User:Eml4500.f08.RAMROD.F/HW3

Monday, September 22 2008
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Derivation of elemental force displacement with respect to the global system.
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$$ K^{(e)}_{4x4} \cdot d^{(e)}_{4x1} = F^{(e)}_{4x1}$$



Statics dictates that this can be defined as a two-force member. Therefore, it can be drawn like this.



The force displacement relationship is as follows for P and q.

$$K^{(e)}_{2x2} \cdot q^{(e)}_{2x1} = P^{(e)}_{2x1}$$

$$K^{(e)}\begin{vmatrix} 1 & -1\\ -1 & 1 \end{vmatrix} \begin{Bmatrix} q^{(e)}_1\\ q^{(e)}_2 \end{Bmatrix}=\begin{Bmatrix} P^{(e)}_1\\ P^{(e)}_2 \end{Bmatrix}$$

$$q^{(e)}_i$$ = the axial displacement of element e at local node i.

$$ P^{(e)}_i$$ = the axial force of element e at local node i.

Goal : Derive equation $$ K^{(e)}_{4x4} \cdot d^{(e)}_{4x1} = F^{(e)}_{4x1}$$ from $$K^{(e)}_{2x2} \cdot q^{(e)}_{2x1} = P^{(e)}_{2x1}$$ . Find the relationship between matrix $$q^{(e)}$$ and $$d^{(e)}$$ and between matrix $$P^{(e)}$$ and $$F^{(e)}$$.

These relations can be expressed in the form:

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

== Consider the displacement vector of local node 1, denoted by $$d^{(e)}_1 $$ == center text



$$ d^{(e)}_1 =  d^{(e)}_1\vec{i} \cdot d^{(e)}_2\vec{j}$$

$$q^{(e)}_1 $$ = the axial displacement of node 1 is the orthogonal projection of the displacement vector of node one. $$d^{(e)}_1 $$ on the zxis of the bar $$\tilde{x}$$ of element e.

$$q^{(e)}_1=\vec{d}^{(e)}_1\cdot \vec{\tilde{i}}$$

$$q^{(e)}_1=(d^{(e)}_1\vec{i}+d^{(e)}_2\vec{j})\cdot \vec{\tilde{i}}$$

$$q^{(e)}_1=d^{(e)}_1(\vec{i}\cdot \vec{\tilde{i}})+d^{(e)}_2(\vec{j}\cdot \vec{\tilde{i}})$$

$$q^{(e)}_1=l^{(e)}d^{(e)}_1+ m^{(e)}d^{(e)}_2$$

$$q^{(e)}_1=\begin{vmatrix} l^{(e)} & m^{(e)} \end{vmatrix}\begin{Bmatrix} d^{(e)}_1 \\d^{(e)}_2

\end{Bmatrix}$$

Consider the displacement vector of local node 2, denoted by $$d^{(e)}_2$$


$$ d^{(e)}_2 =  d^{(e)}_3\vec{i} \cdot d^{(e)}_4\vec{j}$$

$$q^{(e)}_2 $$ = the axial displacement of node 2 is the orthogonal projection of the displacement vector of node two. $$d^{(e)}_2 $$ on the zxis of the bar $$\tilde{x}$$ of element e.

$$q^{(e)}_2=\vec{d}^{(e)}_2\cdot \vec{\tilde{i}}$$

$$q^{(e)}_2=(d^{(e)}_3\vec{i}+d^{(e)}_4\vec{j})\cdot \vec{\tilde{i}}$$

$$q^{(e)}_2=d^{(e)}_3(\vec{i}\cdot \vec{\tilde{i}})+d^{(e)}_4(\vec{j}\cdot \vec{\tilde{i}})$$

$$q^{(e)}_2=l^{(e)}d^{(e)}_3+ m^{(e)}d^{(e)}_4$$

$$q^{(e)}_2=\begin{vmatrix} l^{(e)} & m^{(e)} \end{vmatrix}\begin{Bmatrix} d^{(e)}_3 \\d^{(e)}_4

\end{Bmatrix}$$

The result of the nodal analysis is combined to form this matrix.

$$\begin{Bmatrix} q^{(e)}_1\\ q^{(e)}_2 \end{Bmatrix}=\begin{vmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0& 0 & -l^{(e)} & -m^{(e)} \end{vmatrix}\begin{Bmatrix} d^{(e)}_1\\ d^{(e)}_2 \\ d^{(e)}_3 \\ d^{(e)}_4

\end{Bmatrix} $$