User:Eml4500.f08.ateam.nobrega/Homework Report Five

Proving Hookes Law through the Principle of Virtual Work
The equation $$ F = kd $$ can also be written as $$ kd - F = 0 $$ (Equation 3) or which is true for all values of W if $$ W(kd - F) = 0 $$ (Equation 4). This is a weak form of the Principle of Virtual Work. It is important to note that deriving Equation 4 from Equation 3 is a trivial solution, but deriving Equation 3 from Equation 4 is in fact Not trivial.

From the Force-Displacement relationship,


 * $$ F = kd $$

where F is a 6x1 matrix, k is a 6x6 matrix, and d is a 6x1 matrix, the Principle of Virtual Work can be derived as,


 * $$ W \cdot (kd - F) = 0 $$ (Equation 4)

This is true for all W. For this relationship to work W must be multiplied as a dot product, as 0 is not represented as a matrix. It is desired to show that Equation 3 can be derived from Equation 4 for all W.

To do this the first choice is to select W such that W1 = 1 and W2 through W6 = 0. W is originally defined as a 6x1 matrix but needs to be a 1x6 matrix so,



W^{T} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Equation 4 then becomes,



W \cdot (kd - F) = 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,



\sum_{j=1}^6 k_{ij}d_{j} - F_{1} = 0 $$

The second choice is to select 'W such that W1 equals 0, W2 equals 1, and W3 through W6 also equal 0. So,



W^{T} = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Equation 4 then becomes,



W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,



\sum_{j=1}^6 k_{ij}d_{j} - F_{2} = 0 $$

This repeats as follows.

Choice 3,



W^{T} = \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} $$

Equation 4 then becomes,



W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,



\sum_{j=1}^6 k_{ij}d_{j} - F_{3} = 0 $$

Choice 4,



W^{T} = \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} $$

Equation 4 then becomes,



W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,



\sum_{j=1}^6 k_{ij}d_{j} - F_{4} = 0 $$

Choice 5,



W^{T} = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} $$

Equation 4 then becomes,



W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,



\sum_{j=1}^6 k_{ij}d_{j} - F_{5} = 0 $$

Choice 6,



W^{T} = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$

Equation 4 then becomes,



W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,



\sum_{j=1}^6 k_{ij}d_{j} - F_{6} = 0 $$

Hence $$ F = kd $$ (Equation 1).

To account for the boundary conditions of the two-bar truss where,


 * $$ d_{1} = d_{2} = d_{5} = d_{6} = 0 $$

the weighting coefficient must be "kinematically admissible" or cannot violate the boundary conditions. The weighting coefficients are the virtual displacements where,


 * $$ W_1 = W_2 = W_5 = W_6 = 0 $$

For all $$ W_3 $$ and $$ W_4 $$,



\begin{bmatrix} W_3 \\ W_4 \end{bmatrix} \cdot ( \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} ) = 0 $$

Equation 1 can also be found through Equation 3 by adding F to both sides of the equation as shown below,


 * $$ kd - F = 0 $$

Adding F to both sides,


 * $$ kd - F + F = 0 + F $$

Here F cancels out on the left-hand side of the equation leaving,


 * $$ kd = F $$

The same derivation as shown above can be applied to the axial force displacement relationship. For example...

Deriving :$$ \mathbf k^{(e)} = \mathbf T^{(e)T} * \mathbf \hat k^{(e)} * \mathbf T^{(e)} $$

Recall Force Displacement relation with axial displacement degrees of freedom $$ \mathbf q^{(e)}$$

$$ \mathbf \hat k_{(2x2)}^{(e)} * \mathbf q_{(2x1)}^{(e)} = \mathbf p_{(2x1)}^{(e)} $$

$$ \rightarrow \mathbf \hat k^{(e)} * \mathbf q^{(e)} - \mathbf p^{(e)} = \mathbf 0 $$ : $$ (Eq.1) $$

$$ \mathbf \hat w_{(2x1)} * (\mathbf \hat k^{(e)} * \mathbf q^{(e)} - \mathbf p^{(e)})_{(2x1)} = 0 $$ for all $$ \mathbf \hat w_{(2x1)} $$ : $$ (Eq.2) $$

Recall: $$ \mathbf q_{(2x1)}^{(e)} = \mathbf T_{(2x4)}^{(e)} * \mathbf d_{(4x1)}^{(e)} $$ : $$ (Eq.3) $$

