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Civil - Structural dynamics of earthquake engineering

Free vibration of single degree of freedom systems (undamped) in relation to structural dynamics during earthquakes

   Author :  S. Rajasekaran    Posted On :  26.08.2016 12:19 pm

Abstract: In this page, the governing equations of motion are formulated for free vibration of single-degree-of-freedom (SDOF) (undamped) system. Vibration characteristics are studied by taking an example of a simple pendulum. Free vibration of rigid bodies without damping is also discussed.

Free vibration of single-degree-of-freedom systems (undamped) in relation to structural dynamics during earthquakes

 

Abstract: In this chapter, the governing equations of motion are formulated for free vibration of single-degree-of-freedom (SDOF) (undamped) system. Vibration characteristics are studied by taking an example of a simple pendulum. Free vibration of rigid bodies without damping is also discussed.

 

Key words: frequency, amplitude, phase angle, harmonic motion, NewtonÕs law, Rayleigh method.

 

 

Introduction

The study of vibration deals with oscillatory motion of a machine or a structure about an equilibrium position when it is subjected to shock or an oscillating force. The oscillations may be repeated uniformly, or change with time. Vibration in machines and structures is quite common and undesirable. In most cases its undesirable effects may be classified with respect to human characteristics and damage to engineering structures. An extreme example is a slender skyscraper whose wind-induced oscillations are entirely safe for the structure, yet unpleasant to the occupants of the upper floors. At the other extreme, certain vibrations in aeroplanes may be unnoticeable to the passengers, yet cause damage (fatigue) with catastrophic consequences.

 

The simultaneous occurrence of unpleasant and structurally damaging vibrations such as those in cars and trucks is the most common. Some vibrations are desirable, viz. beating of the heart, planets revolving around the sun and the consolidation of concrete by using vibrators. Damage to the structure occurs through earthquake loadings by forming cracks due to undesirable vibrations. Hence, in such cases, vibrations must be reduced to the minimum or should be eliminated. For elimination or reduction and to produce controlled vibrations, where necessary, a study of basic vibration theory is essential.

 

Formulation of the equation of motion

 

The governing equation of motion can be formulated using

 

       simple harmonic motion theory;

       Newton's second law;

 

       Energy method;

 

       Raleigh method;

 

       DÕAlembertÕs principle.

 

Simple harmonic theory

 

A special kind of motion occurs when the force on the body is proportional to the displacement of the body from equilibrium as shown in Fig. 2.1. If this force acts towards the equilibrium position of the body, a repetitive back and forth motion about this position occurs. Such a motion is an example of periodic or oscillatory motion. F is proportional to (-x).


where ωn is known as natural frequency. The particle moving along x-axis is said to exhibit simple harmonic motion if it satisfies Eq. 2.8, where A, ωn and φ are constants of the motion. In order to give a physical meaning to these constants, the graph in Fig. 2.2 shows x as a function of time t.


 

2.1 Motion of the particle when force is proportional to the displacement.

The constant A is called amplitude of the motion, which is the maximum displacement of the particle in either a positive or negative x-direction. The constant angle φ is called the phase constant (or phase angle) acting along with the amplitude A which can be determined uniquely by the initial displacement and initial velocity of the particle.

 

Note: The function x is periodic and repeats itself when ωnt increases by 2π rad.

 

The period T of the motion is the time it takes for the particle to com

plete one full cycle, i.e.


The inverse of the period is called frequency (f) of the motion. It represents the number of oscillations that a particle makes per unit time.


From Eq. 2.13, we get the maximum displacement, velocity and acceleration as:

xmax = A    2.14a

vmax = ωn2.14b

amax  = ω n2 A   2.14c

Figure 2.3 shows displacement vs. time, velocity vs. time and acceleration vs. time curves. The curves shown in Fig. 2.3 indicate that the phase difference between the velocity and displacement is π/2 radians or 90°, i.e. when x is maximum or minimum, the velocity is zero. Likewise, when x is zero the velocity is maximum. Furthermore, the phase of the acceleration is out of phase by π radians or 180° with displacement, i.e. when x is maximum, acceleration is maximum in the opposite direction.


