Dynamics Course

1. APPROXIMATE METHODS Prof. A. Meher Prasad Department of Civil Engineering Indian Institute of Technology Madras email: prasadam@iitm.ac.in

2. Rayleigh’s Method Background: Consider that an undamped SDF mass-spring system is in free harmonic motion, then x = X sin (pt+α) x = pX cos (pt+α) The strain energy of the system, V, at any time t is given by V = ½ kx 2 = ½ kX 2 sin 2 (pt+α) and its kinetic energy, T, is given by T = ½ mx 2 = ½ mp 2 X 2 cos 2 (pt+α) The principle of conservation of energy requires that, the sum of V and T be the same. Note that when V = V max , T=0, and when T= T max , V =0. Hence V max = T max . . (E1) (E2) (E3) (E4)

3. or ½ kX 2 = ½ mp 2 X 2 From which we conclude that p 2 = k/m This is identical to the expression obtained from the solution of the governing equation of motion. As a second example, consider the SDF system shown Equating V max and T max , we obtain (E5) (E6) (E7) μ m a L k y 0

4. Rayleigh Quotient Consider now a MDF system in free vibration, such that {x} = {X} sin(pt+α) {x} = p {X} cos(pt+α) The maximum strain energy of the system is ½ {F} T [d]{F} V max = ½ {F} T {X} = ½ {X} T [k]{X} In which {F} are the static modal forces corresponding to the displacement amplitudes {X} , and [d] and [d] are the flexibility and stiffness matrices of the system. . (E8) (E9)

5. The maximum Kinetic energy of the system is given by T max = ½ p 2 {X} T [m]{X} = p 2 T max In which T max = ½ {X} T [m]{X} will be referred to as the maximum pseudo-kinetic energy of the system. The principle of conservation of energy requires that V max =T max = p 2 T max where ~ ~ ~ The above equation is known as Rayleigh’s quotient. (E10) (E11) (E12)

6. The Eq.(E12) could also be obtained from the equations of motion of the system as follows: [m] { x } + [k] { x } = 0 Making use of Eq.(E8) and pre multiplying the resulting equation by {X} T , we obtain, - p 2 {X} T [m]{X} + {X} T [k]{X} = 0 V max where .. T max ~

10. If the frequency is computed for several different assumed configurations, the smallest of the computed values will be closest to the exact fundamental frequency, and the associated configuration is closest to the actual fundamental mode. The details of the procedure are illustrated by a series of examples. Example #1 m k k m x 1 x 2 1 1 1 2 1 1.5 (a) (c) (b)

11. V max may be evaluated from the deformations of the stories without having to determine first the stiffness matrix of the system. 2 V max = k [ (1) 2 +0] = k 2 T max = m [ (1) 2 +(1) 2 ] = 2m Assumption 1: Take x 1 =x 2 =1, as shown in Fig. (a) ~

12. Assumption 2: Take x 1 =1 and x 2 = 2, as shown in Fig.(b) 2 V max = k [ (1) 2 +(1) 2 ] = 2k 2 T max = m [ (1) 2 +(2) 2 ] = 5m Assumption 3: Take x 1 =1 and x 2 = 1.5, as shown in Fig.(c) 2 V max = k [ (1) 2 +(0.5) 2 ] = 1.25k 2 T max = m [ (1) 2 +(1.5) 2 ] = 3.25m ~ ~

13. 1. Assumption 3, which leads to the lowest frequency value, is the best of the three approximations considered, and is only slightly off the exact value of, 2. That assumptions 1 and 2 would be poor, could have been anticipated by considering the forces necessary to produce the assumed configurations. The deflection configuration in Fig.(a) is produced by a single concentrated force applied at the first floor level, whereas the configuration in Fig.(b) is produced by a single concentrated force acting at the second floor level. Clearly, neither of these force distributions is a reasonable approximation to the inertia forces associated with the motion of the system in its fundamental mode. Discussion

14. 3. That assumptions 1 and 2 The deflection pattern considered in Fig.(c) is produced by lateral forces which are proportional to the weights of the floors. If these forces are denoted by F, the resulting displacements are as shown. Subject to the justification noted later, this assumption generally leads to an excellent approximation for the fundamental natural frequency of the system. Fig (2) F F 2 F /k 2 F /k + F /k

17. Example # 2 Assume a deflection configuration equal to that produced by a set of lateral forces equal to the weights of the system. 2V max = k [ 2.5+4+ ½(4.5) ] = 8.75 k ( This can also be evaluated from the story deformations as k [ 2.5 2 + 1.5 2 + 0.5 2 ] = 8.75k ) k k x 1 x 2 k x 3 m m 0.5m 2.5 2.5+1.5 = 4 4+0.5 = 4.5 k k 0.5k 1 1.6 1.8

