Dynamics of Structures

Chapter 3. Response of SDOF to Harmonic Loading

3.2 Response of Damped SDOF System to Harmonic Loading

3.2.1 Developing and Solving the Equation of Motion

Figure 3.2.1 Mass-spring of Damped System in Forced Vibration

Figure 3.2.1 shows an undamped mass-spring system subjected to a force excitation. Let's assume that the undamped system is subjected to a harmonic excitation in the form Fo sin(ωˉt)F_o \space sin({\bar{\omega}}t).

The equation of motion can be written as,

mu¨+cu˙+ku=Fosin(ωˉt)m \ddot{u} + c \dot{u} + ku = F_o \sin(\bar{\omega} t)

Eq. 3.2.1

This is also a nonhomogenous differential equation and its solution is two parts, complementary solution ucu_c and and a particular solution upu_p. The complementary solution is the solution of the homogeneous equation which we solved before as,

uc(t)=eζωt(Acos(ωDt)+Bsin(ωDt))u_c(t) =e^{-\zeta \omega t} (A \cos(\omega_D t) + B \sin(\omega_D t))

Eq. 3.2.2

The particular solution will depend on the excitation force. For a sin excitation force, the solution can take the form,

up(t)=(C2sin(ωˉt)+C3cos(ωˉt))u_p(t) = (C_2 \sin(\bar{\omega} t) + C_3 \cos(\bar{\omega} t))

Eq. 3.2.3

Differentiating for the first and second derivative, the constants can be obtained as,

C2=(Fo/k)(1r2)(1r2)2+(2ζr)2C_2 = (F_o/k) \frac{(1-r^2)}{(1-r^2)^2 + (2 \zeta r)^2}

Eq. 3.2.4

C3=(Fo/k)2ζr(1r2)2+(2ζr)2C_3 = (F_o/k) \frac{-2 \zeta r}{(1-r^2)^2 + (2 \zeta r)^2}

Eq. 3.2.5

The complete solution can be give as,

u(t)=eζωt(Acos(ωDt)+Bsin(ωDt))+u(t) = e^{-\zeta \omega t} (A \cos(\omega_D t) + B \sin(\omega_D t)) +
(C2sin(ωˉt)+C3cos(ωˉt))(C_2 \sin(\bar{\omega} t) + C_3 \cos(\bar{\omega} t))

Eq. 3.2.6

The two constants AA and BB can be obtained from the initial conditions, at t=0,u=uot=0, u = u_o and v=vov = v_o .

Using t=0,u=uot=0, u = u_o yields,

A=uoC3A = u_o -C_3

Eq. 3.2.7

Using t=0,v=vot=0, v = v_o yields,

B=vo+ζωAC2ωˉωDB = \frac{v_o + \zeta \omega A - C_2 \bar{\omega} }{\omega_D}

Eq. 3.2.7

The solution can be found as,

u(t)=eζωt(Acos(ωDt)+Bsin(ωDt))Transient Solution +u(t) = \underbrace{e^{-\zeta \omega t} (A \cos(\omega_D t) + B \sin(\omega_D t)) }_{\text{Transient Solution}} \space +
C2sin(ωˉt)+C3cos(ωˉt)Steady State Solution \underbrace{C_2 \sin(\bar{\omega} t) + C_3 \cos(\bar{\omega} t)}_{\text{Steady State Solution}}

Eq. 3.2.8

The transient part of the solution will vansih leaving only the steady state part that goes as long as the applied harmonic excitation exist.

Observations

Damping will make the transient part to vanish after a certain period of time leaving only the steady state park of the solution.

When the frequency ratio r=1r = 1, the forcing frequency is equal to the system frequency and system is said to be at resonance. The amplitude of the steady state solution tends to infinity, however, in real systems, failure will occur long before the system reaches very high amplitudes.

uou_o

vov_o

ω\omega

ωˉ\bar{\omega}

FoF_o

kk

ζ\zeta

tt