User:Eml4500.f08.ateam.shah/Week 3

Homework Report 3 Friday September 25, 2009



\begin{bmatrix} P_1^{(e)}\\ P_2^{(e)}\\ \end{bmatrix} = T^{(e)} \cdot \begin{bmatrix} f_1^{e}\\ f_2^{e}\\ f_3^{e}\\ f_4^{e}\\ \end{bmatrix} $$

Where T(e) is a 2x4 transformation matrix which can be used to transform distance and force from one coordinate system to another.



q^{(e)} = T^{(e)} \cdot d^{(e)} $$



P^{(e)} = T^{(e)} \cdot f^{(e)} $$

Recall



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

Where q and P are axial components with k being a 2x2 overlap stiffness matrix and q being 2x1 distance matrix.

Thus

T^{(e)} \cdot d^{(e)} = q^{(e)} $$



T^{(e)} \cdot f^{(e)} = P^{(e)} $$



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

The goal is to have a FD distance relationship as described by:

k(e) * d(e) = f(e)

So you move the transformation matrix from the RHS to the LHS by premultiplying equation by Y(e)-1. (inverse of the T matrix)

However the T matrix is a rectangular 2x4 matrix, thus it cannot be inverted. However the T matrix can be transposed to achieve similar results:

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

Where transpose of T is a 4x2 matrix and the overall resultant matrix of the transpose of TT * k * T leads to a 4x4 matrix.

The following problem has been worked out to verify the above statement. $$     \begin{bmatrix} &  &   &   &   \\          &   &   &   &   \\          &   & K &   &   \\ &  &   &   &   \\          &   &   &   &      \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix} \cdot \begin{bmatrix} &  &   &   &   \\          &   &   &   &   \\          &   & K &   &   \\ &  &   &   &   \\          &   &   &   &      \end{bmatrix} \cdot \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}^T $$

The justification of this concept can be provided using the principal virtual work theorem introduced later in the semester.


 * See P 10 for the 1st example of PVW/Reduction of global FD relationship.

REMARK The k equation on the exam to determine the displacement could not be solved using: k-1 * F = d because  k  is a singular matrix.

 K  is a singular matrix, which means the determinant of  k  is 0 and thus  k  cannot be inverted.

RECALL: You need to computed (det)-1 to find the inverse of the  k  matrix.


 * A famous quote narrated by the professor:

" Ask what not your country can do for you               ask what you can do for your country." JFK For an unconstrained structure system, there are 3 possible rigid body motions in 2-D (2 truss - 1 rotation).

HW- Find the eigenvalues of the K matrix, and make observation about the number of 0 eigenvalues.

A =

0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.2500   -0.3248   -0.2500         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.2500    2.8248    1.9375    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

>> eig(A)

ans =

8.4194   2.5879   -0.1729   -0.0219   -0.0000   -0.0000

Dynamic Evaluation Problem:
 * k * v = λ * M * V

Where k*v is the stiffness matrix, λ is the eigenvalue related to the vibrational frequency, M is the mass of the system and V is the velocity matrix. Thus zero eigenvalues lead to zero stored elastic energy. This means the system is rigid.