User:Eml4500.f08.gravy.jad/Notes6.g

Continuous PVW Case
Continuing Case: PVW of Dynamics of Elastic Bars

PDE: $$\ \frac{d}{dx} \left \lbrack (EA) \frac{du}{dx} \right \rbrack + f = m \ddot{u} \,$$ Equation(1)

$$\ \Rightarrow \mathbf{m} \mathbf{\ddot{d}} + \mathbf{k} \mathbf{d} = \mathbf{F} \,$$ Equation(2)

Derive (2) from (1): $$\ \int_{x=0}^{x=L} W(x) \left \lbrace \frac{d}{dx} \left \lbrack EA \frac{du}{dx} \right \rbrack + f - m \ddot{u} \right \rbrace \, dx = 0 \,$$ for all possible W(x). Where w(x) is the weighting function.

(1) $$\ \Rightarrow \,$$(3) trivial (3) $$\ \Rightarrow \,$$(1) not trivial (3) rewritten as: $$\ \int W(x)g(x)dx =0 \,$$for all W(x) Since (3) holds for all W(x), select W(x)=g(x), then (3) becomes $$\ \int g^{2}(x)dx = 0 \,$$

Integration by Parts
Integration by Parts: r(x), s(x)

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

$$\ \int (rs)' = rs - \int rs' \,$$ $$\ \Rightarrow \int r's = rs - \int rs' \,$$

Recall cont. PVW (equation 3)

1st term: $$\ r(x)=(E,A) \frac{du}{dx}, s(x)=w(x) \,$$

By integrating by parts: $$\ \int_{x=0}^{x=L} \underbrace{w(x)}_{s} \frac{d}{dx} \underbrace{\left \lbrack (EA) \frac{du}{dx} \right \rbrack}_{r} dx = \left \lbrack W(EA) \frac{du}{dx} \right \rbrack |_{x=0}^{x=L} \,$$

$$\ = W(L) \left \lbrack (EA)(L) \right \rbrack \frac{du}{dx}(L,t) - W(0) \left \lbrack (EA)(O) \right \rbrack \frac{du}{dx} (0,t) \,$$

Now consider model prob:



At x=0, select W(x) such that W(0)=0 (ie. kinetically dismiss-able)

Motivation: Discrete PVW applied to Eq $$\ \mathbf{W}_{6x1} \cdot \underbrace{ \mathbf{k}_{6x1} \mathbf{d}_{2x1}-\mathbf{F}_{6x1} }_{6x1} = 0_{1x1} \,$$

$$\ F^{T} = \left \lfloor F_{1} F_{2} F_{3} F_{4} F_{5} F_{6} \right \rfloor \,$$

Where $$\ F_{1}, F_{2}, F_{5}, F_{6} \,$$ are unknown reactions. Since $$\ \mathbf{W} \,$$ can be selected arbitraily, select $$\ \mathbf{W} \,$$ such that $$\ W_{1}=W_{2}=W_{5}=W_{6}=0 \,$$, so to eliminate equations involving unknown reactions, thus eliminating 1,2,5,6; $$\ \mathbf{\overline{k}} \mathbf{\overline{d}}=\mathbf{\overline{F}} \,$$

Please note: $$\ \overline{\mathbf{W}} \cdot (\mathbf{\overline{kd}} - \mathbf{\overline{F}})=0 \,$$ for all $$\ \mathbf{W} \,$$

This concludes the motivation.

Unknown reaction term $$\ N(0,t) = \left \lbrack (EA)(0) \right \rbrack \frac{du}{dx} (0,t) \,$$

Continuing PVW: $$\ W(L)F(L) - \int_{0}^{L} \frac{dW}{dx}(EA) \frac{du}{dx} dx + \int_{0}^{L} W(x) \left \lbrack f-m \ddot{u} \right \rbrack dx=0 \,$$ for all W(x) such that W(0)=0.

$$\ \Rightarrow \int_{0}^{L} W(m \ddot{u}) dx + \int_{0}^{L} \frac{dW}{dx} (EA) \frac{du}{dx} dx = W(L)F(t) + \int_{0}^{L} Wf dx \,$$ for all W(x) such that W(0)=0.

Comparison between Constant and Discrete Setting
Constant Setting (PVW)

Inertia : $$\ \int_{0}^{L} Wm \ddot{u} dx = \alpha \,$$

Stiffness : $$\ \int_{0}^{L} \frac{dW}{dx} (EA) \frac{du}{dx} dx = \beta \,$$

Applied force : $$\ W(L)F(t) + \int_{0}^{L} Wf dx = \gamma \,$$

For all W(x) such that: $$\ \underbrace{W(0)=0}_{b.c.} \,$$

Discrete setting (PVW)

Inertia : $$\ \overline{\mathbf{W}} \cdot ( \overline{\mathbf{md}}) (1) \,$$

Stiffness : $$\ \overline{\mathbf{W}} \cdot ( \overline{\mathbf{kd}}) (2) \,$$

Applied force : $$\ \overline{\mathbf{W}} \cdot \overline{\mathbf{F}} (3) \,$$

For all W(x) such that: b.c.'s eliminated

Stiffness term:



