User:Eml4500.f08.qwiki.bishop/Lecture 6

Continued from Lecture 5
2. Element Picture

3. Global FD at element level

The global force relationship is given by the equation $$F=KD$$, where $$F$$ is the resultant force matrix, $$K$$ is the global stiffness matrix, and $$D$$ is the global displacement matrix.

In matrix format,

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

Note: The superscript "e" represents the truss element under examination.

$$k^{e}=\frac{E^{e}A^{e}}{L^{e}}$$ This is the axial stiffness of bar element "e"

$$l^{e},m^{e}$$ are director cosines of $$\tilde{x}$$ axis (from local node 1 to local node 2) with respect to the global (x,y) coordinate system.

$$l^{e}=\vec{\tilde{i}}\cdot \vec{i}=cos\theta^{e}$$

$$m^{e}=\vec{\tilde{i}}\cdot \vec{j}=sin\theta^{e}$$

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

Distributing by unit vector "i" it is easily seen that the equation reduces to $$l^{e}$$ because a unit vector dotted with itself equals 1, whereas a unit vector dotted with its perpendicular counter part equals 0.

Note: the director cosines are the components of $$\vec{\tilde{i}}$$

$$\vec{\tilde{i}}=cos\theta ^{e}\vec{i}+sin\theta ^{e}\vec{j}$$

HW 2
Solve: $$\vec{\tilde{i}}\cdot\vec{j}$$

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

Distributing by unit vector "j" it is easily seen that the equation reduces to $$m^{e}$$ because a unit vector dotted with itself equals 1, whereas a unit vector dotted with its perpendicular counter part equals 0.

Solve the element stiffness matrices using the provided information from the problem at the end of Lecture 5.

Element 1:

$$l^{1}=\frac{\sqrt{3}}{2}$$   $$m^{1}=\frac{1}{2}$$     $$k^{1}=\frac{3}{4}$$

$$\textbf{K}^{1}=\frac{3}{4}\begin{bmatrix} \frac{3}{4} & \frac{\sqrt{3}}{4} & -\frac{3}{4} & -\frac{\sqrt{3}}{4}\\ \frac{\sqrt{3}}{4} & \frac{1}{4} & -\frac{\sqrt{3}}{4} & -\frac{1}{4}\\ -\frac{3}{4} & -\frac{\sqrt{3}}{4} & \frac{3}{4} & \frac{\sqrt{3}}{4}\\ -\frac{\sqrt{3}}{4} & -\frac{1}{4} & \frac{\sqrt{3}}{4} & \frac{1}{4} \end{bmatrix}=\begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16}& -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix}$$

$$l^{2}=\frac{\sqrt{2}}{2}$$   $$m^{2}=-\frac{\sqrt{2}}{2}$$  $$k^{2}=5$$

$$\textbf{K}^{2}=5\begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2}\\ -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2}\\ -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \end{bmatrix}=\begin{bmatrix} \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}\\ -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}\\ -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}\\ \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} \end{bmatrix}$$

Calculate the axial forces for the statically determinate case: