linear vibration analysis using screw theory - helix - Georgia Institute ...

31yVEMVxFigure 3.3: Location of Vibration Centers when EM y = 0yMVVExFigure 3.4: Location of Vibration Centers when ! 2 x = !2 yroots of (3.19). Since the constant term is negative, the two roots have opposite signs and thereforelie on opposite sides of M. Thus, one vibration center must lie on each side of ¡¡! EM (see Figure 3.3).Case 2b, ! 2 x = ! 2 y and the chosen coordinate system does not go through M. The vibrationcenters are on the line that connects E and M but (3.6) and (3.7) show that they cannot be betweenE and M. Their location MV x is given by (3.11) which reduces to,(k y EM 2 x + k xEM 2 y )MV 2 x ¡ EM x(m ° ! 2 x ¡ k ° ¡ k y EM 2 x ¡ k xEM 2 y )MV x ¡ ! 2 y m °EM 2 x = 0 (3.21)Since the constant term is negative, the two roots have opposite signs and therefore lie on oppositesides of M. Thus, one vibration center must lie on each side of ¡¡! EM (see Figure 3.4). It should benoted that Figures 3.3 and 3.4 are special cases of Figure 3.2.Case 3, E and M are coincident. This is the trivial case and the single …nite vibration centeris also coincident.The results are summarized by the following theorems.