Unit - 6 Vibrations of Two Degree of Freedom Systems

Unit - 6 Vibrations of Two Degree of Freedom Systems Dr. T. Jagadish. Professor for Post Graduation, Department of Mechanical Engineering, Bangalore Institute of Technology, Bangalore Introduction A two degree of freedom system is one that requires two coordinates to completely describe its equation of motion. These coordinates are called generalized coordinates when they are independent of each other. Thus system with two degrees of freedom will have two equation of motion and hence has two frequencies. A two degree freedom system differs from a single degree of freedom system in that it has two natural frequencies and for each of these natural frequencies there correspond a natural state of vibration with a displacement configuration known as NORMAL MODE. Mathematical terms related to these quantities are known as Eigen values and Eigen vectors. These are established from the two simultaneous equation of motion of the system and posses certain dynamic properties associated. A system having two degrees of freedom are important in as far as they introduce to the coupling phenomenon where the motion of any of the two independent coordinates depends also on the motion of the other coordinate through the coupling spring and damper. The free vibration of two degrees of freedom system at any point is a combination of two harmonics of these two natural frequencies. Under certain condition, during free vibrations any point in a system may execute harmonic vibration at any of the two natural frequencies and the amplitude are related in a specific manner and the configuration is known as NORMALMODE or PRINCIPAL MODE of vibration. Thus system with two degrees of freedom has two normal modes of vibration corresponding two natural frequencies. Free vibrations of two degrees of freedom system: Consider an un-damped system with two degrees of freedom as shown in Figure 6.1a, where the masses are constrained to move in the direction of the spring axis and executing free vibrations. The displacements are measured from the un-stretched positions of the springs. Let x 1 and x be the displacement of the masses m 1 and m respectively at any given instant of time measured from the 1

equilibrium position with x > x 1. Then the spring forces acting on the masses are as shown in free body diagram in Figure 6.1b k 1 x 1 k 1 m 1 k m m k 3 x 1 x m 1 x 1 k (x - x 1 ) m x k 3 x (a) Figure 6.1 Based on Newton s second law of motion ƒ = m x (b) For mass m 1 m 1 x 1 = - k 1 x 1 + k (x x 1 ) m 1 x 1 + k 1 x 1 - k x + k x 1 = 0 m 1 x 1 + (k 1 + k ) x 1 = k x ------------ (1) for mass () m x = - k 3 x - k (x x 1 ) m x + k 3 x + k x - k x 1 = 0 m x + (k + k 3 ) x = k x 1 ------------ () The solution for x 1 and x are obtained by considering that they can have harmonic vibration under steady state condition. Then considering the case when the mass m 1 execute harmonic vibration at frequency ω 1 and the mass m execute harmonic vibration at frequency ω then we have x 1 = X 1 sin ω 1 t, and x = X sin ω t ----------- (3) Where X 1 and X are the amplitudes of vibrations of the two masses under steady state conditions. Substituting equation (3) into equation (1) we have - m 1 ω 1 X 1 sin ω 1 t + (k 1 + k ) X 1 sin ω 1 t = k X sin ω t Therefore X 1 = k sinω t X (k 1 + k ) mω 1 sinω 1 t Since X 1 and X are the amplitude of two harmonic motions, their ratio must be constant and independent of time. Therefore sinω t / sinω 1 t = C a constant.

Consider if C > 1. Then at time t = π/ω 1, sinω 1 t will be sinω 1 x π/ω 1 = sin π/ = 1 Therefore sinω t / sinω 1 t > 1 or sinω t > 1 which is impossible. Hence C > 1 is not possible. Similarly it can be shown that C < 1 is also not possible. Thus the only possibility is that C = 1 Hence sinω t / sinω 1 t = 1 which is only possible if ω = ω 1 = ω. Hence the two harmonic motion have to be of the same frequency. Thus the solution of equation (1) and () can be x 1 = X 1 sin ωt, and x = X sin ωt ----------- (4) x 1 = - ω X 1 sin ωt x = - ω X sin ωt ------------------ (5) Substitute equation (4) and (5) into the equation (1) and () - m 1 ω X 1 sin ωt + (k 1 + k ) X 1 sin ωt = k X sin ωt ----------------- (6) - m ω X sin ωt + (k + k 3 ) X sin ωt = k X 1 sin ωt. ----------------- (7) Canceling the common term sin ωt on both the sides and re arranging the terms we have from equation (6) X 1 /X = k / (k 1 + k m 1 ω ) ---------------- (8) X 1 /X = [(k + k 3 ) m ω ] / k ----------- (9) Thus equating equation (8) and (9) we have X 1 /X = k / (k 1 + k m 1 ω ) = [(k + k 3 ) m ω ] / k -------------- (10) Cross multiplying in equation (10) we have (k 1 + k m 1 ω ) (k + k 3 m ω ) = k on simplification we get m 1 m ω 4 [m 1 (k + k 3 ) + m (k 1 + k )] ω + [k 1 k + k k 3 + k 3 k 1 ] = 0 ---------- (11) The above equation (11) is quadratic in ω and gives two values of ω and therefore the two positive values of ω correspond to the two natural frequencies ω n1 and ω n of the system. The above equation is called frequency equation since the roots of the above equation give the natural frequencies of the system. Now considering m 1 = m = m and k 1 = k 3 = k Then the frequency equation (11) becomes m ω 4 m (k + k ) ω + (k + kk ) = 0 ------------------ (1) Let: ω = λ λ = ω 4, m λ m (k + k ) λ + (k + kk ) = 0 ------------------ (13) The roots of the above equation (13) are as follows: Let a = m, b = - m (k + k ); c = (k + kk ) λ 1, = [- b ± (b 4ac)] / a λ 1, = [- (-m) (k + k ) ± [-m (k+k )] 4 (m ) (k + kk )]/m = [+ m (k +k )] / m ± [ 4m [(k + k + kk ) (k + kk )]/4m 4 = (k+ k ) /m ± (k /m ) = (k +k ) /m ± k /m 3

