x = bulk 1− 2 E x + u1 (F.52) Unlike the half-space solution, however, this solution methodology is similar to that of the traction solution for dissimilar materials and anisotropic materials because the shear and normal tractions must be calculated simultaneously [12]. There is no need to account for singularities, however, because SIEs are not used here. Unfortunately, there are many other convergence issues to be resolved when working within the frequency domain. By knowing the profile, h 0 x, material properties, E , and the total load applied, (P, Q, M, bulk ), the normal and shear tractions, pxqx can be iterated for by assuming . zero, however, A and B are different in this case. A =0 B = ¯q i (F.41) The particular solution of a shear traction on a half-space can now be obtained from the particular solution of G. x = 1 2 − ¯q i −2. size 2c and knowing the total shear load Q applied, the shear traction can be iterated for simultaneously with the normal traction in such a way that an equal and opposite shear traction on each. dissimilar materials and anisotropic materials because the shear and normal tractions must be calculated simultaneously [12]. There is no need to account for singularities, however, because SIEs are
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