Preliminary Study on Application of Axiomatic Design in Tolerance Design
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摘要: 公差设计依赖于设计人员的经验、缺乏科学设计理论指导;产品公差设计通常为多功能需求设计,且功能需求之间往往又存在耦合,需要分析和解耦;功能方程组(功能需求与功能尺寸之间建立的函数方程组)的求解存在不同的求解策略,需确定最佳的求解策略。针对此类问题,引入公理设计理论至公差设计领域,首先将公差设计过程看作是功能域-尺寸与公差域的映射,利用独立公理和信息公理等来理解和分析公差设计问题,提高公差设计的科学性;其次,以轴承座装配为例分析了如何利用独立公理指导公差耦合设计的求解;最后,以齿轮组装配公差设计为例分析了独立公理和信息公理在解耦设计和判断求解顺序中的应用。研究结果表明,公理设计的引入可为公差设计提供理论框架和逐步分解、展开的机制,辅助判断和求解耦合设计问题,为公差设计理论由经验设计变为科学设计提供了有益的启示。Abstract: Tolerance design depends on an experience of designers for its lack of guidance of scientific design theory. Product tolerance design was usually a multi-functional requirements design. There was often a coupling between functional requirements, and it needs to be analyzed and decoupled. Because there existed different solution strategies, it needed to determine the best solution to solve the function equations, which were established to meet the geometrical functional requirements. In view of these problems, the theory of axiomatic design was firstly introduced to the field of tolerance design. The tolerance design process was seen as mapping from function domain to dimensions and tolerances domain. The independent axiom and information axiom were used for understanding and analyzing the tolerance design, which can improve its scientific meaning of tolerance design. Secondly, the bearing seat assembly was taken as an example to explain the solution of the tolerance decoupling design by using the independence axiom. Finally, the decoupling design and the solving sequence of tolerance allocation of gear assembly are taken as an example to explain the application of independence axiom and information axiom. It shows that the introduction of axiomatic design can provide a framework and mechanism for gradually decomposing the tolerance design, and assist in judging and solving coupling design problems. It provides a useful inspiration for tolerance design theory from experience design to scientific design.
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Key words:
- tolerance design /
- tolerance design function /
- axiomatic design /
- coupling design
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图 2 轴承座装配示图[16]
表 1 按策略1求解尺寸链1
名称 基本尺寸 Y1=76 X1=28 X2=40 X3=8 封闭环公差 0.2 比例因子 k=0.2/76 计算公式 kX1 kX2 kX3 组成环公差 0.074 0.105 0.021 表 2 按策略1求解尺寸链2
名称 基本尺寸 Y2=80 X1=28 X4=34 X5=2 X6=16 封闭环公差 0.16 预设公差 0.074 比例因子 k=(0.16-0.074)/(80-28) 计算公式 kX4 kX5 kX6 组成环公差 0.056 0.003 0.027 表 3 按策略2求解尺寸链2
名称 基本尺寸 Y2=80 X1=28 X4=34 X5=2 X6=16 封闭环公差 0.16 比例因子 k=0.16/80 计算公式 kX1 kX4 kX5 kX6 组成环公差 0.056 0.068 0.004 0.032 表 4 按策略2求解尺寸链1
名称 基本尺寸 Y1=76 X1=28 X2=40 X3=8 封闭环公差 0.2 预设公差 0.056 比例因子 k=(0.2-0.056)/(76-28) 计算公式 kX2 kX3 组成环公差 0.120 0.024 -
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