ME 440 Intermediate Vibrations

1. ME 440Intermediate Vibrations Tu, April 28, 2009 Chapter 6: Multi-degree of Freedom (MDOF) Systems ~ The Lumped-Mass Approach ~ http://sbel.wisc.edu/Courses/ME440/2009/index.htm Quote of the Day: An intellectual is a man who takes more words than necessary to tell more than he knows. Dwight D. Eisenhower © Dan Negrut, 2009ME440, UW-Madison

2. Before we get started… • Last Time: • Reviewed approach to obtain the response of forced undamped and underdamped MDOF systems • Covered two examples (falling beam, and then forced longitudinal vibration) • Today: • Finish third example of last time • Lumped mass approach for finding the time evolution of distributed-mass systems • Beam (transverse vibration) • Rod (longitudinal vibration) • HW assigned (due May 5): • 6.74 – forced vibration, three DOF undamped system • 6.80 – bike hitting curb, leads to step function; find response • A look ahead: • On Th and Tu cover Lagrange’s equations • Comprehensive exam: on Th of next week (during regular class hours) • Review session: Wd, May 6th, at 7:15 PM (we’ll meet in this room, 3126ME) 2

3. [AO]Example: Force Response, Undamped System 3

4. [Recall This]Getting Modal Matrix Based on [a] • The characteristic equation: • In other words, first solve for  the equation… • … and then compute the natural frequencies as • Finally, compute the eigenvectors (the modal vectors) that together will make up the modal matrix [u] 4

5. [Cntd]Example • Recall discussion last time: • Instead of getting [k], much easier for beams to determine flexibility matrix [a] • Like last time, use formula for y(x): • Use symmetry of the flexibility matrix [a] to determine all entries based on four “measurements” of y(x) 5

6. I = 1674.7; E = 30e6; L = 40*12; forceVec = [0 ; 3E4; 0]; x = 10*12; b = L - x; a11 = sag(x, 1, E, I, L, b) b = L/2; a12 = sag(x, 1, E, I, L, b) b = 10*12; a13 = sag(x, 1, E, I, L, b) b = L/2; x = L/2; a22 = sag(x, 1, E, I, L, b) function [s]= sag(x, P, E, I, L, b) s = P*b*x; s = s/(6*E*I*L); dummy = L*L - x*x - b*b; s = s*dummy; aMat = [ a11 a12 a13 ; a12 a22 a12; a13 a12 a11] mMat = diag([2.487 ; 2.487 ; 2.487]) [uMat, invOmegaMat] = eig((aMat*mMat)) omega3 = 1/sqrt(invOmegaMat(1,1)) omega2 = 1/sqrt(invOmegaMat(2,2)) omega1 = 1/sqrt(invOmegaMat(3,3)) uMat = [uMat(:,3)/uMat(1,3) uMat(:,2)/uMat(1,2) uMat(:,1)/uMat(1,1)] M_Mat = uMat'*mMat*uMat F_Vec = uMat'*forceVec fHat_Vec = M_Mat\F_Vec; 6

7. [New Topic]Lumped-Mass Modeling of Beams(Transverse Vibration) • What is the general idea? • First, identify points on the rod/beam that are fixed • The amount of mass associated with these points is neglected (this mass doesn’t move, after all…) • Recall beam example we just discussed today • Next example is similar as well… • Relevant as well are masses at the free *end[s]* of the beam • They don’t account for full mass because there is a lack of material on one side… 7

8. [Cntd]Lumped-Mass Modeling of Beams(Transverse Vibration) • Go back to problem at hand: • Given to you: • Need n(i) and [u] • First get n(i). The characteristic equation: 8

9. [Short Detour]Computing the Flexibility Matrix [a] • We are dealing with a fixed-pinned beam with overhang • See below but also consult the handout Disp. at Blue: Disp. at Green (x · l): - not needed anymore (symmetry!) Disp. at Blue: Disp. at Green (x · a): 9

10. [Cntd]Lumped-Mass Modeling of Beams(Transverse Vibration) • Problem at hand: • Mass and flexibility matrices obtained as • End up with the following natural frequencies and modal matrix: 10

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13. [New Topic]Lumped-Mass Modeling of Rods(Axial Vibration) • As the number of “properly selected” lumped masses used to represent the system increases, the accuracy of the frequencies and mode shapes increases as well • What does it mean “properly selected”? Use example in figure at left and below • Maybe counterintuitive in the beginning: • If you use “n” lumped masses, don’t just divide the rod in n equal masses • Rather, first acknowledge that the mass at the right end of rod sees only half as much material as a lumped mass somewhere in the middle of the rod • Also, note that m/(2n) is associated with the left end (lumped mass which doesn’t move) • This effectively means lumped masses add up to 13

14. The “good” model: [Computing k]Lumped-Mass Modeling of Rods(Axial Vibration) • Uniform rod, cross-sectional area A, mass density is , length l • Equivalent spring induced by presence of short rod element in between two consecutive lumped masses • From strength of materials: • The “other” model: 14

15. [AO]Example: Using *Four* Lumped Masses…(Axial Vibration) • Uniform rod, cross-sectional area A, mass density , length l • Modulus of elasticity: E • Use four lumped masses • Compare the “good” model against the “other” model • Incidentally, according to the analytical solution of the wave equation, the natural frequencies are 15

16. [AO]Example: Using *Four* Lumped Masses…(Axial Vibration) • The good case: • The other case: 16

17. [Cntd.]Using Four Lumped Masses…(Axial Vibration) • Notation: • Note: for a five lumped mass model, the error for predicting Mode 1 in the “good” model is 0.4 % 17