User:Eml4500.f08.group/HW6/Lecture Notes

=Lecture Notes=

Continuous PVW for Elastic Bar
Principle of Virtual Work for dynamics of elastic bar

PDE:

$$\frac{\partial }{\partial x}\left[\left(EA \right)\frac{\partial u}{\partial x} \right]+f=m\ddot{u}$$  Equation (1)

Recall:

$$-kd + f=m\ddot{d}$$

$$M\ddot{d}+kd=F$$  Equation (2)

Derive (2) from (1):

$$\int_{x=0}^{x=L}W(x)\left\{\frac{\partial }{\partial x} \left[EA\frac{\partial u}{\partial x} \right] + f-m\ddot{u}\right\}dx$$  Equation (3) = 0 for all possible W(x)-->Weighting Function

(1) => (3) trivial

(3) => (1) not trivial

(3) rewritten as:

$$\int W(x)g(x)dx=0 $$  for all W(x)

Since(3) holds for all W(x), you can select W(x) to be equal to g(x)

$$W(x)=g(x)$$, then (3) becomes:

$$\int g^{2}dx=0$$, where $$g^{2}\geq 0$$ => $$g(x)=0$$

Integration by Parts with Application
Review of Integration by Parts:

Decompose the function into one composed of two separate functions, r(x) and s(x).

$$(rs)' = r's + rs'$$, where $$r' = \frac{dr}{dx}, s' = \frac {ds}{dx}$$

$$\int (rs)' = \int r's + \int rs' = > \int r's = rs - \int rs'$$

Recall the Continuous Principle of Virtual Work and Equation 3 from Lecture 29-1:

The first term: $$r(x) = (EA) \frac{\partial u}{\partial x}$$, where s(x) = w(x)

From integration by parts we see:

$$\int_{x=0}^{x=L} w(x) \frac{\partial}{\partial x}\left[(EA) \frac {\partial u}{\partial x}\right] dx = \left[w(EA) \frac {\partial u}{\partial x}\right]_{x=0}^{x=L} - \int_0^L \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x} dx$$

$$= w(L)(EA)(L)\frac{\partial u}{\partial x}(L,t) - w(0)(EA)(0)\frac{\partial u}{\partial x}(0,t) - \int_0^L \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x} dx$$

Model Problem and PVW
Now consider the following model problem:



At x=0, select w(x) such that w(0)=0 (i.e. the problem is kinematically admissible)

Motivation for the problem: The Discrete Principle of Virtual Work as applied in Lecture 10-1:

$$\boldsymbol w_{6x1} \left ([k]_{6x2} \begin{Bmatrix} d_{3}\\ d_{4}\end{Bmatrix}_{2x1}- \boldsymbol F \right) = 0_{1x1} $$

$$\boldsymbol {F^T} = [F_1 F_2 F_3 F_4 F_5 F_6]$$, where $$F_1, F_2, F_5, F_6$$ are the unknown reactions.

Since w can be selected arbitrarily, select w such that $$w_1 = w_2 = w_5 = w_6$$

So, to eliminate equations involving unknown reactions we need to eliminate rows 1,2,5, and 6, this leaves: $$\boldsymbol k_{2x2} \boldsymbol d_{2x1} = \boldsymbol F_{2x1}$$


 * Note that $$\boldsymbol {w(kd-F)} = 0 $$ for all w.

Back to continuous PVW...

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

From continuous PVW: $$w(L)F(t) - \int_0^L \frac{\partial w}{\partial x}(EA) \frac{\partial u}{\partial x}dx + \int_0^L w(x)[f-m\ddot{u}]dx = 0$$ for all w(x) such that w(0)=0.

$$=> \int_0^L w(m\ddot{u})dx + \int_0^L \frac{\partial u}{\partial x}(EA) \frac{\partial u}{\partial x}dx = w(L)F(t) + \int_0^L wfdx$$ for all w(x) such that w(0)=0.

