User:Eml4500.f08.group/Homework4/Lecture Notes

Lecture 18
connectivity array

Consider the two bar truss system below



The connectivity array:

$$conn=\begin{bmatrix} 1 & 2\\ 2& 3 \end{bmatrix} \begin{matrix} element 1\\ element  2 \end{matrix}$$

Where the numbers 1,2,3 inside the matrix represent the global node number and the columns represent the local node number.

In matlab: conn(e,j)= the global node number of local node j of element e.

Location matrix master array:

$$lmm=\begin{bmatrix} 1 & 2&3&4\\ 3& 4&5&6 \end{bmatrix} \begin{matrix} element 1\\ element  2 \end{matrix}$$

Where the numbers in the matrix represent the global degree of freedom number, or the equation number in the global stiffness matrix K

In matlab: lmm(e,j)= the equation number(global degree of freedom number) for the element stiffness coefficient corresponding to the jth local degree of freedom number.

=Derivation of Element Stiffness Matrix=

Lecture 19: Split, please contribute at least a fifth of the notes
Goal: WE want to find a transformation matrix ~T(e) that transforms one set of local element degrees of freedom, d(e), into another set of local degrees of freedoms, ~d(e), such that the transformation matrix can be inverted.

$$ \tilde{d^{(e)}}_{4x1} = \tilde{T^{(e)}}_{4x4} d^{(e)}_{4x1}$$

Equation (1) $$ \tilde{d_{1}}^{(e)}=\begin{bmatrix} l^{(e)}& m^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}

\end{Bmatrix} $$

Derivation: $$\tilde{d_{1}^{(e)}}=\vec{d}_{1}^{(e)}\cdot \vec{\tilde{i}}$$ $$=(d_{1}^{(e)}\vec{i}+d_{2}^{(e)}\vec{j})\cdot \vec{\tilde{i}}$$ $$=d_{1}^{(e)}(\vec{i}\cdot \vec{\tilde{i}})+d_{2}^{(e)}(\vec{j}\cdot \vec{\tilde{i}})$$ $$ =cos\theta ^{(e)}d_{1}^{(e)} + sin\theta ^{(e)}d_{2}^{(e)} $$ From the definition of l(e) equal to cosθ(e) and m(e) equal to sinθ(e), the above equation reduces to: $$\tilde{d_{1}^{(e)}}=l^{e}d_{1}^{e}+m^{e}d_{2}^{e}$$ In matrix form reduces to: $$\tilde{d_{1}^{(e)}}=\begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)} \\ d_{2}^{(e)} \end{Bmatrix}$$

Equation (2) $$ \tilde{d_{2}}^{(e)}=\begin{bmatrix} -m^{(e)}& l^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}

\end{Bmatrix} $$

Derivation: $$\tilde{d_{2}^{(e)}}=\vec{d}_{1}^{(e)}\cdot \vec{\tilde{j}}$$ $$ =-sin\theta ^{(e)}d_{1}^{(e)} + cos\theta ^{(e)}d_{2}^{(e)} $$ From the definition of l(e) equal to cosθ(e) and m(e) equal to sinθ(e), the above equation reduces to: $$\tilde{d_{2}^{(e)}}=-m^{e}d_{1}^{e}+l^{e}d_{2}^{e}$$ In matrix form reduces to: $$\tilde{d_{2}^{(e)}}=\begin{bmatrix} -m^{(e)} & l^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)} \\ d_{2}^{(e)} \end{Bmatrix}$$

Put equation (1) and equation (2) together in matrix form:

$$ \begin{Bmatrix} \tilde{d_{1}}^{(e)}\\ \tilde{d_{2}}^{(e)}

\end{Bmatrix}=\begin{bmatrix} l^{(e)}& m^{(e)}\\-m^{(e)}& l^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}

\end{Bmatrix} $$

Where $$R^{(e)}= \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix}$$

$$\mathbf{\tilde{d}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{d}^{(e)}_{4x1} \Rightarrow \begin{Bmatrix} \tilde{d^{(e)}_{1}}\\ \tilde{d^{(e)}_{2}}\\ \tilde{d^{(e)}_{3}}\\ \tilde{d^{(e)}_{4}} \end{Bmatrix} = \begin{bmatrix} \textbf{R}^{(e)}_{2x2} & \textbf{0}_{2x2}\\ \textbf{0}_{2x2} & \textbf{R}^{(e)}_{2x2} \end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\\ d^{(e)}_{3}\\ d^{(e)}_{4} \end{Bmatrix}$$



$$\mathbf{\tilde{f}}^{(e)} = k^{(e)}\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\mathbf{\tilde{d}}^{(e)}$$

$$\mathbf{\tilde{f}}^{(e)}_{4x1} = \mathbf{\tilde{k}}^{(e)}_{4x4}\mathbf{\tilde{d}}^{(e)}_{4x1}$$

