Mdof

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

Static Force-Displacement Relationship Let x 1 , x 2 , x 3 ...... x n be the coordinates used to specify the deflection configuration of the system in a dynamic analysis . Assume that at points of definition of these coordinates the system is acted upon by a set of concentrated forces, F 1 , F 2 , F 3 .......F n producing the displacements x 1 ,x 2 , x 3 ...... x n in the direction of the forces. The relationship between the forces F and the displacements x is defined as follows, . . . . . . . . . . . . . . . . (136) (137) or

The quantities ‘d’ are known as flexibility coefficients and the ‘k’ as stiffness coefficients. To establish the meaning of a particular coefficient,consider the equation that contains that coefficient by making the displacement of the term containing the coefficient equal to unity and all the other displacement equal to zero. Then interpret the meaning of the term on left. For example,to determine the meaning of d 21 , we consider the second of the equations in (136) and take F 1 =1 and F 2 = F 3 = ….. = F n = 0. We then conclude that , = displacement at 2 produced by a force of unit magnitude at 1 and no force anywhere else.

Similarly, to determine the meaning of K 21 we consider the second equation in (137) and take x 1 =1 and We conclude that = force at 2 produced by a displacement of unit magnitude at 1 and no displacement anywhere else. In matrix notation,equations 136 and 137 may be written as (138) (139) Where [d] and [k] are square matrices of the stiffness coefficients and the flexibility coefficients, respectively, and {F} and {x} are column matrices of the forces at the nodes and of the associated displacements.

The matrix [d] is known as the flexibility matrix of the system,and [k] is known as the stiffness matrix of the system. Substituting (138) into (139) we obtain {F} = [k] [d] {F} Hence we conclude that [k] [d] =[I] = identity matrix It follows that [k] and [d] are the inverse of one another. From Maxwell’s law reciprocity it follows , d ij= d ji And hence the flexibility matrix is symmetric similarly, from Betti’s work principle it follows that k ij= k ji indicating that [k] is also symmetric (140) (141) (142) (143)

Evaluation of [d] and [k] 1 d 41 d 31 d 21 d 11 2 d 42 d 32 d 22 d 12 Second column of flexibility matrix 3 4 1 1 2 3 4 1 1 k 41 k 31 k 21 k 11 First column of stiffness matrix 1 k 42 k 32 k 22 k 12 Second column of stiffness matrix . . . . . . . . . . . . . . . .

Example #1 For the frame shown, k 1 k 2 k 3 the flexibility matrix is x 1 x 2 x 3

The elements of stiffness matrix are obtained as follows 0 -k 2 k 1 + k 2 -k 3 k 2 +k 3 -k 2 -k 3 0 k 3 Accordingly the stiffness matrix is For systems of this type the stiffness matrix is a banded trapezoidal matrix

Example #2 Consider a prismatic cantilever beam with three nodes spaced at intervals h=L/3 Flexibility Matrix: 3h=L

Stiffness matrix: Formulation is more involved in this case, because a displacement of a mode induces shears not only in the panels connecting to that mode but also in the more distant panels. Such systems are far coupled (have a strong coupling). Show that and verify that

Strain Energy of system For a system deflected in a configuration defined by x 1 , x 2 ,…. x m the strain energy ,V, is equal to the work done by the forces F 1 , F 2….. F n in going through the displacement , x. that is In matrix notation, (144) (145) Substituting Equation (138) into the first of equations (145) we obtain, And substituting Equation (139) into the second of equations (145) we obtain, (146) (147)

Equation (146)expresses V in terms of the flexibility matrix,whereas equations (147) expresses it in terms of the stiffness matrix. If Equation (138) were substituted into Equation (145) we would obtain, hence it follows that or that [d] is symmetric. This is a restatement of Maxwell’s reciprocal relation. (148)

Proceeding in a similar manner and substitute equation 139 into the two equations 145 we obtain, or [k] is symmetric. This result was obtained earlier by the application of Betti’s work principle. Quadratic Form : Equation147 represents an expression of the form. Which is the sum of terms involving the squares of the displacements and the products of pairs of displacements, such an expression is known as a quadratic form and represents a scalar quantity. (149)

