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Example #1: the tritar

The Tritar

Depicted above is a network of three strings called a tritar. We are interested in how the eigenvalues of this simple network vary with changes in the transverse stiffness k i of each string. We assume that the longitudinal stiffnesses σ i are 1 for each string, and we also assume that the lengths of the strings are all 1 for conveninence.

We define our orientation vectors by

v 1 ^ = 1 0 v 2 ^ = 1 2 1 1 v 3 ^ = 1 2 1 - 1

First, we compute the stiffness matrices. For the first string, we obtain

P 1 = K 1 ( I - v 1 ^ v 1 ^ T ) + σ 1 v 1 ^ v 1 ^ T = k 1 0 0 1 - 1 0 0 0 + 1 0 0 0 = 1 0 0 k .

For the second string, we get

P 2 = K 2 ( I - v 2 ^ v 2 ^ T ) + σ 2 v 2 ^ v 2 ^ T = k 1 0 0 1 - 1 2 1 1 1 1 + 1 2 1 1 1 1 = 1 2 k + 1 - k + 1 - k + 1 k + 1 ,

and for string number three, we have

P 3 = K 3 ( I - v 3 ^ v 3 ^ T ) + σ 3 v 3 ^ v 3 ^ T = k 1 0 0 1 - 1 2 1 - 1 - 1 1 + 1 2 1 - 1 - 1 1 = 1 2 k + 1 k - 1 k - 1 k + 1 .

Letting the vector u i = [ u i 1 u i 2 ] T represent the displacements of string i , we obtain the following system of differential equations:

P 1 u 1 ' ' = - λ 2 u 1 P 2 u 2 ' ' = - λ 2 u 2 P 3 u 3 ' ' = - λ 2 u 3 .

Expanding via matrix multiplication, we see that, component-wise, this translates into

u 11 ' ' = - λ 2 u 11 k u 12 ' ' = - λ 2 u 12 s 2 ' ' = - λ 2 s 2 k d 2 ' ' = - λ 2 d 2 k s 3 ' ' = - λ 2 s 3 d 3 ' ' = - λ 2 d 3 ,

where

s i = u i 1 + u i 2 d i = u i 2 - u i 2 .

for i = 1 , 2 . For boundary conditions, we require that the ends be clamped:

u 1 ( 0 ) = 0 u 2 ( 1 ) = 0 u 3 ( 1 ) = 0 ,

that there be continuity at the central node:

u 1 ( 1 ) = u 2 ( 0 ) = u 3 ( 0 ) ,

and that the forces balance at the central node:

P 1 u 1 ' ( 1 ) - P 2 u 2 ' ( 0 ) - P 3 u 3 ' ( 0 ) = 0 .

As with many systems of differential equations, this one can be solved via the time-honored method of guessing. Noting that the differential equations of this form equate the second derivative of a function with a constant multiple of itself, wehypothesize that the solution for each component of displacement is some linear combination of sines and cosines:

u 11 = a 1 , 1 cos λ x + b 1 , 1 sin λ x u 12 = a 1 , 2 cos λ k x + b 1 , 2 sin λ k x u 21 = a 2 cos λ x + b 2 sin λ x + α 2 cos λ k x + β 2 sin λ k x u 22 = a 2 cos λ x + b 2 sin λ x - α 2 cos λ k x - β 2 sin λ k x u 31 = a 3 cos λ k x + b 3 sin λ k x + α 3 cos λ x + β 3 sin λ x u 32 = a 3 cos λ k x + b 3 sin λ k x - α 3 cos λ x - β 3 sin λ x .

where we have guessed

s 2 = 2 a 2 cos λ x + 2 b 2 sin λ x d 2 = 2 α 2 cos λ k x + 2 β 2 sin λ k x s 3 = 2 a 3 cos λ k x + 2 b 3 sin λ k x d 3 = 2 α 3 cos λ x + 2 β 3 sin λ x

and have used the above to translate these into u 21 , u 22 , u 31 , and u 32 . We need to determine the coefficients. Applying the boundary condition that u 1 ( 0 ) = 0 , we get

a 1 , 1 = 0 a 1 , 2 = 0 .

The fact that u 2 ( 1 ) = 0 implies that

a 2 cos λ + b 2 sin λ = 0 α 2 cos λ k + β 2 sin λ k = 0 .

Likewise, for u 3 ( 1 ) = 0 , we get

a 3 cos λ k + b 3 sin λ k = 0 α 3 cos λ + β 3 sin λ = 0 .

Next, we apply the continuity conditions. Since u 1 ( 1 ) = u 2 ( 0 ) , we see that

a 1 , 1 cos λ + b 1 , 1 sin λ - a 2 - α 2 = 0 a 1 , 2 cos λ k + b 1 , 2 sin λ k - a 2 + α 2 = 0 ,

and because u 2 ( 0 ) = u 3 ( 0 ) we have

a 2 - a 3 = 0 α 2 - α 3 = 0 .

