Equations of Motion: Kinetic energy: Potential energy: Sin≈ SYSTEM MODELING AND ANALYSIS Example: Obtain the equations of motion of the simple pendulum shown in the figure. Find the undamped natural frequency of the pendulum. Kinetic energy: Potential energy: There is no external excitation and damping. Sin≈

Kinetic energy: Potential energy: O SYSTEM MODELING AND ANALYSIS Example: Obtain the equations of motion of the compound pendulum shown in the figure. Find the undamped natural frequency. Kinetic energy: G O θ m, L, IO L L1 g Potential energy: Where IO is the mass moment of inertia about point O. For small angular displacements sin θθ

Material: Plain carbon steel L1=0.170299 m SYSTEM MODELING AND ANALYSIS L1=0.170299 m Material: Plain carbon steel

SYSTEM MODELING AND ANALYSIS

Yücel Ercan, İleri Dinamik 2014. SYSTEM MODELING AND ANALYSIS Example: Obtain the equations of motion of the mechanical system shown in the figure. Yücel Ercan, İleri Dinamik 2014. θ x M1 Solution: System has two degree of freedom, namely θ and x. Pendulum rod Bob Tranlational mass Kinetic energy: Potential energy: Virtual work:

Lagrange’s equation for θ: SYSTEM MODELING AND ANALYSIS Potential energy: Kinetic energy: Virtual work: Lagrange’s equation for θ:

Lagrange’s equation for x: (For constant M) In matrix form;

Lagrange’s equation for θ: SYSTEM MODELING AND ANALYSIS Example: Obtain the equations of motion of the mechanical system shown in the figure. x Solution: (Kinetic energy of the beam is zero due to its negligible mass) Lagrange’s equation for θ:

Lagrange’s equation for x: SYSTEM MODELING AND ANALYSIS Lagrange’s equation for x: In matrix form;

Potential energy of pendulum is neglected SYSTEM MODELING AND ANALYSIS Example: Obtain the equations of motion of the mechanical system shown in the figure. G Inputs: f, T, x4 Potential energy of pendulum is neglected

SYSTEM MODELING AND ANALYSIS

SYSTEM MODELING AND ANALYSIS

SYSTEM MODELING AND ANALYSIS In matrix form:

EIGENVALUE EQUATION (Free vibration) Eigenvalue equation Eigenvalue equation:

a=[0.0039,3.75,847,141834,2799360];p=roots(a);vpa(p,4) Eigenvalue equation: m=0.85 kg, L =0.24 m, k=1200 N/m, c=38 Ns/m Matlab code: a=[0.0039,3.75,847,141834,2799360];p=roots(a);vpa(p,4) clc;clear m0=0.85;l0=0.24;k0=1200;c0=38; m=[m0*l0^2/8,0;0,3*m0/4]; c=[27*c0*l0^2/16,-9*c0*l0/4;-9*c0*l0/4,6*c0]; k=[9*k0*l0^2,-3*k0*l0/2;-3*k0*l0/2,4*k0]; syms s;p=solve(det(m*s^2+c*s+k));vpa(p,4) Eigenvalues: -104.3+181.4i, -104.3-181.4i, -22.4, -730.2

Eigenvalues: -104.3+181.4i, -104.3-181.4i, -22.4, -730.2 Form of the free vibration response: A1, φ1, A2 and A3 can be calculated from initial conditions The output reaches zero as time t goes infinity. The value of the steady-state reponse will be zero. If the real parts of all eigenvalues are negative than the system is stable. rad/s ω0 iω -σ φ p=-σ+iω f0=1/T0 For the root -22.4 Δt=0.0142 , t∞=0.28 p=-σ For the root -730.2 için Δt=0.000436 , t∞=0.0086 For the system Δt=0.000436 , t∞=0.28

x(t) t 5 3 1 0.2 0.5 ξ=0.1 t x(t)