User:EML4500.f08.FEABBQ.Hamdan/HW-3

Class Notes: September 26, 2008
Equation [2]

$$ \mathbf{k^{(e)}} \mathbf{q^{(e)}} = \mathbf{P^{(e)}} $$

Fallowing the same argument above we get the relation

$$ \begin{Bmatrix} p_1^{(e)} \\ p_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} l^{(e)} & m^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \end{matrix} \right] \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} $$

Equation [3]

$$ \mathbf{p_{2x1}^{(e)}} = \mathbf{T_{2x4}^{(e)}} \mathbf{f_{4x1}^{(e)}} $$

recalling element axial FD relation

Equation [4]

$$ \mathbf {k_{2x4}^{(e)}} \mathbf{q_{4x1}^{(e)}} =\mathbf{p_{2x1}^{(e)}}  $$

and Equation [5]

$$ \mathbf{q_{2x1}^{(e)}} = \mathbf{T_{2x4}^{(e)}} \mathbf{d_{4x1}^{(e)}} $$

Substituting equations 3 and 5 into equation 4 yields to

$$ \mathbf {k_{2x4}^{(e)}} \mathbf{T_{2x4}^{(e)}} \mathbf{d_{4x1}^{(e)}} =\mathbf{T_{2x4}^{(e)}} \mathbf{f_{4x1}^{(e)}}  $$

The Goal is to have $$ \mathbf{k^{(e)}} \mathbf{d^{(e)}} = \mathbf{f^{(e)}} $$ so move $$ \mathbf{T^{(e)}} $$ from R.H.S. to L.H.S. by premultiplying the equation by $$ \mathbf{{(T^{(e)})}^{(-1)}} $$        ( inverse $$ \mathbf{T^{(e)}} $$)

But unfortunately $$ \mathbf{T^{(e)}} $$ is a rectangular matrix [2x4], can't inverse $$ \mathbf{T^{(e)}} $$

to solve this problem the transpose property was used as follows

$$ \mathbf{{T_{4x2}^{(e)}}^{T}} \mathbf {k_{2x4}^{(e)}} \mathbf{T_{2x4}^{(e)}} \mathbf{d_{4x1}^{(e)}} = \mathbf{f_{4x1}^{(e)}} $$

$$ \mathbf{{T_{4x2}^{(e)}}^{T}} \mathbf {k_{2x4}^{(e)}} \mathbf{T_{2x4}^{(e)}} \mathbf{d_{4x1}^{(e)}} = \mathbf{f_{4x1}^{(e)}} $$

$$ \mathbf {k_{2x4}^{(e)}} \mathbf{d_{4x1}^{(e)}} = \mathbf{f_{4x1}^{(e)}}  $$

Homework

Verify that

$$ \mathbf {I_{4x4}} = \mathbf{{T_{4x2}^{(e)}}^{T}} \mathbf{T_{2x4}^{(e)}}   $$

$$ \mathbf {I_{4x4}} =  \left[ \begin{matrix} l^{(e)} & 0 \\ m^{(e)} & 0 \\ 0 & l^{(e)} \\ 0 & m^{(e)}  \end{matrix} \right]  \left[ \begin{matrix} l^{(e)} & m^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \end{matrix} \right]   $$

Applying the principle of virtual work will result reduction of global FD relation:

$$ \mathbf {k_{2x4}^{(e)}} = \mathbf{{T_{4x2}^{(e)}}^{T}}  \mathbf{\tilde{k}_{4x1}^{(e)}} \mathbf{T_{2x4}^{(e)}}   $$

Remark

why not solve as follows $$ \mathbf{d_{4x1}^{(e)}} = \mathbf{{(T^{(e)})}^{(-1)}}  \mathbf{f_{4x1}^{(e)}}  $$  ?

The answer is that the used methods to compute $$ \mathbf{{(K^{(e)})}^{(-1)}} $$ could not be used since K is singular. this will result a determinants value of zero and thus K is not inverted.

Recall

for an unconstrained structure system there are three possible body motion in 2-D (plane) two translation and one rotation.

Homework

Compute the eigenvalues of K and make observation about the number of zero eigenvalues.

Matlab: Computing eigenvalues
The following is a matlab script and output for the two-bar K example.

K =

0.5625     0.32476      -0.5625     -0.32476            0            0      0.32476       0.1875     -0.32476      -0.1875            0            0      -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5     -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5            0            0         -2.5          2.5          2.5         -2.5            0            0          2.5         -2.5         -2.5          2.5

>> eig(K)

ans =

-0.0000  -0.0000    0.0000    0.0000    1.4706   10.0294

Zero Eigenvalues correspond to zero stored elastic energy, rigid body modes- (shape, eigenvictors)