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FE Analysis

% Script for matrix inversion A = [ 1 1 1 1 0 0 0 0, 1 -1 1 -1 0 0 0 0, 1 -1 -1 1 0 0 0 0, 1 1 -1 -1 0 0 0 0, 0 0 0 0 1 1 1 1, 0 0 0 0 1 -1 1 -1, 0 0 0 0 1 -1 -1 1, 0 0 0 0 1 1 -1 -1 ]; inv(A)

Obtain shape functions for the one-dimensional quadratic element with three nodes using local coordinate system -1 ≤ x ≤ +1.

O--------O--------O 1 2 3 -1--------0--------1------->x

Using shape function, any field (such as displacement, strain, stress) inside the element can be expressed as u(x) = ΣN_{i}u_{i} where i = 1, 2, 3 for 3 nodes of the quadratic element. At nodes the approximated function should be equal to its nodal value. Thus:

From 3 points, we can fit a polynomial of order 2 that is a quadratic polynomial with 3 coefficients. Hence,

Let N_{1} = a_{1} + a_{2} * x + a_{3} * x^{2}

N_{1}(-1) = 1, → a_{1} - a_{2} + a_{3} = 1

N_{1}(0) = 0, → a_{1} = 0

N_{1}(1) = 0, → a_{1} + a_{2} + a_{3} = 0

a_{1} = 0, a_{2} = 1/2, a_{3} = -1/2

N_{1} = 1/2 * x *(x - 1)

Similarly,

Let N_{2} = b_{1} + b_{2} * x + b_{3} * x^{2}

N_{2}(-1) = 0, → b_{1} - b_{2} + b_{3} = 0

N_{2}(0) = 1, → b_{1} = 1

N_{2}(1) = 1, → b_{1} + b_{2} + b_{3} = 0

b_{1} = 1, b_{2} = 0, b_{3} = -1

N_{2} = x *(1 - x)

Similarly,

Let N_{3} = c_{1} + c_{2} * x + c_{3} * x^{2}

N_{3}(-1) = 0, → c_{1} - c_{2} + c_{3} = 0

N_{3}(0) = 0, → c_{1} = 0

N_{3}(1) = 1, → c_{1} + c_{2} + c_{3} = 1

c_{1} = 0, c_{2} = 1/2, c_{3} = 1/2

N_{3} = 1/2 * x *(1 + x)

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