Transcript Process Tolerance Chart

Using Datums for Economic
Process Planning
Dr. R. A. Wysk
ISE789
Spring 2011
Process Planning
• Single datum planning
• Multiple datum plans
Process Tolerance Chart
•Process Boundary Matrices
Values in Process Tolerance Charts
typically represent the BEST attainable
values. They also represent single-feature
relationships. We refer to these intrafeature process planning.
Example #1 - The simplest case; single datum, single feature
4.0
+ .005
- .0
2.0
+ .005
- .005
A
A 2” piece or bar stock needs to be “faced” so
that the required length and surface finish can be
obtained.
Solution:
In checking the work piece, datum -Abecomes the reference plan for the
+ 0.005
length, 4.0 - 0.0 . The OD accuracy is
obtained at the rolling mill, and no OD
turning is required. The length needs
to be faced to final dimension.
Process Plan-Example #1
Oper.
Description
Machine
Tool
10
Retrieve 2’’ Bar
Warehouse
--
20
Cut to 4.25’’ length
Cut-off saw
--
30
Lathe
Facing tool
40
Face backside (remove
1/8 ‘’ stock)
Flip and face front-side
Lathe
Facing tool
50
Remove and inspect
--
--
Example #2 -- Single datum; 2 features.
5  .005
4 + .005
- 0
2.0
+ 0.01
1.0  .005
- 0
A
Sort of like Example #1 but with a 2nd feature related
to the same datum -A-.
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
10
Retrieve 2” bar
Warehouse
20
Cut to 5.25”
Cut-off saw
30
Facing
40
Face backside and
Lathe
invert
Turn 1” Dia. @ .25 in Lathe
depth
(2 passes)
50
Face to 4”
Lathe
Facing
60
Face to 5”
Lathe
Facing
70
Remove and Inspect
Turning
Time
The General Case and Notation.
C23
C12
C4
M12
M13
A
Cij is part specification or
Constraints
Mij is Manufacturing method were
i is the datum feature, and
j is the surface produced
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
C12
 4  .005
0
C23  1 .005
.005
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
Example #4
4 .008
2 holes
.250  .010
 Ø .008 M
C A B
2.0  .01
+
+
+
1
B
1
1
1
Raw Material 4’’ x 2’’ x .5’’
.750  .010
MC A B
 Ø .01 
A
.5  .01
C
All hole features are specified with respect to
datums A-B-C and can be treated as intra-feature
entities.
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
.750  .010
B

.5
1
1




+
.75
A
.01
E
M CDE
 .008
.25±.01
.50 ±.01
C
.25  MAX
D
.25± .01
.5 ±
2 holes
.250  .010
M CDE
 .01 
C23
M12
M13
M14
M15
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-