[精选]孔轴配合精度计算

[精选]孔轴配合精度计算 第一章 尺寸精度及孔轴结合的互换性 ，0.021,3001(已知基准孔与下列三轴相配，试计算配合的极限间隙或极限过盈及配合 公差，画出公差带图，并指明它们各属于哪类配合。 ,0.007，0.028，0.048,30,30,30,0.020，0.016，0.035 (1) (2) (3) maxmaxmin 解:(1) X,D,d=ES,ei=0.021,(,0.02)=0.041mm maxminmin X= D, d=EI,es=0,(,0.007)=0.007mm X,Xfmaxmin T==0.034mm 故属于间隙配合，公差带图略。 maxmaxmin (2) X,D,d=ES,ei=0.021,0.016=0.005mm maxmaxmin Y= D,d=EI,es=0,0.028=,0.028mm X,Yfmaxmax T==0.033mm 故属于过渡配合，公差带图略。 maxmaxmin (3) Y,D, d= EI,es =0,0.048=,0.048mm maxminmin Y= D,d=ES,ei =0.021,0.035=,0.014mm Y,Yfmaxmin T==0.034mm 故属于过盈配合，公差带图略。 ,50TXfmax2( 已知孔轴配合的基本尺寸为mm,配合公差为,0.041mm,=+0.066mm, TH孔公差为,0.025mm,轴下偏差ei=,0.041mm,试求孔轴的极限偏差，画出公差带图，说明配合性质。 fSH解: 轴公差为:T= T,T=0.041,0.025=0.016mm SS 因 T= es,ei 故 es=ei ,T=,0.041+0.016=,0.025mm maxmaxmin 因 X,D,d=ES,ei 即 0.066,ES+0.041 得ES,0.025mm HH 因 T,ES,EI 故 EI=ES,T=0.025,0.025=0mm ，0.025,0.0250,0.041 故 孔为φ50 轴为φ50 maxminmin X= D, d=EI,es,0,(,0.025),0.025mm 属于间隙配合，公差带图略。 TTs1s2213( 已知两根轴，其中:d=φ5mm,=0.005mm, d=φ180mm,=0.025mm,试比较 以上两根轴的加工难易程度。 解:方法一:不查 表 关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf 比较 3*6m11 (1) d=φ5mm,属于3,6尺寸分段，d==4.24 34.241 故 i=0.45+0.001*4.24=0.73mm 120*180m22 d=φ180mm,属于120,180尺寸分段，d==146.97 3146.972 故 i=0.45+0.001*146.97=2.52mm 21 (2) 比较a及a的大小 s111 a,T/ i=5/0.73=6.83 s222 a=T/ i=25/2.52=9.93 211 因为 a,a，所以d的精度高，难加工。 方法二:查表比较 21 由表1,6得，d为IT5，d为IT6 1 所以d的精度高，难加工。 4( 查表画出下列配合的孔轴公差带图;计算其配合的极限间隙或过盈及配合公差;并说明各配合代号的含义及配合性质。 H8H7D9,50,140,100f7h9r6(1) (2) (3) H8K7M6,80,30,15js7h6h6(4) (5) (6) 解:(1) 公差带图略 查表得ES=0.039mm,EI=0,es=,0.025,ei=,0.050 maxmaxmin X,D,d=ES,ei=0.039,(,0.050)=0.089mm maxminmin X= D, d=EI,es=0,(,0.0025)=0.0025mm X,Xfmaxmin T==0.064mm 该配合属于基孔制的配合，配合性质为间隙配合。 (2) 公差带图略 查表得ES=0.207mm,EI=0.120,es=0,ei=,0.087 maxmaxmin X,D,d=ES,ei=0.207,(,0.087)=0.294mm maxminmin X= D, d=EI,es=0.120,0=0.120mm X,Xfmaxmin T==0.174mm 该配合属于基孔制的配合，配合性质为间隙配合 (3) 公差带图略 查表得ES=0.040mm,EI=0,es=0.088,ei,0.063 maxminmin Y= D,d=ES,ei=0.040,(0.063)=,0.023mm maxmaxmin Y,D, d=EI,es=0,0.088)=,0.088mm Y,Yfmaxmin T=,0.065mm 该配合属于基孔制的配合，配合性质为过盈配合。 (4) 公差带图略 查表得ES=0.006mm,EI=,0.015,es=0,ei=,0.013 maxmaxmin X,D,d=ES,ei=0.006,(,0.013)=0.019mm maxmaxmin Y,D, d=EI,es =,0.015,0=,0.015mm X,Yfmaxmax T==0.034mm 该配合属于基孔制的配合，配合性质为过渡配合。 (5) 公差带图略 查表得ES=,0.006mm,EI=,0.015,es=0,ei=,0.008 maxmaxmin,D,d=ES,ei=,0.004,(,0.08)=0.004mm X maxmaxmin Y,D, d=EI,es =,0.015,0=,0.015mm X,Yfmaxmax T==0.019mm 该配合属于基孔制的配合，配合性质为过渡配合。 (6) 公差带图略 查表得ES=0.04mm,EI=0,es=0.015,ei=,0.015 maxmaxmin X,D,d=ES,ei=0.046,(,0.015)=0.061mm maxmaxmin Y,D, d=EI,es =0,0.015=,0.015mm X,Yfmaxmax T==0.076mm 该配合属于基孔制的配合，配合性质为过渡配合。 5( 将下列基孔制配合，转换成配合性质性质相同的基轴制配合，并确定转换前后的极限偏差。 H7H9H6,100,60,40f9g5p6(1) (2) (3) H7H7H7,50,50,30u6k6n6(4) (5) (6) H9F9 f9h6解: (1) ? 转换配合基准 φ60转换为φ60 ? 查表确定转换前孔、轴的极限偏差 φ60属于50,80尺寸分段，IT9,0.074mm SH 即 T=0.074mm T=0.074mm H 对φ60H9，基本偏差EI,0 故ES=EI+ T=+0.074mm S 对φ60f9,基本偏差es=,0.030mm 故ei=es,T=,0.104mm ，0.074H90,0.030f9,0.014 所以 转换前φ60 maxmaxmin X,D,d=ES,ei=0.178mm maxminmin X= D, d=EI,es=0.030mm 属于间隙配合 ? 用换算公式确定转换后的极限偏差 F9属于A,H的范围，所以适用通用规则 '''S对φ60h9 es=,EI=0 故ei= es,T,,0.074mm '''H对φ60F9 EI,,es,+0.030mm 故ES= EI+ T,+0.104mm 0.104F90.0300h9,0.074所以 转换后φ60 ''maxmaxminX,D,d= ES,ei=0.178mm ''maxminmin X= D, d= EI,es=0.030mm 属于间隙配合 即转换前后的配合性质相同。 H6G6 g5h5 (2) ? 转换配合基准 φ40转换为φ40 ? 查表确定转换前孔、轴的极限偏差 φ40属于30,50尺寸分段，IT6,0.016mm IT5,0.011mm SH 即 T=0.016mm T=0.011mm H 对φ40H6，基本偏差EI,0 故ES=EI+ T=+0.016mm S 对φ40g5, 基本偏差es=,0.009mm 故ei=es,T=, 0.020mm ，0.016H60,0.009g5,0.020 所以转换前φ40 maxmaxminX,D,d=ES,ei=0.036mm maxminmin X= D, d=EI,es=0.009mm 属于间隙配合 ? 用换算公式确定转换后的极限偏差 G6属于A,H的范围，所以适用通用规则 '''S对φ40h5 es=,EI=0 故ei= es,T,,0.011mm '对φ40G6 EI,,es,+0.009mm ''H故ES= EI+ T,+0.025mm ，0.025G6，0.0090h5,0.011所以转换后φ40 ''maxmaxminX,D,d= ES,ei=0.036mm ''maxminmin X= D, d= EI,es=0.009mm 属于间隙配合 即转换前后的配合性质相同。 H7P7 p6h6 (3) ? 转换配合基准 φ100转换为φ100 ? 查表确定转换前孔、轴的极限偏差 φ100属于80,120尺寸分段，IT7,0.035mm IT6, 0.022mm SH 即 T=0.035mm T=0.022mm H 对φ100H7，基本偏差EI,0 故ES=EI+ T=+0.035mm 对φ100p6, 基本偏差ei=+0.037mm 故 Ses=ei+T=+0.059mm ，0.035H70，0.059p6，0.037 所以转换前φ100 maxmaxminY,D,d= EI,es =,0.059mm maxminmin Y= D,d=ES,ei =,0.002mm 属于过盈配合 ? 用换算公式确定转换后的极限偏差 P7属于P,ZC(公差等级高于或等于7级)的范围，所以适用特殊规则 ?,IT7,IT6,0.011mm '''S对φ100h6 es=,EI=0 故ei= es,T,,0.022mm '对φ100P7 ES=,ei+?