Considering the case where $$\mathbf{\tilde{d}}^{(e)}_{4} \neq 0 \Rightarrow \mathbf{\tilde{d}}^{(e)}_{1} = \mathbf{\tilde{d}}^{(e)}_{2} = \mathbf{\tilde{d}}^{(e)}_{3} = 0$$

$$\mathbf{\tilde{f}}^{(e)}_{4x1} = \mathbf{\tilde{k}}^{(e)}_{4x4}\mathbf{\tilde{d}}^{(e)}_{4x1} = \textbf{0}_{4x1}$$, the 4th column of the stiffness matrix.

Interpretation of Transverse Degrees of Freedom

 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!HW Force relationship, from team group
 * Using the relationship between distance and T, we can find the relationship between force and T: $$\mathbf{\tilde{d}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{d}^{(e)}_{4x1}$$:
 * Using the relationship between distance and T, we can find the relationship between force and T: $$\mathbf{\tilde{d}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{d}^{(e)}_{4x1}$$:

$$\mathbf{\tilde{f}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{f}^{(e)}_{4x1} \Rightarrow \begin{Bmatrix} \tilde{f^{(e)}_{1}}\\ \tilde{f^{(e)}_{2}}\\ \tilde{f^{(e)}_{3}}\\ \tilde{f^{(e)}_{4}} \end{Bmatrix} = \begin{bmatrix} \textbf{R}^{(e)}_{2x2} & \textbf{0}_{2x2}\\ \textbf{0}_{2x2} & \textbf{R}^{(e)}_{2x2} \end{bmatrix}\begin{Bmatrix} f^{(e)}_{1}\\ f^{(e)}_{2}\\ f^{(e)}_{3}\\ f^{(e)}_{4} \end{Bmatrix}$$

contribution by carbon.w
 * }

In addition, using the force-distance relationship:

$$\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)} \Rightarrow \mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}_{(e)}\mathbf^{(e)} = \mathbf{\tilde{T}}_{(e)}\mathbf^{(e)}$$

If $$\mathbf{\tilde{T}}_{(e)}$$ is invertible, then the relationship becomes:

$$[\mathbf{\tilde{T}}^{(e)-1}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)}]\mathbf^{(e)} = \mathbf^{(e)}$$

$$\mathbf{\tilde{T}}_{(e)}$$ is a block diagonal matrix. The general form of a block diagonal matrix is as follows:

$$\textbf{A} = \begin{bmatrix} \textbf{D}_{1} & & \textbf{0}\\ & ... & \\ \textbf{0} & & \mathbf{D_{s}} \end{bmatrix}$$

Using a simpler example of the diagonal matrix, to display more clearly the format, we can determine A-1:

$$\textbf{B} = \begin{bmatrix} d_{11} & & & \textbf{0}\\ & d_{22} & & \\ & & ... & \\ \textbf{0} & & & d_{nn} \end{bmatrix} = Diag[d_{11}, d_{22},...,d_{nn}]$$

$$\mathbf{B^{-1}} = Diag[\frac{1}{d_{11}}, \frac{1}{d_{22}},...,\frac{1}{d_{nn}}]$$

The above is true if we assume that dii does not equal zero for i=1,...,n and a block diagonal matrix would be:

$$\mathbf{A} = Diag[\mathbf{D_{1}},...,\mathbf{D_{s}}]$$ $$\mathbf{A^{-1}} = Diag[\mathbf{D^{-1}_{1}},...,\mathbf{D^{-1}_{s}}]$$ Therefore $$\mathbf\tilde{T}^{(e)-1} = Diag[\mathbf{R^{(e)-1}},\mathbf{R^{(e)-1}}]$$

