text:
User:EML4500.f08.A-team.VandenBerg/HW5

 Comment: This is HW5 located in the user namespace of EML4500.f08.A-team.VandenBerg 06:35, 30 October 2008 (UTC)

Gobal Force Displacement Relationship vs. Principle of Virtual Work
Justification of eliminating rows 1, 2, 5, 6 to obtain $$\mathbf{K}_{_{2x2}}$$ in original two-bar truss:

Global force displacement relation: $$\mathbf{K}_{_{6x6}} \cdot \mathbf{d}_{_{6x1}} = \mathbf{F}_{_{6x1}} \Rightarrow \mathbf{K} \mathbf{d} - \mathbf{F} = \mathbf{0}_{_{6x1}}$$ Eq.(1)

Using Principle of Virtual Work (PVW), (1) becomes: $$\mathbf{W}_{_{6x1}} \cdot \underbrace{(\mathbf{K}\mathbf{d} - \mathbf{F})}_{6x1} = 0_{1x1}$$Eq.(2) for all $$\mathbf{W}_{_{6x1}}$$ which is a "weighting matrix"

Proving Eq.(1) and Eq.(2) are interchangeable:

A)Eq.(1) $$\Rightarrow$$ (2): Trivial

B)Want to show (2) $$\Rightarrow$$ (1):

We must choose different values for $$\mathbf{W}$$

Choice 1:

Select $$\mathbf{W}$$ so that W1=1, W2 = W3 = W4 = W5 = W6=0

Therefore: $$\mathbf{W^T}= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 1[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j= \mathbf{F}_1 $$

Choice 2:

Select $$\mathbf{W}$$ so that W1=0, W2 = 1, W3 = W4 = W5 = W6=0

Therefore: $$\mathbf{W^T}= \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 1[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j= \mathbf{F}_2 $$

Choice 3:

Select $$\mathbf{W}$$ so that W1= W2 = 0, W3 = 1, W4 = W5 = W6=0

Therefore: $$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 1 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j= \mathbf{F}_3 $$

Choice 4:

Select $$\mathbf{W}$$ so that W1= W2 = W3 = 0, W4 = 1, W5 = W6=0

Therefore: $$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 1 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j= \mathbf{F}_4 $$

Choice 5:

Select $$\mathbf{W}$$ so that W1= W2 = W3 = W4 = 0, W5 = 1, W6=0

Therefore: $$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 1 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j= \mathbf{F}_5 $$

Choice 6:

Select $$\mathbf{W}$$ so that W1= W2 = W3 = W4 = W5 = 0, W6=1

Therefore: $$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 1 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j= \mathbf{F}_6 $$

Conclusion : $$\mathbf{K}_{_{6x6}} \mathbf{d}_{_{6x1}} = \mathbf{F}_{_{6x1}}$$

PVW : Accounting for Boundary Conditions
In the original two-bar truss system: d1 = d2 = d5 = d6 = 0.

Weighting coefficients must be "kinematically admissible", meaning they can not violate the boundary conditions, therefore:

$$\Rightarrow$$W1 = W2 = W5 = W6 = 0

These weighting coefficients represent the virtual displacement by calculus of variations

Now,

$$\mathbf{W} \cdot (\mathbf{Kd}-\mathbf{F})$$

Thus

$$= \begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix} (\mathbf{\overline{K} \overline{d}- \overline{F} })$$ Eq.(3)

and is true for $$\begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix} $$

With the matrices equating to $$\mathbf{\overline{K}}=\begin{bmatrix} K_{33} && K_{34} \\ K_{43} && K_{44}\end{bmatrix} $$

and:

$$\mathbf{\overline{d}}= \begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix} $$

$$\mathbf{\overline{F}} = \begin{Bmatrix}F_{3} \\ F_{4} \end{Bmatrix} $$

being the reduced matrices.

--Eml4500.f08.a-team.robinson 10:47, 3 November 2008 (UTC)

Meeting 25: Wednesday, 22 October 2008. EML4500
No class due to Exam #2

Meeting 26: Monday, 27 October 2008. EML4500
Meeting 26 -- October 27th, 2008

26-1

What was discussed: -Debugging the 2-bar truss Matlab code -Further examination of PVW

Debugging the code:

The code for the two bar truss given to the class by Dr. Vu-Quoc had a slight bug in it since it is a modified version of program in the book for a six bar truss where the Young's Modulus (E) and the area (A) of each element was the same. As this was not the case for the two bar truss, the program had not been corrected to call each different element's properties and thus would not return proper results.

Fixing this code, however, is a fairly simple task and only requires the student to find the function call PlaneTrussResults(e,A) and add a (i) next to the e and the A.

Thus, the homework was given to debug the code, rerun the program, and interpret the results, especially if they differed from the original results. Also another homework assignment was given such that the student should run the code for the six bar truss and interpret those results as well. All Matlab assignments are shown below under the Matlab Section of this report.

After spending the better half of the class on this subject, the discussion returned to grinding through the PVW, which now follows:

26-2

Deriving:

$$\mathbf{k^{(e)}} = \mathbf{T^{(e)^T}}\mathbf{\hat{k}^{(e)}}\mathbf{T^{(e)}}$$

Reference was made to meeting 14 page 3, and also meeting 23 page 3.

Recall FD relation with axial dof's, or q(e).

$$\mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} = \mathbf{p^{(e)}}$$

From this week get:

$$\mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} - \mathbf{p^{(e)}} = \mathbf{0}$$ which we label Eqn. (1)

Using the PVW:

$$\mathbf{\hat{W}} \cdot \left ( \mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} - \mathbf{p^{(e)}} \right ) = \mathbf{0}$$ which we label Eqn. (2)

Which is true for all $$\mathbf{\hat{W}}$$

Earlier it was shown that a similar set of equations where equivalent, and thus by the same logic these equations [(1) and (2)] are also equivalent. Thus:

Proving Eq.(1) and Eq.(2) are interchangeable:

A)Eq.(1) $$\Rightarrow$$ (2): Trivial

B)Want to show (2) $$\Rightarrow$$ (1):

We must choose different values for $$\mathbf{W}$$

Choice 1:

Select $$\mathbf{W}$$ so that W1=1, W2 = W3 = W4 = W5 = W6=0

Therefore: $$\mathbf{W^T}= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{\hat{k}q - p} ) = 1[ \sum^6_{j=1} \mathbf{\hat{k}}_{1j} \mathbf{q}_j- \mathbf{p}_1] + 0[ \sum^6_{j=1} \mathbf{\hat{k}}_{2j} \mathbf{q}_j- \mathbf{p}_2] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{3j} \mathbf{q}_j- \mathbf{p}_3] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{4j} \mathbf{q}_j- \mathbf{p}_4] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{5j} \mathbf{q}_j- \mathbf{p}_5] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{6j} \mathbf{q}_j- \mathbf{p}_6]  $$

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{\hat{k}}_{1j} \mathbf{q}_j= \mathbf{p}_1 $$

Choice 2:

Select $$\mathbf{W}$$ so that W1=0, W2 = 1, W3 = W4 = W5 = W6=0

Therefore: $$\mathbf{W^T}= \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{\hat{k}q-p} ) = 0[ \sum^6_{j=1} \mathbf{\hat{k}}_{1j} \mathbf{q}_j- \mathbf{p}_1] + 1[ \sum^6_{j=1} \mathbf{\hat{k}}_{2j} \mathbf{q}_j- \mathbf{p}_2] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{3j} \mathbf{q}_j- \mathbf{p}_3] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{4j} \mathbf{q}_j- \mathbf{p}_4] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{5j} \mathbf{q}_j- \mathbf{p}_5] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{6j} \mathbf{q}_j- \mathbf{p}_6]  $$

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{\hat{k}}_{2j} \mathbf{q}_j= \mathbf{p}_2 $$

Choice 3:

Select $$\mathbf{W}$$ so that W1= W2 = 0, W3 = 1, W4 = W5 = W6=0

Therefore: $$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{\hat{k}q-p} ) = 0[ \sum^6_{j=1} \mathbf{\hat{k}}_{1j} \mathbf{q}_j- \mathbf{p}_1] + 0[ \sum^6_{j=1} \mathbf{\hat{k}}_{2j} \mathbf{q}_j- \mathbf{p}_2] + 1 [ \sum^6_{j=1} \mathbf{\hat{k}}_{3j} \mathbf{q}_j- \mathbf{p}_3] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{4j} \mathbf{q}_j- \mathbf{p}_4] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{5j} \mathbf{q}_j- \mathbf{p}_5] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{6j} \mathbf{q}_j- \mathbf{p}_6]  $$

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{\hat{k}}_{3j} \mathbf{q}_j= \mathbf{p}_3 $$

Choice 4:

Select $$\mathbf{W}$$ so that W1= W2 = W3 = 0, W4 = 1, W5 = W6=0

Therefore: $$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{\hat{k}q-p} ) = 0[ \sum^6_{j=1} \mathbf{\hat{k}}_{1j} \mathbf{q}_j- \mathbf{p}_1] + 0[ \sum^6_{j=1} \mathbf{\hat{k}}_{2j} \mathbf{q}_j- \mathbf{p}_2] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{3j} \mathbf{q}_j- \mathbf{p}_3] + 1 [ \sum^6_{j=1} \mathbf{\hat{k}}_{4j} \mathbf{q}_j- \mathbf{p}_4] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{5j} \mathbf{q}_j- \mathbf{p}_5] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{6j} \mathbf{q}_j- \mathbf{p}_6]  $$

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{\hat{k}}_{4j} \mathbf{q}_j= \mathbf{p}_4 $$

Choice 5:

Select $$\mathbf{W}$$ so that W1= W2 = W3 = W4 = 0, W5 = 1, W6=0

Therefore: $$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{\hat{k}q-p} ) = 0[ \sum^6_{j=1} \mathbf{\hat{k}}_{1j} \mathbf{q}_j- \mathbf{p}_1] + 0[ \sum^6_{j=1} \mathbf{\hat{k}}_{2j} \mathbf{q}_j- \mathbf{p}_2] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{3j} \mathbf{q}_j- \mathbf{p}_3] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{4j} \mathbf{q}_j- \mathbf{p}_4] + 1 [ \sum^6_{j=1} \mathbf{\hat{k}}_{5j} \mathbf{q}_j- \mathbf{p}_5] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{6j} \mathbf{q}_j- \mathbf{p}_6]  $$

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{\hat{k}}_{5j} \mathbf{q}_j= \mathbf{p}_5 $$

Choice 6:

Select $$\mathbf{W}$$ so that W1= W2 = W3 = W4 = W5 = 0, W6=1

Therefore: $$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{\hat{k}q-p} ) = 0[ \sum^6_{j=1} \mathbf{\hat{k}}_{1j} \mathbf{q}_j- \mathbf{p}_1] + 0[ \sum^6_{j=1} \mathbf{\hat{k}}_{2j} \mathbf{q}_j- \mathbf{p}_2] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{3j} \mathbf{q}_j- \mathbf{p}_3] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{4j} \mathbf{q}_j- \mathbf{p}_4] + 0 [ \sum^6_{j=1} \mathbf{\hat{k}}_{5j} \mathbf{q}_j- \mathbf{p}_5] + 1 [ \sum^6_{j=1} \mathbf{\hat{k}}_{6j} \mathbf{q}_j- \mathbf{p}_6]  $$

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{\hat{k}}_{6j} \mathbf{q}_j= \mathbf{p}_6 $$

