Page 396 - Marks Calculation for Machine Design

P1: Naresh
January 4, 2005
Brown˙C09
Brown.cls
378
k 1 15:28 APPLICATION TO MACHINES
k 1 k 2 k 3
k 2
k 3 (b)
(a)
FIGURE 9.6 Series and parallel springs.

For the three springs in parallel in Fig. 9.6(b), the equivalent spring rate (k eq ) is given by
Eq. (9.32) as

k eq = k 1 + k 2 + k 3 (9.32)
Notice that springs combine completely opposite to resistors in electric circuits.

U.S. Customary SI/Metric
Example 5. Two helical springs are used in Example 5. Two helical springs are used in
series as shown in Fig. 9.6(a). Calculate the series as shown in Fig. 9.6(a). Calculate the
equivalent spring rate (k eq ), where equivalent spring rate (k eq ), where
k 1 = 30 lb/in k 1 = 5,400 N/m
k 2 = 60 lb/in k 2 = 10,800 N/m
solution solution
Step 1. Using Eq. (9.31), determine the equiv- Step 1. Using Eq. (9.31), determine the equiv-
alent spring rate (k eq ) as alent spring rate (k eq ) as
1 1
k eq = k eq =
1 1 1 1
+ +
k 1 k 2 k 1 k 2
1 1
= =
1 1 1 1
+ +
30 lb/in 60 lb/in 5,400 N/m 10,800 N/m
1 1
= =
(0.033 + 0.017) in/lb (0.0001851 + 0.0000925) m/N
1 1
= =
(0.05) in/lb (0.0002776) m/N
= 20 lb/in = 3,600 N/m
Notice that the equivalent spring rate (k eq ) Notice that the equivalent spring rate (k eq )
is less than either of the two individual spring is less than either of the two individual spring
rates. This is because the weaker spring domi- rates. This is because the weaker spring domi-
nates the system. nates the system.