User:Eml4500.f08.Ateam.Slevinski/Homework5

HW 5 - Derivation of FD Relation in Global Coordinates Using the PVW
Derive the transformation matrix based on the lecture for the 2-D truss elements:

$$a_1^{(e)} = d_1^{(e)} \cdot \vec\tilde{i} = ( d_1 \tilde{i} + d_2 \tilde{j} + d_3 \tilde{k} ) \cdot \vec\tilde{i} = d_1 \cdot ( \tilde{i} \cdot \vec\tilde{i} ) + d_2 \cdot ( \tilde{j} \cdot \vec\tilde{i} ) + d_3 \cdot ( \tilde{k} \cdot \vec\tilde{i} )$$

where $$( \tilde{i} \cdot \vec\tilde{i} ) = l^{(e)}, ( \tilde{j} \cdot \vec\tilde{i} ) = m^{(e)}, and ( \tilde{k} \cdot \vec\tilde{i} ) = n^{(e)}$$

$$a_1^{(e)} = l^{(e)}d_1 + m^{(e)}d_2 + n^{(e)}d_3$$

$$a_1^{(e)} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix}\begin{vmatrix} d_1\\ d_2\\ d_3 \end{vmatrix}$$

$$a_2^{(e)} = d_2^{(e)} \cdot \vec\tilde{i} = ( d_4 \tilde{i} + d_5 \tilde{j} + d_6 \tilde{k} ) \cdot \vec\tilde{i} = d_4 \cdot ( \tilde{i} \cdot \vec\tilde{i} ) + d_5 \cdot ( \tilde{j} \cdot \vec\tilde{i} ) + d_6 \cdot ( \tilde{k} \cdot \vec\tilde{i} )$$

where $$( \tilde{i} \cdot \vec\tilde{i} ) = l^{(e)}, ( \tilde{j} \cdot \vec\tilde{i} ) = m^{(e)}, and ( \tilde{k} \cdot \vec\tilde{i} ) = n^{(e)}$$

$$a_2^{(e)} = l^{(e)}d_4 + m^{(e)}d_5 + n^{(e)}d_6$$

$$a_2^{(e)} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix}\begin{vmatrix} d_4\\ d_5\\ d_6 \end{vmatrix}$$

$$\begin{vmatrix} a_1^{(e)}\\ q_2^{(e)} \end{vmatrix} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix} \begin{vmatrix} d_1\\d_2 \\d_3 \\ d_4 \\ d_5 \\ d_6

\end{vmatrix}$$

Note: The transformation matrix is in terms of the three director cosines.

Recall the relation between axial DOF's and element DOF's in the global coordinate system:

$$\underline{q} = \underline{T} \underline{d}$$

and also the relation between axial forces and element forces in the global coordinate system:

$$\underline{P} = \underline{T} \underline{f}$$

Using the principle of virtual work (PVW) we can derive the force-displacement relationship in the global coordinate system.

$$\underline{k} \underline{d} = \underline{F}$$

therefore,

$$\underline{k} \underline{d} - \underline{F} = 0$$

With the PVW, this becomes:

$$\underline{W} \cdot ( \underline{k} \underline{d} - \underline{F} ) = 0$$

where $$\underline{W}$$ is the weighting matrix. For this system we have 6 choices for $$\underline{W}$$:

Choice 1:

$$ W = \begin{vmatrix}1 & 0 & 0 & 0 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{1,j}d_j - F_1 = 0}$$

$$\sum_{j=1}^{6}{k_{1,j}d_j = F_1}$$

Choice 2:

$$ W = \begin{vmatrix}0 & 1 & 0 & 0 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{2,j}d_j - F_2 = 0}$$

$$\sum_{j=1}^{6}{k_{2,j}d_j = F_2}$$

Choice 3:

$$ W = \begin{vmatrix}0 & 0 & 1 & 0 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{3,j}d_j - F_3 = 0}$$

$$\sum_{j=1}^{6}{k_{3,j}d_j = F_3}$$

Choice 4:

