Engineering Tribology 2011 Part 8 pptx

ELASTOHYDRODYNAMIC LUBRICATION 291 It can be noted that for the spheres: R ax = R ay = R A and R bx = R by = R B where: R A and R B are the radii of the spheres ‘A’ and ‘B’ respectively. Substituting into equation (7.2) gives: = 1 R' + 1 R x = 1 R y + 1 R A + 1 R B + 1 R A 1 R B = 2 () + 1 R A 1 R B (7.7) where: = 1 R x = 1 R y + 1 R A 1 R B It should be noted that in some publications the reduced radius for the contact between two spheres is defined as: 1 R' =+ 1 R A 1 R B and consequently the formulae for the contact area dimension ‘a’ and the maximum deflection ‘δ’ are presented in the slightly altered form: a = 3WR' 2E' () 1/3 δ= 1.31 W 2 E' 2 R' () 1/3 , EXAMPLE Find the contact parameters for two steel balls. The normal force is W = 5 [N], the radii of the balls are R A = 10 × 10 -3 [m] and R B = 15 × 10 -3 [m]. The Young's modulus for both balls is E = 2.1 × 10 11 [Pa] and the Poisson's ratio of steel is υ = 0.3. · Reduced Radius of Curvature Since R ax = R ay = R A = 10 × 10 -3 [m] and R bx = R by = R B = 15 × 10 -3 [m] the reduced radii of curvature in the ‘x’ and ‘y’ directions are: = 1 R x + 1 R ax 1 R bx = 1 10 × 10 −3 + 1 15 × 10 −3 = 166.67 ⇒ R x = 6 × 10 −3 [m] = 1 R y + 1 R ay 1 R by = 1 10 × 10 −3 + 1 15 × 10 −3 = 166.67 ⇒ R y = 6 × 10 −3 [m] Note that 1/R x = 1/R y , i.e. condition (7.3) is satisfied (circular contact), and the reduced radius of curvature is: 292 ENGINEERING TRIBOLOGY 1 R' =+ 1 R x 1 R y = 166.67 + 166.67 = 333.34 ⇒ R' = 3 × 10 −3 [m] · Reduced Young's Modulus = 1 E' + 1 −υ A 2 E A [] 1 2 1 −υ B 2 E B =+ 1 − 0.3 2 2.1 × 10 11 [] 1 2 1 − 0.3 2 2.1 × 10 11 ⇒ E' = 2.308 × 10 11 [Pa] · Contact Area Dimensions a = 3WR' E' () 1/3 = 3 × 5 × (3 × 10 −3 ) 2.308 × 10 11 () 1/3 = 5.799 × 10 −5 [m] · Maximum and Average Contact Pressures p max = 3W 2πa 2 = 3 × 5 2π(5.799 × 10 −5 ) 2 = 709.9 [MPa] p average = W πa 2 = 5 π(5.799 × 10 −5 ) 2 = 473.3 [MPa] · Maximum Deflection δ= 1.0397 W 2 E' 2 R' () 1/3 = 1.0397 5 2 (2.308 × 10 11 ) 2 3 × 10 −3 () 1/3 = 5.6 × 10 −7 [m] · Maximum Shear Stress τ max = 1 3 p max = 1 3 709.9 = 236.6 [MPa] · Depth at which Maximum Shear Stress Occurs z = 0.638a = 0.638 × (5.799 × 10 −5 ) = 3.7 × 10 −5 [m] · Contact Between a Sphere and a Plane Surface The contact area between a sphere and a plane surface, as shown in Figure 7.9, is also circular. The contact parameters for this configuration can be calculated according to the formulae summarized in Table 7.1. The radii of curvature of a plane surface are infinite and symmetry of the sphere applies so that R bx = R by = ∞ and R ax = R ay = R A . The reduced radius of curvature according to (7.2) is therefore given by: = 1 R' + 1 R x = 1 R y + 1 R A + 1 ∞ + 1 R A 1 ∞ = 2 R A (7.8) ELASTOHYDRODYNAMIC LUBRICATION 293 Body B R A W Circular contact area Body A W a FIGURE 7.9 Contact between a sphere and a flat surface. where: R x = R y = R A EXAMPLE Find the contact parameters for a steel ball on a flat steel plate. The