Assume displacement u(x) for $$\ x_{i} \le x \le x_{i+1} \,$$

Motivation for linear interpolation of u(x)



Deformed shape is a straight line. ie. there was an implicit assumption of linear interpolation of displacement between 2 nodes

consider the case where there are only axial-displacement (ie. zero transverse displacement)

Question: Express V(x) in terms of $$\ d_{i} = u(x_{i}) \,$$ and $$\ d_{i+1} = u_(x_{i+1}) \,$$ as a linear function in x (ie. linear interpolation

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

Solving Ni(x)
Rube-Goldberg device by (Walter Benjamin which is a German name so the W sounds like a V and the J sounds like a Y much like the Volkswagen where the V sounds like a F and the W sounds like a V) is a device which performs a simple task in a complicated manner. More information can be obtain here

"Honesty, imagination,and ethics" are important when it comes to writing these homework reports. Because everyone post their homework online it is important to remain honest and abide by your ethics and do your own work. When writing up the reports it takes imagination to use narrative in the reports.

From Continuous Principal of Virtual Work to Discrete Principal of Virtual Work
Discrete Principal of Virtual Work can be achieved using Lagrangian interpolation which is the motivation for the formation of Ni(x) and Ni+1(x).

1) Ni(x) and Ni+1(x) are linear straight lines, thus any linear combination of Ni and Ni+1 is also linear and in particular the expression for u(x) shown previously in this report.

$$N_i(x)= \alpha_i + \beta _ix$$, with ($$\alpha_i + \beta _i$$) are numbers

Linear combination of Ni and Ni+1
The Linear combination of Ni and Ni+1 is calculated as the following:

$$\ N_id_i + N_{i+1}d_{i+1}= (\alpha_i + \beta _ix)d_i +(\alpha_{i+1} + \beta _{i+1}x)d_{i+1} = (\alpha_{i}d_{i} +\alpha_{i+1}d_{i+1}) + (\beta _{i}d_{i}+\beta _{i+1}d_{i+1}) \,$$

which is clearly a linear function of x.

2)Recall equation for u(x)(the interpolation of u(x)) from previous in this report

$$\ u(x_i)= \underbrace{N_i(x_i)}_1 d_i+ \underbrace{N_{i+1}(x_i)}_0 d_{i+1}= d_i \,$$

Prove of equation $$u(x_{i+1})= d_{i+1}$$
The same interpolation can be used for w(x) so     $$\ w(x)= N_i(x)w_i+ N_{i+1}(x)w_{i+1} \,$$

The Element Stiffness matrix for element i
The Element Stiffness matrix for element i can be developed using the steps and equations which follow.

$$ \beta= \int_{x_i}^{x_{i+1}} [ \underbrace{N'_iW_i + N'_{i+1}W_{i+1}}_{W'(x)} (EA) \underbrace{N'_id_i + N'_{i+1}d_{i+1}}_{u'(x)}]dx $$

Where $$ N^{\prime}_i:= \frac{d N_i(x)}{dx}$$ Likewise for $$N^{\prime}_{i+1}$$ and written in matrix form for simplification.

$$ u(x)= \underbrace{\begin{bmatrix} N_{i}(x) && N_{i+1}(x) \end{bmatrix}}_{\vec{N}_{1x2}(x)} \begin{Bmatrix} d_i \\ d_{i+1}  \end{Bmatrix}_{2x1}  $$

Recall the element degrees of freedom



$$ \begin{Bmatrix} d_{i} \\ d_{i+1} \end{Bmatrix}= \begin{Bmatrix} d^{(i)}_1 \\ d^{(i)}_{2}  \end{Bmatrix}= \mathbf{d}^{(i)}$$

$$ \begin{Bmatrix} W_i \\ W_{i+1} \end{Bmatrix}= \begin{Bmatrix} W^{(i)}_1 \\ W^{(i)}_{2}  \end{Bmatrix}= \mathbf{W}^{(i)}$$

$$ \beta= \int_{x_i}^{x_{i+1}} \underbrace{(\mathbf{BW}^{(i)}) }_{1x1} \underbrace{(EA)}_{1x1} \underbrace{(\mathbf{Bd}^{(i)}}_{1x1}dx = \mathbf{W}^{(i)} \cdot (\mathbf{k}^{(i)} \mathbf{d}^{(i)}) $$

$$ \beta= \int_{x_i}^{x_{i+1}}(EA) \underbrace{(\mathbf{BW}^{(i)}) }_{1x1}\cdot  \underbrace{(\mathbf{Bd}^{(i)})}_{1x1}dx = \int_{x_i}^{x_{i+1}}(EA)(\mathbf{BW}^{(i)})^T(\mathbf{Bd}^{(i)})dx $$

$$= \int_{x_i}^{x_{i+1}}(EA)(\mathbf{W}^{(i)T}\mathbf{B}^T)(\mathbf{Bd}^{(i)})dx= \int_{x_i}^{x_{i+1}}(EA)(\mathbf{W}^{(i)} \cdot \mathbf{B}^T)(\mathbf{Bd}^{(i)})dx$$

