User:Eml4500.f08.qwiki/Lecture 7

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To continue with modeling the 2-bar truss system, element 1 is further examined to find the stiffness matrix k (1). Element 1 is located at an angle $$\theta ^{(1)}={\color{red}30^{\circ}}$$

Therefore inputting this angle into the following equations of the director cosines yields:

$$l^{(1)}=cos\theta ^{(1)}=cos30^{\circ} =\frac{\sqrt{3}}{2}=0.866$$

$$m^{(1)}=sin\theta ^{(1)}=sin30^{\circ} =\frac{1}{2}=0.50$$

Next, to find the stiffness coefficient the below equation is used.

$$k^{(1)}=\frac{E^{(1)} A^{(1)}}{L^{(1)}}=\frac{3*1}{4}=0.75$$

Element Stiffness Matrix:

$$\mathbf{k^{(1)}}=\begin{bmatrix} \mathbf{k}_{11} & \mathbf{k}_{12} & \mathbf{k}_{13} & \mathbf{k}_{14}\\ \mathbf{k}_{21} & \mathbf{k}_{22} & \mathbf{k}_{23} & \mathbf{k}_{24}\\ \mathbf{k}_{31} & \mathbf{k}_{32} & \mathbf{k}_{33} & \mathbf{k}_{34}\\ \mathbf{k}_{41} & \mathbf{k}_{42} & \mathbf{k}_{43} & \mathbf{k}_{44}\\ \end{bmatrix} $$

Knowing the above element stiffness matrix, is the result of:

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

Where (e) is the element number. An example of finding out what each element of the stiffness matrix is shown below.

$$\mathbf{k_{11}}=k^{(1)}*(l^{(1)})^{2}=\frac{3}{4}*(\frac{\sqrt{3}}{2})^{2}=\frac{9}{16}=0.5625 $$

Doing this for all rows and columns yields an element 1 stiffness matrix of:

$$\mathbf{k^{(1)}}=\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}

=\begin{bmatrix} 0.563 & 0.325 & -0.563 & -0.325\\ 0.325 & 0.188 & -0.325 & -0.188\\ -0.563 & -0.325 & 0.563 & 0.325\\ -0.325 & -0.188 & 0.325 & 0.188 \end{bmatrix}$$

Some observations made from seeing the computed element stiffness matrix are as follows,


 * 1) Only 3 numbers need to be computed, the other values only differ by a sign change.
 * 2) Matrix k(1) is symmetric. Thus

$$k_{ij}=k_{ji} \rightarrow   k_{13}=k_{31}  \rightarrow   -0.563=-0.563$$

Observation 2 also means that the transpose of k is the same as k.