Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 9

Problem 1: Part 9

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Construct a laminar coordinate system at the blue point. (Use equations 9.3.16 and 9.3.17 from the book chapter). Assume that the blue point is at the center of element 5.

The   and   coordinates of the master and slave nodes are

 
 
 

Since there are two master nodes in an element, the parent element is a four-noded quad with shape functions

 

The isoparametric map is

 

Plugging in the numbers, we get

 

Therefore, the derivatives with respect to   are

 

Since the blue point is at the center of the element, the values of   and   at that point are zero. Therefore, the values of the derivatives at that point are

 

Therefore,

 

The local laminar basis vector   is given by

 

The laminar basis vector   is given by

 

A plot of the vectors is shown in Figure 13.

 
Figure 13. Laminar base vectors.

The Maple code for this calculation is shown below.

> #
> # Shape functions
> #
> N1m := 1/4*(1-xi)*(1-eta):
> N2m := 1/4*(1+xi)*(1-eta):
> N1p := 1/4*(1-xi)*(1+eta):
> N2p := 1/4*(1+xi)*(1+eta):
> N := linalg[matrix](1,4,[N1m,N2m,N1p,N2p]);
> #
> # Compute local laminar basis vectors at the blue point
> #
> # Isoparametric map
> #
> x := x1m*N1m + x2m*N2m + x1p*N1p + x2p*N2p;
> y := y1m*N1m + y2m*N2m + y1p*N1p + y2p*N2p;
> #
> # Derivative of Isoparametric map
> #
> dx_dxi := diff(x, xi);
> dy_dxi := diff(y, xi);
> #
> # Jacobian evluated at blue point
> #
> dx_dxi_cen := subs(eta=0, dx_dxi);
> dy_dxi_cen := subs(eta=0, dy_dxi);
> #
> # Local laminar basis vectors at the blue point
> #
> norm_dx_dxi_cen := simplify(sqrt(dx_dxi_cen^2 + dy_dxi_cen^2));
> ehat_x := vector([simplify(dx_dxi_cen/norm_dx_dxi_cen),
> simplify(dy_dxi_cen/norm_dx_dxi_cen), 0]);
> ehat_z := vector([0, 0, 1]);
> ehat_y := crossprod(ehat_z, ehat_x);