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# INTRODUCTION TO ENGINEERING COMPUTATION CAAM 210

Rice University

GPA 3.58

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This 5 page Class Notes was uploaded by Walker Witting on Monday October 19, 2015. The Class Notes belongs to CAAM 210 at Rice University taught by Staff in Fall. Since its upload, it has received 9 views. For similar materials see /class/225002/caam-210-rice-university in Applied Mathematics at Rice University.

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Date Created: 10/19/15

On the Deformation of an Elastic Fiber We consider the case illustrated in Figure l The bold solid line is a fiber in its reference state When we subject its two ends to the two forces f1 f2 and f3 f4 the respective ends are displaced by 1152 and 3164 Figure l The reference solid and deformed dashed ber Our goal is to build a theory that predicts x from knowledge of f The first step is to quantify the associated elongation e E X 7 L where L is the undeformed length and X is the deformed length With respect to Figure l we suppose that the lower left node of the undeformed fiber sits at 0 0 in the Cartesian plane while its upper right node resides at L cos 6 L sin 6 Following Euclid we write Lcos6l 3 7 x12 Lsin l 4 7 2 L2 2Lac3 7 1 cos l x4 7 x2sin6 x3 7 x12 x4 7 2 Lxl 2ac3 7 1 cos l x4 7 2sin6L 103 7 x12 104 7 22L2 To help us here we invoke MacLaurin xl 15 l 152 0052 for small 15 and write L x3 7 1 cos l x4 7 2 sin l 015 7 xj22L Hence assuming that at 7 j22 is small for bothj l and 2 compared with L we find 6 37x1cos6ac4 72sin6 1 In the future we shall write instead of m It will be understood that we are working under the hypothesis that the end displacements are small in comparison to the undeformed length We assume that our fiber is Hookean in the sense that its restoring force y is proportional to its elongation More precisely we presume that Ea 7e 2 y L where E denotes the fibers Young7s modulus 1 denotes its cross sectional area and L denotes its reference length This y positive when the bar is stretched and negative when compressed acts along the reference direction 6 in balance with the applied load f More precisely at the lower node ycos6f10 and ysin6f20 3 and at the upper node ycos7r6f3 0 and ysin7r6f4 0 or 7y cos6l f3 0 and 7 ysin6 f4 0 4 Finally we need only substitute our expression for y in terms of e and e in terms of x and recognize that the 4 equations in 3 and 4 hopefully determine at from f More precisely with k E EaL we find kac3 7 1 cos6 x4 7 2 sin l cos l 7f1 kac3 7 1 cos l x4 7 x2sin6 sin l 7f2 kac3 7 1 cos l 104 7 2 sin l cos l f3 kac3 7 1 cos l 104 7 x2sin6 sin l f4 Let us consider a few concrete examples lf k1 60 f2f40 and f17f3 the above system of four equations dei39lates to 953 i 951 f3 which indeed determines 1 and 3 up to an arbitrary rigid motion stemming from the fact that our fiber is a oater We nail things to a foundation and add a fiber Figure 2 We first compute the two elongations 61 1 cos61 2 sin61 62 1 cos62 2 sin62 we suppose the fibers have stiffnesses k1 and kg and so yi 161 y2 262 while force balance at the only free node yields iyl cos61 7 212 cos62 f1 0 iyl sin61 7 212 sin62 f2 0 Assuming 61 7r4 and 62 37r4 we find 61 101 62 962 7 yen yi y2 f1 m y2 f2 and so at must obey k1 k212 k1 k222 f1 k1 k212 k1 k212 f2 In the case of bars of equal stiffness we find the simple answer that 1 fiki and 2 f2k2 3 For unequal stiffnesses we must solve our 2 linear equations simultaneously Matlab knows Gaussian Elimination We are interested in understanding big nets say m fibers meeting at 11 joints and so we step back and realize that our model was constructed in three easy pieces 1 The fiber elongations are linear combinations of their end displacements 6Ax where A is m by 2n is called the node edge adjacency matrix and encodes the ge ometry7 of the fiber net 2 Each fiber restoring force is proportional to its elongation yK67 where K is m by m and diagonal and encodes the physics7 of the fiber net 3 The restoring forces balance the applied forces at each node ATy f where AT is the transpose exchange rows for columns of A When these steps are combined we arrive at the linear system ATKAx f For the net of Figure 2 we have 7 cos ll smell 7 k1 0 T 7 cos 61 cos lg A7 cos lg sin lg K7 0 k2 A 7 sin ll sin lg Please return to the course page for example demo code And then continue here with a bigger example We now build the adjacency matrix for the net below 1 3 2 6 3 Figure 3 We have numbered the 7 bers and the 3 nodes We shall adopt the convention that the horizontal and vertical displacements of node j are ng and xgj respectively With the fiber angles 6194677T2 6265774 and 93660 the associated elongations are 51 2 62 103 OED 53 3 1 54 4 65 105 65 66 5 7 3 67 6 Which we translate 1 row ie one fiber at a time 010000 008800 7101000 A000100 51x5 000055 0071010 000001 Please return to the lecture page for the associated static and dynamic derno code

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