normal force is W = 5 [N], the radius of the ball is R A = 10 × 10 -3 [m], the Young's modulus for ball and plate is E = 2.1 × 10 11 [Pa] and the Poisson's ratio is υ = 0.3. · Reduced Radius of Curvature Since the radii of the ball and the plate are R ax = R ay = 10 × 10 -3 [m] and R bx = R by = ∞ [m] respectively, the reduced radii of curvature in ‘x’ and ‘y’ directions are: = 1 R x + 1 R ax 1 R bx = 1 10 × 10 −3 + 1 ∞ = 100 ⇒ R x = 0.01 [m] = 1 R y + 1 R ay 1 R by = 1 10 × 10 −3 + 1 ∞ = 100 ⇒ R y = 0.01 [m] Condition (7.3), i.e. 1/R x = 1/R y is satisfied (circular contact), and the reduced radius of curvature is: 1 R' =+ 1 R x 1 R y = 100 + 100 = 200 ⇒ R' = 5 × 10 −3 [m] · Reduced Young's Modulus E' = 2.308 × 10 11 [Pa] 294 ENGINEERING TRIBOLOGY · Contact Area Dimensions a = 3WR' E' () 1/3 = 3 × 5 × (5 × 10 −3 ) 2.308 × 10 11 () 1/3 = 6.88 × 10 −5 [m] · Maximum and Average Contact Pressures p max = 3W 2πa 2 = 3 × 5 2π(6.88 × 10 −5 ) 2 = 504.4 [MPa] p average = W πa 2 = 5 π(6.88 × 10 −5 ) 2 = 336.2 [MPa] · Maximum Deflection δ= 1.0397 W 2 E' 2 R' () 1/3 = 1.0397 5 2 (2.308 × 10 11 ) 2 5 × 10 −3 () 1/3 = 4.7 × 10 −7 [m] · Maximum Shear Stress τ max = 1 3 p max = 1 3 504.4 = 168.1 [MPa] · Depth at which Maximum Shear Stress Occurs z = 0.638a = 0.638 × (6.88 × 10 −5 ) = 4.4 × 10 −5 [m] · Contact Between Two Parallel Cylinders The contact area between two parallel cylinders is circumscribed by a narrow rectangle. The geometry of parallel cylinders in contact is shown in Figure 7.10 and the formulae for the main contact parameters are summarized in Table 7.2. T ABLE 7.2 Formulae for contact parameters between two parallel cylinders. b = 4WR' πlE' () 1/2 rectangle p max = W πbl Elliptical pressure distribution p average = W 4bl δ= 0.319 W E'l () τ max = 0.304p max at a depth of z = 0.786b Contact area dimensions Average contact pressure Maximum shear stress Maximum contact pressure Maximum deflection 2l 2b 2 3 [)] + ln ( 4R A R B b 2 × where: b is the half width of the contact rectangle [m]; ELASTOHYDRODYNAMIC LUBRICATION 295 l is the half length of the contact rectangle [m]; R' is the reduced radius of curvature for the two parallel cylinders in contact [m]. For the cylinders: R ax = R A , R ay = ∞, R bx = R B , R by = ∞ where ‘R A ’ and ‘R B ’ are the radii of the cylinders ‘A’ and ‘B’ respectively. Substituting into equation (7.2) yields: = 1 R' + 1 R x = 1 R y + 1 R A + 1 R B + 1 ∞ 1 ∞ =+ 1 R A 1 R B (7.9) where: 1 R x = 1 R A + 1 R B and 1 R y = 0 The rest of the parameters are as defined for Table 7.1. R A R B W Rectangular contact area Body A W Body B 2b 2l 2l 2b FIGURE 7.10 Geometry of the contact between two parallel cylinders. EXAMPLE Find the contact parameters for two parallel steel rollers. The normal force is W = 5 [N], radii of the rollers are R A = 10 × 10 -3 [m] and R B = 15 × 10 -3 [m], Young's modulus for both rollers is E = 2.1 × 10 11 [Pa] and the Poisson's ratio is υ = 0.3. The length of both rollers is 2l = 10 × 10 -3 [m]. 