If the quadratic form is positive and non-zero for all possible combinations of the unknowns, excluding the trivial case of x 1 =x 2 …x m =0 , the form and associated square matrix are said to be definitely positive or positive definite. The strain energy of the system being a positive quantity, it and the associated matrix are positive. If the system is fixed–base so that the strain energy can be zero only in the trivial case of x 1 =x 2 …=x n =0, the quadratic form and the associated stiffness matrix are positive definite. A quadratic form is said to be positive if its value is positive or zero for all possible combinations of the unknowns. similarly, a square matrix is said to be positive if the associated quadratic form is positive. A system which can undergo rigid body motion can have zero strain energy for finite values of X. Hence V and [k] in this case are positive but not positive definite.

The mass matrix is positive definite because the associated quadratic form represents the kinetic energy of the system, a quantity which is non-zero and positive combinations of , except the trivial case of Mass matrix

Dynamic Equations of motions for MDOF Mass matrix Inertia force vector Damping force vector Elastic force vector Externally applied force vector Damping matrix Stiffnessmatrix A second order non-homogeneous linear ordinary differential equation x 1 x 2 x 3 x i f Ii

Dynamic Equations of motions for MDOF Displacement vector Velocity vector Acceleration vector Are symmetric square matrices Lumped Mass matrix (Diagonal matrix) (Concentrated point masses at each of its corners) Consistent Mass Matrix (Got from assembly of element mass matrices which is based on shape function used in stiffness formulation- more computational work)

Undamped free vibration of MDOF system Equation of motion : can this system undergo purely harmonic motion ? (ie) an admissible solution ? For non-trivial solution: i e Results in n th order polynomial equation in p 2 =0 From fundamental theorem of algebra, this polynomial has roots for p 2 corresponding to each such root of polynomial in p 2 ,there is a mode shape vector defining the vibration configuration required such that purely harmonic sinusoidal motion is possible at circular natural frequency, p i Answer : Yes (i.e) A generalised eigen value problem

Undamped free vibration of MDOF system contd.... Properties : (i) For all i = 1,2,....n (all natural frequencies are +ve real numbers) (ii) Orthogonality conditions More generally, (iii) All n mode shape vectors are linearly independent Vectors and spans complete n-dimensional vector space – i.e. any vector can be expressed as linear combination of modal shape vectors

Orthogonality of Modes The modes corresponding to any two distinct characteristic values (natural frequencies) are orthogonal both with respect to the inertia matrix and the stiffness matrix. Properties of Natural Modes and Frequencies Let and be two distinct natural frequencies and and be the associated modes. The first orthogonality condition is, (157) (158) and the second is,

If [M] is diagonal, as is usually the case, Eq.157 can be written as In which m i is the concentrated mass at the i th coordinate and and are respectively the amplitudes of the i th coordinate when the system vibrates in the p th and s th natural mode. (159) Proof 1: The modes and must satisfy (160) (161)

Next we take the transpose of the two members of Eq.160, obtaining (162) and premultiply the latter equation by . This yields (163) Finally we subtract Eq.163 from 162 and obtain (164) Since p s has been considered to be different from p r , Eq.164 requires the validity of Eq.157. Natural modes corresponding to repeated natural frequencies are orthogonal to all other modes, but are not necessarily orthogonal to each other. premultiply the equation 161 by

However, by appropriate combination of the non-orthogonal modes it is always possible to construct orthogonal points. Therefore, all modes, including those corresponding to repeated characteristic roots, may be considered to be orthogonal in the sense of Eq.157. Give example. Having established that Eq.157 is valid, Eq.158 may be proved by reference to Eq.162 and Eq.163, Since the right-hand member of either of these equalities is zero, the left-hand member must also be zero. It is assumed that that the frequency p r or p s is not equal to zero. Proof 2: Based on structural principles, inertia forces acting on the system while the system is vibrating in the mode: (a)

The corresponding forces for motion in the mode are, (b) Now from Betti’s work principle, the work done by the forces (a) acting through (the displacements produced by (b)) is equal to the work done by the forces (b) acting through , i.e. the displacements produced by (a). The first work quantity is and the second work quantity is hence, which is the same as of Eq 164.

