Question

O CWOIK 2. Submission deadline: 11:59pm on Friday June 7, 2019 Penalty for late submission: 20% per day T3 finite element is defined over AABC (in physical coordinates). The vertices of this triangle have the following coordinates: A(-2, -1), B(3,2), and C(0, 6). Problem 1 Calculate the partial derivatives of T3 basis functions with respect to the physical coordinates x and y. Problem 2. a) Using 1 point and 3 point integration rules, compute f(x, y)ds AABC where f(x,y) =2x2-3xy + y2.

Answer 1

For triangular elements with vertices (x1,y1) = (-2,-1), (x2,y2) = (3,2) and (x3,y3) = (0,6)

the basis or shape functions are given by

N1(x,y) = (a1 + b1*x+c1*y)/(2*A);

N2(x,y) = (a2 + b2*x+c2*y)/(2*A);

N3(x,y) = (a3 + b3*x+c3*y)/(2*A);

where a1 = x2*y3-x3*y2 = 3*6-2*0 = 18; a2 = x3*y1-x1*y3 = -1*-6 = 6 and a3 = x1*y2-x2*y1 = -4+3 = -1

b1 = y2-y3 = -4; b2 = y3-y1 = 7 and b3 = y1-y2 = -3

c1 = x2-x3 = 3; c2 = x3-x1 = 2and c3 = x1-x2 = -5

A = area of triangle = 29/2

Now

differentiation wrt x

dN1/dx = b1/(2*A) =  -8/29

dN2/dx = b2/(2*A) = 14/29

dN3/dx = b3/(2*A) = -6/29

differentiation wrt y

dN1/dy = c1/(2*A) = 6/29

dN2/dy = c2/(2*A) = 4/29

dN3/dy = c3/(2*A) = -10/29

Hope it helps. Cheers!!