=,0.024mm ''H故EI= ES, T,,0.059mm ,0.024P7,0.0590h6,0.022所以转换后φ100 ''maxmaxminY,D,d= EI,es=,0.059mm ''maxminmin Y= D,d= ES,ei=,0.002mm 属于过盈配合 即 转换前后的配合性质相同。 H7U7 u6h6 (4) ? 转换配合基准 φ50转换为φ50 ? 查表确定转换前孔、轴的极限偏差 φ50属于30,50尺寸分段，IT7,0.025mm IT6,0.016mm SH 即 T=0.025mm T=0.016mm H 对φ50H7，基本偏差EI,0 故ES=EI+ T=+0.025mm 对φ50u6, 基本偏差ei=+0.070mm 故 Ses=ei+T=+0.086mm ，0.025H70，0.086u6，0.070 所以转换前φ50 maxmaxminY,D,d= EI,es =,0.086mm maxminmin Y= D,d=ES,ei =,0.045mm 属于过盈配合 ? 用换算公式确定转换后的极限偏差 U7属于P,ZC(公差等级高于或等于7级)的范围，所以适用特殊规则 ?,IT7,IT6,0.09mm '''S对φ50h6 es=,EI=0 故ei= es,T,,0.016mm '对φ50U7 ES=,ei+?=,0.061mm ''H故EI= ES, T,,0.086mm ,0.061U7,0.0860h6,0.016所以转换后φ50 ''maxmaxminY,D,d= EI,es=,0.086mm ''maxminmin Y= D,d= ES,ei=,0.045mm 属于过盈配合 即 转换前后的配合性质相同。 H7K7 k6h6 (5) ? 转换配合基准 φ50转换为φ50 ? 查表确定转换前孔、轴的极限偏差 φ50属于30,50尺寸分段，IT7,0.025mm IT6,0.016mm SH 即 T=0.025mm T=0.016mm H 对φ50H7，基本偏差EI,0 故ES=EI+ T=+0.025mm 对φ50k6, 基本偏差ei=+0.002mm 故 Ses=ei+T=+0.018mm ，0.025H70，0.018k6，0.012 所以转换前φ50 maxmaxminX,D,d=ES,ei =+0.023mm maxmaxmin Y,D, d=EI,es =,0.018mm 属于过渡配合 ? 用换算公式确定转换后的极限偏差 K7属于K、M、N(公差等级高于或等于8级)的范围，所以适用特殊规则 ?,IT7,IT6,0.009mm '''S对φ50h6 es=,EI=0 故ei= es,T,,0.016mm '对φ50K7 ES=,ei+?=+0.007mm ''H故EI= ES, T,,0.018mm ，0.007K7,0.0180h6,0.016所以转换后φ50 ''maxmaxminX,D,d= ES,ei=+0.023mm ''maxmaxmin Y,D, d= EI,es=,0.018mm 属于过渡配合 即 转换前后的配合性质相同。 H6N6 n5h5 (6) ? 转换配合基准 φ30转换为φ30 ? 查表确定转换前孔、轴的极限偏差 φ30属于18,30尺寸分段，IT6,0.013mm IT5=0.009mm SH 即 T=0.013mm T=0.009mm H 对φ30H6，基本偏差EI,0 故ES=EI+ T=+0.013mm 对φ50k6, 基本偏差ei=+0.015mm 故 Ses=ei+T=+0.024mm ，0.013H60，0.024n5，0.015 所以转换前φ30 maxmaxminY,D,d= EI,es =,0.024mm maxminmin Y= D,d=ES,ei =,0.002mm 属于过盈配合 ? 用换算公式确定转换后的极限偏差 N6属于K、M、N(公差等级高于或等于8级)的范围，所以适用特殊规则 ?,IT6,IT5,0.004mm '''S对φ30h5 es=,EI=0 故ei= es,T,,0.009mm '对φ50K7 ES=,ei+?=,0.011mm ''H故EI= ES, T,,0.024mm ,0.011N6,0.0240h5,0.009所以转换后φ30 ''maxmaxminY,D,d= EI,es=,0.024mm ''maxminmin Y= D,d= ES,ei=,0.002mm 属于过盈配合 即 转换前后的配合性质相同。 6(有下列三组孔、轴配合，试根据给定的条件，分别确定其公差等级，并选择适 当的配合(写出代号)、画出公差带图。 XXmaxmin (1)基本尺寸为φ50mm,=+0.050mm,=+0.009mm 解:? 选用基准制:基孔制 X,Xfmaxmin? 