Thus λ 1 = (k + k ) /m k / m = k/m. Then ω n1 = K/m ω n1 = (k/m) ----------------- (14) and λ = (k + k ) /m Thus ω n = (k + k ) /m. Then ω n = [(k + k ) /m] --- (15) ω n1 is called the first or fundamental frequency or 1 st mode frequency, ω n is called the second or nd mode frequency. Thus the number of natural frequencies of a system is equal to the number of degrees of freedom of system. Modes Shapes: From equation (10) we have X 1 /X = k /(k+k ) -mω = (k + k) - mω /k ---(16) Substitute ω n1 = (k/m) in any one of the above equation (16). (X 1 /X ) ωn1 = k / (k+ k m(k/m)) or ((k + k) m(k/m))/k = k /k = 1 (X 1 /X ) ωn1 = 1 ----------------- (17) Similarly substituting ω n = [(k + k ) /m] in any one of the above equation (16). (X 1 /X ) ωn = k / (k + k m(k+ k )/m) or ((k + k) m(k+ k )/m))/k = - k /k = -1 (X 1 /X ) ωn = -1 ----------------- (18) The displacements X 1 and X corresponding to the two natural frequency of the system can be plotted as shown in Figure 6., which describe the mode in which the masses vibrate. When the system vibrates in principal mode the masses oscillate in such a manner that they reach maximum displacements simultaneously and pass through their equilibrium points simultaneously or all moving parts of the system oscillate in phase with one frequency. Since the ratio X 1 /X is important rather than the amplitudes themselves, it is customary to assign a unit value of amplitude to either X 1 or X. When this is done, the principal mode is referred as normal mode of the system. k 1 m 1 k m x 1 x x 1 = 1 x = 1 Node x 1 = 1 x = -1 k 3 (a) (b) (c) Figure 6. ω n1 = (k/m) ω n = [(k + k ) /m] 1 st Mode nd Mode 4

It can be observed from the figure 6.b when the system vibrates at the first frequency, the amplitude of two masses remain same. The motion of both the masses are in phase i.e., both the masses move up or down together, the length of the middle spring remains constant, this spring (coupling spring) is neither stretched nor compressed. It moves rigid bodily with both the masses and hence totally ineffective as shown in Figure 6.3a. Even if the coupling spring is removed the two masses will vibrate as two single degree of freedom systems with ω n = (K/m). When the system vibrates at the second frequency the displacement of the two masses have the same magnitude but with opposite signs. Thus the motions of m 1 and m are 180 0 out of phase, the midpoint of the middle spring remains stationary for all the time. Such a point which experiences no vibratory motion is called a node, as shown in Figure 6.3b which is as if the middle of the coupling spring is fixed When the two masses are given equal initial displacements in the same direction and released, they will vibrate at first frequency. When they are given equal initial displacements in opposite direction and released they will vibrate at the second frequency as shown in Figures 6.3a and 6.3b k 1 k 1 m 1 x 1 m 1 x 1 = - T 1 θ m x m x k 3 k 3 (a) (b) Figure 6.3 ω n1 = (k/m) ω n = [(k + k ) /m] 1 st Mode nd Mode If unequal displacements are given to the masses in any direction, the motion will be superposition of two harmonic motions corresponding to the two natural frequencies. 5

Problems 1. Obtain the frequency equation for the system shown in Figure 6.4. Also determine the natural frequencies and mode shapes when k 1 = k, k = k, m 1 = m and m = m. k 1 x 1 k 1 m 1 x 1 m 1 m k x 1 x k (x - x 1 ) m x Solution (a) Figure 6.4. Consider two degrees of freedom system shown in Figure 6.4a, where the masses are constrained to move in the direction of the spring axis and executing free vibrations. The displacements are measured from the un-stretched positions of the springs. Let x 1 and x be the displacement of the masses m 1 and m respectively at any given instant of time measured from the equilibrium position with x > x 1. Then the spring forces acting on the masses are as shown in free body diagram in Figure 6.4b Based on Newton s second law of motion ƒ = For mass m 1 m 1 x 1 = - k 1 x 1 + k (x x 1 ) m 1 x 1 + k 1 x 1 - k x + k x 1 = 0 m 1 x 1 + (k 1 + k ) x 1 = k x ------------ (1) for mass () m x = - k (x x 1 ) m x + k x - k x 1 = 0 m x + k x = k x 1 ------------ () m x (b) The solution for x 1 and x are obtained by considering that they can have harmonic vibration under steady state condition. Then considering the case when the masses execute harmonic vibration at frequency ω. Thus if x 1 = X 1 sin ωt, and x = X sin ωt ----------- (3) 6

