Machining Operations

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MachiningOperations by Ed Red
POWERPOINT!
JOHN HUERTA
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• 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)

CCHS
C C H S
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• 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

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Machiningoperationson lathe (other than normal
turning)
Facing
Taper
Form
Contour
Threading
Cutoff
Chamfer
Boring
Knurling
Drilling
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Two types of milling operations
Peripheral
Face
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Face milling operations
End milling
Facing
Partial facing
Profiling
Pocketing
Surface contouring
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Face millingmovements
Peripheral milling cutting positions
Face milling cutting positions
Offset face cut
Full face cut
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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
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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
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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
0.150 in. Determine the time to make one pass
across the surface and the metal removal rate
during the cut.
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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)
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Tolerance by process
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Surface finish by process
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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.
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Machinability

• 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

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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.
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Machinability - example
Solution The base material has a machinability
rating 1.0. Its v60 value can be determined
from the Taylor tool life equation as follows
v60 1050/600.28 334 ft/min The cutting speed
at a 60 min. tool life for the test material is
determined similarly v60 1320/600.27 437
ft/min Accordingly, the machinability rating can
be calculated as MR (for the test material)
437/374 1.31 (or 131)
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Optimized machining
Cutting speed can be chosen to maximize the
production rate or minimize the cost per part (or
unit) produced. This trade-off is referred to as
optimized machining. Optimization is possible
because more than one production variable
contributes to the production rate and production
costs. VariablesTh - part handling time Co
(Cg) operator (grinders) cost rate per 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
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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) )
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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/nn Tmax (1 - n)
Tt /n
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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
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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
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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 Cn
Co/(1 n)(Ct CoTt)n Tmin (1 n) (Ct
CoTt)/(n Co)
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Minimum cost per unit - example
Problem statement Suppose a turning operation is
to be performed with HSS tooling on mild steel (n
0125, C 200 from text table). The workpart
has length 20.0 in. and diameter 4.0 in Feed
0.010 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
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Minimum cost per unit - example
Solution Cutting speed for maximum production
rate is vmax C/(1 - n) Tt/nn 200/(.875)
2/0.1250.125 144 ft/min Converting Co from
30/hr to 0.5/min, the cutting speed for minimum
cost is given by vmin Cn Co/(1 n)(Ct
CoTt)n 200(0.125)(0.5)/(0.875)(3.00
(0.5)(2))0.125 121 ft/min
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Machining operations
What did we learn?