User:Eml4500.f08.echo.sell.cm/HW3

Solving for Unknown Displacements
Next, the elimination of known degree's of freedom can create the following matrices.

$$ d_1^{} = d_2^{} = d_5^{} = d_6^{}=0 $$

$$	\mathbf{K}\begin{Bmatrix} d_1=0\\d_2=0\\d_3\\d_4\\d_5=0\\d_6=0 \end{Bmatrix}=\begin{vmatrix} k_{13} & k_{14} \\ k_{23} & k_{24}\\k_{33} & k_{34}\\k_{43} & k_{44}\\k_{53} & k_{54}\\k_{63} & k_{64}\end{vmatrix}\begin{Bmatrix} d_3\\d_4\end{Bmatrix}=\mathbf{F}$$

Applying fixed boundary conditions allows us to delete the corresponding columns in the global stiffness matrix K

By the Principal of Virtual Work (PVW) we're able to delete the corresponding rows (in this case - 1, 2, 5, 6). Corresponding rows of F are also deleted.

The resulting FD relation is: $$ \begin{bmatrix} k_{33} & k_{34} \\ k_{43} & k_{44} \end{bmatrix} \begin{Bmatrix} d_3\\d_4\end{Bmatrix}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}$$

$$\mathbf{F}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}=\begin{Bmatrix} 0\\P\end{Bmatrix}$$

And P =>

What is the inverse of K without using a calculator?

Taking the determinant of K:

$$det \mathbf{K}=k_{33}k_{44}-k_{k34}k_{43}$$

$$\mathbf{K}^{-1}=\frac{1}{det\mathbf{K}}\begin{bmatrix} k_{44} & -k_{34} \\ -k_{43} & k_{33} \end{bmatrix}$$

Verifying:

$$\mathbf{K}\mathbf{K}^{-1}=\mathbf{K}^{-1}\mathbf{K}=\mathbf{I}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

This is essentially:

$$\frac{1}{det(\mathbf{K})}\begin{bmatrix} det(\mathbf{K}) & 0 \\ 0 & det(\mathbf{K}) \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Note: $$\mathbf{K}^T=\begin{bmatrix} k_{33} & k_{43} \\ k_{34} & k_{44} \end{bmatrix}$$

Therefore: $$\mathbf{K}^{-1}\ne \ \frac{1}{det(\mathbf{K})}\mathbf{K}^T$$

Going back to the 2 bar truss problem...

$$\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix}=\mathbf{K}^{-1}\begin{Bmatrix} 0 \\ P \end{Bmatrix}=\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

$$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix}_{2x1} = \mathbf{T}_{2x4}^{(e)} \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix}_{4x1}$$

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

Recall the element axial Force-Displacement relationship

$$\mathbf{\hat{k}}_{2x2}^{(e)} = \mathbf{q}_{2x1}^{(e)} \mathbf{P}_{2x1}^{(e)}$$

where:

$$\mathbf{q}_{2x2}^{(e)} = \mathbf{T}_{2x1}^{(e)} \mathbf{d}_{2x1}^{(e)}$$

$$\mathbf{P}_{2x2}^{(e)} = \mathbf{T}_{2x1}^{(e)} \mathbf{f}_{2x1}^{(e)}$$

therefore:

$$\mathbf{\hat{k}}_{}^{(e)}(\mathbf{T}_{}^{(e)} \mathbf{d}_{}^{(e)}) = (\mathbf{T}_{}^{(e)} \mathbf{f}_{}^{(e)})$$

 Goal:  We want to have $$\mathbf{k}_{}^{(e)} \mathbf{d}_{}^{(e)} = \mathbf{f}_{}^{(e)}$$.

So we need to move $$\mathbf{T}_{}^{(e)}$$ from the right hand side to the left hand side by "pre-multiplying" the equation by $$\mathbf{T}_{}^{(e) -1}$$.

Unfortunately since $$\mathbf{T}_{}^{(e)}$$ is a rectangular matrix it can not be inverted.

