Introduction To Vibrations

Introduction to Vibrations – A Brief Summary Definition: Vibration is defined as oscillatory motion of dynamic systems. The motion could be displacement, velocity, acceleration and so on. Types: Vibration can be categorized in different ways. One of the ways is to classify vibrations into free (natural) and forced (in some cases this is called steady-state) vibrations. The free vibrations are due to the initial conditions and the forced vibrations are due to the external force or excitation acting on the system. In general, a dynamic system has initial conditions and may also be subjected to an external excitation, and hence the system response is found for each free and forced vibration cases and added together to get the total system response. Both free and forced vibrations can be further split into damped and undamped vibrations depending upon whether the system has viscous damping (damped) or not (undamped). Free Vibrations: A typical example of undamped free vibrations is a spring-mass system, as shown below. The system is said to have a single degree-of-freedom because the only variable x(t) is enough to describe the motion.

x(t) k

m

Friction free smooth surface The equation of motion (EOM) of the above system is written as m x + k x = 0 . The above equation can also be written as x + ωn2 x = 0 , where ωn = k / m is known as the natural frequency of the system in rad/s. Physically, it represents the frequency for an undamped system where the system gets into resonance and vibration magnitude becomes high (theoretically infinite). Assuming that the system has the initial conditions (IC) of x(0)=x0 and x (0) = v0, the system response can be found using either the assumed form of solution or the Laplace transform. The assumed form of solution goes with a form of solution of x(t)=Asin(ωnt + φ), where A is the amplitude and φ is the phase angle of vibrations, which can be easily determined from IC as A = x02 + v02 / ωn2 and φ = tan −1 (ωn x0 / v0 ) .

In the case of damped (viscous) free vibrations, the above system is added a dashpot with a viscous damping constant of b, for which the EOM becomes m x + bx + k x = 0 or in a standard form x +2ζωn x + ωn2 x = 0 , where ζ is known as the damping ratio defined as ζ= b /(2 km ) . Depending upon the values of ζ, the damped vibrations are grouped as 1)critically damped (ζ=1), 2) overdamped (ζ>1) and 3) underdamped (0<ζ<1). Among these three cases, only the underdamped case results in oscillations. The critically damped vibrations correspond to the case when the roots of the equation

x +2ζωn x + ωn2 x = 0 are real and same (repeated roots), which results in a response of x(t)=(c1+c2t) e − ωnt , where c1 and c2 are constants determined from the IC as c1=x0 and c2=v0+ωnx0. For the overdamped case, the roots are real and distinct, and the response is calculated assuming a form of solution of x(t)=c e λ t . Substituting this solution in the above equation yields λ2+2ζωnλ+ ωn2 =0, from which two values of λ are found as

λ1,2=ωn(−ζ± ζ 2 − 1 ).

Therefore,

the

response

of

the

system

becomes

λ2t

x(t)=c1 e λ 1 t +c2 e , where c1 and c2 are obtained from the IC. In the case of underdamped vibrations, there are two complex conjugate roots and the response becomes x(t)=Ae−ζ ωnt sin(ωdt + φ), where ωd=ωn 1 − ζ 2 is called the damped natural frequency, and A and φ are found from the IC as A=

(x ζ / 0

1 − ζ 2 + v0 / ωd

) +x 2

2 0

and φ = tan −1 [ωd x0 /(v0 + x0 ζωn )] . Typical curves for these three cases are shown below.

Forced (Steady-State) Vibrations: As sown below, a spring-mass-damper system is now assumed to be acted upon by a sinusoidal force of the form f(t)= f sin(ωt), where ω is the frequency of vibrations of the applied force. An unbalanced force on a rotating machine manifests itself in the form of a sinusoidal force with a magnitude of f =murω2, where mu is the unbalance mass, r is the unbalance distance from the center of rotation and ω is the rotational machine speed in rad/s.

x(t)

b k

m

Friction free smooth surface

f(t)

The EOM of the system in standard form is given as x +2ζωn x + ωn2 x =

1 f (t ) = m

ωn2 f (t ) , for which the transfer function is computed as G(s) = X(s)/F(s) = k ωn2 . The total system response x(t) is composed of responses to the k ( s 2 + 2ζωn s + ωn2 ) IC and to the force f(t). Assuming that the IC are all zero, x(t) becomes same as the steady-state response, which can be easily calculated via sinusoidal transfer function ωn2 G(jω) as x(t)=|G(jω)| f sin(ωt+φ), where |G(jω)|= = k (ωn2 − ω 2 ) 2 + (2ζωn ω) 2 2ζωn ω 1 with Ω=ω/ωn and φ=Im[G(jω)]/Re[G(jω)] = −tan−1 ω 2 − ω 2 = n k (1 − Ω 2 ) 2 + (2ζΩ ) 2 2ζΩ 2ζΩ f 1 −tan−1 1 − Ω 2 . Thus, we obtain x(t)= sin(ωt−tan−1 1 − Ω 2 ), k (1 − Ω 2 ) 2 + (2ζΩ ) 2 f is called the static displacement. The amplitude and phase angle k of x(t) are plotted below for different values of Ω and ζ.

