UNDAMPED FREE VIBRATION
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enotes.pdf (Size: 412.04 KB / Downloads: 199) chapter2.zip (Size: 3.23 MB / Downloads: 94) Presented by:Dr. Chandrashekara Murthy, R V College of Engineering , Bangalore UNDAMPED FREE VIBRATION UNDAMPED FREE VIBRATION 2.1 Introduction Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force. If during vibrations there is no loss of energy, it is known as undamped vibration. The first step in solving a vibration problem is setting up the differential equation of motion. The three approaches to setting up differential equation of motion are as follows SETTING UP THE EQUATION OF MOTION 1.Use D’Alembert’s Principle Forces m x = 0 Torques  I q = 0 m = Mass x = Displacement m x = Inertia Force, q = Angular Displacement I = Mass Moment of Inertia, I q = Inertia Torque D’Alembert’s Principle states that the resultant of all forces acting on a body along with the inertia force is equal to zero. For a rotational system we have to consider torques instead of forces. Using D’Alembert’s Principle we can setup the differential equation of motion. Alternatively, we can get the differential equation of motion by applying Newton’s second law of motion. The third approach to setting up the equation of motion is to apply energy method. Spring mass system Fig 2.1 shows a one degree of freedom simple spring mass system. It represents several practical systems. Free vibrations of a system with a single degree of freedom is one of the most important topics. In fig 2.1 the coordinate x is used to describe the position of the mass. The mass and spring are the basic building blocks for vibrational analysis. Spring stiffness is defined as the force required to elongate or compress the spring by unit length. The differential equation of the spring mass system is set up by considering all the forces and applying D’Alembert’s Principle Spring mass system represents several practical systems Examples: 1. Machine mounted on isolators 2. Mass m attached to the end of a cantilever beam KSpring stiffness in N/m mMass in Kg m xInertia Force Kx = Spring Force, d = Static deflection of spring K d = Force due to static deflection mg = Gravitational pull m x + Kx + K d  mg = 0 m x + Kx = 0 Linear homogeneous second order differential equation X=A sin wn t + B cos wn t = C sin (wn t +f ) wn= Ö K /m = Natural Frequency of vibrations This is the only frequency with which the system vibrates when disturbed and let free. The natural frequency is a characteristic property of the vibrating system. The amplitude of oscillations C and the phase angle F can be determined by applying the initial conditions. Problem A small Pelton wheel rotating at 1500 rpm has a rotor of mass 10 Kg mounted at the centre of a steel shaft which has a span of 0.4 m between bearings. What should be the diameter of the shaft , so that the transverse natural frequency is 50 percent higher than the running speed? Assume E for steel as 2x1011 pa._ 


