1. Machining Operations by
Ed Red

2. Objectives Introduce machining operations terminology
Introduce machining efficiency measures
Reconsider cutting parameters as they apply to efficiency
Review a machining efficiency example
Consider modern machine operations (papers)

3. Machining terms
Chatter – interrupted cutting usually at some frequency
Down milling – cutting speed in same direction as part feed
Up milling – cutting speed in opposite direction as part feed
Peripheral milling – tool parallel to work
Face milling – tool perpendicular to work
Ideal roughness – geometrically determined roughness
Machinability – machining success determined by tool life, surface finish
Optimal machining – parameter choices that increase machining throughput or reduce operational costs

4. Machining operations on lathe
(other than normal turning)
Facing
Taper
Contour
Form
Chamfer
Cutoff
Threading
Boring
Drilling
Knurling

5. Two types of milling operations
Peripheral
Face

6. Face milling operations
Facing
Partial facing
End milling
Profiling
Pocketing
Surface contouring

7. Face milling movements
Peripheral milling cutting positions
Face milling cutting positions
Full face cut
Offset face cut

8. Milling cutter time analysis
Spindle rpm related to cutter diameter and speed:
N (rpm) = v/(p D)
Feedrate in in/min:
fr = N nt f
where f = feed per tooth
nt = number of teeth
MRR is
MRR =w d fr

9. Milling time analysis Slab milling: Approach distance, A : A = d (D-d)
Time to mill workpiece, Tm:
Tm = (L + A)/fr
Face milling:
Allow for over-travel O where A = O:
Full face A = O = D/2
Partial face A = O = w (D – w)
Machining time:
Tm = (L + 2A)/fr

10. Milling time analysis - example
Problem statement:
A face milling operation is performed to finish the top surface of a steel rectangular workpiece 12 in. long by 2 in. wide. The milling cutter has 4 teeth (cemented carbide inserts) and is 3 in. in diameter. Cutting conditions are 500 fpm, f = 0.01 in./tooth, and d = in. Determine the time to make one pass across the surface and the metal removal rate during the cut.

11. Milling time analysis - example
Solution? Numbers?
Full face A = O = D/2
Machining time Tm = (L + 2A)/fr
Metal removal rate MRR = w d fr
Feedrate in in/min fr = N nt f
N (rpm) = v/(p D)

12. Tolerance by process

13. Surface finish by process

14. Surface finish by geometry
Ideal roughness,
Ri = f2/(32 NR)
where NR = tool nose radius
Actual roughness,
Ra = rai Ri (about 2 x Ri )
because of edge effects, chip interactions, surface tearing, etc.

15. Machinability
What is a free machining steel?
Machinability is a measure of machining success or ease of machining.
Suitable criteria:
tool life or tool speed
level of forces
surface finish
ease of chip disposal

16. Machinability - example
Problem statement:
A series of tool life tests is conducted on two work materials under identical cutting conditions, varying only speed in the test procedure. The first material, defined as the base material, yields the Taylor tool life equation
v T0.28 = 1050
and the other material (test material) yields the Taylor equation
v T0.27 = 1320
Determine the machinability rating of the test material using the cutting speed that provides a 60 min. tool life as the basis of comparison. This speed is denoted by v60.

17. Machinability - example
Solution:
The base material has a machinability rating = Its v60 value can be determined from the Taylor tool life equation as follows:
v60 = 1050/ = 334 ft/min
The cutting speed at a 60 min. tool life for the test material is determined similarly:
v60 = 1320/ = 437 ft/min
Accordingly, the machinability rating can be calculated as
MR (for the test material) = 437/374 = (or 131%)

18. Optimized machining
Cutting speed can be chosen to maximize the production rate or minimize the cost per part (or unit) produced. This is referred to as optimized machining because more than one production variable contributes to the production rate and costs.
Variables: Th - part handling time Co (Cg) – operator (grinder’s) cost rate/min
Tm – machining time Ch – cost of part handling time
Tt – tool change time Cm – cost of machining time
np – number of parts cut by Ctc – cost of tool change time tool during tool life
Tc – cycle time per part Ct – cost per cutting edge
T – tool life Ctp = Ct/np - tool cost per part

19. Maximum production rate - turning
Total time per part produced (cycle time):
Tc = Th + Tm + Tt/np
where Tt/np is the tool change time per part.
Consider a turning operation. The machining time is given by
Tm = p D L/(v f)
The number of parts cut per tool is given by
np = T/Tm= f C(1/n)/(p D L v(1/n -1) )

20. Maximum production rate - turning
Substituting, we get the total cutting time
Tc = Th + p D L/(v f) + Tt[ p D L v(1/n -1)/( f C(1/n) )]
Minimizing cycle time (dTc/dv = 0 ) gives optimum (max) cutting speed and tool life:
vmax = C/[(1 - n) Tt/n]n
Tmax = (1 - n) Tt /n

21. Minimum cost per unit - turning
Cost of part handling time:
Ch = CoTh
Cost of machining time:
Cm = CoTm
Cost of tool change time:
Ctc = CoTt /np

22. Minimum cost per unit - turning
Tool cost per part:
Ctp = Ct /np
Tooling cost per edge:
Disposable inserts Ct = Pt /ne ne = num of edges/insert Pt = original cost of tool
Single point grindable Ct = Pt /ng + Tg Cg “includes purchase price” ng = Num tool lives/tool Tg = time to grind tool

23. Minimum cost per unit - turning
Total cost per part:
Cc = Co Th + Co Tm + Co Tt /np + Ct /np
Substituting for Tm and np:
Cc = Co Th + Co p DL/fv + (CoTt + Ct )pDLv(1/n -1)/( f C(1/n) ) Minimizing cost per part (dCc/dv = 0) gives cutting speed and tool life to minimize machining costs per part:
vmin = C{n Co/[(1 – n)(Ct + CoTt)]}n
Tmin = (1 – n) (Ct + CoTt)/(n Co)

24. Minimum cost per unit - example
Problem statement:
Suppose a turning operation is to be performed with HSS tooling on mild steel (n = 0.125, C = 200 from text table). The workpart has length = 20.0 in. and diameter = 4.0 in· Feed = in./rev. Handling time per piece = 5.0 min and tool change time = 2.0 min. Cost of machine and operator = $30.00/hr, and tooling cost = $3.00 per cutting edge. Find (a) cutting speed for maximum production rate and (b) cutting speed for minimum cost

25. Minimum cost per unit - example
Solution:
Cutting speed for maximum production rate is
vmax = C/[(1 - n) Tt/n]n = 200/[(.875) 2/0.125] = 144 ft/min
Converting Co from $30/hr to $0.5/min, the cutting speed for minimum cost is given by
vmin = C{n Co/[(1 – n)(Ct + CoTt)]}n = = 200{(0.125)(0.5)/[(0.875)( (0.5)(2))]} = 121 ft/min

26. Machining operations
What did we learn?