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
TC23 = TM12 + TM13
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
TM12 = .005
TC23 = TM12 + TM13
.010 = .005 + TM13
TM13 = .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
TC12 = ± .01
TC23 = TM12 + TM13
C23m = -M12M + M13m
.008 = -.51 + M13m
From
.518 = M13m
TC23 = TM12 + TM13
.008 = .01 + TM13
TM13 < 0 infeasible
We need to position w.r.t -E-