User:Eml4500.f08.Lulz.js/HW6

PVW for Elastic Bar


Continuing with the discussion of the Principle of Virtual Work (PVW) for the dynamics of the elastic bar shown in Figure ##, consider the partial differential equation (PDE) describing the continuous case equation of motion (EOM) for this bar:


 * $$\frac{\partial}{\partial x}[(EA)\frac{\partial u}{\partial x}] + f = m \ddot{u}$$

Deriving the discrete case EOM for the bar from the continuous EOM yields the following:


 * $$- \mathbf{kd} + \mathbf{F} = \mathbf{M} \mathbf{\ddot{d}} => \mathbf{M}\mathbf{\ddot{d}} + \mathbf{kd} = \mathbf{F}$$

Where $$\mathbf{M}\;$$ is the mass matrix. The presence of the mass matrix indicates a multiple degree of freedom system. A similar equation for the single degree of freedom system is given as:


 * $$m\ddot{d} + kd = F$$

The following integral equation was used in the derivation of the discrete EOM from the continous EOM:


 * $$\int^{1}_{0} w(x) \left\{\frac{\partial}{\partial x} [EA \frac{\partial u}{\partial x}] + f - m \ddot{u}\right\} dx = 0$$ for all possible values of $$w(x)\;$$, the weighting function.

Deriving this equation from the continous EOM is a trivial, but to get the continous EOM from this equation is not. In order to show this, let the integral equation be rewritten as follows:


 * $$\int w(x) g(x) dx = 0$$ for all $$w(x)\;$$

Since this equation should hold for all $$w(x)\;$$, selecting $$w(x)\;$$ to be equal to $$g(x)\;$$ is a valid operation. This results in the following:


 * $$\int g^{2} dx = 0$$

From here, it is obvious that $$g(x) = 0\;$$.


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!Integration by Parts Consider two functions $$r(x)\;\;$$ and $$g(x)\;$$. The derivative of the product of these two functions is as follows:
 * 
 * 


 * $$(rs)^{'} = r^{'}s + rs^{'}\;$$

Where $$r^{'} = \frac{\partial r}{\partial x}$$ and $$s^{'} = \frac{\partial s}{\partial x}$$ Taking the integral of this expression yields the following:


 * $$\int (rs)^{'} =\int r^{'}s + \int rs^{'}\;$$

Noting that the left hand side of this equation is equivalent to $$rs\;$$, with rearranging the following well-known relation can be determined:


 * $$\int r^{'}s = rs - \int rs^{'}\;$$


 * }

Using the integration by parts method on the first term of the continuous EOM, with $$ r(x) = {EA} \frac{\partial u}{\partial x}\;$$ and $$s(x) = w(x)\;$$ yields the following:


 * $$ \int^{L}_{0} w(x) \frac{\partial}{\partial x} \left[(EA) \frac{\partial u}{\partial x}\right] dx = \left[w(EA) \frac{\partial u}{\partial x}\right]^{L}_{0} - \int^{L}_{0} \frac{\partial w}{\partial x}(EA) \frac{\partial u}{\partial x} dx$$


 * $$= \underbrace{w(L)(EA)(L)\frac{\partial u(L,t)}{\partial x}}_{N(L,t)} - \underbrace{w(0)(EA)(0)\frac{\partial u(0,t)}{\partial x}}_{N(0,t)} - \int^{L}_{0} \frac{\partial w}{\partial x}(EA) \frac{\partial u}{\partial x} dx$$



Model Problem
Consider the model problem shown in Figure ##. The boundary conditions for this problem can be found in Homework 5. Additionally, select $$w(x)\;$$ such that $$w(0)=0\;$$; this makes the system "kinematically admissable".

Applying the discrete PVW to a previously studied equation from page 10-1 of the notes yields the following relation:


 * $$\mathbf{W}(\mathbf{K}\begin{Bmatrix} d_{3}\\d_{4}\end{Bmatrix} - \mathbf{F}) = \mathbf{0}$$ for all $$\mathbf{W}$$.