Conclusion : $$\mathbf{\hat{k}}_{_{6x6}} \mathbf{q}_{_{6x1}} = \mathbf{p}_{_{6x1}}$$

Recall also that: $$\mathbf{q^{(e)}} = \mathbf{T^{(e)}}\mathbf{d^{(e)}}$$     Eqn. (3)

is interchangeable with:

$$\mathbf{\hat{W}} = \mathbf{T^{(e)}}\mathbf{W}$$    Eqn. (4)

Meeting 27: Wednesday, 29 October 2008. EML4500
$$ \hat{\underline{w}}_{2x1} {=}$$ virtual axial displacement, corresponding to $$ q^{(e)}_{2x1} $$

$$ \hat{\underline{w}}_{2x1} {=}$$ virtual axial displacement, corresponding to $$ q^{(e)}_{2x1} $$

$$ \underline{w}_{4x1} {=} $$ virtual displacement in global coordinate system, corresponding to $$ d^{(e)}_{4x1} $$

Replace Equations (3) and (4) into (2):

$$ \hat{\underline{w}}_{2x1} \cdot (\hat{\underline{k}}^{(e)} \underline{q}^{(e)} - \underline{p}^{(e)}) {=} 0_{1x1} $$ (2)

$$ \Rightarrow (\underline{T}^{(e)}\underline{w}) \cdot [\hat{\underline{k}}^{(e)}(\underline{T}^{(e)}\underline{d}^{(e)}) - \underline{p}^{(e)}] {=} 0 $$ for all $$ w_{4x1} $$ (5)

Recall: $$ (\underline{A}\underline{B})^T {=} \underline{B}^T\underline{A}^T $$ (6)

Recall: $$ \underline{a}_{nx1} \cdot \underline{b}_{nx1} {=} \underbrace{ \underline{a}^T_{1xn}\underline{b}_{nx1} }_{1x1 (scalar)} $$ (7)

Apply (7) and (6) in (5):

$$ \underbrace{(\underline{T}^{(e)} \underline{w})^T}_{(7)}[\underline{\hat{k}}^{(e)}(\underline{T}^{(e)}\underline{d}^{(e)}) - \underline{p}^{(e)}] {=} 0_{1x1} $$ for all $$ \underline{w}_{4x1} $$

$$ \Rightarrow \underbrace{\underline{w}^T\underline{T}^{(e)T}}_{(6)}[-] {=} 0 $$

$$ \Rightarrow \underbrace{\underline{w}}_{(7)} \cdot [(\underbrace{\underline{T}^{(e)T} \underline{\hat{k}}^{(e)} \underline{T}^{(e)} \underline{d}^{(e)}}_{\underline{k^{(e)}}}) - \underbrace{\underline{T}^{(e)T} \underline{p}^{(e)}}_{\underline{f}^{(e)}}] {=} 0 $$ for all $$ \underline{w}_{4x1} $$

$$ \Rightarrow \underline{w} \cdot [\underline{k}^{(e)}\underline{d}^{(e)} - \underline{f}^{(e)}] {=} 0 $$ for all $$ \underline{w} $$

$$ \Rightarrow \underline{k}^{(e)}\underline{d}^{(e)} {=} \underline{f}^{(e)} $$

So far, only discrete cases have been discussed. Now continuous cases will be discussed.

Motivational model problem: Elastic bar with varying A(x), E(x), subjected to varying axial load (distributed) and concentrated load and inertia force (dynamics). Concentrated load and inertia force are time dependent.

EML4500.f08.A-team.kirley 04:22, 30 October 2008 (UTC)

Meeting 28: Friday, 31 October 2008. EML4500

$$\Sigma$$ Fx = 0 = -N(x,t) + N(x+dx,t) + f(x,t)dx - m(x) $$ \ddot{u} $$, dx

$$\Sigma$$ Fx = $$ \frac{\delta N}{\delta x} $$ dx + h.o.t. + f(x,t) dx - m(x) ,math> \ddot{u}, dx neglect h.o.t.

(h.o.t. = higher order terms)

Recall Taylor series expansion:

f(x+dx) = f(x) + $$ \frac{df(x)}{dx} $$ dx + $$ \underbrace{\frac{1}{2} \frac{d^2f(x)}{dx^2} dx^2 + ...}_{h.o.t.} $$

Eq. (1) $$ \Rightarrow \frac{\delta N}{\delta x} + f = m \ddot{u} $$ (2)

Equation of motion (EOM)

$$ N(x,t) = A(x) \underbrace{\sigma (x,t)}_{E(x) \underbrace{\epsilon (x,t)}_{\frac{\delta u}{\delta x} (x,t)}} $$ (3)

Constitutive relation

(3) in (2) yields

$$ \frac{\delta}{\delta x} [A(x)E(x) \frac{\delta u}{\delta x}] + f(x,t) = m(x) \underbrace{\ddot{u}}_{\frac{\delta^2 u}{\delta t^2}} $$

Need 2 b.c.s (2nd order derivative w.r.t. x)

2 initial conditions (2nd order derivative w.r.t. t)

(initial displacement, initial velocity)

u(0,t) = 0 = u(L,t)

$$ \Rightarrow \frac{\delta u}{\delta x} (L,t) = \frac{F(t)}{A(L)E(L)} $$

1) u(0,t) = 0

2)

$$ \underbrace{N(L,t)}_{A(t)\underbrace{\sigma (L,t)}_{E(L) \underbrace{\epsilon (L,t)}_{\frac{\delta u}{\delta x} (L,t)}}} = F(t)$$

Initial condition at t=0, prescribe

u(x,t=0) = $$\overline{u}$$(x) known function (displacement)

$$ \frac{\delta u}{\delta t} $$ (x,t=0) = $$ \dot{u} $$ (x,t=0) = $$ \overline{\nu} $$ (x)

known function velocity

Matlab and Related Homework
Here is the debugged code for the 2 bar truss:

It can be seen that there are 4 less values in the results matrix, which is appropriate for the 2 bar truss system. Previously the program had printed the correct values, and then spit out garbage for the final four since the code was bugged.

6 Bar Truss from pg. 226
This is the original 6 bar truss code.

Below is a plot of the deformed and un-deformed 6 bar truss. The values for the deformation have been magnified by a factor of 5000 to make the deformation of the truss glaringly obvious.

The following is the code used to create this masterpiece.

Next is a modified code for the 6 bar truss, this time with varying values for the Young's Modulus (E) for each of the elements of the truss

Following is the plot for the modified 6 bar truss. This plot has also had the deformation values magnified by a factor of 5000 for glaring obviousness.

The following is the code used to create the plot.

3 Bar Space Truss From pg 230
Below is the Matlab code from the book that calculates the results for the 3 bar space truss.

Next up are the plots of the truss from several viewpoints of the graph.

This one is from the vantage point of (-2,-2,3)

And now looking at the XY plane:

And looking the XZ plane:

And finally facing the YZ plane:

Below is the code used to generate the graph

Unfortunately, this 3 bar space truss happens to be statically determinate. Personally I'd rather not be bothered to show you how I came up with that other than there are enough equations to cover all of the unknowns. But I suppose I'll prove it out for your benefit (and my own, I suppose) with a section titled....

Thank God For Vector Calculus

First off, let's observe the problem and notice that there are three reaction forces (which from here on out will be know generically as R1, R2, and R3. Also notice that there is a single applied force on this system, which will be known as P.  Next we can divide up each of the three reaction forces into three partial reactions each in the conventional $$\hat{i}$$,  $$\hat{j}$$, and  $$\hat{k}$$ unit vector directions.  Algebraically this looks like:

$$\mathbf{R_1} = \mathbf{R_{1x}\hat{i}} + \mathbf{R_{1y}\hat{j}} + \mathbf{R_{1z}\hat{k}}$$

$$\mathbf{R_2} = \mathbf{R_{2x}\hat{i}} + \mathbf{R_{2y}\hat{j}} + \mathbf{R_{2z}\hat{k}}$$

$$\mathbf{R_3} = \mathbf{R_{3x}\hat{i}} + \mathbf{R_{3y}\hat{j}} + \mathbf{R_{3z}\hat{k}}$$

With a bit of observation of the system, it is seen that $$\mathbf{R_{3x}\hat{i}}$$ and $$\mathbf{R_{3y}\hat{j}}$$ drop out to zero since there are no forces that can be applied in those directions on this element. This is not a necessary step, but greatly simplifies the calculations. Next up we must define each of the partial reactions in order to fully setup the problem. With some knowledge of vectors from calculus, it is quite simple to derive the each of the partial reactions, as seen below:

$$\mathbf{R_{1x}} = \mathbf{R_1}\cos{\phi_1}\cos{\theta_1}$$

$$\mathbf{R_{1y}} = \mathbf{R_1}\cos{\phi_1}\sin{\theta_1}$$

$$\mathbf{R_{1z}} = \mathbf{R_1}\sin{\phi_1}$$

$$\mathbf{R_{2x}} = \mathbf{R_2}\cos{\phi_2}\cos{\theta_2}$$

$$\mathbf{R_{2y}} = \mathbf{R_2}\cos{\phi_2}\sin{\theta_2}$$

$$\mathbf{R_{2z}} = \mathbf{R_2}\sin{\phi_2}$$

$$\mathbf{R_{3z}} = \mathbf{R_3}$$

Note that R3 does not have any x or y direction values since it begins at the origin and extends in the positive z direction. Next thing to be done is to calculate the theta's and phi's in order to finish setting up the problem. Before that happens however, we must the length between node 1 and the origin, which I'll generically call "a" and the length between node 2 and the origin, which I'll generically call "b". Solving for these values is as easy as using the Pythagorean Theorem, such as follows:

$$a = \sqrt{0.96^2 + 1.92^2} = 2.146625$$

$$b = \sqrt{1.44^2 + 1.44^2} = 2.036468$$

Now we are in a position to calculate all of the angles that we will need. Thus,

$$\tan{\theta_1} = \frac{1.92}{0.96}$$

$$\tan{\theta_2} = \frac{1.44}{-1.44}$$

$$\tan{\phi_1} = \frac{2}{2.146625}$$

$$\tan{\phi_2} = \frac{2}{2.036468}$$

Therefore,

$$\theta_1 = 63.435^\circ$$

$$\theta_2 = -45^\circ$$

$$\phi_1 = 42.975^\circ$$

$$\phi_2 = 44.482^\circ$$

Setting up the 3 summation of the force equations we get:

$$\mathbf{F_x} = \mathbf{R_1}\cos{\phi_1}\cos{\theta_1} + \mathbf{R_2}\cos{\phi_2}\cos{\theta_2} = 0$$

$$\mathbf{F_y} = \mathbf{R_1}\cos{\phi_1}\sin{\theta_1} + \mathbf{R_2}\cos{\phi_2}\sin{\theta_2} - \mathbf{P} = 0$$

$$\mathbf{F_z} = \mathbf{R_1\sin{\phi_1}} + \mathbf{R_2\sin{\phi_2}} + \mathbf{R_3} = 0$$

And after a small bit of rearranging and plugging in (or plugging and using a bit of linear algebra) which I'll leave up to the reader, you will get:

$$\mathbf{R_1} = -20374.65877$$

$$\mathbf{R_2} = -13214.4037$$

$$\mathbf{R_3} = 23148.11723$$

These values match up nearly perfectly to the Matlab's code, and are really only off by about a few hundreths of a point (due to rounding of the decimals on the lengths and the angles). However, you may notice that the values seem to have the opposite signs than what the code gives. You must remember that the code solves for the axial forces in the element, and while the absolute values of these numbers are the same, the signs are not. Simply put, the reactions with negative (-) signs are in tension and conventionally listed as positive. Also, the reaction with the positive (+) sign is in compression, which is conventionally listed as negative.

