Friction and Lubrication in Mechanical Design Episode 1 Part 5 pptx

25 330 0
Friction and Lubrication in Mechanical Design Episode 1 Part 5 pptx

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

Thông tin tài liệu

80 0.6 : Region 2 0.5 0.4 Chapter 3 0.9 - I 0.7: o.o, , , , , , , , , ,.,,,., , 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 (4 Total Applied Force (x 103 Ibf) = 0.32 z 0.2: 6 -0.3: -0.5 i -0.4 1 = 0.3 0.1 z 0.2 6 -0.3: -0.5 i -0.4 1 Region 1 A-A-A-~A-A-A-~~A-A-A-A-A-A-~ (b) Total Applied Force (x 10 Ibf) Figure 3.19 (a) Tangential load sharing between the two contact zones at differ- ent applied loads. (b) Distance between the line of action of the resultant frictional resistance and centroid at different applied loads. Traction Distribution and Microslip in Frictional Contacts 81 f2= 0.12 p2= 10000 psi / f2= 0.12 p2= 20000 psi 500 600 700 800 900 650 1000 1Ox) 1100 1150 1200 1250 1280 1296 Figure 3.20 Development of slip regions on a discrete contact area with increas- ing tangential load. 3.4.3 Iterative Procedure The modified linear programming formulation is first implemented with an initial guess for the center of rotation in order to find the discretized traction forces whose directions are predetermined by the preprocessor. Now the residual forces (the rectangular components of the sum of the traction forces) can be calculated. These residual forces must be equal to zero when the real center of rotation is found, since no tangential forces are applied. The residual forces are then used to modify the center of rotation and the process is repeated until the residual forces vanish. The real center of rotation, the traction force distribution, the microslip region, and the angle of rigid body rotation are determined by this iterative procedure, as depicted in the flow chart (Fig. 3.22). 82 Chapter 3 0.9 l.O, 0.8 1 1 LL- 5 0.4- ‘J. 5 0.3- 0.2- F* ,., , , ’.“(.’ ., 0.0 0.5 1 .o 1.5 2.0 2.5 3.0 RigM Body Movement (xlW in.) (a) 1.4 1.2- E. & 1.0- ‘3 t 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 Rigid Body hV8ltWnt (Xfw in.) (b) Figure 9.21 (a) Joint compliance under same tangential loads before gross slip (Case 1). (b) Joint compliance under different tangential loads before gross slip (Cases 2 and 3). Traction Distribution and Microslip in Frictional Contacts b 83 forces negligible? Print the results Find the traction distribution using the modified linear programming Calculate the residual force c Assume the new center of rotation 1 using a linear interpolation scheme Figure 3.22 problem subjected to a twisting moment. Flow chart for the iterative procedure to solve frictional contact 3.4.4 Illustrative Examples EXAMPLE 1: Circular Hertzian Contact. The contact between two steel spheres of 1 in. radius (Fig. 3.23) is first considered in order to compare the result from the developed procedure with the analytical solution by Lukin [23]. The normal load is taken as 21601bf, the twisting moment is 2.45in. -lbf, and the coefficient of friction is 0.1. A grid with 80 square elements is used to discretize the circular contact area of 0.36628 x 10-' in. radius. A comparison between Lubkin's theory (solid line) and the numerical results (symbol s) is plotted in Fig. 3.24 and very good agreement can be seen. The angle of rigid rotation (0.10641 x 10-* rad) is also found to com- pare favorably with Lubkin's theory (0.1 11 19 x 10-* rad) with a deviation of 4.30%. The center of rotation is at the centroid. 84 Chapter 3 P Ip Figure 3.23 Contact of spherical bodies subjected to a twisting moment. x 10’ Figure 3.24 Lubkin’s theory. Traction distribution on the circular contact as compared with Traction Distribution and Microslip in Frictional Contacts 85 EXAMPLE 2: Elliptical Hertzian Contact. The elliptical Hertzian con- tact area with an aspect ratio of 2 is assumed to occur when a normal load of 21601bf is applied on two steel bodies. The pressure distribution is assumed to be Hertzian in this case. The coefficient of friction is taken to be equal to 0.1 and a twisting moment of 3.8in lbf is applied on the interface. A grid with 80 rectangular elements of the side ratio of 2 is used to discretize the elliptical contact area. The contours of the magnitude and the direction of the tractions are plotted