User:Eml4500.f08.FEABBQ/HW7

=Class Notes November 17, 2008=

Model Frame Problem with 2 Elements:
In this lecture, the two element system below was used to introduce the concept of frames.



Frames include beam elements which involve both translational and rotational degrees of freedom (dofs). Figure 2 below shows the free-body diagrams of both elements.

Element FBD's:


Note that each dof has a corresponding force. Also to be noted is that $$d^{(e)}_3$$ and $$d^{(e)}_6$$ correspond to rotational dofs. Similarly,$$f^{(e)}_3$$ and $$f^{(e)}_6$$ correspond to moments.

2-D Frame Showing Global dofs:
Figure 3 below shows the global FBD for the system. Notice the three dofs at each node compared to two dofs at each node for the similar two bar truss example.



For this example, the stiffness matrices for each element are square matrices with six rows and six columns: $$ \underline{K}^{(e)}_{6x6}$$, e=1,2

Similarly, the global stiffness matrix is a square matrix with nine rows and nine columns: $$ \underline{K}_{9x9}= {A^{e=2}_{e=1}}$$    $$\underline{K}^{(e)}_{6x6}$$

Assembly of Global Stiffness Matrix
The final portion of this lecture included the "recipe" for assembling the global stiffness matrix and is shown in figure 4 below.



=Class Notes November 19, 2008=



$${\tilde{k}}^{(e)}_{6x6} {\tilde{d}}^{(e)}_{6x1}={\tilde{f}}^{(e)}_{6x1}$$

$${\tilde{d}}^{(e)}_{6x1}=\begin{Bmatrix} {\tilde{d}}^{(e)}_{1} \\ {\tilde{d}}^{(e)}_{2}\\ {\tilde{d}}^{(e)}_{3}\\ {\tilde{d}}^{(e)}_{4}\\ {\tilde{d}}^{(e)}_{5}\\ {\tilde{d}}^{(e)}_{6} \end{Bmatrix} $$   ,    $${\tilde{f}}^{(e)}_{6x1}=\begin{Bmatrix} {\tilde{f}}^{(e)}_{1} \\ {\tilde{f}}^{(e)}_{2}\\ {\tilde{f}}^{(e)}_{3}\\ {\tilde{f}}^{(e)}_{4}\\ {\tilde{f}}^{(e)}_{5}\\ {\tilde{f}}^{(e)}_{6} \end{Bmatrix}$$

$$\tilde{z}=z $$ ( Rotation about Z-axis)

Moments about the Z-axis are equal in both coordinates: $$\tilde{f}^{(e)}_{3}=f^{(e)}_{3}, \tilde{f}^{(e)}_{6}=f^{(e)}_{6}$$

$$

\tilde{k}=\begin{bmatrix} \frac{EA}{L} &0 &0  &\frac{-EA}{L}  &0  &0 \\ 0 &\frac{12EI}{L^{2}} &\frac{6EI}{L^2} &0 &\frac{-12EI}{L^3}  &\frac{6EI}{L^2} \\ 0&0 &\frac{4EI}{L}  &0  & \frac{-6EI}{L^2} &\frac{2EI}{L} \\ \frac{-EA}{L}& 0 & 0 & \frac{EI}{L}&  0& 0\\ 0&\frac{-12EI}{L^3} &\frac{-6EI}{L^2}  & 0 &  \frac{12EI}{L^3}& \frac{-6EI}{L^2}\\ 0&\frac{6EI}{L^2} & \frac{2EI}{L} &  0& \frac{-6EI}{L^2} &\frac{4EI}{L} \end{bmatrix}_{(6X6)}$$


 * Dimension Analysis:

For Displacement the dimension $$\begin{bmatrix} \tilde{d_1} \end{bmatrix}=L= \begin{bmatrix} \tilde{d_i} \end{bmatrix} \rightarrow   i= 1,2,4,5 $$

$$L\rightarrow$$ Lenght

$$\begin{bmatrix} \tilde{d_i} \end{bmatrix} \rightarrow$$ Dimension of $$\tilde{d}_i$$

