User:Eml4500.f08.team.guan/spacetrusses

Space Truss Matrix Relations
Knowledge of the 2D trusses can be projected into 3 dimensional trusses as well. The following is the derivation of some matrix relations of trusses in the 3 dimensional space.

Axial Force Displacement Relationship
Here, the axial force displacement relationship is the exact same as the 2D truss. This is because the no matter if the truss is in 2D or 3D space, there can only be two axial displacements so the size of the matrix doesn't change. Hence, the following relation. $$ \frac{EA}{L} \begin{bmatrix} 1 & -1\\ -1 & 1\\ \end{bmatrix}

\begin{bmatrix} q_1\\ q_2\\ \end{bmatrix}= \begin{bmatrix} P_1\\ P_2\\ \end{bmatrix} $$ Where q is the axial displacement and P is the force. As seen in the bottom figure, the truss in 3D space is subject to 6 displacement degrees of freedom and equivalently 2 axial displacements. 

Element Dofs and Element Forces in Global (x,y,z) Coordinates
3D trusses introduce 2 extra degrees of freedom in each element. This results in a 6 by 1 elemental displacement matrix and consequently a 6 by 1 elemental force matrix. This is evident from the above figure. These matrices are listed below. $$ d= \begin{bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)}\\ \end{bmatrix}$$    $$ f= \begin{bmatrix} f_1^{(e)}\\ f_2^{(e)}\\ f_3^{(e)}\\ f_4^{(e)}\\ f_5^{(e)}\\ f_6^{(e)}\\ \end{bmatrix} $$

Transformation Matrix
In order to change from local displacement degrees of freedom in global coordinates to axial displacements in local coordinates, a new transformation matrix must be implemented. A new direction cosine must be added. The figure below shows how the new direction cosine can be calculated. P_1 is the location of the node 1 and P_2 is the location of node 2. The new direction cosine is similar to the other two direction cosines and describes the displacement of a local node in a the z direction. The equations for the direction cosines can also be found below.  $$l_t=\frac{x_2-x_1}{L}$$ ,   $$m_t=\frac{y_2-y_1}{L}$$   ,   $$n_t=\frac{z_2-z_1}{L}$$ The transformation matrix derivation is as follows. Consider the axial displacement of node 1 on the below diagram.  This displacement is essentially the orthogonal projection of the displacement vector $$d_{[1]}^{(e)}$$ of node 1 on the axis of the element e described by the unit vector $$\tilde {w}$$. In order to describe this relationship mathematically, the dot product is taken between the $$d_{[1]}^{(e)}$$ displacement vector and the unit vector $$\tilde {w}$$. $$ q_1^{(e)}=\vec {d_{[1]}^{(e)}} \cdot \tilde {w}= (d_1^{(e)} \vec{i} +d_2^{(e)} \vec{j} +d_3^{(e)} \vec{k} )\cdot \tilde{w}= d_1^{(e)}\cdot \tilde{w}+d_2^{(e)}\cdot \tilde{w}+d_3^{(e)}\cdot \tilde{w} $$ Node 1 is located at the origin (0,0,0) and node 2 is located at (x_2,y_2,z_2). $$\tilde {w}=\frac{x_2 \vec{i}+y_2 \vec{j}+z_2 \vec{k}}{\sqrt{x_2^2+y_2^2+z_2^2}}=\frac{x_2 \vec{i}+y_2 \vec{j}+z_2 \vec{k}}{L}$$ $$q_1^{(e)}=\frac{x_2}{L}d_1^{(e)}+\frac{y_2}{L}d_2^{(e)}+\frac{z_2}{L}d_3^{(e)}=l_td_1^{(e)}+m_td_2^{(e)}+n_td_3^{(e)}$$ In matrix form, the above  equation yields to the following. $$q_1^{(e)}= \begin{bmatrix} l_t & m_t & n_t\\ \end{bmatrix} \begin{bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ \end{bmatrix} $$ $$q_2^{(e)}= \begin{bmatrix} l_t & m_t & n_t\\ \end{bmatrix} \begin{bmatrix} d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)}\\ \end{bmatrix} $$ Combining these two equations yields the following relationship. $$ \begin{bmatrix} q_1^{(e)}\\ q_2^{(e)}\\ \end{bmatrix}= \begin{bmatrix} l_t & m_t & n_t &0 &0 &0\\ 0 &0 &0 &l_t & m_t & n_t\\ \end{bmatrix} \begin{bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)}\\ \end{bmatrix} $$ Where the transformation matrix is $$T= \begin{bmatrix} l_t & m_t & n_t &0 &0 &0\\ 0 &0 &0 &l_t & m_t & n_t\\ \end{bmatrix}$$

The Principle of Virtual Work
The principle of virtual work also applied in the same way in 3 dimensional space and is pervasive in the these matrix relationships as well. The force displacement relationship in global coordinates is derived below using the principle of virtual work. $$\displaystyle Kd=F$$ Unlike the elemental stiffness, displacement, and force matrices, the size of the global force displacement matrices depend on the number of truss elements there are. The sizes are not fixed like the elemental case. Give this relation, we can subtract the force matrix from both sides and apply the PVW. Then, we have $$w \cdot (Kd-F)=0$$ for all $$w$$. To prove this, let us select some choices for the weighting function matrix. Choice 1: We select the weighting function such that, $$\displaystyle w_1=1, w_2=0, w_3=0,...,w_n=0$$ where "n" is the number of global displacement degrees of freedom we have. $$w^T= \begin{bmatrix} 1 & 0& 0& ... &...&...\\ \end{bmatrix} $$