Then we have x 1 = - ω X 1 sin ωt, x = - ω X sin ωt ------------------ (4) Substitute equation (3) and (4) into the equation (1) and () we get - m 1 ω X 1 sin ωt + (k 1 + k ) X 1 sin ωt = k X sin ωt ----------------- (5) - m ω X sin ωt + k X sin ωt = k X 1 sin ωt. ----------------- (6) From equation (5) we have X 1 /X = k /[(k 1 + k ) m 1 ω ] -------------- (7) From equation (6) we have X 1 /X = [ k m ω ] / k -------------- (8) Equating (7) and (8) k / (k 1 + k m 1 ω ) = [k m ω ] /k k = (k 1 + k m 1 ω ) (k m ω ) k = (k 1 + k ) k m 1 ω k m ω (k 1 + k ) + m 1 m ω 4 m 1 m ω 4 - ω [m 1 k + m (k 1 + k )] + k 1 k = 0 --------------- (9) letting ω = λ m 1 m λ λ [m 1 k + m (k 1 + k )] + k 1 k = 0 ---------------------- (10) Equation (10) is the frequency equation of the system which is quadratic in λ and hence the solution is λ = [[m 1 k + m (k 1 + k )] ± [[{m 1 k + m (k 1 +k )} ]- 4 m 1 m k 1 k ]] / m 1 m To determine the natural frequencies Given k 1 = k, k = k and m 1 = m, m = m λ = [mk + m (k +k) ± [mk + 6mk) 4m mk k]] / m. m = [7mk ± [(7mk) 4 (4m k )]] / 4m = [7mk ± (49m k 16m k ] / 4m λ = [7mk ± 5.744 mk] /4m Thus λ 1 = [7mk - 5.744 mk]/4m and λ = [7mk + 5.744 mk] /4m λ 1 = ω n1 = [7 mk 5.744 mk] /4m = 1.55 mk /4m = 0.3138 k/m Thus ω n1 = 0.56 (k/m) λ = ω n = [7mk + 5.744 mk] /4m = 3.186 k/m. Thus ω n = 1.784 (k/m) Substituting the values of frequencies into the amplitude ratio equation as given by equation (7) and (8) one can determine the mode shapes: FOR THE FIRST MODE: Substituting ω n1 = 0.3138 K/m into either of the equation (7) or (8) we get first mode shape: I.e. X 1 /X = k /[(k 1 + k ) m 1 ω ] -------------- (7) X 1 /X = [ k m ω ] / k -------------- (8) X 1 /X = k / [(k + k m ω ] = k/ [3k m. ω n1 ] X 1 /X = k/ [3k m. 0.3138k/m] = 1/(3 0.3138) =1/.686 = 0.374 Thus we have X 1 /X = 0.374. Then If X 1 = 1, X =.685 7

FOR THE SECOND MODE: Substituting ω n = 3.186 K/m into either of the equation (7) or (8) we get first mode shape: I.e. X 1 /X = k /[(k 1 + k ) m 1 ω ] -------------- (7) X 1 /X = [ k m ω ] / k -------------- (8) X 1 /X = k / [(k + k m ω ] = k/ [3k m. ω n1 ] X 1 /X = k/ [3k m. 3.186k/m] = 1/(3 3.186) = 1/ (-0.186) = - 5.37 or X 1 /X = (k m.3.186 k/m)/k = [1-(3.186)] = - 5.37 Thus we have X 1 /X = -5.37. Then If X 1 = 1, X = - 0.186 m 1 k 1 k x 1 x 1 = 1 Node x 1 = 1 m x x =.685 x = -0.186 MODE SHAPE FOR FIRST MODE SECOND MODE ω n1 = 0.3138 K/m ω n = 3.186 K/m Figure 6.5. Derive the frequency equation for a double pendulum shown in figure.6.6.. Determine the natural frequency and mode shapes of the double pendulum when m 1 = m = m l 1 = l = l Figure 6.6 Consider two masses m 1 and m suspended by string of length l 1 and l as shown in the figure 6.6.. Assume the system vibrates in vertical plane with small amplitude under which it only has the oscillation. Let θ 1 and θ be the angle at any given instant of time with the vertical and x 1 and x be the horizontal displacement of the masses m 1 and m from the initial vertical position respectively. 8

For small angular displacement we have sinθ 1 = x 1 / l 1 and sinθ = (x x 1 ) / l --- (1) Figure.6.7 Free body diagram Figure 6.7hows the free body diagram for the two masses. For equilibrium under static condition the summation of the vertical forces should be equal to zero. Thus we have At mass m 1 T 1 cosθ 1 = mg + T cosθ ----- () At mass m T cosθ = mg ------- (3) For smaller values of θ we have cosθ = 1. Then the above equations can be written as T = m g ----- (4) and T 1 = m 1 g + m g T 1 = (m 1 + m )g ----- (5) When the system is in motion, the differential equation of motion in the horizontal direction can be derived by applying Newton Second Law of motion. Then we have for mass m 1 m 1 x 1 +T 1 sin θ 1 = T sin θ m 1 x 1 = -T 1 sin θ 1 + T sin θ ----- (6) For mass m m x - T sin θ = m x + T sin θ = 0 ----- (7) Substituting the expression for T and T 1 from equation (4) and (5) and for sinq 1 and sinq from equation (1) into the above equation (6) and (7) we have Equation (6) becomes m 1 x 1 + [(m 1 + m ) g](x 1 /l 1 ) = m g[(x x 1 )/l m 1 x 1 + [{(m 1 + m )/l 1 } + m /l ]gx 1 = (m g/l ) x --------- (8) 9

Equation (7) becomes m x + m g(x x 1 )/l m x + (m g/l ) x 1 = (m g/l ) x ------ (9) Equations (8) and (9) represent the governing differential equation of motion. Thus assuming the solution for the principal mode as x 1 = - ω Asinωt Substitute in (10) into equation (8) and (9) and cancelling the common term sinωt we have [-m 1 ω +{(m 1 +m )/l 1 +m /l }g]a=(m g /l )B --(11) [-m ω +(m g/l )]B = (m g/l )A ---------- (1) From equation (11) we have A/B=(m g/l )/[{(m 1 +m )/l 1 +m /l }g]-m 1 ω -- (13) From equation (1) we have and A/B = [(m g / l ) -m ω ] / (m g / l ) ------ (14) Equating equation (13) and (14) we have x = - ω Bsinωt -------- (10) A/B = (m g/l )/[{(m 1 +m )/l 1 + m /l }g-m 1 ω ] = [(m g/l ) -m ω ]/(m g/l ) [{(m 1 + m )/l 1 +m /l }g-m 1 ω ][(m g/l ) -m ω ] = (m g / l ) ------- (15) Equation (15) is a the quadratic equation in w which is known as the frequency equation. Solving for ω we get the natural frequency of the system. Particular Case: When m 1 = m = m and l 1 = l = l Then equation (13) will be written as A/B = (mg/l)/[(3mg/l)-mω ] = (g/l)/[(3g/l) - ω ] A/B= 1/[3 (ω l/g)] ------ (16) and equation (14) will be written as A/B = [1 (ω l/g)] ------ (17) Equating equation (16) and (17) we get A/B = 1/[3- (ω l/g)] = [1- (ω l/g)] [3- (ω l/g)] * [1- (ω l/g)] = 1 or (3g- ω l)*(g- ω l) = g 3g 3glω glω +l ω 4 = g or l ω 4 4glω + g = 0 or ω 4 (4g/l) ω + (g /l ) = 0 ------- (18) letting λ = ω in equation (18) we get λ (4g/l)λ + (g /l ) = 0 ----- (19) Which is a quadratic equation in l and the solution for the equation (19) is 10