Interpolation of Deformed Bar
Stiffness term:





In the figure above, assuming the displacement u(x) for $$x_{i} \leq x \leq x_{i+1}$$ (for example, $$x\in [x_{i}, x_{i+1}]$$, where the symbol $$\in $$ means "belongs to". The figure shows the interpolation of u(x) where there is only axial displacement (or, no transverse displacement)



The motivation for linear interpolation of u(x)
In a 2-bar truss the deformation can be shown as:



The deformed shape is a straight line, meaning that there is an implicit asumption of linear interpolation of displacement between two nodes.

u(x) can be expressed in terms of $$d_{i} = u(x_{i})$$ and $$d_{i+1} = u(x_{i+1})$$ as a linear function in x (for example, linear interpolation) as follows:

$$u(x) = N_{i}(x)d_{i} + N_{i+1}(x)d_{i+1}$$

where $$N_{i}(x)$$ and $$N_{i+1}$$ are linear functions in x.

$$N_{i} = \frac {x-x_{i+1}}{x_{i} - x_{i+1}}$$

$$N_{i+1} = \frac {x - x_{i}}{x_{x+1} - x_{i}}$$

Therefore, $$u(x) = \frac {x-x_{i+1}}{x_{i} - x_{i+1}}d_{i} + \frac {x - x_{i}}{x_{x+1} - x_{i}}d_{i+1}$$



Lagrangian Interpretation
Motivation for form of $$\displaystyle N_i(x)$$ and $$\displaystyle N_{i + 1}(x)$$

1) $$\displaystyle N_i(x)$$ and $$\displaystyle N_{i + 1}(x)$$ are linear (straight lines), thus any linear combination of $$\displaystyle N_i$$ and $$\displaystyle N_{i + 1}$$ is also linear, and in particular the expression for $$\displaystyle u(x)$$ on p. 31-3: $$\displaystyle u(x) = N_i (x) d_i + N_{i + 1} (x) d_{i + 1}$$

$$\displaystyle N_i(x) = \alpha_i + \beta_i x$$ $$\displaystyle N_{i + 1}(x) = \alpha_{i + 1} + \beta_{i + 1} x$$ with $$\displaystyle \alpha_i$$ and $$\displaystyle \beta_i$$ as numbers

Linear combo of $$\displaystyle N_i(x)$$ and $$\displaystyle N_{i + 1}(x)$$: $$\displaystyle N_i(x) d_i + N_{i + 1} d_{i +1}(x) = (\alpha_i + \beta_i x) d_i + (\alpha_{i +1} + \beta_{i + 1} x) d_{i + 1}$$ $$\displaystyle => N_i(x) d_i + N_{i + 1} d_{i +1}(x) = (\alpha_i d_i + \alpha_{i +1} d_{i +1}) + (\beta_i d_i+ \beta_{i + 1}d_{i +1})x$$

2) Recall equation for $$\displaystyle u(x)$$ on p. 31-3: $$\displaystyle u(x_i) = N_i (x_i) d_i + N_{i + 1} (x_i) d_{i + 1}$$

set $$\displaystyle N_i (x_i) = 1$$ and $$\displaystyle N_{i + 1} (x_i) = 0$$

therefore $$\displaystyle u(x_i) = d_i$$

FEM via PVW
Finite Element Method via the Principal of Virtual Work. (cont'd)

Our goal here is to calculate the Element stiffness matrix $$ \beta $$, which is defined as follows.



By examining the figure drawn during lecture 31, we can interpolate to find the values of u(x) and w(x). u(x) = Ni(x)*di+Ni+1(x)*di+1 w(x) = Ni(x)*wi+Ni+1(x)*wi+1

After differentiating both u(x) and w(x), we can plug into the above equation to yield the following element stiffness matrix for a generic element i.



$$\frac{\partial u(x)}{\partial x}=\begin{bmatrix} N^{'}_{i}(x) & N^{'}_{i+1}(x) \end{bmatrix} \begin{Bmatrix} d_{i}\\ d_{i+1} \end{Bmatrix} $$

$$B(x)=\begin{bmatrix} N^{'}_{i}(x) & N^{'}_{i+1}(x) \end{bmatrix}$$

Similarly: $$W(x)=N(x)\begin{Bmatrix} w_{i}\\ w_{i+1}

\end{Bmatrix}$$

$$\frac{\partial W}{\partial x}(x)=B(x)\begin{Bmatrix} W_{i}\\ W_{i+1}

\end{Bmatrix}$$

Recall the element dof(S):