Equation 2.3 is a second order differential equation and therefore two initial conditions are required to solve the equation. Let x t = 0 = x 0 ; v t =0 be the initial conditions. Substituting t = 0 in Eq. 2.13,

x0 = A sin φ

v0 = + ωn A cos φ    2.15a

a0 = ω n2 A sin φ    2.15b

From Eq. 2.15a and 2.15b, we get    2.15c


Thus we see that φ and A can be calculated if x 0 ,ω n and v0 are known. The following important properties are to be noted if a particle is moving

 

in simple harmonic motion.

 

The displacement, velocity and acceleration shown in Fig. 2.3 vary sinusoidally with time but are not in phase.               

   The acceleration of the particle is proportional to the displacement but in    the opposite direction.

                The frequency and period of motion are independent of amplitude.

 

In this book, we use force unit of newton, length unit of metres and mass unit of kilogram.

 

Example 2.1

 

A body oscillates with a simple harmonic motion along the x-axis. Its displacement varies with time according to x = 8 cos (πt + π/4), where t is in seconds and the angle is in radians.

 

(a)  Determine the amplitude, frequency and period of motion.

 

Calculate the velocity and acceleration of the body at any time t.

 

(c)  Using the results of (b), determine the position, velocity and acceleration of the body at t = 1 second.

 

(d) Determine the maximum speed and acceleration.

 

(e)  Find the displacement of the body between t = 0 to t = 1 second.

 

Solution


At t = 1

 

x = 8 cos (π + π/4) = 8 cos (5π/4) = Ð5.66 m

 

v = Ð8π sin (5π/4) = 17.78 m/s2

 

a = Ð8π2 cos (π + π/4) = 55.8 m/s2

 

vmax = 8π m/s, amax = 8π 2 m/s2

 

At t = 0

 

x0 = 8 cos (0 + π/4) = 5.66 m

 

At t = 1 s

 

x = -2.83 × 2 = -5.66 m

 

Hence displacement from t = 0 to t = 1 second is

 

x = x - x0 = -5.66 - 5.66 = -11.32 m

 

Since the particle's velocity changes sign during the first second, the magnitude of x is not the same as the distance travelled in the first second.

 

Newton’s second law

 

Spring–mass system

 

Consider a physical system consisting of mass attached to the end of a spring as shown in Fig. 2.4 where it is free to move. Due to the self-weight of the mass, the spring elongates by x0 and this position is called the equilibrium position. Considering the free body diagram


kxg = mg    2.17

Assume that the spring oscillates back and forth when it is disturbed from equilibrium position. From Newton's second law,


which is the same equation as obtained in Eq. 2.3

 

We see that the solution must be that of simple harmonic motion. Wherever the force acting on a particle is linearly proportional to the displacement and acts in the opposite direction, the particle is said to be in simple harmonic motion.


In Eq. 2.18, self-weight is cancelled with kxg and usually it is not considered during the analysis. Since period T = 2π/ωn and frequency is inversely proportional to period, we can express the period and frequency of the system as


 

2.6 Displacement, velocity and acceleration time curves.

Case II

 

Suppose if the mass is given an initial velocity v0 in the downward direction from the equilibrium position so that at t = 0, v = v0 and x0 = 0 at t = 0, we get




Example 2.2

 

A car of mass 1300 kg is constructed using a frame supported by four springs. Each spring has a force constant 20 000 N/m. If the combined mass of two people in a car is 160 kg, find the frequency of vibration when it is driven over a pothole on the road. Also determine the period of execution of two complete vibrations.

 

Solution

 

From angular frequency,


Period of vibration T = 1/f = 0.847 seconds. Time for one complete vibration = 0.847 seconds.

Time taken for two complete vibrations = 1.694 seconds.

 

Example 2.3

 

A mass of 400 g shown in Fig. 2.7 is connected to a light spring whose force constant is 5 kN/m. It is free to oscillate on a horizontal frictionless track. If the mass is displaced 10 cm from equilibrium and released from rest, find (a) period of motion, (b) maximum speed of the mass, (c) maximum acceleration of the mass, and (d) equations for displacement, speed and acceleration as function of time.