296 ENGINEERING TRIBOLOGY · Reduced Radius of Curvature Since the radii of the cylinders are R ax = R A = 10 × 10 -3 [m], R ay = ∞ and R bx = R B = 15 × 10 -3 [m], R by = ∞ respectively, the reduced radii of curvature in the ‘x’ and ‘y’ directions are: = 1 R x + 1 R ax 1 R bx = 1 10 × 10 −3 +=166.67 1 15 × 10 −3 ⇒ R x = 6 × 10 −3 [m] = 1 R y + 1 R ay 1 R by =+ 1 ∞ = 0 1 ∞ ⇒ R y = ∞[m] Since 1/R x > 1/R y condition (7.3) is satisfied and the reduced radius of curvature is: 1 R' = 1 R x = 166.67 ⇒ R' = 6 × 10 −3 [m] · Reduced Young's Modulus E' = 2.308 × 10 11 [Pa] · Contact Area Dimensions = 4 × 5 × (6 × 10 −3 ) π × (5 × 10 −3 ) × (2.308 × 10 11 ) () 1/2 = 5.75 × 10 −6 [m]b = 4WR' πlE' () 1/2 · Maximum and Average Contact Pressures p max = W πbl = 5 π × (5.75 × 10 −6 ) × (5 × 10 −3 ) = 55.4 [MPa] p average = W 4bl = 5 4 × (5.75 × 10 −6 ) × (5 × 10 −3 ) = 43.5 [MPa] · Maximum Deflection = 2.40 × 10 −8 [m] δ= 0.319 W E'l [] 2 3 [)] + ln ( 4R A R B b 2 = 0.319 [] 2 3 [)] + ln ( 4 × (10 × 10 −3 ) × (15 × 10 −3 ) (5.75 × 10 −6 ) 2 5 (2.308 × 10 11 ) × (5 × 10 −3 ) · Maximum Shear Stress τ max = 0.304p max = 0.304 × 55.4 = 16.8 [MPa] · Depth at which Maximum Shear Stress Occurs z = 0.786b = 0.786 × (5.75 × 10 −6 ) = 4.5 × 10 −6 [m] ELASTOHYDRODYNAMIC LUBRICATION 297 · Contact Between Two Crossed Cylinders With Equal Diameters The contact area between two cylinders with equal diameters crossed at 90° is bounded by a circle. This configuration is frequently used in wear experiments since the contact parameters can easily be determined. The contacting cylinders are shown in Figure 7.11 and the contact parameters can be calculated according to the formulae summarized in Table 7.1. Circular contact area a R B W Body B W R A Body A FIGURE 7.11 Geometry of the contact between two cylinders of equal diameters with axes perpendicular. Since R A = R B then in this configuration R ax = ∞, R ay = R A , R bx = R B and R by = ∞. The reduced radius according to (7.2) is given by: = 1 R' + 1 R x = 1 R y + 1 ∞ + 1 R B + 1 R A 1 ∞ = 2 R A (7.10) which is the same as for a sphere on a plane surface. If the cylinders are crossed at an angle other than 0° or 90°, i.e. their axes are neither parallel nor perpendicular, then the contact area is enclosed by an ellipse. Examples of the analysis of such cylindrical contacts can be found in the specialized literature [14]. The formulae for evaluation of parameters of elliptical contacts are described next. EXAMPLE Find the contact parameters for two steel wires of the same diameter crossed at 90°. This configuration is often used in fretting wear studies. The normal force is W = 5 [N], radii of the wires are R A = R B = 1.5 × 10 -3 [m], the Young's modulus for both wires is E = 2.1 × 10 11 [Pa] and the Poisson's ratio is υ = 0.3. 