1. 337: Materials & Manufacturing Processes
IE 337 Lecture 13: Machining - I
337: Materials & Manufacturing Processes
Lecture 6:
Machining Operations and Machinability
Chapter 22 and 24
S.V. Atre

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 d mm3/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)
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
= (103) m/min
= m/min

10. Power and Energy Relationships
Power requirements to perform machining can be computed from:
Pc = Fc v N-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 (= 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 =  %
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 specific energy (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 = kW

16. What if feed changes?

17. Facing
Tool is fed
radially inward

18. Contour Turning
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

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

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

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

22. Engine Lathe

23. IE 337 Lecture 13: Machining - I
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
S.V. Atre

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 vs. Blind Holes
Through‑holes - drill exits the opposite side of work
Blind‑holes – drill does not exit work on opposite side
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. Slab Milling
The basic form of peripheral milling in which the cutter width extends beyond the workpiece on both sides
(tool axis parallel to machined surface)

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

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. IE 337 Lecture 14: Machining 2
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.
S.V. Atre

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 = ( )/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