(d) Equations as a function of time

 

x = A cos ω n t = 0.1 cos 3.53 t

 

v = ÐωnA sin ωnt = Ð3.53 × 0.1 sin (3.53t) = Ð 0.353 sin (3.53t)

 

a = ω n2 A cos ω n t = 1.246 cos 3.53t

 

 

Simple pendulum

 

The simple pendulum is another mechanical system that moves in an oscillatory motion. It consists of a point mass 'm' suspended by means of light inextensible string of length L from a fixed support as shown in Fig. 2.8. The motion occurs in a vertical plane and is driven by a gravitational force. The forces which are acting on the mass are shown in the figure. The tangential component of the gravitational force, mg sin θ, always acts towards the mean position θ = 0 opposite to the displacement, restoring force acting tangent to the arc.


For small displacement sin θ  == θ and the motion of the bob is along the arc


From the above equation, it is seen that the period and frequency of a simple pendulum depend only on the length of the string and the value g.

 

Since the period is independent of the mass, a pendulum of equal length at the same location oscillates with equal periods. The analogy between the simple pendulum is the massÐspring system as shown in Fig. 2.9. The displacement, velocity, acceleration, kinetic energy and potential energy are given in Table 2.1 for various positions of the pendulum.

 

Example 2.4

 

A man wants to measure the height of a tower. He notes that a long pendulum extends from the ceiling almost to the floor and that its period is 24 s. Determine (a) the height of the tower and (b) the period when the pendulum is taken to the moon where g = 1.67 m/s2.


 

2.6     Comparison of simple harmonic motion and uniform circular motion

Consider a particle at a point P moving in a circle of radius A with constant angular speed ωn as shown in Fig. 2.10. We refer to the circle as reference


circle. As the particle rotates its position, the vector rotates about the origin 0 and at t = 0, 0P makes an angle of φ. At time t, θ = ωnt + φ.

This expression shows that point M moves with simple harmonic motion along the y-axis. Therefore we conclude that:

 

Simple harmonic motion along a straight line can be represented by the projection of uniform circular motion along diameter of a reference circle.

 

Similarly, we can show that point θ exhibits simple harmonic motion. Therefore:

 

Uniform circular motion can be considered as a combination of two simple harmonic motions.

 

Energy method

 

Energy of simple harmonic oscillator

 

Consider a massÐspring system discussed in Section 2.3 (see Fig. 2.4). Assuming the system to be conservative, we expect the total mechanical energy as constant. We can express kinetic energy as


Plots of kinetic and potential energy versus time are shown in Fig. 2.11a for φ = 0. The variations of T and V with displacements are plotted in Fig. 2.11b. Energy is continuously being transformed between potential energy stored in the spring and the kinetic energy of the mass. The kinetic energy and potential energy for the pendulum and spring mass system are shown in Fig. 2.9.

 

Example 2.5

 

A mass 0.5 kg is connected to a light spring of stiffness 20N/m, and oscillates on a horizontal frictionless track.

 

(a)  Calculate the total energy of the system and the maximum speed of the mass if the amplitude of motion is 3cm.

 

(b) Calculate the velocity of the mass when the displacement is equal to 2 cm.

(c)  Compute kinetic and potential energies of the system when the displacement is equal to 2cm.


 

Rayleigh method

 

E = V + T = VMAX = TMAX             2.41a

When strain energy is maximum, kinetic energy is zero and vice versa. From Eq. 2.37 and 2.38


 

 

D’Alembert’s principle

D’AlembertÕs principle of dynamic equilibrium is a convenient method for establishing the equation of motion for simple single-degree-of-freedom (SDOF) and multiple-degree-of-freedom (MDOF) systems. It essentially involves invoking Newton’s second law of motion to the system.

 

Considering Eq. 2.20, introduce the appropriate inertia force and it can be reasoned that applied force on the mass is in equilibrium with inertia force, i.e. inertia force is acting in the opposite direction to the applied force. Therefore dynamic problem is reduced to equivalent static problem.

 

D Alembert’s principle states that a system may be set in a state of dynamic equilibrium by adding to the external forces a fictitious force which is commonly known as the inertial.

Applying the equation of equilibrium for the free body shown in Fig. 2.12 we get

Mx + kx =0  ----- -- 2.42

 

Free vibration of rigid bodies without damping

                            The basic concepts of analysing a vibrating system that were developed up to Section 2.6 are a particular class of problem. The following list gives the characteristics of problems and some practical considerations.

