10.3 The physics of springs  (Page 2/11)

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For this system, we have 16 strings and 7 seven nodes. So we will be working with vectors x , e , y , and f which contain the following information:

x : displacement of nodes

e : elongation of springs

y : restoring force in each spring

f : load force at each node

Since there are 7 nodes, each with a horizontal and vertical displacement, vectors x and f with be vectors of length 14. The 2n-1 entry will correspond to the horizontal displacement or load force on node n and the 2n entry will correspond to the vertical displacement or load force on node n. Since there are 16 springs, y and e will be vectors of length 16 with each entry corresponding to the restoring force or displacement of that spring.

$x=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ ⋮\\ {x}_{13}\\ {x}_{14}\end{array}\right]$

$\begin{array}{cc}& {x}_{1}:\phantom{\rule{4.pt}{0ex}}\text{horizontal}\phantom{\rule{4.pt}{0ex}}\text{displacement}\phantom{\rule{4.pt}{0ex}}\text{at}\phantom{\rule{4.pt}{0ex}}\text{Node}\phantom{\rule{4.pt}{0ex}}\text{1}\hfill \\ & {x}_{2}:\phantom{\rule{4.pt}{0ex}}\text{vertical}\phantom{\rule{4.pt}{0ex}}\text{displacementd}\phantom{\rule{4.pt}{0ex}}\text{at}\phantom{\rule{4.pt}{0ex}}\text{Node}\phantom{\rule{4.pt}{0ex}}\text{1}\hfill \\ \\ & {x}_{13}:\phantom{\rule{4.pt}{0ex}}\text{horizontal}\phantom{\rule{4.pt}{0ex}}\text{displacement}\phantom{\rule{4.pt}{0ex}}\text{at}\phantom{\rule{4.pt}{0ex}}\text{Node}\phantom{\rule{4.pt}{0ex}}\text{7}\hfill \\ & {x}_{14}:\phantom{\rule{4.pt}{0ex}}\text{vertical}\phantom{\rule{4.pt}{0ex}}\text{displacement}\phantom{\rule{4.pt}{0ex}}\text{at}\phantom{\rule{4.pt}{0ex}}\text{Node}\phantom{\rule{4.pt}{0ex}}\text{7}\hfill \end{array}$

As for the matrices translating us from one vector to the next, we will be working with A and K . A will have fourteen columns relating to the 14 degrees of freedom of the nodes of system, and it will have 16 rows, corresponding to the springs of the system. K will be a square matrix, 16x16. Diagonal elements of K will correspond to the stiffnesses of each spring

$K=\left[\begin{array}{ccccc}{k}_{1}& 0& 0& \cdots & 0\\ 0& {k}_{2}& 0& \cdots & 0\\ 0& 0& {k}_{3}& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & {k}_{16}\end{array}\right],\phantom{\rule{1.em}{0ex}}{k}_{i}=\text{spring}\phantom{\rule{4.pt}{0ex}}\text{constant}\phantom{\rule{4.pt}{0ex}}\text{of}\phantom{\rule{4.pt}{0ex}}\text{spring}\phantom{\rule{4.pt}{0ex}}\text{i}$

To construct A , we use a pattern of three linear approximations. The first two assume that the elongation of horizontal springs will be approximately equal to the horizontal displacements of the two adjacent nodes. The same will hold true for vertical nodes. The third recurring equation in the adjacency matrix approximates the diagonal elongation with a first order Maclaurin expansion. This elongation is equal to $\sqrt{2}/2$ multiplied by the sum of the horizontal and vertical displacements. The specific adjacency matrix for this set up is the following:

$A=\left[\begin{array}{cccccccccccccc}0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ \alpha & -\alpha & 0& 0& -\alpha & \alpha & 0& 0& 0& 0& 0& 0& 0& 0\\ 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& \alpha & -\alpha & 0& 0& -\alpha & \alpha & 0& 0& 0& 0& 0& 0\\ -1& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& -1& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& -1& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& \alpha & -\alpha & 0& 0& -\alpha & \alpha & 0& 0\\ 0& 0& 0& 0& -1& 0& 1& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& -1& 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0& 0& 0& 0& -\alpha & \alpha & 0& 0& \alpha & -\alpha \\ 0& 0& 0& 0& 0& 0& -1& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& -1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& -1& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& -1& 0\end{array}\right],\phantom{\rule{4pt}{0ex}}\alpha =\frac{\sqrt{2}}{2},$

Now that we have created our adjacency matrix for the system and know the relation $f={A}^{T}KAx$ , we can create the matrix defined by ${A}^{T}KA$ . If this matrix is invertible, then we can solve what is referred to as the forward problem, finding x given f . This relationship is simply done by finding the inverse of our matrix and solving $x={\left({A}^{T}KA\right)}^{-1}f$ .

An inverse problem

We move now to our inverse problem: given the applied forces and the displacements of nodes in our network, can we determine the stiffnesses of the springs? Although the problem may at first appear trivially similar to the forward problem, we can see that this is not the case.From the previous section, we know that our calculations involve force vector f and displacement vector x ,

$f=\left[\begin{array}{c}{f}_{1}\\ {f}_{2}\\ ⋮\\ {f}_{13}\\ {f}_{14}\end{array}\right],\phantom{\rule{1.em}{0ex}}x=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ ⋮\\ {x}_{13}\\ {x}_{14}\end{array}\right]$

for 14 degrees of freedom. However, the spring constants involved are

$K=\left[\begin{array}{ccccc}{k}_{1}& 0& 0& \cdots & 0\\ 0& {k}_{2}& 0& \cdots & 0\\ 0& 0& {k}_{3}& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & {k}_{16}\end{array}\right],\phantom{\rule{1.em}{0ex}}{k}_{i}=\text{spring}\phantom{\rule{4.pt}{0ex}}\text{constant}\phantom{\rule{4.pt}{0ex}}\text{of}\phantom{\rule{4.pt}{0ex}}\text{spring}\phantom{\rule{4.pt}{0ex}}i$

for 16 springs. This dimensional mismatch leads to an underdetermined system when we attempt to solve for K .

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