Page 402 - Marks Calculation for Machine Design

P1: Naresh
15:28
January 4, 2005
Brown.cls
Brown˙C09
384
APPLICATION TO MACHINES
Substituting the expression for the the number of active coils (N a ) from Eq. (9.25) in
Eq. (9.44) for the mass (m) of the spring gives
2 2 2 2 4 2 6
ρπ d DN a ρπ d D d G ρπ d G
m = = = (9.45)
2
3
4 4 8D k 32D k
Substitute the expression for the mass (m) from Eq. (9.45) into the expression for the
critical frequency ( f cr ) in Eq. (9.43) to give

2 2
1 k k 1 32D k
f cr = = 1 2 6 = 2 6
2 m 2 ρπ d G 2 ρπ d G

2
32D k
(9.46)

Dk 8
=
πd 3 ρG
As stated earlier, multiply the critical frequency by 15 to 20 to avoid resonance with the
frequency of the cyclic loading on the spring.
U.S. Customary SI/Metric
Example 9. Determine the limiting frequency Example 9. Determine the limiting frequency
( f limiting ) of the cyclic loading on a helical ( f limiting ) of the cyclic loading on a helical
spring, where spring, where
D = 1.5 in D = 4cm = 0.04 m
d = 0.125 in d = 0.3 cm = 0.003 m
k = 50 lb/in k = 9,000 N/m
3
ρ = 15.2 slug/ft = 8.8 × 10 −3 slug/in 3 ρ = 7,850 kg/m 3
2
9
2
6
G = 11.5 × 10 lb/in (steel) G = 80.5 GPa = 80.5 × 10 N/m (steel)
solution solution
Step 1. As the units are somewhat awkward, Step 1. As the units are somewhat awkward,
first calculate the term (ρG) as first calculate the term (ρG) as
slug lb kg N
ρG = 8.8 × 10 −3 11.5 × 10 6 ρG = 7,850 80.5 × 10 9
in 3 in 2 m 3 m 2
slug · lb kg · N
= 1.012 × 10 5 = 6.32 × 10 14
in 5 m 5
2
2
(lb · s /ft) · lb (N · s /m) · N
= 1.012 × 10 5 = 6.32 × 10 14
in 5 m 5
2
2
s · lb 2 1ft s · N 2
= 1.012 × 10 5 × = 6.32 × 10 14
ft · in 5 12 in m 6
2
2
lb · s 2 14 N · s 2
= 8.43 × 10 3 = 6.32 × 10
in 6 m 6