Shape Function for a Rectangular Isoparametric Element

Shape functions are ubiquitous concept present in every Finite Element simulations of elastic components. This is an attempt to demystify the concept of shape functions by describing the step-by-step approach to get the function as they are used. The work is based on concepts presented in Zienkiewicz and Bathe where the authors have been [a bit extra] succinct in the textbook to describe the way shape functions are evolved.
Shape Function and Jacobian of Isoparametric Elements
Shape Function and Jacobian of Isoparametric Elements
Shape Function and Jacobian of Isoparametric Elements
Shape Function and Jacobian of Isoparametric Elements

OCTAVE Scrip for Matrix Inversion

% 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)

Shape Function Example for 1D Quadratic Element

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) = ΣNiui 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:
u(-1) = u1, u(0) = u2, u(1) = u2

From 3 points, we can fit a polynomial of order 2 that is a quadratic polynomial with 3 coefficients. Hence, Let N1 = a1 + a2 * x + a3 * x2
N1(-1) = 1, → a1 - a2 + a3 = 1
N1(0)  = 0, → a1 = 0
N1(1)  = 0, → a1 + a2 + a3 = 0
Thus:
a1 = 0, a2 = 1/2, a3 = -1/2
Finally, N1 = 1/2 * x *(x - 1)

Similarly,
Let N2 = b1 + b2 * x + b3 * x2
N2(-1) = 0, → b1 - b2 + b3 = 0
N2(0)  = 1, → b1 = 1
N2(1)  = 1, → b1 + b2 + b3 = 0
Thus:
b1 = 1, b2 = 0, b3 = -1
Finally, N2 = x *(1 - x)

Similarly,
Let N3 = c1 + c2 * x + c3 * x2
N3(-1) = 0, → c1 - c2 + c3 = 0
N3(0)  = 0, → c1 = 0
N3(1)  = 1, → c1 + c2 + c3 = 1
Thus:
c1 = 0, c2 = 1/2, c3 = 1/2
Finally, N3 = 1/2 * x *(1 + x)


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