$$ W = \begin{vmatrix}0 & 0 & 0 & 1 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{4,j}d_j - F_4 = 0}$$

$$\sum_{j=1}^{6}{k_{4,j}d_j = F_4}$$

Choice 5:

$$ W = \begin{vmatrix}0 & 0 & 0 & 0 & 1 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{5,j}d_j - F_5 = 0}$$

$$\sum_{j=1}^{6}{k_{5,j}d_j = F_5}$$

Choice 6:

$$ W = \begin{vmatrix}0 & 0 & 0 & 0 & 0 & 1\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{6,j}d_j - F_6 = 0}$$

$$\sum_{j=1}^{6}{k_{6,j}d_j = F_6}$$

$$\begin{vmatrix} k_{1,1} & k_{1,2} & k_{1,3} & k_{1,4} & k_{1,5} & k_{1,6}\\ k_{2,1} & k_{2,2} & k_{2,3} & k_{2,4} & k_{2,5} & k_{2,6}\\ k_{3,1} & k_{3,2} & k_{3,3} & k_{3,4} & k_{3,5} & k_{3,6}\\ k_{4,1} & k_{4,2} & k_{4,3} & k_{4,4} & k_{4,5} & k_{4,6}\\ k_{5,1} & k_{5,2} & k_{5,3} & k_{5,4} & k_{5,5} & k_{5,6}\\ k_{6,1} & k_{6,2} & k_{6,3} & k_{6,4} & k_{6,5} & k_{6,6}\\ \end{vmatrix} \begin{vmatrix} d_1\\d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6

\end{vmatrix} = \begin{vmatrix} F_1\\F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6

\end{vmatrix}$$

HW 5 - Derivation of Elemental Stiffness Matrix in Global Coordinates Using PVW
The following is an expression of the element stiffness matrix $$\underline{K}_{6x6}^{(e)}$$ in terms of the transformation matrix $$\underline{T}^{(e)}$$ and the element stiffness matrix $$\underline\hat{k}^{(e)}$$ from the axial force displacement relationship. We will show that by using the priciple of virtual work, we can derive the equation for the element stiffness matrix in global coordinates.

$$\underline{K}^{(e)} = \underline{T}^{(e)T} \underline\hat{k}^{(e)} \underline{T}^{(e)}$$

Recall the force displacement relationship with axial DOF's $$ q^{(e)}$$

$$\underline\hat{k}_{2x2}^{(e)} \underline{q}_{2x1}^{(e)} = \underline{p}_{2x1}^{(e)}$$

Rearranging this equation yields:

$$ \underline\hat{k}^{(e)} \underline{q}^{(e)} - \underline{p}^{(e)} = \underline{0}_{2x1}$$    (1)

We can now apply the principle of virtual work (PVW) to get:

$$\underline\hat{W} \cdot ( \underline\hat{k}^{(e)} \underline{q}^{(e)} - \underline{p}^{(e)} ) = 0^{'}_{1x1}$$   (2)  for all $$\underline\hat{W}_{2x1}$$

We now can recall the equation:

$$\underline{q}_{2x1}^{(e)} = \underline{T}_{2x6}^{(e)} \underline{d}_{6x1}^{(e)}$$    (3)

Similarly,

$$\hat{W}_{2x1} = \underline{T}^{(e)}_{2x6} \underline{W}_{6x1}$$ (4)

Now, substitute equations (3) & (4) into equation (2) which yields:

$$( \underline{T}^{(e)} \underline{W} ) \cdot [ \underline\hat{k}^{(e)} ( \underline{T}^{(e)} \underline{d}^{(e)} ) - \underline{p}^{(e)} ] = 0$$  for all $$ \underline{W}_{6x1}$$   (5)

Recall that:

$$ ( \underline{A}_{pxq} \underline{B}_{qxr} )^{T} = \underline{B}^{T} \underline{A}^{T}$$  (6)

