FREE VIBRATIONS OF SDOF SYSTEMS
3.4 UNDERDAMPED FREE VIBRATIONS
The constants of integration are determined by applying the initial conditions, Equation (3.8) and (3.9), resulting in
(3.28) An alternative form of the solution is developed by using the trigonometric identity, Equation (3.20)
(3.29) where
(3.30) (3.31) and
(3.32) Equation (3.29) is plotted in Figure 3.7. Once free oscillations of a viscously damped system commence, the nonconservative viscous damping force continually dissipates energy. Since no work is being done on the system, this leads to a continual decrease in the sum of the potential and kinetic energies. For underdamped free vibrations, the system oscillates about an equilibrium position. However, each time it reaches equilibrium, the system’s total energy level is less than at the previous time. The maximum displacement on each cycle of motion is continually decreasing. Equation (3.29) and Figure 3.7 show that the amplitude decreases exponentially with time.
The free vibrations of an underdamped system are cyclic but not periodic. Even though the amplitude decreases between cycles, the system takes the same amount of time to exe- cute each cycle. This time is called the period of free underdamped vibrationsor the damped periodand is given by
(3.33) Td =
2p vd
vd = vn21 - z2 fd = tan-1a x0vd
x#
0 + zvnx0b A =
Ax20 + ax#
0 + zvnx0 vd b2 x(t) = Ae-zvntsin(vdt + fd)
x(t) = e-zvnt cx0cos1vn21 - z2t2 + x#
0 + zvnx0
vn21 - z2sin1vn21 - z2t2 d
t Ae–ζωnt
1
x(t)/x(0)
FIGURE 3.7
Free vibrations of an under- damped SDOF system decay exponentially.
Thus, dis called the damped natural frequency. Note that d n and Td T. This is due to the viscous friction which resists the motion of the system and slows it down.
Consider a mass-spring and viscous-damper system with and . Then (3.34)
Hence, cos d⫽ , and
(3.35) The total energy present in an underdamped system at time tis
(3.36) The total energy in the system at the end of the nth cycle, , is
(3.37) The energy dissipated as the system executes one cycle of motion is
(3.38) The ratio of the energy dissipated over a cycle compared to the total energy at the begin- ning of the cycle is
(3.39) Equations (3.38) and (3.39) show that the energy dissipated per cycle of motion is con- stant, and thus, it has a constant ratio. The sequence of energies at the beginning of each cycle is a geometric sequence with ratio . For example, if
. The percentage of energy at the end of the nth cycle is (0.717)ntimes the initial energy. The larger the damping ratio, the smaller the ratio, and a larger fraction of energy is dissipated per cycle. Since the sequence of energies is a geometric sequence, the energy is never completely dissipated, thus indicating that the free vibrations of an under- damped system continues indefinitely with exponentially decreasing amplitude.
Taking the limit of the energy ratio as the damping ratio approaches one, lim . All of the energy would be dissipated within the first cycle. This is the origin of the term underdamped; the damping force is not large enough to ever dissipate all of the energy.
The logarithmic decrement, , is defined for underdamped free vibrations as the natu- ral logarithm of the ratio of the amplitudes of vibration on successive cycles.
d
z:1
¢En
En = 1
¢En
En = 0.717
z = 0.1, 1 - e-4pz>21-z2
¢En
En = 1 - e4pz>21-z2
= 1
2kx20e-4nzp>11-z2(1 - e-4pz>11-z2)
⌬En = En - En+1
En = E(nTd) = 1
2kx20e-4nzp>21-z2
t = 2np
vd
cos(vdt + fd ) + (1 - z2)cos2(vdt + fd)
D
- 2z21 - z2sin(vdt + fd)
= 1 2
kx20e-2zvnt
(1 - z2)
C
(1 + z2)sin2(vdt + fd) E = 12kx2 + 1 2mx#
2
A = x0 21 - z2
z sinfd = 21 - z2, f
fd = tan-1a21 - z2 z b
x# (0) = 0 x(0) = x0
v 7 v 6 v
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(3.40) For small ,
(3.41) The logarithmic decrement is often measured by experiment and damping ratio deter- mined from
(3.42) It can be shown that the following equations can also be used to calculate the logarithmic decrement:
(3.43) for any integer nand
(3.44) (3.45) Equation (3.43) implies that the logarithmic decrement can be determined from ampli- tudes measured on nonsuccessive cycles, while Equations (3.44) and (3.45) imply that velocity and acceleration data can also be used to determine the logarithmic decrement.
The free vibrations of an underdamped system decay exponentially with time. When the ini- tial conditions are x(0)⫽x0and x# , the response of the system is shown in Figure 3.8.
