User:Eml4500.f08.jamama.justin/HW2

The Global Forces are determined by the following matrix

$$\begin{Bmatrix} f1\\ f2\\ f3\\ f4\\ f5\\ f6\\ \end{Bmatrix} = \begin{bmatrix} & &  &  &  & \\  &  &  &  &  & \\  &  &  K&  &  & \\ & &  &  &  & \\  &  &  &  &  & \\  &  &  &  &  & \end{bmatrix} \times \begin{Bmatrix} d1\\ d2\\ d3\\ d4\\ d5\\ d6\\ \end{Bmatrix} $$ 6X1      6X6           6X1

Global Force      Global Stiffness           Global Displacement

Column Matrix      Matrix           Column Matrix

Before the global matrix can be constructed, you first need to find the displacements in the elements themselves. To do this for each element, the following equation may be used:

$$K^{\left(e \right) } \times D^{\left(e \right)} = F^{\left(e \right)} $$

4X4      4X1           4X1

where: $$K^{\left(e \right) }$$=Element stiffness matrix for element e

$$D^{\left(e \right)}$$=Element displacement matrix of element e

$$F^{\left(e \right)}$$=Element force matrix

Element Stiffness Matrix:

The element stiffness matrix $$K^{\left(e \right)}$$ for the equation above is calculated with the matrix shown (below).

$$k^{\left(e \right)}\times \begin{bmatrix} \left(l^{\left(e \right)} \right)^2 & l^{\left(e \right)}m^{\left(e \right)} & -\left(l^{\left(e \right)} \right)^2 & -l^{\left(e \right)}m^{\left(e \right)}\\ l^{\left(e \right)}m^{\left(e \right)} & \left(m^{\left(e \right)} \right)^2 & -l^{\left(e \right)}m^{\left(e \right)} & -\left(m^{\left(e \right)} \right)^2 \\ -\left(l^{\left(e \right)} \right)^2& -l^{\left(e \right)}m^{\left(e \right)} & \left(l^{\left(e \right)} \right)^2 & l^{\left(e \right)}m^{\left(e \right)} \\ -l^{\left(e \right)}m^{\left(e \right)}& -\left(m^{\left(e \right)} \right)^2 & l^{\left(e \right)}m^{\left(e \right)} & \left(m^{\left(e \right)} \right)^2 \end{bmatrix} $$

where: $$k^{\left(e \right)}= \frac{E^{\left(e \right)}A^{\left(e \right)}}{L^{\left(e \right)}}$$ Axial stiffness of bar element "e"

and $$l^{\left(e \right)}, m^{\left(e \right)}$$=Director cosines of the $$\tilde{x}$$ axis with respect to the (x,y) coordinates for element e.

Director Cosines

Director cosines relate the angle at which the element is on to the traditional x,y axis. These values are needed to calculate the stiffness matrix k.

$$l^{\left(e \right)}=\vec{\tilde{i}}\bullet \vec{i}=\cos \theta^{\left(e \right)}$$

$$m^{\left(e \right)}=\vec{\tilde{i}}\bullet \vec{j}=\cos \left(\frac{\pi }{2} - \theta^{\left(e \right)}\right)=\sin \theta ^{\left(e \right)}$$



$$\vec{\tilde{i}}=\cos \theta ^{\left(e \right)} \vec{i} + \sin \theta ^{\left(e \right)}\vec{j}$$

$$\vec{\tilde{i}}\bullet \vec{i}=\cos \theta ^{\left(e \right)} \vec{i}\bullet \vec{i} + \sin \theta ^{\left(e \right)}\vec{j}\bullet \vec{i}$$

$$\vec{\tilde{i}}\bullet \vec{j}=\cos \theta ^{\left(e \right)} \vec{i}\bullet \vec{j} + \sin \theta ^{\left(e \right)}\vec{j}\bullet \vec{j}$$