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Using Datums for Economic
Process Planning
Dr. R. A. Wysk
IE550
Fall 2008
Process Planning
• Single datum planning
• Multiple datum plans
Process Tolerance Chart
Values in Process Tolerance Charts
typically represent the BEST attainable
values. They also represent single-feature
relationships. We refer to these intra-
feature process planning.
•Process Boundary Matrices
A 2” piece or bar stock needs to be “faced” so
that the required length and surface finish can be
obtained.
4.0
+ .005
- .0
2.0 + .005
- .005
A
Example #1 - The simplest case; single datum, single feature
Solution:
In checking the work piece, datum -A-
becomes the reference plan for the
length, 4.0 . The OD accuracy is
obtained at the rolling mill, and no OD
turning is required. The length needs
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 (remove
1/8 ‘’ stock)
Lathe Facing tool
40 Flip and face front-side Lathe Facing tool
50 Remove and inspect -- --
Process Plan-Example #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
to the same datum -A-.
Example #2 -- Single datum; 2 features.
Solution:
- 4” segment is the same as in Example #1
-Addition segment requires that:
-OD is reduced to 1”
-Length needs to be reduced to 5  .005
Process Plan for Example #2
OP# Description Machine Tool Time
10 Retrieve 2” bar Warehouse
20 Cut to 5.25” Cut-off saw
30 Face backside and
invert
Lathe Facing
40 Turn 1” Dia. @ .25 in
depth
Lathe Turning
(2 passes)
50 Face to 4” Lathe Facing
60 Face to 5” Lathe Facing
70 Remove and Inspect
A
C12
C23
C4
M12
M13
Cij is part specification or
Constraints
Mij is Manufacturing method were
i is the datum feature, and
j is the surface produced
The General Case and Notation.
From the part, you can see that
C12 @ M12
This reads, “C12 comes directly from process
M12 (our facing operation).”
Also from the drawing, one can see that
TC23 = TM12 + TM13
This reads, “the tolerance for feature C23 can be
as large as the sum of the tolerance for
producing M12 and the tolerance for producing
M13”  Tolerance Stacking
Notation: subscript m implies minimum
M implies maximum
C23m = -M12M + M13m
Let’s suppose
0
005
.
4 

12
C 005
.
005
.
1

23
C
Then
TM12 = .005
TC23 = TM12 + TM13
.010 = .005 + TM13
TM13 = .005
If a negative value results then the process
specification is unfeasible
Since C23m = - M12M + M13m
.995 = -4.005 + M13m
5.000 = M13m
 Set the process specifications for M13 at
5.000 - 5.005
.750  .010
All hole features are specified with respect to
datums A-B-C and can be treated as intra-feature
entities.
 Ø .01  C A B
M
Example #4
+ +
+
4 .008
2 holes
.250  .010
 Ø .008 C A B
M
2.0  .01
A
1 1 1
B
1
.5  .01
C
Raw Material 4’’ x 2’’ x .5’’
Process Plan for Example #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 #5
 

B
.5 1 1
.750  .010
  .01  C D E
M
+
.25  MAX
.75
D
.25± .01 A
.5 ±
.01
E
2 holes
.250  .010
  .008 C D E
M
.50 ±.01
C
M12
M13
C23
M14 M15
.25±.01
Raw Material 4’’ x 2’’ x
.5’’
C12 @ M12
TC12 = ± .01
TC23 = TM12 + TM13
C23m = -M12M + M13m
.008 = -.51 + M13m
From 
.518 = M13m
TC23 = TM12 + TM13
.008 = .01 + TM13
TM13 < 0  infeasible
We need to position w.r.t -E-

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ie550cngdtm.ppt

  • 1. Using Datums for Economic Process Planning Dr. R. A. Wysk IE550 Fall 2008
  • 2. Process Planning • Single datum planning • Multiple datum plans
  • 3. Process Tolerance Chart Values in Process Tolerance Charts typically represent the BEST attainable values. They also represent single-feature relationships. We refer to these intra- feature process planning. •Process Boundary Matrices
  • 4. A 2” piece or bar stock needs to be “faced” so that the required length and surface finish can be obtained. 4.0 + .005 - .0 2.0 + .005 - .005 A Example #1 - The simplest case; single datum, single feature
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  • 7. Solution: In checking the work piece, datum -A- becomes the reference plan for the length, 4.0 . The OD accuracy is obtained at the rolling mill, and no OD turning is required. The length needs to be faced to final dimension. + 0.005 - 0.0
  • 8. Oper. Description Machine Tool 10 Retrieve 2’’ Bar Warehouse -- 20 Cut to 4.25’’ length Cut-off saw -- 30 Face backside (remove 1/8 ‘’ stock) Lathe Facing tool 40 Flip and face front-side Lathe Facing tool 50 Remove and inspect -- -- Process Plan-Example #1
  • 9. 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 to the same datum -A-. Example #2 -- Single datum; 2 features.
  • 10. Solution: - 4” segment is the same as in Example #1 -Addition segment requires that: -OD is reduced to 1” -Length needs to be reduced to 5  .005
  • 11. Process Plan for Example #2 OP# Description Machine Tool Time 10 Retrieve 2” bar Warehouse 20 Cut to 5.25” Cut-off saw 30 Face backside and invert Lathe Facing 40 Turn 1” Dia. @ .25 in depth Lathe Turning (2 passes) 50 Face to 4” Lathe Facing 60 Face to 5” Lathe Facing 70 Remove and Inspect
  • 12. A C12 C23 C4 M12 M13 Cij is part specification or Constraints Mij is Manufacturing method were i is the datum feature, and j is the surface produced The General Case and Notation.
  • 13. From the part, you can see that C12 @ M12 This reads, “C12 comes directly from process M12 (our facing operation).” Also from the drawing, one can see that TC23 = TM12 + TM13 This reads, “the tolerance for feature C23 can be as large as the sum of the tolerance for producing M12 and the tolerance for producing M13”  Tolerance Stacking
  • 14. Notation: subscript m implies minimum M implies maximum C23m = -M12M + M13m Let’s suppose 0 005 . 4   12 C 005 . 005 . 1  23 C Then TM12 = .005 TC23 = TM12 + TM13 .010 = .005 + TM13 TM13 = .005
  • 15. If a negative value results then the process specification is unfeasible Since C23m = - M12M + M13m .995 = -4.005 + M13m 5.000 = M13m  Set the process specifications for M13 at 5.000 - 5.005
  • 16. .750  .010 All hole features are specified with respect to datums A-B-C and can be treated as intra-feature entities.  Ø .01  C A B M Example #4 + + + 4 .008 2 holes .250  .010  Ø .008 C A B M 2.0  .01 A 1 1 1 B 1 .5  .01 C Raw Material 4’’ x 2’’ x .5’’
  • 17. Process Plan for Example #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
  • 18. Example #5    B .5 1 1 .750  .010   .01  C D E M + .25  MAX .75 D .25± .01 A .5 ± .01 E 2 holes .250  .010   .008 C D E M .50 ±.01 C M12 M13 C23 M14 M15 .25±.01 Raw Material 4’’ x 2’’ x .5’’
  • 19. C12 @ M12 TC12 = ± .01 TC23 = TM12 + TM13 C23m = -M12M + M13m .008 = -.51 + M13m From  .518 = M13m TC23 = TM12 + TM13 .008 = .01 + TM13 TM13 < 0  infeasible We need to position w.r.t -E-