第3题:
用查表法确定下列配合的孔、轴的极限偏差、计算极限间隙或过盈,并画出公差带图。ф20H8/f7
(1)φ20H8( 0 +0.033 ) φ20f7( -0.041 -0.020 ) X max =[+0.033-(一0.041)]mm=+0.074mm X min =[0一(一0.020)]mm=+0.020mm X av =[(+0.074+0.020)/2]mm=+0.047mm T f =[+0.074一(+0.020)]mm=0.054mm 间隙配合 (2)φ14H7( 0 +0.018 ) φ14r6( +0.023 +0.034 ) Y max =(0—0.034)mm=一0.034mm Y min =(+0.018—0.023)mm=0.005mm Y av =[((一0.034)+(一0.005))/2]mm=一0.0195mm T f =[一0.005一(一0.034)]mm=0.029mm 过盈配合 (3)φ30M8( -0.029 +0.004 ) φ30h7( -0.021 0 ) Y max =(一0.029一0)mm=一0.029mm X max =[+0.004一(一0.021)]mm=+0.025mm Y av =[((一0.029)+(+0.025))/2]mm=一0.002mm T f =[+0.025一(一0.029)]mm=0.054mm 过渡配合 (4)φ45JS6(±0.008) φ45h5( -0.011 0 ) Y max =(一0.008—0)mm=一0.008mm X max =[+0.008一(一0.01 1)]mm=+0.019mm Y av =[(一0.008)+(+0.019)]/2mm=+0.0055mm T f =[+0.019一(一0.008)]mm=0.027mm 过渡配合 (1)φ20H8(0+0.033)φ20f7(-0.041-0.020)Xmax=[+0.033-(一0.041)]mm=+0.074mmXmin=[0一(一0.020)]mm=+0.020mmXav=[(+0.074+0.020)/2]mm=+0.047mmTf=[+0.074一(+0.020)]mm=0.054mm间隙配合(2)φ14H7(0+0.018)φ14r6(+0.023+0.034)Ymax=(0—0.034)mm=一0.034mmYmin=(+0.018—0.023)mm=0.005mmYav=[((一0.034)+(一0.005))/2]mm=一0.0195mmTf=[一0.005一(一0.034)]mm=0.029mm过盈配合(3)φ30M8(-0.029+0.004)φ30h7(-0.0210)Ymax=(一0.029一0)mm=一0.029mmXmax=[+0.004一(一0.021)]mm=+0.025mmYav=[((一0.029)+(+0.025))/2]mm=一0.002mmTf=[+0.025一(一0.029)]mm=0.054mm过渡配合(4)φ45JS6(±0.008)φ45h5(-0.0110)Ymax=(一0.008—0)mm=一0.008mmXmax=[+0.008一(一0.011)]mm=+0.019mmYav=[(一0.008)+(+0.019)]/2mm=+0.0055mmTf=[+0.019一(一0.008)]mm=0.027mm过渡配合