Page 395 - Marks Calculation for Machine Design

P1: Naresh
15:28
January 4, 2005
Brown˙C09
Brown.cls
U.S. Customary MACHINE ENERGY SI/Metric 377
Example 4. Calculate the work done to com- Example 4. Calculate the work done to com-
pressahelicalspringthatisalreadycompressed, pressahelicalspringthatisalreadycompressed,
where where
L o = 2.0 in L o = 5.0 cm = 0.05 m
L i = 1.75 in L i = 4.5 cm = 0.045 m
L f = 1.25 in L f = 3.5 cm = 0.035 m
k = 100 lb/in k = 18,000 N/m
solution solution
Step 1. Using Eq. (9.29), calculate the dis- Step 1. Using Eq. (9.29), calculate the dis-
placements (x 1 ) and (x 2 ) as placements (x 1 ) and (x 2 ) as
x 1 = L i − L o = (1.75 in) − (2.0in) x 1 = L i − L o = (0.045 m) − (0.05 m)
=−0.25 in =−0.005 m
x 2 = L f − L o = (1.25 in) − (2.0in) x 2 = L f − L o = (0.035 m) − (0.05 m)
=−0.75 in =−0.015 m
Step 2. Substitute the displacements (x 1 ) and Step 2. Substitute the displacements (x 1 ) and
(x 2 ) found in step 1 in Eq. (9.28) to give the (x 2 ) found in step 1 in Eq. (9.28) to give the
work done as work done as
1 2 2 1 2 2
Work = k x − x 2 Work = k x − x 2
1
1
1→2 2 1→2 2
1 1
= (100 lb/in) = (18,000 N/m)
2 2
2
2
2
2
×((−0.25 in) − (−0.75 in) ) ×((−0.005 m) − (−0.015 m) )
1 1
= (100 lb/in) = (18,000 N/m)
2 2
2
2
×((0.0625 − 0.5625) in ) ×((0.000025 − 0.000225) m )
1 2 1 2
= (100 lb/in)(−0.5in ) = (18,000 N/m)(−0.0002 m )
2 2
=−25 in · lb =−1.8N · m =−180 N · cm
The negative sign on the work done means The negative sign on the work done means
work was done on the spring. work was done on the spring.
9.2.4 Series and Parallel Arrangements
When more than one spring is being used in a design, they are either in series, meaning
one after another, or in parallel, meaning side by side, or a combination of both. These
two arrangements are shown for three springs in Fig. 9.6, combined in series in (a) and
combined in parallel in (b).
Using the spring rate (k) of each spring, an equivalent spring rate (k eq ) can be determined
depending on whether the springs are in series or parallel. For the three springs in series in
Fig. 9.6(a), the equivalent spring rate (k eq ) is given by Eq. (9.31) as
1
k eq = (9.31)
1 1 1
+ +
k 1 k 2 k 3