Multiple Degree of Freedom Systems The Millennium bridge required many degrees of freedom to model and design with.
The first step in analyzing multiple degrees of freedom (DOF) is to look at DOF DOF: Minimum number of coordinates to specify the position of a system Many systems have more than DOF Examples of DOF systems car with sprung and unsprung mass (both heave) elastic pendulum (radial and angular) motions of a ship (roll and pitch) D.J. Inman /58 Mechanical Engineering at Virginia Tech
4. Two-Degree-of-Freedom Model (Undamped) A degree of freedom system used to base much of the analysis and conceptual development of MDOF systems on.
Free-Body Diagram of each mass Figure 4. k (x -x ) k x m m x x
Summing forces yields the equations of motion: m x ( t) k x ( t) k x ( t) x ( t) m x ( t) k x ( t) x ( t) (4.) Rearranging terms: m x ( t) ( k k ) x ( t) k x ( t) 0 m x ( t) k x ( t) k x ( t) 0 (4.)
Note that it is always the case that A Degree-of-Freedom system has Two equations of motion! Two natural frequencies (as we shall see)!
The dynamics of a DOF system consists of homogeneous and coupled equations Free vibrations, so homogeneous eqs. Equations are coupled: Both have x and x. If only one mass moves, the other follows Example: pitch and heave of a car model In this case the coupling is due to k. Mathematically and Physically If k = 0, no coupling occurs and can be solved as two independent SDOF systems
Initial Conditions Two coupled, second -order, ordinary differential equations with constant coefficients Needs 4 constants of integration to solve Thus 4 initial conditions on positions and velocities x (0) x, x (0) x, x (0) x, x (0) x 0 0 0 0
Solution by Matrix Methods The two equations can be written in the form of a single matrix equation (see pages 7-75 if matrices and vectors are a struggle for you) : x( t) x( t) x( t) x( t ), ( t ), ( t) x ( t) x x ( t) x x ( t) (4.4), (4.5) m 0 k k k M, K 0 m k k (4.6), (4.9) Mx Kx 0 m x m x ( t) ( k ( t) k k x ) x ( t) k ( t) k x x ( t) 0 ( t) 0
Initial Conditions IC s can also be written in vector form x x (0), and x(0) x x 0 0 x 0 0
The approach to a Solution: For DOF we assumed the scalar solution ae lt Similarly, now we assume the vector form: Let x( t) ue jt (4.5) j, u 0,, u unknown - M K ue 0 jt - M K u 0 (4.6) (4.7)
This changes the differential equation of motion into algebraic vector equation: - M K u0 (4.7) This is two algebraic equation in 3 uknowns ( vector of two elements and scalar): u u u =, and
The condition for solution of this matrix equation requires that the the matrix inverse does not exist: If the inv - M K exists u 0: which is the static equilibrium position. For motion to occur u 0 - M K does not exist or det - M K 0 (4.9) The determinant results in equation in one unknown (called the characteristic equation)
Back to our specific system: the characteristic equation is defined as det- M K 0 det m k k k k 0 m k (4.0) m m 4 (m k m k m k ) k k 0 (4.) Eq. (4.) is quadratic in so four solutions result: and and
Once is known, use equation (4.7) again to calculate the corresponding vectors u and u This yields vector equation for each squared frequency: and ( M K) u 0 (4.) ( M K) u 0 (4.3) Each of these matrix equations represents equations in the unknowns components of the vector, but the coefficient matrix is singular so each matrix equation results in only independent equation. The following examples clarify this.