λ 1, = (g/l) + [(4g /l ) (g /l ) or λ 1, = (g/l)( ± ) -------------------- (0) λ 1 = (g/l)( - ) = 0.5858(g/l) ----- (1) and λ = (g/l)( + ) = 3.414(g/l) ------ () Since λ=ω then the natural frequency ω n1 = l 1 = 0.7654 (g/l) thus ω n1 = 0.7654 (g/l) --- (3) and ω n = l = 1.8478 (g/l) thus ω n = 1.8478 (g/l) ---------- (4) Substituting ω n1 and ω n from equation (3) and (4) into either of the equation (16) or (17) we get the mode shape FOR THE FIRST MODE: Mode shapes for the first natural frequency ω n1 = 0.7654 (g/l) or ω n1 = (g/l)( - ) I mode from equation (16) A/B= 1/[3 (ω l/g)] (A/B) 1 =1/[3-ω n1 l/g] =1/[(3-{(g/l)(- )*l/g}] = 1/(3-+ ) =1/(1+ ) = 1/.414 = 0.414 Thus when A =1 B =.414 Also from equation (17) A/B = [1 (ω l/g)] For ω n1 = 0.7654 (g/l) or ω n1 n1 = (g/l)( - ) (A/B) 1 = (1- ω n1 l/g) = [1- {(g/l)(- )}l/g] or (A/B) 1 = (1 + ) = -1 = 0.414 Thus when A =1 B =.414 Modes shape is shown in figure-6.8 Figure- 6.8 FOR THE SECOND MODE: Mode shapes for second natural frequency ω n = 1.8478 (g/l) or ω n = (g/l)(+ ) II mode from equation (16) is given by A/B=1/[3 (ω l/g)] (A/B) =1/[3-ω n1 l/g] =1/[(3-{(g/l)(+ {(g/l)(+ )*l/g}] = 1/(3-- ) =1/(1- ) (A/B) = 1/(- 0.414) = -.414 or Thus when A =1 B = -0.414 Also from equation (17) A/B = [1 (ω l/g)] ω n = 1.8478 (g/l) or ω n = (g/l)(+ ) (A/B) = (1- ω n1 l/g) = [1- {(g/l)(+ )}l/g] (A/B) = (1 - ) = -(1+ ) = -.414 Thus when A =1 B = -0.414 Modes shape is shown in figure-6. 6.9 Figure- 6.9 11

Determine the natural frequencies of the coupled pendulum shown in the figure 6.10. Assume that the light spring of stiffness k is un-stretched and the pendulums are vertical in the equilibrium position. ka(θ 1 - θ ) a k ka(θ 1 - θ ) Solution: Figure 6.10. Considering counter clockwise angular displacement to be positive and taking the moments about the pivotal point of suspension by D.Alembert s principle we have ml θ 1 ml θ Equation (1) and () can also be written as ml θ ----------------- (1) ----------------- () ----------------- (3) ----------------- (4) Equation (3) and (4) are the second order differential equation and the solution for θ 1 and θ are obtained by considering that they can have harmonic vibration under steady state condition. Then considering the case when the masses execute harmonic vibration at frequency ω Thus if θ 1 = Asin ωt, and θ = Bsin ωt ----------- (5) Substitute equation (5) into the equation (3) and (4) and canceling the common terms we get (- ml ω + mgl + ka )A = ka B ----------------- (6) (- ml ω + mgl + ka )B = ka A ----------------- (7) From equation (6) we have A/B = ka / [mgl + ka ml ω ] -------------- (8) From equation (7) we have A/B = [mgl + ka ml ω ] / ka -------------- (9) Equating (8) and (9) l m = - mglθ 1 ka(θ 1 θ ) = - mglθ + ka(θ 1 θ ) ml θ 1 + (mgl + ka)θ 1 = kaθ + (mgl + ka)θ = kaθ 1 A/B = ka / [mgl + ka ml ω ] = [mgl + ka ml ω ] / ka [mgl + ka ml ω ] = [ka ] ------------------- (10) or θ m mg mgl + ka ml ω = + ka ω = ( mgl + ka + ka ) / ml -------- (11) m θ 1 m mg 1

ω 1, = [( mgl + ka + ka ) / ml ] -------- (1) ω 1 = [( mgl + ka - ka ) / ml ] = (g/l) -------- (13) ω = [( mgl + ka + ka ) / ml ] = [(g/l) + (ka /ml )] -------- (14) Substituting the values of frequencies into the amplitude ratio equation as given by equation (8) and (9) one can determine the mode shapes: FOR THE FIRST MODE: Substituting ω n1 = g/l into either of the equation (8) or (9) we get first mode shape: A/B = ka / [mgl + ka ml ω ] = ka / [mgl + ka ml g/l] = ka / [mgl + ka mlg] = ka / ka A/B = 1 FOR THE SECOND MODE: Substituting ω n = [(g/l) + (ka /ml )] into either of the equation (8) or (9) we get second mode shape: A/B = ka / [mgl + ka ml ω ] = ka / [mgl + ka ml [(g/l) + (ka /ml )]] = ka / [mgl + ka mlg - ka ] = (ka / -ka ) = -1 Thus A/B = -1 Mode shapes at these two natural frequencies are as shown in figure- 6.10 MODE SHAPES AT TWO DIFFERENT FREQUENCIES FIRST MODE SECOND MODE ω n1 = g/l A/B = 1 ω n = [(g/l) + (ka /ml )] A/B = -1 Figure-6.10 Mode Shapes at first frequency Figure-6.11 Mode Shapes at second frequency 13

