互换性与技术测量第六版重点课后习题答案PPT参考课件.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,*,习题,0,1,R10,系列：,250 315 400 500 630 800 1000 1250 1600 2000 2500 3150,1,2,孔或轴,最大极限尺寸,最小极限尺寸,上偏差,下偏差,公差,尺寸标注,孔,:,10,9.985,9.97,-0.015,-0.03,0.015,孔,:,18,18.018,18,+0.018,0,0.018,孔,:,30,30.012,29.991,+0.012,-0.009,0.021,轴,:,40,39.95,39.888,-0.050,-0.112,0.062,轴,:,60,60.041,60.011,+0.041,+0.011,0.030,轴,:,85,85,84.978,0,-0.022,0.022,习题,1-1,填表,2,习题,1,2,0,+,-,20,孔,轴,+0.033,-0.065,-0.098,0,+,-,55,孔,+0.030,轴,+0.060,+0.041,0,+,-,35,孔,+0.007,轴,-0.018,-0.016,（,1,）间隙配合,X,max,=ES-ei=+0.033-(-0.098)=+0.131 mm,X,min,=EI-es=0-(-0.065)=+0.065 mm,T,f,=,X,max,-X,min,=,0.131-0.065,=0.066 mm,（,2,）过渡配合,X,max,=ES-ei=0.007+0.016=+0.023 mm,Y,max,=EI-es=-0.018 mm,T,f,=X,max,Y,max,=+0.023-(-0.018)=0.041 mm,（,3,）过盈配合,Y,max,=EI-es=0-(+0.060)=-0.060mm,Y,min,=ES-ei=+0.03-(+0.041)=-0.011mm,T,f,=,min,-,max,=-0.011-(-0.060)=0.049 mm,3,习题,1,3,0,+,-,50,孔,轴,+0.039,-0.025,-0.050,0,+,-,80,孔,+0.01,轴,+0.13,-0.12,0,+,-,30,孔,+0.006,轴,-0.015,-0.013,（,1,）基孔制间隙配合,X,max,=0.089 mm,X,min,=0.025 mm,T,f,=0.064 mm,（,2,）基轴制过渡配合,X,max,=+0.019 mm,Y,max,=-0.015 mm,T,f,=0.034 mm,（,3,）基轴制间隙配合,X,max,=+0.25mm,X,min,=+0.01mm,T,f,=0.24 mm,4,习题,1,3,0,+,-,180,孔,+0.04,轴,+0.235,+0.21,0,+,-,18,孔,-0.004,轴,-0.015,-0.008,（,4,）基孔制过盈配合,（,6,）基轴制过渡配合,X,max,=+0.004 mm,Y,max,=-0.015 mm,T,f,=0.019 mm,（,5,）基孔制过盈配合,Y,max,=-0.235mm,Y,min,=-0.17mm,T,f,=0.065 mm,0,+,-,140,孔,+0.063,轴,+0.126,Y,max,=-0.126mm,Y,min,=0mm,T,f,=0.126 mm,5,习题,1,4,（,1,）,（,2,）,（,3,）,（,4,）,（,5,）,（,6,）,6,习题,1,5,（,1,）,解：,选择基准制,。无特殊要求，选择基孔制，孔的基本偏差代号为,H,，,EI=0,。,确定公差等级,。,假设孔、轴同级，,T,D,T,d,T,f,/2,33,m,查,表,1-8,孔轴公差等级均为,IT8,之间，,IT8=33,m.,选择配合种类,确定轴的基本偏差,轴,0,+,-,25,孔,+0.033,由于间隙配合，轴的基本偏差为,es,。,由于,X,min,=EI-es,es=EI-X,min,0-0.020,-0.020,mm,基孔制，孔为,H8,，,EI=0,ES=EI+T,D,=0.033mm,即,7,查表,1,10,得：轴的基本偏差代号为,f,，,es=-0.020,mm,ei=es-T,d,=-0.020-0.033=-0.053,mm,即,配合为,验算：,满足要求。,8,习题,1,5,（,2,）,解：,选择基准制,。无特殊要求，选择基孔制，孔的基本偏差代号为,H,，,EI=0,。,确定公差等级,。,查,表,1-8,IT6=16,m.IT7=25,m.,选择配合种类,确定轴的基本偏差,轴,0,+,-,25,孔,+0.025,由于过盈配合，轴的基本偏差为,ei,。,由于,Y,min,=ES-ei,ei=ES-Y,min,0.025+0.035,+,0.060,mm,基孔制，孔为,H7,，,EI=0,ES=EI+T,D,=0.025mm,即,根据工艺等价原则，孔取,IT7,，轴取,IT6,。,9,查表,1,10,得：轴的基本偏差代号为,u,，,ei=+0.060,mm,es=ei+T,d,=0.060+0.016=0.076,mm,即,配合为,验算：,满足要求。,10,习题,1,5,（,3,）,解：,选择基准制,。无特殊要求，选择基孔制，孔的基本偏差代号为,H,，,EI=0,。,确定公差等级,。,查,表,1-8,IT8=46,m.IT7=30,m.,选择配合种类,确定轴的基本偏差,轴,0,+,-,25,孔,+0.046,由于过渡配合以及极限间隙过盈值，得出轴的基本偏差为,ei,。