298 ENGINEERING TRIBOLOGY · Reduced Radius of Curvature Since the radii of the wires are R ax = ∞, R ay = R A = 1.5 × 10 -3 [m], and R bx = R B = 1.5 × 10 -3 [m], R by = ∞ respectively, the reduced radii of curvature in the ‘x’ and ‘y’ directions are: = 1 R x + 1 R ax 1 R bx =+ 1 ∞ 1 1.5 × 10 −3 = 666.67 ⇒ R x = 0.0015 [m] = 1 R y + 1 R ay 1 R by =+ 1 ∞ = 666.67 1 1.5 × 10 −3 ⇒ R y = 0.0015 [m] Since 1/R x = 1/R y condition (7.3) is satisfied and the reduced radius of curvature is: 1 R' =+ 1 R x 1 R y = 666.67 + 666.67 = 1333.34 ⇒ R' = 7.5 × 10 −4 [m] · Reduced Young's Modulus E' = 2.308 × 10 11 [Pa] · Contact Area Dimensions a = 3WR' E' () 1/3 = 3 × 5 × (7.5 × 10 −4 ) 2.308 × 10 11 () 1/3 = 3.65 × 10 −5 [m] · Maximum and Average Contact Pressures p max = 3W 2πa 2 = 3 × 5 2π(3.65 × 10 −5 ) 2 = 1791.9 [MPa] p average = W πa 2 = 5 π(3.65 × 10 −5 ) 2 = 1194.6 [MPa] · Maximum Deflection δ= 1.0397 W 2 E' 2 R' () 1/3 = 1.0397 5 2 (2.308 × 10 11 ) 2 × (7.5 × 10 −4 ) () 1/3 = 8.9 × 10 −7 [m] · Maximum Shear Stress τ max = 1 3 p max = 1 3 1791.9 = 597.3 [MPa] · Depth at which Maximum Shear Stress Occurs z = 0.638a = 0.638 × (3.65 × 10 −5 ) = 2.3 × 10 −5 [m] ELASTOHYDRODYNAMIC LUBRICATION 299 · Elliptical Contact Between Two Elastic Bodies, General Case Elliptical contacts are found between solid bodies which have different principal relative radii of curvature in orthogonal planes. Examples of this are encountered in spherical bearings and gears. The contact area is described by an ellipse. An illustration of this form of contact is shown in Figure 7.5 and the formulae for the main contact parameters are summarized in Table 7.3. T ABLE 7.3 Formulae for contact parameters between two elastic bodies; elliptical contacts, general case. a = k 1 3WR' E' () 1/3 ellipse p max = 3W 2πab p average = W πab δ= 0.52k 3 τ max = k 4 p max at a depth of z = k 5 b Contact area dimensions Average contact pressure Maximum shear stress Maximum contact pressure Maximum deflection a b b = k 2 3WR' E' () 1/3 Elliptical pressure distribution W 2 E' 2 R' () 1/3 ≈ 0.3p max where: a is the semimajor axis of the contact ellipse [m]; b is the semiminor axis of the contact ellipse [m]; R' is the reduced radius of curvature [m]; k 1 , k 2 , k 3 , k 4 , k 5 are the contact coefficients. The other parameters are as defined previously. Contact coefficients can be found from the charts shown in Figures 7.12 and 7.13 [13]. In Figure 7.12 the coefficients ‘k 1 ’, ‘k 2 ’ and ‘k 3 ’ are plotted against the ‘k 0 ’ coefficient which is defined as: k 0 = − 1 R ax 1 R ay [( ) − 1 R bx 1 R by () + 22 + 2 − 1 R ax 1 R ay () − 1 R bx 1 R by () cos2φ ] 1/2 + 1 R ax ( + 1 R ay + 1 R bx 1 R by ) where: φ is the angle between the plane containing the minimum principal radius of curvature of body ‘A’ and the plane containing the minimum principal radius of curvature of body ‘B’. For example, for a wheel on a rail contact φ = 90° while for parallel cylinders in contact φ = 0°. The remaining contact coefficients ‘k 4 ’ and ‘k 5 ’ are plotted against the k 2 /k 1 ratio as shown in Figure 7.13. A very useful development in the evaluation of contact parameters is due to Hamrock and Dowson [7]. The method of linear regression by the least squares