Derive the equation of motion of the system shown in figure 6.1. Assume that the initial tension T in the string is too large and remains constants for small amplitudes. Determine the natural frequencies, the ratio of amplitudes and locate the nodes for each mode of vibrations when m 1 = m = m and l 1 = l = l 3 = l. m m l 1 l Figure 6.1. At any given instant of time let y 1 and y be the displacement of the two masses m 1 and m respectively. The configuration is as shown in the figure 6.13. θ 1 sin θ 3y 1 θ y θ 3 θ 3 Tcosθ Figure 6.13. Figure 6.14 The forces acting on the two masses are shown in the free body diagram in figure 6.14(a) and (b) From figure 6.13 we have sinθ1 = (y 1 /l 1 ) sinθ = [(y 1 y )/l ] and sinθ 3 = (y /l 3 ) For small angle we have sinθ 1 = θ 1 = (y 1 /l 1 ), sinθ = θ = [(y 1 y )/l ] and sinθ 3 = θ 3 = (y /l 3 ) and cosθ 1 = cosθ = cosθ 3 = 1.0 Thus the equation of motion for lateral movement of the masses T θ 1 y 1 m 1 (a) θ Tcosθ T T T(sinθ 1 + Tcosθ θ y m Tsinθ (b) θ 3 Tcosθ T For the mass m 1 m 1 y 1 = - (Tsinθ 1 + Tsinθ ) = - T (θ 1 + θ ) m 1 y 1 = - T [(y 1 /l 1 ) + (y 1 - y ) /l ] or m 1 y 1 + [(T/l 1 ) + (T/l )]y 1 = (T/l )y ---- (1) For the mass m m y = (Tsinθ Tsinθ 3 ) m y = T[(y 1 - y ) /l - (y /l 3 )] or y + [(T/l ) +(T/l 3 )]y = (T/l )y 1 ----- () 14

Assuming harmonic motion as y 1 =Asinωt and y = Bsinωt ------- (3) and substituting this into equation (1) and () we have [-m 1 ω + (T/l 1 ) + (T/l )] A = (T/l ) B ----- (4) [-m ω + (T/l ) + (T/l 3 )] B = (T/l ) A ----- (5) Thus from equation (4) we have A/B = (T/l ) / [(T/l 1 ) + (T/l ) - m 1 ω ] ----- (6) and from equation (5) we have A/B = [(T/l ) + (T/l 3 ) - m ω ] / (T/l ) ----- (7) Equating equation (6) and (7) we have A/B=(T/l )/[(T/l 1 )+(T/l )- m 1 ω ]/[(T/l )+(T/l 3 )-m ω ]/(T/l ) Thus we have [(T/l 1 )+(T/l ) m 1 ω ][(T/l )+(T/l 3 )-m ω ] = (T /l ) -------------- (8) Equation (8) is the equation on motion which is also known as frequency equation. Solving this equation gives the natural frequencies of the system. Particular Case: When m 1 = m = m and l 1 = l = l 3 = l then equation (6) can be written as A/B = (T/l)/[(T/l)+(T/l)-mω ] = (T/l)/[(T/l)-mω ] ---------- (9) and equation (7) can be written as A/B = [(T/l)+(T/l) -mω ]/(T/l) = [(T/l) - mω ]/(T/l) ---- (10) Equating equation (9) and (10) we have [(T/l - mω ] = (T/l) -------- (11) Thus T/l - mω = + (T/l) ------- (1) Therefore we have ω = [(T+T)]/ml ------ (13) ω n1 = [(T-T]/ml] = (T/ml) -------- (14) and ω n = [(T+T]/ml] = (3T/ml) --------- (15) Substituting equation (14) and (15) into either of the equation (9) or (10) we have the ratio of amplitudes for the two natural frequencies. For the first natural frequency ω n1 = (T/ml) then from equation (9) we have (A/B) ωn1 = (T/l)/[(T/l)-mω ] = (T/l)/[(T/l) m(t/ml)] = (T/l)/(T/l) = +1 or from equation (10) we have (A/B) ωn1 = [(T/l) - mω ]/(T/l) = [(T/l) m(t/ml)]/(t/l) Thus (A/B) ωn1 = (T/l)/(T/l) = +1 For the second natural frequency ω n = (3T/ml) then from equation (9) we have (A/B) ωn = (T/l)/[(T/l)-mω ] = (T/l)/[(T/l) m(3t/ml)] = (T/l)/(-T/l) = -1 Thus (A/B) ωn = (T/l)/(-T/l) = -1 Then the mode shape will be as shown in figure 6.15(a) and (b) +1 m 1 m m 1 +1 +1-1 m Figure 6.15(a) Figure 6.15(b) First Mode ω n1 = (T/ml), (A/B)ω n1 = +1 Second Mode ω n = (3T/ml), (A/B)ω n1 = -1 15

Torsional Vibratory systems Derive the equation of motion of a torsional system shown in figure 6.16. Let J 1 and J be the mass moment of inertia of the two rotors which are coupled by shafts having torsional stiffness of K t1 and K t. Jθ K t1 θ 1 1 K t1 Disc-1 θ 1 J 1 θ 1 K t (θ - θ 1 ) Jθ K t θ J θ Disc- Figure 6.16 Two Degree of Freedom Figure 6.17 Free Body Diagram torsional system If θ 1 and θ are the angular displacement of the two rotors at any given instant of time, then the shaft with the torsional stiffness K t1 exerts a torque of K t1 θ 1 and the shaft with the torsional stiffness K t exerts a torque of K t (θ - θ 1 ) as shown in the free body diagram figure 6.17 Then by Newton second law of motion we have for the mass m 1 J 1 θ 1 = - K 1 θ 1 + K (θ θ 1 ) or J 1 θ 1 +( K 1 + K )θ 1 = K θ ------ (1) for the mass m J θ = - K (θ θ 1 ) or J θ + K θ = K θ 1 -------- () Equation (1) and () are the governing Equations of motion of the system. Equivalent Shaft for a Torsional system In many engineering applications we find shaft of different diameters as shown in Figure 6.18 are in use. J a d 1 d d 3 d 4 J 1 θ J J 3 J 4 1m 3 4 L 1 L L 3 L 4 J B Figure-6.18 Stepped shaft 16

