Steady-state vibration requires $D1a=D1\gamma =0$ in Eq. (25) and $D2a=D2\gamma =0$ in Eq. (30). Combination of the real and imaginary parts of Eqs. (25) and (30) according to the reconstitution method [41,42], substitution of $D1a=D1\gamma =0$ into Eq. (25) and $D2a=D2\gamma =0$ into Eq. (30), and use of Eqs. (18) and (19) in that sequence give the frequency-amplitude relationsDisplay Formula

(31)$a\sigma -\Delta 02a-3\alpha 3,0a38-\alpha 0,12cos\gamma +q12cos(\gamma -\rho 1)+{14\mu \alpha 0,1sin\gamma -14\mu q1sin(\gamma -\rho 1)+[\alpha 0,1cos\gamma -q1cos(\gamma -\rho 1)](332\alpha 3,0a2+14\sigma )\u2003\u2003\u2003-38\Delta 0\alpha 0,1cos\gamma -12\alpha 0,0\Delta 1cos(\nu 1-\gamma )-14\Delta 2acos(\nu 2-2\gamma )-38\alpha 2,1a2cos(\Phi 2,1-\gamma )+18\Delta 02a-316\alpha 3,0\Delta 0a3+12\mu 2a+15256a5\alpha 3,02\u2003\u2003\u2003+a\alpha 2,0\alpha 0,0\Delta 0+512a3\alpha 2,02}=0-\mu a-12\alpha 0,1sin\gamma +12q1sin(\gamma -\rho 1)+{-14\mu \alpha 0,1cos\gamma +14\mu q1cos(\gamma -\rho 1)\u2003\u2003\u2003+[\alpha 0,1sin\gamma -q1sin(\gamma -\rho 1)](932\alpha 3,0a2+14\sigma )-38\Delta 0\alpha 0,1sin\gamma +38\mu \alpha 3,0a3+12\alpha 0,0\Delta 1sin(\nu 1-\gamma )+14\Delta 2asin(\nu 2-2\gamma )\u2003\u2003\u2003+18\alpha 2,1a2sin(\Phi 2,1-\gamma )}=0(31)$