x 1 To help us here we invoke MacLaurin, 1 + t = 1 + t/2 + O(t 2 ) for small t, and write

Transcription

1 On the Deormation o an Elastic Fiber We consider the case illustrated in Figure. The bold solid line is a iber in its reerence state. When we subject its two ends to the two orces, (, ) and (, ) the respective ends are displaced by (, ) and (, ). Figure. The reerence (solid) and deormed (dashed) iber. Our goal is to build a theory that predicts rom knowledge o. The irst step is to quantiy the associated elongation, e l L, where L is the undeormed length and l is the deormed length. With respect to Figure, we suppose that the lower let node o the undeormed iber sits at (0, 0) in the Cartesian plane, while its upper right node resides at (L cos, L sin ). Following Euclid, we write l = (L cos + ) + (L sin + ) = L + L{( ) cos + ( ) sin} + ( ) + ( ) = L + {( ) cos + ( ) sin}/l + {( ) + ( ) }/L. To help us here we invoke MacLaurin, + t = + t/ + O(t ) or small t, and write l = L + ( ) cos + ( ) sin + O(( j j+ ) /L). Hence, assuming that ( j j+ ) is small (or both j = and ) compared with L, we ind e ( ) cos + ( ) sin. () In the uture we shall write = instead o. It will be understood that we are working under the hypothesis that the end displacements are small in comparison to the undeormed length.

2 We assume that our iber is Hookean, in the sense that its restoring orce, y, is proportional to its elongation. More precisely, we presume that y = Ea L e () where E denotes the iber s Young s modulus, a denotes its cross sectional area, and L denotes its reerence length. This y, positive when the bar is stretched and negative when compressed, acts along the reerence direction,, in balance with the applied load. More precisely, at the lower node and at the upper node y cos + = 0 and y sin + = 0 () y cos(π + ) + = 0 and y sin(π + ) + = 0 or y cos() + = 0 and y sin() + = 0 () Finally, we need only substitute our epression or y in terms o e and e in terms o and recognize that the equations in () and () (hopeully) determine rom. More precisely, with k Ea/L, we ind k{( ) cos + ( ) sin} cos = k{( ) cos + ( ) sin} sin = k{( ) cos + ( ) sin} cos = k{( ) cos + ( ) sin} sin = Let us consider a ew concrete eamples. I k =, = 0, = = 0 and = the above system o our equations delates to = which indeed determines and up to an arbitrary rigid motion, stemming rom the act that our iber is a loater.

3 We nail things to a oundation and add a iber. Figure. We irst compute the two elongations e = cos( ) + sin( ) e = cos( ) + sin( ) we suppose the ibers have stinesses k and k and so y = k e y = k e while orce balance at the only ree node yields y cos( ) y cos( ) + = 0 y sin( ) y sin( ) + = 0 Assuming = π/ and = π/, we ind e = ( + )/ e = ( )/ and (y y )/ = (y + y )/ = and so must obey (k + k ) / + (k k ) / = (k k ) / + (k + k ) / = In the case o bars o equal stiness we ind the simple answer that = /k and = /k.

4 For unequal stinesses we must solve our linear equations simultaneously. Matlab knows Gaussian Elimination. We are interested in understanding big nets (say m ibers meeting at n joints) and so we step back and realize that our model was constructed in three easy pieces. [] The iber elongations are linear combinations o their end displacements, e = A, where A is m-by-n, is called the node-edge adjacency matri, and encodes the geometry o the iber net. [] Each iber restoring orce is proportional to its elongation, y = Ke, where K is m-by-m and diagonal and encodes the physics o the iber net. [] The restoring orces balance the applied orces at each node, A T y = where A T is the transpose (echange rows or columns) o A. When these steps are combined, we arrive at the linear system A T KA =. For the net o Figure, we have cos sin A = cos sin k 0 K = 0 k A T cos cos = sin sin Please return to the course page or eample demo code. And then continue here with a bigger eample.

5 We now build the adjacency matri or the net below Figure. We have numbered the 7 ibers and the nodes. We shall adopt the convention that the horizontal and vertical displacements o node j are j and j respectively. With the iber angles, = = 7 = π/, = 5 = π/, and = 6 = 0, the associated elongations are e = e = ( + )/ e = e = e 5 = ( )/ e 6 = 5 e 7 = 6. which we translate, row, i.e., one iber, at a time s s A = s s s = / Please return to the lecture page or the associated static and dynamic demo code. 5