Advanced Structural Dynamics ProjectThe dynamic response and stability analysis of the beam under vertical excitationInstructor：Dr. Li WeiName：Student ID：1.Problem description and the purpose of the project1.1 calculation modelAn Eular beam subjected to an axial force. Please build the differential equation of motion and use a proper difference method to solve this differential equation. Study the dynamic stability of the beam related to the frequency and amplitude of the force. As shown in the Fig 1.1.Fig1.11.2 purpose and process arrangementa.learning how to create mathematical model of the continuoussystem and select proper calculation method to solve it.b.learning how to build beam vibration equation and solve Mathieuequation.ing Floquet theory to judge vibration system’s stability andanalyze the relationship among the frequency and amplitude of the force and dynamic response.This project will introduce the establishment of the mathematical model of the continuous system in section 2, the movement equation and the numerical solution of using MATLAB in section 3, Applying Floquent theory to study the dynamic stability of the beam related to the frequency and amplitude of the force in section 4. In the last of the project, we get some conclusions in section 5.2. the mathematical model of the systemThe geometric model of the beam and force-simplified diagram is shown in Fig.2.1.We assume that its stiffness(EI) is constant and the deflection of the beam is small, and the boundary conditions is simply support. Now the beam subjected to an axial force. We assume the force is equal to 0cos P t ω. F=f 0coswt yxFig.2.1We select the length of x ∆ in any position of the beam, the free-body diagram is shown in Fig.2.2.Fig.2.2Using equations of movement equilibrium, that is to say:+↑()y y F m a ∑=∆ （1）0G M +∑= （2） From equation (1), we will get:22),(),(ty x A t x x S t x S ∂∂∆=∆+-ρ （3）Divide equation (3) by x ∆ and take the limit:22xy A x S ∂∂=∂∂-ρ （4） Then synthesize equation (2),we can get:0),()],(),([),(),(=∆∆+--∆++-∆+x t x x S t x y t x x y F t x M t x x M （5）Divide equation (5) by x ∆ and take the limit:S xy F x M =∂∂+∂∂ （6） Combine equation (4) with (6):0222222=∂∂+∂∂+∂∂ty A x y F x M ρ （7） And 22),(xy EI t x M ∂∂= （8） Combine equation (7) with (8):0)(22222222=∂∂+∂∂+∂∂∂∂ty A x y F x y EI x ρ （9） We know EI is a constant, so0)(222244=∂∂+∂∂+∂∂ty m x y t F x y EI （10） In equation (10), m is the mass of unit length. Now we will use assumed-modes method. Named lx n t T t x y n πsin )(),(=,so: 0sin )(22244422=⎥⎦⎤⎢⎣⎡-+l x n T l n t F T l n EI dt T d m n n n πππ （11） 0))(1(222=-+n non n T F t F P dt T d n=1,2,...... （12） In the equation (12)222222,l EI n F m EI l n P n on ππ==And t F t F ϖcos )(= ,so0)cos 1(222=-+n non n T t F F P dt T d ϖ n=1,2,...... （13） 0)cos (22=-+n n T t dtT d ϖεδ （14） In the equation (12)4)(L n A EI πρδ= 22)(Ln F πε= Equation (14) is the Mathieu equation. it is difficult to solve the analytical solution directly, thus, we use the approximate derivative namely an average acceleration method to get the numerical solution from the reference.3. Numerical solution3.1 using MATLAB to solve equationWe will use the Newmark-β method [1] to solve equation (14). We can use the initial condition 00u u 和to integrate the move equation:m 0u cu ku ++= （15）Fig.3.1As shown in Fig.3.1))(2(11+++∆+=i i i i i u u t u u （16） )(4121+++∆+∆+=i i i i i i i u u t t u u u （17） ()0cos =-+i i u wt u εδ （18）From equation (16), (17) and (18), we will get:()i i u t wn u ∆∆--=∆cos εδ （19）i i ii u u t u 2)2(-∆∆=∆ （20） ()()22]cos [44]cos [)(2tt wn t u t wn u t u i i i ∆∆-+∆+∆-∆-=∆εδεδ （21） When applying the MATLAB, we need discrete the processing time t, get time step 02.0=∆t .When solving the vibration stability interval, there are three variables to participate in the discussion, namely w c ,,δ. So take a particular w first and