User:Eml4500.f08.echo.mott/HW2

This is the drafting page for homework report 2 - due Friday, September 26.

The link to the FINAL SUBMITTAL PAGE is HERE
 * This time I will leave everything on this page and just copy it to the final page. Chris 15:11, 26 September 2008 (UTC)

Mon, Sept 8
2 Force body example:



Solution to this problem:

$$\sum F_x = 0; R_1cos( \theta_1) - R_2cos( \theta_2) + Pcos( \theta_P) = 0$$

$$\sum F_y = 0; R_1sin( \theta_1) - R_2sin( \theta_2) + Psin( \theta_P) = 0$$

$$\sum M_1 = 0; R_2sin( \theta_2)*(l_2cos( \theta_2) + l_1cos( \theta_1)) + Psin( \theta_P)*(l_1cos( \theta_1)) = 0$$

Three equations and three unknowns, therefore it is statically determinate.

Hey are there values for this? I know he wanted us to solve it using statics but I don't know if he gave us values - --70.185.100.27 00:28, 26 September 2008 (UTC)Eml4500.f08.echo.bruce.jp

I don't have any values either. I don't think he gave us any values, just wanted us to do it symbolically or something. I did add Step 1) below (it was from the previous lecture) Eml4500.f08.echo.sell.br 04:54, 26 September 2008 (UTC)

I thought he wanted us to solve the problem that had P just being a vertical force but that sounds like maybe it would be too easy so i just wanted to see what everyone else had. I have time today Friday 9/26 to fix it but i wasn't sure if that was what he was asking for.- --70.185.100.27 00:28, 26 September 2008 (UTC)Eml4500.f08.echo.bruce.jp

I don't really know for sure. The rest of this section should be finished now. Eml4500.f08.echo.sell.br 06:44, 26 September 2008 (UTC)

I just realized the picture is slightly wrong. I'm going to update it and add the R1, R2,theta's, etc. Eml4500.f08.echo.lacroix 15:22, 26 September 2008 (UTC)

Finite Element Method

Step 1) Draw and label Global Picture

Step 2) Draw and label Element Pictures

Step 3) Create global FD at element level


 * Drawing and Labeling the Global Picture:




 * Breaking it down into two elements:




 * Statically indeterminate examples:



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

$$	\mathbf{k}^{(e)}$$ = 4x4 element stiffness matrix for element e where e = 1,2

$$	\mathbf{d}^{(e)}$$ = 4x1 element displacement matrix of element e

$$	\mathbf{f}^{(e)}$$ = 4x1 element force matrix of element e

The element stiffness matrix can be broken down as follows:

$$\mathbf{k}^{(e)}=k^{(e)}\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & (m^{(e)})^2 & -l^{(e)}m^{(e)} & -(m^{(e)})^2\\ -(l^{(e)})^2 & -l^{(e)}m^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)}\\-l^{(e)}m^{(e)} & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2\end{bmatrix}$$

The scalar value of the axial stiffness of bar element e can be found using the following equation:

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

$$l^{(e)}$$ is the x-length of the element in reference to global coordinates and $$m^{(e)}$$ is the y-length of the element from a global perspective. They can be thought of as the length of the element projected onto global coordinates from local coordinates.



$$l^{(e)}= \mathbf{\tilde{i}} \cdot \mathbf{i}=cos( \theta^{(e)})$$

$$m^{(e)}= \mathbf{\tilde{i}} \cdot \mathbf{j}=cos( \frac{ \pi}{2} - \theta^{(e)})=sin( \theta^{(e)})$$

The direction cosines are the components of $$\mathbf{\tilde{i}}$$ (unit vector along $$\mathbf{x}$$ axis) with respect to basis $$(\mathbf{i},mathbf{j})$$ $$\mathbf{\tilde{i}}=cos( \theta^{(e)})\mathbf{i}+sin( \theta^{(e)})\mathbf{j}=l^{(e)}+m^{(e)}$$

$$\mathbf{\tilde{i}}=cos( \theta^{(e)})\mathbf{i}+sin( \theta^{(e)})\mathbf{j}$$

$$\mathbf{\tilde{i}} \cdot \mathbf{i}=cos( \theta^{(e)})$$

$$\mathbf{\tilde{i}} \cdot \mathbf{j}=sin( \theta^{(e)})$$

Eml4500.f08.echo.sell.br 06:44, 26 September 2008 (UTC)

