CE573 – Structural Dynamics Homework #9ce573/Homeworks/Homework_2012/CE573... · 1 CE573 –...
Transcript of CE573 – Structural Dynamics Homework #9ce573/Homeworks/Homework_2012/CE573... · 1 CE573 –...
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CE573 – Structural Dynamics Homework #9
1. Using the interstory drifts (story distortions) as the coordinates, say 1 and 2 , the equations of motion for
free vibration can be written as (remember the free-body diagrams and force balances we did in class):
11
22
02 0
0 0
story
story
kM M
M M k
Casting the problem of finding modal frequencies and modeshapes as an eigenvalue problem, the modal
frequencies 1 and 2 will be those unique frequencies that will satisfy the characteristic equation, i.e.
2 2 2 2
2 2 2 2
2 20det
story story
story story
k M M k M M
M k M M k M
2 2 2 2
2 4 2 2
2 2 2 2
21 2 2
2 0
3 0
3 9 4 3 5
2 2,
story
story story
story story
story
story story
k M k M M M
M Mk k
Mk M k M k k
M M
ordering the modal frequencies in ascending order, the roots of the characteristic equation are
1 0 618. storyk
M and 2 1 618. storyk
M
It should be no surprise that the natural frequencies are identical to those we found using absolute positions of the masses as the coordinates. Remember, Nature is independent of how you model Her. (Aside: If you are wrong, it will demonstrate in the first opportunity that you are wrong.)
How about the modeshapes? They are the eigenvectors corresponding to the modal frequencies found above.
For the first mode, with 1 0 618. storyk
M , we have
2 21 1 1
2 221 1
1 12 1 1
2 2
2 0
0
0 236 0 382 0 10 618
0 382 0 618 0 0 618
. ..
. . .
story
story
story
k M M
M k M
k
Note that the two equations expressed in matrix form are linearly dependent. That’s why the matrix of coefficients is singular; its rows are not linearly independent. Accordingly, there is only one “useful” equation to work with (pick either one) even though we have two unknowns. The best we could do is to express one of
the components (say, 2 ) in terms of the other one (say, 1 ). That’s why we find a “shape” and not fixed
numbers, for the modeshape (characteristic shape).
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For the second mode, with 2 1 618. storyk
M :
2 22 2 1
2 222 2
1 12 1 1
2 2
2 0
0
4 236 2 618 0 11 618
2 618 1 618 0 1 618
. ..
. . .
story
story
story
k M M
M k M
k
The modeshapes are (let’s give unity for the first degree of freedom) 1
0 618.
and 1
1 618.
, for the first
(fundamental) and second modes, respectively. Do these modeshapes make sense when compared with the modeshapes we found in class using absolute
positions as the coordinates, 1
1 618.
and 1
0 618.
, respectively? Yes, they do. If you express these
modeshapes in terms of interstory drifts, you get what we have found above.
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There is another approach to illustrate the forces acting on this system. In this second approach --called the D’Alembert’s approach—t he inertial forces (tangential and centrifugal) are shown on the figure. They are shown as acting in the direction opposite to associated accelerations.
and the rest follows just as done above.