Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 8

Problem 1: Part 8

Consider the curved composite beam shown in Figure 12.

Figure 12. Composite Beam.

Assume that the beam has been shaped into an arc of a circle. The material of the beam is a transversely isotropic fiber composite material with the fibers running along the length of the beam. The rate constitutive relation of the material is given by

${\cfrac {D}{Dt}}{\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{31}\\\sigma _{12}\end{bmatrix}}={\begin{bmatrix}C_{11}&C_{12}&C_{13}&0&0&0\\C_{12}&C_{11}&C_{13}&0&0&0\\C_{13}&C_{13}&C_{33}&0&0&0\\0&0&0&C_{44}&0&0\\0&0&0&0&C_{44}&0\\0&0&0&0&0&(C_{11}-C_{12})/2\end{bmatrix}}{\begin{bmatrix}D_{11}\\D_{22}\\D_{33}\\D_{23}\\D_{31}\\D_{12}\end{bmatrix}}~.$

The problem becomes easier to solve if we consider numerical values of the parameters. Let the local nodes numbers of element $5$ be $1$ for node $5$, and $2$ for node $6$.

Let us assume that the beam is divided into six equal sectors. Then,

$\theta ={\cfrac {\pi }{4}}~;~~\theta _{1}={\cfrac {\pi }{3}}~;~~\theta _{2}={\cfrac {\pi }{6}}~.$

Let $r_{1}=1$ and $r_{2}=1.2$. Since the blue point is midway between the two, $r=1.1$.

Also, let the components of the stiffness matrix of the composite be

$C_{11}=10;~~C_{33}=100;~~C_{12}=6;~~C_{13}=60;~~C_{44}=30~.$

Let the velocities for nodes $1$ and $2$ of the element be

$\mathbf {v} _{1}={\begin{bmatrix}v_{1}^{x}\\v_{1}^{y}\\\omega _{1}\end{bmatrix}}={\begin{bmatrix}1\\2\\1\end{bmatrix}}~;~~\mathbf {v} _{2}={\begin{bmatrix}v_{2}^{x}\\v_{2}^{y}\\\omega _{2}\end{bmatrix}}={\begin{bmatrix}2\\1\\1\end{bmatrix}}$

The Maple code for these initializations is shown below

> with(linalg):
> #
> # Input geometry
> #
> theta:= evalf(Pi/4):
> theta1 := evalf(theta + Pi/12):
> theta2 := evalf(theta - Pi/12):
> r1:= 1; r2 := 1.2; r:= (r1+r2)/2;
> #
> # Input material properties
> #
> C11 := 10; C33:= 100; C12:= 6; C13 := 60; C44 := 30;
> CC := (C11-C12)/2;
> #
> # Input velocities
> #
> vx1 := 1; vy1 := 2; w1 := 1;
> vx2 := 2; vy2 := 1; w2 := 1;
> v1master := linalg[matrix](3,1,[vx1,vy1,w1]);
> v2master := linalg[matrix](3,1,[vx2,vy2,w2]);

Nonlinear finite elements/Homework 6/Solutions/Problem 1