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Dynamics Course

### Transcript of Approximate Methods

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

2. Rayleighs Method Background: Consider that an undamped SDF mass-spring system is in free harmonic motion, then x = X sin (pt+) x = pX cos (pt+) The strain energy of the system, V, at any time t is given by V = kx 2= kX 2sin 2 (pt+) and its kinetic energy, T, is given by T = mx 2= mp 2 X 2cos 2 (pt+) 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 kX 2= mp 2 X 2 From which we conclude thatp 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 maxand T max , we obtain (E5) (E6) (E7) m a L k y 0 4. Rayleigh Quotient Consider now a MDF system in free vibration, such that {x} = {X} sin(pt+) {x} = p {X} cos(pt+) The maximum strain energy of the system is {F} T [d]{F}V max= {F} T {X} = {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= p 2 {X} T [m]{X}= p 2T maxIn whichT max= {X} T [m]{X}will be referred to as the maximum pseudo-kinetic energy of the system. The principle of conservation of energy requires thatV max=T max= p 2T maxwhere ~ ~ ~ The above equation is known asRayleighs 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 maxwhere .. T max~ 7. Properties of Rayleigh quotient

• Ifandin Eq.(E12) are evaluated for {X} = to the j thmode {X j }, then the value of p 2will be precisely p j 2 .
• If Eq.(E12) is evaluated for a vector {X} which does not correspond to a natural mode, then the resulting value of p 2willnotbe 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 2in Eq.(E12) defines a surface in a space having (n-1) dimensions.
• It can be shown that

V max(1) The natural frequencies of the system, i.e., the values of p 2obtainedwhen {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.

• The values of p 2obtained 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 ).
• 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 .

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:

• 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.
• Compute the values of V maxandcorresponding to the estimated mode.
• Evaluate p 2from,
• 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.

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 #1m k k m x 1 x 2 1 1 1 2 1 1.5 (a) (c) (b) 11. V maxmay 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 byF,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.

• In the following diagram, the value of p 2determined by application of Rayleighs 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 .

X 2 / X 1 3 2 1 2.618 -0.618 1.618 1 -2 -1 0 2 3 0.38197 16.

• 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 computeV maxas the product of these forces and the resulting displacements. For example, for the forces considered in Fig.2,
• 2 V max=F F /k[2+3] = 5 ( F 2 /k)
• 2 T max= mF /kF /k[2 2 +3 2 ] = 13m ( F 2 /k 2)
• p 2= 0.3846 (k / m)
• which is the same as the answer obtained above.

~ 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+ (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 1.61.8 18. The exact value ofp 2 is0.2680k/m,and theerrorin p is only0.42%. In the following figure, the value of p 2is plotted as a function of the displacement ratios X 2 / X 1and X 3 / X 1in the range between 3 and 3. As would be expected, there are three stationary points:(1)aminimum pointofp 2 =0.268k/mcorresponding to the fundamentalfrequency;(2) amaximum valueofp 2 =3.732k/mcorresponding to the third natural frequency;(3) asaddle pointofp 2 =2k/mcorresponding to the second natural frequency. The values X 2 / X 1and X 3 / X 1of the associated modes can be read off the figure. T max ~ 2= m [ 2.5 2+ 4 2+ (4.5 2 ) ] = 32.375m 19. Dimensional representation of Rayleighs 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 nbe the inertia forces corresponding to the assumed mode and y 1 ,y 2 y j ,..y nbe 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 jto 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 jas the inertia forces corresponding to the assumed configuration.It is important to note that y jin 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.

• Illustration:
• 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)
• 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 exac