(0) = 0 d = lna x$
(t) x$
(t + Td)b d = lna x#
(t) x#
(t + Td)b d = 1
nlna x(t) x(t + nTd)b z =
d 24p2 + d2 d = 2pz
z
= zvnTd = 2pz 21 - z2 d = lna x(t)
x(t + Td)b = lna Ae-zvntsin(vdt + fd) Ae-zvn(t+Td)sin3vd(t + Td) + fd4 b
1.2
1
0.8
0.6
0.4
0.2
0
–0.2
0 0.5 1 1.5
t (s) x/x0
2 2.5 3
FIGURE 3.8
Underdamped response due to initial conditionsx(0)⫽x0 and (0)⫽0. The overshoot is the amplitude at the end of the first half-period.
x#
The absolute value of the displacement after the first half-cycle is called the overshoot. The over- shoot is calculated by
(3.46) The percent overshoot is 100xh .
0
= 100 e-zp>21-z2
= x0e-zp>21-z2 h = -xaTd
2b = - x0
21 - z2e-zp>21-z2sin(p + fd)
Determine (a) the response of the accelerometer of Example 2.20 if it has an initial veloc- ity of 30 m/s and an initial displacement of 0 m. (b) What is the value of the displacement at t⫽1 s?
S O L U T I O N
(a) The differential equation governing the free response of the accelerometer is
(a) Putting the equation in standard form, we have
(b) The natural frequency is
(c) and the damping ratio is determined as
(d) The system is underdamped and the response for the given initial conditions is
(e) where
(f) Thus,
(g) (b) At t⫽1 s,
x(10-6 s) = 1.04 * 10-4e-5.36*104(10-6)sin32.82 * 105(10-6)4 = 3.07 * 10-5 m (h) m
= 1.04 * 10-4e-5.36*104t(sin2.82 * 105t) m x(t) = 30m/s
2.82 * 105rad/se-0.187(2.87*105)tsin(2.82 * 105t)
vd = vn21 - z2 = 2.87 * 105 rad/s 21 - (0.187)2 = 2.82 * 105 rad/s x(t) =
x#
0
vde-zvntsinvdt z = 1.07 * 105
2(2.87 * 105) = 0.186
vn = 28.26 * 1010 = 2.87 * 105 rad/s x$
+1.07 * 105x#
+ 8.26 * 1010x = 0 4.6 * 10-12x$
+ 4.93 * 10-7x#
+ 0.380x = 0 m
E X A M P L E 3 . 6
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E X A M P L E 3 . 7
The slender bar of Figure 3.9(a) has a mass of 31 kg and a length of 2.6 m. A 50 N force is statically applied to the bar at Pthen removed. The ensuing oscillations of Pare moni- tored, and the acceleration data is shown in Figure 3.9(b) where the time scale is calibrated but the acceleration scale is not.
(a) Use the data to find the spring stiffness kand the damping coefficient c.
(b) Calibrate the acceleration scale.
S O L U T I O N
FBDs of the system at an arbitrary instant are shown in Figure 3.9(c). Applying (∑MO)ext⫽(∑MO)effto these FBDs leads to the differential equation of motion:
x$ (a) +
3c 7mx#
+ 27k 7mx = 0
c = ?
k = ? 50 N
m = 31 kg
1.95 m 0.65 m
(a)
4 3 2 1 0 –1 –2 –3
0 0.05 0.1
Time (s)
a(t) (scale not calibrated)
0.15 0.2 0.25
(b)
Ry mg k[x(0) + ∆st] 50 N
(c) (d)
External forces Effective forces R
kx
(
3x .)
2m x.
3 c
冢
3L/4x¨ 冣
1 12mL2 mxø
3
FIGURE 3.9
(a) System of Example 3.7. (b) Accelerometer data for free vibration response. (c) FBD when system is in equilibrium. (d) FBDs of system at an arbitirary instant.
The natural frequency and damping ratio are determined by comparing the preced- ing equation with the standard form of the differential equation for damped free vibra- tions as
(b) (c) The period of damped free vibrations is determined from the accelerometer data as 0.1 s.
The value of the logarithmic decrement is determined from the accelerometer data and Equation (3.45) as
(d) The damping ratio is calculated using Equation (3.42) as
(e) The damped natural frequency is
(f) from which the natural frequency is calculated as
(g) (a) The stiffness is calculated from Equation (b) as
(h) and the damping coefficient is calculated from Equation (c) as
(i) (b) A static analysis of the equilibrium position in Figure 3.9(c) provides the initial dis- placement from equilibrium as
(j) The initial acceleration is calculated using the governing differential equation as
x$ (k)
(0) = -2zvnx#
(0) - v2nx(0) = -(63.0)2(0.0016 m) = -6.22 m/s2 x(0) =
F k =
50 N 3.19 * 104 N/m
= 1.6 mm c =
14mvnz
3 =
14(31 kg)(63.0 rad/s)(0.0643)
3 = 585.7 N
#
s/mk = 7mv2n
27 =
7(31 kg)(63.0 rad>s)2
27 = 3.19 * 104 N/m vn =
vd 21 - z2
=
62.8rad/s 21 - (0.0644)2
= 63.0 rad/s vd = 2p
Td = 2p
0.1 s = 62.8 rad/s
z = 0.406
24p2 + (0.406)2
= 0.0644 d = lnc x$
(0) x$
(0.1 s)d = ln3
2 = 0.406 2zvn = 3c
7mQz = 3c 14mvn vn =
A 27k
7m
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