#### Transcript response of a damped system under the harmonic

RESPONSE OF A DAMPED SYSTEM UNDER THE HARMONIC MOTION OF THE BASE y(t) = Displacement of the base From the figure, the equation of motion is mx c( x y ) k ( x y ) 0 (3.64) If y (t ) Y sin t , mx cx kx ky cy kY sin t cY cos t A sin( t ) where c A Y k (c ) and tan k 2 2 1 (3.65) The steady-state response of the mass can be expressed as Y k 2 (c ) 2 x p (t ) sin( t 1 ) 2 2 2 1/ 2 (k m ) (c ) where (3.66) c 2 k m 1 tan 1 OR using trigonometric identities, x p (t ) X sin( t ) (3.67) where Displacement transmissibility X k (c ) Y (k m 2 ) (c ) 2 2 2 1/ 2 1 (2r ) 2 2 2 (1 r ) (2r ) 2 mc 3 2r 3 1 tan tan 2 2 2 2 k ( k m ) ( c ) 1 ( 4 1 ) r 1 1/ 2 (3.68) (3.69) If expressed in complex form, response of the system can be expressed as 1 i 2r it Ye x p (t ) Re 2 1 - r i 2r and (3.70) the displacement transmissibility as X 2 1/ 2 Td 1 (2r ) H (i ) Y (3.71) Displacement transmissibility : The ratio of the amplitude of the response x p (t ) to that of the motion y(t). The variations of displacement transmissibility is shown in the figure below. EXAMPLE VEHICLE MOVING ON A ROUGH ROAD The figure below shows a simple model of a motor vehicle that can vibrate in the vertical direction while traveling over a rough road. The vehicle has a mass of 1200kg. The suspension system has a spring constant of 400 kN/m and a damping ratio of ζ = 0.5. If the vehicle speed is 20 km/hr, determine the displacement amplitude of the vehicle. The road surface varies sinusoidally with an amplitude of Y = 0.05m and a wavelength of 6m. SOLUTION The frequency can be found by v 1000 1 2f 2 0.290889v rad/s 3600 6 For v = 20 km/hr, ω = 5.81778 rad/s. The natural frequency is given by, k 400 10 n m 1200 3 1/ 2 18.2574 rad/s Hence, the frequency ratio is 5.81778 r 0.318653 n 18.2574 7 SOLUTION CON’T The amplitude ratio can be found from Eq.(3.68): 8 1/ 2 X 1 (2r ) Y (1 r 2 ) 2 (2r ) 2 1.469237 2 1/ 2 1 (2 0.5 0.318653) 2 2 ( 1 0 . 318653 ) ( 2 0 . 5 0 . 318653 ) 2 Thus, the displacement amplitude of the vehicle is given by X 1.469237Y 1.469237(0.05) 0.073462 m This indicates that a 5cm bump in the road is transmitted as a 7.3cm bump to the chassis and the passengers of the car. VIBRATION UNDER GENERAL FORCING CONDITIONS INTRODUCTION A general forcing function may be periodic (nonharmonic) or nonperiodic. A nonperiodic forcing function may be acting for a short, long, or infinite duration. Shock is defined as the small forcing function or excitation as compared to the natural time period of the system. Some examples of general forcing functions include the motion imparted by a cam to the follower; the vibration felt by an instrument when its package is dropped from a height; etc. The transient response of a system can be found by using what is known as the convolution integral. RESPONSE UNDER A GENERAL PERIODIC FORCE The equation of motion can be expressed as a0 mx cx kx F (t ) a j cos jt b j sin jt 2 j 1 j 1 (4.4) The steady-state solution of the equation is derived as: (a j / k ) a0 x p (t ) cos( jt j ) 2 2 2 2 2k j 1 (1 j r ) (2jr) j 1 (b j / k ) (1 j r ) (2jr ) 2 2 2 2 sin( jt j ) (4.13) RESPONSE UNDER A NONPERIODIC FORCE When the exciting force F(t) is nonperiodic, such as that due to the blast from an explosion, a different method of calculating the response is required. Various methods can be used to find the response of the system to an arbitrary excitation. Some of these methods are as follows: 1. 2. 3. 4. 5. Representing the excitation by a Fourier integral Using the method of convolution integral Using the method of Laplace transforms First approximating F(t) by a suitable interpolation model and then using a numerical procedure Numerically integrating the equation of motion. CONVOLUTION INTEGRAL Impulse Ft mx2 mx1 (4.17) By designating the magnitude of the impulse by F, we can write, in general, F t t t ~ Fdt (4.18) A unit impulse ( f ) is defined as ~ f lim ~ t t t 0 t Fdt Fdt 1 (4.19) CONVOLUTION INTEGRAL • Response to an impulse For an undamped system, the solution of the (4.20) equation of motion mx cx kx 0 is given by Eq.