User:Eml4500.f08.ateam.boggs.t/Lecture 6

An example of a statically indeterminate problem is given below:



The Force Equation as seen in matrix form is as follows:



Where F is the global force column matrix, K is the global stiffness matrix, and d is the global displacement matrix.

It was pointed out that the previous example, a truss system with two elastic bars, was in fact a solvable system correcting that it was originally stated to be indeterminate. The class was then told that the system was to be solved and turned in with the next homework report. As a reminder it is important that when the system is broken up into it's individual elements to place any forces shared by two or more elements in the global force diagram on only one of the individual elements.



To begin solving the above problem one must compute the the k matrix for each element, a depiction of what is to go in to each k matrix is given below:



K^{(e)}= k^{(e)} \cdot \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} $$ Here l(e) and m(e) are director cosines for the x'-axis and the equation for k(e) is:



k^{(e)} = \frac{E^{(e)}A^{(e)}}{L^{(e)}} $$



l(e) = î • i = cosθ(e)

m(e) = î • j = cos(π/2 - θ(e)) = sinθ(e)

î = cosθ(e) i + sin θ(e) j

î • i = (cosθ(e) i + sinθ(e) j) • i

î • i = cosθ(e) i • i + sinθ(e) j • i

î • i = cosθ(e)

Therefore, î • j also equals sinθ(e)