$$ \beta=\mathbf{W}^{(i)} \cdot (\int_{x_i}^{x_{i+1}}\mathbf{B}^T(EA)\mathbf{B}dx) d^{(i)}$$

$$ k^{(i)}_{2x2}= \int_{x_i}^{x_{i+1}} \underbrace{\mathbf{B}^T(x)}_{2x1} \underbrace{(EA)(x)}_{1x1} \underbrace{\mathbf{B}  (x)}_{1x2} dx$$

$$B(x)= \begin{bmatrix} 1^{st} Term && \frac{1}{L^i} \end{bmatrix}$$ where $$L^{(i)}= x_{i+1} -x_i $$ (which is the length of element i).

Considering EA is a constant
Consider EA = constant, then the equation for k simplifies to:

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

Transformation of variable coordinates from x to $$ \tilde{x}$$
$$\tilde{x}:= x-x_i$$ and $$d\tilde{x}=dx$$

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

The tapered bar can be drawn and solved using the tilde axis



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

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

$$k^{(i)}_{2x2}$$ derived
$$N_1^{(i)}(\tilde{x}) = N_1^{(i)}(x)$$ which was solve eariler in this report.

$$N_2^{(i)}(\tilde{x})= \frac{\tilde{x}}{L^{(i)}}= \bigg\{ \begin{matrix} 0 && at, \tilde{x}=0 \\ 1 && at, \tilde{x}=L^{(i)} \end{matrix}$$

k(1)
Another way to solve the tapered bar is using the average area and modulus of elasticity.

$$\underbrace{\frac{E(A_1+A_2)}{L}}_{Average Area} \begin{bmatrix}1&&-1 \\1&& 1 \end{bmatrix}= \mathbf{k}^{(i)}$$



Remark
Recall the mean value theorem (MVT) and its relation to the centroid:

MVT : $$\int_{x=a}^{x=b} f(x) dx= f(\bar{x})[b-a]$$

for $$\bar{x}\in[a,b]$$

$$\in$$ symbol means in so $$ a \le \bar{x} \ge b$$

centroid$$= \int_A xdA= \bar{x}\int_A = \bar{x}A$$



$$\int_{a}^{b}f(x)g(x)dx= f(\bar{x})g(\bar{x})[b-a]$$

where $$a \le \bar{x} \ge b$$

but $$f(\bar{x}) \ne \underbrace{\frac{1}{b-a} \int_{a}^{b}f(x)dx}_{Average value of f}$$

and $$g(\bar{x}) \ne \underbrace{\frac{1}{b-a} \int_{a}^{b}g(x)dx}_{Average value of g}$$

Modification of 2-bar truss code to accommodate general k(i)


The first modification to the 2-bar truss code was adding the values for E1(i), E2(i), A1(i), and A2(i) to the matrices at the beginning of the code. This was done with a 2x2 matrix for both E and A. The columns represent the elements, and the rows represent the nodes. Therefore, e(1,2) is E1(2). These values were then used to caculate k(i) using the following general formula:

$$k_{2x2}(\tilde{x})^{(i)} = \frac{\begin{bmatrix} 2 \left ( A_1 * E_1 \right ) +  \left ( A_2 * E_1 \right ) + \left ( A_1 * E_2 \right ) + 2\left ( A_2 * E_2 \right ) \end{bmatrix}}{6*L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

Then, the values of E1(i), E2(i), A1(i), and A2(i) were fed into the modified versions of PlaneTrussElement.m and PlaneTrussResults.m that are shown below.

The results were then plotted on the same graph as the deformed two bar truss without tapering. The blue dashed line represents the initial conditions, the green line is the non-tapered deformation, and the red line is the deformation of the tapered bars.

Electric pylon with the FE mesh
The following matlab code was used to solve the electric pylon. The functions: PlaneTrussElement, PlaneTrussResults,and NodalSoln were obtained from the Fundamental Finite Element Analysis and Applications book website. Here

The element number for the element with the highest tensile stress is represented by Tbn and equals 81. The highest tensile stress is represented by Ts and equals 9.0511e+005

The element number for the element with the highest compressive stress is represented by Cbn and equals 55. The highest compressive stress is represented by Cs and equals -8.6957e+005

This problem is statically indeterminate. One reaction was given but there are too many reactions remaining to solve using the static equations.

Contributing Team Members
The following students contributed to this report:

Jessica Dyer Eml4500.f08.gravy.jad 18:09, 21 November 2008 (UTC)

Brian Eggeman Eml4500.f08.gravy.bje 23:46, 20 November 2008 (UTC)

Sean Sullivan Eml4500.f08.gravy.sms 03:42, 21 November 2008 (UTC)

Mikal McFarlane Eml4500.f08.gravy.mmm 15:01, 21 November 2008 (UTC)

Patrick Clayton Eml4500.f08.gravy.psc 12:14, 21 November 2008 (UTC)

Justin Tennant Eml4500.f08.gravy.JMT 19:45, 21 November 2008 (UTC)