The locations of the first and last mass are fixed

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Intermediate Algebra
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Chapter 8 / Exercise 65
Intermediate Algebra
Mckeague
Expert Verified
The locations of the first and last mass are fixed. The equilibrium location of the other masses is the one that minimizes E . (a) Show how to find the equilibrium positions of the masses 2 ,...,n 1 using convex optimization. Be sure to justify convexity of any functions that arise in your formulation (if it is not obvious). The problem data are m i , k i , l i , g , x 1 , y 1 , x n , and y n . (b) Carry out your method to find the equilibrium positions for a problem with n = 10, m i = 1, k i = 10, l i = 1, x 1 = y 1 = 0, x n = y n = 10, with g varying from g = 0 (no gravity) to g = 10 (say). Verify that the results look reasonable. Plot the equilibrium configuration for several values of g . 14.3 Elastic truss design. In this problem we consider a truss structure with m bars connecting a set of nodes. Various external forces are applied at each node, which cause a (small) displacement in the node positions. f R n will denote the vector of (components of) external forces, and d R n will denote the vector of corresponding node displacements. (By ‘corresponding’ we mean if f i is, say, the z -coordinate of the external force applied at node k , then d i is the z -coordinate of the displacement of node k .) The vector f is called a loading or load . The structure is linearly elastic, i.e. , we have a linear relation f = Kd between the vector of external forces f and the node displacements d . The matrix K = K T 0 is called the stiffness matrix of the truss. Roughly speaking, the ‘larger’ K is ( i.e. , the stiffer the truss) the smaller the node displacement will be for a given loading. We assume that the geometry (unloaded bar lengths and node positions) of the truss is fixed; we are to design the cross-sectional areas of the bars. These cross-sectional areas will be the design variables x i , i = 1 ,...,m . The stiffness matrix K is a linear function of x : K ( x ) = x 1 K 1 + · · · + x m K m , where K i = K T i followsequal 0 depend on the truss geometry. You can assume these matrices are given or known. The total weight W tot of the truss also depends on the bar cross-sectional areas: W tot ( x ) = w 1 x 1 + · · · + w m x m , 127
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Intermediate Algebra
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Chapter 8 / Exercise 65
Intermediate Algebra
Mckeague
Expert Verified
where w i > 0 are known, given constants (density of the material times the length of bar i ). Roughly speaking, the truss becomes stiffer, but also heavier, when we increase x i ; there is a tradeoff between stiffness and weight. Our goal is to design the stiffest truss, subject to bounds on the bar cross-sectional areas and total truss weight: l x i u, i = 1 ,...,m, W tot ( x ) W, where l , u , and W are given. You may assume that K ( x ) 0 for all feasible vectors x . To obtain a specific optimization problem, we must say how we will measure the stiffness, and what model of the loads we will use. (a) There are several ways to form a scalar measure of how stiff a truss is, for a given load f . In this problem we will use the elastic stored energy E ( x,f ) = 1 2 f T K ( x ) 1 f to measure the stiffness. Maximizing stiffness corresponds to minimizing E ( x,f ).

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