For vibration analysis it is required to have an equivalent system. In this section we will study how to obtain the torsionally equivalent shaft. Let θ be the total angle of twist in the shaft by application of torque T, and θ 1, θ, θ 3 and θ 4 be twists in section 1,, 3 and 4 respectively. Then we have θ = θ 1 + θ + θ 3 + θ 4 From torsion theory we have, T = Gθ J Thus L Where J = pd4/3 Polar moment of inertia of shaft. θ = θ 1 + θ + θ 3 + θ 4 will be θ = TL 1 + TL + TL 3 + TL 4 J 1 G 1 J G J 3 G 3 J 4 G 4 If material of shaft is same, then the above equation can be written as θ = 3T [ L 1 + L + L 3 + L 4 ] 4 4 4 πg [ d 1 d d 3 d 4 4 ] If de and Le are equivalent diameter and lengths of the shaft, then: L e = [ L 1 + L + L 3 + L 4 ] 4 4 4 d e [ d 1 d d 3 d 4 4 ] L e = L 1 [ d e ] + L [ d e ] + L 3 [ d e ] + L 4 [ d e ] [d 1 4 ] [d 4 ] [d 3 4 ] [d 4 4 ] Equivalent shaft of the system shown in Figure- 6.19 J A d e J B L e Figure 6.19 Equivalent shaft of the system shown in figure 6.18 Definite and Semi-Definite Systems Definite Systems A system, which is fixed from one end or both the ends is referred as definite system. A definite system has nonzero lower natural frequency. A system, which is free from both the ends, is referred as semi-definite system. For semi-definite systems, the first natural frequency is zero. Various definite linear and a torsional systems are shown in figure-6.19 17

k 1 x 1 x K 1 m 1 x 1 K K K m 1 m J 1 θ 1 k K J θ m x Figure-6.19 Various definite systems Semi Definite or Degenerate System Systems Systems for which one of the natural frequencies is equal to zero are called semi definite systems. Various definite linear and a torsional systems are shown in figure-6.0 l x 1 K m m x K θ Figure-6.0 Semi-Definite systems Problem to solve Derive the equation of motion of a torsional system shown in figure 6.1. J 1 J θ 1 K θ J 1 J Figure-6.1 Two Rotor System 18

Vibration of Geared Systems Consider a Turbo-generator geared system is shown in the figure 6.. Turbine Gears K t1 Shaft-1 J g1 K t Generator J 1 J g Shaft- J Figure-6.: Turbo-Generator Geared System. The analysis of this system is complex due to the presence of gears. Let i be the speed ratio of the system given by i = Speed of Turbine Speed of Generator First step in the analysis of this system is to convert the original geared system into an equivalent rotor system. Which is done with respect to either of the shafts. When the Inertia of Gears is Neglected The basis for this conversion is to consider the energies i.e. the kinetic and potential energy for the equivalent system should be same as that of the original system. Thus if θ 1 and θ are the angular displacement of the rotors of moment of inertia J 1 and J respectively then neglecting the inertia of the gears the Kinetic and Potential energy of the original system are given by Since θ = iθ 1.. T =1/ J 1 θ 1 + 1/ J θ Then the above equations can be written as Thus the above equation shows that the original system can be converted into equivalent system with respect to the first shaft as shown in figure- 6.3 U = 1/ k t1 θ 1 +1/ k t θ.... T =1/ J 1 θ 1 + 1/ J (iθ ) = 1/ J 1 θ 1 + 1/ (i J )θ 1 U = 1/ k t1 θ 1 +1/ k t (iθ ) = 1/ k t1 θ 1 +1/ (i k t )θ 1 k t1 i k t J 1 Turbine J Generator Figure-6.3 Turbo-generator geared system neglecting the inertia of gears 19

Which is obtained by multiplying the inertia and stiffness of the second shaft by i and keeping this part of the system in series with the first part. Thus the stiffness of this equivalent two rotor system is k te = i k t1 k t / (k t1 + i k t ) Thus the frequency of the system is given by ω n = k te (J 1 + i J ) / i J 1 J rad/sec When the Inertia of Gears is Considered If the inertia of the gears is not negligible then the equivalent system with respect to the first shaft can be obtained in the same manner and finally we have the three rotor system as shown in figure- 6.4 k t1 i k t J 1 J Turbine J g1 +i J g Generator Figure-6.4 Considering the inertia of gears CO-ORDINATE COUPLING AND PRINCIPAL COORDINATES. Consider a two degree of freedom system as shown in the figure- 6.5. The vibration is restricted in plane of paper. m, J G K 1 K l 1 Figure-6.5. Two degree of freedom system If m is the mass of the system, J is the Mass Moment of Inertia the system and G is the centre of gravity. k1 and k are the stiffness of the springs which are at a distance l1 and l from the line passing through the centre of gravity of the mass. Then the system has two generalized co-ordinates, x is in Cartesian and θ is in Polar co-ordinate systems when it is vibrating. 0

