FREE VIBRATIONS OF SDOF SYSTEMS
3.3 FREE VIBRATIONS OF AN UNDAMPED SYSTEM
When the system is undamped, the roots of the characteristic equation given by Equation (3.12) are purely imaginary, as ni.The general solution is a linear combina- tion of all possible solutions, thus
(3.14) where B1and B2are constants of integration.
Euler’s identity states
(3.15) Application of Euler’s identity to Equation (3.14) leads to
(3.16) or
(3.17) where C1⫽B1⫹B2and C2⫽i(B1–B2) are redefined constants of integration. As defined, C1and C2are real, while B1and B2are complex conjugates. Substituting the initial condi- tions, Equations (3.8) and (3.9), into Equation (3.17) leads to
(3.18) An alternate and more instructive form of Equation (3.18) is
(3.19) Expanding Equation (3.19) using the trigonometric identity for the sine of the sum of angles
(3.20) gives
(3.21) x(t) = Acosfsinvnt + Asinfcosvnt
sin(a + b) = sinacosb + cosasinb x(t) = Asin(vnt + f)
x(t) = x0cosvnt + x#
0
vnsinvnt x(t) = C1cosvnt + C2sinvnt
x(t) = B1(cosvnt + isinvnt) + B2(cosvnt - isinvnt) eiu = cosu + isinu
x(t) = B1eivnt + B2e-ivnt
⫾v z
a = vn(-z ⫾ 2z2 - 1) z 7 1
a = -vn z = 1,
a = vn(-z ⫾ i21 - z2) z 6
6
a = ⫾ivn z = 0,
Equating coefficients of like trigonometric terms of Equations (3.18) and (3.21) leads to (3.22) and
(3.23) Equation (3.19) is an example of the simple harmonic motion discussed in Section 1.6.
The amplitude of the motion is A, the frequency is n, its phase is , and its period is . The parameter nis called the natural frequency, because it is the frequency at which the undamped free response occurs naturally.
The undamped motion of a SDOF system is simple harmonic motion. The initial con- ditions determine the energy initially present in the system. Potential energy is converted to kinetic energy and vice versa without dissipation. Since energy is conserved, the system eventually returns to its initial state with the original potential and kinetic energies, com- pleting one full cycle of motion. The subsequent cycle duplicates the first cycle. The system takes the same amount of time to execute the second cycle as it does the first. Since no energy is dissipated, it executes subsequent cycles in the same amount of time. Thus, the motion is cyclic and periodic. Figure 3.2 illustrates simple harmonic motion of an undamped SDOF system.
The amplitude A, defined by Equation (3.22), is the maximum displacement from equi- librium. The amplitude is a function of the system parameters and the initial conditions.
The amplitude is a measure of the energy imparted to the system through the initial con- ditions. For a linear system
(3.24) where Eis the sum of kinetic and potential energies.
A = A
2E keq v
2p vn
f v
f = tan-1avnx0 x#
0
b A =
Ax20 + ax#
0
vnb2
T =
t 2π
ωn
π/2 – φ ωn
x0
x(t)
–x0 0
A
FIGURE 3.2
Illustration of free response of an undamped system. The motion is cyclic and periodic.
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The phase angle , calculated from Equation (3.23) is an indication of the lead or lag between the response and a pure sinusoidal response. The response is purely sinusoidal with
⫽0 if x0⫽0. The response leads a pure sinusoidal response by /2 rad if . The system takes a time of
(3.25)
to reach its equilibrium position from its initial position.
t = d p - f
vn f 7 0
- f
vn f … 0
x#
0 = 0 p
f
f
E X A M P L E 3 . 1
An engine of mass 500 kg is mounted on an elastic foundation of equivalent stiffness 7⫻105N/m. Determine the natural frequency of the system.
S O L U T I O N
The system is modeled as a hanging mass-spring system. Equation (3.3) with ceq⫽0 gov- erns the displacement of the engine from its static-equilibrium position. The natural fre- quency is determined by using Equation (3.5)
(a) or expressed in Hz.
(b)
E X A M P L E 3 . 2
f = vn
2p = 37.4rad/s
2prad/cycle = 5.96Hz vn =
A k m =
A
7 * 105N/m
500kg = 37.4rad/s
E X A M P L E 3 . 2
A wheel is mounted on a steel shaft (G⫽83⫻109N/m2) of length 1.5 m and radius 0.80 cm. The wheel is rotated 5 and released. The period of oscillation is observed as 2.3 s.
Determine the mass moment of inertia of the wheel.
S O L U T I O N
The oscillations of the wheel about its equilibrium position are modeled as the torsional oscillations of a disk on a massless shaft, as illustrated in Figure 3.3. The differential equa- tion for such a system is derived in Example 2.17 as
(a) Equation (a) is written in the standard form by dividing by I, giving
(b)
$ u +
JG IL u = 0 I
$ u +
JG L u = 0
°
The natural frequency is obtained from Equation (b) as
(c) The natural frequency is calculated from the observed period by
(d) The moment of inertia of the wheel is calculated using Equation (c) as
I = (e) JG Lv2n =
p
2(0.008 m)4(83 * 109 N/m2)
(1.5 m)(2.73rad/s)2 = 47.7 kg
#
m2vn = 2p T =
2prad/cycle
2.3s/cycle = 2.73rad/s vn =
A JG IL
FIGURE 3.3
System of Example 3.2. A wheel is mounted on a shaft, and the period of oscillations is observed, which is used to calculate the moment of inertia of the wheel.
G = 83 × 109 N/m2 r = 8 mm
θ (t) 1.5 m
E X A M P L E 3 . 3
A mass of 5 kg is dropped onto the end of a cantilever beam with a velocity of 0.5 m/s, as shown in Figure 3.4(a). The impact causes vibrations of the mass, which sticks to the beam.
