me 482 - manufacturing systems machining operations by ed red machining operations by ed red

ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems

MachiningOperations

by

Ed Red

MachiningOperations

by

Ed Red


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)

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)


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

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


Machining operations on lathe(other than normal turning)

Machining operations on lathe(other than normal turning)

Chamfer

Taper Contour

FormFacing

Cutoff Threading

Boring Drilling

Knurling


Two types of milling operationsTwo types of milling operations

Peripheral Face


Face milling operationsFace milling operations

Facing Partial facing

End milling

Profiling

Pocketing Surface contouring


Face millingmovementsFace millingmovements

Peripheral milling cutting positions

Face milling cutting positions

Full face cut

Offset face cut


Milling cutter time analysis 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

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


Milling time analysis 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

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:


Partial face A = O = w (D – w)

Machining time:

Tm = (L + 2A)/fr


Milling time analysis - example 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.

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.


Milling time analysis - example Milling time analysis - example

Solution? Numbers?Solution? Numbers?


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)


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)


Tolerance by processTolerance by process


Surface finish by processSurface finish by process


Surface finish by geometrySurface finish by geometry

Ideal roughness,

Ri = f2/(32 NR)

where

NR = tool nose radius

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.

Actual roughness,

Ra = rai Ri (about 2 x Ri )

because of edge effects, chip

interactions, surface tearing, etc.


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 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

MachinabilityMachinability What is a free machining steel?

http://www.sandmeyersteel.com/303.html#1


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.

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 - exampleMachinability - 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%)

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 - exampleMachinability - example


Optimized machiningOptimized 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

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


Maximum production rate - turningMaximum 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) )

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) )


Maximum production rate - turningMaximum production rate - turning

Substituting, 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

Substituting, 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


Minimum cost per unit - turningMinimum 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

Cost of part handling time:

Ch = CoTh

Cost of machining time:

Cm = CoTm

Cost of tool change time:

Ctc = CoTt /np



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

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



Total 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)

Total 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)


Minimum cost per unit - exampleMinimum 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 = 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

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 = 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


Minimum cost per unit - exampleMinimum 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

Solution:

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


Machining operationsMachining operations

What did we learn?What did we learn?

me 482 - manufacturing systems machining operations by ed red machining operations by ed red

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