And also that:

$$\underline{a}_{nx1} \cdot \underline{b}_{nx1} = \underline{a}^{T}_{1xn} \underline{b}_{nx1}$$   (7)

Now we can apply equations (6) and (7) into (5) to get:

$$ ( \underline{T}^{(e)} \underline{W} )^{T} [ \underline\hat{k}^{(e)} ( \underline{T}^{(e)} \underline{d}^{(e)} ) - ( \underline{p}^{(e)} ] = 0 $$ for all $$\underline{W}_{6x1}$$

In this equation, $$\underline\hat{W}_{2x1}$$ represents the virtual axial displacement and from eq. (4) and the matrix $$\underline{W}_{6x1}$$ represents the virtual displacement in the global coordinate system corresponding to $$\underline{d}^{(e)}$$

The previous equation now becomes:

$$\underline{W}^{T} \underline{T}^{(e)T} [ \underline\hat{k}^{(e)} ( \underline{T}^{(e)} \underline{d}^{(e)} ) - \underline{p}^{(e)} ] = 0$$

$$\underline{W} \cdot [ ( \underline{T}^{(e)T} \underline\hat{k}^{(e)} \underline{T}^{(e)} ) \underline{d}^{(e)} - ( \underline{T}^{(e)T} \underline{p}^{(e)}) ] = 0$$ for all $$\underline{W}_{6x1}$$

In the equation above, the element stiffness matrix $$\underline{K}_{6x6}^{(e)}$$ is represented by $$( \underline{T}^{(e)T} \underline\hat{k}^{(e)} \underline{T}^{(e)} )$$and the force matrix $$\underline{f}_{6x1}^{(e)}$$ is represented by $$\underline{T}^{(e)T} \underline{p}^{(e)}$$

The equation now becomes:

$$\underline{W} \cdot [ \underline{K}^{(e)} \underline{d}^{(e)} - \underline{f}^{(e)} ] = 0$$ for all $$\underline{W}$$

We have the original expression for the elemental stiffness matrix in terms of the axial force displacement relationship.

$$\underline{K}^{(e)} \underline{d}^{(e)} = \underline{f}^{(e)}$$

HW 5 - Element DOF's and Element Forces in Global xyz-Coordinate System for 6x1 Matrices


Using the previous diagram, we can write the following relationships.

$$\underline{k}^{(e)} \underline{q}^{(e)} = \underline{p}^{(e)}$$

$$\begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix} \begin{bmatrix}q_{1}^{(e)}\\ q_{2}^{(e)}\end{bmatrix} = \begin{bmatrix}p_{1}^{(e)}\\ p_{2}^{(e)}\end{bmatrix}$$



From the diagram above, we can write the element DOF's and forces for a 3-D system.

$$\underline{f}^{(e)} = \begin{vmatrix}F_1\\F_2 \\F_3 \\ F_4\\ F_5\\ F_6\end{vmatrix}_{6x1}$$ and $$\underline{d}^{(e)} = \begin{vmatrix} d_1\\d_2\\ d_3\\ d_4\\ d_5\\ d_6\end{vmatrix}_{6x1}$$

We can now use the transformation matrix $$\underline{T}$$ to transform the 6 element DOF's in the global xyz-coordinate system to the 2 axial DOF's $$p$$ and $$q$$.

$$\underline{q}^{(e)} = \underline{T} \underline{d}^{(e)}$$

This equation expands to:

$$\begin{vmatrix}q_{1}^{(e)}\\q_{2}^{(e)} \end{vmatrix}_{2x1} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix}_{2x6} \begin{vmatrix}d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\end{vmatrix}_{6x1}$$

Similarly, this can be applied to the equation:

$$\underline{p} = \underline{T} \underline{f}$$

From either of the above equations, to solve for the element forces $$\underline{f}$$ or displacements $$\underline{d}$$ we can use the transpose of the transformation matrix $$\underline{T}^{T}$$.

$$\underline{f} = \underline{T}^{T} \underline{p}$$

$$\underline{d} = \underline{T}^{T} \underline{q}$$