... the subscripts 1, 2, 3, etc indicate the positions 1, 2, 3, etc in Figure 6 .15 at which the quantities are calculated, or else the subscripted quantities are defined in the figure. Further details ... the point of view of chip formation. Z′ is parallel to z and z′, still in the cutting direction, but X′ is normal to, and Y′ is in, the plane containing the cutting and chip velocities. In terms ... introduced in Figure 6 .15 . In Figure 6 .15 (a), for example, there are four intervals indicated by A 1 , A 2 , A 3 , A 4 . In Table 6 .1, the values of lim 1 ,i , lim 2 ,i , A q ,i , t 1e ,i and the appropriate...
Ngày tải lên: 21/07/2014, 17:20
... Annals CIRP 26 (1) , 21 26. Kakino, K. (19 84) Monitoring of metal cutting and grinding processes by acoustic emission. J. Acoustic Emission 3, 10 8 11 6. Miwa, Y., Inasaki, I. and Yonetsu, S. (19 81) In-process ... slip-line field is the complete set of orthogonal curvilinear slip-lines existing in a plastic region. Slip- line field theory provides rules for constructing the slip-line field in particular ... effects, is introduced in the next section. 6.3 Introducing variable flow stress behaviour Slip-line field modelling investigates the variety of chip formation allowed by equilibrium and flow conditions...
Ngày tải lên: 21/07/2014, 17:20
Metal Machining - Theory and Applications Episode 1 Part 8 ppsx
... the similarity of slope between this and the sintered steel observations is striking. Figure 4 .16 is an example of tool wear and wear distribution influenced by the machine tool. It gives the ... quartz is high, the amplifier must itself have high input impedance: 10 5 MW is not unusual. Figure 5 .11 shows the piezoelectric equivalent of the dynamometers of Figure 5.8. The stiffness is basically ... specified point can be measured, Temperatures in machining 15 1 Fig. 5 .17 A detail of the hot junction and the associated measurement circuit Fig. 5 .18 Calibration test results for P10 carbide and a...
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Metal Machining - Theory and Applications Episode 1 Part 7 ppt
... temperature 0 .1 1 10 10 0 Chipping Micro chipping Abrasion Fracture Attrition Damage size (àm) Fig. 4.4 Classification of mechanical damages Childs Part 1 28:3:2000 2: 41 pm Page 12 1 in drilling, particularly ... to find uses in interrupted turning and light milling oper- ations. Considering the thicknesses of both the coatings and modified substrate surface layers, the composition (and hence thermal and ... understanding of the interactions between the graded surface compositions and the mechanical and thermal fields generated in machining, leading to still further improvements in tool design, continues...
Ngày tải lên: 21/07/2014, 17:20
Metal Machining - Theory and Applications Episode 1 Part 6 pdf
... calcium, to form alumina, silica or calcium oxides. Alumina is hard and abrasive and is certainly detrimental to tool life in machining. The addition of silicon and calcium can result in softer inclusions. ... state temperature rise in machining. In transient conditions, heat capacity is also important because, with conductivity, it determines thermal diffusivity k and the rate of penetration of heat into the tool. ... ceramics based on alumina and silicon nitride, and the super- hard materials polycrystalline diamond and cubic boron nitride (single crystal diamonds are also used for the finishing of IT mirror and...
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Metal Machining - Theory and Applications Episode 1 Part 5 potx
... intermediate thermal conductivities and diffusivities result in their spanning the range with respect to tempera- ture rise per unit feed and also cutting speed. Work material characteristics in machining ... understood. 2.4.2 Lubrication in metal cutting The previous section has emphasized the high friction conditions that exist between a chip and tool, in the absence of solid lubricants. The conditions that ... 84 The initial wear region is the running-in regime of Figure 2.28. Surface smoothing occurs until the contacting asperities deform mainly elastically. If the surface adhesion is small (mild...
Ngày tải lên: 21/07/2014, 17:20
Metal Machining - Theory and Applications Episode 1 Part 4 pot
... it is clear that significant temperature rises may occur in the chip. This is without considering the additional heating due to friction between the chip and tool. It is important to understand ... the heating. Given that the primary shear acts on the workpiece, these simple considerations point to the possibility of workpiece thermal damage when machin- ing aluminium and titanium alloys, ... flow into the tool. The first question in considering the heating of the chip is what is the value of a*? The answer comes from recognizing that the contact area is common to the chip and the tool....
Ngày tải lên: 21/07/2014, 17:20
Metal Machining - Theory and Applications Episode 1 Part 3 pot
... formed is thick and straight. Adding a lubricating fluid causes the chip to become thinner and curled. In this case, adding the lubricant caused the friction coefficient between the chip and tool ... by the rate of investment in the new, more powerful and stiffer machine tools needed for their potential to be realized. Instead, it is a growth in ceramic (titanium nitride, titanium carbide and ... the tool shape alone. In fact, determining the chip shape is the grand challenge for mechanics. Once the shape is known, determining the cutting forces is relatively simple; and determining the...
Ngày tải lên: 21/07/2014, 17:20
Metal Machining - Theory and Applications Episode 1 Part 2 pps
... C t is the initial price of the tool, plus all the reconditioning costs, divided by the number of times it is reconditioned. It is less than the purchase price (if it were more, reconditioning ... (1. 13) can be used with a = 1. For the example concerned, the time and cost of removing material by drilling is negligible. It is the loading and unloading time and cost that dominates. It is ... Dimensions are given in Figure 1. 25. The part is created by turning the external diameter, milling the keyway, and drilling four holes. The turning operation will be consid- ered first. 1. 4.1...