18. The exact value of p 2 is 0.2680k/m, and the error in p is only 0.42%. In the following figure, the value of p 2 is plotted as a function of the displacement ratios X 2 / X 1 and X 3 / X 1 in the range between –3 and 3. As would be expected, there are three stationary points: (1) a minimum point of p 2 =0.268k/m corresponding to the fundamental frequency; (2) a maximum value of p 2 =3.732k/m corresponding to the third natural frequency; (3) a saddle point of p 2 =2k/m corresponding to the second natural frequency. The values X 2 / X 1 and X 3 / X 1 of the associated modes can be read off the figure. T max ~ 2 = m [ 2.5 2 + 4 2 + ½(4.5 2 ) ] = 32.375m

19. Dimensional representation of Rayleigh’s quotient for a 3 DOF system **20 squares to the inch

20. Assumption # 1: The exact value of , Example # 3 m 1.5 m L/3 L/3 L/3 x

21. Assumption #2 Which is in excellent agreement with the exact value W 1.5W

22. Let F 1 ,F 2 , F j,… F n be the inertia forces corresponding to the assumed mode and y 1 ,y 2 ……y j ,…..y n be the deflections induced by these forces. Then, m 1 m 2 m j m n . . . . . . . . Application to systems with Lumped masses: . . . . . . . . F 1 F 2 F j F n y j

23. As already demonstrated, good accuracy is achieved by taking the forces F j to be equal to, or proportional to the weights W j . It should be realized, however that these forces are not the exact inertia forces.Rather they represent the inertia forces associated with a uniform (rigid body) motion of the system. An improved approximation may be achieved by assuming a configuration for the mode and taking F j as the inertia forces corresponding to the assumed configuration.It is important to note that y j in Eq. E13 are the deflections produced by the forces F j, not the deflections assumed for the purpose of estimating F j Selection of F j

27. Applications to continuous systems EI, μ , L associated with the assumed deflection

28. y(x) = y o sin( π x/L) This force may be determined from the differential equation for beams. Recalling that , We obtain, Hence, Which is the same as the result found by procedure (a)

30. This is 0.8% too high This is the exact frequency – Explain why? If we had assumed as y (x) the deflection produced by a single concentrated force at the center, we would have found that,

31. Dunkerley’s Method Equation of motion : Let [m] be the diagonal matrix, set Det

32. Given a n th order polynomial equation, (1/p 2 ) Sum of the roots of characteristic equation, d ii is the flexibility coefficient equal to deflection at i resulting from a unit load of i, its reciprocal must be the stiffness coefficient k ii , equal to the force per unit deflection at i.

33. By neglecting these terms(1/p 2 2 , …1/p n 2 ) ,1/p 1 2 is larger than its true value and there fore p 1 is smaller than the exact value of the fundamental frequency The estimate to the fundamental frequency is made by recognizing p 2 ,p 3 etc are natural frequencies of higher modes and larger than p 1 .

34. Dunkerley’s Approximation It provides a lower bound estimate for the fundamental frequency. Let p = natural frequency of system p A , p B , p C , …….. p N = exact frequencies of component systems Then or The frequency so determined can be shown to be lower than the exact.

35. Example # 1 If natural modes of component systems A, B, C are close of each other, then the value of p determined by this procedure can be shown to be close to the exact. k k k m m 0.5m m m 0.5m

36. Consider the cantilever beam shown for which the component systems A,B,C are indicated . Since the natural modes of the system are in closer agreement in this case than for the system of the shear beam type considered in the previous example, the natural frequency computed by Dunkereley’s method can be expected to be closer to the exact value than with case before. m m m/2 m m m/2 A B C Example # 2 L/3 2L/3 2L/3 L/3 L L/3 L/3 L/3

37. As expected the agreement is excellent in this case.

38. Example # 3 Upper bound: Determined by Rayleigh’s method with y(x) = y 0 sin( π x/L) is, Lower bound: Determined by Dunkerley’s approximation If we consider one mode, m μ

39. Consider all modes, m μ dx x

42. 5. It can further be shown that if, at the end of a cycle, we compute the values of which will make the elements or components of the derived and assumed vectors equal, the highest characteristic value lies between the largest and smallest of these values. 4. In general, the vector computed in 2 will not be proportional to {X}. Now if we take as our next assumption the result of step 2 and repeat the process, the procedure will converge to the characteristic vector associated with the largest characteristic value .

