User:Eml4500.f08.Ateam.Slevinski/Homework3

HW#3: 9/26/08 Lecture Notes (Meeting 14)
From the relationship betweeen qi(e) and di(e):

$$ \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} l^{(e)} & m^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \end{matrix} \right] \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} $$  where  q(e) = T(e)×d(e)

we can use the same argument to solve for the axial displacements, qi(e), and the axial forces, pi(e).

$$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} T^{(e)} \end{matrix} \right] \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} $$         where  P(e) = T(e)×f(e)

We now can recall the force-displacement relationship:

 $$ k^{(e)}(T^{(e)}d^{(e)}) = (T^{(e)}f^{(e)}) $$

The goal here is to have the original force-displacement relationship: $$ k^{(e)}d^{(e)} = f^{(e)} $$. To accomplish this we would need to move matrix $$ T^{(e)} $$ from the right hand side to the left hand side. This could be achieved by multiplying the equation by the inverse $$ T^{-1} $$ of the matrix $$ T^{(e)} $$. As seen above though, the matrix $$ T^{(e)} $$ is a rectangular matrix and therefore cannot be inverted.

To overcome this problem, we will use the transpose of $$ T^{(e)} $$ and replace $$ k^{(e)} $$ with $$ [T^{(e)T}k^{(e)}T^{(e)}] $$. The force-displacement relationship now becomes:

$$ [T^{(e)T}k^{(e)}T^{(e)}]d^{(e)} = f^{(e)} $$

We can recall from previous lectures that $$ \theta^{(1)} = 30^{o} $$, $$ \theta^{(2)} = 45^{o} $$ and $$ l^{(e)} = cos(\theta^{(e)}) $$, $$ m^{(e)} = sin(\theta^{(e)}) $$. From these values and equations we can determine the matrix $$ T^{(e)} $$.

$$ \begin{Bmatrix} T^{(1)} \end{Bmatrix} = \left[ \begin{matrix} l^{(1)} & m^{(1)} & 0 & 0 \\ 0 & 0 & l^{(1)} & m^{(1)} \end{matrix} \right] = \left[ \begin{matrix} 0.866 & 0.5 & 0 & 0 \\ 0 & 0 & 0.866 & 0.5 \end{matrix} \right] $$

And the transpose of $$ T^{(1)} $$ is:

$$\left[ \begin{matrix} 0.866 & 0 \\ 0.5 & 0 \\ 0 & 0.866 \\ 0 & 0.5 \end{matrix} \right]$$

From lecture 12 we know that, $$ k^{(1)} = \left[ \begin{matrix} 1 & -1 \\ -1 & 1 \end{matrix} \right] $$

So therefore,

$$ [T^{(1)T}k^{(1)}T^{(1)}] $$ = $$\left[ \begin{matrix} 0.866 & 0 \\ 0.5 & 0 \\ 0 & 0.866 \\ 0 & 0.5 \end{matrix} \right]$$'$$ \left[ \begin{matrix} 1 & -1 \\ 1 & -1 \end{matrix} \right] $$'$$\left[ \begin{matrix} 0.866 & 0.5 & 0 & 0 \\ 0 & 0 & 0.866 & 0.5 \end{matrix} \right] $$ = $$\left[ \begin{matrix} 0.7499 & 0.433 & -0.7499 & -0.433 \\ 0.433 & 0.25 & -0.433 & -0.25 \\ -0.7499 & -0.433 & 0.7499 & 0.433 \\ -0.433 & -0.25 & 0.433 & 0.25 \end{matrix} \right] $$

And if we recall from previous lectures, we will see that these are the coefficients of the 4x4 matrix $$ \underline k^{(1)} $$.

Calculating the eigenvalues of this matrix leads to values of:

$$ \lambda = \left[ \begin{matrix} 0 & 2 & 0 & 0 \end{matrix} \right] $$

The three zero eigenvalues in this matrix correspond to the three possible rigid body motions. Two of these being translational and the other rotational.

