F2 r F2=F 1 F e =4 2 .1 6 94 2N a d+ a
∵ F1 /F1 =1 39 1/1 0 5 .4 8 0=00 5 .8 5<e=03 .5 a r ∴ X =1 1 Y =0 1
∵F 2/F2 =4 2 .1 / 8 0=25 > 6 94 2 1 0 .7 e=03 .5 a r ∴ X =04 . 2 Y =17 . 2
2 Fe r 1
F2 d
Fe a F2 r F1 r
F1 d
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2
Fe r
1
F2 d
Fe a F2 r
F1 d
F=06 F, .8 r d F 1 .8 1 0 N 1 N 2 0 =8 6 , d =06 F 2 .8 7 0 =4 6 N 00 70 , d =06
F1 r
F 1 a =8 6 1 0 9 6 > d ,轴有向右窜 1 +4 0 =4 1 N F 2 d +F e 动的趋势,所以轴承2 被压紧,轴承1 被放松. F1 F1 1 N F2 F1 a =4 1 N 96 a = d =8 6 a = d +Fe 轴的直径为7 ,初选型号为7 1 A 的轴承 0 24C
1 Fe a F1 r F2 r 2
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1 F1 d Fe a F1 r
2 F2 d
F2 r
解:F 1 r / y=1 0 / 8 0 (2 17 .)=5 94 2 , 2 .1 N d =F1 2 方向向右 F 2 r / y=5 0 ( 17) 5 .4 N 2 / 2 . =1 29 1 , d =F2 2 方向向左 注:计算内部轴向力用的Y a F> 时的Y 指F / r e . ∵F 1 F e = 5 94 2 4 0 2 .1 + 1 0=4 2 .1 F 2 = 6 94 2> d d+ a 1 39 1N 5 .4 ∴轴有向右窜动趋势 .轴承2被压紧, 轴承1被放松
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图示,某轴轴颈的直径d 0 =7 mm,轴承的径向反力F r 1 =10 N,F =7 0 N,轴的轴向载荷Fe 10 , 20 00 r 2 a=40 N 方向如图示,轴的转速n 4 0/ n =17 r ,轴承的使用寿 mi 命要求为L=8 0 h 0 0 ,取载荷系数f . =1 ,轴承工作温 1 h p 度一般,请选择一对接触角为2 o 5的角接触球轴承的型 号.
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轴的直径为7 ,初选型号为7 1 A 的轴承 0 24C 查得,该轴承基本额定动载荷 C=5 2 0 30 r
例:图示,某轴采用一对正装的型号为3 3 9 0 0 的圆锥 滚子轴承,已知轴承的径向反力F =10 N,F = r 1 80 r 2 50 2 N,轴的轴向载荷Fe 10 ,方向如图示,轴的 a=40 N 转速n 4 0/ n =14 r ,轴承的使用寿命要求为L= mi h 1 0h 1 0 ,取载荷系数f .,请分析这对轴承能否满 0 =1 3 p 足寿命要求.
10 C ε 10 98500 Lh = ( ) = ( ) =51794 .08h 60 n P 60 ×1470 7861 .612
6 6 10 3
说明7 1 A 轴承能满足寿命要求 34C
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3 ,计算当量动载荷 P =p X r + F 1 = .(2 0 0= 3 0 ( 1 f F1 Y a )111 0 + )1 2 N P =p X r + F 2 = .(.1 7 0 +.7 4 1 ) ( 00 8 9 6= 2 f F2 Y a )1104 7 6 .1 N 8 16 2 P< 2 的受力大小验算 1 P ,所以按轴承2
F F F F
a 1 r1 a 2 r 2
816 = 1200 4916 = 7000
= 0 . 68 = 0 . 7023 > 0 . 68
由表1 -5 3 ,取X =1 1 , ,Y =0 1 X =04 ,Y =08 .1 1 .7 2
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1 F1 d Fe a F1 r ∴F 1 =F 1=1 39 1N 5 .4 a d
2 F2 d
10 C ε 10 53200 Lh = ( ) = ( ) =6645.559 h 60n P 60 ×1470 7861.612
6 6 10 3
可见所选轴承不满足寿命要求,重新选择7 1 A 34C 查得,该轴承基本额定动载荷 C = 8 0 N r950
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