However through usage of the Principle of Virtual Work we find $$\mathbf{k}_{}^{(e)} = [\mathbf{T}_{}^{(e)T} \mathbf{\hat{k}}_{}^{(e)}\mathbf{T}_{}^{(e)}]$$

Which yields answer of $$[\mathbf{T}_{4x2}^{(e)T} \mathbf{\hat{k}}_{2x2}^{(e)}\mathbf{T}_{2x4}^{(e)}]_{4x4} \mathbf{d}_{4x1}^{(e)} = \mathbf{f}_{4x1}^{(e)}$$

and gives us $$\mathbf{k}_{}^{(e)} \mathbf{d}_{}^{(e)} = \mathbf{f}_{}^{(e)}$$

 HW:  Verify $$\mathbf{k}_{}^{(e)} = [\mathbf{T}_{}^{(e)T} \mathbf{\hat{k}}_{}^{(e)}\mathbf{T}_{}^{(e)}]$$

Using the Principle of Virtual Work $$\mathbf{\hat{k}}_{}^{(e)} = \mathbf{d}_{}^{(e)}) = \mathbf{f}_{}^{(e)}$$ because the transformation matrix $$\mathbf{T}_{}^{(e)}$$ is an arbitrary matrix. However through further evaluation of the equation $$\mathbf{\hat{k}}_{}^{(e)} = \mathbf{d}_{}^{(e)}) = \mathbf{f}_{}^{(e)}$$ we recognize that the sizes of the matrices prevent this operation from being carried out. We can re-size $$\mathbf{\hat{k}}_{}^{(e)}$$ by multiplying it by and identity matrix where the matrix transpose is multiplied by the matrix. So, $$\mathbf{T}_{}^{(e)T}\mathbf{T}_{}^{(e)}$$ is our identity matrix and by definition when multiply the identity matrix by the unit matrix we are left with a matrix the re-sized matrix. i.e. $$\mathbf{k}_{}^{(e)} = [\mathbf{T}_{}^{(e)T} \mathbf{\hat{k}}_{}^{(e)}\mathbf{T}_{}^{(e)}]$$

We could not solve as $$\mathbf{d}_{}^{} = \mathbf{K}_{}^{-1} = \mathbf{F}_{}^{}$$ because of the singulrity associated with matrix $$\mathbf{K}_{}^{}$$. ($$det\mathbf{K}_{}^{} = 0$$)

For an unconstrained strut system there are 3 possible rigid body motions in 2-D.(2trans. and 1 rat.)

These 3 motions are Axial Deformation, Rotation at Nodes and Translation at Nodes.

 HW:  Find the Eigenvalues of $$\mathbf{K}_{6x6}^{}$$ and make observations about the number of zero eigenvalues

dynamic evaluation problem $$\mathbf{K}_{}^{} \mathbf{v}_{}^{} = \lambda_{}^{} \mathbf{M}_{}^{} \mathbf{v}_{}^{}$$

where:

K = Stiffness Matrix

$$\mathbf{\lambda}$$ = eigenvalues(~vibration frequency)

M = Mass Matrix

$$(\mathbf{K}_{}^{} - \lambda_{}^{} \mathbf{M}_{}^{}) \mathbf{v}_{}^{} = 0$$

$$\begin{bmatrix} .5625 & .3248 & -.5625 & -.3248 & 0 & 0\\ .3248 & .1875 & -.3248 & -.1895 & 0& 0\\ -.5625 & -.3248 & 3.0625 & -2.1752 & -2.5 & 2.5\\ -.3248 & -.1875 & -2.1752 & 2.6875 & 2.5 & -2.5\\ 0 & 0 & -2.5 & 2.5 & 2.5 & -2.5\\ 0 & 0 & 2.5 & -2.5 & -2.5 & 2.5 \end{bmatrix}v = \mathbf{\lambda_{}^{}M_{}^{}}v$$

Solving yields the following eigenvalues: $$\mathbf{\lambda_{1}}$$ = 10.03, $$\mathbf{\lambda_{2}}$$ = 1.471, $$\mathbf{\lambda_{3}}$$ = 0, $$\mathbf{\lambda_{4}}$$ = 0, $$\mathbf{\lambda_{5}}$$ = .2983, $$\mathbf{\lambda_{6}}$$ = 0

Zeros occur in rows 3, 4 and 6 of lambda.

The eigenvector Matrix is: $$\begin{bmatrix} 0.0139 & 0.6174 & -0.6455 & 0.4494 & -0.6131 & 0.5997\\ 0.0080 & 0.3565 & -0.1291 & -0.6754 & 0.0754 & 0.1199\\ -0.5123 & -0.5409 & -0.6454 & -0.1682 & -0.6131 & 0.5997\\ 0.4904 & -0.4330 & -0.1291 & 0.3942 & 0.0754 & 0.1199\\ 0.4984 & -0.0765 & -0.2582 & -0.2812 & -0.3442 & 0.5014\\ -0.4984 & 0.0765 & 0.2582 & 0.2812 & 0.3442 & 0.0216 \end{bmatrix}$$

The zero eigenvalues correspond with zero stored elastic energy yields rigid body nodes

--C. Sell 23:30, 7 October 2008 (UTC)