Examples 4..5 & 4..6:calculating u and m =9 kg,m =kg, k =4 N/m and k =3 N/m The characteristic equation becomes 4-6 +8=( -)( -4)=0 = and =4 or,3 rad/s,,4 rad/s Each value of yields an expression or u:
Computing the vectors u For =, denote u then we have u (- M K) u 0 u 7 9() 3 u 0 3 3 () u 0 9u 3u 0 and 3u u 0 equations, unknowns but DEPENDENT! (the nd equation is -3 times the first)
Only the direction of vectors u can be determined, not the magnitude as it remains arbitrary u u u u 3 3 results from both equations: only the direction, not the magnitude can be determined! This is because: det( M K) 0. The magnitude of the vector is arbitrary. To see this suppose that u satisfies ( M K) u 0, so does au, a arbitrary. So ( M K) au 0 ( M K) u 0
Likewise for the second value of : For = 4, let u then we have u (- M K) u 0 u 7 9(4) 3 u 0 3 3 (4) u 0 9u 3u 0 or u u 3 Note that the other equation is the same
What to do about the magnitude! Several possibilities, here we just fix one element: Choose: Choose: u u u u 3 3
Thus the solution to the algebraic matrix equation is:,3, has mode shape u 3,4, has mode shape u Here we have introduce the name mode shape to describe the vectors u and u. The origin of this name comes later 3
Return now to the time response: We have computed four solutions: x( t) u e, u e, u e, u e jt jt jt jt Since linear, we can combine as: x() t au e bu e cu e du e jt jt jt jt j t jt jt jt x() t ae be u ce de u A sin( t ) u A sin( t ) u (4.4) (4.6) where A, A,, and are constants of integration determined by initial conditions. Note that to go from the exponential form to to sine requires Euler s formula for trig functions and uses up the +/- sign on omega
Physical interpretation of all that math! Each of the TWO masses is oscillating at TWO natural frequencies and The relative magnitude of each sine term, and hence of the magnitude of oscillation of m and m is determined by the value of A u and A u The vectors u and u are called mode shapes because the describe the relative magnitude of oscillation between the two masses
What is a mode shape? First note that A, A, and are determined by the initial conditions Choose them so that A = = =0 Then: x(t) x (t) x (t) A Thus each mass oscillates at (one) frequency with magnitudes proportional to u the st mode shape u u sin t A u sin t
A graphic look at mode shapes: If IC s correspond to mode or, then the response is purely in mode or mode. Mode : x k m x =A/3 k x x =A m u 3 Mode : x k m x =-A/3 k x x =A m u 3
Example 4..7 given the initial conditions compute the time response cos cos cos 3 cos 3 ) ( ) ( sin sin sin 3 sin 3 ) ( ) ( 0 0 (0) mm, 0 = (0) consider t A t A t A t A t x t x t A t A t A t A t x t x x x
cos cos cos 3 cos 3 0 0 sin sin sin 3 sin 3 0 mm A A A A A A A A At t=0 we have
4 equations in 4 unknowns: 3 A sin A sin 0 A sin A sin 0 A cos A cos 0 A cos A cos Yields: A.5 mm, A.5 mm, rad
The final solution is: x (t) 0.5cos t 0.5cost x (t).5cos t.5cost (4.34) These initial conditions gives a response that is a combination of modes. Both harmonic, but their summation is not. Figure 4.3
Solution as a sum of modes x(t) a u cos t a u cos t Determines how the first frequency contributes to the response Determines how the second frequency contributes to the response
Things to note Two degrees of freedom implies two natural frequencies Each mass oscillates at with these two frequencies present in the response and beats could result Frequencies are not those of two component systems k m.63, k m.73 The above is not the most efficient way to calculate frequencies as the following describes
Some matrix and vector reminders A T xx a b d b c c ad cb c a A x x m 0 M x Mx m x m x 0 T m T M 0 x Mx 0 for every value of x except 0 Then M is said to be positive definite
4. Eigenvalues and Natural Frequencies Can connect the vibration problem with the algebraic eigenvalue problem developed in math This will give us some powerful computational skills And some powerful theory All the codes have eigen-solvers so these painful calculations can be automated
Some matrix results to help us use available computational tools: A matrix M is defined to be symmetric if M M T A symmetric matrix M is positive definite if x T Mx 0 for all nonzero vectors x A symmetric positive definite matrix M can be factored M LL T Here L is upper triangular, called a Cholesky matrix
If the matrix L is diagonal, it defines the matrix square root The matrix square root is the matrix M / such that M / M / M If M is diagonal, then the matrix square root is just the root of the diagonal elements: m L M / 0 (4.35) 0 m
A change of coordinates is introduced to capitalize on existing mathematics For a symmetric, positive definite matrix M: m 0 m 0 M, M, M / 0 0 m m 0 m / / Let x( t) M q( t) and multiply by M : / / / / M MM t M KM t q( ) q( ) 0 (4.38) I identity K symmetric or q( t) Kq( t) 0 where K M KM / / k K is called the mass normalized stiffness and is similar to the scalar m used extensively in single degree of freedom analysis. The key here is that K is a SYMMETRIC matrix allowing the use of many nice properties and computational tools m 0
How the vibration problem relates to the real symmetric eigenvalue problem Assume q( t) ve in q( t) Kq( t) 0 jt ve Kve 0, v 0 or jt jt Kv v Kv lv v 0 vibration problem real symmetric eigenvalue problem (4.40) (4.4) Note that the martrix as does n K contains the same type of information in the single degree of freedom case.