(2.27) as x(t ) e where nt x0 n x0 sin d t x0 cos d t d (4.21) c 2mn d n n k m k c 2 1 m m (4.22) 2 (4.23) (4.24) 14 CONVOLUTION INTEGRAL Figure 4.3: A single degree of freedom system subjected to an impulse 15 CONVOLUTION INTEGRAL If the mass is at rest before the unit impulse is applied, we obtain, from the impulsemomentum relation, Impulse f 1 mx (t 0) mx (t 0 ) mx0 (4.25) ~ Thus the initial conditions are given by x(t 0) x0 0 1 x (t 0) x0 m (4.26) Hence, Eq.(4.21) reduces to e nt x(t ) g (t ) sin d t md (4.27) 16 CONVOLUTION INTEGRAL which is also known as the impulse response function. The Eq.(4.27) is shown in Fig.4.3(c). If the magnitude of the impulse is F instead of ~ x unity, the initial velocity 0 is F/ m and the ~ response of the system becomes x(t ) Fe ~ n t md sin d t F g (t ) ~ (4.28) If the impulse is applied at an arbitrary time t = τ, it will change the velocity at t = τ, shown in Fig.4.4(a). Thus, x(t ) F g (t ) ~ (4.29) 17 CONVOLUTION INTEGRAL Figure 4.4: Impulse Response 18 EXAMPLE 4.4 RESPONSE OF A STRUCTURE UNDER IMPACT In the vibration testing of a structure, an impact hammer with a load cell to measure the impact force is used to cause excitation, as shown in Fig.4.5(a). Assuming m = 5kg, k = 2000 N/m, c = 10 N-s/m and F~ = 20 N-s, find the response of the system. 19 SOLUTION From the known data, we can compute, k 2000 c c 10 20 rad/s, 0.05, m 5 cc 2 km 2 2000(5) 20 n d 1 2 n 19.975 rad/s Assuming that the impact is given at t = 0, we find the response of the system as e nt x1 (t ) F sin d t ~ m d 20 e 0.05( 20)t sin 19.975t (5)(19.975) 0.20025e t sin 19.975t m (E.1) CONVOLUTION INTEGRAL • Response to General Forcing Condition Consider the response of the system under an arbitrary external force, shown in Fig.4.6. Hence, the response is given by x(t ) F ( )g (t ) Fig.4.6: An arbitrary (nonperiodic) forcing function (4.30) 21 CONVOLUTION INTEGRAL The total response at time t can be found by summing all the responses due to the elementary impulses acting at all times τ: x(t ) F ( ) g (t ) (4.31) Letting 0 and replacing the summation by integration, we obtain 1 t n ( t ) x(t ) F ( ) e sin d (t )d (4.33) md 0 which is called the convolution or Duhamel integral. 22 TWO DEGREE OF FREEDOM SYSTEMS INTRODUCTION As is evident from the systems shown in Figs.5.1 and 5.2, the configuration of a system can be specified by a set of independent coordinates termed as generalized coordinates, such as length, angle, or some other physical parameters. Principle coordinates is defined as any set of coordinates that leads a coupled equation of motion to an uncoupled system of equations. 24 EQUATIONS OF MOTION FOR FORCED VIBRATION Consider a viscously damped two degree of freedom spring-mass system, shown in Fig.5.3. Figure 5.3: A two degree of freedom spring-mass-damper system EQUATIONS OF MOTION FOR FORCED VIBRATION m1x1 (c1 c2 ) x1 c2 x2 (k1 k2 ) x1 k 2 x2 F1 m2 x2 c2 x1 (c2 c3 ) x2 k2 x1 (k2 k3 ) x2 F2 26 The application of Newton’s second law of motion to each of the masses gives the equations of motion: (5.1) (5.2) Both equations can be written in matrix form as [m]x (t ) [c]x (t ) [k ]x (t ) F (t ) (5.3) where [m], [c], and [k] are called the mass, damping, and stiffness matrices, respectively, and are given by EQUATIONS OF MOTION FOR FORCED VIBRATION c1 c2 c2 [c ] c c c 2 2 3 k1 k 2 k 2 [k ] k k k 2 2 3 And the displacement and force vectors are given respectively: x1 (t ) F1 (t ) x (t ) F (t ) x2 (t ) F2 (t ) It can be seen that the matrices [m], [c], and [k] are symmetric: 27 m1 0 [ m] 0 m 2 EQUATIONS OF MOTION FOR FORCED VIBRATION [m]T [m], [c]T [c], [k ]T [k ] where the superscript T denotes the transpose of the matrix. The solution of Eqs.(5.1) and (5.2) involves four constants of integration (two for each equation). We shall first consider the free vibration solution of Eqs.(5.1) and (5.2). 28