公差等级的确定:T==0.041mm T,T，TfHs又 反查表1-6得: TsH=0.025mm(孔公差等级为7级)，T,0.016mm(轴公差等级为6 级) T,T，TfHs 此时满足 ? 确定配合:由题知该孔轴配合为间隙配合，在基孔制条件下 es,Xmin=0.009mm 故 es=,0.009mm 反查表1,8，确定轴基本偏差代号为g s 而 ei=es,T=,0.009,0.016=,0.025mm TH ES==+0.025mm,EI=0 ，0.025,0.009,50,50,0.0250? 检验:φ50H7为，φ50g6为 max此时 X,ES,ei=0.025+0.025=+0.050mm min X=EI,es=0,(,0.009)=+0.009mm H7,50g6 所以 满足题意。 XYmaxmax (2)基本尺寸为φ25mm,=+0.013mm,=,0.021mm 解:? 选用基准制:基孔制 X,Yfmaxmax? 公差等级的确定:T==0.034mm T,T，TfHs又 反查表1-6得: TsH=0.021mm(孔公差等级为7级)，T,0.013mm(轴公差等级为6 级) T,T，TfHs 此时满足 ? 确定配合:由题知该孔轴配合为过渡配合，在基孔制条件下 TmaxHei=,X,0.021,0.013,+0.008mm 反查表1,8得,轴基本偏差代号为m s故 ei=+0.008mm es=ei+ T=+0.021mm THES==+0.025mm,EI=0 ，0.021，0.021,25,25，0.0080? 检验:φ25H7为，φ25m6为 max此时 X,ES,ei=0.025+0.025=+0.050mm max Y= EI,es=0,0.028=,0.021mm H7,25m6满足题意。 所以 YYmaxmin(3)基本尺寸为φ18mm,= 0,=,0.029mm 解:? 选用基准制:基孔制 X,Yfmaxmax? 公差等级的确定:T==0.029mm T,T，TfHs又 反查表1-6得: TsH=0.018mm(孔公差等级为7级)，T,0.011mm(轴公差等级为6 级) T,T，TfHs 此时满足 ? 确定配合:由题知该孔轴配合为过盈配合，在基孔制条件下 YTminHei=,,0.018+0,+0.018mm s反查表1,8得,轴基本偏差代号为p , es=ei+ T=+0.029mm THES==+0.025mm,EI=0 ，0.018，0.029,18,18，0.0180? 检验:φ18H7为，φ18m6为 max此时 Y,EI,es =0,0.029=,0.029mm min Y= ES,ei =0.018,0.018=0 H7,18p6所以 满足题意。 ttsH7(活塞与汽缸体的配合，其配合尺寸为φ150mm,工作温度=100?，,180?， ,6,6,,SH线膨胀系数=12*10/?,=24*10/?,实验确定其工作时的间隙量应该在 0.1,0.3mm内，若设计者选择配合为φ150H9/d9，问是否合适,若不合适，应选那种配合, 解: (1) 对于φ150H9/d9的配合 查表得ES,0.1mm EI=0 es=,0.145mm ei=,0.245mm 所以在不考虑热变形得影响时候，极限间隙为: maxmaxminX,D,d=ES,ei=0.1,(,0.245)=0.345mm maxminmin X= D, d=EI,es=0,(,0.145)=0.145mm 而由热变形引起得间隙变化量为: ,,D(,*,t,,*,t)HHss ,6,6 ,150【12*10*(100,20),24*10*(180,20)】 ,,0.432mm 所以实际的极限间隙为: Xmax实max ,X+?,,0.087mm Xmin实min = X+?= ,0.287mm 故 不合格 (2)重新进行设计 考虑到热变形的影响，实际应该满足的极限间隙为 maxX,0.3,?,0.732mm min X= 0.1,?,0.532mm ? 选用基准制:基孔制 X,Xfmaxmin? 公差等级的确定:T==0.2mm T,T，TfHs又 反查表1-6得: TsH=0.1mm(孔公差等级为9级)，T,0.1mm(轴公差等级为9级) T,T，TfHs 此时满足 ? 确定配合:该孔轴配合为间隙配合，在基孔制条件下 es,Xmin=0.532mm 故 es=,0.532mm 反查表1,8，确定轴基本偏差代号为a 此时es=,0.520mm s 而 ei= es ,T,0.520,0.100=,0.620mm TH ES==+0.100mm,EI=0 ，0.100,0.520,150,1500,0.620? 检验:φ150H9为，φ150a9为 max此时 X,ES,ei=0.1,(,0.620)=+0.720mm min X= EI,es=0,(,0.520)=+0.520mm 即工作时候的间隙量在0.088~0.288之间，基本满足题意。 所以应选择的配合为φ150H9/a9