Similarly: $$ \mathbf \hat w_{(2x1)} = \mathbf T_{(2x4)}^{(e)} * \mathbf w_{(4x1)} $$ : $$ (Eq.4) $$

Note: $$ \mathbf \hat w_{(2x1)} $$ is the equivalent of the axial virtual displacement

If one were to apply all the possible $$ \mathbf \hat w $$ one derive the axial force displacement relationship as shown below...

$$ \mathbf \hat k_{(2x2)}^{(e)} * \mathbf q_{(2x1)}^{(e)} = \mathbf p_{(2x1)}^{(e)} $$

^Everything should go above this. This was the last day of notes

Meeting 27 Wednesday October 29th 2008
Today’s lecture was a continuation of previous lecture and further understanding and proof of the Principle of Virtual Work. The following variables must be defined for the proof and will be used for rest of this section. The real axial displacement is $$q_{2x1}^{(e)}$$

$$\hat{W}_{2x1}$$ is the virtual axial displacement that corresponds to the axial displacement $$q_{2x1}^{(e)}$$.

$$W_{4x1}$$ is the virtual axial displacement in the global coordinate system that corresponds to the real displacements in global coordinates $$d^{(e)}_4x1$$.

The following equations correspond to the real axial and virtual axial displacements. Equation 3 $$ q_{2x1}^{(e)}=T_{2x4}^{(e)}d_{4x1}^{(e)} $$

Equation 4 $$ \hat{W}_{2x1}=T_{2x4}^{(e)}W_{4x1} $$

The following equation can then be obtained by replacing equations 3 and 4 into equation 2 given in lecture meeting 26

$$ \left( T^{(e)}W\right)\cdot \left[\hat{k}_{2x2}^{(e)}\left(T^{(e)}d^{(e)} \right)-P_{2x1}^{(e)} \right]=0 $$ for all $$ W_{4x1}$$

The Equation above can then be simplified and solved using principles of Matrix multiplication and Matrix transposes. See the following proof and example for further explanation.

The transpose of two matrices that are multiplied together is equal to the transpose of each individual matrix multiplied together.

Given: $$ \left(A_{pxq}B_{qxr} \right)^{T}=B_{rxq}^{T}A_{qxp}^{T} $$

A and B are given as the following matices

$$ A_{2x3} $$ and $$ B_{3x3} $$

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

$$ B=\begin{bmatrix} 7 & 8 & 9\\ 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} $$

$$ AB=\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 7 & 8 & 9\\ 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}=\begin{bmatrix} 21 & 27 & 33\\ 57 & 72 & 87 \end{bmatrix} $$

$$ \left( AB\right)^{T}=\begin{bmatrix} 21 & 27 & 33\\ 57 & 72 & 87 \end{bmatrix}^{T}=\begin{bmatrix} 21 & 57\\ 27 & 72\\ 33 & 87 \end{bmatrix} $$

$$ B^{T}A^{T}=\begin{bmatrix} 7 & 1 & 4\\ 8 & 2 & 5\\ 9 & 3 & 6 \end{bmatrix}\begin{bmatrix} 1 & 4\\ 2 & 5\\ 3 & 6 \end{bmatrix}=\begin{bmatrix} 21 & 57\\ 27 & 72\\ 33 & 87 \end{bmatrix} $$

Elimination of the dot product in equation 2 can conducted using the following priciple:

$$ a_{nx1}\cdot b_{nx1}=a_{1xn}^{T}b_{nx1} $$

This simplifies equation 2 into:

$$ \left( T^{(e)}W\right)^{T}\left[\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-P^{(e)} \right]=0_{1x1} $$

Which can be further simplified by using the fact that transpose of two matrices that are multiplied together is equal to the transpose of each individual matrix multiplied together resulting in:

$$ W^{T}T^{(e)T}\left[\hat{k}{(e)}\left(T^{(e)}d^{(e)} \right)-P^{(e)} \right]=0_{1x1} $$

By applying the reverse of the principle above the final manipulation of equation 2 becomes:

$$ W \cdot \left[\left(T^{(e)T}\hat{k}^{(e)}T^{(e)} \right)d^{(e)} - \left(T^{(e)T}P^{(e)} \right) \right]= 0_{1x1}$$ for all $$ W_{4x1} $$

By realizing that $$T^{(e)T}\hat{k}^{(e)}T^{(e)} $$ equals element stiffness matrix, $$k^{(e)}$$, and that $$T^{(e)T}p^{(e)} $$ equals to the element force matrix, $$f^{(e)}$$. This relation in equation 2 simplifies

$$ W \cdot \left[k^{(e)}d^{(e)}-f^{(e)}\right]= 0$$

Since this has been proven for all $$ W_{4x1}$$ in previous lectures the equation above reduces to the original force-displacement relation discussed in previous lectures about the Finite Element Method FEM

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

Proving the principle of virtual work.

This is final section for FEM method using the discrete case with matrices.

The next section will use partial differential equations for the non continuous case. The model motivation for this problem is an elastic bar with varying cross section area A(x), Young’s Modulus E(X) (such as composite materials) which are subjected to varying axial load distributions, concentrated load distributions, and inertia forces. See the image below and following lecture notes for further explanation.



Composite Materials
Composite materials consist of a strengthening fiber and matrix substrate. The addition of these fibers to a substrate is termed 'doping'. Due to the mixed material properties, composite materials often contain a varying elastic modulus. In other words, the Young's modulus is a function of position in the material.

Some examples of composite materials with this characteristic:


 * Carbon fiber
 * Fiber Glass

The Elastic Modulus is much higher in these materials in the fiber direction (longitudinal direction). However, perpendicular to the fibers (transverse direction) the Elastic Modulus is smaller as the fibers lose their strength in shear.

Free body diagram of a rigid body cross section


$$ \sum{F_x}=0=-N(x,t) + N(x+dx,t)+f(x,t)dx-m(x)\ddot{u}dx $$

$$ \sum{F_x}=0=\frac{\delta N}{\delta x}(x,t)dx+H.O.T.+f(x,t)dx-m(x)\ddot{u}dx $$

Cancel dx and neglect the higher order terms (H.O.T.)

(Eqn. 1): $$ \sum{F_x}=0=\frac{\delta N}{\delta x}(x,t)+f(x,t)-m(x)\ddot{u} $$

Recall Taylor Series Expansion:

$$f(x+dx)=f(x)+\frac{df(x)}{dx}dx+[\frac{d^2f(x)}{2dx^2}+...] $$

Where the content inside the brackets represent higher order terms.

(Eqn. 2):  $$\frac{\delta N}{\delta x}+ f =m\ddot{u} $$

Where Equation 2 represents the equation of motion(EOM) of the elastic bar.

$$ N(x,t)=A(x)\sigma(x,t) $$ where: $$ \sigma(x,t)=E(x)\epsilon (x,t)$$ and: $$ \epsilon (x,t)=\frac{\delta u}{\delta x }(x,t)$$ (Eqn. 3)

Substituting Eqn. 3 into Eqn. 2:

$$ \frac{\delta }{\delta x}(A(x)E(x)\frac{\delta u}{\delta x})+f(x,t)=m(x)\ddot{u} $$, where: $$ \ddot{u}=\frac{\delta ^2u}{\delta t^2}$$

Partial Differential Equation of Motion
In order to integrate differential equations of motion, boundary conditions are needed to evaluate the constants of integration. For the problem above, this translates to two initial conditions (due to the 2nd order nature of the equations).

Boundary Conditions:



$$u(0,t)=0=u(L,t)$$



$$u(0,t)=0$$

$$N(L,t)=F(t)$$

TWO-BAR Truss De-Bugged
Below is the debugged version of the two bar truss example. The problem was found at the end of the code for the results matrix. Initially the code did not take into account that the elements had different cross section areas (A) and Young's modulus of elasticy (e). This was corrected in the for statement and we were able to obtain the correct results.

SIX BAR Truss Plot




Derivation of Matrix Relations and Matrices
Section 2.4 in the Fundamental Finite Element Analysis and Applications textbook is concerned with space trusses. We can use our FEA knowledge and use the 2D plane truss concepts to derive expressions for a 3D space truss. It is important to note that the FD relations remain the same for each element.


 * $$\begin{bmatrix}

F_{1}\\ F_{2} \end{bmatrix} = \begin{bmatrix} k & -k\\ -k & k \end{bmatrix} \begin{bmatrix} d_{1}\\ d_{2} \end{bmatrix} = \begin{bmatrix} P_{1}\\ P_{2} \end{bmatrix} $$

The figure above depictsthe local and global coordinates for an axially loaded bar in three dimensions. The nodal degrees of freedom are u1, v1, w1, u2, v2 and w2.

