User:Eml4500.f08.team.Bengtson/teamHW3

Meeting 12)
To find the derivation of element FD wrt global coordinate system please consult page 6-1 of the notes.

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

K (e)is a 4x4 matrix

d(e) is a 4x1 matrix

f(e) is both multiplied together to form a 4x1 matrix.



Reference from page 4-5 in the notes

$$ \boldsymbol{K}^{(e)} \begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix} $$$$\begin{Bmatrix} q_{1}^{(e)} \\ q_{1}^{(e)}\\ \end{Bmatrix} $$ =$$\begin{Bmatrix} P_{1}^{(e)}\\ P_{2}^{(e)} \end{Bmatrix}$$

$$ \boldsymbol{K}^{(e)}$$is a 2x2 matix $$ q_{i}^{(e)} $$ is an axial displacement of element e at local node i

$$P_{i}^{(e)}$$ is the axial force at local node i.

Goal: Derive eq(1) from eq(2)

Want to find relation (already done in meeting 4)

d(e) 2x1 and d(e) 4x1

p(e) 2x2 and f(e) 4x1

The relationship between q and d is $$ q^{(e)}=d^{e}_{i}\bullet \tilde{\imath }$$

The relationship between p and f is axis on which it works is different. One works on the normal x-y axis, the other works on the $$\tilde{x}$$ axis

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

$$\boldsymbol{K}^{(e)}*\boldsymbol{q}^{(e)}=\boldsymbol{P}^{(e)}$$

So multiply by the inverse of K and you get the distance and axial distance. You can then find the relation by using the previous relationship between q and d.

These relationships can be expressed in for $$q^{(e)}=\boldsymbol{T}^{(e)}\boldsymbol{d}^{(e)}$$

Consider the displacement vector of the load mode [1], denoted by d[1](2)



$$d_{[1]}^{(e)}=d_{1}^{e}\imath +d_{2}^{e}\jmath $$

$$q_{1}^{(e)}$$= axial displacement of node [1] is the orthogonal projection of the displacement vector math>d_{[1]}^{(e)} of node 1 on the axis ~χ of element e

$$\Rightarrow q_{1}^{(e)}=\vec{d_{[1]}^{(e)}}\bullet \vec{\imath} $$

d(e) can be separated into a horizontal and vertical component

$$=\left(d_{1}^{(e)}\vec{\imath} +d_{2}^{(e)}\vec{\jmath} \right)\bullet \vec{\iota} $$

The dot product can be distributed and you get

$$=d_{1}^{(e)}\left(\vec{\imath} \bullet \vec{\iota} \right) + d_{2}^{(e)}\left(\vec{\jmath} \bullet \vec{\iota} \right)$$

The first part of the equation above is equal to $$\cos \theta ^{(e)}$$ or l(e)

The second part of the eqution above is equal to $$\sin \theta ^{(e)}$$ or M(e)

$$q^{(e)}_{1}= l^{(e)}d_{1}^{(e)}+ m^{(e)}d_{2}^{(e)}$$

$$q=\begin{bmatrix} l^{(e)} & M^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{Bmatrix}$$

q is a 1x1 matrix (otherwise known as a scalar)

It is also the displacement along the $$\tilde{x}$$ axis.

Similarly for node [2]:

$$q_{2}^{(e)}=\begin{bmatrix} l^{(e)} & M^{(e)} \end{bmatrix} \begin{Bmatrix} d_{3}^{(e)}\\ d_{4}^{(e)} \end{Bmatrix}$$

$$\begin{Bmatrix} q_{1}^{(e)} \\ q_{2}^{(e)} \end{Bmatrix}= \begin{bmatrix} l^{(e)} & M^{(e)}& 0 & 0 \\ 0 & 0 & l^{(e)} & M^{(e)}\end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\d_{2}^{(e)}\\d_{3}^{(e)}\\d_{4}^{(e)}\end{Bmatrix} $$

This equation shows how you can find the axial displacement of the two nodes by multiplying the displacements by the proper m(e) and l(e)