# Process Tolerance Chart - Penn State Engineering

Using Datums for Economic Process Planning Dr. R. A. Wysk IE550 Fall 2008 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 2.0 + .005 - .0 + .005 128 - .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. Oper. 10 Process PlanExample #1 Description Machine Tool 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 A M13 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 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 Lets suppose C12 4 .005 0 C 23 1.005 .005 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 Example #4 4 .008 2 holes .250 .010 .008 MC A B 2.0 .01 + + + 1 1

.750 .010 .01 MC A B A 1 B 1 Raw Material 4 x 2 x .5 .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 2 holes .250 .010 1 .01 MC D E .25 MAX

+ .75 D .25 .01 .5 A .0 1 E .008M C D E .25.01

.50 .01 C C23 M12 M13 M14 M15 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

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