where the factor

The vertical axis for the amplitude graph stands for the magnification factor 1 and the vertical axis for the phase angle represents the negative 2 2 (1 − Ω ) + (2ζΩ ) 2 phase angle in rad. It is clear from the amplitude graph that for small values of damping ratio ζ, the vibration magnitude becomes very large when Ω=1, i.e. ω=ωn. Furthermore, it is evident from the phase angle graph that regardless of the values of ζ, for Ω=1 the phase angle becomes around −1.57 rad or −90o, indicating that when ω=ωn, then φ = −90o. This observation can be used to identify resonance problems in dynamic systems. Free Vibrations of Two-Degree-of-Freedom Systems: Many times, it may happen that a dynamic system has more than one-degree-of-freedom, i.e. the system has multi degree-of-freedom with more than one variable to describe the system behavior. A two-degree-of-freedom system with two masses is shown below.

x1(t) k

m

x2(t) k

m

k

Friction free smooth surface The EOM for this system can be written as mx1 + 2kx1 − kx2 = 0 mx2 + 2kx2 − kx1 = 0 The solutions are assumed as x1(t)=Asin(ωnt + φ) and x2(t)=Bsin(ωnt + φ). Substituting these in the above equations yields

(−mA ωn2 +2kA−kB) sin(ωnt + φ) = 0 (−mB ωn2 +2kB−kA) sin(ωnt + φ) = 0 To satisfy the above equations, we should have the following −mA ωn2 +2kA−kB = 0

(1)

−mB ωn2 +2kB−kA = 0

(2)

In matrix form, ⎡2k − mωn2 ⎢ ⎣ −k

− k ⎤ ⎡ A⎤ ⎡0⎤ ⎥⎢ ⎥=⎢ ⎥ 2k − mωn2 ⎦ ⎣ B ⎦ ⎣0⎦

For the non-trivial solution, we should have the determinant of the above matrix to be zero, i.e. 2k − mωn2 −k

−k =0 2k − mωn2

Or, (2k − mωn2 ) 2 − k 2 = 0 , from which we obtain two natural frequencies (or modes) computed as ωn1 = k / m and ωn 2 = 3k / m When we substitute these natural frequencies in equation (1) or (2) above, we obtain two relations between A and B or two mode shapes as

ωn = ωn1 → A/B = 1 ωn = ωn2 → A/B = −1 The general solutions (responses) for arbitrary initial conditions are given by x1(t)=A1 sin(ωn1t + φ1) + A2 sin(ωn2t + φ2) x2(t)=A1 sin(ωn1t + φ1) − A2 sin(ωn2t + φ2) where A1, A2, φ1 and φ2 are found from four IC of x1(0)=x10, x1 (0) = v10 , x2(0)=x20 and x 2 (0) = v 20 . Forced Vibrations of Two-Degree-of-Freedom Systems: When the two masses of the above system are acted upon by sinusoidal external forces of f1(t)= f1 sin(ωt) and f2(t)= f 2 sin(ωt), the EOM become

mx1 + 2kx1 − kx2 = f1 (t ) mx2 + 2kx 2 − kx1 = f 2 (t )

Assuming solutions of the form x1(t)=Asin(ωt) and x2(t)=Bsin(ωt) and substituting them in the above equations yields the following (in matrix form) ⎡2k − mω 2 ⎢ ⎣ −k

− k ⎤ ⎡ A⎤ ⎡ f1 ⎤ ⎥⎢ ⎥=⎢ ⎥ 2k − mω 2 ⎦ ⎣ B ⎦ ⎣ f 2 ⎦

which can be solved for A and B as A=

(2k − mω 2 ) f1 + kf 2

B=

kf1 + (2k − mω 2 ) f 2

(2k − mω 2 ) 2 − k 2 (2k − mω 2 ) 2 − k 2

It is clear that values of A and B become infinite when ω=ωn1 or ω=ωn2, i.e. when the external frequency ω reaches either of ωn1 or ω=ωn2, then the magnitudes of x1(t) and x2(t) become theoretically infinite (since there is no damping in the system).