1. ME 440 Intermediate Vibrations
April 26, 2017
ME 440 Intermediate Vibrations
Tu, April 28, 2009
Chapter 6: Multi-degree of Freedom (MDOF) Systems
~ The Lumped-Mass Approach ~
M1051 ECB 2, 2hr EES seminars:
2-4 pm on Jan 20
4-6 pm on Jan 20
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, 2009 ME440, UW-Madison

2. Before we get started… Last Time: Today: A look ahead: 2
April 26, 2017
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
April 26, 2017
[AO] Example: Force Response, Undamped System 3

4. [Recall This] Getting Modal Matrix Based on [a]
April 26, 2017
[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. Recall discussion last time:
April 26, 2017
[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. function [s]= sag(x, P, E, I, L, b) s = P*b*x; s = s/(6*E*I*L);
April 26, 2017
I = ;
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)
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])
[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)
April 26, 2017
[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)
April 26, 2017
[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]
April 26, 2017
[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 · a):
Disp. at Blue:
Disp. at Green (x · l):
- not needed anymore (symmetry!) 9

10. [Cntd] Lumped-Mass Modeling of Beams (Transverse Vibration)
April 26, 2017
[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

11. April 26, 2017 11

12. April 26, 2017 12

13. [New Topic] Lumped-Mass Modeling of Rods (Axial Vibration)
April 26, 2017
[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. [Computing k] Lumped-Mass Modeling of Rods (Axial Vibration)
April 26, 2017
[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 “good” model:
The “other” model: 14

15. [AO] Example: Using *Four* Lumped Masses… (Axial Vibration)
April 26, 2017
[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)
April 26, 2017
[AO] Example: Using *Four* Lumped Masses… (Axial Vibration)
The good case:
The other case: 16

17. [Cntd.] Using Four Lumped Masses… (Axial Vibration)
April 26, 2017
[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