in Figs. 3.25 and 3.26. The border line between the no-slip region and the slip region is also shown as a broken line. The centroid in this case is the center of rotation. Interpreted Magnitude of Stress Figure 3.25 Contour plot for the magnitude of traction on the elliptical contact area. 86 Chapter 3 Direction of Predirected Stress (N=80) Figure 5.26 Contour plot for the direction of traction on the elliptical contact. EXAMPLE 3: Disconnected Contact Areas on Semi-Infinite Bodies. Two disconnected square areas of the same size (0.6in. x 0.6in.) on a semi- infinite steel body are assumed to be in contact with another semi-infinite steel body. The centroids of the two squares are located 1 in. apart (Fig. 3.27). Uniform pressure is assumed on each contact region and the coeffi- cient of friction is 0.12 for both regions. Two cases of normal loading are considered here. Case 1. PI = 10,OOOpsi and P2 = 10,OOOpsi; Case 2. PI = 20,OOOpsi and P2 = 10,OOOpsi. Each region is discretized with 36 square elements to have 72 elements for the entire contact area. The contours of the magnitude and the direction of the tractions are plotted in Figs 3.27 and 3.28 for Case 1 with a twisting moment of 500 in lbf and in Figs 3.29 and 3.30 for Case 2 with a twisting moment of 700 in lbf. It 500 in-Ibfe f, = 0.12 P, = 1000 psi 7 I I I I I I Figure 3.27 Contour plot for the magnitude of traction on the contact I area of disconnected squares with I A4 = 500in lbf and normal loading of Case 1. I I f, = 0.12 P, = 10000 p8i Figure 3.28 Contour plot of the direction of traction on the contact area of disconnected squares with M = 500in lbf and normal loading of Case I. -1 70 600 inJM @ f, = 0.12 P, = IOOOO psi 87 i I All Slip I I I I I I 700 in-lbfQ f, = 0.12 P, 20000 psi -1 I I I I I I I Figure 3.29 Contour plot for the magnitude of traction on the contact I area of disconnected squares with I M = 700in lbf and normal loading of Case 2. I f, = 0.12 Pz = 10000 psi + 700 in-Ibfo f, t 0.12 P, = 20000 pai Figure 3.30 Contour plot for the direction of traction on the contact area of disconnected squares with M = 700in lbf and normal loading of Case 2. 88 Traction Distribution and Microslip in Frictional Contacts 89 can be seen that the traction contours and the slip patterns for both regions 1 and 2 are identical and the center of rotation is the centroid for the symmetric normal loading. As would be expected, the case of asymmetric normal loading shows different traction distributions in the two discon- nected contact areas and the center of rotation is consequently found to be displaced from the centroid. Also notice that region 2 reaches the state of total slip for Case 2, with a twisting moment of 700in lbf, and circumfer- ential tractions are assumed for region 2. The center of rotation always occurred on the line connecting the centroids of two disconnected squares. The x-distance between the center of rotation and the centroid for Case 2 versus the applied twisting moment is plotted in Fig. 3.31. The development of the slip region with the increasing twisting moment is shown in Fig. 3.32 for Case 2. It can be seen that region 2 reaches a state of total slip at a twisting moment of 700 in lbf, and that gross slip occurs at 770 in lbf. Some slip is also shown to occur in region 1 below 700 in lbf. The compliance curve relating the angle of rigid rotation and the twist- ing moment is plotted in Fig. 3.33a for Case 1 and in Fig. 3.33b for Case 2. 0.40 0.0 0.1 0.2 0.3 0.4 015 0:s 017 0.8 1) Twisting Moment (xl Oin-lbf) Figure 3.31 twisting moments on the contact area of disconnected squares for Case 2. Locations of the center of rotation from the centroid versus applied [...]