User:EML4500.F08.RAMROD.B/Homework 5.1

=3-D Truss Relationships=

This process is very similar to that of the 2-D case. Therefore we will be taking the same steps to derive the Force-Displacement relationships. The elemental axial force displacement relationship is identical to the 2-D case and is shown below.

$$\displaystyle (k^e)(q^e)=p^e$$

Since the 3-D case will have three displacements at each node, the element displacement matrix will be a six by one (6x1) as will the element force matrix. These matrices are shown below.

 $$\displaystyle d^e = \begin{Bmatrix}d_1^e\\ d_2^e\\ d_3^e\\ d_4^e\\ d_5^e\\ d_6^e \end{Bmatrix}$$ $$\displaystyle f^e = \begin{Bmatrix}f_1^e\\ f_2^e\\ f_3^e\\ f_4^e\\ f_5^e\\ f_6^e \end{Bmatrix}$$

For the 3-D case the two by six (2x6) transformation matrix $$T^e$$ that relates the axial degrees of freedom to the global degrees of freedom is shown below.

 $$\displaystyle T^e = \begin{bmatrix} l^e & m^e & n^e & 0 & 0 & 0 \\ 0 & 0 & 0 & l^e & m^e & n^e \end{bmatrix}$$

This transformation matrix can be used to relate the axial and global degrees of freedom and forces as shown below.

$$\displaystyle \begin{bmatrix}l^e & m^e & n^e & 0 & 0 & 0 \\ 0 & 0 & 0 & l^e & m^e & n^e\end{bmatrix} \begin{Bmatrix}d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\end{Bmatrix}=\begin{Bmatrix}q_1\\ q_2\end{Bmatrix}$$

It can also be used to relate the global forces to the local axial forces. This is shown below.

$$\displaystyle \begin{bmatrix}l^e & m^e & n^e & 0 & 0 & 0 \\ 0 & 0 & 0 & l^e & m^e & n^e\end{bmatrix} \begin{Bmatrix}f_1\\ f_2\\ f_3\\ f_4\\ f_5\\ f_6\end{Bmatrix}=\begin{Bmatrix}p_1\\ p_2\end{Bmatrix}$$