产品综合样本 release 8.5 - page 1053

附录
02
a/3.05
计算：
EK = mv2 = 50 . 1 2 = 25 Nm
2 2
考虑到总能 ET 至少要大于 25 Nm:
可选用 SA 2015，S = 0.015 m
ED = F . S = ( D2 . π . P . g/100 – m . g) . S = ( 632 . π 6 . 9.81/100 – 50 . 9.81) . 0.015 = 20.1 Nm
4 4
ET = EK + ED = 25 + 20.1 = 45.7 Nm
ETC = ET . C = 45.1 . 600 = 27060 Nm/h
Me = 2ET = 2 . 45.7 = 91.4 kg
v2 12
根据 SA 2015 缓冲器的技术参数可知满足要求：
ET (max) = 59 Nm，E TC (max) = 38000 Nm/h，Me (max) = 120 kg
计算：
EK = mv2 = 50 . 1 2 = 25 Nm
2 2
ED = F . S = (m . g + D2 . π . P . g/100 ) . S = (50 . 9.81 + 63 . π . 6 . 9.81/100) . 0.025 = 58.1 Nm
4 4
ET = EK + ED = 25 + 58.1 = 83.1 Nm
ETC = ET . C = 83.1 . 600 = 49860 Nm/h
Me = 2ET = 2 . 84 = 168 kg
v2 1 2
可选用 SA 2725 缓冲器，其技术参数为：
ET (max) = 147 Nm，ETC (max) = 72000 Nm/h，Me (max) = 270 kg
计算：
v = √ (2g . h) = √ (2 . 9.81 . 0.35) = 2.6 m/s
EK = m . g . h = 5 . 9.81 . 0.35 = 17.2 Nm
考虑到总能 ET 至少要大于 17.2 Nm:
可选用 SA 1412，S = 0.012 m
ED = F . S = m . g . S = 5 . 9.81 . 0.012 = 0.6 Nm
ET = EK + ED = 17.2 + 0.6 = 17.8 Nm
ETC = ET . C = 17.8 . 1500 = 26700 Nm/h
Me = 2ET = 2 . 17.5 = 5 kg
v2 2.62
根据 SA 1412 缓冲器的技术参数可知满足要求：
ET (max) = 20 Nm，ETC (max) = 33000 Nm/h，Me (max) = 40 kg
例 3：自由落体的撞击
h = 0.35 m
m = 5 kg
S = 0.01 m
C = 1500 /h
例 4：有附加推力垂直撞击（一）
m = 50 kg
S = 0.025 m
P = 6 bar
D = 63 mm
C = 600 /h
v = 1.0 m/s
例 5：有附加推力垂直撞击（二）
m = 50 kg
h = 0.3 m
S = 0.025 m
P = 6 bar
D = 63 mm
C = 600 /h
v = 1.0 m/s
附录 > 与产品相关的一些技术信息 产品综合样本 > 8.5 版本
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