Problem 1: A simple 3-dof shear-building model has the following equation: =

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Archibald Cobb
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1 MEEN 6 E: Normalization and Damped Response Fall 24 Problem : A simple -dof shear-building model has the following equation: m u k+ k2 k2 u p( t) m u k k k k + + u = p ( t) m u k k u p( t) where m = 4 lb-s 2 /in, m 2 = 2 lb-s 2 /in, m = 6 lb-s 2 /in, k = lb/in, k 2 = lb/in and k = 2 lb/in. he modal properties for this system are: = Φ λ = (a) Mass Orthonormalize the above mode Shapes and Find the corresponding frequencies in rad/sec. (b) If the system has initial conditions of: u ( t = ) = inches, u2 ( t = ) = -.5 inches and u ( t = ) = -.2 inches find the free-vibration response using modal superposition. Use the mode shapes from part (a). (c) In order to reduce the size of the problem, only mode shapes believed to contribute significantly to the response will be used. In this case, the response will be approximated by using only the first mode contribution. What is the resulting equation for the free-vibration response? (d) Compare the response found from parts (b) and (c). How accurate is the approximation? Problem 2: Now the system has % damping in the first mode, 5% damping in the second mode, and 2% damping in the third mode. (a) Find modal damping matrix with all the damping information given. Find the corresponding system damping matrix in the original coordinate system. (b) If the system has initial conditions of u ( t = ) =.5in, u ( t = ) =.75in, ( ) the free-vibration response using modal superposition 2 u t = = ; find (c) Now find the damping matrix using Rayleigh Damping and the damping given in the first and third modes only. What does the modal damping matrix look like? What happened to the damping in the second mode?

2 MEEN 6 E: Normalization and Damped Response Fall 24 Problem : phi = lambda = Part (a) Frequencies: omega = Scaling Constants: c = Mass-Orthonormalized Mode Shapes: Phi_star = For forces that can be described as proportional forces: P( t ) = s p ( t ) = M{ } u g ( t ) s Participaton Factors: the equation of motion for the i th mode due to base excitation can be written as: ({ φi} [ K]{ φi} ) ( φi s ) { q i} + q i = x g =Γ ixg φ M φ φ M φ ({ } [ ]{ }) { } { } { } i i { i} [ ]{ i} ( ) Where Γ i is the Seismic Participation Factor for the i th mode. his is a special parameter that: Implies the measure of the degree to which the i th mode participates in the response. Factor is NO independent on how the mode is normalized Factor also NO a DIREC measure of how mode contributes to the response. ake a look at participation factors calculated using both the original mode shapes (in problem statement) as well as the factors found using the mass-orthonormalized mode shapes found in part a. For the first mode, the equation when using the original mode shapes is:

3 MEEN 6 E: Normalization and Damped Response Fall 24 4 { } 2 ({ φi} { s} ) Γ = = = =.282 ({ φ} [ ]{ }) i M φ i { } Repeating for all modes, we get the following: Mode Symbol Original φ Orthonormalized φ Γ Γ Γ Strict Sum of Participation Factors Sum of Absolute Values Note that in one case the factors add up to one and in another they do not. ake a look at the ratio of contributions of each participation factor as a percent of the sum of the absolute values: Mode Symbol Original φ Orthonormalized φ Γ Γ Γ Notice that depending on your normalization, you would expect different modes to have the higher contributions to the seismic response. ypically, building codes will ask you to include modes to add up to at least 9% contribution. In this case, either method would result in using both modes and 2. Part (b) Inverse of Mass Orthonormalized Phi Matrix. iphi_star = Initial Conditions: Original Coordinate System Uo = Initial Conditions: Modal Coordinate System Qo =

4 MEEN 6 E: Normalization and Damped Response Fall 24 Response equations: Modal Coordinate System q(t) = cos(.277t) q2(t) = cos(.9464t) q(t) = -.64 cos(.648t) Response equations: Original Coordinate System u(t) = cos(.277t) +.75 cos(.9464t) cos(.648t) u2(t) = -.75 cos(.277t) cos(.9464t) cos(.648t) u(t) = cos(.277t) cos(.9464t) +.49 cos(.648t) Compare these equations with the resulting plots shown below. Response time histories are given in both Modal and Original Coordinate Systems Free Vibration: Modal Coordinates Modal Coordinate Modal Coordinate 2 Modal Coordinate Free Vibration: Original Coordinates Degree of Freedom Degree of Freedom 2 Degree of Freedom

5 MEEN 6 E: Normalization and Damped Response Fall 24 Part (c) Response Using Only Mode u(t) = cos(.277t) u2(t) = -.75 cos(.277t) u(t) = cos(.277t) 2 Free Vibration: Modal Coordinates Modal Coordinate Free Vibration Using Mode Only: Original Coordinates Degree of Freedom Degree of Freedom 2 Degree of Freedom You can clearly see that only frequency is present in the response of all degrees of freedom, and that comes from Mode.