Contributions
--EML4500.f08.A-team.kirley 04:22, 30 October 2008 (UTC)

--Eml4500.f08.a-team.robinson 10:47, 3 November 2008 (UTC)

--EML4500.f08.A-team.melvin 18:22, 6 November 2008 (UTC)

--EML4500.f08.A-team.rieth 04:25, 7 November 2008 (UTC)

--EML4500.f08.A-team.morford 20:06, 7 November 2008 (UTC)

--EML4500.f08.A-team.VandenBerg 21:57, 7 November 2008 (UTC)

title: User:EML4500.f08.A-team.VandenBerg/HW5


text: File:ABC.JPG

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: File:ABC.JPG


title: File:Mmproblem1.jpg


text:
User:Eas4200c.f08.carbon.orear/HW5

=Homework 5=

 * Lecture Notes (current version) Lecture Notes (archived version)
 * MATLAB Motivation (current version) MATLAB Motivation (archived version)
 * MATLAB Code (current version) MATLAB Code (archived version)
 * HW5 Contributing Team Members (current version) Contributing Team Members (archived version)

title: User:Eas4200c.f08.carbon.orear/HW5


text:
User:Eas4200c.f08.carbon.orear/HW5/Lecture Notes

=Lecture Notes=

Kinematic Assumptions and Stress Strain Relations (Lecture 25)
There are four zero strain components.

εxx = εyy = εzz = γxy = 0

And four zero stress components, from our stress-strain relations.

σxx = σyy = σzz = τxy = 0

And these equations become important in analyzing last weeks results in the displacement field (warping) of a non-uniform cross-section of a bar. We can see that τyz and τxz are the only stress components remaining that are non-zero. Both of these values do not depend on z! Future examination of the Prandtl Stress function will utilize these important results!

At this point we shift our focus, however. We want to look more at the stress-strain relationship and make some comments about matrix notation for our convenience.

We will examine the stress-strain relations for the case of isotropic elasticity.

What is isotropic elasticity? Isotropic Materials (from Wikipedia)

Also, here is an additional note from our Team.

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

!Composite Materials and non-isotropic elasticity, from team Carbon When we think about composite materials, the properties of the fibers in the matrix are only optimal along the fiber direction. The tension strength, bending stiffness, or other beneficial properties we may desire are direction-dependant. The transverse fiber direction and the out-of-plane direction in laminar composites do not have the same poisson ratio or elastic modulus as the fiber-axial direction.

In our matrix notation, the numbers 1, 2, and 3 refer to row and column matrix position. This is not to be confused with "Material Coordinate Systems" that denote 1 as the axial fiber direction, 2 as the transverse in-plane direction, and 3 as the out-of-plane normal direction.

Isotropic materials are presumed to have properties independant of direction in the material on a macro-scale.

Contribution by carbon.orear
 * }

We can look at either tensorial or engineering strain, as discussed in previous notes. Here we use tensorial strain for our relationship. If for some matrix shown below, where A, B , and 0 are each 3x3 sub-matrices (note that 0 is a zero matrix):

{εij} = $$\begin{bmatrix} A & 0\\0 & B \end{bmatrix}\,$$ {σij}

then it holds that, by inversion,

{εij} = $$\begin{bmatrix} A^{-1} & 0\\0 & B^{-1} \end{bmatrix}\,$$ {σij}

Note that for an invertible matrix, the forward product of a matrix and itself yields the identity matrix [ I ].

[ C ][ C -1] = [ I ]

The components of matrix A above are given in Chapter 2 of the Sun book along with the components for matrix B.

HOMEWORK SOLUTION: The homework problem given was to find a relationship for σ12. From our matrices we can see that

σ12 * {1 / G} = ε12

Equilibrium Equation for Stresses in a Non-Uniform Stress Field (Lecture 25 cont'd)
Considering the 1-dimensional case as our model:

The diagram below shows uniform axial force on the left (as in 2.3) and non-uniform on the right (as in 2.4). The model will set up our examination for analysis of a stress-field after this lecture.

Bidirectional Bending (Lecture 26)
Bidirectional Bending (see Section 4.2 and Example 4.1 in the book)

Bidirectional bending "recipe:"

$$M_y = \int_A z\sigma_{xx}dA$$

$$M_z = \int_A y\sigma_{xx}dA$$

Moment of Inertia Tensor: $$I_y, I_z, I_{yz}$$ (sometimes referred to by $$I_{yy},I_{zz},I_{yz}$$ or $$I_{22},I_{33},I_{23}$$), whose components are given by:

$$I_y = \int_A z^2dA$$

$$I_z = \int_A y^2dA$$

$$I_{yz} = \int_A yzdA$$

Remember, $$\sigma_{xx}=E\epsilon_{xx}$$

$$=>\sigma_{xx}=\frac{I_yM_z-I_{yz}M_y}{I_yI_z-(I_{yz})^2}y+\frac{I_zM_y-I_{yz}M_z}{I_yI_z-(I_{yz})^2}z$$

$$\boldsymbol I = \begin{bmatrix} I _{11} & I _{12} & I _{13}\\ I _{21} & I _{22} & I _{23}\\ I _{31} & I _{32} & I _{33} \end{bmatrix}$$

Note: The denominator in the above equation for $$\sigma_{xx}$$ is equivalent to the determinant of the "sub-matrix" of the above matrix consisting of:

$$\begin{bmatrix} I _{22} & I _{23}\\ I _{32} & I _{33}\\ \end{bmatrix}$$

$$D=I_{22}I_{33}-(I_{23})^2$$

The Neutral Axis is where $$\sigma_{xx}=0$$

$$=>\sigma_{xx}=m_yy+m_zz=0$$

$$=>z=(-\frac{m_y}{m_z})y=(tan\beta)y$$

Equation of Equilibrium in Terms of σ (Lecture 27)
The goal is to arrive at equation 3.14:

$$\frac{\partial \sigma _{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z} = 0$$

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

!Contribution Permutation and Indicial Notation, from team Carbon
 * To convert to Indicial Notation, follow the form below (standard → indicial):
 * To convert to Indicial Notation, follow the form below (standard → indicial):
 * To convert to Indicial Notation, follow the form below (standard → indicial):

Numerator: x → 1, y → 2, z → 3 Denominator: x → x1, y → x2, z → x3

Look at the Wikipedia article below to find out more about the indicial or "Einstein" notation:

http://en.wikipedia.org/wiki/Summation_convention

 * To convert from the book notation to the class notation follow the form below (book → class):

x → y → z → x... (cyclic permutation)

 * To convert from the class notation to the book notation follow the form below (class → book):

x → z → y → x... (cyclic permutation)

Contribution by carbon.w
 * }

Using Indicial Notation, the above equation becomes:

$$\frac{\partial \sigma _{21}}{\partial x_{2}} + \frac{\partial \sigma _{31}}{\partial x_{3}} = 0$$

In addition,

(1) $$\frac{\partial \sigma _{yy}}{\partial y} + \frac{\partial \sigma _{zy}}{\partial z} +  \frac{\partial \sigma _{xy}}{\partial x} = 0$$

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

!HW Equations of Equilibrium, from team Carbon
 * Equation (1) can be deduced from the following equation:
 * Equation (1) can be deduced from the following equation:

1: $$\frac{\partial \sigma _{yy}}{\partial y} + \frac{\partial \tau _{zy}}{\partial z} +  \frac{\partial \tau _{xy}}{\partial x} = 0$$

And in indicial notation, Eqn. (1) becomes:

(1)IND $$\frac{\partial \sigma _{22}}{\partial x_{2}} + \frac{\partial \tau _{32}}{\partial x_{3}} +  \frac{\partial \tau _{12}}{\partial x_{1}} = 0$$

Similarly, the following equations (2) and (3) can be deduced from the original equations 2: and 3: and subsequently converted to indicial notation (2)IND and (3)IND:

2: $$\frac{\partial \tau _{yz}}{\partial y} + \frac{\partial \sigma _{zz}}{\partial z} +  \frac{\partial \tau _{xz}}{\partial x} = 0$$

(2) $$\frac{\partial \sigma _{yz}}{\partial y} + \frac{\partial \sigma _{zz}}{\partial z} +  \frac{\partial \sigma _{xz}}{\partial x} = 0$$

(2)IND $$\frac{\partial \sigma _{23}}{\partial x_{2}} + \frac{\partial \sigma _{33}}{\partial x_{3}} +  \frac{\partial \sigma _{13}}{\partial x_{1}} = 0$$

And finally,

3: $$\frac{\partial \tau _{yx}}{\partial y} + \frac{\partial \tau _{zx}}{\partial z} +  \frac{\partial \sigma _{xx}}{\partial x} = 0$$

(3) $$\frac{\partial \sigma _{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z} +  \frac{\partial \sigma _{xx}}{\partial x} = 0$$

(3)IND $$\frac{\partial \sigma _{21}}{\partial x_{2}} + \frac{\partial \sigma _{31}}{\partial x_{3}} +  \frac{\partial \sigma _{11}}{\partial x_{1}} = 0$$

Contribution by carbon.w
 * }

All of the equations above can be combined in a simple form:

$$\sum_{i=1}^{3}{\frac{\partial \sigma_{ij}}{\partial x_{i}}} = 0$$ for j=1,2,3

This simplified form was essential to the development of Einstein's Theory of Relativity. Relativity (wikipedia article)

3D Non-Uniform Stress Field (Lecture 28)
Using the kinematic assumptions and stress strain relations,

σxx = σyy = σzz = τxy = 0

The equation, $$\sum_{i=1}^{3}{\frac{\partial \sigma_{ij}}{\partial x_{i}}} = 0$$ for j=1,2,3, can be rewriten to

$$\frac{\partial \sigma_{21}}{\delta x_2} + \frac{\partial \sigma_{31}}{\partial x_3}=0$$

$$\sum F_x=0=-\sigma(x)A + \sigma(x+dx)A+f(x)dx$$

$$\displaystyle f(x)$$ is the distributed load

$$\displaystyle 0=A[\sigma(x+dx)-\sigma(x)]+f(x)dx$$

using a taylor series expansion on the section within the square brackets to give us

$$\frac{d\sigma(x)dx}{dx}+h.o.t.$$  h.o.t. - higher order terms

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

!Taylor Series Expansion, from team Carbon A Taylor Series Expansion is a way to solve a function as a infinate sum of terms. The Taylor series is represented as $$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+\cdots\,,$$ or as $$\sum_{n=0}^{\infin} \frac{f^{(n)}(a)}{n!} (x-a)^{n}\,,$$. If the function converges then as the term number increases the term amount approaches zero and the higher order terms become so small that they make an insignifiacant different on the total and can be ignored.

For more information on Taylor Series Expansions see the Wikipedia article [Taylor Series]

Contribution by carbon.booth
 * }

Recall: $$f(x+dx) = f(x) + \frac{df(x)dx}{dx} + \frac{1d^{2}f(x)dx}{2dx^{2}} + ...$$ and we neglect the h.o.t.

Now neglecting the higher order terms $$\frac{d\sigma}{dx}+\frac{f(x)}{A}=0$$ where A is the applied load

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

!Body Force, from team Carbon Body force, $$F_{body}=\frac{F}{V} = \frac{Force}{Volume}$$, is the force on an object per unit volume. An example, as mentioned in class, is the Earth's gravitational pull on the moon. This is represented as the gravitational force of the two masses divided by the volume of the moon.

