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# Matrix Methods for Mechanical Systems: The General Planar Truss

Module by: Doug Daniels, Steven J. Cox. E-mail the authors

Summary: In this module we generalize our analysis of trusses to include any planar structure. Considering a vein model, we figure the elongation of its fibers, taking into account both vertical and horizontal displacement. After determining a matrix to represent these values, we show how to automate our data entry into MATLAB in the form of an M-file. Finally, we show how to take a pseudo-inverse in MATLAB, and display the results of the process.

Let us now consider something that resembles the mechanical prospection problem introduced in the introduction to matrix methods for mechanical systems. In the figure below we offer a crude mechanical model of a planar tissue, say, e.g., an excised sample of the wall of a vein.

Elastic fibers, numbered 1 through 20, meet at nodes, numbered 1 through 9. We limit our observation to the motion of the nodes by denoting the horizontal and vertical displacements of node jj by x 2j-1 x 2j-1 (horizontal) and x 2j x 2j (vertical), respectively. Retaining the convention that down and right are positive we note that the elongation of fiber 1 is e 1 = x 2 x 8 e 1 x 2 x 8 while that of fiber 3 is e 3 = x 3 x 1 . e 3 x 3 x 1 . As fibers 2 and 4 are neither vertical nor horizontal their elongations, in terms of nodal displacements, are not so easy to read off. This is more a nuisance than an obstacle however, for noting our discussion of elongation in the small planar truss module, the elongation is approximately just the stretch along its undeformed axis. With respect to fiber 2, as it makes the angle π4 4 with respect to the positive horizontal axis, we find e 2 = x 9 cosπ4 x 10 sinπ4= x 9 x 1 + x 2 x 10 22 . e 2 x 9 x 1 4 x 10 x 2 4 x 9 x 1 x 2 x 10 2 2 . Similarly, as fiber 4 makes the angle 3π4 3 4 with respect to the positive horizontal axis, its elongation is e 4 = x 7 cos3π4 x 8 sin3π4= x 3 x 7 + x 4 x 8 22 . e 4 x 7 x 3 3 4 x 8 x 4 3 4 x 3 x 7 x 4 x 8 2 2 . These are both direct applications of the general formula

e j = x 2 n - 1 cos θ j x 2n sin θ j e j x 2 n - 1 x 2 m - 1 θ j x 2n x 2m θ j
(1)
for fiber jj, as depicted in Figure 2 below, connecting node mm to node nn and making the angle θ j θ j with the positive horizontal axis when node mm is assumed to lie at the point (0,0). The reader should check that our expressions for e 1 e 1 and e 3 e 3 indeed conform to this general formula and that e 2 e 2 and e 4 e 4 agree with ones intuition. For example, visual inspection of the specimen suggests that fiber 2 can not be supposed to stretch (i.e., have positive e 2 e 2 ) unless x 9 > x 1 x 9 x 1 and/or x 2 > x 10 . x 2 x 10 . Does this jive with Equation 1?

Applying Equation 1 to each of the remaining fibers we arrive at e=Ax e A x where AA is 20-by-18, one row for each fiber, and one column for each degree of freedom. For systems of such size with such a well defined structure one naturally hopes to automate the construction. We have done just that in the accompanying M-file and diary. The M-file begins with a matrix of raw data that anyone with a protractor could have keyed in directly from Figure 1:


data =     [                                 % one row of data for each fiber, the
1       4       -pi/2           % first two columns are starting and ending
1       5       -pi/4           % node numbers, respectively, while the third is the
1       2       0               % angle the fiber makes with the positive horizontal axis
2       4       -3*pi/4
...and so on...                  ]


This data is precisely what Equation 1 requires in order to know which columns of AA receive the proper cos or sin. The final AA matrix is displayed in the diary.

The next two steps are now familiar. If KK denotes the diagonal matrix of fiber stiffnesses and ff denotes the vector of nodal forces then y=Ke        and        ATy=f y K e        and        A y f and so one must solve Sx=f S x f where S=ATKA S A K A . In this case there is an entire three--dimensional class of zz for which Az=0 A z 0 and therefore Sz=0 S z 0 . The three indicates that there are three independent unstable modes of the specimen, e.g., two translations and a rotation. As a result SS is singular and  x = S\f in MATLAB will get us nowhere. The way out is to recognize that SS has 183=15 18 3 15 stable modes and that if we restrict SS to 'act' only in these directions then it should' be invertible. We will begin to make these notions precise in discussions on the Fundamental Theorem of Linear Algebra. For now let us note that every matrix possesses such a pseudo-inverse and that it may be computed in MATLAB via the pinv command. Supposing the fiber stiffnesses to each be one and the edge traction to be of the form f=( -110111-100010-1-10-11-1 )T , f -110 111 -100 010 -1-10 -11-1 , we arrive at xx via x=pinv(S)*f` and offer below its graphical representation.

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