尺寸链计算方法

Xmini is negative so there is the possibility to have an interference.
Conclusion (preliminary) The proposed tolerances and dimensions are NOK .
9
Example. Quadratic method.
级别: 级别 :
0
目标公差: 目标公差 : 计算公差: 计算公差 :
Nominal 名义值: Nominal名义值 : + 0,5 + 0,2
C2011
C
C2009
Repère序号 序号
级别
Designation des maillons尺寸描述 图纸 尺寸描述 Plan图纸 The Rivet step length
CHAINES DE COTES 公差配合
Designation:The gap between fixing brkt and lower welding brkt Observation:备注 备注
Hiérarchisation特性 : 特性 N° Effet Client 客户要求 :
CC0001
CU MU L PROB AB ILIST E IT ²P FAISABILITE PREVISIONNELLE IT/D(Quadratique)能力预测值 能力预测值 N o minal 名义值 CONCLUSION结论 结论: Arithmetique 结论 0,20 mm 0,50 -0,10 mm 0 Sce部门 : R&D Auteur analyse编辑 : DATE : 部门 编辑
3
The theoretical part
The real part
4
Traditional stack up
Nominal value : b Tolerance : tb
= =
5
Quadratic method
tX <
X Nominal + 1/2
>
•This method can be used if the stack up has a minimum of 5 dimensions . •It assumes the production processes are capable . •It is not recommend to use this method for key safety dimensions
= X mini
9 / 2) = 1,5
Xmini is positive so we have always a freeplay. Conclusion (preliminary) The proposed tolerances and dimensions are OK . The tolerance can even be larger.
版本
名义值
TOL. + 0,1 + 0,0 + 0,1 - 0,1 + 0,0 + 0,0 IT BE = Dq =
公差值
IT²平方 平方 0,01 0,04 0 0,05
C2009 C2011 -
+ -
CL0000457 Tolérance marche appui flasque
R/S/B Lower pivot bracket thinkness Profondeur embouቤተ መጻሕፍቲ ባይዱi flasque CL0000461 0
Quadratic method
Nominal value : b Tolerance : tb
X Nominal - 1/2
113-110-( a= 113 +/- 1 b= 20 +/- 0,5 c= 40 +/- 0,5 d= 10 +/- 0,5 e= 20 +/- 0,5 f= 20 +/- 0,5
8
Example . Traditional stack up
Nominal value : b Tolerance : tb
Traditional stack up
= = 113-110-(7/2)= -0,5
a= 113 +/- 1 b= 20 +/- 0,5 c= 40 +/- 0,5 d= 10 +/- 0,5 e= 20 +/- 0,5 f= 20 +/- 0,5
6
Probabilistic method
>
7
Example
Nominal value : b Tolerance : tb
Proposed dimensions and tolerances to be checked a= 113 +/- 1 b= 20 +/- 0,5 Working conditions: we need a free play , X must be c= 40 +/- 0,5 bigger than zero . d= 10 +/- 0,5 e= 20 +/- 0,5 Are the proposed tolerances and dimensions OK ? f= 20 +/- 0,5
Traditional stack up 113-110-(7/2)= -0,5
Result NOK
Quadratic method 113-110-(
Result OK
9 / 2) = 1,5
Probabilistic method 113 - 110 - (
Result OK
9 / 2) = 0,4
10
Example. Probabilistic method.
Probabilistic method
Nominal value : b Tolerance : tb
X Nominal 113 - 110 - ( a= 113 +/- 1 b= 20 +/- 0,5 c= 40 +/- 0,5 d= 10 +/- 0,5 e= 20 +/- 0,5 f= 20 +/- 0,5 9 / 2) = 0,4
= > X mini
X mini is positive so we have always a freeplay. Conclusion (preliminary) The proposed tolerances and dimensions are OK .
11
Comparison of the different methods.
Stack up tolerances
1
Summary: 1) 2) 3) 4) 5) Stack up tolerances usefulness Traditional calculation Quadratic and probabilistic method Stack up tolerances calculation sheet DFSS tools
0,39 Dp = 1,34 IT / Dq = Quadratique 0,16 mm -0,06 mm 00/01/1900
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Design for Six Sigma
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Q&A
26
A A 0
+ 3,0 + 2,5 + 0,0
0,1 0,2 0 0,30 0,22
+ 0 CUMUL ARITHMETIQUE 和值 IT²A
CUMUL QUADRATIQUE平方根 平方根 IT²Q
00000
Maxi 0,70
Probabilist (Pour info) 0,24 -0,14 Visa签名 : 签名
When possible the use of the quadratic or probabilistic method allow to accept larger tolerance, or to have smaller free play.
12
Stack up tolerance calculation sheet
2
Introduction
Stack up tolerances usefulness
Any physical part has some variations in its dimensions, whatever the production process utilized . The 3D part shown on the computer screen will in fact never physically exist. The cad model shows only the theoretical nominal of the design. The real part will be always different. The study of the impact the dimension tolerances have on the design’s main functions is integrally part of the engineering design activity. Late or no stack up during the design phase can drive to complete concept redesign and or tool scrap/modification. Tolerance management is a key thing in mechanical engineering.