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337: Materials & Manufacturing Processes
Lecture 6:
Machining Operations and Machinability
Chapter 22 and 24
This Time
Parameters Material Removal Rate Power Requirements Surface Finish Machinability
2
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) Do = outer diameter
Df = final diameter
Machining Time - Tm (min) L = length of cut
Mat’l Removal Rate - MRR (mm3/min -or- in3/min)
oDπ
vN
2fo DD
d
rm f
LT
fNfr
dfvMRR
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 = DoL/vf
v = DoL/fTm
= (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 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 (= 33000 ft-lb/min)
11
Power and Energy Relationships
The Gross machine power (Pg) available is:
Pc = Pg• Ewhere E = mechanical efficiency of machine tool
Typical E for machine tools = 80 - 90%
Note: Textbook relationship is same -
EP
P cg
EHP
HP cg
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 21.14where MRR = material removal rate
MRRP
P cu
MRRHP
HP cu
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
wt
F
MRR
PPU
o
ccu
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
MRRP
P cu
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
What if feed changes?
16
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
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 vs. Blind Holes
Two hole types: (a) through‑hole, and (b) blind hole
Through‑holes - drill exits the opposite side of work
Blind‑holes – drill does not exit work on opposite side
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)
Dπ
vN
rm f
dT
fNfr
4
2rfDπ
MRR
r
m f
DtT 22
1 90tan
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
Two forms of milling:
(a) peripheral milling, and (b) face milling
Milling Parameters Illustrated
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)
Dπ
vN
fnNf tr
rfdwMRR
r
m f
dDdLT
rm f
DLT
rm f
wDwLT
2-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
Dπ
vN fnNf tr
rm f
DLT
35
Example: Solution
N = (3000 mm/s)/150= 6.37 rev/s
fr = 6.37(4)(.27) = 6.88 mm/s
Tm = (400 + 150)/6.88 = 80 s = 1.33 min.
Dπ
vN
fnNf tr
rm f
DLT
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