User:Eml4500.f08.Lulz.js/HW7

Frame Elements
A basic frame element consists of a truss element combined with a beam element, which gives the frame element the capacity to handle both axial and transverse deformations. Consider Figure 1, which shows a model problem consisting of a system of two frame elements. The point of connection between the two elements, where the force $$\;P$$ is applied, is a rigid connection. This means that the angle between the two elements remains constant after deformation.



The free body diagrams for both elements are shown in Figures 2 and 3. Generally speaking, the degrees of freedom for such a system correspond directly to the general forces, or $$d_i^{(e)} \Rightarrow f_i^{(e)}$$. However, for this problem, two new degrees of freedom have been introduced for both elements. These degrees of freedom correspond as follows from the force vector:


 * $$\underbrace{\left\{\begin{matrix} d_3^{(e)} \\ d_6^{(e)} \end{matrix}\right\}}_{\mbox{rotational degrees of freedom}} \Rightarrow \underbrace{\left\{\begin{matrix} f_3^{(e)} \\ f_6^{(e)} \end{matrix}\right\}}_{\mbox{bending moments}}$$

These new degrees of freedom correspond to rotations, and the associated forces are actually bending moments. Figure 4 shows the global view of the system, with the global degrees of freedom. Once element stiffness matrices $$\mathbf{k}^{(e)}\;$$ are generated, they can be assembled into the global stiffness matrix as follows:


 * $$\mathbf{K} = A\mathbf{k}^{(e)}$$

$$A\;$$ is the assembly operator. An overview of the global stiffness matrix is given in Figure 5.



In order to generate the element stiffness matrices, consider the local degrees of freedom for element 1, as shown in Figure 6. The stiffness matrices can be determined from the following familiar relationship:


 * $$\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)}$$

Where $$\mathbf{\tilde{d}}^{(e)}$$ and $$\mathbf{\tilde{f}}^{(e)}$$ are given as:


 * $$\mathbf{\tilde{d}}^{(e)} = \begin{Bmatrix} \tilde{d}^{(e)}_1\\ \vdots \\\tilde{d}^{(e)}_6 \end{Bmatrix}$$


 * $$\mathbf{\tilde{f}}^{(e)} = \begin{Bmatrix} \tilde{f}^{(e)}_1\\ \vdots \\\tilde{f}^{(e)}_6 \end{Bmatrix}$$

For the rotational degrees of freedom, the following relationships are true:


 * $$\tilde{f}^{(e)}_3 = f^{(e)}_3$$, $$\tilde{f}^{(e)}_6 = f^{(e)}_6$$

Using these considerations, the general element stiffness matrix $$\mathbf{\tilde{k}}^{(e)}$$ can be constructed as follows:


 * $$\mathbf{\tilde{k}}^{(e)} = \begin{bmatrix} \frac{EA}{L}& 0 & 0 & \frac{-EA}{L} & 0 & 0\\ & \frac{12EI}{L^2} & \frac{6EI}{L^2} & 0 & \frac{-6EI}{L^2} & \frac{6EI}{L^2}\\  &  & \frac{4EI}{L} & 0 & \frac{-6EI}{L^2} & \frac{-2EI}{L}\\  &  &  & \frac{EA}{L} & 0 & 0\\  &  &  &  & \frac{12EI}{L^2} & \frac{-6EI}{L^2}\\ sym. &  &  &  &  & \frac{4EI}{L}\end{bmatrix}$$

Dimensional Analysis
It is useful to look at the dimensions of some of the variables involveed in the frame system. Consider first the non-rotational degrees of freedom:


 * $$[\tilde{d}_1] = L = [\tilde{d}_i]$$, for $$i=1,2,4,5\;$$

Conversely, consider the rotational degrees of freedom:


 * $$[\tilde{d}_3] = 1 = [\tilde{d}_6]$$

Figure 7 displays the arc length relationship for circular sections. This relationship is summarized as:


 * $$\widehat{AB} = R\theta \Rightarrow \theta = \frac{\widehat{AB}}{R}$$

Performing a dimensional analysis on this concept results in:


 * $$[\theta] = \frac{[\widehat{AB}]}{[R]} = \frac{L}{L} = 1$$

Furthermore, consider the stress strain relation:


