User:Eml4500.f08.gravy.mmm/HW6

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

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

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
So the stiffness matrix becomes

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

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 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.