43. Stiffness Matrix Flexibility Matrix Converges to highest natural frequency and mode Converges to fundamental natural frequency and mode

44. Using Flexibility formulation first,we obtain Example Frame k k x 1 x 2 k x 3 m m 0.5m

46. ξ

47. Obviously it is diverging from the fundamental mode

49. Rayleigh Quotient Combining iteration with Rayleigh Method Stodola Method: From Rayleigh’s Quotient,

50. If convergence is incomplete, the Rayleigh Quotient gives the better approximation.Any error in the first mode frequency computed by the Rayleigh Quotient is always on high side

51. Stodola convergence Assumed mode can be expressed as, Highest characteristic value On Iteration,

52. After ‘s’ Iterations,

53. Stodola process for the second mode The vector, has a zero first mode component Any Vector,

54. Then premultiplying any arbitrary vector {X} by the sweeping matrix [S 1 ] removes the first mode component [H] 2 {X}= λ 2 [X] Define sweeping matrix [S 1 ] to be, Matrix iteration is carried out for,

55. 0.5m m m k k k x 3 x 2 x 1

60. Execute the first few natural frequencies and the associated modes of the beam shown , and study the rate of convergence of the results as a function of the stiffness of the spacing ,i.e. or an appropriate dimensionless measure of it. Application of Rayleigh Ritz Procedure L/4 3L/4

61. The coefficients ‘a’ in this expression must be such that the value of p 2 ,determined from When expressed in terms of the dimensionless distance The expression for y(x) becomes, is stationary. This requires that,

62. Consider first only Two Terms in the Series,

64. Application of the above Equation for a j =a 1 and a j =a 2 leads to , or,

65. Canceling the factor ½ on the two sides of these equations , and Introducing the dimensionless frequency parameter We obtain after rearrangement of term: Expanding,we obtain the following quadratic equation in λ 0

66. λ 0 ²-(17+3ρ 0 )λ 0 +16+18ρ 0 =0 λ 0 =1/2 {17+3ρ 0 ± } The modes are defined by the ratio a 2 /a 1 this is given by, a 2 / a 1 = -

67. Consider Three Terms in Series or,

68. or, Application of the Equation for a j =a 1 and a j =a 2 leads to,

69. Expanding we obtain the following cubic equation in  : Modes: These are defined by the ratios a 2 / a 1 and a 3 / a 1 . Considering the first two equations (13) and eliminating a 3 we obtain: Considering the first and third of equation (13) , and eliminating a 2 we obtain: (16) (17) (15) This leads to the determinantal equation,

70. Note that Equations (16) and (17) are independent of  o . However  o enters in these equations indirectly through  o .The equations are valid irrespective of the order of  o considered (i.e. for all three modes) In considering the second mode ,it is more convenient to express it in terms of the ratios a 1 / a 2 and a 3 / a 2 (i.e normalize it with respect to a 2 ).These ratios are given by, and (18) (19)

71. Convergence of natural frequencies and modes 1 0.1238 0.0723 -0.1456 1 0.3007 -0.0344 -0.3050 1 87.237 27.097 3.6661 3 1 0.3178 -0.3178 1 28.247 3.7526 2 1 6 1 (b) For  o =5 1 0.0224 0.0129 -0.0235 1 0.0864 -0.0109 -0.0867 1 82.044 18.089 1.8665 3 1 0.0877 -0.0877 1 18.124 1.8760 2 1 2 1 (a) For  o =1 a 3 a 2 a 1 a 3 a 2 a 1 a 3 a 2 a 1  1  1  1 Third mode Second mode Fundamental mode Frequency coefficient No of terms used

77. Starting Vectors (1) When some masses are zero, for non zero d.o.f have one as vector entry. (2) Take ratio .The element that has minimum value will have 1 and rest zero in the starting vector.

79. Eqn. 2 are not true. Eigen values unless P = n If [  ] satisfies (2) and (3),they cannot be said that they are true Eigen vectors. If [  ] satisfies (1),then they are true Eigen vectors. Since we have reduced the space from n to p. It is only necessary that subspace of ‘P’ as a whole converge and not individual vectors.

80. Algorithm: Pick starting vector X R of size n x p For k=1,2,…..   k+1 { X } k+1 -  k   static p x p p x p Smaller eigen value problem, Jacobi

81. Factorization Subspace Iteration Sturm sequence check (1/2)nm 2 + (3/2)nm nq(2m+1) (nq/2)(q+1) (nq/2)(q+1) n(m+1) (1/2)nm 2 + (3/2)nm 4nm + 5n nq 2

82. Total for p lowest vector. @ 10 iteration with nm 2 + nm(4+4p)+5np q = min(2p , p+8) is 20np(2m+q+3/2) This factor increases as that iteration increases. N = 70000,b = 1000, p = 100, q = 108 Time = 17 hours

84. (2) Neq x m [ I ] k,k+1 - with # of rows = # of attachment d.o.f. between k and k+1 = # of columns Ritz analysis: Determine [ K r ] = [R] T [k] [R] [ M r ] = [R] T [M] [R] [k r ] {X} = [M] r +[X] [ ] - Reduced Eigen value problem Eigen vector Matrix, [  ] = [ R ] [ X ]

85. Use the subspace Iteration to calculate the eigen pairs (  1 ,  1 ) and (  2 ,  2 ) of the problem K  =  M  ,where Example