Where $$\mathbf{F}^{T} = \begin{bmatrix}F_1 & F_2 & F_3 & F_4 & F_5 & F_6\end{bmatrix}$$. Since $$\mathbf{W}$$ can be arbitrarily selected, select it such that $$W_1 = W_2 = W_5 = W_6 = 0\;$$. This will eliminate equations involving unknown reactions, eliminating rows 1,2,5, and 6 from the two-bar truss system ans yielding the following relation:


 * $$\mathbf{\bar{K}}\underbrace{\mathbf{\bar{d}}}_{\begin{Bmatrix} d_{3}\\d_{4}\end{Bmatrix}}=\underbrace{\mathbf{\bar{F}}}_{\begin{Bmatrix} F_{3}\\F_{4}\end{Bmatrix}}$$

Note also that $$\mathbf{\bar{W}} \cdot \mathbf{\bar{K}}\mathbf{\bar{d}}=\mathbf{\bar{F}}$$ for all $$\mathbf{\bar{W}}$$, where $$\mathbf{\bar{W}} = \begin{Bmatrix} W_{3}\\W_{4}\end{Bmatrix}$$.

Returning to the continuous case, consider the following unknown reaction:


 * $$N(0,t) = (EA)(0)\frac{\partial u}{\partial x} (0,t)$$

This equation can be written as:


 * $$W(L)F(t) - \int^{L}_{0} \frac{\partial w}{\partial x}(EA) \frac{\partial u}{\partial x} dx + \int^{L}_{0}w(x)[f-m\ddot{u}]dx = 0$$ for all $$w(x)\;$$ such that $$w(0) = 0\;$$.

The weak form of continuous PVW can be expressed as:


 * $$\int^{L}_{0}W(m\ddot{u})dx + \int^{L}_{0} \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx = W(L)F(t) + \int^{L}_{0} Wfdx$$ for all $$w(x)\;$$ such that $$w(0) = 0\;$$.

Comparison Between Continuous and Discrete PVW



 * {| border=1


 * bgcolor="#99ccff"|
 * bgcolor="#99ccff" style="text-align: center;" |Continuous Setting (PVW)
 * bgcolor="#99ccff" style="text-align: center;" |Discrete Setting (PVW)
 * bgcolor="#05fa67"|Inertia
 * bgcolor="#ffffff" style="text-align: center;" | $$\int^{L}_{0}wm\ddot{u}dx$$
 * bgcolor="#ffffff" style="text-align: center;" | $$\mathbf{\bar{W}} \cdot (\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}})$$
 * bgcolor="#05fa67"|Stiffness
 * bgcolor="#ffffff" style="text-align: center;" | $$\int^{L}_{0} \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx$$
 * bgcolor="#ffffff" style="text-align: center;" | $$\mathbf{\bar{W}} \cdot (\mathbf{\bar{k}}\mathbf{\bar{d}})$$
 * bgcolor="#05fa67"|Applied Force
 * bgcolor="#ffffff" style="text-align: center;" | $$W(L)F(t) + \int^{L}_{0}wfdx$$
 * bgcolor="#ffffff" style="text-align: center;" | $$\mathbf{\bar{W}} \cdot \mathbf{\bar{F}}$$
 * bgcolor="#05fa67"|Conditions
 * bgcolor="#ffffff" style="text-align: center;" | all $$w(x)\;$$ such that $$w(0)=0\;$$
 * bgcolor="#ffffff" style="text-align: center;" | for all $$\mathbf{\bar{W}}\;$$
 * }
 * bgcolor="#05fa67"|Conditions
 * bgcolor="#ffffff" style="text-align: center;" | all $$w(x)\;$$ such that $$w(0)=0\;$$
 * bgcolor="#ffffff" style="text-align: center;" | for all $$\mathbf{\bar{W}}\;$$
 * }
 * }

Figures ## and ## describe the interpolation of displacements along the elastic bar. Figure ## shows the elastic bar segmented by nodes, whereas Figure ## shows a close-in view of the ith segment of the bar. Assume the displacement $$u(x)\;$$ for $$x_i \leq x \leq x_i + 1$$, which can also be written as $$(x \in [x_i,x_i+1])$$.

Motivation for Linear Interpolation of u(x)


Consider the deformed truss element seen in Figure ##. The deformed shape is a straight line; i.e., there was an implicit assumption of linear interpolation of displacement between 2 nodes.

Next, consider the case where there are only axial displacements. Expressing $$u(x)\;$$ in terms of $$d_i = u(x_i)\;$$ and $$d_{i+1} = u(x_{i+1})\;$$ as a linaer function of $$\;x$$ (i.e, performing linear interpolation) is as follows:

$$u(x) = N_i(x)d_i + N_{i+1}(x)d_{i+1}\;$$, where $$N_i(x)\;$$ and $$N_{i+1}(x)\;$$ are linear functions in x.



