User:Eml4500.f08.ateam.mcnally/Week 7

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.