337: Materials & Manufacturing Processes

1. 337: Materials & Manufacturing Processes Lecture 6: Machining Operations and Machinability Chapter 22 and 24

2. This Time • Parameters • Material Removal Rate • Power Requirements • Surface Finish • Machinability

3. Turning Single point cutting tool removes material from a rotating workpiece to form a cylindrical shape

4. Turning • A single point cutting tool removes material from a rotating workpiece to generate a rotationally symmetric shape • Machine tool is called a lathe • Types of cuts: • Facing • Contour turning • Chamfering • Cutoff • Threading • Workholding methods: • Holding the work between centers • Chuck • Collet • Face plate

5. Turning Parameters Illustrated

6. Primary Machining Parameters • Cutting Speed – (v) • Primary motion • Peripheral speed m/s ft/min • Feed – (f) • Secondary motion • Turning: mm/rev in/rev • Milling: mm/tooth in/tooth • Depth of Cut – (d) • Penetration of tool below original work surface • Single parameter mm in • Resulting in Material Removal Rate –(MRR) MRR = v f dmm3/s in3/min where v = cutting speed; f = feed; d = depth of cut

7. Machining Calculations: Turning • Spindle Speed - N(rpm) • v = cutting speed • Do = outer diameter • Feed Rate - fr(mm/min -or- in/min) • f = feed per rev • Depth of Cut - d(mm -or- in) • Do = outer diameter • Df = final diameter • Machining Time - Tm(min) • L = length of cut • Mat’l Removal Rate - MRR(mm3/min -or- in3/min)

8. Example • In a production turning operation, the foreman has decided that a single pass must be completed on a cylindrical workpiece in 5.0 min. The piece is 400 mm long and 150 mm in diameter. Using a feed = 0.30 mm/rev and a depth of cut = 4.0 mm, what cutting speed must be used to meet this machining time requirement?

9. Example: Solution • Tm = L/fr = L/Nf = pDoL/vf • v = pDoL/fTm = p(0.4)(0.15)/(0.30)(10-3)(5.0) = 0.1257(103) m/min = 125.7 m/min

10. Power and Energy Relationships • Power requirements to perform machining can be computed from: Pc = Fc vN-m/s (W) ft-lb/min where: Pc = cutting power; Fc = cutting force; and v = cutting speed • Customary U.S. units for power are Horsepower (= 33000 ft-lb/min)

11. Power and Energy Relationships • The Gross machine power (Pg) available is: Pc = Pg• E where E = mechanical efficiency of machine tool • Typical E for machine tools =  80 - 90% Note: Textbook relationship is same -

12. Unit Power in Machining • Useful to convert power into power per unit volume rate of metal cut • Called the unit power, Pu or unit horsepower, HPu or • Tool sharpness is taken into account multiply by 1.00 – 1.25 • Feed is taken into account by multiplying by factor in Figure 21.14 where MRR = material removal rate

13. Specific Energy in Machining • Unit power(Pu) is also known as the specificenergy (U), or the power required to cut a unit volume of material: where t0 = un-deformed chip thickness; w = width of the chip; and Fc = cutting force • Units for specific energy are typically N‑m/mm3(J/mm3) or in‑lb/in3 • Table 21-2 (p. 497) in the text approximates specific energy for several materials based on est. hardness

14. Example • In a turning operation on stainless steel with hardness = 200 HB, the cutting speed = 200 m/min, feed = 0.25 mm/rev, and depth of cut = 7.5 mm. How much power will the lathe draw in performing this operation if its mechanical efficiency = 90%. • From Table 21.2, U = 2.8 N-m/mm3 = 2.8 J/mm3

15. Example: Solution • MRR = vfd = (200 m/min)(103 mm/m)(0.25 mm)(7.5 mm) = 375,000 mm3/min = 6250 mm3/s • Pc = (6250 mm3/s)(2.8 J/mm3) = 17,500 J/s = 17,500 W = 17.5 kW • Accounting for mechanical efficiency, Pg = 17.5/0.90 = 19.44 kW

16. What if feed changes?

17. Facing Tool is fed radially inward

18. Instead of feeding the tool parallel to the axis of rotation, tool follows a contour that is other than straight, to create a contoured form Contour Turning

19. Cutting edge cuts an angle on the corner of the cylinder, forming a "chamfer" Chamfering

20. Tool is fed radially into rotating work at some location to cut off end of part Cutoff

21. Pointed form tool is fed linearly across surface of rotating workpart parallel to axis of rotation at a large feed rate to create threads Threading

22. Engine Lathe

23. Boring • Difference between boring and turning: • Boring is performed on the inside diameter of an existing hole • Turning is performed on the outside diameter of an existing cylinder • In effect, boring is an internal turning operation • Boring machines • Horizontal or vertical - refers to the orientation of the axis of rotation of machine spindle

24. Drilling Used to create a round hole, usually by means of a rotating tool (drill bit) that has two cutting edges

25. Through‑holes - drill exits the opposite side of work Blind‑holes – drill does not exit work on opposite side Through Holes vs. Blind Holes Two hole types: (a) through‑hole, and (b) blind hole

26. Machining Calculations: Drilling • Spindle Speed - N(rpm) • v = cutting speed • D = tool diameter • Feed Rate - fr(mm/min -or- in/min) • f = feed per rev • Machining Time - Tm(min) • Through Hole : • t = thickness •  = tip angle • Blind Hole : • d = depth • Mat’l Removal Rate - MRR(mm3/min -or- in3/min)

27. Milling Rotating multiple-cutting-edge tool is moved slowly relative to work to generate plane or straight surface • Two forms: peripheral milling and face milling

28. Milling • Machining operation in which work is fed past a rotating tool with multiple cutting edges • Axis of tool rotation is perpendicular to feed direction • Creates a planar surface; other geometries possible either by cutter path or shape

29. Milling Parameters Illustrated Two forms of milling: (a) peripheral milling, and (b) face milling

30. The basic form of peripheral milling in which the cutter width extends beyond the workpiece on both sides (tool axis parallel to machined surface) Slab Milling

31. Cutter overhangs work on both sides (tool axis perpendicular to machined surface) Conventional Face Milling

32. Machining Calculations: Milling • Spindle Speed - N(rpm) • v = cutting speed • D = cutter diameter • Feed Rate - fr(mm/min -or- in/min) • f = feed per tooth • nt = number of teeth • Machining Time - Tm(min) • Slab Milling: • L = length of cut • d = depth of cut • Face Milling: • w = width of cut • 2nd form is multi-pass • Mat’l Removal Rate - MRR(mm3/min -or- in3/min) -or-

33. Example • A face milling operation is used to machine 5 mm from the top surface of a rectangular piece of aluminum 400 mm long by 100 mm wide. The cutter has four teeth (cemented carbide inserts) and is 150 mm in diameter. Cutting conditions are: v = 3 m/s, f = 0.27 mm/tooth, and d = 5.0 mm. Determine the time to make one pass across the surface.

34. Example: Solution

35. Example: Solution N = (3000 mm/s)/150p = 6.37 rev/s fr = 6.37(4)(.27) = 6.88 mm/s Tm = (400 + 150)/6.88 = 80 s = 1.33 min.

36. You should have learned • Parameters • Material Removal Rate • Power Requirements • Surface Finish • Machinability

37. Assignment • HW 2 (Due Tuesday): • CH 21,22 and 24 Problems • In Assignments folder

38. Next Time • Casting Chapter 10