At any given instant of time for a small disturbance the system occupy the position as shown in figure-6.6(a). Static equilibrium l 1 l (x l 1 θ) x (x + l m,j G θ) G θ K 1 K l 1 l Figure-6.6 (a) system under vibration k 1 (x-l 1 θ) k (x + l θ) - k (x+l θ) (b) displacements at the springs If x is the displacement at the center of gravity of the system. Then the amount of displacements that take place at the left spring is (x l 1 θ)and at the right spring is (x+ l θ) which is as shown in figur-6.6(b). At any given instant of time when the body is displaced through a rectilinear displacement x and an angular displacement θ from its equilibrium position. The left spring with the stiffness k 1 and the right spring with the stiffness k are compressed through (x l 1 θ) and (x+l θ) from their equilibrium position, The forces acting on the system is as shown in the free body diagram in figure-6.6. The differential equation of motion of the system in x and θ direction are written by considering the forces and moments in their respective direction. Thus we have the equation of motion mx = - k 1 (x - l 1 θ) - k (x + l θ) ----- (1) Jθ =+k 1 (x-l 1 θ)l 1 - k (x+l θ)l ----- () Rearranging the above two equation we have mx + (k 1 + k )x = ( k 1 l 1 - k l )θ ----- (3) Jθ + (k 1 l 1 + k l ) θ = (k 1 l 1 - k l ) x ----- (4) Since J = mr The above two equation can also be written as x + [(k 1 + k )/m]x = [( k 1 l 1 - k l )/ m]θ ----- (5) θ + [(k 1 l 1 + k l )/ mr ] θ = [(k 1 l 1 - k l )/ mr ] x ----- (6) Letting [(k 1 + k )/m = a, (k 1 l 1 - k l )/ m = b and (k 1 l 1 + k l )/ mr = c Thus substituting these into equation (5) and (6) we have 1

x + ax = bθ ----- (7) θ + c θ = (b/r ) x ----- (8) The above two differential equation (7) and (8) are coupled with respect to the coordinates in which b is called the coupling coefficient or coordinate coupling. Since if b=0 the two coordinate coupling equations (7) and (8) are independent of each other. The two equations are then decoupled and each equation may be solved independently of the other. Such a coordinate are called PRINCIPAL COORDINATE OR NORMAL COORDINATES. Therefore the two i.e. rectilinear and angular motions can exists independently of each other with their natural frequency as a and c. Thus for the case of decoupled system when b=0 then (k 1 l 1 k l )/m = 0 or k 1 l 1 k l = 0 or k 1 l 1 = k l. Then the natural in rectilinear and angular modes are ω nl = a and ω na = c ω nl = a = (k 1 + k )/m and ω na = c = (k 1 l 1 + k l )/ mr In general for a two degree of freedom under damped free vibration the equation of motion can be written in the matrix form as m 11 m 1 x 1 c 11 c 1 x 1 k 11 k 1 x 1 0 +. + = m 1 m x c 1 c x k 1 k x 0. Which reveal the type of coupling present in the system as Dynamic or Mass Coupling exist if the mass matrix is non diagonal matrix. Where as stiffness or static Coupling exist if the stiffness matrix is non diagonal. Where as damping Coupling exist if the damping matrix is non diagonal. The system is dynamically decoupled when the mass matrix exists is a diagonal matrix. m 11 0 x 1 c 11 c 1 x 1 k 11 k 1 x 1 0 + + =. 0 m x c 1 c x k 1 k x 0. The system is damped decoupled when the damping matrix exists is a diagonal matrix. m 11 m 1 x 1 c 11 0 x 1 k 11 k 1 x 1 0 +. + = m 1 m x 0 c x k 1 k x 0.

The system is statically decoupled when the stiffness matrix exists is a diagonal matrix. m 11 m 1 x 1 c 11 c 1 x 1 k 11 0 x 1 0 + + =. m 1 m x c 1 c x 0 k x 0. Dynamic or Mass Coupling: If there is some point C in the system along which a force is applied to the system produces pure translation along the line of action of force as shown in figure-6.7 l 3 l 4 Static equilibrium line l 3 l 4 M,J C G e (x c -l 3 θ) C e G θ x c (x c + l 4 θ) k 1 k k 1 (x-l 3 θ) k (x+l 4 θ) Figure-6.7 (a) system under vibration Then the equation of motion is l 1 l Mx c + meθ = -k 1 (x c -l 3 θ) - k (x c +l 4 θ) Jθ + mex c = -k 1 (x c -l 3 θ) - k (x c +l 4 θ) (b) displacements at the springs Rearranging the above two equation we have Mx c + meθ + (k 1 + k )x c + (k l 4 - k l 3 )θ = 0 Jθ + mex c + (k l 4 - k 1 l 3 )x c + (k 1 l 3 + k l 4 ) θ = 0 The above equation can be written in matrix form as M me me J x c θ + (k 1 + k ) (k l 4 - k 1 l 3 ) (k l 4 - k 1 l 3 ) (k 1 l 3 + k l 4 ) x c θ = 0 0 When k l 4 k 1 l 3 = 0 or k l 4 = k 1 l 3 then the system is statically decoupled but dynamically coupled in which the equation of motion will be which was 3

M me me J x c θ + (k 1 + k ) 0 0 (k 1 l 3 + k l 4 ) x c θ = 0 0 Static/Stiffness and Dynamic/Mass Coupling: If there is a point C in the system along which a displacement produces pure translation along the line of action of spring force as shown in figure-6.8 Static equilibrium line l C G M,J x c C G θ (x c + lθ) k 1 k l k 1 x c l 1 l k (x c + lθ) l 1 l Figure-6.8 (a) system under vibration (b) displacements at the springs Then the equation of motion is Mx c + ml 1 θ = - k 1 x c - k (x c +lθ) and Rearranging the above two equation we have Mx c + ml 1 θ + (k 1 + k )x c + k l = 0 Jθ + ml 1 x c + k lx c + k 1 l θ = 0 Jθ + m l 1 x c = -k (x c + lθ)l The above equation can be written in matrix form as M ml 1 ml 1 J x c θ + (k 1 + k ) k l k l k l x c θ = 0 0 In which both the mass matrix and stiffness matrix are non-diagonal matrix hence the system is both statically and dynamically coupled. 4