Eml4500.f08.echo.lacroix 23:57, 22 September 2008 (UTC)

Wed, Sept 10
I'll show how to compute the k matrix from given values of the sample problem. I might need a little more information from the last part of the lecture. *Chris 15:56, 15 September 2008 (UTC)

The element stiffness matrix takes the form shown below.

$$\mathbf{k^{(e)}}=k^{(e)} \begin{bmatrix} l^{(e)^2} & l^{(e)}m^{(e)} & -l^{(e)^2} & -l^{(e)}m^{(e)} \\l^{(e)}m^{(e)} & m^{(e)^2} & -l^{(e)}m^{(e)} & -m^{(e)^2} \\-l^{(e)^2} & -l^{(e)}m^{(e)} & l^{(e)^2} & l^{(e)}m^{(e)} \\-l^{(e)}m^{(e)} & -m^{(e)^2} & l^{(e)}m^{(e)} & m^{(e)^2}\end{bmatrix}$$

The scalar $$k^{(e)}$$ can be found from the properties of the structure where E is the Elastic Modulus, A is the cross-sectional area of the element, and L is the length of the element.

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

$$l^{(e)}$$ is the x-length of the element in reference to global coordinates and $$m^{(e)}$$ is the y-length of the element from a global perspective. They can be thought of as the length of the element projected onto global coordinates from local coordinates.

The x-length can be found by using: $$l_{s}= \frac{x_{2}-x_{1}}{L}$$

The y-length of the element can be found by using: $$l_{s}= \frac{y_{2}-y_{1}}{L}$$

With the inclined angle referenced to the horizontal, the horizontal and vertical lengths can be found by:

lsx = cos θ

lsy = sin θ

After reviewing the stiffness matrix two observations can be made.

1) Of all the numbers in the matrix, only three need to be computed. The other coefficients have the same absolute values.
 * $$l_s^{} m_s$$
 * $${l_s}^2$$
 * $${m_s}^{2}$$

2) Matrix k is symmetric. That is $$k_{i j}^{1}=k_{j i}^{1}$$.

Fri, Sept 12
I'll do the global and the element matrices assembly process.--C. Sell 18:28, 16 September 2008 (UTC) Still working on this days notes decided to post what i had for now.--C. Sell 16:24, 21 September 2008 (UTC)

Process to Assemble a Global Stiffness Matrix using Element Stiffness Matrix:

Consider the Global Force Displacement Relationship for a Truss with 2 elastic bars (6 dof's in Global picture.

Written in compact notation we have the following:

$$\begin{bmatrix}K_{ij}^{}\end{bmatrix}_{6x6} \left\{d_{ij}\right\}_{6x1} = \left\{F_{i}\right\}_{6x1}$$

In the more general form (nxn):

$$\sum_{j=1}^{6} K_{ij}d_{j}= F_{i}$$; i = 1, 2, ..., 6 $$K_{nxn} = \begin{bmatrix}K_{ij}^{}\end{bmatrix}_{nxn}$$ = global stiffness matrix $$d_{nx1} = \left\{d_{j}^{}\right\}_{nx1}$$ = global displacement matrix $$F_{nx1} = \left\{F_{i}^{}\right\}_{nx1}$$ = global force matrix

Now recall the Element Force-Displacement relationship:

$$k_{4x4}^{(e)} d_{4x4}^{(e)} = f_{4x1}^{(e)}$$ $$k_{4x4}^{(e)} = \begin{bmatrix}k_{ij}^{(e)}\end{bmatrix}_{4x4}$$ = element stiffness matrix $$d_{4x1}^{(e)} = \left\{d_{j}^{(e)}\right\}_{4x1}$$ = element displacement matrix $$f_{4x1}^{(e)} = \left\{f_{i}^{(e)}\right\}_{4x1}$$ = element force matrix

Use an assembly process to go from element matrices(stiffness, displacement, force) to global matrices.