Ngày tải lên: 21/07/2014, 17:20
Metal Machining - Theory and Applications Episode 1 Part 1 potx
... retrieval system, without either prior permission in writing from the publishers or a licence permitting restricted copying. In the United Kingdom such licences are issued by the Copyright Licensing Agency: ... machines was holding steady, with roughly twice the investment in turning as in milling machines. At the same 10 Introduction Fig. 1. 10 Mass/power and price/mass relationships for turning machines Childs ... and its finite element formulation 328 A1 .1 Yielding and flow under triaxial stresses: initial concepts 329 A1.2 The special case of perfectly plastic material in plane strain 332 A1.3 Yielding...
Ngày tải lên: 21/07/2014, 17:20
Wiley the official guide for GMAT Episode 1 Part 10 pps
... summarizes the relationship between Arizona v. California in lines 38–42, and the criteria citing the Winters doctrine in lines 10 –20? (A) Arizona v. California abolishes these criteria and ... 3 911 1_449745-ch07.indd 3 91 2/23/09 11 :40: 51 AM2/23/09 11 :40: 51 AM The Of cial Guide for GMAT đ Review 12 th Edition 392 Line (5) (10 ) (15 ) (20) (25) (30) In 19 55 Maurice Duverger published The Political ... patient. 10 6. The author mentions patients’ ages (line 33) primarily in order to (A) identify the most critical variable differentiating subgroups of patients (B) cast doubt on the advisability...
Ngày tải lên: 22/07/2014, 14:20
Wiley the official guide for GMAT Episode 1 Part 9 potx
... sufficient. 10 9. In triangle ABC, point X is the midpoint of side AC and point Y is the midpoint of side BC. If point R is the midpoint of line segment XC and if point S is the midpoint of line segment ... SUFFICIENT. (2) When 14 0 is divided by 11 0, the quotient is 1 R30. However, 250 divided by 11 0 yields a quotient of 2 R30, and 360 divided by 11 0 gives a quotient of 3 R30. Since there is ... 12 th EditionThe Of cial Guide for GMAT đ Review 12 th Edition 17 1. What is the tens digit of positive integer x ? (1) x divided by 10 0 has a remainder of 30. (2) x divided by 11 0 has a remainder...
Ngày tải lên: 22/07/2014, 14:20
Wiley the official guide for GMAT Episode 1 Part 8 potx
... D 11 6. A 11 7. D 11 8. E 11 9. E 12 0. B 12 1. E 12 2. C 12 3. A 12 4. A 12 5. D 12 6. E 12 7. D 12 8. B 12 9. C 13 0. C 13 1. D 13 2. C 13 3. B 13 4. B 13 5. D 13 6. D 13 7. A 13 8. E 13 9. ... SUFFICIENT. 10 _449745-ch06.indd 29 410 _449745-ch06.indd 294 2/23/09 11 :37 :11 AM2/23/09 11 :37 :11 AM 316 The Of cial Guide for GMAT đ Review 12 th EditionThe Of cial Guide for GMAT đ Review 12 th Edition Arithmetic Statistics Given ... right of the median and the list in ascending order is either 6, k, n, 12 , 17 or k, 6, n, 12 , 17 . In either case, n is the middle number, and since the median is 10 , n = 10 ; SUFFICIENT. e correct...
Ngày tải lên: 22/07/2014, 14:20
Wiley the official guide for GMAT Episode 1 Part 7 ppsx
... x increases from 16 5 to 16 6, which of the following must increase? I. 2x – 5 II. 1 – 1 x III. 1 x 2 – x (A) I only (B) III only (C) I and II (D) I and III (E) II and III Algebra Simplifying ... Solving Answer Explanations (A) I only (B) II only (C) I and II only (D) II and III only (E) I, II, and III Arithmetic Properties of numbers Since a, b, and c are consecutive positive integers ... 16 5 16 5 16 6 1 165 16 6 − = − ()() = ()() , which is positive, and thus the function increases from x = 16 5 to x = 16 6. III. For x = 16 5, the denominator is 16 5 2 – 16 5 = (16 5) (16 5 – 1) = (16 5) (16 4),...
Ngày tải lên: 22/07/2014, 14:20
Wiley the official guide for GMAT Episode 1 Part 6 ppt
... 224 The Of cial Guide for GMAT đ Review 12 th EditionThe Of cial Guide for GMAT đ Review 12 th Edition 11 9. Which of the following is equivalent to the pair of inequalities x + 6 > 10 and ... 10 = 560 5x = 550 solve for x x = 11 0 e fi rst integer in the sequence is 11 0, so the next integers are 11 1, 11 2, 11 3, and 11 4. From this, the last 5 integers in the sequence, and thus their ... Review 12 th Edition (10 10 )(0.0 012 ) = 10 ⁴(0.0 012 ) – 10 ²(0.0 012 ) = 10 ,000(0.0 012 ) – 10 0(0.0 012 ) = 12 .12 – 0 .12 multiply by multiples of 10 to move the decimals distribute the (0.0 012 ) 10 ⁴...
Ngày tải lên: 22/07/2014, 14:20