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


 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!In-class Team Exercise, from team group l^{(e)} & -m^{(e)}\\ m^{(e)} & l^{(e)} \end{bmatrix} * \begin{bmatrix} l^{(e)} & m(e)\\ -m^{(e)} & l^{(e)} \end{bmatrix} = [(l^{(e)})^{2}  (m^{(e)})^{2}][(-l^{(e)}m^{(e)})   (m^{(e)}l^{(e)}]$$
 * $$\mathbf{R^{(e)T}}*\mathbf{R^{(e)}} = \begin{bmatrix}
 * $$\mathbf{R^{(e)T}}*\mathbf{R^{(e)}} = \begin{bmatrix}

$$\mathbf{R^{(e)T}}*\mathbf{R^{(e)}} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}_{2x2} = \mathbf{\bar{I}_{2x2}}$$ (the identity matrix)

contribution by carbon.w
 * }

$$\Rightarrow \mathbf{R^{(e)-1}} = \mathbf{R^{(e)T}}$$

$$\mathbf{\tilde{T}^{(e)-1}} = Diag[\mathbf{R^{(e)T}}, \mathbf{R^{(e)T}}] = (Diag[\mathbf{R^{(e)}}, \mathbf{R^{(e)}}])^{T}$$

$$\Rightarrow \mathbf{\tilde{T}^{(e)-1}} = \mathbf{\tilde{T}^{(e)T}}$$


 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!HW Force relationship, from team group
 * Using the relationship shown above,
 * Using the relationship shown above,

$$[\mathbf{\tilde{T}}^{(e)-1}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)}]\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{T}}_{(e)}\mathbf{\tilde{f}}^{(e)}$$,

and plugging in $$\mathbf{\tilde{T}^{(e)-1}} = \mathbf{\tilde{T}^{(e)T}}$$, we arrive at:

$$[\mathbf{\tilde{T}}^{(e)T}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)}]\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)}$$

$$\mathbf{k}^{(e)} = \mathbf{\tilde{T}}^{(e)-1}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)} $$

contribution by carbon.w
 * }

Lecture 22
Justification of assembling local element stiffness matrices into a global stiffness matrix. Consider the example of the 2-bar truss system discussed during the 9th lecture.









First we note the global displacements and forces that correspond to the element displacements and forces as shown below.

$$\begin{bmatrix} k^{(1)}_{11} & k^{(1)}_{12} & k^{(1)}_{13} & k^{(1)}_{14}\\ k^{(1)}_{21} & k^{(1)}_{22} & k^{(1)}_{23} & k^{(1)}_{24}\\ k^{(1)}_{31} & k^{(1)}_{32} & k^{(1)}_{33} & k^{(1)}_{34}\\ k^{(1)}_{41} & k^{(1)}_{42} & k^{(1)}_{43} & k^{(1)}_{44} \end{bmatrix}*\begin{Bmatrix} D1\\ D2\\ D3\\ D4 \end{Bmatrix}=\begin{Bmatrix} F1\\ F2\\ F3\\ F4 \end{Bmatrix}$$ $$\begin{bmatrix} k^{(2)}_{11} & k^{(2)}_{12} & k^{(2)}_{13} & k^{(2)}_{14}\\ k^{(2)}_{21} & k^{(2)}_{22} & k^{(2)}_{23} & k^{(2)}_{24}\\ k^{(2)}_{31} & k^{(2)}_{32} & k^{(2)}_{33} & k^{(2)}_{34}\\ k^{(2)}_{41} & k^{(2)}_{42} & k^{(2)}_{43} & k^{(2)}_{44} \end{bmatrix}*\begin{Bmatrix} D3\\ D4\\ D5\\ D6 \end{Bmatrix}=\begin{Bmatrix} F3\\ F4\\ F5\\ F6 \end{Bmatrix}$$

Next we can fill in the global stiffness matrix by observing that the columns of this stiffness matrix correspond to global displacements and the rows correspond to the global forces as shown below.

$$\begin{bmatrix} & (D1) & (D2) & (D3) & (D4) & (D5) & (D6)\\ (F1) & K11 & K12 & K13 & K14 & K15 & K16\\ (F2) & K21 & K22 & K23 & K24 & K25 & K26\\ (F3) & K31 & K32 & K33 & K34 & K35 & K36\\ (F4) & K41 & K42 & K43 & K44 & K45 & K46\\ (F5) & K51 & K52 & K53 & K54 & K55 & K56\\ (F6) & K61 & K62 & K63 & K64 & K65 & K66 \end{bmatrix}$$

My plugging the local stiffness values into the above matrix, we achieve the global stiffness matrix as shown below. This matrix exactly matches the global stiffness matrix for this 2-bar truss system which was given in meeting 9. $$\begin{bmatrix} k_{11}^{(1)}&k_{12}^{(1)}&k_{13}^{(1)}&k_{14}^{(1)} &0  &0 \\ k_{21}^{(1)}&k_{22}^{(1)} &k_{23}^{(1)}  &k_{24}^{(1)}  &0  &0 \\ k_{31}^{(1)}&k_{32}^{(1)} &k_{33}^{(1)}+k_{11}^{(2)}  &k_{34}^{(1)}+k_{12}^{(2)}  &k_{13}^{(2)}  &k_{14}^{(2)} \\ k_{41}^{(1)}&k_{42}^{(1)} &k_{43}^{(1)}+k_{21}^{(2)}  &k_{44}^{(1)}+k_{22}^{(2)}  &k_{23}^{(2)}  &k_{24}^{(2)} \\ 0& 0&  k_{31}^{(2)}&k_{32}^{(2)}  &k_{33}^{(2)}  &k_{34}^{(2)} \\ 0& 0&  k_{41}^{(2)}&k_{42}^{(2)}  &k_{43}^{(2)}  &k_{44}^{(2)} \end{bmatrix}$$

Equilibrium of Node 2

Based on the above element diagrams, the equilibrium of node 2 can be shown as follows.



$$\sum{F_{x}}=0=-f^{(1)}_{3}-f^{(2)}_{1}=0$$

$$\sum{F_{y}}=0=P-f^{(1)}_{4}-f^{(2)}_{2}=0$$

Lecture 23
From equations (1) and (2), we find:

Eq. (1) $$=> f_3^{(1)} + f_1^{(2)} = 0$$

Eq. (2) $$=> f_4^{(1)} + f_2^{(2)} = P$$

Futhermore using the relation f=kd, we find:

(1) $$[k_{31}^{(1)}d_1^{(1)} + k_{32}^{(1)}d_2^{(1)} + k_{33}^{(1)}d_3^{(1)} + k_{34}^{(1)}d_4^{(1)}] + [k_{11}^{(2)}d_1^{(2)} + k_{12}^{(2)}d_2^{(2)} + k_{13}^{(2)}d_3^{(2)} + k_{14}^{(2)}d_4^{(2)}] = 0 $$

(2) $$[k_{41}^{(1)}d_1^{(1)} + k_{42}^{(1)}d_2^{(1)} + k_{43}^{(1)}d_3^{(1)} + k_{44}^{(1)}d_4^{(1)}] + [k_{21}^{(2)}d_1^{(2)} + k_{22}^{(2)}d_2^{(2)} + k_{23}^{(2)}d_3^{(2)} + k_{24}^{(2)}d_4^{(2)}] = P $$

Next, we want to transform from local to global dofs:

(1) $$[k_{31}^{(1)}d_1 + k_{32}^{(1)}d_2 + k_{33}^{(1)}d_3 + k_{34}^{(1)}d_4] + [k_{11}^{(2)}d_3 + k_{12}^{(2)}d_4 + k_{13}^{(2)}d_5 + k_{14}^{(2)}d_6] = 0 $$

(2) $$[k_{41}^{(1)}d_1 + k_{42}^{(1)}d_2 + k_{43}^{(1)}d_3 + k_{44}^{(1)}d_4] + [k_{21}^{(2)}d_3 + k_{22}^{(2)}d_4 + k_{23}^{(2)}d_5 + k_{24}^{(2)}d_6] = P $$

Thus, we have obtained the 3rd and 4th rows of the global stiffness matrix, K, as seen on P. 9-1.

Next, let's look at the Assembly of $$\boldsymbol k^{(e)}$$, where e=1,...,nel (the number of elements), into the global stiffness matrix K:

$$\boldsymbol K = \overset{e=nel}{\underset{e=1}{A}} \boldsymbol k^{(e)}$$

where,

n: is the total number of global dofs before eliminating boundary conditions.

ned: is the number of elemental dofs (ned< kd-F=0$$

(4) $$ <=> w(kd-F)=0$$ for all w.

Proof:

(3) => (4) is trivial.

(4) => (3) is not trivial. Since (4) is valid for all w, you can select and arbitrary w, say w=1, then (4) becomes:

1. $$(kd-F)=0 =>$$ (3)