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

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

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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

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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

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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)

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Conventional Face Milling

Cutter overhangs work on both sides

(tool axis perpendicular to machined surface)

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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-

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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.

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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