 * First, Identify the correspondence between the element displacement degrees of freedom's and the global displacement degrees of freedom's.

Global Level: $$\left\{d_1, d_2, \dots, d_6\right\}$$

Element level:
 * $$\left\{d_1^{(1)}, d_2^{(1)}, d_3^{(1)}, d_4^{(1)}\right\}$$
 * $$\left\{d_1^{(2)}, d_2^{(2)}, d_3^{(2)}, d_4^{(2)}\right\}$$

The relationship between global-local(element) dof's are as follows:

$$\text{Global Node 1} = \begin{cases} d_1 = d_1^{(1)} \\ d_2 = d_2^{(1)}\end{cases}$$

$$\text{Global Node 2} = \begin{cases} d_3 = d_3^{(1)} = d_1^{(2)} \\ d_4 = d_4^{(1)} = d_2^{(2)}\end{cases}$$

$$\text{Global Node 3} = \begin{cases} d_5 = d_3^{(2)} \\ d_6 = d_4^{(2)}\end{cases}$$

Conceptual step of the assembly:



Mon, Sept 15
Taking the global stiffness matrix and substituting in the appropriate values:

$$\mathbf{K}= \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)}+k_{11}^{(2)} & k_{34}^{(1)}+k_{12}^{(2)} & k_{13}^{(2)} & k_{14}^{(2)}\\k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)}+k_{21}^{(2)} & k_{44}^{(1)}+k_{22}^{(2)} & k_{23}^{(2)} & k_{24}^{(2)}\\0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)}\\0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)}\end{bmatrix}= \begin{bmatrix} K_{11} & K_{12} & K_{13} & K_{14} & K_{15} & K_{16}\\K_{21} & K_{22} & K_{23} & K_{24} & K_{25} & K_{26}\\K_{31} & K_{32} & K_{33} & K_{34} & K_{35} & K_{36}\\K_{41} & K_{42} & K_{43} & K_{44} & K_{45} & K_{46}\\K_{51} & K_{52} & K_{53} & K_{54} & K_{55} & K_{56}\\K_{61} & K_{62} & K_{63} & K_{64} & K_{65} & K_{66}\end{bmatrix}$$

$$K_{11}=k_{11}^{(1)}=\frac{9}{16}=0.5625$$

$$K_{12}=K_{21}=k_{12}^{(1)}=\frac{3\sqrt{3}}{16}=0.3248$$

$$K_{13}=K_{31}=k_{13}^{(1)}=-k_{11}^{(1)}=-0.5625$$

$$K_{14}=K_{23}=K_{32}=K_{41}=k_{14}^{(1)}=-k_{12}^{(1)}=-0.3248$$

$$K_{22}=k_{22}^{(1)}=\frac{3}{16}=0.1875$$

$$K_{24}=K_{42}=k_{24}^{(1)}=-k_{22}^{(1)}=-0.1875$$

$$K_{33}=k_{33}^{(1)}+k_{11}^{(2)}=	\begin{matrix} \frac{9}{16} \end{matrix}+\begin{matrix} \frac{5}{2} \end{matrix}=3.0625$$

$$K_{34}=k_{34}^{(1)}+k_{12}^{(2)}=	\begin{matrix} \frac{3\sqrt{3}}{16} \end{matrix}+\begin{matrix} \frac{-5}{2} \end{matrix}=-2.1752$$