method has been applied to 300 ENGINEERING TRIBOLOGY derive simplified expressions for the elliptic integrals required for the stress and deflection calculations in Hertzian contacts. The derived formulae apply to any contact and eliminate the need to use numerical methods or charts such as those shown in Figures 7.12 and 7.13. The formulae are summarized in Table 7.4. Although they are only approximations, the differences between the calculated values and the exact predictions from the Hertzian analysis are very small. This can easily be demonstrated by applying these formulae to the previously considered examples, with the exception of the two parallel cylinders. In this case the contact is described by an elongated rectangle and these formulae cannot be used. In general, these equations can be used in most of the practical engineering applications. 1.5 2.0 0.5 1.0 1 2 5 10 0 0 0.2 0.4 0.6 0.8 1.00.1 0.3 0.5 0.7 0.9 k 0 k 3 k 2 k 1 FIGURE 7.12 Chart for the determination of the contact coefficients ‘k 1 ’, ‘k 2 ’ and ‘k 3 ’ [13]. 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0 0.2 0.4 0.6 0.8 1.00.1 0.3 0.5 0.7 0.9 k 2 /k 1 k 4 & k 5 k 4 k 5 (line contact) (point contact) FIGURE 7.13 Chart for the determination of contact coefficients ‘k 4 ’ and ‘k 5 ’ [13]. [...]... × 10−3 16.67 83 .33 ) ) ] 1/2 = 0.2 From Figure 7.12, for k0 = 0.2: k1 = 1.17, k2 = 0 .88 and k3 = 1. 98 and from Figure 7.13 where k 2 /k 1 = 0 .88 /1.17 = 0.75, the other constants have the following values: k4 = 0.33 and k5 = 0.54 · Contact Area Dimensions a = k1 ( ) = 1.17 ( ) = 0 .88 3W R' E' b = k2 3W R' E' 1/3 1/3 ( 3 × 50 × 0.012 2.3 08 × 1011 ) = 2.32 × 10−4 [m] ( 3 × 50 × 0.012 2.3 08 × 1011 ) =... 1.0339 0.636 = 1.3 380 Simplified Elliptical Integrals ε = 1.0003 + 0.59 68 Rx 0.59 68 × 0.02 = 1.0003 + Ry 0.03 ξ = 1.5277 + 0.6023 ln · ( ) 0.03 0.02 ( ) = 1.3 982 ( ) Ry 0.03 = 1.5277 + 0.6023ln Rx 0.02 = 1.7719 Contact Area Dimensions a= ( ) ( b= ( ) ( ) 6k2 εWR' 1/3 6 × 1.3 380 2 × 1.3 982 × 50 × 0.012 1/3 = πE' π(2.3 08 × 1011) 6 εWR' πkE' 1/3 = ) 6 × 1.3 982 × 50 × 0.012 1/3 π × 1.3 380 × (2.3 08 × 1011) = 2.32... the formula [7]: ^ [ ^ Hmin = Hc = 1 28 αa λb2 0.131 tan−1 () ] αa + 1. 683 2 2 where: H min is the non-dimensional minimum film thickness; Hc is the non-dimensional central film thickness; α a and λb are coefficients which can be calculated from: αa = RB ≈ 0.955 k RA ( λb = 1 + k 0.6 98 k ) −1 is the ellipticity parameter as previously defined (7. 28) 3 18 ENGINEERING TRIBOLOGY It can be seen that the minimum... 1011) = 2.32 × 10−4 [m] = 1.73 × 10−4 [m] 304 ENGINEERING TRIBOLOGY · Maximum and Average Contact Pressures pmax = 3W 3 × 50 = 2 πab 2 π(2.32 × 10−4) × (1.73 × 10−4) = 594 .8 [MPa] W 50 = πab π(2.32 × 10−4) × (1.73 × 10−4) = 396.5 [MPa] paverage = · Maximum Deflection [( )( ) ] [( )( 4.5 W 2 1/3 εR' πkE' 2 4.5 50 = 1.7719 11 1.3 982 × 0.012 π1.3 380 × (2.3 08 × 10 ) δ=ξ )] 1/3 = 1.6 × 10−6 [m] When comparing... Contact Pressures pmax = 3 × 50 3W = 2 πab 2 π(2.32 × 10−4) × (1.75 × 10−4) = 588 .0 [MPa] 50 W = πab π(2.32 × 10−4) × (1.75 × 10−4) = 392.0 [MPa] paverage = · Maximum Deflection δ = 0.52k3 · ( ) ( ) 1/3 W2 1/3 502 = 0.52 × 1. 