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ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
MachiningOperations
by
Ed Red
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
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)
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
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
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Machining operations on lathe(other than normal turning)
Chamfer
Taper Contour
FormFacing
Cutoff Threading
Boring Drilling
Knurling
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Two types of milling operations
Peripheral Face
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Face milling operations
Facing Partial facing
End milling
Profiling
Pocketing Surface contouring
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Face millingmovements
Peripheral milling cutting positions
Face milling cutting positions
Full face cut
Offset face cut
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Milling cutter time analysis Spindle rpm related to cutter diameter and speed:
N (rpm) = v/( D)
Feedrate in in/min:
fr = N nt f
where f = feed per tooth
nt = number of teeth
MRR is
MRR =w d fr
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
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
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
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.
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
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/( D)
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Tolerance by process
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Surface finish by process
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Surface finish by geometryIdeal 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.
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
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
Machinability What is a free machining steel?http://www.sandmeyersteel.com/303.html#1
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
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.
Machinability - example
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
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%)
Machinability - example
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Optimized machiningCutting 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
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
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 = D L/(v f)
The number of parts cut per tool is given by
np = T/Tm= f C(1/n)/( D L v(1/n -1) )
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Maximum production rate - turningSubstituting, we get the total cutting time
Tc = Th + D L/(v f) + Tt[ 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
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
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
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
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
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Minimum cost per unit - turningTotal cost per part:
Cc = Co Th + Co Tm + Co Tt /np + Ct /np
Substituting for Tm and np:
Cc = Co Th + Co DL/fv + (CoTt + Ct )DLv(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)
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Minimum cost per unit - exampleProblem 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 = 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
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Minimum cost per unit - exampleSolution:
Cutting speed for maximum production rate is
vmax = C/[(1 - n) Tt/n]n = 200/[(.875) 2/0.125]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)(3.00 + (0.5)(2))]}0.125
= 121 ft/min
ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems
Machining operations
What did we learn?
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