Buoyancy force is also often represented as a body force.

For more information on Body Forces see the Body Force Wikipedia article

Contribution by carbon.guidry
 * }

Now the equilibrium equation of a 3-d non-uniform stress field will be shown, but without an applied load. Focusing on the x-direction to avoid clutter and make a more appealing figure. Each side of the cube in the figure is of length dx, dy, and dz along their respective axis. The shear forces on the front and rear faces are equal, as are the shear stresses on the top and bottom. The normal stress in the x direction can be seen on the right and left faces.

$$ \sum{F_{x}} = 0= (dydz)[-\sigma_{xx}(x,y,z) + \sigma_{xx}(x+dx,y,z)]\, $$ Face normal to X axis.

$$ + (dzdx)[\sigma_{yx}(x,y,z) + \sigma_{yx}(x,y+dy,z)]\,$$ Face normal to Y axis.

$$+ (dxdy)[\sigma_{zx}(x,y,z) + \sigma_{zx}(x,y,z+dz)]\,$$ Face normal to Z axis.

Resulting in

$$0=(dxdydz)[\frac{d\sigma_{xx}}{dx} + \frac{\sigma_{yx}}{dy} + \frac{\sigma_{zx}}{dz}]$$

$$0=[\frac{d\sigma_{xx}}{dx} + \frac{\sigma_{yx}}{dy} + \frac{\sigma_{zx}}{dz}]$$

Which is therefore, equivalent to equation (1) in Lecture 27.


text:
User:Eas4200c.f08.carbon.orear/HW5/MATLABMotivation

=MATLAB Motivation=

This homework adds T-brackets to the previous assignments and exams how it affects the airfoil. Given cross-sectional areas, the new centroid, area, inertia, highest bending moment and highest bending stress will be computed. For the second part, the same is computed but now including bending moment effects (given). Then the highest bending moments are observed in the stringers (not the skin) and the ultimate bending moment is found and compared.

contribution by carbon.w


text:
User:Eas4200c.f08.carbon.orear/HW5/Contributors

Jeff Wright Eas4200c.f08.carbon.w 04:18, 30 October 2008 (UTC)

Brian Taylor Eas4200c.f08.carbon.taylor 19:39, 4 November 2008 (UTC)

Aaron Clausen Eas4200c.f08.carbon.clausen 02:07, 7 November 2008 (UTC)

Jeff O'Rear Eas4200c.f08.carbon.orear 02:36, 7 November 2008 (UTC)

Abby Booth Eas4200c.f08.carbon.booth 07:42, 7 November 2008 (UTC)

Beau Guidry Eas4200c.f08.carbon.guidry 17:45, 7 November 2008 (UTC)


text:
User:Eas4200c.f08.carbon.orear/HW5/MATLAB

=HW5 MATLAB Code=

Case 1 Results
Neglect the bending effects of the skin and the partition walls (spar webs).

location of the centroid of the stringers; plot the stringers with black solid circles, the centroid as a crosshair, with the skin and spar webs in dotted lines.

 * Stringer Centroid = (0 , 0.2083   , 0.0081) m

all components of the moment of inertia tensor: I_22, I_33, I_23

 * I_22 = 6.3913e-007
 * I_33 = 8.3333e-006
 * I_23 = -7.6055e-008

slope beta = - alpha of the neutral axis (see classnotes); plot the neutral axis.

 * TanAlpha = -0.0216 m/m

highest normal bending stress sigma_11 and location

 * stress_11Max = 85,134,000 Pa

 * LocationMax = (0, 0.1538, 0.0470) m (See Red Cross Hair)

let the ratio between M_y and M_z to be constant based on the above data, set the maximum bending stress sigma_11 to the ultimate stress for steel 300 M (see Table 1.1, p.14), find the ultimate bending moments M_y and M_z.

 * stress_11Ult =  300,000,000 Pa

 * M_y_ult = -7,176.2 N m

 * M_z_ult = 2,870.5 N m

Case 1 Results
location of the centroid; plot the NACA airfoil in solid lines, with the stringers in black solid circles, this centroid and the centroid in Part I (as crosshairs with different colors and line styles) in the same figure. You may need to magnify the figure in case the two centroids are very close to each other.

 * Entire Centroid  = (0  , 0.2344   , 0.0071) m (See Black Cross Hair)
 * Stringer Centroid = (0 , 0.2083   , 0.0081) m (See Blue X)

all components of the moment of inertia tensor: I_22, I_33, I_23

 * I_22 = 1.0893 x 10-4
 * I_33 = 1.7658 x 10-4
 * I_23 = -3.4076 x 10-7

slope beta = - alpha of the neutral axis (see classnotes); plot the neutral axis.

 * TanAlpha = -0.2450 m/m

highest bending stress sigma_11 and location.

 * stress_11Max = 836,720 Pa

 * LocationMax = (0,  0.5000, 0) m (See Red Cross Hair)

'''compare the results of the above full-blown NACA airfoil to the results obtained in Part I, and draw some conclusions. '''

Next, since the thin skin would buckle under high normal bending stress acting in the plane of the skin, and thus the skin would not participate in resisting bending at high bending moments; to estimate the ultimate bending moments as done in Part I, do the following:

find the highest normal bending stress sigma_11 in any of the 4 stringers, and identify that stringer(s).

 * stress_11B = 763,870 Pa
 * stress_11F = 170,880 Pa
 * stress_11H = 630,850 Pa
 * stress_11E = 108,530 Pa

Stringer B has the highest bending stress.

let the ratio between M_y and M_z to be constant based on the above data, set the maximum bending stress sigma_11 in the above stringer(s) to the ultimate stress for steel 300 M (see Table 1.1, p.14), find the ultimate bending moments M_y and M_z for the full-blown NACA 2415 airfoil.

 * stress_11_B_Ult =  300,000,000 Pa

 * M_y_ultB = -490,920 N m

 compare the results to those obtained in Part I.
 * M_z_ultB = 196,370 N m


text:
User:Jade Knight/Tools

To help me find stuff:
 * Blocking policy
 * Category:User warning templates
 * Template:Nothanks-sd
 * Maintenance
 * Special:PrefixIndex

title: User:Jade Knight/Tools


text:
Talk:Consensus in a wiki environment/An introduction to Consensus

I really dislike the current section of "What consensus isn't" as it currently stands. I think it either needs a significant rewrite, or should be removed entirely: it is too likely to be a source of wiki-lawyering, I fear. The Jade Knight (d'viser) 05:30, 30 October 2008 (UTC)
 * It doesn't make much sense to the page. This isn't a Wikipedia policy, it's a learning project.  It doesn't help to start off by making it sound like a Wikipedia policy. Bastique 01:52, 7 November 2008 (UTC)


text:
User:Spacebirdy/monobook.js

mw.loader.load('https://meta.wikimedia.org/w/index.php?title=User:Spacebirdy/altækt.js&action=raw&ctype=text/javascript');

title: User:Spacebirdy/monobook.js


text:
User talk:Mhz1363

'''

title: User talk:Mhz1363


text: Brezhoneg Unan/Lesson 4/Quiz on mutations after the article


text:
Internet and Project Competence/group5

 Interaction between children and nature

The group members

 * Robbins, Lindsay M.
 * Hendrix, Dennis C.
 * Bolick, Megan L.
 * Stacy, Ashley E
 * Jones, Whitney S.
 * ElliA Elisabeth Altmann
 * Gergana Georgieva
 * Raia
 * Eva
 * Desislava Maneva
 * Michelle Hemraj

final powerpoint
http://upload.wikimedia.org/wikiversity/en/8/89/Interaction_with_Nature.pdf

Working area: the questionnaire
Everyone needs to ask these questions to 2 or 3 students from their country and then create a slide that shows your results...this slide doesnt have to be the exact answers from your survey just an overall view of how your kids interact with nature.. then we can put them all together and make a few final slids comparing the different countries.

1. Personal Questions

- How old are you?

-what grade are you in?

- Mark with a cross:           Girl o 			Boy o

- Describe where you live:city/village/town

2. How would you describe nature?

3. What animals do you offen see in nature?

4. Have you ever planted any plant?

Describe how you have taken care of it?

5. how much time during the week do you spend outside at home? what activities do you participate in?

6. How much time do you spend outdoors during the school day? What types of activities do you do?

7.What natural phenomenons have you seen? Describe them? (Suggestions: mist, rainbow, hail, dew, frost, a solar eclipse, etc.)

8. How do you think pollution affects our enviornment?

9. How much time do you spend a week watching tv/playing video games or using a computer?

what types of thing to you watch or play while watching tv or using a computer?

 * edited by HemrajM
 * edited by Elisabeth A.

 * i wrote what i thought about the questions so that we can get them finalized and deside what questions we are asking... we really need to know by sometime this weekend so we can do this with our students next week....it is our last week in our schools.....i think we need to keep it somewhat simple...Thanks Megan

 * HemrajM- I think we should take off the question about the dreamhouse, making it 10 questions; add an age/school/country/male-female section, and finalize.
 * i canged some of the questions to see more how children interact with nature. i think through these we can see how children really interact with nature....i think we should stick to 2 or 3 students per person so we really have time to analyze the results. i hope this is what we can finalize the questions as???? we just need to know what questions we will be asking our students....because it need to be done in the next week or so....so we have time to compare our results.......Thanks Megan

Conclusion Questionnaire
okay, we have to work on it :)

title: Internet and Project Competence/group5


text:
Internet and Project Competence/group6a

The Teachers Emotional Influence on First Graders

group members
Group 6a
 * Graham, Ashley M.
 * Valentine, Casey D.
 * Murrell, Genevieve N.
 * Sain, Jennifer L.
 * Goodson, Jordan C.
 * Daniela Widdmann
 * Lilia Popova

Group 6b
 * Endaya, Verna F.
 * Register, Cayton P.
 * Grimm, Erica R.
 * Surette, Megan E.
 * Mariana Igova
 * Kathrin Lehnert
 * Lindsay Robbins

Group Power Point Project http://upload.wikimedia.org/wikiversity/en/f/f4/International_project_slide_show.pdf

KWL CHART
Please click on "edit" (here above), copy the code, paste it on your personal user site and insert Your text in the columns.

Project Outline/Questions to be Discussed
I. How does the teacher's work motivate the children? II. How does the teacher affect the development of the childrens interests/ of the childrens creativity? III. When does the friendship between teacher and students begin? IV. What is the role that teachers play in affecting children emotions? V. What triggers the emotional influence between the child and the teacher. VI. How long after the school year beings does the student realize their emotional attachement with their teacher( I am sure this varies of course)? VII. Does every first grade teacher have an emotional influence on all of their students? VIII. What do teachers and parents think about the first teachers' influence on the first graders? IX. How a child is affected by positive vs. negative criticism?

4. Links
This link gives a teacher's input on how emotions affect the children's learning, and ways that the teacher can help those students who are not in the mood to learn. However, this is from a teacher in high school, but I'm sure it can be adapted for a first grade class. Emotions affect a student's learning Verna Endaya. Topic question number 4.

This link briefly describes a study that involved the results between the affects of positive and negative criticism on students. [] Ashley Graham. Topic question 9

title: Internet and Project Competence/group6a


text:
Internet and Project Competence/group6b

group wiki 6

Group Power Point project http://upload.wikimedia.org/wikiversity/en/f/f4/International_project_slide_show.pdf

KWL CHART
Please click on "edit" (here above), copy the code, paste it on your personal user site and insert Your text in the columns.