2. Independence of modes The natural modes are linearly independent, in the sense that some mode cannot be expressed as a linear combination of all the other modes. Expressed differently, the equation is satisfied if and only if all the coefficients are zero. Proof Pre-multiplying Eq. 165 by we obtain hence (165) (164 )

But since [m] is positive definite, the triple product cannot be zero and hence C r must be zero. By a similar reasoning it is concluded that all the other coefficients must also vanish. An important consequence of this property of independence is that any n-dimensional vector can be expressed as a linear combination of n characteristics vectors. 3. Reality and Positiveness of Natural Frequency The natural frequency are real and positive quantities. Proof of Reality We start by assuming that the frequencies are not real and show that this assumption leads to a contradictions Assume that (165 )

Since both [k] and [m] are real, it follows that the associated natural modes must also be complex. i.e. Further, the complex conjugates of p r and {  } r must also represent a solution. Letting, (166 ) (167) (168 ) (169)

Now it can be verified that the triple product of the matrices in this equation is a real quantity, and since [m] is positive definite, the product is a non-zero positive number. It follows that  must be zero. Hence p r cannot be complex as assumed in Eqn. 165 but must be real. Proof of Positiveness: Pre-multiplying Therefore, P r is either zero or positive . However, if system cannot undergo rigid body motion, the left-handed member of the equation is non-zero and positive. Hence P r is non-zero and positive is used such cases. To summarize, provided the system is stable and cannot undergo rigid body motion, the natural frequencies are non-zero positive quantities (170)

Solution for General Forced Vibration of MDOF systems in Time Domain a)Direct Integration method b)Modal Superposition method Modal superposition method: Both mass matrix,[m] and Stiffness method, [k] can be simultaneously made into diagonal matrices by proper choice of coordinates(known as modal coordinates) Transformation of coordinates results in a set of uncoupled SDOF equations in terms of modal coordinates Useful for many practical problems where the response can be approximated very well by using few eigen modes of the problem

Equations of motion: Modal Superposition Method… b) Find mode shape and natural frequencies by solving Eigen Value Problem, obtain for i=1,2,…..n where, = i th mode shape vector c) Express (2) (1) Substitute eqn (2) in (1) and then premultiply by and using Orthogonality Properties, one obtains for j=1,2,……..n (3)

Let {  i } T [C] {  j } = 0 for i  j then [C] is said to be classically damped matrix Special case satisfying orthogonality , [C] =  [M] +  [K] = Rayleigh Damping “ FOSS” condition [C] [M] -1 [K] = [K] [M] -1 [C] Then division by {  i } T [M] {  j } of Equation (3) leads to , For j=1,2,..n Where  j = j th modal damping ratio, c j = max Participation factor (4)

(d) Express initial conditions By pre multiplying above expression with {  j } T [m] and using orthogonality of modes, a j ,b j - participation factors

(e) Solve q i (t) of the uncoupled SDOF systems (f) Express

In general take first ‘r’ modes only Elastic forces: Had the load been static , {f s (t) = {P(t)} and (IAF) j = 1 i.e Implementation issues

Considering static force correction for the neglected modes r+1, ….n

From Equation (4) Mode Acceleration Method

Equations of motion are obtained by application of procedure described earlier which may be summarized as follows: 1. Consider system in a deflected configuration and identify all forces acting on in, including the d’Alembert inertia forces 2. Draw force body diagram for the individual masses and write the equations expressing the equilibrium of the forces Response of a MDOF System - An Example m 1 m 2 m 3 k 1 k 2 k 3 x 1 x 2 x 3 P 3 (t) P 2 (t) P 1 (t) k 3 ( x 3 -x 2 ) k 2 (x 2 -x 1 ) k 1 x 1

Application of this procedure leads to the following, For Mass 1 For Mass 2 For Mass 3 Response of a MDOF System Or (a) (b) (c) (d)

Consider special case in which, k 1 =k 2 =k 3 and m 1 =m 2 =m and m 3 = ½ m (132) (133) Natural Frequencies and Modes

Let we get For a non-trivial solution Expanding, we get the frequency equation (134)

which, upon factoring the term , may be rewritten as The roots are Hence natural frequencies are (135)

From the first of these equations, With the roots (natural frequencies) evaluated, the natural modes are determined from above as follows: Normalize by taking X 1 =1, then