The transofrmation betwwen local and global degrees of freedom:

d=Td_{l} $$


 * $$\begin{bmatrix}

d_{1}\\ d_{2} \end{bmatrix} = \begin{bmatrix} l_{s} & m_{s} & n_{s} & 0 & 0 & 0\\ 0 & 0 & 0 & l_{s} & m_{s} & n_{s} \end{bmatrix} \begin{bmatrix} u_{1}\\ v_{1}\\ w_{1}\\ u_{2}\\ v_{2}\\ w_{2} \end{bmatrix} $$

$$ l_{s}=\frac{x_{2}-x_{1}}{L}::: m_{s}=\frac{y_{2}-y_{1}}{L}::: l_{s}=\frac{z_{2}-z_{1}}{L} $$

$$ L=((x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} + (z_{2}-z_{1})^{2}))^{0.5} $$

Now we must transform the elemental equations in local coordinates to the global coordinates relation. $$ k_{l}d_{l}=r_{l} $$ or $$ k_{l}Td=r_{l} $$ Mutliply by the transpose of T:

$$ T^{T}k_{l}Td=T^{T}r_{l} $$ Which simplifies to: kd=r



\frac{EA}{L} \begin{bmatrix} l_{s}^{2} & m_{s}l_{s} & n_{s}l_{s} & -l_{s}^{2} & -m_{s}l_{s} & -n_{s}l_{s}\\ m_{s}l_{s} & m_{s}^{2} & m_{s}n_{s} & -m_{s}l_{s} & -m_{s}^{2} & -m_{s}n_{s}\\ n_{s}l_{s} & m_{s}n_{s} & n_{s}^{2} & -n_{s}l_{s} & -m_{s}n_{s} & -n_{s}^{2}\\ -l_{s}^{2} & -m_{s}l_{s} & -n_{s}l_{s} & l_{s}^{2} & m_{s}l_{s} & n_{s}l_{s}\\ -m_{s}l_{s} & -m_{s}^{2} & -m_{s}n_{s} & m_{s}l_{s} & m_{s}^{2} & m_{s}n_{s}\\ -n_{s}l_{s} & -m_{s}n_{s} & -n_{s}^{2} & n_{s}l_{s} & m_{s}n_{s} & n_{s}^{2}\\ \end{bmatrix} \begin{bmatrix} u_{1}\\ v_{1}\\ w_{1}\\ u_{2}\\ v_{2}\\ w_{2}\\ \end{bmatrix} = \begin{bmatrix} F_{1x}\\ F_{1y}\\ F_{1z}\\ F_{2x}\\ F_{2y}\\ F_{2z}\\ \end{bmatrix} $$

If we assume that the concentrated loads are added directly to the global equations at the beginning of the assembly, then the F matrix is simply a 1x6 matrix of zeros. Using this knowledge we may now compute the nodal displacemnets. The axial displacments can be computed by transforming the global displacemnts into local coordinates. For example:
 * $$\begin{bmatrix}

d_{1}\\ d_{2} \end{bmatrix} = \begin{bmatrix} l_{s} & m_{s} & n_{s} & 0 & 0 & 0\\ 0 & 0 & 0 & l_{s} & m_{s} & n_{s} \end{bmatrix} \begin{bmatrix} u_{1}\\ v_{1}\\ w_{1}\\ u_{2}\\ v_{2}\\ w_{2} \end{bmatrix} $$ Therefore the axial displacement over the element is simply: $$ u(s)=(1-\frac{s}{L}\frac{s}{L})\begin{bmatrix} d_{1}\\ d_{2} \end{bmatrix} $$ This relation is true for s values greater than or equal to zero and less than or equal to L. Note that the axial strain can easily be computed by taking the derivative of this function with respect to time. Yielding: $$ strain=(-d_{1}+ d_{2})\frac{1}{L} $$

Contributing Team Members
Eml4500.f08.ateam.nobrega 01:01, 7 November 2008 (UTC)

Eml4500.f08.ateam.shah 04:41, 7 November 2008 (UTC)

Eml4500.f08.ateam.mcnally 04:45, 7 November 2008 (UTC)

Eml4500.f08.ateam.boggs.t 05:18, 7 November 2008 (UTC)

Eml4500.f08.ateam.carr 17:35, 7 November 2008 (UTC)

Eml4500.f08.ateam.rivero 17:49, 7 November 2008 (UTC)