Project Outline/Questions to be Discussed
I. How does the teacher's work motivate the children? II. How does the teacher affect the development of the childrens interests/ of the childrens creativity? III. When does the friendship between teacher and students begin? - Mariana Igova IV. What is the role that teachers play in the education and the influence on the children emotions? V. What triggers the emotional influence between the child and the teacher. VI. How long after the school year beings does the student realize their emotional attachement with their teacher( I am sure this varies of course)? VII. Does every first grade teacher have an emotional influence on all of their students? VIII. what teachers,pupils and their parents think about the first teachers' influence on the first graders> IX. How a child is affected by positive vs. negative criticism?

title: Internet and Project Competence/group6b


text:
User talk:Michelle Gingras

 Hello Michelle Gingras, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --The Jade Knight (d'viser) 09:37, 30 October 2008 (UTC)

Good luck with learning about the sanshin! The Jade Knight (d'viser) 09:37, 30 October 2008 (UTC)

title: User talk:Michelle Gingras


text:
User talk:Celenetolentino

 Hello Celenetolentino, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --The Jade Knight (d'viser) 09:40, 30 October 2008 (UTC)

Good luck with your t-shirt project! The Jade Knight (d'viser) 09:40, 30 October 2008 (UTC)

title: User talk:Celenetolentino


text:
User talk:ErinShipton

 Hello ErinShipton, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --The Jade Knight (d'viser) 09:48, 30 October 2008 (UTC)

Good luck with your script! You'll find that Wikiversity and Wikibooks both have several resources on Writing. I hope you manage to reach Kripke, but if not, several other writers wander around the web and can be reached easily. I could also recommend a few good Folklorists, if you'd like some expert advice finding myths/urban legends/etc. The Jade Knight (d'viser) 09:48, 30 October 2008 (UTC)

title: User talk:ErinShipton


text:
User talk:PeterTran

 Hello PeterTran, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --The Jade Knight (d'viser) 10:09, 30 October 2008 (UTC)

Good luck learning how to play Mahjong! The Jade Knight (d'viser) 10:09, 30 October 2008 (UTC)

title: User talk:PeterTran


text:
User talk:Kibs~enwikiversity

Welcome to wiki the best way to stay informed

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Kibs. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Kibs~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:41, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:13, 19 April 2015 (UTC)

title: User talk:Kibs~enwikiversity


text:
User:WikimediaNotifier/template

 * … do not reply to this automatically delivered notification here ; answer/discuss together on Meta, centrally!

title: User:WikimediaNotifier/template


text:
User:Eml4500.f08.ramrod.D/HW5

Debugged 2 Bar Truss System
Below is the code to the debugged two bar truss system. Since the modulus of elasticity and the areas for the given example were not the same the code had to be modified in order for it to run properly. Before the correction the code was running as if the all of the elements had the same modulus of elasticity and the same area. Below is the corrected code and the results from the corrected code. Functions PlaneTrussElemtens.m, NodalSoln.m and PlaneTrussResults.m were used in this code and the discription for these functions can be found in the functions section.

Results from the debugged two bar truss code

Comparison of Results
Once the code was corrected the same reaction results were obtained when comparing it to the statics methods that have been done in class. Below are the results to the statics method.

Finite Element Method:

Using this equation, we can now solve for the f(1) matrix and thus we find the values for the reactions: $$f^{(1)}_{1} = F_{1} = -4.4378$$ $$ f^{(1)}_{2} = F_{2} = -2.5622$$

Using this same procedure applied to element 2, we can solve for the values of the other two unknown reactions giving us: $$f^{(2)}_{3} = F_{5} = 4.4378$$ $$f^{(2)}_{4} = F_{6} = -4.4378$$

Statics

Using the Euler cut method, you find the following components from the resultant P forces.

$$P_x^1 = -4.4378$$ $$P_y^1=-2.5622$$ $$P_x^2=4.4378$$ $$P_y^2=-4.4378$$

Six Bar Truss Code
Below is the code for the six bar truss system found on page 226 in the the textbook. All six of the bars have the same modulus of elasticity (e) and the same area (A). The MATLAB functions of PlaneTrussElement.m, NodalSoln.m, and PlaneTrussResults.m are all found in the code. The descriptions and functions can be found below the results of the six bar truss system.

Results of the above code.

Functions
The functions PlaneTrussElement.m, NodalSoln.m, PlaneTrussResults.m were all called in the code for the 5 truss system. They are all necessary for the completion of the truss problem.

PlaneTrussElement.m

This function take the Young's Modulus (e), the Area of the element (A) and the coordinates of the element ends (coord) and generates the stiffness matrix for a plane truss element. The function first calculates the lengths of the the elements. Once those are calculated the stiffness matrix is calculated.

NodalSoln.m

The function takes the global coefficient matrix (K), the global right hand side vector (R), the list of degrees of freedom with specified values, and the specified values to determine the displacements and reactions at each node. The dof of the the system is first found by using the  command which finds the longest dimension of R. df then finds the difference between the dofs that have known values (a value of zero) and the dof that were found in the previous line. The displacements and the reactions are then calculated.

PlaneTrussResults.m

This function computes the plane truss element results. It takes in the modulus of elasticity (e), the area of the cross-section (A), the coordinates at the element ends (coord), and the displacements at the elemtent ends (disps) and calculates the axial strain (eps), stress(sigma) and force (force). The results are stored in the variable  and sent back to the main program.

Plot of Six Bar Truss System
Below is the code and the plot of the six bar truss system. The plot was made from the results that were found in the above code. The results demonstrate that only node 2 and node 5 had a displacement of (0.2131, 0.2500) and (-0.0061, 0.0122) respectively. The undeformed system is plotted with the dotted lines and the deformed system is plotted with solid lines. The deformed displacements have been scaled up by a factor of 1.2 in order to enhance in the plot.

Three Bar Space Truss Example
Below is the code for the the 3D three bar truss system found in the textbook on page 230. There are four nodes in the system with each node containing 3 degrees of freedom. The code uses three additional functions for it to be run. SpaceTrussElement.m, NodalSoln.m, and NodalSoln.m were all used for the code. Below this code a description of each function can be found.

Functions
The functions SpaceTrussElement.m, NodalSoln.m, SpaceTrussResults.m were all called in the code for the 5 truss system. They are all necessary for the completion of the truss problem.

SpaceTrussElement.m This function creates the stiffness matrix of a space truss element. It takes in the modulus of elacticity (e), the truss area (A), and the corresponding coordinates for the truss. The function first calculates the length of each element and then calculates the stiffness matrix (k).

NodalSoln.m

The function takes the global coefficient matrix (K), the global right hand side vector (R), the list of degrees of freedom with specified values, and the specified values to determine the displacements and reactions at each node. The dof of the the system is first found by using the  command which finds the longest dimension of R. df then finds the difference between the dofs that have known values (a value of zero) and the dof that were found in the previous line. The displacements and the reactions are then calculated.

SpaceTrussResults.m

This function takes in the modulus of elasticity (e), the area of each truss (A), the coordinate of the nodes and the displacements (disps) at the element ends. It calculates the axial strain (eps), stress(sigma) and force (force). The results are stored in the variable  and sent back to the main program.

3 Bar Space Truss Results
Below are the results from the three bar space truss example.

Code for the lot of the 3 Bar Space Truss
Below is the code for the 3 Bar Space Truss.

Plot of 3 Bar Space Truss
Below is the plot of the 3 bar Space truss with the viewing mode of x = 1m, y = 1m, and z= 1m. The dotted lines represent the undeformed truss system and the solid lines represent the deformed truss system.

]

Below is the same plot but in plane view down the x axis, plane view down the y axis, and plane view down the z axis.

title: User:Eml4500.f08.ramrod.D/HW5


text:
User:Dlohcierekim~enwikiversity

I was looking for a list of German verbs, and came to the page here.

title: User:Dlohcierekim~enwikiversity


text:
Talk:High School Biology/Teachers

I need some space to put links for my biology class. I will help with the course if it is OK with you. Rozzychan 15:35, 30 October 2008 (UTC)

title: Talk:High School Biology/Teachers


text:
User:Eml4500.f08.bottle.vitello/HW 5

Principle of Virtual Work
In order to use the Principle of Virtual Work method, we must show justification for the eliminations of rows 1, 2, 5, and 6 to obtain $$ \textbf{K}_{2x2} $$ in a two-bar truss.

The force displacement relation for a two-bar truss system is:

$$ \textbf{Kd}-\textbf{F}=\textbf{0}_{6x1}\ $$

For the Principle of Virtual Work, this is derived to:

$$ \textbf{W}_{6x1}\cdot (\textbf{Kd}-\textbf{F})_{6x1}=0_{1x1}\ $$

with W being the weighing matrix, for all $$ \textbf{W}_{6x1}\ $$

In order to prove that these two equations are both equivalent to each other, we use the following proof:

Proof: Select W at $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}_{1x6}\ $$ plugging W back into the Principle of Virtual Work equation gives: $$ \begin{bmatrix}1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 1 \cdot \bigg[ \sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1} \bigg]+ 0\cdot \bigg[\sum_{j=1}^{6}{K_{2-6j}d_{j}} - F_{2-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{1j}d_{j}=F_{1}}\ $$ Following through as before for the rest of the matrices gives these results: $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{2j}d_{j}} - F_{2} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{3-6j}d_{j}} - F_{3-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{2j}d_{j}=F_{2}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-2j}d_{j}} - F_{1-2} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{3j}d_{j}} - F_{3} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{4-6j}d_{j}} - F_{4-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{3j}d_{j}=F_{3}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-3j}d_{j}} - F_{1-3} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{4j}d_{j}} - F_{4} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{5-6j}d_{j}} - F_{5-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{4j}d_{j}=F_{4}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-4j}d_{j}} - F_{1-4} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{5j}d_{j}} - F_{5} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{5j}d_{j}=F_{5}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-5j}d_{j}} - F_{1-5} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{6j}d_{j}=F_{6}}\ $$ Hence, adding up all of these results gives us $$ \textbf{Kd}=\textbf{F} $$.

