The system of mono-degree of freedom with some type of aerodynamic force and the analysis of solution for the case has been considered in [1, 2]. In [3] we have conducted a survey in the non-resisting equation system of two-degree of freedom with the aerodynamic force as in f5]. In this article, we will continue to study the abovementioned mathematical problem with the regard to the resisting element of environment make some suggestions for the solution and survey its stability. | T~p chi Ccr h9c Journal of Mechanics, NCNST of Vietnam T. XIX, 1997, No 4 (73 - 78) ON THE INSTABILITY OF A MECHANICAL SYSTEM OF 2 - DEGREE OF FREEDOM TAKING INTO ACCOUNT THE RESISTING FORCE Vu Quae TRU, NGUYEN DANG BICH Institute for Building Sciences and Technology High projects, soft structure are destroyed due to aerodynamic instability. The action mechanism of aerodynamic force is very complicated, which makes the mathematical problem of multi-degree freedom system more difficult to be solved. The system of mono-degree of freedom with some type of aerodynamic force and the analysis of solution for the case has been considered in [1, 2]. In [3] we have conducted a survey in the non-resisting equation system of two-degree of freedom with the aerodynamic force as in f5]. In this article, we will continue to study the abovementioned mathematical problem with the regard to the resisting element of environment make some suggestions for the solution and survey its stability. 1. Formulation of the problem The mechanical system is considered as a high column, the mass quantity M is concentrated on the column top end, the vibration of the system with the regard to the resisting force is described in the following equation: i' + 2vr + w2 r- rp 2 = (1- a/' r () where: v, a-· resisting coefficient, w2 = _:_ M c - corresponding rigidity of the system; a, b - coefficient of aerodynamic force. The second equation of system (} has an initial integral: () where: C 1 - integral constant. Substitute () into (), the first equation of () reduces to: r + 2vr +. w2 r- C 1 r2 b+ 1e- 2 at = •2 (1- a):_ () r Using variable change method: () s, >. are arbitrary coefficients, which will be chosen in the solution process, from () we have: p+2(v->.a)p+s(w 2 +a>. 2 -2v>.)p= 73 (1-;)P: +C (sp)"/'+1e1 2 (aHb)t () We choose s and A providing that: a+ ,\b = 0 2b and () -+1=0, s then the equation () will be taken of the .