Free Vibration Write expression for the motion of the system due to an initial displacement Without any initial velocity where,

Free Vibration In this case Consider next, the displacement induced by an excitation of the second mass, that is an initial velocity without an initial displacement Where,

Forced Vibration Consider first the effect of a force P 2 (t)=P 2 f(t) applied on the second mass Noting that: P is given as: Making use of the values of obtained earlier we obtain,

Forced Vibration Maximum values of (I.A.F) j are obtained from the response spectrum applicable to those particular forcing functions considered

Discuss relative importance of higher models. Note that it depends on response gravity under consideration and on the relative values of (A.F) 3 and (A.F.) 1 . The latter depend, in turn on the value of f 1 t 1 and on the ratio of f 3 /f 1 . Finally, note that the participation factors are functions of the spatial distribution of the forcing functions. If the force were applied at the first rather than the second floor level, the expression for the response would be as follows:

Finally if the system were subjected to a set of forces Hence,

The participation factors would be,

Note: The contribution of the higher modes is significantly less in this phase than when the system is excited by a single force on the first floor

Analysis of n-DF Damped Systems In the following analysis of the response of damped systems, damping will be assumed to be viscous. The damping forces, {F d }, are then a linear function of the velocities of the coordinates, and can be expressed in the form, (204) In which [c] is the damping matrix of the system. The element c ij corresponding to the i th row and the j th column represents the force at node i induced by a unit velocity of mode j. The equations of motion of the system can then be written as, or, in terms of the flexibility of the matrix, [d], as (206) (205)

The form of the matrix [C] is obviously a function of the distribution of the damping resistance throughout the structure. However, unlike the elements of [M], [K] and [D] which can be determined from the basic properties of the structural elements, those of [C] can not be so evaluated, as the relative magnitudes and distributions of the various factors which contribute to the damping of the structure are generally not known. Factors contributing to the overall damping of the system include, the ‘resistance of the fluid’ (air or water) in which the structure is oscillating, energy dissipation due to the straining of the elements composing the structure, losses at the faying surfaces of the cracked sections, friction at joints and supports, and energy absorption in components or elements that may be superimposed on the main structure.

The difficulties involved may be illustrated by reference to the 3DF of the shear-beam type considered earlier. It is clear that the damping resistance of the system in this, as in any other case can be modeled by a series of dashpots, but how these should be attached to the various floors is not clear.

(a) External or absolute damping (b) Internal or relative damping (c) Combination of External and Internal damping For example, if we consider a set of three dashpots and attach one end of each to a separate floor and the other to the ground, we obtain the arrangement shown in Fig.(a). Alternatively, if we attach the ends to consecutive floors, we obtain the arrangement shown in Fig.(b) m 2 m 1 m 3 c 1 c 2 c 3 k 1 k 2 k 3 c 1 c 2 c 3 c 1 c 2 c 3

For case(a), the damping forces would be of the form (207) and the damping matrix would be of the same form as the mass matrix. Furthermore, if the ratio (c j / m j ) for all the elements were the same, the matrix [c] would be proportional to the mass matrix [m]. For case(b), the damping forces would be of the form, (208) and the damping matrix would be of the same form as the stiffness matrix, [k]. Furthermore, if the ratio c j /m j for all the dashpots and springs were the same, the matrix [c] would be proportional to the mass matrix [k].

In general, the form of [c] may be different from either [m] or [k]. An arrangement of dashpots leading to such a matrix is shown in Fig.(c). If the damping in a structure is due exclusively to the resistance of the fluid in which it is oscillating, it would be reasonable to consider [c] to be of the same form as [m]. Similarly, if damping is exclusively due to energy dissipation due to the straining of the members, [c] can be considered to be of the same form as [k]. A mass-proportional or stiffness-proportional [c] are special cases of the class referred to above.