Going back to the Principle of virtual work and the two-bar truss, we had originally removed the displacements at d1, d2, d5, and d6. These results for the weighing coefficient also result in $$ W_{1}=W_{2}=W_{5}=W_{6}=0 $$. Using only the virtual displacement at d3 and d4 of the two-bar truss system gives:

$$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\begin{Bmatrix} w_{3}\\w_{4}

\end{Bmatrix}\cdot (\textbf{K}_{2x2}\textbf{d}_{2x1}-\textbf{F}_{2x1})=0 $$

In which case the results for K, d, and F matrices gives:

$$ \textbf{K}=\begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44} \end{bmatrix} $$

$$ \textbf{d}=\begin{Bmatrix} d_{3}\\d_{4}

\end{Bmatrix} $$

$$ \textbf{F}=\begin{Bmatrix} F_{3}\\F_{4}

\end{Bmatrix} $$

We derive $$ k^{(e)}_{4X4}=T^{(e)^{T}}_{4X2}\hat{k}^{(e)}_{2X2}T^{(e)}_{2X4} $$ Recall the FD relationship with axial dof's $$q^{(e)}: \! $$ $$ \hat{k}^{(e)}_{2X4}q^{(e)}_{2X1}=p^{(e)}_{2X1} $$ $$ \Rightarrow equation (1) \hat{k}^{(e)}q^{(e)}-p^{(e)}= 0_{2X1}\! $$ Therefore: $$ \hat{W}_{2X1}\circ \underbrace{(\hat{k}^{(e)}q^{(e)}-p^{(e)})}_{2X1}=0_{1X1}$$ for all $$ \hat{W}_{2X1} \! $$

Recall that $$ q^{(e)}_{2X1}=T^{(e)}_{2X4}d^{(e)}_{4X1} $$ similarly, $$ \hat{w_{2X1}}=T^{(e)}_{2X4}w_{4X1}$$ where $$\hat{w}_{2X1}$$ is the virtual displacement corresponding to $$q^{(e)}_{2X1}$$ and $$w_{4X1} \!$$ is the virtual displacement in the global coordinate system corresponding to $$d^{(e)}_{4X1}$$

If we place equation $$ (3) \! $$ and $$ (4) \! $$ in equation $$ (2) \! $$ then we get $$ equation(5) = \! $$$$ (T^{(e)}w)\circ [\hat{k}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}] =0 $$ for all $$ w_{4X1} \!$$

Recall: $$ equation (6) = \! $$$$ (AB)^{T}=B^{T}A^{T}\! $$  (see below for proof) and $$ equation (7) = \!$$ $$a\circ b=a^{T}b$$ Substituting equations $$ (6) \!$$ and $$ (7) \!$$ into $$ (5) \!$$ we get $$(T^{(e)}w)^{T}[\hat{k}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}]=0_{1X1} $$ for all $$ w_{4X1} \! $$ $$ \Rightarrow w^{T}T^{(e)}[\hat{k}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}]-0 $$ $$ \Rightarrow w\circ [\underbrace {(T^{(e)^{T}}\hat{k}^{(e)}T^{(e)})}_{k^{(e)}}d^{(e)}-\underbrace {(T^{(e)^{T}}P^{(e)})}_{f^{(e)}}]$$ for all $$ w_{4X1} \! $$  $$ \Rightarrow w\circ [k^{(e)}d^{(e)}-f^{(e)}]=0 $$ for all $$ w \! $$  therefore $$ k^{(e)}d^{(e)}=f^{(e)} \!$$

Continuous Case with PDE's
Motivational Model Problem: Elastic Bar with varying $$ A(x) \! $$ and $$ E(x) \! $$, subject to a varying axial load (distributed load), concentrated loading, and inertial forces (damping).

$$ \sum{F_{x}}=0=-N(x,t)+N(x +dx, t)+f(x,t)dx-m(x)\ddot{u} \! $$ $$ equation (1) \! $$$$=\frac{\partial N}{\partial x}+h.o.t.+f(x,t)-m(x)\ddot{u}dx \! $$ neglecting the higher order terms, we recall the Taylor series expansion: $$f(x+dx)=f(x)+\frac{df(x)}{dx}+ \underbrace {\frac{1}{2}\frac{d^{2}f(x)}{dx^{2}}dx^{2}+\cdots}_{h.o.t.} $$

Equation $$ (1) \! $$ becomes $$ equation (2) = \! $$$$\frac{\partial N}{\partial x}+f=m\ddot{u}\! $$ the equation for the motion of the elastic bar $$ equation (3)= \! $$ $$ N(x,t)=A(x)\underbrace {\sigma (x)}_{E(x)\underbrace {\varepsilon (x,t)}_{\frac{\partial u (x,t)}{\partial x}}}$$ equation $$ (3) \! $$ into equation $$ (2) \! $$ yields: $$ \frac{\partial }{\partial x}[A(x)E(x)\frac{\partial u}{\partial x}]+f(x,t)=m(x)\underbrace {\ddot{u}}_{\frac{\partial ^{2}u}{\partial t^{2}}} $$ In order to solve this problem, we need 2 boundary conditions and 2 initial conditions.

6-Bar Truss: Case 1
For the first case, all of the values for the modulus of elasticity and cross sectional area are the same.

The following code was compiled to produce the plot for the deformed and undeformed six bar truss.

6-Bar Truss: Case 2
For the second case, all of the values for the modulus of elasticity are different, while the values for the cross sectional area remain the same. The values for the different moduli can be found in the code below.

The following code was compiled to produce the plot for the deformed and undeformed six bar truss, including the deformed truss with the varying moduli of elasticity.

Truss Systems in 3D
The next section of HW 5 will be discussing truss systems in three dimensional space, which had previously not been considered. The axial force displacement relation, commonly referred to as the FD relation, is very similar when compared to the 2-D truss element. Note that the equation of the 2-D truss element is: $$\mathbf{kd}=\mathbf{F}$$ We also recall that the FD relation for axial displacement of each element in the notation used in class is as follows: $$\frac{EA}{L}\begin{pmatrix} 1 & -1\\ -1 & 1 \end{pmatrix} \begin{pmatrix} d_{1}\\d_{2} \end{pmatrix} = \begin{pmatrix} P_{1}\\P_{2} \end{pmatrix} \Rightarrow \mathbf{k^{(e)}q^{(e)}}=\mathbf{p^{(e)}}$$ The main difference for the 3-D truss is that each matrix is a bit larger to account for the movement in the z-direction. The local degrees of freedom become the 6x1 matrix $$\mathbf{d^{(e)}}=\begin{Bmatrix} d_{1}^{(e)} \\ d_{2}^{(e)} \\ d_{3}^{(e)} \\ d_{4}^{(e)} \\ d_{5}^{(e)} \\ d_{6}^{(e)} \end{Bmatrix}$$ and the local forces become the 6x1 matrix below. $$\mathbf{f^{(e)}}=\begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \\ f_{3}^{(e)} \\ f_{4}^{(e)} \\ f_{5}^{(e)} \\ f_{6}^{(e)} \end{Bmatrix} $$ Now the transform matrix T(e) must contain two extra columns, but the axial displacement q(e) and axial load p(e) will remain as 2x1 matrices. The transform matrix will become: $$\mathbf{T^{(e)}}=\begin{pmatrix} l_{s} & m_{s} & n_{s} & 0 & 0 & 0\\ 0 & 0 & 0 & l_{s} & m_{s} & n_{s} \end{pmatrix}$$ The transform matrix for the 3-D truss system was derived similarly to the 2-D transform matrix in lecture 12. The figure below displays the unit vectors in the normal x-y coordinate axis along with the rotated coordinate axis. The displacement vector for a local node is defined as follows. $$ \bar{d}_{1}^{(e)}=d_{1}^{(e)}\hat{i}+d_{2}^{(e)}\hat{j}+d_{3}^{(e)}\hat{k} $$ It is possible to take the dot product of the displacement vector and the rotated unit vector and carry out the steps listed below. $$ q_{1}^{(e)}=(d_{1}^{(e)}\hat{i}+d_{2}^{(e)}\hat{j}+d_{3}^{(e)}\hat{k})\hat{\tilde{i}} $$ $$ q_{1}^{(e)}=d_{1}^{(e)}(\hat{i}\cdot\hat{\tilde{i}})+d_{2}^{(e)}(\hat{j}\cdot\hat{\tilde{i}})+d_{3}^{(e)}(\hat{k}\cdot\hat{\tilde{i}}) $$ $$ q_{1}^{(e)}=l^{(e)}d_{1}^{(e)}+m^{(e)}d_{2}^{(e)}+n^{(e)}d_{3}^{(e)} $$ $$ q_{1}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)} \end{Bmatrix} $$ And Similarly, for local node 2, the following can also be said. $$ q_{2}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} d_{4}^{(e)}\\ d_{5}^{(e)}\\ d_{6}^{(e)} \end{Bmatrix} $$ Combining the axial displacements into one expression results in the following expression: $$ \begin{Bmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{Bmatrix}= \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ d_{5}^{(e)}\\ d_{6}^{(e)} \end{Bmatrix} $$ This expression can be written in its most simple form as the following: $$\mathbf{q}^{(e)}=\mathbf{T}^{(e)}\mathbf{d}^{(e)}$$ Similarly, the local forces can be multiplied by the same transform matrix to equal the axial forces. $$\mathbf{p}^{(e)}=\mathbf{T}^{(e)}\mathbf{f}^{(e)}$$

Knowing the definition of T, k and r can be expressed as $$k_l d_l = r_l$$ $$k_l T d = r_l$$ $$k d = r$$ $$k = T^T k_l T$$ $$r = T^T r_l$$ Where the element equation for a space truss element can be expressed as:

$$ (EA/L) \begin{bmatrix} l_{s}^{2} & m_{s}l_{s} & n_{s}l_{s} & -l_{s}^{2} & -m_{s}l_{s} & -n_{s}l_{s}\\ m_{s}l_{s}& m_{s}^{2} & m_{s}n_{s} & -m_{s}l_{s} & -m_{s}^{2} & -m_{s}n_{s}\\  n_{s}l_{s}& m_{s}n_{s} & n_{s}^{2} & -n_{s}l_{s} & -m_{s}n_{s} & -n_{s}^{2}\\  -l_{s}^{2}& -m_{s}l_{s} & -n_{s}l_{s} & l_{s}^{2} & m_{s}l_{s} & n_{s}l_{s}\\  -m_{s}l_{s}& -m_{s}^{2} & -m_{s}n_{s} & m_{s}l_{s} & m_{s}^{2} & m_{s}n_{s}\\  -n_{s}l_{s}& -m_{s}n_{s} & -n_{s}^{2} & n_{s}l_{s} & m_{s}n_{s} & n_{s}^{2} \end{bmatrix} \begin{bmatrix} u_1\\v_1\\w_1\\u_2\\v_2\\w_2 \end{bmatrix} = \begin{bmatrix} F_1x\\F_1y\\F_1z\\F_2x\\F_2y\\F_2z \end{bmatrix} $$

Running the following code from the course textbook:

Results in:

Knowing that

$$\sum{F_{x}}= R_{1}l^{(1)} + R_{2}l^{(2)} + R_{3}l^{(3)} = 0$$ $$ \sum{F_{y}}= R_{1}m^{(1)} + R_{2}m^{(2)} + R_{3}m^{(3)} = -P$$ $$ \sum{F_{z}}= R_{1}n^{(1)} + R_{2}n^{(2)} + R_{3}n^{(3)} = 0$$

and from the MATLAB solution (which corresponds to the forces defined on page 231 of the course textbook)

$$ R_{1} = 20375$$ $$ R_{2} = 13214$$ $$ R_{3} = -23149$$

With $$ A_{1} = 200$$ $$ A_{2} = 200$$ $$ A_{3} = 600$$

It follows that $$ \sigma_{1} = \frac{R_{1}}{A_{1}} = \frac{20375}{200} = 101.87$$ $$ \epsilon_{1} = \frac{\sigma_{1}}{E_{1}} = \frac{101.875}{200\times10^{3}} = 5.0936\times10^{-4}$$ $$ \sigma_{2} = \frac{R_{2}}{A_{2}} = \frac{13214}{200} = 66.072$$ $$ \epsilon_{2} = \frac{\sigma_{2}}{E_{2}} = \frac{66.07}{200\times10^{3}} = 3.3036\times10^{-4}$$ $$ \sigma_{3} = \frac{R_{3}}{A_{3}} = \frac{-23149}{600} = -38.58$$ $$ \epsilon_{3} = \frac{\sigma_{3}}{E_{3}} = \frac{-38.58}{200\times10^{3}} = -1.929\times10^{-4}$$

Which were already determined through calculations via the MATLAB code above, and hence proving that the problem is statically determinate as solved by both statics and FEA.

The following is MATLAB code written to model the deformed and undeformed 3-bar 3D truss system:

The following plots correspond to the deformed and undeformed 3-Bar 3D truss.