                               One degree of freedom. A degree of freedom is defined as the independent coordinate with which we define the displaced shape of the structure. A system with a single coordinate function is said to be a one-degree-of- freedom system.

                               Free vibration. Equation 2.3 is valid when a disturbing force is applied only once to give a mass on initial displacement. The mass is in free vibration when only two kinds of forces are acting on it: (a) an elastic- restoring force within the system and (b) gravitational or other constant forces that cause no displacement from the equilibrium position of the system.

                               Undamped vibration. In the absence of dissipative forces acting on a vibrating mass, the amplitude of vibration is constant, and the motion is said to be undamped.

                               Natural frequency. Each mass spring system vibrates at a characteristic frequency in free vibration. This is known as natural frequency of the system.


 

       Lumped parameters. Strictly speaking, Eq. 2.3 is valid only for a particle of mass m and a spring of no mass and spring constant k. In practice, the mass of a translating rigid body is assumed to be concentrated as a particle and the mass of the spring is completely ignored.

 

Example 2.6

 

Find the natural frequency of the system shown in Fig. 2.13.

 

Solution

 

Taking moment at A


 

Example 2.7

 

A single one storey reinforced concrete (RC) building idealized as a massless frame is shown in Fig. 2.14 supporting a dead load of 50 kN at the roof level. The frame is 8 m wide and 4 m high. Each column and beam is 250 mm square. Assume YoungÕs modulus of concrete as 30 × 106 kN/m2 determine the natural frequency and period of the system. Assume stiffness of an equivalent SDOF system is k = 96EI/7h3.

 

Solution

 

Moment of inertia of the cross-section


 

Program 2.1: MATLAB program to draw displacement, velocity and acceleration with respect to time

 

Consider the springÐmass system shown in Fig. 2.4 with mass of 2kg m and stiffness of 8 N/m. We can write the following MATLAB program to draw the displacementÐtime, velocityÐtime and accelerationÐtime curves. We can solve symbolically the second order differential equation as shown in the listing. Initial displacement and velocity may be assumed as 3 m and 5 m/s.

 

1  Listing of MATLAB program

 

clc; close all; m=2; k=8; dt=0.02;

 

w=sqrt(k/m); y=dsolve(ÔD2y=-2^2*yÕ,Ôy(0)=3Õ,ÔDy(0)=5Õ,ÔxÕ); simplify(y)

 

for i=1:1500 t(i)=(i-1)*dt;

 

z(i)=3*cos(w*t(i))+5*sin(w*t(i))/w; v(i)=-w*3*sin(w*t(i))+5*cos(w*t(i)); a(i)=-3*w^2*cos(w*t(i))-5*w*sin(w*t(i));

 

end figure(1)

 

plot(t,z,ÔkÕ) xlabel(ÔtÕ) ylabel(ÔuÕ)

 

title(Ô Displacement Time CurveÕ) figure(2)

 

plot(t,v,ÔkÕ) xlabel(ÔtÕ) ylabel(ÔvÕ)

 

title(ÔVelocity time curveÕ) figure(3)

 

plot(t,a,ÔkÕ) xlabel(ÔtÕ) ylabel(ÔaÕ)

title(ÔAcceleration time curveÕ)

 

Figures 2.15, 2.16 and 2.17 represent displacementÐtime, velocityÐtime and accelerationÐtime curves.

 


 

Program 2.2: MATHEMATICA program to draw displacement, velocity and acceleration with respect to time

 

2.12.1  Listing of MATHEMATICA program

The listing of the program in MATHEMATICA is shown below. Using MATHEMATICA we can solve the second order differential equation and plot the displacementÐtime, velocityÐtime and accelerationÐtime curves.


 

 

Free vibration of structural systems

 

1  Laterally loaded elastic system

 

Consider the portal frame as shown in Fig. 2.18.

 

 

Case (a) When the beam stiffness is infinitely rigid (shear frame)

 

From Fig. 2.19a:

 


 


Example 2.9

 

Find the natural frequency of the system (Fig. 2.24) having mass less rigid rod






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