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- 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 ~
- 7. Properties of Rayleigh quotient <ul><li>If and in Eq.(E12) are evaluated for {X} = to the j th mode {X j }, then the value of p 2 will be precisely p j 2 . </li></ul><ul><li>If Eq.(E12) is evaluated for a vector {X} which does not correspond to a natural mode, then the resulting value of p 2 will not be a natural frequency. Furthermore, to each assumption of {X} there corresponds a different value of p 2 . In fact, if we recall that (n-1) displacement ratios are necessary to define the configuration of a MDF system, we may conclude that p 2 in Eq.(E12) defines a “surface” in a space having (n-1) dimensions. </li></ul><ul><li>It can be shown that </li></ul>V max (1) The natural frequencies of the system, i.e., the values of p 2 obtained when {X} is equal to any of the natural modes, correspond to stationary (maximum, minimum or saddle) points of this surface. It follows that an error in estimating the mode will produce an error in frequency which is of smaller order (since the surface is ‘flat’ in the neighbourhood of the stationary point). T max ~
- 8. <ul><li>The values of p 2 obtained for an arbitrary {X} lies between the lowest and highest natural frequencies of the system(i.e., p 1 2 < p 2 < p n 2 ). </li></ul><ul><li>It follows that if one assumes an {X} which approximates the fundamental natural mode of the system, then the resulting value of p will be greater than (will represent an upper bound estimate for) p 1 . Similarly, if one assumes an {X} that approximates the highest natural mode, the resulting value of p will be lower than (will represent a low bound for) the true p n . </li></ul>
- 9. Because the fundamental natural mode of a system can normally be estimated with reasonable accuracy, the procedure is ideally suited to the computation of the fundamental natural frequency.Application of the procedure requires the following steps: <ul><li>Estimate the fundamental mode of vibration. This may be done either by assuming directly the displacements of the nodes, or the associated forces and computing the displacements. </li></ul><ul><li>Compute the values of V max and corresponding to the estimated mode. </li></ul><ul><li>Evaluate p 2 from, </li></ul><ul><li>The value thus determined is higher than the true fundamental natural frequency of the system, and, unless the assumption regarding the mode was quite poor, it will be close to the actual frequency. </li></ul>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
- 15. <ul><li>In the following diagram, the value of p 2 determined by application of Rayleigh’s method is plotted as a function of the displacement ratio x 2 /x 1 . Note that the two natural frequencies correspond to the maximum and minimum points of the curve, and that in the vicinity of these extremum points the frequency values are insensitive to variations in the displacement ratio x 2 /x 1 . </li></ul>X 2 / X 1 3 2 1 2.618 -0.618 1.618 1 -2 -1 0 2 3 0.38197
- 16. <ul><li>The first step in the solution that has been presented was to estimate the mode of vibration of the system. Alternatively, we could have assumed the distribution of the inertia forces associated with the mode of interest and compute V max as the product of these forces and the resulting displacements. For example, for the forces considered in Fig.2, </li></ul><ul><li>2 V max = F F /k [2+3] = 5 ( F 2 / k) </li></ul><ul><li>2 T max = m F /k F /k [2 2 +3 2 ] = 13m ( F 2 / k 2) </li></ul><ul><li> p 2 = 0.3846 (k / m) </li></ul><ul><li>which is the same as the answer obtained above. </li></ul>~
- 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
- 24. <ul><li>Illustration: </li></ul><ul><li>As a guide in the selection of the forces { f }, assume that the fundamental mode varies as a sine curve , as shown in fig (a) </li></ul><ul><li>The inertia forces corresponding to this assumption are shown in fig.(b). For the example considered, these turn out to be the exact forces, and hence the frequency computed from these forces will be exact in this case. </li></ul>(a) Assumed Mode (b) Inertia Forces 0.5 3/2 1 0.5F 3/2F 0.5F 0.5F 3/2F 0.5F m m 0.5m x 1 x 2 x 3
