User:Eml4500.f08.bottle.butler/HW 3

For this homework assignment I typed all of the formulas in the notes in Matlab code and added to the notes with Jonathon Hipps. I have included the entire note section here in hopes that it is easier for the reader to follow.

Initially in the previous lectures, we derived an element force diagram with $$f \! $$ representing element force and $$ d \! $$ representing element displacement on a truss element. We replace these unknowns with $$ P \! $$ representing the element force and $$ q \! $$ representing the element displacement through the axis of the truss element.

With:

1.   $$q_i^{(e)}$$=axial displacement of element e at local node i   2.    $$P_i^{(e)}$$=axial force of element e at local node i

To derive $$ k \!$$ $$* \!$$$$d \!$$$$= \!$$$$f \!$$ from $$k \!$$$$* \!$$$$q \!$$$$=$$$$P \!$$, we need to find the relation between $$q\!$$ and $$d\!$$, and $$P\!$$ and $$f\!$$. For displacement, we derive that:

$$ q_{2X1}^{(e)}$$ $$ = \!$$ $$T_{2X4}^{(e)}$$ $$ * \!$$ $$d_{4X1}^{(e)}$$

In order to derive $$T \!$$, we must consider the displacement vector of the nodes in each truss element. Considering the previous truss system derived in the lecture notes, we derive vector $$ d \! $$ of node 1 on the axis x of element e, giving us:

$$q_1^{(e)}$$ $$=$$ $$d_1^{(e)}$$ $$ *$$ $$ \bar{i} $$ $$ = $$ $$d_1^{(e)}$$ $$ \tilde{i} $$ $$ + $$ $$ d_2^{(e)} \tilde {j} $$ $$ $$ = $$ $$d_1^{(e)}$$ $$ ($$ $$\tilde{i}$$$$*$$ $$\bar{i}$$$$)$$ $$ + $$ $$ d_2^{(e)}$$ $$ ( $$ $$ \tilde{j}$$ $$*$$$$ \bar{i}) $$ $$=$$$$ d_1^{(e)}$$$$*$$ $$\cos (\theta^{(e)}) \! $$ $$ + $$ $$ d_2^{(e)}$$$$ * $$ $$ \sin (\theta^{(e)}) \!$$ $$q_1^{(e)}$$$$= \!$$$$ d_1^{(e)}$$$$ * \!$$ $$ l^{(e)} \!$$ $$+ \!$$ $$d_2^{(e)} $$ $$ * \!$$ $$ m^{(e)} \!$$
 * $$ $$ \bar {i} $$

Homework Problem Using the same procedure as before for element 2 of the truss system gives us:

$$q^{(e)}_2= (d^{(e)}_3\bar{i}+d^{(e)}_4\bar{j}) \cdot \bar{\tilde{i}}$$

$$= d^{(e)}_3(\bar{i}\cdot \bar{\tilde{i}})+d^{(e)}_4(\bar{j} \cdot\bar{\tilde{i}}) $$

$$= d_3^{(e)}\cdot cos(\theta^{(e)})+d_4^{(e)}\cdot sin(\theta^{(e)})$$

$$q_2^{(e)}$$ $$ = \!$$ $$d_3^{(e)}$$$$ *\! $$$$ l^{(e)}\!$$ $$ + \!$$ $$d_4^{(e)}$$ $$ *\!$$ $$ m^{(e)}\!$$

Combining these two equations gives us the result for the $$ T \! $$ matrix:

$$ \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix} $$ $$ = \! $$$$ \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\  0  &  0  & l^{(e)} &  m^{(e)} \end{bmatrix} $$ $$ * \! $$ $$ \begin{Bmatrix} d_1^{(e)} \\d_2^{(e)} \\d_3^{(e)} \\ d_4^{(e)}\end{Bmatrix} $$

The same argument can also be applied to the axial force on the truss system, using the same procedure used to find the axial displacement of the truss system:

$$P^{(e)} \!$$ $$ = \!$$ $$T^{(e)}\!$$ $$* \!$$ $$f^{(e)}\!$$

Gives:

$$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix} $$ $$ = \! $$$$ T_{2X4}^{(e)} $$ $$ * \! $$ $$ \begin{Bmatrix} f_1^{(e)} \\f_2^{(e)} \\f_3^{(e)} \\ f_4^{(e)}\end{Bmatrix} $$

