# Of the system the force transmitted to the foundation

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Unformatted text preview: into standard form yields, m1 x1 + ( k1 + k2 ) x1 − k2 x2 = f1 + m1 g m2 x2 + k2 x2 − k2 x1 = f 2 + m2 g m2 g Lecture Notes -56- 06/16/06 12:25 PM Mechanical Vibrations I Or expressed in matrix form as, m1 0 x1 k1 + k2 0 m x + −k 2 2 2 − k2 x1 f1 m1 g = + k2 x2 f 2 m2 g Recall that when the gravitational term ( mi g ) ends up on the force side of the equation, the associated coordinate ( xi ) can be redefined from the static equilibrium position thereby eliminating the mi g term from the equation. In this case, both terms are eliminated. m1 0 x1 k1 + k2 0 m x + −k 2 2 2 − k2 x1 f1 = k2 x2 f 2 Equations of Motion – Lagrange’s Method As has been shown, Lagrange’s Method involves identifying the kinetic, potential and dissipative energies for the entire system, then evaluating the Lagrange Equation for each degree of freedom. d ∂T ∂T ∂U ∂D + + = Fi − dt ∂qi ∂qi ∂qi ∂qi First, identifying the kinetic energy term, 2 T = 1 m1 x12 + 1 m2 x2 2 2 And the potential energy term, U = 1 k1 x12 + 1 k2 ( x2 − x1 ) − m1 gx1 − m2 gx2 2 2 2 Provides all the information necessary to determine the equations of motion. Note that there are no dissipative terms for this problem, so D = 0 . Second, evaluating the various partial derivatives for each degree of freedom, yields, (First degree of freedom) d d ∂T d ∂ 1 ( 2 m1x12 + 12 m2 x22 ) = dt ( m1x1 ) = m1x1 = dt ∂x1 dt ∂x1 Lecture Notes -57- 06/16/06 12:25 PM Mechanical Vibrations I ∂U ∂ = ∂x1 ∂x1 ( 1 2 k1 x12 + 1 k2 ( x2 − x1 ) − m1 gx1 − m2 gx2 = k1 x1 + k2 ( x2 − x1 )( −1) − m1 g 2 2 ) (Second degree of freedom) d d ∂T d ∂ 1 ( 2 m1x12 + 12 m2 x22 ) = dt ( m2 x2 ) = m2 x2 = dt ∂x2 dt ∂x2 ∂U ∂ = ∂x2 ∂x2 ( 1 2 k1 x12 + 1 k2 ( x2 − x1 ) − m1 gx1 − m2 gx2 = k2 ( x2 − x1 ) − m2 g 2 2 ) Finally, collecting the various terms for each degree of freedom yields the equations of motion. m1 x1 + ( k1 + k2 ) x1 − k2 x2 = f1 + m1 g m2 x2 + k2 x2 − k2 x1 = f 2 + m2 g Or expressed in matrix form as, m1 0 x1 k1 + k2 0 m x + −k 2 2 2 − k2 x1 f1 m1 g = + k2 x2 f 2 m2 g Again recall, when the gravitational term ( mi g ) ends up on the force side of the equation, then the associated coordinate ( xi ) can be redefined from the static equilibrium position thereby eliminating the mi g term from the equation. In this case, both terms are eliminated. m1 0 x1 k1 + k2 0 m x + −k 2 2 2 − k2 x1 f1 = k2 x2 f 2 Notice that the actual equations of motion identified are identical, regardless of whether Newton’s Method or Lagrange’s Method was used. Natural Frequency and Mode Shape In order to identify the natural frequency and mode shape for a set of simultaneous differential equations of motion, it is necessary to solve an eigenvalue problem. By transforming the set of equations to the Laplace Domain ( s ), the differential equations of motion are transformed into a set of simultaneous algebraic equations. Lecture Notes -58- 06/16/06 12:25...
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## This note was uploaded on 09/29/2013 for the course MECHANICAL ME taught by Professor Regalla during the Fall '11 term at Birla Institute of Technology & Science, Pilani - Hyderabad.

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