... Reaming Milling Mold casting Drilling Chemical milling Elect discharge machining Planing, shaping Sawing Forging Snagging Hot rolling Flame cutting Sand casting Average surface roughness (pin.) 2-8 2 -16 4 -16 4-32 4-63 8 -16 8-32 16 - 250 32-63 32 -12 5 32 -1 25 32 -12 5 32- 250 63 -1 25 63- 250 63- 250 63- 250 63 -50 0 63 -10 00 12 5- 500 250 -1 000 50 0- 10 00 50 0- 10 00 50 0- 10 00 literature [ 1- 18] Although various factors... theoretical and experimental studies on surface roughness during turning are available in the 10 0 10 I The Contact Between Rough Surfaces Table 4 .1 Average Surface Roughness for Some Common Manufacturing Processes ~~~~~~ Manufacturing process Super finishing Lapping Polishing Honing Grinding Electrolytic grinding Barrel finishing Boring, turning Die casting Cold rolling, drawing Extruding Reaming Milling Mold... iterative procedure are used as initial guesses REFERENCES 1 2 Hertz, H., “Miscellaneous Papers” translated by Jones, D E., and Schott, G A., Macmillan, New York, NY, 18 96, pp 14 616 2, 16 3 -18 3 Lundberg, G., “Elastische Beruhrung Ingenieurw., 19 39, Vol 10 , pp 2 01- 21 1 Zweier Halbraume,”Forsch 98 7 8 9 10 11 12 13 14 15 16 17 18 19 Chapter 3 Cattaneo, C., “Teoria del contatto elasiico in seconda approssimazione,”... Uniformstr~s -12 ~psi I I I : J I 97 Traction Distribution and Microslip in Frictional Contacts f, = 0 .12 P, = 20000 psi 4oo in, M -6OOIW f, = 0 .12 P, = 10 000 psi Figure 3. 41 Contour plot for the direction of traction on the contact area of disconnected squares for Case 2 (using iterative procedure) programming technique (0 .16 50 2 x 10 -4 in and 0 .12 263 x 10 -4 rad for Case 1 and 0. 312 57 x 10 - 4in and 0.30892 x 10 -4rad... moment is a highly nonlinear problem The problem is piecewisely linearized using an iterative method and a modified linear programming technique is utilized at each iteration The procedure followed in the iterative method is shown in Fig 3.34 91 Traction Distribution and Microslip in Frictional Contacts 0.7 b I I 1 l 1 1 1 1’7 0.0 0 .5 1. 0 0.8 1. 5 2.0 2 .5 3.0 3 .5 4.0 A g d Twist (x10’ rat.) nk (a) - m... corresponding results for Case 2 are shown in Figs 3.40 and 3. 41 In this case, region 1 is found to be in a state of total slip The rigid body motions from the iterative procedure (0 .16 436 x 10 F 4in and 0 .12 323 x 10 -4rad for Case 1 and 0.3 050 9 x 10 - 4in and 0.327 21 x 10 -4rad for Case 2) compare favorably with those from the nonlinear f, = 0 .12 e, = loo00 f, - 0 .12 P, = 20000 p i Figure 3.38 Contour plot... deviations of 2.39% and 0.49% for Case 1, and 2. 31% and 5. 92% for Case 2, respectively The elapse CPU times on a Harris 800 to obtain the above results by the iterative procedure are 2 min for Case 1, and 8 min for Case 2, whereas those necessary to obtain the results from the nonlinear programming technique are 18 min for Case 1, and 46 min for Case 2, respectively, when the solutions obtained by the iterative... 3 f,= 0 .12 p2= 10 000 psi f , 300 400 50 0 600 700 750 = 0 .12 p, = 20000 psi 770 Figure 3.32 Progression of slip with increasing twisting for contact area of disconnected squares 35 3 .5 .1 FRICTIONAL CONTACTS SUBJECTED T O A COMBINATION OF TANGENTIAL FORCE A N D TWISTING MOMENT Iterative Procedure The analysis of the frictional contact problem under a combination of tangential force and twisting moment... 3 21- 3 35 Francavilla, A., and Zienkiewicz, 0 C., “Note on Numerical Computation of Elastic Contact Problems,” Int J Numer Meth Eng., 19 75, Vol 9(4), pp 9 13 -924 Haug, E., Chand, R., and Pan, K., “Multibody Elastic Contact Analysis by Quadratic Programming,” J Optim Theory Appl., Feb 19 77, Vol 21( 2), pp 18 9 -19 8 Kravchuk, A S., “On the Hertz Problem for Linearly and Non-Linearly Elastic Bodies of Finite... Dec 19 71, Vol 7 (12 ), pp 16 13 -16 26 Tsai, K C., Dundurs, J., and Keer, L M., “Contact between an Elastic Layer with a Slightly Curved Bottom and a Substrate,” J Appl Mech., Trans ASME, Sept 19 72, Ser E., Vol 39(3), pp 8 21- 823 “Minimum Principle for Frictionless Elastic Kalker, J J., and Van Randen, Y., Contact with Application to Non-Hertzian Contact Problems,”J Eng Math., April 19 72, Vol 6(2), pp 19 3-206 . ’.“(.’ ., 0.0 0 .5 1 .o 1. 5 2.0 2 .5 3.0 RigM Body Movement (xlW in. ) (a) 1. 4 1. 2- E. & 1. 0- ‘3 t 0.0 0 .5 1. 0 1. 5 2.0 2 .5 3.0 3 .5 4.0 4 .5 5. 0 5. 5 6.0 Rigid Body. Halbraume,”Forsch. Ingenieurw., 19 39, Vol. 10 , pp. 2 01- 21 1. 98 Chapter 3 7. 8. 9. 10 . 11 . 12 . 13 . 14 . 15 . 16 . 17 . 18 . 19 . Cattaneo, C., “Teoria del contatto elasiico in seconda. ASME, 19 67, Vol. 24, pp. 623-624. Vol. 7 (12 ), pp. 16 13 -16 26. 19 71, pp. 387-392. Dec. 19 74, Vol. 30(3), pp. 3 21- 3 35. 9 1 3-924. 18 9 -19 8. 5 31- 54 8. Traction Distribution and Microslip in Frictional

Ngày đăng: 05/08/2014, 09:20

Từ khóa liên quan

Tài liệu cùng người dùng

Tài liệu liên quan