Problem Determine the normal mode of vibration of an automobile shown in figure-6.9 simulated by a simplified two degree of freedom system with the following numerical values m = 1460 kg, L 1 = 1.35m, L =.65 m, K 1 = 4.x10 5 N/m, K = 4.55x10 5 N/m and J=mr where r= 1. m L1 L K1 x K θ Figure-6.9 Automobile can be modeled as shown in figure -6.30 Static equilibrium line l 1 l m,j G (x l 1 θ) G x θ (x + l θ) K 1 K l 1 l Figure-6.30 (a) system under vibration k 1 (x-l 1 θ) k (x + l θ) - k (x+l θ) (b) displacements at the springs Let at any given instant of time the translatory displacement be x and an angular displacement be θ from its equilibrium position of the automobile. Then the left spring with the stiffness k 1 and the right spring with the stiffness k are compressed through (x l 1 θ) and (x+l θ) from their equilibrium position, The forces acting on the system are as shown in the free body diagram in figure-6.30(b). The differential equation of motion of the automobile in x and θ direction are written by considering the forces and moments in their respective direction. Thus we have the equation of motion mx = - k 1 (x - l 1 θ) - k (x + l θ) ----- (1) Jθ =+k 1 (x-l 1 θ)l 1 - k (x+l θ)l ----- () 5

Rearranging the above two equation we have mx + (k 1 + k )x = ( k 1 l 1 - k l )θ ----- (3) Jθ + (k 1 l 1 + k l ) θ = (k 1 l 1 - k l ) x ----- (4) Equation (3) and (4) are the second order differential equation and the solution for x and θ are obtained by considering that they can have harmonic vibration under steady state condition. Then considering the case when the system execute harmonic vibration at frequency ω Thus if x = Asinωt, and θ = Bsin ωt ---- (5) Substitute equation (5) into the equation (3) and (4) and canceling the common term sinωt we get [- mω + (k 1 + k )] A = (k 1 l 1 k l ) B ------- (6) [- Jω + (k 1 l 1 + k l )] B = (k 1 l 1 k l ) A ---- (7) From equation (6) we have A/B = (k 1 l 1 -k l )/ [(k 1 + k ) mω ] ---------- (8) From equation (7) we have A/B = [(k 1 l 1 + k l ) - Jω ] / (k 1 l 1 k l ) ---- (9) Equating (8) and (9) A/B = (k 1 l 1 -k l )/ [(k 1 + k ) mω ] = [(k 1 l 1 + k l ) - Jω] / (k 1 l 1 k l ) [(k 1 +k ) mω ][(k 1 l 1 +k l )-Jω ]=(k 1 l 1 -k l ) Further Simplification will give mjω 4 [J(k 1 +k ) + m(k 1 l 1 +k l )]ω + k 1 k (l 1 +l ) = 0 --------- (10) Substituting the value of m, J, k 1, k, l 1, l into the above equation (10) we have 3.173x10 6 ω 4 4.831x10 9 ω +1.7x10 1 =0 --- (11) or ω 4 1.53x103ω +5.49x10 5 =0 ---- (1) Letting ω =λ we have λ 1.53x10 3 λ+5.49x10 5 = 0 ---- (13) Equation (13) is quadratic equation in λ. Thus solving equation (13) we get two roots which are λ 1 = 569.59, λ = 953.13 Since ω =λ we have ω = λ. Thus ω 1 =3.86 rad/sec and ω =30.87 rad/sec Thus f n1 = 3.797 Hz and f n = 4.911 Hz Un-damped Dynamic Vibration Absorber Consider a two degree of freedom system with a forcing function F 1 = F o sinωt as shown in figure-6.31(a). k 1 x 1 k 1 m 1 F 1 = F o sinωt F 1 m 1 k x 1 k (x x ) m x m (a) Figure- 6.31(a) Two degree of freedom system with forcing function F 1 on mass 1 (b) 6

Let x 1 and x be the displacement of the masses m 1 and m respectively at any given instant of time measured from the equilibrium position with x > x 1. Then the spring forces acting on the masses are as shown in free body diagram in Figure 6.31(b) Based on Newton s second law of motion ƒ = For mass m 1 we have m 1 x 1 m 1 x 1 + k 1 x 1 - k x + k x 1 = F o sinωt ------------ (1) for mass () = - k 1 x 1 + k (x x 1 ) + F o sinωt m x = - k (x x 1 ) m x + k x - k x 1 = 0 m x + k x = k x 1 ------------ () The solution for x 1 and x are obtained by considering that the masses execute harmonic vibration at frequency w. Thus if x 1 = X 1 sin ωt, and x = X sin ωt ----- (3) Then we have x 1 = - ω X 1 sinωt and x = - ω X sinωt --------- (4) Substituting equation (3) and (4) into the equation (1) and () we get -m 1 ω X 1 sinωt + (k 1 + k )X 1 sinωt = k X sin ωt + F o sinωt ----------------- (5) - m ω X sinωt + k X sinωt = k X 1 sin ωt -------------- (6) Canceling the common term sinwt on both the sides of equation 95) and (6) we have [(k 1 + k ) - m 1 ω ]X 1 - k X = F o ----------- (7) k X 1 - [k - m ω ]X = 0----- (8) Solving for X 1 and X by cramer s rule F 0 K 0 K mω K 0 X -------- (9) -------- (10) 1 = X = where is the determinant of characteristic equations. m x (K 1 + K ) mω 1 F 0 (K1 + K = K ) mω K 1 K mω 0 -------- (11) { }{ } Solving the above determinant we get = (K + K ) mω K mω ------- (1) 1 1 K 7

If the two vibratory masses are considered separately as shown in Figure- 6.3, the mass 1 is a main system and mass is an secondary system. This system can be used as Dynamic vibration absorber or Tuned damper by using the amplitude Equations (9) and (10). k 1 F 1 m 1 x 1 Main System k m x Secondary System Figure- 6.3 If the system has to be used as a Dynamic vibration absorber, then the amplitude of vibration of mass m 1 should be equal to zero, i.e X 1 =0. X F 0 0 K K m ω 1 = = 0 -------- (13) Then we have F 0 0 K K mω = 0 F o (k m ω ) = 0 since F o cannot be equal to zero we have k m ω = 0 ω = k /m or ω = k /m rad/sec -------- (14) The above Eqn. is the natural frequency of secondary or absorber system. 8