We can then replace the axial forces for the equations derive, representing the element forces:

$$k_{2X1}^{(e)}$$ $$*$$ $$ q_{2X1}^{(e)}$$$$=\!$$$$ P_{2x1}^{(e)}$$ and $$k^{(e)} \!$$$$ * \!$$ $$\underbrace{(T^{(e)}\! d^{(e)}) \!}_{q^{(e)}}=\!$$ $$ \underbrace{(T^{(e)} \!  f^{(e)})\!}_{P^{(e)}}$$ Using the inverse of $$ T \! $$, we can get the following solution for the element forces: $$ \begin{bmatrix} (T^{(e)})^{T}k^{(e)}I^{(e)} \end{bmatrix}*d^{(e)} = f^{(e)}$$ However, we cannot invert $$ T \!$$ with itself since it is a rectangular matrix with a two by two matrix $$ k \!$$; we instead use the Principal of Virtual Work (PVW) to find $$f_1 \! $$, $$ f_2 \!$$, $$ f_5 \!$$, and $$ f_6 \!$$. There is no need to find $$ f_3 \! $$ and $$ f_4 \!$$ since they are the applied load on the truss system, which is already known.

The following loop used for the two bar truss system to compare between the Finite Element Method (FEM) and the statics method went as follows, starting with the FEM method:

FEM Compute displacements Compute reactions

Then the loop goes back using the statics method:

Statics Compute reactions Compute displacements By using the statics method, the reactions $$ f_1 \! $$, $$ f_2 \! $$, $$ f_5 \! $$, and $$ f_6 \! $$ are known and with it, the forces $$ P \! $$ are then known. This knowledge can then be used to find the axial displacement $$ q \! $$ at each node on the truss system:

$$q_2^{(1)} $$ $$ =\frac{P_{2}^{(1)}}{k^{(1)}} \! $$$$ = \!$$ $$AC \!$$

$$q_1^{(1)} $$$$ = \!$$ $$0 \!$$ ,fixed node $$1 \!$$

$$q_1^{(2)}$$ $$ =\frac{P_{2}^{(2)}}{k^{(2)}} $$ $$= \!$$ $$ -AB \! $$

$$q_2^{(2)} $$$$=\! $$ $$0 \! $$ ,fixed node $$ 2 \! $$

Another method used to find the displacement of global node 2 of the two body truss system is to find the deformed shape of the truss system itself.

The first part is to find the deformed lengths of $$ AC \! $$ and $$ AB \! $$:

$$ AC = \! $$ $$ \frac{P_{2}^{(1)}}{k^{(1)}} = \frac{5.1243}{.75} = 6.8324$$

$$ AB = \! $$ $$\frac{P_{1}^{(2)}}{k^{(2)}}= \frac{6.2760}{5} = 1.255 \!$$

Then the x-y coordinates of points B and C are determined, leaving only two unknowns $$ x_D \! $$ and $$ y_D \! $$. To find this displacement, the formula used to determine said point is:

$$ x-x_{p}= (PQ)cos \theta \!$$ $$ y-y_{p}= (PQ)sin\theta \!$$

$$ y-y_{p}= tan(\theta + \frac{\pi }{2})(x-x_{p})$$

with $$ x_p \!$$ and $$ y_p \! $$ representing points B and C on the displacement diagram. However, we can simply use point A as our reference point since it is already on the point of origin on the x-y axis.

$$AD = (x_{d}-x_{a})\bar{i} + (y_{d}-y_{a})\hat{j} \!$$ $$AD = (x_{d} -0)\bar{i} + (y_{d} -0)\hat{j} \! $$

With the resulting definition becoming:

$$AD = d_{3}*\bar{i} + d_{4}*\hat{j} \! $$

This ultimately closes the loop for the two bar truss system. The same procedure can be used to find the displacement for a three bar truss system