The rotation has a dimension of 1 $$\begin{bmatrix} \tilde{d_3} \end{bmatrix}=1= \begin{bmatrix} \tilde{d_6} \end{bmatrix} $$



$$AB=R\cdot \theta$$

$$AB\rightarrow$$ Arc Length

$$\theta \rightarrow$$ Angle

$$ \theta =\frac{AB}{R}$$

$$\left[\theta \right]=\frac{\left[AB \right]}{\left[R \right]}=\frac{L}{L}=1$$

=Class Notes November 21, 2008= Continuation of Dimensional Analysis

$$[ \varepsilon ] = \frac{du}{dx} = \frac{L}{L}=1$$

$$[ \sigma ] = [E] = \frac{F}{L^2}$$

$$\left[ \frac{EA}{L}\right] = \left[ \tilde{k_{11}}\right]=\frac{(F/L^2)(L^2)}{L}=\frac{F}{L}$$

$$\left[ \tilde{k_{11}} \tilde{d_1} \right] = [\tilde{k_{11}}][\tilde{d_1}]=\frac{[6][E][I]}{[L^2]} = \frac{1(F/L^2)L^4}{L^2} = F$$

Element F-d relation in global coordinates from element F=d relations in local coordinates, i.e. obtain

$$\mathbf{k}_{6x6}^{(e)}\mathbf{d}_{6x1}^{(e)}=\mathbf{f}_{6x1}^{(e)}$$

$$\mathbf{k}_{6x6}^{(e)}=\tilde{\mathbf{T}}_{6x6}^{(e)T}\tilde{\mathbf{k}}_{6x6}^{(e)}\tilde{\mathbf{T}}_{6x6}^{(e)}$$

=Class Notes November 24, 2008=

Deform Shape of Truss Element:Interpolation of Transverse Displacement $$ v(\tilde{x})$$ element
PVW for beams

$$ \int_{0}^{L}{w(\tilde{x}) \left[-\frac{\delta ^2}{\delta x^2} \left((EI)\frac{\delta^2 v}{\delta x^2} \right)  f_{t} - m\ddot{v}     \right] dx} = 0 $$ for all possible $$ w(x) $$

Now, Integrate by parts of 1st Term:

$$ \int_{0}^{L}{w(\tilde{x}) \left[-\frac{\delta ^2}{\delta x^2} \left((EI)\frac{\delta^2 v}{\delta x^2} \right)\right] dx} $$

where, $$ r(x) = \left( \frac{\partial }{\partial x}\left\{ (EI)\frac{\delta^2 v}{\delta x^2}\right\}\right) $$

$$ r^2(x) = \frac{\partial }{\partial x} \left( \frac{\partial }{\partial x}\left\{ (EI)\frac{\delta^2 v}{\delta x^2}\right\}\right) $$

$$ = \left[w \frac{\partial }{\partial x} \left\{(EI) \frac{\delta ^2v}{\delta x^2} \right\} \right]^{L}_{0} - \int_{0}^{L}{\frac{dw}{dx}\left( \frac{\partial }{\partial x}\left\{ (EI)\frac{\delta^2 v}{\delta x^2}\right\}\right)dx } $$

for simplicity, $$ \beta _{1} = \left[w \frac{\partial }{\partial x} \left\{(EI) \frac{\delta ^2v}{\delta x^2} \right\} \right]^{L}_{0} $$

Integrate again,

$$ = \beta _{1} - \left[\frac{dw}{dx}\left\{ (EI)\frac{\delta^2 v}{\delta x^2}\right\} \right]^{L}_{0} + \int_{0}^{L}{\frac{\delta ^2w}{\delta x^2} (EI) \frac{\delta^2v }{\delta x^2}dx} $$

to simplify further,

$$ \beta _2 = \left[\frac{dw}{dx}\left\{ (EI)\frac{\delta^2 v}{\delta x^2}\right\} \right]^{L}_{0} $$

$$ \gamma = \int_{0}^{L}{\frac{\delta ^2w}{\delta x^2} (EI) \frac{\delta^2v }{\delta x^2}dx} $$

Therefore the base equation (the boxed equation above) becomes,

$$ = - \beta _1 + \beta _2 - \gamma + \int_{0}^{L}{wf_t dx} - \int_{0}^{L}{wm\ddot{v}dx} $$

Now for further understanding focus on the stiffness term $$ \gamma $$ to dervie the beam stiffness matrix and also identify the beam beam shape functions.