Proof: As before, the solution is expressed in the form of Eq.189, i.e., as a linear combination of the natural modes of vibration of the associated undamped systems . Furthermore, we limit ourselves to forces of the form of Eq.179. Substituting Eqs.189 and 179 into Eq.205, and following the steps used in the analysis of elastic systems, we obtain the following equations, which is the counterpart of Eq.192. (210) We shall now investigate if the modal superposition method which we have used for the analysis of undamped systems can also be used for damped systems. It will be shown that this is indeed possible provided the damping matrix, [c], is of a form that satisfies the orthogonality relationship , (209)

Now, if it is assumed that Eq.209 is valid, namely, that pairs of natural modes are also orthogonal with respect to the damping matrix, then all but the r th term of the series in the second term of Eq.210 vanish, and Eq.210 simplifies to, (211) or (212) In which , and are defined by Eqs.196 to 199, respectively, and the effective damping for the r th mode, defined as , (213) Eq.211 , or its equivalent Eq.212, permits the values of q r (t) to be evaluated independently of each other. In other words, the use of q r (t) uncouples the governing equations of motion, as for undamped systems.

If Eq.211 is now divided by the coefficient of the acceleration term, the resulting coefficient of the displacement term is equal to the square of the r th circular natural frequency of the system, p p 2 (see Eq.170), and the coefficient of the time function on the right hand member of the equation becomes p p 2 c r , in which c r is the participation factor defined by Eq.191. If we also let, (214) we obtain (215)

In which  r represents the percent of the critical coefficient of damping for motion in the r th natural mode. Finally, if we compare this equation with the equation of motion of a SDF damped system, (216) And recall that the solution of this equation can be expressed symbolically as (217) in which I.A.F. is the Instantaneous Amplification Factor for the particular damped system and excitation considered. We conclude that q r (t) can also be stated as (218)

The time function (I.A.F.) r is the I.A.F. for a SDF system which has a natural frequency, p r , a damping factor,  r , and is subjected to a forcing function, the time variation of which is defined by f(t). Substituting Eq.218 into Eq.189, we obtain again Eq.188. It should be emphasized, however that the (I.A.F.) r in this equation refer to damped rather than undamped systems. The other quantities are the same as for the associated undamped systems.

Rayleigh Damping If [c] is proportional to either [M] or [K], or is a linear combination of [m] and [k] as follows [c] =  [M] +  [K] In which  and  are constants, it is clear that [c] will satisfy the orthogonality relation defined by Eq.209. Indeed, substituting Eq.219 into Eq.209 we obtain, (219) which is zero because each of the triple products is zero. Damping of the form of Equation 21a is known as Rayleigh damping ,after Lord Rayleigh who was the first to demonstrate the applicability of the modal superposition method of analysis in this case.

36 Substitution of Equation 21a we find that, (220) Note: If the damping is mass proportional ,  =0,and the damping factor for a mode is inversely proportional to the frequency of the mode, If the damping matrix is stiffness proportional ,the damping factor ,  p is proportional to the natural frequency of the mode under consideration. Implications of Rayleigh Damping

For a given excitation ,it follows that the higher modes will be damped significant more in the second case than the first (see the sketch). For example ,if  1 for a system is ,say,0.03 and the first three natural frequencies are in the ratio of 1:3:5 ,the assumption of a mass proportional damping will lead to effective damping factors for the second and third modes of 0.01 and 0.03  (1/5) = 0.006, respectively. By contrast, the assumption of stiffness proportional damping will lead to effective damping factors of 0.09 and 0.15 for the second and third modes, respectively. for [c] =  [M] for [c] =  [K]  p p p

Unless the contribution of these two modes to the overall response of the system is significant, the response values computed in the two cases will be substantially different. It is important, therefore, that great care be exercised in the selection of the form of the damping matrix. For a two degree of freedom system with arbitrary modal damping factors,  1 and  2 ,it is always possible to determine the damping matrix of the Rayleigh type. Indeed, satisfying Eqn. 220 for p=1 and p=2 ,we obtain, (221)

which, upon solving simultaneously, yields (222) In the special case in which  2 =  1 (p 1 / p 2 ) ,  =0 ,indicating that [c] =  [M] , and  = 2p 1  1 similarly ,when  2 =  1 (p 2 / p 1 ) ,  = 0 (i.e [c] =  [K]) , and  =2  1 /p 1. Fore systems with more than two degree of freedom for which the modal damping factors are arbitrary, it is generally not possible to obtain a matrix [c] of the Rayleigh type.

Mdof

More Related Content

What's hot (20)

Similar to Mdof (20)

More from Teja Ande (20)

Recently uploaded (20)

Mdof