D R " % 5 A 6 Y U K L command Final project 1: Single-Degree-of-Freedom Vibratory Systems Instructor: Xiaolei Yang MEC 102: Engineering Computing and Problem Solving April 4, 2019 1 Introduction Consider a spring-mass-damper system with a mass m, a spring with linear spring constant k, and a damper with damping coefficient c. In addition, we include a general nonlinear element g(x, z) to represent a general nonlinearity that is a function of displacement and velocity . The mass is subjected to an excitation F(t) = Fof(t). This system, which is illustrated in Figure 1, is a prototypical model used to study mechanical systems ranging from washing machines to vehicles. The governing equation of motion is of the form d'y +24 +y+9(y.y) f(T) مولی dy dr (1) where y=x/st, st = Fo/k, wat is the nondimensional time, and the damping factor ( is given by ! 2mwa In Eq. (3), the natural frequency w,, is given by √k/m (2) (3) When the system is subjected to an initial displacement X, and an initial velocity Vo, then in terms of the nondimensional coordinates, we have y(0) = Xo/ost = yo dy(0) == Vo/(0) = =20 dT (4) (5) x(1) F(1) = Fof(t) ни m g(x,x) 2 Figure 1: Spring-mass-damper system with a general nonlinear element g(x, z). Example: System with nonlinear spring and forced oscilla- tions For this system, there is a nonlinear cubic spring with spring constant, thus, where 9(3.9)=aoy", (6) Q= (7) k The initial conditions are zero. The harmonic forcing at frequency w is given by F(T) Fo cos(wt) = Fa cos(r) (8) where = w/w is the nondimensional excitation frequency ratio. Substitute Eqs. (6) and (8) into Eq. (1), the governing equation is obtained as follows: d²y dy dr² +26+y+aoy³ = cos(fr) Introducing the following variables Y₁ = y dy 32. (9) (10) -velocity (11) Eq. (9) becomes dy (12) = y2 dT dy2 -2y-1-a0 + cos(T), (13) dr and 91 (0) = 0 92 (0) = 0, (14) (15) 9 10 11 12 13 ONDERST 14 15 16 17 18 19 At a given excitation frequency, we shall determine the steady-state response of the system given by Eq. (9)-that is, after the transients have died out-and examine the spectral content of this steady-state response employing fft for the following cases: (1) a = 0 (linear system) and (2) 6,250 (nonlinear system). Equation (9) is numerically integrated using ode45. In addition, we assume that (=0.2 and ?= 3.0. The excitation frequency 2 has been chosen to be three times the natural frequency of the system. We shall examine the time histories for 0≤30 and take 12,000 samples in this region. This is equivalent to the data being acquired every T, = 30/12,000 = 0.0025. Therefore, the (dimensionless) sampling frequency is 2/7T,800. This is far in excess of what is necessary to sample the response based on the excitation frequency = 3.0. The consequence of this over- sampling is that we have to truncate the spectrum plot; thus, we shall display only the first forty 3,200 and the number of samples used by fit is N = 2138, 192. values. Also, we let Natart The script is as follows: 1 function Example01 0.2; alphao = [0, 62501; 3; M =12000; linspace (0, 30, M); N = 2 13; Nstart = 3200; Fs = M/30; 2 zeta = 3 Omega E 4 tspan = 5 6 7 8 f = (Fs (0:N-1)/N) *2.0*pi; for m = 1:2 [t, y] = ode 45 (@ForcedNLOscillator, tspan, [0 0]'); figure (m); plot (t, y(:,1), 'k-'); axis ([0, 25, -0.2, 0.46]); xlabel(’\tau); ylabel(’y(\tau)); axes ('position', [0.55, 0.63, 0.25, 0.25]) Amp = abs (fft (y (Nstart: Nstart+N, 1), N))/N; plot (f (1:40), 2*Amp (1:40), 'k-'); v = axis; v (2) = f (40); v(4) = max (2*Amp (1:40)); axis (v) xlabel('\Omega'); ylabel('Amplitude'); 21 end function xdot = ForcedNLOscillator (t, x) 22 23 22 24 end end xdot = [x (2); -2*zeta*x (2) -x (1) - alphao (m) *x (1) ^3+cos (Omega*t)]; Execution of the program results in Figure 2, where it is seen that the responses of both the linear and nonlinear systems reach steady state when r≥ 8. This time corresponds to an index Natart =3,200. Although both of the steady-state responses have a period equal to the period of the harmonic forcing function, they have different characteristics that can be more clearly distinguished in the frequency domain. The corresponding spectral response appears in the upper right-hand cor- ner of each figure where it is seen that the amplitude spectrum of the displacement response in the nonlinear case shows spectral peaks at the forcing frequency 2 and at integer multiples of it. The additional peaks are due to the cubic nonlinearity of the spring. In the linear case, there is only one spectral peak, which corresponds to the excitation frequency. This example illustrates that the response of a nonlinear system can have spectral components different from the excitation (input) frequency. 3 Problem for final project 1 An overhead crane's trolley is carrying, via a cable, a load of mass m as shown in Figure 3. When the trolley is moved with an acceleration b(t), the governing equation of motion of the crane load is d20 L- dt2 +g sin(0)-b(t) cos(0) (16) where 9 = 9.8m/s2 is the gravity constant. If the cable length is L = 2 m, then graph the swing motion (t) for the following accelerations of the trolley over the time interval 0 <t<10 s: 1. b(t) = 10u(t) m/s², 0(0) = 0.4 rad, and de(0)/dt = 0.1 rad/s, where u(t) is the unit step function. 2. b(t)=0.2u(t) m/s², 0(0) = 0.4 rad, and de(0)/dt = 0.1 rad/s, where u(t) is the unit step function. 1 You need to submit: 1) the MATLAB code, and 2) the images. 0.4- 0.1 0.3 0.2- 0.1 -0.1- 0.05 5 10 9 -0.2 10 15 20 25 0.4- (a) 0.04 0.3- 0.02 0.2 0.1- -0.1- 5 10 .. -0.25 10 15 20 25 T (b) Figure 2: (a) Linear system subjected to a harmonic forcing at 2 = 3.(b) Nonlinear system sub- jected to a harmonic forcing at = 3 and 0 = 6,250. Trolley • b(t) m L k IC = XCQ)=xo J 9C44)-8048 -FCT) =Focos (7) Crane boom Figure 3: Trolley on an overhead crane carrying a swinging load m.