Using Datums for Economic Process Planning

Dr. R. A. Wysk

IE550

Fall 2008

Process Planning

• Single datum planning

• Multiple datum plans

Process Tolerance ChartProcess Tolerance Chart

Values in Process Tolerance Charts Values in Process Tolerance Charts typically represent the BEST attainable typically represent the BEST attainable values. They also represent single-feature values. They also represent single-feature relationships. We refer to these intra-relationships. We refer to these intra-feature process planning.feature process planning.

•Process Boundary MatricesProcess Boundary Matrices

A 2” piece or bar stock needs to be “faced” so A 2” piece or bar stock needs to be “faced” so that the required length and surface finish can be that the required length and surface finish can be obtained.obtained.

4.0+ .005

- .0

2.0 + .005

- .005

A

128

Example #1 - The simplest case; single datum, single feature

Solution:Solution:

In checking the work piece, datum -A- In checking the work piece, datum -A- becomes the reference plan for the becomes the reference plan for the length, 4.0 . The OD accuracy is length, 4.0 . The OD accuracy is obtained at the rolling mill, and no OD obtained at the rolling mill, and no OD turning is required. The length needs turning is required. The length needs to be faced to final dimension.to be faced to final dimension.

+ 0.005

- 0.0

Oper. Description Machine Tool

10 Retrieve 2’’ Bar Warehouse --

20 Cut to 4.25’’ length Cut-off saw --

30 Face backside (remove1/8 ‘’ stock)

Lathe Facing tool

40 Flip and face front-side Lathe Facing tool

50 Remove and inspect -- --

Process Plan-Process Plan-Example #1Example #1

1.0 .005

4 + .005

- 0

2.0 + 0.01

- 0

A

5 .005

Sort of like Example #1 but with a 2nd feature related Sort of like Example #1 but with a 2nd feature related to the same datum -A-.to the same datum -A-.

Example #2 -- Single datum; 2 features.

Solution:Solution:

- 4” segment is the same as in Example #1- 4” segment is the same as in Example #1

-Addition segment requires that:-Addition segment requires that:

-OD is reduced to 1”-OD is reduced to 1”

-Length needs to be reduced to 5 -Length needs to be reduced to 5 .005 .005

Process Plan for Example Process Plan for Example #2#2

OP# Description Machine Tool Time

10 Retrieve 2” bar Warehouse

20 Cut to 5.25” Cut-off saw

30 Face backside andinvert

Lathe Facing

40 Turn 1” Dia. @ .25 indepth

Lathe Turning

(2 passes)

50 Face to 4” Lathe Facing

60 Face to 5” Lathe Facing

70 Remove and Inspect

A

C12C23

C4

M12

M13

CCijij is part specification or is part specification or

ConstraintsConstraints

MMijij is Manufacturing method were is Manufacturing method were

i is the datum feature, and i is the datum feature, and

j is the surface producedj is the surface produced

The General Case and Notation.

From the part, you can see thatFrom the part, you can see that

CC12 12 M M1212

This reads, “CThis reads, “C1212 comes directly from process comes directly from process

MM1212 (our facing operation).” (our facing operation).”

Also from the drawing, one can see thatAlso from the drawing, one can see that

TTCC2323 = T = TMM1212 + T + TMM1313

This reads, “the tolerance for feature CThis reads, “the tolerance for feature C2323 can be can be

as large as the sum of the tolerance for as large as the sum of the tolerance for producing Mproducing M1212 and the tolerance for producing and the tolerance for producing

MM1313” ” Tolerance Stacking Tolerance Stacking

Notation: subscript Notation: subscript mm implies minimum implies minimum

MM implies maximum implies maximum

CC2323mm = -M = -M1212MM + M+ M1313mm

Let’s supposeLet’s suppose

0005.4 12C 005.

005.123C

ThenThen

TTMM1212 = .005 = .005

TTCC2323 = T = TMM1212 + T + TMM1313

.010 = .005 + T.010 = .005 + TMM1313

TTMM1313 = .005 = .005

If a negative value results then the process If a negative value results then the process specification is unfeasible specification is unfeasible

Since CSince C2323mm = - M = - M12M12M + M + M1313mm

.995 = -4.005 + M.995 = -4.005 + M1313mm

5.000 = M5.000 = M1313mm

Set the process specifications for MSet the process specifications for M1313 at at

5.000 - 5.005 5.000 - 5.005

.750 .010

All hole features are specified with respect to All hole features are specified with respect to datums A-B-C and can be treated as intra-feature datums A-B-C and can be treated as intra-feature entities.entities.

Ø .01 C A B M

Example #4Example #4

+ ++

4 .0082 holes

.250 .010

Ø .008 C A B M

2.0 .01

A

1 1 1

B1

.5 .01C

Raw Material 4’’ x 2’’ x .5’’

Process Plan for Process Plan for Example #4Example #4

OP# Description Machine Tool

10 Load part in vise Fadal CNC

20 Drill 1st small hole Fadal CNC .25 drill

30 Drill 2nd small hole Fadal CNC .25 drill

40 Drill large hole Fadal CNC .25 drill

50 Unload and inspect

Example #5Example #5

B

.5 1 1

.750 .010

.01 C D EM

+

.25 MAX

.75D

.25± .01 A.5 ± .01

E

2 holes.250 .010

.008 C D EM

.50 ±.01

CM12

M13

C23

M14 M15

.25±.01

Raw Material 4’’ x 2’’ x .5’’

CC1212 M M1212

TTCC1212 = ± .01 = ± .01

TTCC2323 = T = TMM1212 + T + TMM1313

CC2323mm = -M = -M12M12M + M + M1313mm

.008 = -.51 + M.008 = -.51 + M1313mm

From From

.518 = M.518 = M1313mm

TTCC2323 = T = TMM1212 + T + TMM1313

.008 = .01 + T.008 = .01 + TMM1313

TTMM1313 < 0 < 0 infeasibleinfeasible

We need to position w.r.t -E-We need to position w.r.t -E-