The first three plots correspond to views along the x-, y-, and z- axes.

Also included are views from [-2, -2, 3] and for clarity [3, 3, 3].

Note that the magnifying factor for these images is 1.0.

Prove AB Transpose = B Transpose A Transpose
Given:

$$\textbf{A} = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}

\textbf{B} = \begin{bmatrix} 7 & 8 & 9\\ 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}

$$ Such that $$

\mathbf{AB} = \begin{bmatrix} 21 & 27 & 33\\ 57 & 72 & 87 \end{bmatrix}

\mathbf{AB}^T = \begin{bmatrix} 21 & 57 \\ 27 & 72 \\ 33 & 87 \\ \end{bmatrix}

$$ While $$

\textbf{A}^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \\ \end{bmatrix}

\textbf{B}^T = \begin{bmatrix} 7 & 1 & 4\\ 8 & 2 & 5\\ 9 & 3 & 6 \end{bmatrix}

\mathbf{B}^T\mathbf{A}^T = \begin{bmatrix} 21 & 57 \\ 27 & 72 \\ 33 & 87 \\ \end{bmatrix} $$

Hence, $$\mathbf{B}^T\mathbf{A}^T = \mathbf{(AB)}^T$$

Research of Composite Materials
 Composite materials  are materials created from at least two constituent materials, generally with differing physical or chemical characteristics such that the resulting material contains properties each contributing element. Consider it as equivalent to a holistic Gestalt approach to fabrication synthesis, such that the final synergetic compound reflects the invested engineering aimed at deriving the resultant requisite material properties.

Historically, basic composite materials included the combination of mud and straw to provide both structure, strength, and formability. More advanced and modern composites include carbon fiber, plywood, fiberglass, and asphalt concrete. Common to most composites is the existence of a matrix element and a reinforcement element. The matrix is considered the composite constituent which encases and provides form for the reinforcement material. The reinforcement, in turn, provide enhancements to the matrix via its own complementary material properties.

Typically, the combination of the matrix and reinforcement material(s) occurs in what's considered a melding event. This results in the finite combination of the two or more materials, hence creating the composite. Further, the physical properties of composites are usually not isotropic, implying that the direction of load will affect the material in different ways. If particular reinforcing fibers are arranged in such a way that a load is applied parallel to their axis of orientation, the composite will fair much better than if the load was applied perpendicular to the fiber arrangement axis. This can sometimes result in failure, which can also be experienced if the reinforcing material comes separated from the matrix.

For more information, it is recommended that the reader consider using the following resource as endorsed by Professor Vu-Quoc's Vision:

http://en.wikipedia.org/wiki/Composite_materials

Weighting Coefficients
Given that $$K d - F = 0 $$ and $$ w ( K d - F) = 0 $$ If w refers to any set of weighting coefficients, it is possible to choose the set which can be defined as $$ w_1 = 0, $$ $$ w_2 = w_3 = ... w_6 = 0 $$ Thus $$ w^{T} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$ Which can be expressed as $$w ( K d - F) = 1 * \left[{\sum_{j=1}^{6}K_{1j}d_{j}-F_1} \right] + 0 * \left[{\sum_{j=1}^{6}K_{2j}d_{j}-F_2} \right] + 0 * \left[{\sum_{j=1}^{6}K_{3j}d_{j}-F_3} \right] + ... 0 * \left[{\sum_{j=1}^{6}K_{6j}d_{j}-F_6} \right] = 0 $$ Hence $$w ( K d - F) = 1 * \left[{\sum_{j=1}^{6}K_{1j}d_{j}-F_1} \right] $$ It follows that the choice of w element as a value of 1 will result in the corresponding sum per that chosen w element if all other elements are chosen as a value of 0, as observed through a similar pattern as seen above.

Axial Displacements
From above, if $$ w^{T} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Then the quotient w (K q - p) for axial displacement can be written as

$$w (K q - p) = 1 * \left[{\sum_{j=1}^{6}K_{1j}q_{j}-p_1} \right] + 0 * \left[{\sum_{j=1}^{6}K_{2j}q_{j}-p_2} \right] + 0 * \left[{\sum_{j=1}^{6}K_{3j}q_{j}-p_3} \right] + ... 0 * \left[{\sum_{j=1}^{6}q_{6j}d_{j}-p_6} \right]$$

Such that

$$w (K q - p) = 1 * \left[{\sum_{j=1}^{6}K_{1j}q_{j}-p_1} \right]$$

It follows that this relationship is true per w element for all w elements selected as value of 1 where all other w elements are selected as 0.

Debugging of Two Bar Truss Code
The code given in conjunction with the second homework report contained an error which was to be corrected. The previous code returned an error when attempting to generate a results matrix. This error was caused by the lines:

for i=1:elems results = [results; PlaneTrussResults(e, A, ...           nodes(conn(i,:),:), d(lmm(i,:)))]; end

This section of code attempted to pass two arrays as scalar values.

To fix this problem, the code was modified to read:

for i=1:elems results = [results; PlaneTrussResults(e(i), A(i), ...   nodes(conn(i,:),:), d(lmm(i,:)))]; end

The code in its entirety is as follows:

% Debugged Two-Bar Truss System Code clear all; e = [3 5]; A = [1 2]; P = 7; L=[4 2]; alpha = pi/3; beta = pi/4; nodes = [ 0, 0; % x,y position of node 1 L(1)*cos(pi/2-alpha), L(1)*sin(pi/2-alpha); % x,y position of node 2 L(1)*cos(pi/2-alpha)+L(2)*sin(beta),L(1)*sin(pi/2-alpha)-L(2)*cos(beta) % x,y position of node 3 ]; dof=2*length(nodes); conn=[1,2; 2,3]; lmm = [1, 2, 3, 4; 3, 4, 5, 6]; elems=size(lmm,1); K=zeros(dof); R = zeros(dof,1); debc = [1, 2, 5, 6]; ebcVals = zeros(length(debc),1); %load vector R = zeros(dof,1); R(4) = P; % Assemble global stiffness matrix K=zeros(dof); for i=1:elems lm=lmm(i,:); con=conn(i,:); k_local=e(i)*A(i)/L(i)*[1 -1; -1 1] k=PlaneTrussElement(e(i), A(i), nodes(con,:)) K(lm, lm) = K(lm, lm) + k; end K R % Nodal solution and reactions [d, reactions] = NodalSoln(K, R, debc, ebcVals) results=[]; for i=1:elems results = [results; PlaneTrussResults(e(i), A(i), ...   nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g results

This correction yields the following results:

k_local = 0.750  -0.750   -0.750    0.750 k = 0.5625   0.3248   -0.5625   -0.3248    0.3248    0.1875   -0.3248   -0.1875   -0.5625   -0.3248    0.5625    0.3248   -0.3248   -0.1875    0.3248    0.1875 k_local = 5   -5    -5     5 k = 2.5000  -2.5000   -2.5000    2.5000   -2.5000    2.5000    2.5000   -2.5000   -2.5000    2.5000    2.5000   -2.5000    2.5000   -2.5000   -2.5000    2.5000 K = 0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.1875   -0.3248   -0.1875         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000 R = 0    0     0     7     0     0 d = 0        0    4.3520    6.1271         0         0 reactions = -4.4378  -2.5622    4.4378   -4.4378 results = 1.7081      5.1244       5.1244       0.6276        3.138        6.276

Individual Contributions
Eml4500.f08.bottle.vitello 15:51, 30 October 2008 (UTC) Eml4500.f08.bottle.butler 22:29, 3 November 2008 (UTC) Eml4500.f08.bottle.hipps 11:28, 4 November 2008 (UTC) Eml4500.f08.bottle.loschak 09:04, 5 November 2008 (UTC) Eml4500.f08.bottle.ranto 05:10, 6 November 2008 (UTC) Eml4500.f08.bottle.barnes 18:30, 6 November 2008 (UTC) Eml4500.f08.bottle.brockmiller 21:57, 6 November 2008 (UTC)

title: User:Eml4500.f08.bottle.vitello/HW 5


text:
User:WikimediaNotifier/config.js

{$$}

title: User:WikimediaNotifier/config.js


text: File:Halloween intro.OGG

title: File:Halloween intro.OGG


text: User:WikimediaNotifier/self.js

title: User:WikimediaNotifier/self.js


title: File:Drawing1jn.svg


title: File:MATLABHW5.svg


text:
Echinoderm Nervous System Evolution Project

Echinoderm Nervous System Evolution Project

Introduction
The broad objective of this project is to help understand the homology of the urbilaterian nervous system, a question that has puzzled researchers for more than a century. Modern tools and techniques offer new ways of searching for missing pieces to this puzzle. This research focuses on the origin of cell types in the echinoderm nervous system as compared to the nervous system of other Bilateria and Eumetazoa, a promising avenue of exploration that may elucidate aspects of homology not apparent from developmental and gene-expression studies.

Echinoderms are of particular interest because their pentaradial (five-fold) symmetry is so enigmatic. It is likely that, at an early stage in their evolutionary history, they lost their cephalic ganglion. One class, Holothuroidea (sea cucumbers) has regained a cephalic ganglion. The unusual echinoderm body shape may have evolved from the translocation and inversion of the gene segment containing Hox1-Hox3, and the loss (at an unknown period in time) of Hox4. It is not known whether other taxa besides echinoids also possess this unusual Hox gene order.

Research

 * Research Proposal

 * Progress Record

Overview
This area of research involves learning about invertebrate zoology, neuroscience, and molecular lab techniques.

Recommended Coursework

 * Comparative Neuroscience

Other Materials

 * History of Life

 * List of Animal Phyla

Researchers

 * La comadreja formerly AFriedman RESEARCH (talk)


text:
User talk:Acey~enwikiversity

Welcome
 Hello and Welcome to Wikiversity Acey~enwikiversity! You can contact us with questions at the colloquium or me personally when you need help. Please remember to sign and date your finished comments when participating in discussions. The signature icon above the edit window makes it simple. All users are expected to abide by our Privacy, Civility, and the Terms of Use policies while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations.
 * Discuss Wikiversity issues or ask questions at the colloquium.

You do not need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --MathXplore (discuss • contribs) 05:44, 2 March 2024 (UTC)

62.56.124.53 18:30, 30 October 2008 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Acey. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Acey~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:26, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:01, 19 April 2015 (UTC)

title: User talk:Acey~enwikiversity


text:
G-Protein Coupled Receptor Gene Duplication Project

G-Protein Coupled Receptor Gene Duplication Project

Introduction
The broad aim of this project is to analyze genomic sequence data with hidden Markov models (HMMs). The HMM analysis program used is called HMMER and can be downloaded here.

Research

 * Progress Record
 * Accession numbers of the Rhodopsin sequences used by Fredriksson and Schioth (2005) to construct Hidden Markov Models

Overview
This area of research requires training in bioinformatics. As a result, it involves an academic background which is a combination of biology and computer science.

Recommended Coursework

 * Introduction to Computer Science

Researchers

 * La comadreja formerly AFriedman RESEARCH (talk) 23:07, 27 August 2010 (UTC)

Acknowledgments
La comadreja would like to thank the following people:
 * David B., for help in using the HMMER software
 * User:Graeme E. Smith, for finding the locations of complete genomes relevant to the study


text:
Portal:Automotive Engineering/To-do

Kinematics

title: Portal:Automotive Engineering/To-do


text:
Template:Automotive Engineering/To-do

Kinematics

title: Template:Automotive Engineering/To-do


text:
User:La comadreja/Courses

These courses provide some of the background for the research projects. Within each subject, they are arranged roughly in order of increasing difficulty. I am actively involved with course content development and course mentoring.