 * $$\sigma = E\varepsilon \Rightarrow [\sigma] = [E]\underbrace{[\varepsilon]}_{1} = [E]$$


 * $$[\sigma] = [E] = \frac{F}{L^2}$$

Consider the axial displacement $$q\;$$:


 * $$[q] = \frac{[du]}{[dx]} = \frac{L}{L} = 1$$

For some more fundamental variables, consider the area and moment of inertia:


 * $$[A] = L^2\;$$


 * $$[I] = L^4\;$$

With the dimensions of these more basic terms defined, it is possible to analyze the dimensions of more complex variables. For example, consider the stiffness factor $$k\;$$:


 * $$\left[\frac{EA}{L}\right] = [\tilde{k}_1] = \frac{(\frac{F}{L^2})(L^2)}{L} = \frac{F}{L}$$

Analyzing the product of the global stiffness coefficient and the global degree of freedom proves that this product generates the global force components:


 * $$[\tilde{k}_{11}\tilde{d}_1] = [\tilde{k}_{11}][\tilde{d}_1] = (\frac{F}{L})(L) = F$$


 * $$[\tilde{k}_{23}\tilde{d}_3] = [\tilde{k}_{23}]\underbrace{[\tilde{d}_3]}_1 = \frac{[6][E][I]}{[L^2]} = \frac{(1)(\frac{F}{L^2})(L^4)}{L^2} = F$$

Element Displacement Relation
Returning to the local degrees of freedom, the following relation can be written to convert $$\mathbf{d}^{(e)}\;$$ to $$\mathbf{\tilde{d}}^{(e)}\;$$:


 * $$\begin{Bmatrix} \tilde{d}_1\\ \tilde{d}_2\\ \tilde{d}_3\\ \tilde{d}_4\\ \tilde{d}_5\\ \tilde{d}_6 \end{Bmatrix} = \underbrace{\begin{bmatrix} l^{(e)}& m^{(e)} & 0 & 0 & 0 & 0\\ -m^{(e)} & l^{(e)} & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & 0\\ 0 & 0 & 0 & -m^{(e)} & l^{(e)} & 0\\ 0 & 0 & 0 & 0 & 0 & 1\end{bmatrix}}_{\mathbf{\tilde{T}^{(e)}}} \begin{Bmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6 \end{Bmatrix}$$

Where $$\begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)}\end{bmatrix}$$ is the previously discussed $$\mathbf{R}\;$$.

Derivation of Element Stiffness Matrix from the Principle of Virtual Work
Consider the Principle of Virtual Work (PVW) for beams given as Equation 1 below:


 * $$\int_0^L w(\underbrace{x}_\tilde{x}) [\frac{-\partial^2}{\partial x^2}[(EI)\frac{\partial v^2}{\partial x^2}] + f_t(x) - m(x)\ddot{v}] = 0\;,\;\mbox{for all possible} \;w(x)\;\;(1)$$

Isolating the first term and performing an integration by parts yields the following:


 * $$\alpha := \int_0^L \underbrace{w(x)}_{s(x)} \underbrace{\frac{\partial^2}{\partial x^2}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}}_{\underbrace{\frac{\partial}{\partial x} \underbrace{\left(\frac{\partial}{\partial x}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}\right)dx}_{r(x)}}_{r'(x)}}$$


 * $$\alpha := \underbrace{\left[w(x)\frac{\partial}{\partial x}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}\right]^L_0}_{\beta_1} - \int_0^L \underbrace{\frac{\partial w}{\partial x}}_{s'(x)}\underbrace{\left(\frac{\partial}{\partial x}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}\right)dx}_{r(x)}$$

Performing another integration by parts on the second term of the above yields the following:


 * $$\alpha := \underbrace{\left[w(x)\frac{\partial}{\partial x}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}\right]^L_0}_{\beta_1} - \int_0^L \underbrace{\frac{\partial w}{\partial x}}_{s'(x)}\underbrace{\left(\frac{\partial}{\partial x}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}\right)dx}_{r(x)}$$


 * $$\alpha := \beta_1 - \underbrace{\left[\frac{-\partial w}{\partial x}(EI)\frac{\partial v^2}{\partial x^2}\right]^L_0}_{\beta_2} + \int_0^L \underbrace{\frac{\partial^2 w}{\partial x^2}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}dx}_{\gamma}$$