$$\begin{Bmatrix} d_{i}\\ d_{i+1}

\end{Bmatrix} = \begin{Bmatrix} d^{(i)}_{1}\\ d^{(i)}_{2}

\end{Bmatrix} =d^{(i)}$$

$$\begin{Bmatrix} w_{i}\\ w_{i+1}\\ \end{Bmatrix} = \begin{Bmatrix} w^{(i)}_{1}\\ w^{(i)}_{2}\\ \end{Bmatrix} =w^{(i)}$$

$$\displaystyle \beta = \int_{x_i}^{x_{i+1}} (\boldsymbol B w^{(i)})_{1x1} (EA)_{1x1} (\boldsymbol B d^i)_{1x1} dx$$ $$\displaystyle => \beta = \boldsymbol w^{(1)} * (k^{(1)} d^{(1)})$$ This is the Goal from p. 31-3

$$\displaystyle \beta = \int_{x_i}^{x_{i+1}} (EA) (\boldsymbol B w^{(i)})_{1x1} * (\boldsymbol B d^i)_{1x1} dx$$

$$\displaystyle (\boldsymbol B w^{(i)})_{1x1} * (\boldsymbol B d^i)_{1x1} = (\boldsymbol B w^{(i)})^T (\boldsymbol B d^i)$$

where $$\displaystyle (\boldsymbol B w^{(i)})^T = w^{(i) T} \boldsymbol B^T = w^{(i)} * \boldsymbol B^T$$

$$\displaystyle => \beta = w^{(i)} * (\int B^T (EA) B dx) d^{(i)}$$

$$\boldsymbol k^{(i)} = \int_{x_i}^{x_{i+1}} \boldsymbol B(x) (EA) B(x) dx$$

$$\boldsymbol B(x)=\begin{bmatrix} \frac{-1}{L^{(i)}} & \frac{1}{L^{(i)}} \end{bmatrix}$$, where $$L^{(i)}=x_{i+1}-x_i$$

$$\boldsymbol k^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

Transformation of variables (coordinates) from $$x$$ to $$\tilde x$$.

$$\tilde x:= x-x_i, \tilde {dx} = dx$$

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


 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!HW Expression for $$\mathbf{k^{(i)}}$$, from team group
 * The expression for $$\mathbf{k^{(i)}}$$ can be solved as such:
 * The expression for $$\mathbf{k^{(i)}}$$ can be solved as such:

$$\mathbf{k^{(i)}} = \int_{x=0}^{\tilde{x} = L^{(i)}}{\mathbf{B^{T}}(\tilde{x})(EA)(\tilde{x})\mathbf{B}(\tilde{x})d\tilde{x}} = \frac {EA}{\tilde{x}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}^{L^{(i)}}_{0} = \frac {EA}{L^{(i)}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

The equation of the area is: $$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

The equation of the modulus of elasticity is: $$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$



Substituting the equations above into the stiffness matrix, is as follows:

$$\mathbf{k^{(i)}} = \frac {[N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2][N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2]}{L^{(i)}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

contribution by carbon.w
 * }

Stiffness Matrices Application
Refering to the figures drawn during lecture 31, we find the shape functions N(i)1 and N(i)2. N(i)2 was given in class as the following.



HW: Find an expression for N(i)1

Noting that the value for N(i)1 has to equal 1 at x = 0 and it must also equal 0 at x = 1, we find the linearized value for N(i)1 to be:



HW: obtain expressions for k(i) using the average values for A and E as well as Linearized values for A and E

Using the average values for A $$\frac{A_{1} + A_{2}}{2}$$ into the equation $$ k = \frac {EA} {L} $$, we find the element stiffness matrix for element i to be:

Next, we use the following linearized equation for A to solve for the same stiffness matrix.





HW: Compare stiffness matrices obtained by using average values and linearized values for both E and A

Average vales of A and E,, yield the following stiffness matrix.



While the linearized values of A and E,, yield this stiffness matrix.

$$k^{(i)} = \frac {[(L^{(i)} - \tilde{x}) E_1 + (\tilde{x}) E_2] [(L^{(i)} - \tilde{x}) A_1 + (\tilde{x}) A_2]} {L^{(i)^3}} \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} $$

Next, the mean value theorem (MVT) is recalled. This theorem is to help determine the location of a centroid of an area. For example, take the area shown below.





Recall that to find the centroid we use the equation

In our case, when calculating a stiffness matrix, there could be two different functions of x in the same equation. In that case, the mean value theorem yields.

where it is important to note that, in general: