1 2K      α11 α21 α12 α31 α13 3 2K    α22 α32 α23 5 2K  α33 Deflection at point 1 is: y    α11m1g α12m2g α13m3g mg 5mg 5g 2 2 1      2K 2000 2K Deflection at point 2 is: y2    α21m1g α22m2g α23m3g g 11g 2 y2 6 3     2K 2000 2K Deflection at point 3 is:    α31m1g α32m2g y3 α33m3g mg 13mg 13g 2 6 5      2K 2K 2000 Rayleigh’s formula is: n i 1  ω2 n ∑mi yi2 i 1 5 11 13 2 2x  2x  2x 2000 2000  ω2 11 2 52 13 2 2 2 2 2 g   2000 2000 2000 12.41 rad/s  ω Examples: 2 Find the lowest natural frequency of transverse vibrations of the system shown in Fig.9 by Rayleigh’s method. E=196 GPa, I=10-6 m4, m1=40 kg, m2=20 kg
1 α11  α21  α12  α31  α13  2K 3 α22  α32  α23  2K 5 α33  2K Deflection at point 1 is:  α11m1g  α12m2g  α13m3g  mg 2  2  1  5mg  5g 2K 2000 2K Deflection at point 2 is: y2  α21m1g  α22m2g  α23m3g y2 2  6  3   11g 2K 2000 2K Deflection at point 3 is: y3  α31m1g  α32m2g  α33m3g  mg 2  6  5  13mg  13g 2K 2000 2K Rayleigh’s formula is: n ω2  i 1 n ∑mi yi2 i 1 5 11 13 2 2x  2x  2x g ω2  2000 2000 52 11 2 13 2 2 2  2  2 g 2000 2000 2000 ω  12.41 rad/s Examples: 2 Find the lowest natural frequency of transverse vibrations of the system shown in Fig.9 by Rayleigh’s method. E=196 GPa, I=10-6 m4, m1=40 kg, m2=20 kg y
Unit-7_Lect-3.docx