Plugging this column matrix into equation 2 will yield the following. $$w \cdot (Kd-F)= 1[\sum_{j=1}^n{K_{1j}d_j-F_1}]+0[\sum_{j=1}^n{K_{2j}d_j-F_2}]+0[\sum_{j=1}^n{K_{3j}d_j-F_3}]+...+0[\sum_{j=1}^n{K_{nj}d_j-F_n}]=0 $$ Multiplying it all out gives $$\sum_{j=1}^n{K_{1j}d_j}=F_1 $$ We continue to choose choices for the weighting matrix strategically for "n" amount of times. Putting all the resulting equations in matrix form will give you $$\displaystyle Kd=F$$

From the 2D case, we derived a square transformation matrix that is invertible. Recall that it equals: $$

T^{(e)}= \begin{bmatrix} R^{(e)}_{2x2} & 0_{2x2} \\ 0_{2x2} & R^{(e)}_{2x2} \\ \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\ -m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \\ \end{bmatrix}

$$ In the 3D case, we must account for the extra directional cosine (n_t). It is necessary to derive the transformation matrix in 3D to derive the elemental matrix stiffness matrix in global coordinates in terms of the transformation matrix. Now, lets consider the bottom figure to assist in derive this square, invertible, 3D transformation matrix.  The top truss indicates the displacements in an arbitrary coordinate system whereas the bottom truss indicates the displacements in a specific coordinate system. This coordinate system has x tilda as the axis of the element and the y tilda and z tilda axes are mutually orthogonal with x tilda and each other. In other words, the element is normal to the y tilda and z tilda plane. Thus, the following equation is true.  $$ \begin{bmatrix} q_1\\ q_2\\ \end{bmatrix}= \begin{bmatrix} \tilde {d_1}\\ \tilde {d_4}\\ \end{bmatrix} $$ These are the axial displacements and the rest of the displacements are transverse displacements. The force displacement relationship in this form is as follows. $$

\tilde{f}^{(e)}_{6x1}=\tilde{k}^{(e)}_{6x6}\tilde{d}^{(e)}_{6x1}=

\begin{bmatrix} \tilde{f}^{(e)}_{1}\\ \tilde{f}^{(e)}_{2}\\ \tilde{f}^{(e)}_{3}\\ \tilde{f}^{(e)}_{4}\\ \tilde{f}^{(e)}_{5}\\ \tilde{f}^{(e)}_{6}\\ \end{bmatrix} =k^{(e)} \begin{bmatrix} 1 &0&0&-1&0&0\\ 0&0&0&0&0&0\\ 0 &0&0&0&0&0\\ -1 &0&0&1&0&0\\ 0&0&0&0&0&0\\ 0 &0&0&0&0&0\\ \end{bmatrix} \begin{bmatrix} \tilde{d}^{(e)}_{1}\\ \tilde{d}^{(e)}_{2}\\ \tilde{d}^{(e)}_{3}\\ \tilde{d}^{(e)}_{4}\\ \tilde{d}^{(e)}_{5}\\ \tilde{d}^{(e)}_{6}\\ \end{bmatrix}$$ The new transformation matrix is given below. $$H= \begin{bmatrix} l_t&m_t&n_t\\ l_s&m_s&n_s\\ l_r&m_r&n_r\\ \end{bmatrix} $$     ,       $$ \tilde{T}= \begin{bmatrix} H&0\\ 0&H\\ \end{bmatrix}= \begin{bmatrix} l_t&m_t&n_t&0&0&0\\ l_s&m_s&n_s&0&0&0\\ l_r&m_r&n_r&0&0&0\\ 0&0&0&l_t&m_t&n_t\\ 0&0&0&l_s&m_s&n_s\\ 0&0&0&l_r&m_r&n_r\\ \end{bmatrix}$$ Where l_s is the cosine of the angle between the s and y axes and l_r is the cosine of the angle between the r and z axes. t is the axis defined by the unit vector w and r and s are the axes mutually orthogonal to t. This gives: <p style="text-align:center;">$$ k^{(e)}=\tilde{T}^{(e)T}\tilde{k}^{(e)}\tilde{T}^{(e)}...(1) $$ $$ f=kd=[\tilde{T}^{(e)T}\tilde{k}^{(e)}\tilde{T}^{(e)}][\tilde{T}^T\tilde{d}$$] The derivation of the transformation matrix relationship 1 is as follows. <p style="text-align:center;"> (5) $$(undefined^{(e)}) \begin{bmatrix} \tilde^{(e)}(undefined^{(e)}^{(e)})-^{(e)} \end{bmatrix} =0$$ (6)  $$\displaystyle (_{pxq}_{qxr})^{T}= ^{T}^{T} $$ (7)  $$_{nx1}_{nx1}= ^{T}_{1xn}_{nx1} $$  Note:  The right side of equation (7) is scalar: $$==(Scalar)

$$

When you apply equations (7) & (6) into equation (5) it yields:

$$(undefined^{(e)})^{T} \begin{bmatrix} \tilde^{(e)}(undefined^{(e)}^{(e)})-^{(e)} \end{bmatrix} =o $$  for all $$_{6x1}$$

 Note:  $$(undefined^{(e)})^{T}$$ comes from equation (7). Applying the transpose rule for matrix multiplication from equation (6) yields:

$$(^{(T)}undefined)^{(e)T} \begin{bmatrix}\tilde^{(e)}(undefined^{(e)}^{(e)})-^{(e)} \end{bmatrix} =o $$

When you replace the transpose with the dot product it yields:

$$ \begin{bmatrix} (undefined^{(e)}\tilde^{(e)}undefined^{(e)})^{(e)}-(undefined^{(e)T}^{(e)} \end{bmatrix} =0 $$ Now we can choose the choices for the weighting function that would strategically yield the desired result.