MODELING OF SDOF SYSTEMS
2.3 SPRINGS IN COMBINATION
2.3.2 SERIES COMBINATION
Torsional oscillations occur in the system of Figure 2.7. A thin disk of mass moment of inertia I is attached to a circular shaft of length L, shear modulus G, and polar moment of inertia J. When the disk is rotated through an angle from its equilibrium position, a moment
(2.24) develops between the disk and the shaft. Thus, if the polar mass moment of inertia of the shaft is small compared with I, then the shaft acts as a torsional spring of stiffness
(2.25)
changes in length of the springs in the series combination. If xiis the change in length of the ith spring, then
(2.29) Since the force is the same in each spring, xiF/k and Equation (2.29) becomes
(2.30) Since the series combination is to be replaced by a spring of an equivalent stiffness, Equation (2.26) is used in Equation (2.30), leading to
(2.31)
Electrical circuit components also can be placed in series and parallel and the effect of the combination replaced by a single component with an equivalent value. The equivalent capac- itance of capacitors in parallel or series is calculated like that of springs in parallel or series. The equivalent resistance of resistors in series is the sum of the resistances, whereas the equivalent resistance of resistors in parallel is calculated by using an equation similar to Equation (2.31).
keq = 1 a
n i=1
1 ki x = a
n i=1
F ki
x = x1 + x2 + Á
+ xn = a
n i=1
xi
k1 k2 k3 kn
m
FIGURE 2.10
The springs in the series combination each develop the same force, but the total displace- ment of the combination is the sum of the indi- vidual changes in length.
E X A M P L E 2 . 4
Model each of the systems of Figure 2.11 by a mass attached to a single spring of an equivalent stiffness. The system of Figure 2.11(c) is to be modeled by a disk attached to a torsional spring of an equivalent stiffness.
S O L U T I O N
(a) The steps involved in modeling the system of Figure 2.11(a) by the system of Figure 2.8 are shown in Figure 2.12. Equation (2.28) is used to replace the two parallel springs by an equivalent spring of stiffness 3k.The three springs on the left of the mass are then in series, and Equation (2.31) is used to obtain an equivalent stiffness.
If the mass in Figure 2.11(a) is given a displacement xto the right, then the spring on the left of the mass will increase in length by x, while the spring on the right of the mass will decrease in length by x.Thus, each spring will exert a force to the left on the mass. The spring forces add; the springs behave as if they are in parallel. Hence Equation (2.28) is used to replace these springs by the equivalent spring shown in Figure 2.12(c).
(b) The deflection of the simply supported beam due to a unit load at x 2 m is cal- culated using Table D.2
v(z = 2 m) = va2L (a)
3 b = 4L3 243EI
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k 3k 2k
2k
k
m
(a)
(b)
(c)
(d) k m
2 m 1 m E = 210 × 109 N/m2
I = 5 × 10–4 m4 k = 1 × 108 N/m
m
h2 = 20 mm h1 = 25 mm b = 13 mm E = 210 × 109 N/m2 2 m
b
h2 h1
A B C
Gst = 80 × 109 N/m2 Gal = 40 × 109 N/m2 r1 = 20 mm
r2 = 25 mm
r3 = 18 mm r4 = 30 mm
20 cm AB: Steel shaft
with aluminum core BC: Hollow steel shaft 30 cm
r2 r3 r4
r1
FIGURE 2.11
Systems for Example 2.4.
k 3k 3k 2k
m x
(a)
(b)
(c) 13k/5
m
3k/5 2k
m x
FIGURE 2.12
Steps in replacing the combination of springs in Figure 2.11 (a) using a single spring of an equivalent stiffness.
from which the equivalent stiffness is obtained
(b) The displacement of the block of mass mequals the displacement of the beam at the location where the spring is attached plus the change in length of the spring. Hence the beam and spring act as a series combination. Equation (2.31) is used to calculate their equivalent stiffness
(c) (c) The aluminum core of shaft AB is rigidly bonded to the steel shell. Thus the angu- lar rotation at B is the same for both materials. The total resisting torque transmitted to section BC is the sum of the torque developed in the aluminum core and the torque devel- oped in the steel shell. Thus the aluminum core and steel shell of shaft AB behave as two torsional springs in parallel. The resisting torque in shaft AB is the same as the resisting torque in shaft BC. The angular displacement at Cis the angular displacement of B plus the angular displacement of Crelative to B. Thus shafts AB and BC behave as two torsional springs in series. In view of the preceding discussion and using Equations (2.28) and (2.31), the equivalent stiffness of shaft ACis
(d)
where the torsional stiffness of a shaft is ktJG/L and
(e)
(f)
(g) Substitution of these values into the equation for keqgives
(h) (d) Under the assumption that the rate of taper of the bar is small the following mechanics of materials equation is used to calculate the change in length of the bar due to a unit load applied at its end:
(i)
¢ = L
L
0
dz AE
kt,eq = 1.01 * 105N
#
m/radkt
BC
= = p
323(0.06m)4 - (0.036m)44a80 * 109N m2b
0.2m = 4.43 * 105 N . m
rad kt
ABst
= p
323(0.05 m)4 - (0.04 m)44a80 * 109N m2b
0.3m = 9.66 * 104 N . m
rad kt
ABal
= p
32(0.04 m)4a40 * 109 N m2b
0.3m = 3.35 * 104 N . m rad kt
eq
=
1 1 kt
ABal
+ kt
ABst
+ 1 kt
BC
keq = 1
1
2.36 * 108 N/m + 1 1 * 108 N/m
= 7.03 * 107 N/m k1 = 243EI
4L3 =
243(210 * 109 N/m2)(5 * 10-4m4)
4(3m)3 = 2.36 * 108N/m
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The area varies linearly over the length of the bar . The change in length is
(j) Thus, the equivalent stiffness of the shaft is
(k)