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[求助] 帮忙解答关于公差尺寸链的疑惑

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发表于 2010-2-22 17:53:25 | 只看该作者 回帖奖励 |正序浏览 |阅读模式
图中是配合的两个零件(示意图不必较真),目标是在列出的尺寸基础上,计算图示间隙的尺寸及公差水平。对于定位间隙的处理,请大家伙各抒己见。

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发表于 2010-3-2 18:23:10 | 只看该作者
我认为可以有这样的思路:图中间隙受两个条件影响,柱孔的gap C和A,B,D,E的制造公差。
1.不考虑公差,也不考虑柱孔组装偏差,则gap=A+B+C-D-E=(A-E)+(B+C-D)=5。
2.增加考虑组装偏差,4.75<gap<5.25
3增加考虑制造公差,分两种情况极限公差和统计公差,极限公差=(0.3+0.2)+(0.1/2+0.1/2+0)=0.6,此时gap的范围是4.15<gap<5.85。统计公差=根号(0.3^2+0.2^2+0.05^2+0.05^2)=0.367,此时gap范围4.383<gap<5.617
(其中C公差为0)
21
发表于 2015-1-20 15:02:40 | 只看该作者
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20
发表于 2015-1-20 14:00:45 | 只看该作者
本帖最后由 yakeke 于 2015-1-20 14:03 编辑

关于本问题,个人觉得应该是传动链上公差标注重复了。
我认为可以有这样的思路:图中间隙受两个条件影响,柱孔的gap C和A,B,D,E的制造公差。
1.不考虑公差,也不考虑柱孔组装偏差,则gap=A+B+C-D-E=(A-E)+(B+C-D)=5。
2.增加考虑组装偏差,4.75<gap<5.25
3增加考虑制造公差,分两种情况极限公差和统计公差,极限公差=(0.3+0.2)+(0.1/2+0.1/2+0)=0.6,此时gap的范围是4.15<gap<5.85。统计公差=根号(0.3^2+0.2^2+0.05^2+0.05^2)=0.367,此时gap范围4.383<gap<5.617
(其中C公差为0)
这个解答应该是准确的,但在3。极限公差=(0.3+0.2)+(0.1/2+0.1/2+0)=0.6,感觉这里不应该是1/2.
19
发表于 2011-1-10 14:32:03 | 只看该作者
iDFMA 发表于 2010-2-24 12:52
1.决定定位柱与孔的间隙。

定位柱与孔的间隙、定位柱的尺寸、和孔的尺寸是一个尺寸链。定位柱与孔的间隙需要保证定位柱在装进孔的时候不会发生干涉,至于具体是多少取决于定位柱和孔的所能达到的尺寸精度以及公差 ...

个人认为装配偏移,基本尺寸为0
公差应为+/-[(孔基本尺寸+上偏差)-(轴基本尺寸+下偏差)]/2较为合适。
孔和轴都是在最小实体状态下的尺寸

本帖最后由 feiyue_2001 于 2011-1-10 14:53 编辑
18
发表于 2011-1-10 11:56:57 | 只看该作者
是一份好料
17
发表于 2010-10-21 13:30:11 | 只看该作者
学习,学习
16
发表于 2010-10-21 13:22:57 | 只看该作者
好复杂, 能总结一下到底应该咋做是标准答案吗?
15
发表于 2010-8-9 17:07:08 | 只看该作者
谁能说一下正确答案呀?
14
发表于 2010-8-9 16:36:20 | 只看该作者
说到这个cetol
就用他来做了一下分析,软件对间隙的处理是:把公称尺寸定为0,公差为0.350,但是他提供几种间隙方式,见图片

分析结果见

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13
发表于 2010-8-9 16:03:28 | 只看该作者
发的帖子,不能显示图,怎么能显示图?
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