ME 563 Mechanical Vibrations Lecture #12 Multiple Degree of Freedom Free Response + MATLABFree Response 1"We can solve for the homogeneous solution to a coupled set of equations in a multiple degree of freedom linear system by: - Identifying the initial conditions on all the states - Assuming a solution of the form {x(t)}={A}est What does this last assumption imply about the response?Two DOF System 2"Consider the two degree of freedom system of equations: If we make a solution of the form, , as we did for the single DOF case, we obtain: Non-trivial solutions satisfy: € Ms2+ 2Cs + 2K( )⋅ Ms2+ Cs + K( )− Cs + K( )2= 0Characteristic Equation 3"There are four solutions that satisfy the characteristic equation and these solutions are expressed as follows when the modal frequencies are underdamped: The solution to the homogeneous equation is then written as follows (for the first two complex conjugate roots): s1=σ1+jω1, s2=σ1-jω1, s3=σ2+jω2, and s4=σ2-jω2 Real part: decay rate Imaginary part: oscillation frequencyFree Response Form 4"The free response is usually written in the following form for a multiple degree of freedom system: Four constants  Four initial conditions are required. Modal vector (can be scaled) Decaying co-sinusoid (common to both degrees of freedom)Example Solution 5"For K=1 N/m, C=0.1 Ns/m, and M=1 kg, the solution is given by:MATLAB Solution 6"To obtain solutions for the free response in MATLAB, the procedure explained before for a single DOF system is used: OrMATLAB Solution 7"We use the following formulation to define the outputs, {y}, if we are only interested in the displacement variables: Or: € y{ }2×1= C[ ]2×4q{ }4×1+ D[ ]2×2f (t){ }2×1Free Response (Eigen-Analysis) 8"We can also solve the homogeneous equations of motion by: - Identifying the initial conditions on all the states - Identifying the modal frequencies, s, and vectors, {X}, using eigen-analysis. Do you remember what an eigenvalue problem looks like?Eigenvalue Problem 9"Consider the case when we have no damping: We can write this set of equations of motion as follows: Now assume a solution of the form {x(t)}={A}estEigenvalue Problem 10"Consider the case when we have no damping: This last equation is the standard eigenvalue problem. Matlab solves this equation with the command “eig(A)”: [v,d]= eig(A);Matlab eig 11"How do we interpret what Matlab gives us with this command? is a diagonal matrix of -s1,22 is a matrix of modal vectors How do we find the modal frequencies and vectors? d v d = 2.6180 0 0 0.3820 v = -0.8507 -0.5257 0.5257 -0.8507Matlab eig 12"First, we take the negative square root of the diagonal entries of the d matrix providing the modal frequencies in rad/s: Second, we can scale the columns of v to produce easy to interpret modal vectors: √-d = 0+1.6180i 0 0 0+0.6181i v = 1.0000 0.6180 -0.6180 1.0000Matlab Form of Response 13"The form of the free response can then be written in the form: In summary, the eigenvalue problem for an undamped system is given by: Where does the damping fit?Damped Eigenvalue Problem 14"To obtain solutions for the free response in a damped system, the state variable form of the equations of motion are used: and then the eigenvalues and eigenvectors of the state matrix are calculated using eig. This approach works because the assumed solution {q}est is also used for the 1st order system: [v,d]= eig(A);Matlab eig 15"How do we interpret what Matlab gives us with this command in the first order form of the state variable model? is a diagonal matrix of modal frequencies s1,2 and s3,4 (1:2,1:4) is a matrix of modal vectors {X}1 and {X*}1 d v diag(d) = -0.1309 + 1.6127i -0.1309 - 1.6127i -0.0191 + 0.6177i -0.0191 - 0.6177i s3,4 s1,2 {X}2 and {X*}2Matlab eig 15"Eigenvectors are not the modal vectors of the 2nd order system – eigenvectors involve both position and velocity states. We look to the position states to obtain the scaled modal vectors. v = [ -0.0362 - 0.4457i -0.0362 + 0.4457i 0.0224 + 0.2755i 0.0224 - 0.2755i 0.7236 0.7236 -0.4472 + 0.0000i -0.4472 - 0.0000i -0.4472 + 0.0000i -0.4472 - 0.0000i -0.7236 -0.7236 0.0085 - 0.2763i 0.0085 + 0.2763i 0.0138 - 0.4470i 0.0138 + 0.4470iDamping in MDOF Systems 16"Note that for the previous example, the damping matrix [C] corresponded to a proportionally viscously damped system: For this type of damping, note that the modal vectors were entirely real, i.e., the phase between DOFs was either 0 or 180o.Damping in MDOF Systems 16"What if the system is not proportionally damped? The modal vectors in this case are complex but the response is not: For this type of damping, note that the modal vectors are complex, so the phase between the DOFs is not 0 or 180o. What does this mean?Damping in MDOF Systems 17"What if the system is not proportionally damped? The modal vectors in this case are complex but the response is not: Real Imag Mode #2 Mode #1 2nd order EOMS cannot be uncoupled using real modal