discuss the remaining two parameters.From Floquent theory [2]，we can use parameter A to judge stability.Equation 0)()(22=++y t dt dy t dt y d ξξ（22） Take two sets of special solution:1)0(,0)0(0)0(,1)0(2211====yy yy （20） Parameter [2] )]()([2121T y T y A +=（21） If abs (A) is less than 1, the system is stability. And if abs (A) is greater than 1, the system is instability. When abs(A) is equal to 1, the system is critical state.We use MATLAB Codes to solve equations. We use ω=2 Math ieu Equation to judge the validity of the codes. From Fig.3.2 and Fig.3.3, wecan consider the codes are correct.In these follow figures, ω=2, the horizontal axis is δ, vertical axis is ε.Fig.3.2. stable domain in reference [3]and[2]Fig.3.3. stable domain in MATLAB solutionCompared Fig.3.2 with Fig.3.3, we can see that the stability domain of numerical solutions applying average acceleration method are consistent with the standard solutions. it can concluded that when the system have solution whose cycle is equal to π or 2π, )3,2,1(2 ==n n δThis chapter discusses the accuracy of the vibration stability determination with Floquent theory. The next chapter will discuss the numerical solution and stable domain and two parameters ’ influences on the stability for this question.4. Parameters influenceIn this part, we only consider two parameters, namely the frequency and amplitude of the force.4.1the influence of the force’s frequency4.1.1the stability of the systemWhen we discuss the stability of the system related to the frequency of the force, we should select some different frequencies, so we choose ω=1,2,4,6,8 and10. Using MATLAB codes, we can obtain the figs of the stability. We can know the stable region is bigger with the increase of the frequency in Fig.4.1.ω=1 ω=2ω=4 ω=6ω=8 ω=10Fig.4.1 stable domain with different ω4.1.2the response of the systemWhen we discuss the response of the system, the system should be stable. So we choose 7δ=，1ε=-，ω=2,4 and 6. In Fig.4.2, the cycle of the response increase and the range of the reactive amplitude is smaller with the increase of the frequency.Fig.4.2 responses of the system with different ω4.2the influence of the force’s amplitudeThe εis related to the force’s amplitude P. The cycle of the response a little increase and the range of the reactive amplitude is bigger with the increase of the force’s amplitude, in Fig.4.3.and Fig4.4.Fig.4.3 vibration response curve with different δThe red curve is w=2, δ=12,ε=1; the blue curve is w=2, δ=14, ε=1.This figure state that the vibration cycle is smaller and the amplitude have a little change with the increase of the δ.Fig.4.3 vibration response curve with different εThe red curve is w=2, δ=12,ε=1; the blue curve is w=2, δ=12, ε=5. This figure state that the amplitude is smaller and the vibration cycle have a little change with the increase of the ε.5. Conclusion(1) With the increase of the frequency, the stable region and the cycle of the response are bigger, but the range of the reactive amplitude is smaller.(2)With the incr ease of the force’s amplitude, the cycle of the response a little increase and the range of the reactive amplitude is much bigger. (3)Vibration in the stable region, the vibration cycle is smaller with the increase of the δ; the amplitude is smaller with the increase of the ε.AcknowledgementsThe author is grateful for upperclassman Li Yong, he give me much assistance. And the author is also grateful for Doctor Li Wei. In his classes, I felt very happy and can understand his class effectively. At last, the author is also grateful for classmates in the same laboratory, they give me much guidance and encourage. I gain a lot of knowledge through this study and I will work harder in the future.References1、ROY R.CRSIG, Jr. STRUCTURAL DYNAMICS. New York: John Wiley & Sons.2、王海期. 非线性振动. 北京: 高等教育出版社, 19923、顾志平. 非线性振动. 北京: 中国电力出版社, 2012。