 * Picture****

$$ v(\tilde{x}) = N_2(\tilde{x} )\tilde{d_2} +N_3(\tilde{x} )\tilde{d_3} + N_5(\tilde{x} )\tilde{d_5} + N_6(\tilde{x} )\tilde{d_6} $$

Recall: $$ u(\tilde{x}) = N_1(\tilde{x} )\tilde{d_1} N_4(\tilde{x} )\tilde{d_4} $$


 * Picture Set ****

HMWK ??? pg 246

=Class Notes December 1, 2008=

Computing $$U(\tilde{x}), V(\tilde{x)}$$ :



Local system: $$U(\tilde{x})=U(\tilde{x})\vec{\tilde{i}}+V(\tilde{x})\vec{\tilde{j}}$$

Global System:$$ U(\tilde{x})=U_{x}(\tilde{x})\vec+V_{y}(\tilde{x})\vec $$

Computing $$U(\tilde{x}), V(\tilde{x)}$$ using
 * $$V(\tilde{x)}=N_{2}(\tilde{x})\tilde{d_2}+N_{3}(\tilde{x})\tilde{d_3} +N_{5}(\tilde{x})\tilde{d_5}+N_{6}(\tilde{x})\tilde{d_6}$$
 * $$U(\tilde{x)}=N_{1}(\tilde{x})\tilde{d_1}+N_{4}(\tilde{x})\tilde{d_4}$$

Compute $$U_x(\tilde{x}), V_y(\tilde{x)}$$ from $$U(\tilde{x}), V(\tilde{x)}$$

$$\begin{Bmatrix} U_x(\tilde{x)}\\ U_y(\tilde{x)} \end{Bmatrix}= R^T \begin{Bmatrix} U(\tilde{x})\\V(\tilde{x})

\end{Bmatrix}$$

$$\begin{Bmatrix} U(\tilde{x})\\V(\tilde{x})

\end{Bmatrix}=\begin{bmatrix} N_1 &0 & 0 &N_4  &0  & 0\\ 0 &N_2 &N_3  & 0 & N_5 & N_6 \end{bmatrix}\begin{Bmatrix} \tilde{d_1^{(e)}}\\ \tilde{d_2^{(e)}}\\ \tilde{d_3^{(e)}}\\ \tilde{d_4^{(e)}}\\ \tilde{d_5^{(e)}}\\ \tilde{d_6^{(e)}} \end{Bmatrix}$$

let: $$ \begin{bmatrix} N_1 &0 & 0 &N_4  &0  & 0\\ 0 &N_2 &N_3  & 0 & N_5 & N_6 \end{bmatrix} =\mathbb{N(\tilde{\mathrm{x})}}$$

$$\begin{Bmatrix} U_x(\tilde{x)}\\ U_y(\tilde{x)} \end{Bmatrix}=R^T \mathbb{N(\tilde{\mathrm{x})}}\tilde{T}^{(e)}d^{(e)}$$

=Class Notes December 5, 2008=

$$\begin{Bmatrix} u_x(\tilde{x})\\ u_y(\tilde{x}) \end{Bmatrix} = \mathbf{R}^T\begin{Bmatrix} u(\tilde{x})\\ v(\tilde{x}) \end{Bmatrix}$$  (1)

$$\begin{Bmatrix} u_x(\tilde{x})\\ u_y(\tilde{x}) \end{Bmatrix} = \begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0\\ 0 & N_2 & N_3 & 0 & N_5 & N_6 \end{bmatrix} \begin{Bmatrix} \tilde{d_1}^{(e)}\\ \vdots\\ \tilde{d_1}^{(e)} \end{Bmatrix}$$  (2)

Perform dimensional analysis of equation (2):

$$[u]=L$$
 * First, a look at the normal linear degrees of freedom:

$$[N_1]=[N_4] = 1$$

$$[N_1][\tilde{d_1}],[N_4][\tilde{d_4}]$$

$$1\cdot L,1\cdot L = L$$

The dimensional analysis for DOF's 2 and 5


 * Now a look at the dimensional analysis of a rotational DOF:

$$[v]=L$$

$$[\tilde{d_3}]=[1]$$

$$[N_3][\tilde{d_3}],[N_6][\tilde{d_6}]$$

$$L\cdot 1,L\cdot 1 = L$$

Derivation of Shape Functions

Recall: Governing PDE for beams. Look at governing equations without internal forces of transverse forces or distributed truss loads. This is the static case:

$$\frac{\delta ^2}{\delta x^2} \left\{(EI) \frac{\delta v^2}{\delta x^2}\right\}=0$$

If EI is constant along the beam: $$\frac{\delta ^2}{\delta x^2}v=0$$

After integrating this equation four times, you arrive at: $$\Rightarrow v(x)=C_0+C_1x^1+C_2x^2+C_3x^3$$

Now the boundary conditions can be applied to find $$C_0, C_1, C_2, C_3$$ for each shape function.