Finally, from the force-balance equation, we have

- a 1 , 1 λ sin λ + b 1 , 1 λ cos λ - 1 2 ( k + 1 ) b 2 λ + β 2 λ k - 1 2 ( k + 1 ) b 2 λ - β 2 λ k - 1 2 ( k - 1 ) b 3 λ k + β 3 λ - 1 2 ( k + 1 ) b 3 λ k - β 3 λ = 0

and

- a 1 , 2 λ k sin λ k + b 1 , 2 λ k cos λ k - 1 2 ( 1 - k ) b 2 λ + β 2 λ k - 1 2 ( k + 1 ) b 2 λ - β 2 λ k - 1 2 ( k - 1 ) b 3 λ k + β 3 λ - 1 2 k + 1 b 3 λ k - β 3 λ = 0 .

These twelve equations can be represented by a single matrix equation with the following coefficient matrix:

1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 cos λ sin λ 0 0 0 0 0 0 0 0 0 0 0 0 cos λ k sin λ k 0 0 0 0 0 0 0 0 0 0 0 0 cos λ k sin λ k 0 0 0 0 0 0 0 0 0 0 0 0 cos λ sin λ cos λ 0 sin λ 0 - 1 0 - 1 0 0 0 0 0 0 cos λ k 0 sin λ k - 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 - 1 0 0 0 0 0 0 0 0 0 1 0 0 0 - 1 0 - λ sin λ 0 λ cos λ 0 0 - λ 0 λ k 0 ( 1 - k ) λ k 0 - λ 0 - λ k sin λ k 0 λ k cos λ k 0 - λ 0 λ k 0 - λ k 0 λ .

The determinant of this matrix is

- sin λ k sin λ l 2 6 cos λ sin λ k k 2 cos λ k sin λ + 4 cos λ 2 sin λ k 2 k 3 / 2 + 3 cos λ sin λ k k cos λ k sin λ + sin λ 2 cos λ k 2 k 3 / 2 - sin λ 2 cos λ k 2 k + 2 sin λ 2 cos λ k 2 k 5 / 2 k - 3 / 2 ,

and by substituting in the desired value of k and setting this determiniant to zero we can then solve for the eigenvalues λ of our tritar net. A plot of the first seven eigenvalues as a function of k is displayed below:

First seven planar eigenvalues for a tritar net, computed with fzero().

As expected, the eigenvalues increase as k increases; however for eigenvalues beyond the first, we observe some rather strange behavior, which suggests that something has gone wrong in the above process. The eigenvalues in the above plot werecomputed using MATLAB's fzero() function at a tolerance of 1e-10. Using a more naive bisection method (which is less likely to lock onto the wrong root) at the same tolerance, we obtain the following plot:

First seven planar eigenvalues for a tritar net, computed with the bisection method.

This seems to have fixed some of the erratic behavior, but neither tightening the tolerance nor increasing the fine-ness of the mesh along which the determinant is evaluated provides much further improvement. On the other hand, it is apparent thatthe solver's structure and parameters impacts the shape of the plots. Perhaps a better solver of some sort (e.g. Newton's method, but adapted to search only in a given interval) can fix more of the problem.

Example #2: the quintar

A depiction of the quintar, with the end angles at 90 degrees.

Rather than develop all of the mathematical relations as in the previous example, it should suffice to say that the same procedure is followed. The solutions to the differential equations are still sums of sines and cosines, but more equations have been added to the system. The function obtained by setting the determinant equal to zero is not enlightening and longer than that for the quintar, and so only the final plots will be presented here. By iteratively increasing the angle at the ends of the network, a plot of the angle versus the eigenvalues is obtained in which traces are formed as the eigenvalues change. It is interesting to note how some of the eigenvalues increase in magnitude while others decrease.

A second plot is presented in which the first nine eigenvalues are plotted versus the transverse stiffness parameter for our analytic model. A region of stiffnesses was chosen where the root-finding algorithm is able to successfully lock onto the zeroes of the determinant as they change. It is clear from this plot that for this range of stiffnesses, increasing stiffness results in the modes of vibration increasing in frequency. This result is expected, since in general stiffer members possess higher vibrational frequencies.

A plot of the opening angle at the ends versus the eigenvalues.
A plot of the first nine eigenvalues for planar modes versus the transverse stiffness.

Questions & Answers

how do they get the third part x = (32)5/4
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ninjadapaul
20/(×-6^2)
Salomon
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Source:  OpenStax, The art of the pfug. OpenStax CNX. Jun 05, 2013 Download for free at http://cnx.org/content/col10523/1.34
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