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

Problem 1: Part 8
Consider the curved composite beam shown in Figure 12.

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