User:Eml4500.f08.Ateam.philbin/HW1

September 8, 2008 - Lecture 6

Statically Determinate versus Statically Indeterminate
A statically determinate problem is one that can be solved using simple statics relationships. There are three main equations, summing forces in the x and the y directions and solving for the moment about a point. With these three equations, any problem with up to three unknowns can be solved. Problems with multiple beams or more than three reaction forces are statically indeterminate since there are insufficient equations to solve for all the unknowns.

An exception to the indeterminate rules is shown in the following example.



Here there are four reaction forces, two at each support and a known applied force at the top. It can be noted that the beams in this problem are both two-force members. By looking at each element separately and summing the forces in any direction it is found that they must be both equal and opposite. Summing the moment about either end shows that both forces' line of action must go through each end. Otherwise the moments would not sum to zero.



Therefore the two forces are constrained in one direction. With these simple straight beams, the line of action is along the beam itself. If the beam were curved or had an angle in it, the same logic would apply and the line of action of the two forces will be along the invisible line going through the two attachment points.

This simplification of the problem lowers the degrees of freedom (unknowns) from four to three. Since the direction of the reaction forces are now known, only the magnitude is left to be found.

Lecture Notes
2) Element picture

Here the element degrees of freedom and element forces are shown in either the global system or in a local coordinate system. The element number is in the parentheses and the forces (and displacements) are numbered from node 1 to 3, each starting with the x and then y axes.



3) Global force / displacement relationship

The next step is to create a relationship between the forces and displacements in the global coordinate system. The matrices will be in the following form.

$$\underline k^{(e)}*\ \underline d^{(e)}= \underline f^{(e)}$$

Where k (e) is a 4x4 matrix and d (e) and f (e) are 4x1 matrices. In order to find the values of k (e), the formula found on page 225 of Fundamental Finite Element Analysys and Applications must be used. It states that the scalar, $$K = \frac{EA}{L}$$, is multiplied by a matrix composed of functions of the director of cosines.



The director of cosines is a function of the angle of the element and given as, $$l^{(e)}= \vec \tilde{i} \centerdot \vec i = \cos \ \Theta^{(e)}$$ and  $$m^{(e)}= \vec \tilde{i} \centerdot \vec j = \sin \ \Theta^{(e)}$$. The completed matrix for K (e) is as follows.

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

The k (e) matrices for elements 1 and 2 were calculated based on the given data and are as follows.

$$ \underline K^{(1)}= \begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & \frac{-9}{16} & \frac{-3\sqrt{3}}{16} \\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & \frac{-3\sqrt{3}}{16} & \frac{-3}{16} \\ \frac{-9}{16} & \frac{-3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16} \\ \frac{-3\sqrt{3}}{16} & \frac{-3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix}=\begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 \\ -0.5625 & -0.3248 & 0.5625 & 0.3248 \\ -0.3248 & -0.1875 & 0.3248 & 0.1875 \end{bmatrix} $$

$$ \underline K^{(2)}= \begin{bmatrix} \frac{5}{2} & \frac{-5}{2} & \frac{-5}{2} & \frac{5}{2} \\ \frac{-5}{2} & \frac{5}{2} & \frac{5}{2} & \frac{-5}{2} \\ \frac{-5}{2} & \frac{5}{2} & \frac{5}{2} & \frac{-5}{2} \\ \frac{5}{2} & \frac{-5}{2} & \frac{-5}{2} & \frac{5}{2} \end{bmatrix}=\begin{bmatrix} 2.5 & -2.5 & -2.5 & 2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ 2.5 & -2.5 & -2.5 & 2.5 \end{bmatrix} $$