Question: Note that, the iso-parametric shape functions are obtained using Lagrange Multipliers. For an Nth degree polynomial approximation, there should be N + 1 shape

Note that, the iso-parametric shape functions are obtained using Lagrange Multipliers. For

Note that, the iso-parametric shape functions are obtained using Lagrange Multipliers. For an Nth degree polynomial approximation, there should be N + 1 shape functions. Note that, the shape functions should have a value unity (e.g., 1) at the node that they belong to and zero (e.g. O) at the other nodes. For example, a first-order approximation would be given as: u(x) = $(x) u(x-1) + 2(x) u(x-2) where u(x-1) and u(x-2) represents the nodal values of the function u at the first and second nodes. Ibilaxs 25 X-X2 Here, $(x) = *-*2 and (x) = *-*1. x1-x X2-X1 a) If the element extends between coordinates -1 x+1, what are the shape functions in simplified forms (e.g., x = -1 and x = +1)? b) Show that (-1) = 1 and (+1) = 0. Also show that (-1) = 0 and (+1) = 1 c) If the approximation is extended to second order (quadratic) polynomials as u(x) = (x) u(x-) + (x) u(x-2) + 3(x) u(x-3) selection of the base functions as: X-X1 X-X2 3 (x) = Show that with above discussion $(x) = x-x2 x-x3 X1 X2 X1 X3 ; $(x) = would be a proper choice (e.g., X-X X-X3 X2-X1 X2 X3 (x) = 1 and - and 1515700 X3-X1 X3-X2 (x) = 1 (x3) = 0 and for the others, the similar rule applies). d) For the quadratic approximation given above and with node locations x = -1, x = 0 and x3 = +1 find explicit functions for , 2 and 3 1306

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