HW #3
From the relationship betweeen qi(e) and di(e):

$$ \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} l^{(e)} & m^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \end{matrix} \right] \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} $$  where  q(e) = T(e)×d(e)

we can use the same argument to solve for the axial displacements, qi(e), and the axial forces, pi(e).

$$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} T^{(e)} \end{matrix} \right] \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} $$         where  P(e) = T(e)×f(e)

We now can recall the force-displacement relationship:

 $$ k^{(e)}(T^{(e)}d^{(e)}) = (T^{(e)}f^{(e)}) $$

The goal here is to have the original force-displacement relationship: $$ k^{(e)}d^{(e)} = f^{(e)} $$. To accomplish this we would need to move matrix $$ T^{(e)} $$ from the right hand side to the left hand side. This could be achieved by multiplying the equation by the inverse $$ T^{-1} $$ of the matrix $$ T^{(e)} $$. As seen above though, the matrix $$ T^{(e)} $$ is a rectangular matrix and therefore cannot be inverted.

To overcome this problem, we will use the transpose of $$ T^{(e)} $$ and replace $$ k^{(e)} $$ with $$ [T^{(e)T}k^{(e)}T^{(e)}] $$. The force-displacement relationship now becomes:

$$ [T^{(e)T}k^{(e)}T^{(e)}]d^{(e)} = f^{(e)} $$

We can recall from previous lectures that $$ \theta^{(1)} = 30^{o} $$, $$ \theta^{(2)} = 45^{o} $$ and $$ l^{(e)} = cos(\theta^{(e)}) $$, $$ m^{(e)} = sin(\theta^{(e)}) $$. From these values and equations we can determine the matrix $$ T^{(e)} $$.

$$ \begin{Bmatrix} T^{(1)} \end{Bmatrix} = \left[ \begin{matrix} l^{(1)} & m^{(1)} & 0 & 0 \\ 0 & 0 & l^{(1)} & m^{(1)} \end{matrix} \right] = \left[ \begin{matrix} 0.866 & 0.5 & 0 & 0 \\ 0 & 0 & 0.866 & 0.5 \end{matrix} \right] $$

And the transpose of $$ T^{(1)} $$ is:

$$\left[ \begin{matrix} 0.866 & 0 \\ 0.5 & 0 \\ 0 & 0.866 \\ 0 & 0.5 \end{matrix} \right]$$

From lecture 12 we know that, $$ k^{(1)} = \left[ \begin{matrix} 1 & -1 \\ -1 & 1 \end{matrix} \right] $$

So therefore,

<p style="text-align:center;">$$ [T^{(1)T}k^{(1)}T^{(1)}] $$ = $$\left[ \begin{matrix} 0.866 & 0 \\ 0.5 & 0 \\ 0 & 0.866 \\ 0 & 0.5 \end{matrix} \right]$$'$$ \left[ \begin{matrix} 1 & -1 \\ 1 & -1 \end{matrix} \right] $$'$$\left[ \begin{matrix} 0.866 & 0.5 & 0 & 0 \\ 0 & 0 & 0.866 & 0.5 \end{matrix} \right] $$ = $$\left[ \begin{matrix} 0.7499 & 0.433 & -0.7499 & -0.433 \\ 0.433 & 0.25 & -0.433 & -0.25 \\ -0.7499 & -0.433 & 0.7499 & 0.433 \\ -0.433 & -0.25 & 0.433 & 0.25 \end{matrix} \right] $$

And if we recall from previous lectures, we will see that these are the coefficients of the 4x4 matrix $$ k^{(1)} $$.

Given the 2x2 matrix $$ k^{(1)} = \left[ \begin{matrix} 1 & -1 \\ -1 & 1 \end{matrix} \right]$$ we can calculate the eigenvalues of this matrix as follows:

<p style="text-align:center;"> det$$ \left[ \begin{matrix} (1-\lambda) & -1 \\ -1 & (1-\lambda) \end{matrix} \right]$$ = $$ (1-\lambda)^{2} - 1 = 0 $$.

Therefore, the eigenvalues (or roots) of this equation are:

<p style="text-align:center;">$$ \lambda_{1} = 0 $$ and $$ \lambda_{2} = 2 $$