$$K_{35}=K_{53}=k_{13}^{(2)}=-k_{11}^{(2)}=\frac{-5}{2}=-2.5$$

$$K_{36}=K_{45}=K_{54}=K_{63}=k_{14}^{(2)}=-k_{12}^{2}=2.5$$

$$K_{43}=K_{34}=k_{43}^{(1)}+k_{21}^{(2)}=k_{34}^{(1)}+k_{12}^{(2)}=-2.1752$$

$$K_{44}=k_{44}^{(1)}+k_{22}^{(2)}=\frac{3}{16}+\frac{5}{2}=2.6875$$

$$K_{46}=K_{64}=k_{24}^{(2)}=\frac{-5}{2}=-2.5$$

$$K_{55}=k_{33}^{(2)}=\frac{5}{2}=2.5$$

$$K_{56}=K_{65}=k_{34}^{(2)}=\frac{-5}{2}=-2.5$$

$$K_{66}=k_{44}^{(2)}=\frac{5}{2}=2.5$$

$$K_{15}^{}=K_{16}=K_{25}=K_{26}=K_{51}=K_{52}=K_{61}=K_{62}=0$$

$$\mathbf{K}=\begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0\\0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0\\-0.5625 & -0.3248 & 3.0625 & -2.1752 & -2.5 & 2.5\\-0.3248 & -0.1875 & -2.1752 & 2.6875 & 2.5 & -2.5\\0 & 0 & -2.5 & 2.5 & 2.5 & -2.5\\0 & 0 & 2.5 & -2.5 & -2.5 & 2.5\end{bmatrix}$$

Wed, Sept 17
Next, the elimination of known degree's of freedom can create the following matrices.

$$ d_1^{} = d_2^{} = d_5^{} = d_6^{}=0 $$

$$	\mathbf{K}\begin{Bmatrix} d_1=0\\d_2=0\\d_3\\d_4\\d_5=0\\d_6=0 \end{Bmatrix}=\begin{vmatrix} k_{13} & k_{14} \\ k_{23} & k_{24}\\k_{33} & k_{34}\\k_{43} & k_{44}\\k_{53} & k_{54}\\k_{63} & k_{64}\end{vmatrix}\begin{Bmatrix} d_3\\d_4\end{Bmatrix}=\mathbf{F}$$

Applying fixed boundary conditions allows us to delete the corresponding columns in the global stiffness matrix K

By the Principal of Virtual Work (PVW) we're able to delete the corresponding rows (in this case - 1, 2, 5, 6). Corresponding rows of F are also deleted.

The resulting FD relation is: $$ \begin{bmatrix} k_{33} & k_{34} \\ k_{43} & k_{44} \end{bmatrix} \begin{Bmatrix} d_3\\d_4\end{Bmatrix}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}$$

$$\mathbf{F}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}=\begin{Bmatrix} 0\\P\end{Bmatrix}$$

And P =>

What is the inverse of K without using a calculator?

Taking the determinant of K:

$$det \mathbf{K}=k_{33}k_{44}-k_{k34}k_{43}$$

$$\mathbf{K}^{-1}=\frac{1}{det\mathbf{K}}\begin{bmatrix} k_{44} & -k_{34} \\ -k_{43} & k_{33} \end{bmatrix}$$

Verifying:

$$\mathbf{K}\mathbf{K}^{-1}=\mathbf{K}^{-1}\mathbf{K}=\mathbf{I}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

This is essentially:

$$\frac{1}{det(\mathbf{K})}\begin{bmatrix} det(\mathbf{K}) & 0 \\ 0 & det(\mathbf{K}) \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Note: $$\mathbf{K}^T=\begin{bmatrix} k_{33} & k_{43} \\ k_{34} & k_{44} \end{bmatrix}$$

Therefore: $$\mathbf{K}^{-1}\ne \ \frac{1}{det(\mathbf{K})}\mathbf{K}^T$$

Going back to the 2 bar truss problem...

$$\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix}=\mathbf{K}^{-1}\begin{Bmatrix} 0 \\ P \end{Bmatrix}=\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

Now we must compute the reactions:


 * Method 1:

Use element FD relations.

$$\mathbf{k}^{(1)}\mathbf{d}^{(1)}=\mathbf{f}^{(1)}$$

Taking a look at element 1 again and applying the knowns:



$$d^{(1)}=\begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}$$

Taking a look at element 2:



$$\mathbf{k}^{(2)}\mathbf{d}^{(2)}=\mathbf{f}^{(2)}$$

$$d^{(1)}=\begin{Bmatrix} 4.352 \\ 6.1271 \\ 0 \\ 0 \end{Bmatrix}=\begin{Bmatrix} d_1^{(2)} \\ d_2^{(2)} \\ d_3^{(2)} \\ d_4^{(2)} \end{Bmatrix}$$


 * Method 2: Not really sure when he phased into this....

Fri, Sept 19






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