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Equivalent Dynamics system

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raj0305
9 diapositives · 156 Views · Aug 15, 2020

Unit 1
Dynamics of Machine

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Equivalent Dynamics system
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  1. EQUIVALENT DYNAMICAL SYSTEM

  2. In order to determine the motion of a rigid body, under the action of external forces, it is usually convenient to replace the rigid body by two masses placed at a fixed distance apart, in such a way that, • 1. the sum of their masses is equal to the total mass of the body ; • 2. the centre of gravity of the two masses coincides with that of the body • 3. the sum of mass moment of inertia of the masses about their centre of gravity is equal to the mass moment of inertia of the body. • When these three conditions are satisfied, then it is said to be an equivalent dynamical system. Consider a rigid body, having its centre of gravity at G

  3. Let m = Mass of the body, KG = Radius of gyration about its centre of gravity G, m1 and m2 = Two masses which form a dynamical equivalent system, l1 = Distance of mass m1 from G, l2 = Distance of mass m2 from G,

  4. This equation gives the essential condition of placing the two masses, so that the system becomes dynamical equivalent. The distance of one of the masses (i.e. either l1 or l2) is arbitrary chosen and the other distance is obtained from equation (vi).

  5. Formulas

  6. A connecting rod is suspended from a point 25 mm above the centre of small end, and 650 mm above its centre of gravity, its mass being 37.5 kg. When permitted to oscil- late, the time period is found to be 1.87 seconds. Find the dynamical equivalent system constituted of two masses, one of which is located at the small end centre. Solution. Given : h = 650 mm = 0.65 m ; l1 = 650–25 = 625 mm = 0.625 m ; m = 37.5 kg ; tp = 1.87 s First of all, let us find the radius of gyration (kG) of the connecting rod (considering it is a compound pendulum), about an axis passing through its centre of gravity, G. We know that for a compound pendulum, time period of oscillation (tp),

  7. It is given that one of the masses is located at the small end centre. Let the other mass is located at a distance l2 from the centre of gravity G, as shown in Fig. 15.19. We know that, for a dynamicallyequivalentsystem, l1.l2 = (kG)2