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ME4189_F12_9_18_Examples

This is accomplished by finding the unit vectors in

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Unformatted text preview: t be quantified. This is accomplished by finding the unit vectors in the direction of each spring, shown below as ̂ and ̂ . ̂ ̂ Using the frame shown in the figure, ̂ ̂ cos cos ̂ ̂ sin In vector form, the deflection of the mass ̅ is sin ̅ ̂ ̂ ̂ Therefore, Δ ̂ ∙ ̅ cos ̂ ∙ ̅ Δ cos The potential energy is then simplified to be cos Lagrange’s equation is 0 21 cos 0 Thus, interesting enough, the system behaves as if it were restrained by a single spring of magnitude 21 cos Example 2: The example below is significantly more complicated than that discussed above. The L-shaped bar is pinned at point , and is free to rotate about that point. Hence, the single degree of freedom required for solution of the problem (obtaining the equation of motion) is , as shown in the figure. There are two springs, each of stiffness , and two dampers, each of magnitude . The mass of the bar is , though you will see that it is not exactly relevant here. Relevant points on the body are , which is the L-beam weld, , where the force acts, and , the end of the bar. Both springs are inclined an angle from the bottom portion of the beam. Relevant lengths are also demonstrated in the figure. The goal of the problem is to find the equation of motion of the system, using Lagrange’s equation: where signifies Rayleigh’s dissipation function. The kinetic energy of the system is very easy to obtain, because the whole body purely rotates about point : 1 2 For the sake of simplicity, the inertia term of the body about knowledge of the mass is not entirely relevant). will be left as (hence, the Unit vectors are shown designating the direction of each spring and damper. The subscript on the unit vector indicates the designation of the component. Therefore, the general forms of and are 1 Δ 2 1 Δ 2 1 Δ 2 1 Δ 2 The symbol Δ designates a displacement (or stretching/compressing), whereas the symbol Δ designates the velocity of one end of the compon...
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