Thus, Equation 1 can be rewritten as:


 * $$ -\beta_1 + \beta_2 - \gamma + \int^L_0 w f_t dx - \int^L_0 w m \ddot{v} dx = 0$$

Consider the beam diagram in Figure 7. Focusing on the $$\gamma\;$$ term for now, the beam stiffness matrix and shape functions can be derived as follows. First, consider the definitions for the deformation components $$v(\tilde{x})\;$$ and $$u(\tilde{x})\;$$:


 * $$ v(\tilde{x}) = N_2(\tilde{x})\tilde{d}_2 + N_3(\tilde{x})\tilde{d}_3 + N_5(\tilde{x})\tilde{d}_5 + N_6(\tilde{x})\tilde{d}_6$$


 * $$ u(\tilde{x}) = N_1(\tilde{x})\tilde{d}_1 + N_4(\tilde{x})\tilde{d}_4$$

Figure 8 shows the drawings for the various shape functions. The functions themselves are defined as:


 * $$ N_2(\tilde{x}) = 1 - \frac{3\tilde{x}^2}{L^2} - \frac{2\tilde{x}^3}{L^3}\;\;\tilde{d}_2$$


 * $$ N_3(\tilde{x}) = \tilde{x} - \frac{2\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2}\;\;\tilde{d}_3$$


 * $$ N_5(\tilde{x}) = \frac{3\tilde{x}^2}{L^2} - \frac{2\tilde{x}^3}{L^3}\;\;\tilde{d}_5$$


 * $$ N_6(\tilde{x}) = - \frac{\tilde{x}^2}{L} - \frac{\tilde{x}^3}{L^2}\;\;\tilde{d}_6$$

Recall the general element transformation relation:


 * $$\mathbf{\tilde{d}^{(e)}} = \mathbf{\tilde{T}^{(e)}}\mathbf{d^{(e)}}$$



Consider Figure 9, the deformation components, and then the following relation for computing them:


 * $$\mathbf{u}(\tilde{x}) = u(\tilde{x}) \vec{\tilde{i}} + v(\tilde{x}) \vec{\tilde{j}} = u_x(\tilde{x}) \vec{\tilde{i}} + u_y(\tilde{x}) \vec{\tilde{j}}$$

Using the above and the previously discussed shape-function relationships, the following relationships can be derived:


 * $$\begin{Bmatrix} u_x(\tilde{x})\\u_y(\tilde{x})\end{Bmatrix} = \mathbf{R}^T \begin{Bmatrix} u(\tilde{x})\\v(\tilde{x})\end{Bmatrix}$$


 * $$\begin{Bmatrix} u(\tilde{x})\\v(\tilde{x})\end{Bmatrix} = \underbrace{\begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & N_3 & 0 & N_5 & N_6 \end{bmatrix}}_{\mathbb{N}(\tilde{x})} \mathbf{\tilde{d}^{(e)}}$$

Combining the results of this section, the following overall relation can be determined:


 * $$\begin{Bmatrix} u_x(\tilde{x})\\u_y(\tilde{x})\end{Bmatrix} = \mathbf{R}^T \mathbb{N}(\tilde{x}) \mathbf{\tilde{T}^{(e)}}\mathbf{d^{(e)}}$$

Dimensional Analysis
As before, it is useful to revisit the dimensional analysis of some of the results for the previous section:


 * $$[u] = L\;$$


 * $$ [N_1] = \cdots = [N_6] = 1 $$


 * $$[N_1 \tilde{d}_1] = [N_1][\tilde{d}_1] = L$$


 * $$[N_4 \tilde{d}_4] = [N_4][\tilde{d}_4] = L$$


 * $$[v] = L\;$$


 * $$[N_2][\tilde{d}_2] = [N_5][\tilde{d}_5] = L$$


 * $$[N_3][\tilde{d}_3] = [N_6][\tilde{d}_6] = 1$$

Solving the Beam PDE
Consider again the PDE for beams:


 * $$\frac{\partial^2}{\partial x^2} \left\{(EI)\frac{\partial^2 v}{\partial x^2}\right\} = 0$$

Making the assumption that $$EI\;$$ is a constant leads to the following:


 * $$\frac{\partial v^4}{\partial x^4} = 0$$

This is an easily integrable equation which yields the following relationship and its derivative:


 * $$v(x) = C_0 + C_1x + C_2x^2 + C_3x^3\;$$


 * $$v'(x) = C_1 + 2C_2x + 3C_3x^2\;$$

In order to solve for the coefficients of integration, certain boundary conditions must be used. The following set are used to obtain $$N_2(x)\;$$, where $$\tilde{x} = x\;$$ for simplicity:


 * $$v(0) = 1\;$$


 * $$v'(0) = 1\;$$


 * $$v(L) = 1\;$$


 * $$v'(L) = 0\;$$

Application of the first two boundary condition leads directly to $$C_0\;$$ and $$C_1\;$$


 * $$v(0) = 1 = C_0\;$$


 * $$v'(0) = C_1 = 0\;$$

Evaluating the equation and its derivative at $$x = L$$ yields the following:


 * $$v(L) = \underbrace{C_0}_1 + \underbrace{C_1L}_0 + C_2L^2 + C_3L^3 = 0\;$$


 * $$v'(L) = \underbrace{C_1}_0 + 2C_2L + 3C_3L^2 = 0\;$$

Evaluating for the determined coefficients allows the above to be rewritten as:


 * $$C_2L^2 + C_3L^3 = -1\;$$


 * $$2C_2L + 3C_3L^2 = 0\;$$

These equations form a simple system of two equations, which can be easily solved to yields expressions for the remaining two coefficients of integration:


 * $$C_2 = \frac{-3}{L^2}$$


 * $$C_3 = \frac{2}{L^3}$$

Thus, the original equation can be rewritten as follows:


 * $$v(x) = 1 - \frac{3}{L^2} + \frac{2}{L^3}$$

Obtain $$N_3(x)\;$$, $$N_5(x)\;$$, and $$N_6(x)\;$$ can be achieved with an identical solution method and the following boundary conditions (the order of sets corresponds to the order of the list):


 * $$v(0) = 0\;$$


 * $$v'(0) = 1\;$$


 * $$v(L) = 0\;$$


 * $$v'(L) = 0\;$$


 * $$v(0) = 0\;$$


 * $$v'(0) = 1\;$$


 * $$v(L) = 0\;$$


 * $$v'(L) = 0\;$$


 * $$v(0) = 0\;$$


 * $$v'(0) = 0\;$$


 * $$v(L) = 0\;$$


 * $$v'(L) = 1\;$$

Derive Coefficients in Element Stiffness Matrix
A sample coefficient of the element stiffness matrix can be written as follows:


 * $$\tilde{k}_{23} = \frac{6EI}{L^2} = \int_0^L \frac{\partial^2 N_2}{\partial x^2}(EI)\frac{\partial^2 N_3}{\partial x^2} dx $$

In general, the following relationship is true:


 * $$\tilde{k}_{ij} = \int_0^L \frac{\partial^2 N_i}{\partial x^2}(EI)\frac{\partial^2 N_j}{\partial x^2} dx $$

Elastodynamics (Vibrations via FEA)
Consider the following discrete PVW equation:


 * $$\mathbf{\bar{w}} \cdot [\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{k}}\mathbf{\bar{d}} - \mathbf{\bar{F}}] = 0\; \mbox{for all}\; \mathbf{\bar{w}}$$

This can easily be rewritten as Equation 1, with the associated initial conditions:


 * $$\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{k}}\mathbf{\bar{d}} = -\mathbf{\bar{F}}(t)\;\;(1)$$


 * $$\mathbf{\bar{d}}(0) = \bar{d}_0$$


 * $$\mathbf{\dot{\bar{d}}}(0) = \bar{v}_0$$

Solving Equation 1
Consider the following "unforced" vibrations problem, given as Equation 2. The problem can be considered unforced because the right hand side is the zero vector:


 * $$\mathbf{\bar{M}}\mathbf{\ddot{v}} + \mathbf{\bar{k}}\mathbf{v} = \mathbf{0}\;\;(2)$$

Assume that $$\mathbf{v}(t) = (\sin \omega t) \boldsymbol{\phi}$$, where $$\boldsymbol{\phi}$$ is not time dependent. Applying this relationship to Equation 2 yields:


 * $$-\omega^2 (\sin \omega t)\mathbf{\bar{M}} \boldsymbol{\phi} + (\sin \omega t)\mathbf{\bar{k}}\boldsymbol{\phi} = \mathbf{0}$$

This leads to the following generalized eigenvalue problem given as Equation 3:


 * $$\mathbf{\bar{k}}\boldsymbol{\phi} = \omega^2 \mathbf{\bar{M}} \boldsymbol{\phi}\;\;(3)$$


 * {| class="collapsible collapsed"

!Standard Eigenvalue Problems
 * 
 * 

The general form of the eigenvalue problem is as follows:

$$\mathbf{A}\mathbf{x} = \lambda \mathbf{B}\mathbf{x}$$

This can be reduced to the standard eigenvalue problem by making the assumption that $$\mathbf{B} = \mathbf{I}\;$$, where $$\mathbf{I}\;$$ is the identity matrix:

$$\mathbf{A}\mathbf{x} = \lambda \mathbf{x}$$

$$\mathbf{I} = \left[\begin{matrix} 1\\ 0 \end{matrix}\right. \ddots \left.\begin{matrix} 0\\ 1 \end{matrix}\right]$$
 * }


 * $$\lambda = \omega^2\;$$ is defined as the eigenvalue, and the eigenpair is given as $$(\lambda_i,\phi_i)\;$$, for $$i = 1, \cdots, n\;$$. A general eigen mode is defined as:


 * $$\mathbf{v}_i(t) = (\sin \omega_i t) \boldsymbol{\phi_i}$$, for $$i = 1, \cdots, n\;$$

Now, considering the following method based on superposition. The orthogonal project of eigenpairs is given as:


 * $$\boldsymbol{\phi_i}^T \mathbf{\bar{M}}\boldsymbol{\phi_j} = \delta_{ij} = \left\{ \begin{matrix} 1, \mbox{if}\; i = j \\ 0, \mbox{if}\; i \neq j \end{matrix} \right.$$


 * $$\delta_{ij}\;$$ refers to the Kronecker delta. The mass orthogonality of eigenvectors is given as:


 * $$\mathbf{\bar{M}}\boldsymbol{\phi_j} = \lambda_j^{-1} \mathbf{\bar{k}}\boldsymbol{\phi_j}$$


 * $$\underbrace{\boldsymbol{\phi_i}^T \mathbf{\bar{M}}\boldsymbol{\phi_j}}_{\delta_{ij}} = \lambda_j^{-1} \boldsymbol{\phi_i}^T \mathbf{\bar{k}}\boldsymbol{\phi_j}$$

This leads to the following relationship:


 * $$\boldsymbol{\phi_i}^T \mathbf{\bar{k}}\boldsymbol{\phi_j} = \lambda_j \delta_{ij}$$

From Equation 1, the following can be derived:


 * $$\mathbf{\bar{d}}(t) = \Sigma \xi_i(t) \boldsymbol{\phi_i}$$

Combining these results yields the following:


 * $$\mathbf{\bar{M}} \underbrace{\left(\sum_j \ddot{\xi_j} \boldsymbol{\phi_j}\right)}_{\mathbf{\ddot{\bar{d}}}} + \mathbf{\bar{k}} \underbrace{\left(\sum_j \xi_j \boldsymbol{\phi_j}\right)}_{\mathbf{\bar{d}}} = \mathbf{\bar{F}}$$

Rearranging and multiplying by $$\boldsymbol{\phi_i}^T\;$$ yields:


 * $$\sum_j \ddot{\xi_j} \underbrace{\left(\boldsymbol{\phi_i}^T \mathbf{\bar{M}}\boldsymbol{\phi_j}\right)}_{\delta_{ij}} + \sum_j \xi_j \underbrace{\left(\boldsymbol{\phi_i}^T \mathbf{\bar{k}}\boldsymbol{\phi_j}\right)}_{\lambda_j \delta_{ij}} = \boldsymbol{\phi_i}^T \mathbf{\bar{F}}$$

Finally, the overall result can be written as:


 * $$\ddot{\xi_j} + \lambda_j\xi_j = \boldsymbol{\phi_i}^T \mathbf{\bar{F}} i = 1, \cdots, n\;$$