The force balance equations can be expressed in the

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Unformatted text preview: r 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 xi , i = 1, . . . , m. The stiffness matrix K is a linear function of x: K (x) = x1 K1 + · · · + xm Km , where Ki = KiT 0 depend on the truss geometry. You can assume these matrices are given or known. The total weight Wtot of the truss also depends on the bar cross-sectional areas: Wtot (x) = w1 x1 + · · · + wm xm , 117 where wi > 0 are known, given constants (density of the material times the length of bar i). Roughly speaking, the truss becomes stiffer, but...
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This note was uploaded on 09/10/2013 for the course C 231 taught by Professor F.borrelli during the Fall '13 term at University of California, Berkeley.

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