Important Properties of the n x n Real Symmetric Eigenvalue Problem There are n eigenvalues and they are all real valued There are n eigenvectors and they are all real valued The set of eigenvectors are orthogonal The set of eigenvectors are linearly independent The matrix is similar to a diagonal matrix
Square Matrix Review Let a ik be the ik th element of A then A is symmetric if a ik = a ki denoted A T =A A is positive definite if x T Ax > 0 for all nonzero x (also implies each l i > 0) The stiffness matrix is usually symmetric and positive semi definite (could have a zero eigenvalue) The mass matrix is positive definite and symmetric (and so far, its diagonal)
Normal and orthogonal vectors x x M, y x n y M y n n, inner product is x T y x i y i i x orthogonal to y if x T y 0 x is normal if x T x if a the set of vectores is is both orthogonal and normal it is called an orthonormal set The norm of x is x x T x (4.43)
Normalizing any vector can be done by dividing it by its norm: x x T x has norm of (4.44) To see this compute x x T x T x x T x x x T x x T x x T x
Examples 4.. through 4..4 / / 3 0 7 3 3 0 K M KM 0 3 3 0 3 so K which is symmetric. 3 3-l - - 3-l det( K li) det l 6l 8 0 which has roots: l and l 4
( K l I) v 0 3 v 0 3 v 0 v v v v 0 v ( ) The first normalized eigenvector
Likewise the second normalized eigenvector is computed and shown to be orthogonal to the first, so that the set is orthonormal v, v T v ( ) 0 v T v ( ) v T v ( ()()) v i are orthonormal
Modes u and Eigenvectors v are different but related: u v and u v x M / q u M / v (4.37) Note M / u 3 0 3 0 v
This orthonormal set of vectors is used to form an Orthogonal Matrix P T P P v v called a matrix of eigenvectors (normalized) T T v v v 0 v I T T 0 v v v v P is called an orthogonal matrix P KP P Kv Kv P lv l v T T T T T lv v lv v l 0 diag(, ) T T l 0 l (4.47) v v lv v P is also called a modal matrix
Example 4..3 compute P and show that it is an orthogonal matrix From the previous example: P v v P T P 0 0 I
Example 4..4 Compute the square of the frequencies by matrix manipulation T P KP 3 3 4 4 4 0 0 0 0 8 0 4 0 rad/s and rad/s In general: T P KP diag l diag( ) (4.48) i i
Example 4..5 The equations of motion: m x ( k k ) x k x 0 m x k x ( k k ) x 0 3 (4.49) Figure 4.4 In matrix form these become: m 0 0 m x k k k k k k 3 x 0 (4.50)
Next substitute numerical values and compute P and m kg, m 4 kg, k k 3 0 N/m and k = N/m 0 M, K 0 4 / / K M KM det K li det l 5l 35 0 l l l.890 and l.098.7 rad/s and.098 rad/s
Next compute the eigenvectors For l equation (4.4 ) becomes: Normalizing -.890 v 3-.890 v 9.089v yields v v v v (9.089) v v v v v 0. 09, and v 0.9940 0.09 0.9940, likewise 0.9940 v 0.09 0
Next check the value of P to see if it behaves as its suppose to: P 0.09 0.9940 v v 0.9940 0.09 T P KP 0.09 0.9940 0.09 0.9940.840 0 0.9940 0.09 3 0.9940 0.09 0.098 T PP 0.09 0.99400.09 0.9940 0 0.9940 0.09 0.9940 0.09 0 Yes!
A note on eigenvectors In the previous section, we could have chosed v to be v 0.9940 0.09 instead of v -0.9940 0.09 because one can always multiple an eigenvector by a constant and if the constant is - the result is still a normalized vector. Does this make any difference? No! Try it in the previous example
All of the previous examples can and should be solved by hand to learn the methods However, they can also be solved on calculators with matrix functions and with the codes listed in the last section In fact, for more then two DOF one must use a code to solve for the natural frequencies and mode shapes. Next we examine 3 other formulations for solving for modal data
Matlab commands To compute the inverse of the square matrix A: inv(a) or use A\eye(n) where n is the size of the matrix [P,D]=eig(A) computes the eigenvalues and normalized eigenvectors (watch the order). Stores them in the eigenvector matrix P and the diagonal matrix D (D=)
More commands To compute the matrix square root use sqrtm(a) To compute the Cholesky factor: L= chol(m) To compute the norm: norm(x) To compute the determinant det(a) To enter a matrix: K=[7-3;-3 3]; M=[9 0;0 ]; To multiply: K*inv(chol(M))
An alternate approach to normalizing mode shapes From equation (4.7) M K u 0, u 0 Now scale the mode shapes by computing such that i u i T M i u i i u i T u i w u i i i is called mass normalized and it satisfies: Mw Kw 0 w Kw, i, T i i i i i i (4.53)
There are 3 approaches to computing mode shapes and frequencies (i) Mu Ku (ii) u M Ku (iii) v M KM v (i) Is the Generalized Symmetric Eigenvalue Problem easy for hand computations, inefficient for computers (ii) Is the Asymmetric Eigenvalue Problem very expensive computationally (iii) Is the Symmetric Eigenvalue Problem the cheapest computationally