 * $$\mathbf{N_2}$$

$$v(0)=1=C_0$$

$$v(L)=0=1+C_1L+C_2L^2+C_3L^3$$

$$v'(x)=C_1+2C_2x+C_3x^2$$

$$v'(0)=0=C_1$$

$$v'(L)=0=2C_2L+3C_3L^2$$

Now solve for $$C_2, C_3$$:

$$C_3={-2C_2}/{3L}$$

plugging $$C_3$$ back into the equation for v(L): $$0=1+C_2L^2+{-2C_2}/{3L}L^3$$

$$C_2=-3/L, C_3=2/L^3$$

$$N_2(x)=1-\left( \frac{3}{L^2}\right)x+\left( \frac{2}{L^3}\right)x^3$$


 * $$\mathbf{N_3}$$

change... $$v(0)=0=C_0$$

$$v'(0)=1=C_1$$

$$v(L)=0=L+C_2L^2+C_3L^3$$

solving for $$C_3=-1/L^2-C_2/L$$

$$v'(L)=0=1+2C_2L+3C_3L^2$$

solving for $$C_3=-1/3L^2-2C_2/3L$$

Now solve for $$C_2, C_3$$ by first setting $$C_3$$ equal to itself.

$$C_3=-1/3L^2-2C_2/3L=-1/L^2-C_2/L$$

$$C_2=-2/L$$

$$C_3=1/L^2$$

$$N_2(x)=x-\left( \frac{2}{L}\right)x^2+\left( \frac{1}{L^2}\right)x^3$$


 * $$\mathbf{N_5}$$

$$v(0)=0=C_0$$

$$v'(0)=0=C_1$$

$$v(L)=1=0+C_2L^2+C_3L^3$$

$$v'(x)=C_1+2C_2x+C_3x^2$$

$$v'(L)=0=2C_2L+3C_3L^2$$

Now solve for $$C_2, C_3$$:

$$C_3={1-C_2L^2}/{L^3}$$ from v(L)

$$C_3={-2C_2}/{3L}$$ from v'(L)

Setting $$C_3$$ equal to itself: $${(1-C_2L^2)}/{L^3}={-2C_2}/{3L}$$

$$C_2={3}/{L^2}, C_3={-2}/{L^3}-2L/3$$

$$N_5(x)=\left( \frac{3}{L^2}\right)x^2-\left( \frac{2}{L^3}\right)x^3$$


 * $$\mathbf{N_6}$$

$$v(0)=0=C_0$$

$$v'(0)=0=C_1$$

$$v(L)=0=C_2L^2+C_3L^3$$

Solve for $$C_3$$... $$C_3=-C_2/L$$

$$v'(L)=1-2C_2L+3C_3L^2$$

Solving again for $$C_3$$... $$C_3 = 1/3L^2-{2C_2}/{3L}$$

Setting $$C_3$$ equal to itself yields $$C_2$$: $$C_2=-1/L$$

$$C_3=1/L^2$$

$$N_6(x)=-\left( \frac{1}{L}\right)x^2+\left( \frac{1}{L^2}\right)x^3$$

=Matlab Matters=

2 Bar Truss Problem With Original Conditions
The code used to calculate the deformations for the original two bar truss problem is shown below. Please note that code was appended to the end to create the plot of the deformed elements.

The figure below is a plot of the undeformed and deformed elements.



2 Bar Truss Problem With Modified Conditions
For this homework it was also necessary to modify the 2 bar truss code to accommodate varying cross-sectional areas and elasticities. To do this, two different formulas were used to find the stiffness matrices of each element:

$$ \hat{k}_{gen}^{(e)}=\frac{(2A_{1}E_{1}+A_{2}E_{1}+A_{1}E_{2}+2A_{2}E_{2})}{6L}\begin{bmatrix} 1 &-1 \\ -1 &1 \end{bmatrix} $$

$$ \hat{k}_{avg}^{(e)}=\frac{(A_{1}+A_{2})(E_{1}+E_{2}))}{4L}\begin{bmatrix} 1 &-1 \\ -1 &1 \end{bmatrix} $$

The following Matlab code was used to calculate the deformations for the given values.

The figure below is a plot comparing the results of the two methods. Notice that due to the choice of given values, the results of from using the average values is the same as the original two bar truss.



Frame-Bar Truss
This function was used in the frame-bar truss code above

The following code is provided for solving problems that are purely frames (no bars as above)

=Wikiversity vs. E-learning=

=Contributing members:=
 * Stark 21:14, 9 December 2008 (UTC)
 * Esmail Hadman 23:07, 8 December 2008 (UTC)
 * Nik Vitt 00:04, 9 December 2008 (UTC)
 * Andy Koby 18:58, 9 December 2008 (UTC)
 * J.Jayma 19:48, 9 December 2008 (UTC)