User:EML4500.f08.FEABBQ.Jayma/HW3

Class Notes: Friday, October 3, 2008
There is a clarification/correction to the angle for element 3, $$\theta^{(3)}$$, as shown by the following image and equation...



$$ \theta^{(3)}=-90^{\circ}-45^{\circ}=-135^{\circ} $$

Now that the angle for element 3 is clear, we can examine the three-bar truss system with respect to the reaction forces and the external force P...



We know the following equations to be true for this three-bar truss...

$$\Sigma{F}_{x}=0\, $$

$$\Sigma{F}_{y}=0\, $$

$$\Sigma{M}_{A}=0\, $$

where $$\Sigma{M}_{A}=0$$ is trivial, due to the fact that all force vectors pass through this point.

But what about point B ~ $$\Sigma{M}_{B}\, $$?

We will use the following image and subsequent equations to explain the moment about point B. Note: Points A and B are unrelated to the original three-bar truss figure...



The following is true for all $$A'\,$$ on the line of action of $$\vec{F}$$... $$ \Sigma{\vec{M}}_{B}=\vec{BA}\times\vec{F}=\vec{BA'}\times\vec{F}$$ $$ \vec{BA'}=\vec{BA}+\vec{AA'}$$ $$ \Sigma{\vec{M}}_{B}=(\vec{BA}+\vec{AA'})\times\vec{F}=\vec{BA}\times\vec{F}+\vec{AA'}\times\vec{F}$$

where we know that $$\vec{AA'}\times\vec{F}=\vec{0}\, $$. Now we go back to the three-bar truss: Node $$A$$ is in equilibrium: $$ \sum_{i=0}^3\vec{F}=\vec{0}\, $$

And we know that $$ \sum_{i}\vec{M}_{B,i}=\sum_{i}\vec{BA'_{i}}\times\vec{F_{i}}\, $$

$$A'_{i}\, $$ is any point on the line of action of $$\vec{F_{i}}\, $$

$$ \sum_{i}\vec{M}_{B,i}=\sum_{i}\vec{BA'_{i}}\times\vec{F_{i}}=\vec{BA}\times\sum_{i}\vec{F_{i}}=\vec{0}\, $$, therefore proving $$ \Sigma{M}_{B}=\vec{0}\, $$.

Next, we move on to the global stiffness matrix for the three-bar truss. A general outline of the overlapping nature of the three individual element stiffness matrices is shown below...



Next, we fill this matrix...