- 25. <ul><li>If the mode were assumed to increase linearly along the height, the forces and the deflections would be as follows: </li></ul><ul><li>In this case </li></ul>F 2F 1.5F 4.5 F/k 8 F/k 9.5 F/k
- 26. <ul><li>Taking the {F} proportional to the masses will not lead to satisfactory results. Directions important </li></ul><ul><li>Take </li></ul>m 1 m 2 Another example : F 1 F 2
- 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
- 40. <ul><li>Limitation of procedures: </li></ul><ul><li>One cannot improve the accuracy of the solution (depends on the deflected shape of structure) in a systematic manner. </li></ul><ul><li>Extension of procedure : Rayleigh - Ritz </li></ul>
- 41. <ul><li>Applicable to systems governed by [H]{X} = {X} , where [H] is not necessarily symmetric. </li></ul><ul><li>Meaning to a solution : Finding an {X} which when operated by [H] will give a vector is proportional to itself.Then {X}=characteristic vector and = the associated characteristic value. </li></ul><ul><li>Procedure: </li></ul><ul><li>Assume an {X}. </li></ul><ul><li>Compute [H]{X} </li></ul><ul><li>If step 2 results in a vector which is proportional to {x}, then {X} is a characteristic vector, and the factor of proportionality is the associated characteristic value. </li></ul>Stodola Method – (Method of Iteration)
- 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
- 72. Mass Condensation or Guyan Reduction <ul><li>Extensively used to reduce the number of D.O.F for eigen value extraction. </li></ul><ul><li>Unless properly used it is detrimental to accuracy </li></ul><ul><li>This method is never used when optimal damping is used for mass matrix </li></ul>
- 73. <ul><li>Assumption: Slave d.o.f do not have masses – only elastic forces are important </li></ul>
- 74. <ul><li>Choice of Slave d.o.f </li></ul><ul><ul><li>All rotational d.o.f </li></ul></ul><ul><ul><li>Find ratio, neglect those having large values for this ratio </li></ul></ul><ul><ul><li>If [ M ss ] = 0, diagonal, [K r ] = same as static condensation then there is no loss of accuracy </li></ul></ul>
- 75. Subspace Iteration Method <ul><li>Most powerful method for obtaining first few Eigen values/Eigen vectors </li></ul><ul><li>Minimum storage is necessary as the subroutine can be implemented as out-of core solver </li></ul><ul><li>Basic Steps </li></ul><ul><ul><li>Establish p starting vectors, where p is the number of Eigen values/vectors required P<<n </li></ul></ul><ul><ul><li>Use simultaneous inverse iteration on ‘p’ vectors and Ritz analysis to extract best Eigen values/vectors </li></ul></ul><ul><ul><li>After iteration converges, use STRUM sequence check to verify on missing Eigen values </li></ul></ul>
- 76. <ul><li>Method is called “Subspace” iteration because it is equivalent </li></ul><ul><li>to iterating on whole of ‘p’ dimension (rather that n) and not </li></ul><ul><li>as simultaneous iteration of “p’ individual vectors </li></ul><ul><li>Starting vectors </li></ul><ul><li>Strum sequence property </li></ul>For better convergence of initial lower eigen values ,it is better if subspace is increased to q > p such that, q = min( 2p , p+8) Smallest eigen value is best approximated than largest value in subspace q.
- 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.
- 78. <ul><li>Starting vectors can be generated by Lanczos algorithm- converges fast. </li></ul><ul><li>In dynamic optimisation , where structure is modified previous vectors could be good starting values. </li></ul>Eigen value problem (1) (2) (3)
- 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
- 83. <ul><li>Aim: Generate (neq x m) modal matrix (Ritz vector). </li></ul><ul><li>Find k and { u } k for the k th component </li></ul><ul><li>Let [ ] k = substructure Modal matrix </li></ul><ul><li>which is nk x n , nk = # of interior d.o.f </li></ul><ul><li>n = # of normal modes take determined for that structure </li></ul><ul><li>Assuming ‘l’ structure, </li></ul>
- 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

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