The beam is made of steel (E⫽210⫻109N/m2), is 2.1 m long, and has a moment of inertia I⫽3⫻10–6m4. Neglect inertia of the beam and determine the response of the mass.
S O L U T I O N
Let x(t) represent the displacement of the mass, which is measured positive downward from the equilibrium position of the mass after it is attached to the beam. As shown in Figure 3.4(b), the system is modeled as a 5 kg mass hanging from a spring of stiffness
(a) The natural frequency of free vibration is
(b) The beam is in equilibrium at t⫽ 0 when the particle hits. However, xis measured from the equilibrium position of the system with the particle attached. Thus,
x(0) = - ¢ (c)
st = - mg keq = -
(5 kg)(9.81m/s2)
2.04 * 105N/m = -2.40 * 10-4m vn =
A keq
m = A
2.04 * 105N/m
5 kg = 202.0 rad/s keq =
3EI L3 =
3(210 * 109 N/m2)(3 * 10-6 m4)
(2.1 m)3 = 2.04 * 105 N/m
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The initial velocity is .The time history of motion is calculated using Equation (3.19) as
(d) where the amplitude Aand the phase are determined using Equations (3.22) and (3.23), respectively:
(e) (f) f = tan-1c(202.0rad/s)(-2.40 * 10-4m)
0.5m/s d = -0.0968 rad = -5.59°
A =
A(-2.40 * 10-4 m)2 + a 0.5 m/s
202.2 rad/sb2 = 2.48 mm f
x(t) = Asin(202.0t + f) x#
(0) = 0.5 m/s
E = 210 × 109 N/m2 I = 3 × 10–6 m4
2.04 × 105 N/m 5 kg
x
Velocity = 0.5 m/s 2.1 m
(a)
5 kg (b)
FIGURE 3.4
(a) System of Example 3.3. A mass is dropped onto a fixed-free beam. (b) The system is modeled as a mass hanging from a spring of equivalent stiffness. Since xis measured from the equilibrium position of the system, the initial displacement is the negative of the static deflection of the beam.
E X A M P L E 3 . 4
An assembly plant uses a hoist to raise and maneuver large objects. The hoist shown in Figure 3.5 is a winch attached to a beam that can move along a track. Determine the nat- ural frequency of the system when the hoist is used to raise a 800-kg machine part at a cable length of 9 m.
S O L U T I O N
The beam is modeled as a pinned-pinned beam. If the hoist is at its midspan, its stiffness is kb = 48EI (a)
L3 =
48(200 * 109 N/m2)(3.5 * 10-4 m4)
(3.1 m)3 = 1.13 * 108 N/m
The stiffness of the cable is
(b) The beam and the cable act as springs in series with an equivalent stiffness of
(c)
The system’s natural frequency is vn = (d)
A keq
m = A
9.71 * 107 N/m
800 kg = 3.48 * 102 rad/s keq =
1 1 kb +
1 kc
=
1 1
1.13 * 108 N/m +
1 6.98 * 108 N/m
= 9.71 * 107 N/m kc = AE
L =
p(0.1 m)2(200 * 109 N/m2)
9 m = 6.98 * 108 N/m
Cable
(a) (b)
Beam
Beam: L = 3.1 m E = 200 × 109 N/m2 I = 3.5 × 10–4 m4 Cable: E = 200 × 109 N/m2
r = 10 cm L = 9 m
m
kb
ks
FIGURE 3.5
(a) System of Example 3.4 in which a hoisting mechanism consists of a cable attached to an overhead beam. (b) The system is modeled as a SDOF system with the stiffness of the beam and the stiffness of the cable acting as springs in series.
E X A M P L E 3 . 5
The pendulum of a cuckoo clock consists of a slender rod on which an aesthetically designed mass slides. If the clock gains time, should the mass be moved closer to or farther away from the support to correct the tuning?
S O L U T I O N
The pendulum is modeled as a particle of mass mon a rigid, massless rod. The particle is assumed to be a distance lfrom its axis of rotation. Summing moments about the point of support on the free-body diagrams of Figure 3.6 leads to
(a) Application of the small-angle assumption yields the linearized equation of motion
(b)
$ u +
g l u = 0
$ u +
g
l sinu = 0
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from which the natural frequency is calculated as
The period of oscillation is
Since the clock is running fast, the period of the pendulum needs to be increased. Thus l should be increased and the mass moved farther away from the axis of rotation.
T = 2pA l g vn =
A g l
FIGURE 3.6
(a) System of Example 3.5 in which the pendulum of a cuckoo clock is a massless rod with a particle attached. (b) FBDs at an arbitrary instant.
l
mg Ox Oy
mlθ˙2
mlθ¨
=
External forces Effective forces
The nonlinear differential equation derived in Example 3.5 is linearized by assuming small and replacing sin by . The exact nonlinear pendulum equation, Equation (a) of Example 3.5, is one of the few nonlinear equations for which an exact solution is known.
The solution subject to and is developed in terms of elliptic integrals, which are well-known tabulated functions.
The period of motion of a nonlinear system is dependent upon the initial conditions, while the period of a linear system is independent of initial conditions. One method of assessing the validity of the small-angle approximation for a given amplitude is to compare the period calculated using the exact solution to the period calculated using the linearized differential equations for different initial displacements. This comparison is given in Table 3.1, which shows that the small angle approximation leads to accurate prediction of the period for amplitudes as large as 40 . For an initial angular displacement of 40 , the error in the period from using the small angle approximation is only 3.1 percent.
The success of the use of the small-angle approximation in the pendulum example should give confidence to its use in other problems, where an exact solution is not available.
°
°
# u(0) = 0 u(0) = u0
u u u