Page 198 - Dynamics of Mechanical Systems

0593_C06_fm Page 179 Monday, May 6, 2002 2:28 PM

Forces and Force Systems 179

θ
k k
1 2

F F

+ + δ
M 1 2
FIGURE 6.7.3
A representation of a torsional spring sub- FIGURE 6.7.4
jected to a turning moment. Two springs in series.
Because the springs are in series, the force is the same in each spring. Therefore, from
Eq. (6.7.1) δ and δ are:
1
2
δ = Fk and δ = Fk (6.7.4)
1 1 2 2
Hence, from Eq. (6.7.3) δ is:

 1 1 
δ= Fk 1 + Fk 2 = F  +  (6.7.5)
 k 1 k 2 

Solving for F we obtain:

 kk 
F =  12  δ (6.7.6)
 k + k 
1
2
where the ratio k k /(k + k ) is the equivalent modulus k for the spring series.
1 2
1
2
Observe that Eq. (6.7.6) also holds if the spring series is compressed. Also, by repeated
use of Eq. (6.7.6) we can obtain equivalent moduli for any number of springs in a series.
Consider next a pair of springs parallel as in Figure 6.7.5. As before, let k and k be the
1
2
moduli of the springs. Let F be a tension force on the spring combination, let be the
unstretched length of the springs, and let δ be the elongation of the springs. In this case,
the springs elongate the same amount, whereas the forces in the springs are different.
That is,
δ = δ = δ (6.7.7)
1 2
thus,

Fk = F k = δ (6.7.8)
1 1 2 2

where F and F are the individual spring forces. From equilibrium considerations of the
1
2
spring ends we have:
F = F + F (6.7.9)
1 2