User:Eml4500.f08.team.guan/hw3

Elemental Axial Displacements
The derivation of the elemental force displacement relationship in a specified local coordinate system from global coordinates can be obtained as follows. Essentially, as the displacements of the global nodes are solved for and obtained, the information can be used to determine the normal forces and stresses that each element is subject to. In order to do this, the axial displacement of each element must be obtained. This can be done by utilizing a transformation matrix to change the displacements from global coordinates into axial displacements that will have the local coordinates of the element intrinsic to the magnitude. The following image is a picture representation of the described process and ultimate goal.  Here, "P" designates the axial force and "q" designates the axial displacement. The coordinate system (s,t) is the local coordinate with s being in the direction of the axis of the element and t being in the direction perpendicular to it. This coordinate system is rotated an angle theta (in this case, 30 degrees) from the global coordinate system (x,y). The force displacement relation for the element equation in local coordinates is shown below:  $$ k^{(e)} \begin{bmatrix} 1 & -1 \\ -1 &  1 \\ \end{bmatrix} \begin{bmatrix} q_1^{(e)} \\ q_2^{(e)} \\ \end{bmatrix}= \begin{bmatrix} P_1^{(e)} \\ P_2^{(e)} \\ \end{bmatrix} $$ $$\displaystyle \hat k^{(e)}q^{(e)}=P^{(e)}$$ "k" is the axial stiffness, $$q_i^{(e)}$$ is the axial displacement of element "e" at local node "i", and $$P_i^{(e)}$$ is the axial force of element "e" at the local node "i". $$k_{hat}$$ in the above equation is a two by two matrix, q is a two by one matrix, and P is a two by one matrix. The goal is to find the relationship between $$q_i^{(e)}$$, and $$d_i^{(e)}$$ and $$P_i^{(e)}$$ and $$f_i^{(e)}$$. First of all let's examine the relationship between axial displacement in local coordinates and nodal displacement in global coordinates. This relation can be expressed in the form: $$\displaystyle q^{(e)}=T^{(e)}d^{(e)}$$ where q is a two by one matrix, T is the transformation matrix (a two by four matrix), and d is a four by one matrix which is essentially the displacement matrix of the element in global coordinates. Consider the displacement matrix of local node 1, denoted by $$d_{1box}^{(e)}$$.  $$\displaystyle d_{1box}^{(e)}=d_{1}^{(e)} \hat i + d_{2}^{(e)} \hat j$$ $$q_1^{(e)}$$ is the axial displacement of local node one and is the orthogonal projection of the displacement vector $$d_{1box}^{(e)}$$ on the axis $$x_{tilda}$$ of element "e". The following math processes result in the transformation matrix. Element 1: $$q_1^{(e)}= \vec d_{1box}^{(e)} \bullet \vec i_{tilda}=(d_{1}^{(e)} \vec i + d_{2}^{(e)} \vec j) \bullet \vec i_{tilda}=d_{1}^{(e)}(\vec i \bullet \vec i_{tilda})+d_{2}^{(e)}(\vec j \bullet \vec i_{tilda})$$ $$=d_{1}^{(e)}cos(\theta^{(e)})+d_{2}^{(e)}sin(\theta^{(e)})=d_{1}^{(e)}l^{(e)}+d_{2}^{(e)}m^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} \\ \end{bmatrix} \begin{bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ \end{bmatrix} $$ Element 2: $$q_2^{(e)}= \vec d_{2box}^{(e)} \bullet \vec i_{tilda}=(d_{3}^{(e)} \vec i + d_{4}^{(e)} \vec j) \bullet \vec i_{tilda}=d_{3}^{(e)}(\vec i \bullet \vec i_{tilda})+d_{4}^{(e)}(\vec j \bullet \vec i_{tilda})$$ $$=d_{3}^{(e)}cos(\theta^{(e)})+d_{4}^{(e)}sin(\theta^{(e)})=d_{3}^{(e)}l^{(e)}+d_{4}^{(e)}m^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} \\ \end{bmatrix} \begin{bmatrix} d_3^{(e)}\\ d_4^{(e)}\\ \end{bmatrix} $$ So, the final matrix equation for the relation between axial displacement and displacement in global coordinates can be written as follows: $$ \begin{bmatrix} q_1^{(e)}\\ q_2^{(e)}\\ \end{bmatrix}= \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & l^{(e)} \\ \end{bmatrix} \begin{bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\\ \end{bmatrix} $$ where $$ \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & l^{(e)} \\ \end{bmatrix}$$ is $$\displaystyle T^{(e)}$$