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Unformatted text preview: out for 2-D and 3-D vector motions. • Be sure to draw the appropriate free body diagrams for each mass (or combination of masses) in the system. o Whenever the system is separated in order to draw a free body diagram, replace the separation with the appropriate internal forces/moments (equal and opposite forces/moments on each side of the separation). o Do not move forces/moments arbitrarily from one mass to another. The internal forces account for the effects of one mass on another. • Develop one equation of motion for each degree of freedom of the system using Newton's ( -or- d'Alembert’s ) method. Be sure to watch for moving reference frame issues. Also, check that the units are the same for each term in an equation (Forces + Moments: NOT!) • If necessary, once the exact equations of motion have been determined, linearize the equations of motion by neglecting nonlinear terms in the equations of motion. Note that the linear equations of motion may not adequately describe the original equations of motion if some of the terms that have been neglected are not insignificant. Lecture Notes -10- 06/16/06 12:25 PM Mechanical Vibrations I Newton Example A
In order to solve Newton’s Equation, m x(t) f(t) ∑F i = mx , for the given single degree-of-freedom k c system, it is first necessary to draw the complete free-body diagram. This involves identifying all internal and external forces acting upon the degree-of-freedom. (Recall that the internal force resulting from the spring is due to the actual relative motion of the ends of the spring. Therefore the coordinate x is measured from the free length of the spring.) All the internal and external forces are combined to yield the actual equation of motion. f − kx − mg − cx = mx collecting all the coordinate based terms to one side of the equation and all the externally applied forces to the other yields f mx m
cx m mg mx + cx + kx = f − mg Since the mg term ends up on the right hand side with the external force, the coordinate x can be redefined to be from the static equilibrium position. As a result, the mg term can be eliminated as shown. Evaluating the static equilibrium equation (i.e. x = 0 , x = 0 & f = 0 .) yields, kx = −mg
Therefore the static deflection ( ∆ ) can be evaluated as ∆=− mg k The coordinate system reference can be shifted to the static equilibrium position by defining Lecture Notes -11- 06/16/06 12:25 PM Mechanical Vibrations I xs = x − ∆
Since the static deflection ( ∆ ) is constant, obviously the velocity and acceleration is unchanged by the coordinate shift. xs = x and xs = x
Substituting into the original solution,
mxs + cxs + k ( xs + ∆ ) = f − mg But recall that k ∆ = − mg , therefore mxs + cxs + kxs = f
Note: Often the solution solved about the static equilibrium position will eliminate the ‘mg’ term when there are springs to carry the load. When this occurs, the ‘mg’ term will end up on the right hand side of the equation...
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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.
- Fall '11