,由于,X,max,=ES-ei,ei=ES-X,max,0.046-0.044,+,0.002 m,m,基孔制，孔为,H8,，,EI=0,ES=EI+T,D,=0.046mm,即,根据工艺等价原则，孔取,IT8,，轴取,IT7,。,11,查表,1,10,得：轴的基本偏差代号为,k,，,ei=+0.002,mm,es=ei+T,d,=0.002+0.030=0.032,mm,即,配合为,验算：,满足要求。,12,习题,1,6,通过查表,1-8,得知：,d,1,=100mm,，,T,d1,=35m,的轴为,IT7,级,；,d,2,=10mm,，,T,d2,=22m,的轴为,IT8,级；,由于精度等级越高加工难度越大，因此,d,2,=10mm,，,T,d2,=22m,的轴加工更容易。,13,习题,2,1,试从,83,块一套的量块中，同时组合下列尺寸：,29.875,，,48.98,，,40.79,，,10.56,？,29.875:1.005/1.37/7.5/20,48.98,：,1.48/,7.5,/40,40.79,：,1.29/9.5/30,10.56,：,1.06/,9.5,应尽量减少量块组的量块数目。,14,习题,2,2,仪器读数在,20mm,处的示值误差为,+0.002mm,，当用它测量工件时，读数正好是,20mm,，问工件的实际尺寸是多少？,15,习题,2,3,16,弦长,L=95mm,，,弓高,h,30mm,，,U,L,2.5m,，,U,h,2m,习题,2,7,17,习题,2,8,：,：,18,习题,2,8,：,由于,U,2,U,1,U,3,所以,第三种方案最优。,19,习题,2,9,查表,2-7,：,A=12,m u,1,=10,m,上验收极限,200-0.012=199.988mm,下验收极限,200-0.115+0.012=199.897mm,设工件尺寸为,200h9,，试按,光滑工件尺寸的检验,标准选择计量器具，并确定检验极限,。,选用分度值为,0.01,的外径千分尺,u=0.0070.01,。,20,3-3,第,3,章 几何公差及检测,圆锥面圆度公差,0.006,；圆锥面直线度公差,0.002(+);,圆锥面的斜向圆跳动公差,0.012,；,左端面相对于孔轴线基准,B,的垂直度公差,0.015,；,右端面相对于左端面基准的平行度公差,0.005,；,内孔素线相对于孔轴线基准,B,的平行度公差,0.01,；,21,3-4,a,：端面相对于基准轴线,A,的垂直度公差,0.05,；公差带定义是与基准轴线,A,垂直距离为,0.05,的两个平行平面之间的区域；,b,：端面相对于基准轴线,A,的端面圆跳动公差,0.05,；公差带定义是与基准轴线,A,同轴的任一直径位置的测量圆柱面上距离为,0.05,的两圆之间的区域；,c,：端面相对于基准轴线,A,的端面全跳动公差,0.05,；公差带定义是与基准轴线,A,垂直距离为,0.05,的两个平行平面之间的区域；,22,3-6.,23,3-7.,24,3-8.,25,3-9.,26,3-11.,公差原则中，独立原则和相关要求的主要区别何在？包容要求和最大实体要求有何异同？,独立原则中尺寸公差和形位公差相互无关；相关要求中尺寸公差和形位公差相互有关。,包容要求和最大实体要求都是相关要求，包容要求遵循最大实体边界，仅用于形状公差；最大实体要求遵循最大实体实效边界。,27,图例,采用的公差原则,遵守的理想边界,边界尺寸,给定的垂直度公差,允许的最大垂直度误差,a,最大实体要求的零几何公差,最大实体边界,20,0,0.13,b,可逆的最大实体要求,最大实体实效边界,19.95,0.05,0.18,b,最小实体要求,最小实体实效边界,20.18,0.05,0.18,c,独立原则,0.05,0.05,3-12,28,习题,5,1,T=2.4 Z=2.8,T=3 Z=4,-41,T,Z,-29,-38.6,-26.6,0,50,+,-,+25,T,+2.5,+5.5,Z,+22,-25,H7,f6,TT,TS,ZT,-26.2,-27.8,-39.8,29,习题,5,3,孔合格条件：,轴合格条件：,30,习题,6,3,5,级,210,P137,表,6,2,，,6,3,表,1,8,，,1,10,1-11,0,+,-,50,轴,孔,+5.5,-8,-5.5,0,+,-,90,孔,轴,+0.016,-10,-0.006,X,max,=5.5,m,Y,max,=-13.5,m,Y,av,=-4,m,X,max,=26,m,Y,max,=-6,m,X,av,=10,m,31,A,1,/2,A,2,A,3,/2,A,4,习题,7,4,A,4,是封闭环，如图所示判断出,A,2,和,A,3,/2,为增环，,A,1,/2,为减环。,32,A,2,的中间公差,33,A,2,的公差,A,2,的上、下偏差,A,2,的尺寸,34,习题,9,1,外螺纹,M20X2-5g6g,P182,表,9-4,，,9-5,9-7,中径,5g,es=-38,m ei=-38-125=-163m,大径,6g,es=-38,m ei=-38-280=-318m,内螺纹,M20X2-6H,中径,6H,EI=0,ES=212m,小径,6H,EI=0,ES=375m,35,习题,10-1,0.04 A,35,-0.2,0,40,A,3.2,3.2,12.5,1.6,12N9,（）,0,-0.043,对称度数值查,P115,表,3-14,36,习题,10-1,对称度数值查,P115,表,3-13,40,A,3.2,3.2,12JS9,（）,0.0215,0.04 A,43.3,+0.2,0,1.6,37,习题,10-2,内花键,外花键,N=6,NdDB,38,