6 MEEN 6 E: Normalization and Damped Response Fall 24 Part (d) Estimate the peak response of each dof using both ABSSUM and SRSS methods described in H2. Compare peak response from part (b) and part (c) Response Found Using All Modes: ABSSUM: Absolute Sum We assume the worst response in each mode all happen at the exact same time. So take a look at the response equations for each mode. he worst case happens when the COS term is equal to + or -. hen the resulting peak values (independent of sign) are given in the table below. Mode Response Equation Peak Value (Independ of Sign) q(t) = cos(.277t) q2(t) = cos(.9464t) 2.47 q(t) = -.64 cos(.648t).64 Looking at the response equation for Mode, we can clearly see the terms coming from each mode. u( t) = cos(.277 t) +.75 cos(.9464 t) cos(.648 t) Mode Contribution Mode 2 Contribution Mode Contribution Again, the worst case scenario would be for the COS terms to equal + or -. Imposing that condition and summing the absolute value of each modal term, we get: u ( t ) = =.454 Doing this for each degree of freedom, we can fill in the table with estimates of the peak value. SRSS: Square-Root of the Sum of the Squares his method is based on random vibration theory. While not guaranteed to be conservative, this method is generally conservative for systems with well spaced frequencies (which is the case for this example). o estimate the peak response, the worst case for each mode is squared, added together, and then we take the square root of the resulting sum. So for the first degree of freedom: ( ) ( ) ( ) u ( t ) = =.572 he results of the estimates using both methods are shown in the table below. he actual peak values are also given for comparison purposes. Degree of Freedom Actual Peak Value Using ABSSUM Using SRSS.652 inches.454 inches.572inches inches.45 inches.958 inches.2 inches.298 inches.88 inches

7 MEEN 6 E: Normalization and Damped Response Fall 24 Response Found Using Mode Only: Since only one mode is now used, the two modal combination strategies result in the same numerical answer for the peak estimates. Again, the peak values of the response for the actual system are given on this same table. Degree of Freedom Actual Peak Value Using ABSSUM Using SRSS.652 inches.227 inches.227 inches inches.75 inches.75 inches.2 inches.9674 inches.9674 inches In this case, using only mode does not result in good estimates of peak response. Problem 2: Part (a) Using Mass Orthonormalized Mode Shapes: Cm_Modal = C = Part (b) Initial conditions: Qo_b = Qdo = Damped Natural Frequencies: omega_d = Coeff. In Front of COS term: A_b = Coeff. In Front of SIN term: B_b =

8 MEEN 6 E: Normalization and Damped Response Fall 24 While explicit response equations are not provided, the resulting response is shown below Damped Free Vibration: Modal Coordinates Modal Coordinate Modal Coordinate 2 Modal Coordinate Damped Free Vibration: Original Coordinates Degree of Freedom Degree of Freedom 2 Degree of Freedom

9 MEEN 6 E: Normalization and Damped Response Fall 24 Part (c) Rayleigh Damping Recall that the coefficients for Rayleigh Damping can be found using the following: ω ω i i α ζ i 2 ω ω = j j α ζ j Substituting values from Modes and as required by the problem: α = α.2 α.48 = α.2 Now can substitute to solve for damping matrix in original coordinate system. m k+ k2 k2 C = α m2 α k2 k2 k k + + m k k Resulting in the following damping matrix: C = his matrix can now be converted to modal space. Since it is proportional to the mass and stiffness matrices, then the resulting matrix must also be diagonal: Cm = Note that this is NO the same modal damping matrix we had before. Knowing that the diagonal m elements must be: 2ζω i imi then we can determine the resulting damping ratios for all modes. In this example, we get zeta = We do get the exact damping ratios in modes and as specified. All other modes follow the trend described in the damping handout. As expected, the damping ratio for mode two is smaller than that of the two specified modes as it lies within the frequency range of the specified modes. his results in a MUCH smaller damping ratio than originally specified (5%).

Design of Structures for Earthquake Resistance

NATIONAL TECHNICAL UNIVERSITY OF ATHENS Design of Structures for Earthquake Resistance Basic principles Ioannis N. Psycharis Lecture 3 MDOF systems Equation of motion M u + C u + K u = M r x g(t) where: