User:Eml4500.f08.gravy.jad/Notes4.g

FEA Notes and Homework 4

Solving Truss systems in Matlab
Connectivity Array

Here we are using a connectivity array "conn" (from 2 bar truss system)

$$\ conn = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} \,$$

Where the columns represent the local node number, but the numbers inside the matrix state the global node number. The rows tell the element number. (ie. row 1 represents the first element, row 2 represents the second element, ect.)

conn(e,j)= global node # of local node j of element e.

Location Matrix Master Array

Here we are using a location matrix master array "lmm" (from 2 bar truss system)

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

Where the columns represent the local dof's, but the numbers inside the matrix state the global dof's or equation number in matrix k. The rows tell the element number. (ie. row 1 represents the first element, row 2 represents the second element, ect.)

lmm(i,j)= eq # (global dof's #) for the elem. stiffness coeff. corresponding to the jth local dof #.

Solving Truss systems in the $$\tilde{x}$$ and $$\tilde{y}$$ coordinate axis
Back to Derivation of element Force Displacement with respect to global coordinate system: Method 2 to desire k(e)4x4 : transform a system with 4 dofs into a system with also 4 dofs (instead of only 2 dofs) so that the transformation matrix is now 4x4 and inverted. We want a system of equations that follows a coordinate system which correlates to the axial direction of the bar.

Goal : Want to find $$\ T^{(e)}_{4x4} \,$$ that transform the set of local (elem) dofs $$\ d^{(e)}_{4x4} \,$$ to another set of local (elem) dofs $$\ \underline{\tilde{d}}^{(e)}_{4x4} \,$$ such that $$\ \underline{\tilde{T}}^{(e)} \,$$  is invertible.



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

Continuing the example from Force Displacement matrices calculations:

$$\ \tilde{d}^{(e)}_{1} = \left \lfloor \ l^{e} m^{e} \right \rfloor \begin{Bmatrix} d^{(e)}_{1} \\ d^{(e)}_{2} \end{Bmatrix} \,$$  where $$\ \underline{\tilde{ d }}^{(e)}_{1} \,$$ also equals $$\ q^{(e)}_{1} \,$$ (Equation 1)

$$\ \tilde{d}^{(e)}_{2} = \tilde{d}^{(e)}_{1} = \vec{d}^{(e)}_{1} \cdot \vec{\tilde{j}} \,$$ (component of $$\ \tilde{d}^{(e)}_{1} \,$$ along $$\ \vec{\tilde{j}} \,$$ j, ie. $$\ \tilde{y}-axis \,$$) (of the local nodes)

$$\ = - \sin\theta^{(e)} d^{(e)}_{1} + \cos \theta^{(e)} d^{(e)}_{2} \,$$

$$\ = \left \lfloor \ -m^{e} l^{e} \right \rfloor \begin{Bmatrix} d^{(e)}_{1} \\ d^{(e)}_{2} \end{Bmatrix} \,$$ (Equation 2)

Deriving the $$\tilde{d}^{(e)}$$ equation
Put (Equation 1) and (Equation 2) in matrix form:

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

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

A complete matrix form of all dofs

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



$$\ \underline {\tilde{f}}^{(e)}_{4x1}= k^{(e)} \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \underline{\tilde{d}}^{(e)} \,$$

In conclusion:

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

Special Case

$$\ \tilde{d}^{(e)}_{4} \ne 0 \,$$, $$\ \tilde{d}^{(e)}_{1} =\tilde{d}^{(e)}_{2} =\tilde{d}^{(e)}_{3} = 0 \,$$

Therefore: $$\ \underline{\tilde{f}}^{(e)}_{4x1} = \underline{\tilde{k}}^{(e)}_{4x4} \underline{\tilde{d}}^{(e)}_{4x1} = \underline{0}_{4x1} \,$$ where the zero matrix is the 4th column of $$\ \underline{\tilde{k}}^{(e)} \,$$

Now interpret the transverse of the dofs: $$\ \underline{\tilde{d}}^{(e)} = \underline{\tilde{T}}^{(e)} \tilde{d}^{(e)} \,$$

Similarly: $$\ \underline{\tilde{f}}^{(e)} = \underline{\tilde{T}}^{(e)} \tilde{f}^{(e)} \,$$

Prove of $$ \mathbf{\tilde{f}}^{(e)}$$ equation
Also: $$\ \underline{\tilde{k}}^{(e)} \underline{\tilde{d}}^{(e)} = \tilde{f}^{(e)} \Rightarrow \underline{\tilde{k}}^{(e)} \underline{\tilde{T}}^{(e)} {d}^{(e)} \,$$

Hence, if $$\ \underline{\tilde{T}}^{(e)} \,$$ is invertible then: $$\ \left [ {\underline{\tilde{T}}^{(e)}}^{-1} \underline{\tilde{k}}^{(e)} \underline{\tilde{T}}^{(e)} \right ] \underline{\tilde{d}}^{(e)} = \underline{f}^{(e)} \,$$

$$\ \underline{\tilde{T}}^{(e)} \,$$ is a block diagonal matrix (from earlier in this report).

Now consider the general block-diagonal matrix: $$\ A = \begin{bmatrix} \underline{D}_{1} & & \underline{0} \\ & \ddots &  \\ \underline{0} &  & \underline{D}_{s} \end{bmatrix} \,$$

Question: Then what is $$\ A^{-1} \,$$?

Answer: Lets make a simpler case- $$\ B = \begin{bmatrix} d_{11} & & & \underline{0} \\ & d_{22} & &  \\ & & \ddots & \\ \underline{0} &  & & D_{s} \end{bmatrix} = Diag \left [ d_{11},d_{22}, \ldots, d_{nn} \right ] \,$$

$$\ B^{-1}= Diag \left [ \frac{1}{d_{11}},\frac{1}{d_{22}}, \ldots, \frac{1}{d_{nn}} \right ] \,$$ assuming that $$\ d_{ij} \ne 0 \,$$ for $$\ i=1 \ldots n \,$$.

$$\ B \cdot B^{-1} = I \,$$

Now going back to the A-matrix:

$$\ A= Diag \left [ \underline{D}_{1},\underline{D}_{2}, \ldots, \underline{D}_{s} \right ] \,$$

$$\ A^{-1}= Diag \left [ \frac{1}{\underline{D}_{1}},\frac{1}{\underline{D}_{2}}, \ldots, \frac{1}{\underline{D}_{s}} \right ] = Diag \left [ \underline{D}^{-1}_{1},\underline{D}^{-1}_{2}, \ldots, \underline{D}^{-1}_{s} \right ] \,$$

Therefore: $$\ {\underline{\tilde{T}}^{(e)}}^{-1} = Diag \left [ {\underline{R}^{(e)}}^{-1}, {\underline{R}^{(e)}}^{-1} \right ] \,$$

Recall from earlier in this report $$\ \underline{R}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix} \,$$

Now: $$\ {\underline{R}^{(e)}}^{T} = \begin{bmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \end{bmatrix} \,$$

Therefore: $$\ {\underline{R}^{(e)}}^{T} \underline{R}^{(e)} = \begin{bmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \end{bmatrix}_{2x2} \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix}_{2x2} = \begin{bmatrix} {l^{(e)}}^{2} + {m^{(e)}}^{2} & 0 \\ 0 & {l^{(e)}}^{2} + {m^{(e)}}^{2} \end{bmatrix} \,$$

since $$\ {l^{(e)}}^{2} + {m^{(e)}}^{2} = 1 \,$$, then $$\ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I_{2x2} \,$$ the Identity matrix.

In conclusion: $$\ \Rightarrow {R^{(e)}}^{-1} = {R^{(e)}}^{T} \,$$ and $$\ \Rightarrow {\underline{\tilde{T}}^{(e)}}^{-1} = {\underline{\tilde{T}}^{(e)}}^{T} \,$$

Prove of $$ \mathbf{k}^{(e)}$$ equation
Verify that the equation: $$ \mathbf k^{(e)} = \tilde \mathbf T^{(e)^{T}}\tilde \mathbf k^{(e)}\tilde \mathbf T^{(e)}$$ is true.

Solving for the eigenvalues of the system
Closing the loop, infinitesimal displacement, eigenvalues problems, zero eigenvalues, rigid body mechanisms.

Solving for the eigenvalues of a 2 bar truss system using Matlab
With the Equation

$$\ \underline{k} = \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}_{6x6}\,$$

$$\ \ = \begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{49}{16} & \frac{3\sqrt{3} - 40}{16} & -\frac{5}{2} & \frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3} - 40}{16} & \frac{43}{16} & \frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2} \\ 0 & 0 & \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$

The following Matlab code was created

To evaluate the problem $$\  \mathbf{kv} = \boldsymbol{\lambda}  \mathbf{v}\,$$

Let {u1,u2,u3,u4} be the pure eigenvalues corresponding to four zero eigenvalues.

$$\ \mathbf{k}_{6x6} \mathbf{u}_i =0 \cdot \mathbf{u}_i \,$$, i=1,2,3,4

where $$ 0 \cdot u_i $$ is the zero vector of size 6x1.

Linear combination of {u_i, i=1,2,3,4}

$$ \sum_{i=1}^4 \alpha_i u_i =: \mathbf{W}_{6x1}$$

=: means equal by definition and $$\alpha$$ is a real number so it is 1x1 and u_1 is 6x1.

W is also an eigenvalue vector corresponding to a zero eigenvalue:

$$\ \mathbf{KW}= \mathbf{K}( \sum_{i=1}^4 \boldsymbol{\alpha}_i \mathbf{W}_i ) =  \sum_{i=1}^4 \boldsymbol{\alpha}_i (\mathbf{Ku}_i) \,$$ where $$(\mathbf{Ku}_i)$$ is a zero matriz of size 6x6. $$\ =\mathbf{0}_{6x1} = 0 \cdot \mathbf{W}_{6x1} \,$$

Solving for the eigenvalues of square bar truss system (a) using Matlab


$$\ \mathbf{kv} = \boldsymbol{\lambda}  \,$$

a=b=1, E=2,  A=3

Assembling the global stiffness matrix
Justification of assembly of element stiffness matrix $$k^{(e)} e=1, number of elements$$ into global stiffness matrix K

consider example of 2 bar truss from previous lecture.

Recall element force displacement relation

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

from previous lecture: Euler cut principal method 2 (equilibrium of global node 2) Euler cut principal method

from previous lecture: Free body diagram of element 1 and element 2 element degrees of freedom d(e) Trusses, matrix method

from previous lecture: for node 2, identify global degrees of freedom to element degrees of freedom for both element 1 and 2. Global_F-D_Relationship



$$\ \sum F_x= 0-f_3^{(1)}- f_1^{(2)}= 0 \,$$  (Equation 1)

$$ \sum F_y= P-f_4^{(1)}- f_2^{(2)}= 0$$  (Equation 2)

Next use element Force Displacement relation $$\mathbf{k}^{(e)} \mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

Equation 1 becomes $$\ f_3^{(1)} + f_1^{(2)}=0 \,$$

Equation 2 becomes $$\ f_4^{(1)} + f_2^{(2)}=P \,$$

So substituting for the f's in equation 1 gives

$$\ \underbrace{[ k_{31}^{(1)} d_{1}^{(1)} + k_{32}^{(1)} d_{2}^{(1)} + k_{33}^{(1)} d_{3}^{(1)} + k_{34}^{(1)} d_{4}^{(1)} ]}_{f_3^{(1)}} + \underbrace{[k_{11}^{(2)} d_{1}^{(2)} + k_{12}^{(2)} d_{2}^{(2)} + k_{13}^{(2)} d_{3}^{(2)}+ k_{14}^{(2)} d_{4}^{(2)}]}_{f_1^{(2)}}=0 \,$$

Replacing the local d's with the global d's as explained in Global_F-D_Relationship You end up with

$$[ 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 $$

Which is the third row of the $$\ \underline{K} \underline{d} = \underline{F} \,$$ equation.

Combine like d's, and convert to resulting equation into a multiplication of matrices to form:

$$\ \begin{bmatrix} K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)}   \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix} = 0 \,$$

Derivation of equation 2 and the assembly of row 4 of the global stiffness matrix
The assembly of K(e) e=1 to the number of elements, into the global stiffness matrix K can be represented as

$$\ K_{nxn}=Ak^{(e)}_{mxm}\,$$

Where A is the Assembly operator, n is the total number of global degrees of freedom, m is the number of element degrees of freedom and m is much less than n. Which is the principal of virtual work insert link here

Elimination of rows corresponding to the boundary conditions lets you obtain $$\mathbf{\bar{q}}_{2x2}$$

Remember that:

$$\ \mathbf{q}^{(e)}_{2x1} = \mathbf{T}^{(e)}_{2x4} d^{(e)} _{4x1} \,$$

$$\ k^{(e)} = \mathbf{T}^{(e)T} \hat{k}^{(e)}  \mathbf{T}^{(e)} \,$$

Deriving FEM of Partial differential Equations

Force Displacement relation of a bar or spring: Kd=F implies that Kd-F=0  (Equation 3)

which is equivalent to w(Kd-F)=0 for all w  (Equation 4)

Equation 4 is a "weak form" which is Principal of virtual work

Prove:

A) Equation 3 implies equation 4 which is a trivial conclusion

B) Equation 4 implies equation 3 which is not a trivial conclusion

Since equation 4 is valid for all w, first select w=1 and equation 4 becomes 1(Kd-F)=0 which is equation 3

Contributing Team Members
The following students contributed to this report:

Jessica Dyer Eml4500.f08.gravy.jad 15:43, 24 October 2008 (UTC)

Patrick Clayton Eml4500.f08.gravy.psc 23:19, 23 October 2008 (UTC)

Mikal McFarlane Eml4500.f08.gravy.mmm 02:32, 24 October 2008 (UTC)

Justin Tennant Eml4500.f08.gravy.JMT 02:37, 24 October 2008 (UTC)

Brian Eggeman Eml4500.f08.gravy.bje 03:47, 24 October 2008 (UTC)

Sean Sullivan Eml4500.f08.gravy.sms 04:32, 24 October 2008 (UTC)