互换性技术习题.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,标注,30K7,和,50M7,采用包容原则。,底面,F,的平面度公差为,0,.,02mm,；,30K7,孔和,50M7,孔的内端面对它们的公共轴线的圆跳动公差为,0,.,04 mm,。,30K7,孔和,50M7,孔对它们的公共轴线的同轴度公差为,0,.,03mm,。,6-11H10,对,50M7,孔的轴线和,F,面的位置度公差为,0,.,05mm,，,基准要素的尺寸和被测要素的位置度公差应用最大实体要求。,E,E,0,.,02,0,.,04,A,B,A-B,0,.,03,A-B,0,.,05,B,M,M,C,C,210h7,70H7,4-,20H8,0.01,0.04,A,A,0.02,A,0.03,B,B,0.15,A,B,形,位,公差标注举例,将下列技术要求标注在图上。,（,1,）,100h6,圆柱表面的圆度公差为,0.005mm,。,（,2,）,100h6,轴线对,40P7,孔轴线的同轴度公差为,0.015,。,（,3,）,40P,孔的圆柱度公差为,0.005mm,。,（,4,）,左端的凸台平面对,40P7,孔轴线的垂直度公差为,0.01 mm,。,（,5,）,右凸台端面对左凸台端面的平行度公差为,0.02mm,。,0.005,0.015,C,0.005,0.01,C,0.02,A,A,标注的解释,说明右图中标注的形位公差的含义。,如图所示销轴的三种形位公差标注，它们的公差带有何不同？,图,a,为给定方向上素线的直线度，,其公差带为宽度等于公差值,0,02mm,的两平行平面间的区域。,图,b,为轴线在任意方向的直线度，,其公差带为直径等于公差值,0,02mm,的圆柱体内的区域。,图,c,为给定方向上被测素线对基准素线的平行度,，其公差带为宽度等于公差值,0,02mm,且平行于基准,A,的两平行平面间的区域。,(a),(b),(c),改正下图图样上形位公差的标注错误（形位公差项目不变）。,试将以下要求用形位公差代号标注在图中,1,、,90h6,轴的轴心线对,30H7,孔的轴心线的同轴度公差为,0.025mm,；,2,、,60h7,轴的圆柱度公差为,0.02mm,；,3,、,B,面对,30H7,孔轴心线的端面圆跳动公差为,0.04mm,；,4,、,C,面对,B,面的平行度公差为,0.04mm,；,5,、,30H7,孔的轴心线对,B,面的垂直度公差为,0.03mm,；,什么是公差原则？,确定,形位公差与尺寸公差,之间相互关系所遵循的原则,公差原则的分类？,采用公差原则,标记,遵循边界及边界尺寸,mm,给定的尺寸公差,给定的形位公差,MMC,下允许的最大形位误差值,LMC,下允许的最大形位误差值,尺寸合格,独立原则,无,无,IT,t,t,t,D,min,D,a,D,max,包容要求,在单一要素尺寸公差带后加注,E,最大实体边界,MMS(dmax,，,Dmin,),IT,0,0,IT,D,min,D,a,D,max,最大实体要求,在形位公差框格相应格公差值后加,M,最大实体实效边界,MMVS,MMSt,IT,t,t,t+IT,D,min,D,a,D,max,0.008,A,图例,采用公差原则,边界及边界尺寸,mm,给定的形位公差,mm,可能允许的最大形位误差值,mm,a,独立原则,无,0.008,0.008,b,包容要求,最大实体边界,20,0,0.021,c,最大实体要求,最大实体实效边界,39.9,0.1,0.2,例题：,a,b,c,E,M,0.1,A,练习,下列配合属于哪种基准制的哪种配合，确定其配合的极限间隙（过盈）和配合公差。并画出其公差带图。,50H8/f7,，,30K7/h6,，,30H7/p6,0,0,0,+,+,+,-,-,-,50,30,30,+0.039,-0.025,-0.050,+0.006,+0.021,+0.035,+0.022,-0.015,-0.013,4,、正计算（校核计算）举例,校核计算的步骤是：,根据装配要求确定封闭环；寻找组成环；画尺寸链线图；判别增环和减环；由各组成环的基本尺寸和极限偏差验算封闭环的基本尺寸和极限偏差。,如图,a,所示的结构，已知各零件的尺寸：,A,1,=30 mm,，,A,2,=A,5,=5 mm,，,A,3,=43 mm,，,A,4,=3 mm,，设计要求间隙,A,0,为,0,1,0,45mm,，,试做校核计算。,解,（,l,）,确定封闭环为要求的间隙,A,0,；,寻找组成环并画尺寸链线图（上图,b,）；,判断,A,3,为增环，,A,1,、,A,2,、,A,4,和,A,5,为减环。、,（,2,）封闭环的基本尺寸,A,0,A,3,（,A,1,+A,2,+A,4,+A,5,）,=43,（,30+5+3+5,）,=0,即要求封闭环的尺寸为,0 mm,。,（,3,）,计算封闭环的极限偏差,ES,。,ES,3,（,EI,1,+EI,2,+EI,4,+EI,5,）,+0,18,（,0,130,0750,040,075,）,=+0,50,EI,。,EI,3,（,ES,1,+ES,2,+ES,4,+ES,5,）,0,02mm,（,0,0,0,0,）,mm,0,02mm,（,4,）,计算封闭环的公差,T,。,T,1,+T,2,+T,3,+T,4,+T,5,=0,13,0,075,十,0,16,0,04,十,0,075=0,48mm,校核结果表明，封闭环的上、下偏差及公差均已超过规定范围。,