例题轴孔间隙配合
例题一:图示为某机器活塞销与连杆小头孔的装配示意图,
其中:
,0.0100,0.0070销φ28mm,孔φ28mm; ,0.0125,0.0095
?试问销、孔配合间隙的范围是多少,
?按分组互换法装配并确定各组尺寸。(建议分组数为4组)
解:?销、孔尺寸的配合间隙范围是0.0005,0.0055mm;
,0.0055,0.0005 ?由题得:A=0mm,即T0=0.0050mm,0
按等公差分配得:
T1=T2=0.0025mm,此公差相当于IT2级,显然制造
起来很困难也不经济,因此将它们的公差扩大4倍
(下偏差不动,变动上偏差)
,0.0005,0.0095,即:T1’=T2’=40.0025=0.01mm,即:孔,φ28mm,
,0.0025,0.0125销,φ28mm; 这样销可用无心磨床加工,孔可用金刚镗床加工,将它们分为4组:
,0.0005,0.0020孔尺寸:第一组φ28mm
,0.0020,0.0045 第二组φ28mm
,0.0045销,0.0070孔 第三组φ28mm 销
,0.0070,0.0095第四组φ28mm。
确定销尺寸,以第一组孔为例,用极值解法得:
,0.0025,0.0050第一组φ28mm,同理可得:
,0.0050,0.0075 第二组φ28mm;
,0.0075,0.0100 第三组φ28mm;
,0.0100,0.0125 第四组φ28mm;
,0.003例题二:某一轴与孔的配合为间隙配合,孔:Φ30,轴:0
,0.002Φ30,试确定: ,0.005
(1) 轴与孔的配合间隙的范围是多少,
(2)若按分组装配法,将公差扩大4倍,问:轴与孔按什么样的尺寸及偏差来进行加工,为达到装配精度要求,每组的尺寸范围又是多少,
解:(1)轴、孔配合间隙为0.002,0.008mm(装配精度) (2)按分组装配法将轴、孔公差扩大四倍后得:
0.003×4=0.012mm;
若上偏差不动,改变下偏差:
,0.003,0.002则孔变为:φ30mm;轴变为:φ30mm。 ,0.009,0.014
这样,孔可用金刚镗加工,轴可用无心磨加工,相应的精度
约为IT6。
,0.0030第一组孔为:φ30mm;第二组孔为:φ30mm; ,0.0030
,0.006,0.003,0.009第三组孔为:φ30mm;第四组孔为:φ30mm。 ,0.006
由计算得:
,0.002,0.005第一组轴为:φ30mm;第二组轴为:φ30mm; ,0.005,0.008
,0.008,0.011第三组轴为:φ30mm;第四组轴为:φ30mm; ,0.011,0.014
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