98 = 1.6 × 10−6 [m] 2 11 2 E' R' (2.3 08 × 10 ) 0.012 Maximum Shear Stress τmax = k4 pmax = 0.33 × 588 .0 · 303 = 194.0 [MPa] Depth at which Maximum Shear Stress Occurs z = k5 b =... pressure p* = 0.0020 0.0015 Dimensionless speed parameter U = 5.0500 × 10-11 Maximum Hertzian Stress 0 .84 16 × 10-11 0 .84 16 × 10-12 0.0010 0.0005 0 h −6 R' 100 × 10 Dimensionless film thickness H = Dimensionless speed parameter U = 5.0500 × 10-11 −6 80 × 10 −6 60 × 10 −6 40 × 10 0 .84 16 × 10-11 20 × 10−6 0 .84 16 × 10 0 -2 -1 0 -12 1 x = x b FIGURE 7.17 Effects of speed parameter ‘U’ on the pressure and film... mixed or partial lubrication prevails The theory describing the mechanism of partial elastohydrodynamic lubrication was developed by Johnson, Greenwood and Poon [30] It was found that during partial lubrication, the average surface separation between two rough surfaces is about the same as predicted for smooth surfaces It has also been found that the average asperity pressure 324 ENGINEERING TRIBOLOGY. .. effectively predicted at a particular velocity and load There was, however, some discrepancy concerning the height 314 ENGINEERING TRIBOLOGY of the pressure peak since the measured peak was very much smaller than that predicted by theory This was eventually rectified by introducing the lubricant compressibility into the calculations which resulted in a reduction in the pressure spike [ 18] p E' Dimensionless... is necessary to transpose the directions of the coordinates, so ‘Rx’ and ‘Ry’ become: Rx = 0.02 [m] and Ry = 0.03 [m] 302 ENGINEERING TRIBOLOGY and the reduced radius of curvature is: 1 1 1 = 50.0 + 33.33 = 83 .33 = + R' Rx Ry · ⇒ R' = 0.012 [m] Reduced Young's Modulus E' = 2.3 08 × 1011 [Pa] · Contact Coefficients The angle between the plane containing the minimum principal radius of curvature of the... thickness is determined The derivation of the film thickness equation for elastohydrodynamic contacts begins with the 1-dimensional form of the Reynolds equation without squeeze effects (i.e 4.27): 3 08 ENGINEERING TRIBOLOGY ( ) dp h−h = 6Uη dx h3 where the symbols follow the conventions established in Chapter 4 and are: p is the hydrodynamic pressure [Pa]; U is the surface velocity [m/s]; η is the lubricant . Modulus E' = 2.3 08 × 10 11 [Pa] 294 ENGINEERING TRIBOLOGY · Contact Area Dimensions a = 3WR' E' () 1/3 = 3 × 5 × (5 × 10 −3 ) 2.3 08 × 10 11 () 1/3 = 6 .88 × 10 −5 [m] · Maximum. 10 −3 1 −30 × 10 −3 ) 1 = 0.2 = 16.67 83 .33 From Figure 7.12, for k 0 = 0.2: k 1 = 1.17, k 2 = 0 .88 and k 3 = 1. 98 and from Figure 7.13 where k 2 /k 1 = 0 .88 /1.17 = 0.75, the other constants. Maximum Shear Stress τ max = 1 3 p max = 1 3 504.4 = 1 68. 1 [MPa] · Depth at which Maximum Shear Stress Occurs z = 0.638a = 0.6 38 × (6 .88 × 10 −5 ) = 4.4 × 10 −5 [m] · Contact Between Two Parallel