This relationship can be seen graphically in Figure ##. $$N_{i+1}(x)\;$$ can be expressed as $$\frac{x - x_i}{x_{i+1} - x_i}$$.

Lagrange Interpolation
The motivation for the form of $$N_i(x)\;$$ and $$N_{i+1}(x)\;$$ is as follows:

1) $$N_i(x)\;$$ and $$N_{i+1}(x)\;$$ are linear; thus, any linear combination of $$N_i\;$$ and $$N_{i+1}$$ is also linear, and in particular the previously discussed expression for $$u(x)\;$$. This relationship can be further expressed as:


 * $$N_i(x) = \alpha_i + \beta_ix\;$$


 * $$N_{i+1}(x) = \alpha_{i+1} + \beta_{i+1}x\;$$

Where $$\alpha_i,\alpha_{i+1},\beta_i,\;$$ and $$\beta_{i+1}\;$$ are real numbers. A linear combination of these numbers can be written as follows:


 * $$N_id_i + N_{i+1}d_{i+1} = (\alpha_i + \beta_ix)d_i + (\alpha_{i+1} + \beta_{i+1}x)d_{i+1}\;$$


 * $$= (\alpha_i d_i + \alpha_{i+1} d_{i+1}) + (\beta_i d_i + \beta_{i+1} d_{i+1})x\;$$

It is clear that this relation is a function in x.

2) Recall the previously discussed equation for $$u(x_i)\;$$:


 * $$u(x_i) = \underbrace{N_i(x_i)}_{1}d_i + \underbrace{N_{i+1}(x_i)}_{0}d_{i+1}\;$$
 * $$ = d_i\;$$

Determination of Stiffness Matrix for Element i
Applying the interpolation process for w(x) yields the following relation:


 * $$w(x) = N_i(x)W_i + N_{i+1}(x)W_{i+1}\;$$

This equation can be employed to determine the elemental stiffness matrix for the element i. Consider the following integral equation for $$\beta \;$$:


 * $$\beta = \int^{x_{i+1}}_{x_i}\left[N_i^'W_i + N_{i+1}^'W_{i+1}\right](EA)\left[N_i^'d_i + N_{i+1}d_{i+1}\right]dx$$

Where $$N_i^' := \frac{\partial N_i(x)}{\partial x}$$, and similarly for $$N_i^'d_i\;$$. Note also the following two definitions:


 * $$u(x) = \underbrace{\begin{bmatrix}N_i(x) & N_{i+1}(x)\end{bmatrix}}_{\mathbf{N}(x)}\begin{Bmatrix}d_i\\d_{i+1}\end{Bmatrix}$$


 * $$\frac{du(x)}{dx} = \underbrace{\begin{bmatrix}N_i^'(x) & N_{i+1}^'(x)\end{bmatrix}}_{\mathbf{B}(x)}\begin{Bmatrix}d_i\\d_{i+1}\end{Bmatrix}$$

Similarly,


 * $$W(x) = \mathbf{N}(x)\begin{Bmatrix}W_i\\W_{i+1}\end{Bmatrix}$$


 * $$\frac{dW(x)}{dx} = \mathbf{B}(x)\begin{Bmatrix}W_i\\W_{i+1}\end{Bmatrix}$$



Recall the element degrees of freedom as shown in Figure ##. These degrees of freedom can be summarized as:


 * $$\begin{Bmatrix}d_i \\ d_{i+1}\end{Bmatrix} = \begin{Bmatrix}d_1^{(i)} \\ d_2^{(i)}\end{Bmatrix} = \mathbf{d}^{(i)}$$

Similarly for $$\mathbf{W}^{(i)}\;$$,


 * $$\begin{Bmatrix}W_i \\ W_{i+1}\end{Bmatrix} = \begin{Bmatrix}W_1^{(i)} \\ W_2^{(i)}\end{Bmatrix} = \mathbf{W}^{(i)}$$

Substituting the above relations into the integral equation for $$\beta \;$$ yields the following:


 * $$\beta = \int^{x_{i+1}}_{x_i}\left(\mathbf{BW}^{(i)}\right)(EA)\left(\mathbf{Bd}^{(i)}\right)dx = \mathbf{W}^{(i)} \cdot (\mathbf{k}^{(i)}\mathbf{d}^{(i)})$$

Since all of the terms in the integral are constants (1x1 matrices), they can be rearranged as follows:


 * $$\beta = \int^{x_{i+1}}_{x_i}(EA)\left(\mathbf{BW}^{(i)}\right) \cdot \left(\mathbf{Bd}^{(i)}\right)dx = \underbrace{\left(\mathbf{BW}^{(i)}\right)^T}_{\underbrace{\mathbf{W}^{(i)T}\mathbf{B}^T}_{\mathbf{W}^{(i)} \cdot \mathbf{B}^T}}\left(\mathbf{Bd}^{(i)}\right)$$

Finally, $$ \beta \;$$ can be written as:


 * $$\beta = \mathbf{W}^{(i)} \cdot \left(\int \mathbf{B}^T (EA) \mathbf{B}dx\right)\mathbf{d}^{(i)}$$

This leads to the following expression for the element stiffness matrix $$\mathbf{k^{(i)}}\;$$:


 * $$\mathbf{k}^{(i)} = \int^{x_{i+1}}_{x_i} \mathbf{B}^T(x) (EA)(x) \mathbf{B}(x)dx$$

Note also the following definition for $$L^{(i)}$$:


 * $$L^{(i)} = x_{i+1} = x_{i}\;$$

Also, note the transformation of coordinates from $$x\;$$ to $$\tilde{x}\;$$:


 * $$\tilde{x} := x - x_i\;$$


 * $$d\tilde{x} = dx\;$$

This leads to the following version of the element stiffness matrix:


 * $$\mathbf{k}^{(i)} = \int^{\tilde{x}=L^{(i)}}_{\tilde{x}=0} \mathbf{B}^T(\tilde{x}) (EA)(\tilde{x}) \mathbf{B}(\tilde{x})d\tilde{x}$$



Consider the an ith element as shown in Figure ##. The relations for area and Young's modulus as a function of $$\tilde{x}\;$$ can be written as follows:


 * $$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$


 * $$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

Comparison of Element Stiffness Matrix Determination Methods
The shape function $$N_2^{(i)}(\tilde{x})\;$$ can be written as follows:


 * $$N_2^{(i)}(\tilde{x}) = \frac{\tilde{x}}{L^{(i)}}= \left\{\begin{matrix}0\;at\;\tilde{x} = 0 \\ 1\;at\;\tilde{x} = L^{(i)}\end{matrix}\right.$$



Following the book on page 159, setting $$E_1 = E_2 = E\;$$ and letting $$A(\tilde{x})\;$$ be linear, it is possible to obtain $$\mathbf{k}^{(i)}$$ and compare to the equation in the book:


 * $$\frac{E}{L^{(i)}}\frac{(A_1 + A_2)}{2}\begin{bmatrix}1&-1\\-1&1\end{bmatrix} = \mathbf{k}^{(i)}$$

Where the quantity $$\frac{(A_1 + A_2)}{2}\;$$ is the average area. A graphical view of this concept is given in Figure ##. Next, it is useful to compare the general $$\mathbf{k}^{(i)}\;$$ to a stiffness matrix determined using $$.5(A_1+A_2)\;$$ and $$.5(E_1+E_2)\;$$, where $$E_1 \neq E_2\;$$. This yields the following relation for $$\mathbf{k}^{(i)}_{ave}\;$$:


 * $$\frac{(E_1+E_2)(A_1+A_2)}{4}\begin{bmatrix}1&-1\\-1&1\end{bmatrix} = \mathbf{k}^{(i)}_{ave}$$

The comparison in question is simply the difference $$\mathbf{k}^{(i)} - \mathbf{k}^{(i)}_{ave}$$.

[[Image:Lulz6.9.png|thumb|right|200px|Fig #. Centroid of Area]


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!Mean Value Theorem Recall the Mean Value Theorem(MVT) and its relation to the centroid:
 * 
 * 


 * $$\int^{x=a}_{x=b}f(x)dx = f(\bar{x})[b-a]\;$$

for $$\bar{x} \in [a,b]$$. For the centroid of an area, as given in Figure ##, the MVT can be expressed as follows.


 * $$\int_A xdA = \bar{x} \int_A dA = \bar{x}A$$

Applying the MVT to a product of two functions produces a similar result:


 * $$\int^{x=a}_{x=b}f(x)g(x)dx = f(\bar{x})g(\bar{x})[b-a]\;$$

However, it should be noted that $$f(\bar{x})$$ and $$g(\bar{x})$$ are not equivalent to the average values of those functions. This property is shown in equation form below:


 * $$f(\bar{x}) \neq \frac{1}{b-a} \int^a_b f(x)dx$$


 * $$g(\bar{x}) \neq \frac{1}{b-a} \int^a_b g(x)dx$$


 * }