Computer Science

 * How to Use the Wiki Markup Language, a workshop from WikiEducator
 * Introduction to Computer Science
 * Introduction to Programming
 * Introduction to Programming in Java
 * C++
 * HTML
 * Introductory Discrete Mathematics for Computer Science
 * Discrete Mathematics for Computer Science

Biology

 * Cell Biology
 * Introduction to Genetics
 * Introduction to Non-Genetic Darwinism
 * Introduction to Taxonomy
 * Fundamentals of Neuroscience
 * Comparative Neuroscience

title: User:La comadreja/Courses


text:
Portal:Automotive Engineering/Intro

Automotive/Automobile engineering is a branch of mechanical engineering, incorporating elements of mechanical engineering, electrical engineering, electronic engineering, software engineering and safety engineering as applied to the design, manufacture and operation of automobiles, buses and trucks and their respective engineering subsystems. Used to design automotive vehicles. This also includes working on vehicles.

title: Portal:Automotive Engineering/Intro


text: File:McBeth intro.OGG

title: File:McBeth intro.OGG


text:
Organic chemistry/Naming

There are thousands of known organic compounds and millions of others yet to be discovered. Because of the wide variety of compounds a naming system was developed by International Union of Pure and Applied Chemistry (IUPAC) to give a unique descriptive name to each compound. In this system the name of the compound directly shows its structure through an elaborate system of suffixes and prefixes. Common names are used for compounds that were either around before IUPAC or who have a more widely known name than the IUPAC. For example, trichloromethane is IUPAC name for chloroform, though it is almost universally referred to by students and scientists alike as chloroform.

Basic Nomenclature
The IUPAC nomenclature of organic molecules is based on the longest carbon chain in the molecule. For alliphatics, which are hydrocarbons containing chains of C, the following prefixes are used:

Alkanes
Alkanes are a compound in which the carbon backbone contains only single bonds. They are named according to the number of carbons they have. Alkanes combust to produce water and carbon dioxide in air. Some familiar alkanes are methane (natural gas), propane (gas used in homes), and butane (blue-flamed cigarette lighters). Alkanes containing four carbons or more can form different chain isomers; that is, they can become "branched," forming, for example, iso-butane.

General Alkane Formula: CnH2n+2

Suffix: "ane"

Example: CH3CH2CH3 - Propane

To name branched molecules, find the longest continuous chain first. This chain forms the base name. Number the carbon atoms, starting at the end closest to the branching. Locate the branch by the number of the C stem to which it is attached on the main chain. Name the branch, which is called an alkyl group, by using the chart above. If more than one of the same alkyl groups are present as branches, the number of these branches is indicated by prefix multipliers (di-, tri-, etc.). The location of each of these alkyl groups is indicated by a number, using the lowest combination possible.

Example: CH3CH(CH3)CH3
 * Name As: 2-methyl propane or common name is iso-butane

Example: CH3CH(CH3)CH2CH(CH3)CH3
 * Names As: 2,4-dimethyl pentane, as we start naming from the end closest to a branch (in this case, the back end)

For IUPAC names, the following table of prefixes is important. Each prefix corresponds to a different number of carbons.

The simplest systematic name is given to straight chain alkanes. An alkane is a compound containing solely hydrogen and carbon, held together only by single bonds. Straight chain implies that there are no branches or rings in the structure. To name simple alkanes, you need to know the appropriate base part for the number of carbons in a continuous chain. Say for example, you had the following molecule:

This molecule contains a continuous chain of five carbons. By referencing the chart, we see that five carbons merits the prefix "pent-". Because this molecule contains only carbon and hydrogen held together by single bonds, it is an alkane, which merits the suffix "-ane". Therefore the name of this molecule is pentane.

Alkenes
Alkenes, like Alkanes, have carbon backbones but contain at least one carbon-carbon double bond. Double bonds are reactive and can be reduced to single bonds using hydrogen gas, pressure, and a catalyst, such as Pd or Pt. Alternatively, Alkenes can be reduced to Alkanes by reacting with Hydrogen gas over a Nickel catalyst. Nickel acts as a heterogeneous surface catalyst, weakening the H-H bonds in hydrogen, thus speeding up the reaction. This also leads to syn addition, where both of the hydrogens are added to the same side of the double bond.

General Alkene Formula: CnH2n

Suffix: "ene"

Example: CH2CH2 - Ethene

See the alkane naming scheme for branched molecules, as the same principal applies to alkenes we simply retain the "ene" rather than the "ane" ending. However, the placement of the double bond matters when naming.

Example: CH3(CH)2CH3 - 2-butene

Example: CH2CHCH2CH3 - 1-butene

IUPAC rules actually specify that alkenes should be named as but-1-ene, but-2-ene, and so on; but the practice of putting the number before the name remains widespread in the US.

To figure out which is which, number the carbon backbone starting with end closest to the double bond. Whichever number carbon the double bond starts on is the prefix which you add to the name.

Another problem in naming is when there is more than one double bond.

Example: CH2(CH)2CH2 - 1,3-butadiene

Example: CH2CCCH2 - 1,2,3-butatriene

To name these types of compounds, again, number the carbons starting from the end closest to a double bond. The number prefixes will correspond to the carbons at which each double bond starts, and an additional suffix (di, tri, tetra, etc.) will be added in front of the 'ene' suffix.

Alkynes
Alkynes have carbon-carbon triple bonds. Like alkenes, Alkynes are reactive and can be brought down to either double or single bonds. If excess hydrogen is used, and they are allowed to react, they will become single bonds. If a poison catalyst (such as Pd/CO3, palladium with carbonate) is used instead of a regular catalyst, the alkynes will be selectively reduced to alkenes.

General Alkyne Formula: CnH2n-2

Suffix: "yne"

Example: CHCH - Ethyne

See the alkane naming scheme for branched molecules, as the same principal applies to alkynes we simply retain the "yne" rather than the "ane" ending.

Alcohols
General Alcohol Formula: CnH2n+1OH

Suffix: "ol"

Example: CH3CH2OH - Ethanol (IUPAC)

Example: CH3CH2OH - Ethyl Alcohol (Common Name)

Example: CH3CH2CH2OH - 1-Propanol (IUPAC)

Example: CH3CH2CH2OH - Propyl Alcohol (Common)

Example: CH3CHOHCH3 - Propan-2-ol (IUPAC)

Example: CH3CHOHCH3 - Isopropyl Alcohol (Common)

The numbered prefix is indicative of an isomer, not the location of a double bond.

Ketones
Ketones are carbonyl compounds. Thus they contain the C=O group. This group is responsible for the typical peak present in the mass spectrometer profiles for carbonyl compounds. The suffix "-one" is added to the alkane name to indicate a ketone. Ketones, if formed from alcohols, must be formed from secondary alcohols as you need to have two groups either side of the "carbonyl" carbon.

General Ketone Formula: CnH(2(n-1)+2)O

Suffix: "one"

Examples of ketones include:

CH3COCH3 Name: propanone. This is the lowest member of the series.

CH3COCH2CH3 Name: butanone.

CH3CH2COCH2CH3 Name: pentan-3-one. The "-3-" indicates the position of the C=O (carbonyl) group. It shows that the carbonyl group is present on the third carbon atom in the chain. Such naming is essential in ketones (except propanone and butanone). For example, an isomer of pentan-3-one is pentan-2-one, which has a different structural formula.

Ethers
In Ethers the functional group is -O-. The General Formula of ethers is R-O-R or R-O-R'. R-O-R are simple ethers. Whereas R-O-R' represents mixed ethers. R and R' are different alkyl groups. According to IUPAC they are called Alkoxy Alkanes (O-R is alkoxy group).

For Example,

CH3OCH3 is a simple ether whose common name is Dimethyl Ether and the IUPAC name is Methoxy Methane.

C2H5OCH3 is a mixed ether whose common name is Ethyl Methyl ether and the IUPAC name is Methoxy Ethane. You should name with the lowest group first: methyl, then ethyl, etc... which explains the IUPAC difference in naming, as the methyl come before the ethyl.

Esters
Carbon Compounds consisting of a carbonyl carbon adjacent to an ether (-COO-). Esters are often characterized by a unique smell or taste. Esters are used by manufacturers due to their unique smells and tastes. Suffix is -oate.

Aldehydes
Aldehydes are carbonyl compounds with a C=O and a H bonded to the same carbon. The suffix attached to an aldehyde is "al". The quickest way to form an aldehyde is to oxidise a primary alcohol using potassium dichromate.

Suffix: "al"

General formula: CnH2n+1CHO

Functional Group: CHO

Examples of aldehydes:

CH2O Formaldehyde (common name) or Methanal (IUPAC), is the simplest aldehyde.

CH3CH2CHO Propanal

CH3CH2CH2CHOButanal

Carboxylic Acids
Carboxylic acids are carbonyl compounds containing a C=O and an -OH group bonded to the same carbon. They can be obtained by oxidising a aldehyde or a primary alcohol using potassium dichromate (K2Cr2O7) with concentrated sulphuric acid.

Suffix: "oic acid"

General formula: CnH2n-1COOH

Functional Group: COOH

Examples of carboxylic acids:

CH3COOH              Ethanoic Acid (usually called Acetic acid)

CH3CH2COOH Propanoic Acid

Benzenes
Benzenes, part of the aromatic group, are all based on the benzene ring (C6H6). In it, there are three double bonds that create a resonance structure. Resonance theory states that we can draw two different structures for benzene where the double bonds are in different places, but neither structure adequately represent the molecule, so instead we think of the molecule as a resonance hybrid, a combination of the two. Benzene is incredibly stable, and is also planar. Although the term aromatic should refer to a smell, is really refering to a Huckel number of pi electrons.

General formula: C6H6

Derivatives of benzene are named by numbering the C atoms in a clockwise manner, starting at one for the first group. Alkyl monosubstituted benzenes are names as alkyl benzenes.

Example: C6H5CH3 - Methylbenzene, a common name for methylbenzene is Toluene.

If the alkyl branch is attached to the benzene ring by a carbon other than an end carbon, the benzene is then considered a branch of the carbon chain, and is called phenyl.

Phenyls
C6H5

Allyls
Allyl as a group is represented as -CH2-CH=CH2. Allyl is considered to be a common name. The term allyl deals with a carbon attached to a carbon in a double bond (that is sp2 hybridized). This is not to be confused with allylic as allylic refers to just a carbon attached to the carbon in the double bond, not the entire group of three carbons. An example of Allyl groups with its corresponding systematic names is

H2C=CHCH2Br  allyl bromide or 3-bromopropene.

Note it is allyl because the bromine is attached to the CH2 carbon which is sp2 hybridized, double bonded.

Vinyls
The vinyl group is similar to the allyl group. However, vinyl refers to a group attached directly to an sp2 hybridized (double bonded) carbon. The vinyl group is represented as -CH=CH2. This is not to be confused with vinyllic carbons which are only the carbons participating in the double bond, not the entire group. An example of naming a vinyl compound with its systematic name is shown below

H2C=CHCl   vinyl chloride or chloroethene

To relate vinylic and allylic (not vinyl and allyl) consider the following part of a carbon chain:

R-CH2-CH=CH-CH2-R

The first and last carbons shown are allylic (CH2), the middle two carbons participating in the double bond are vinylic (CH).

title: Organic chemistry/Naming