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User:Eas4200c.f08.carbon.orear/HW7/LectureNotes

=Homework 7 Lecture Notes=

Roadmap
We have developed a Mini-plan as follows:

For single-cell sections, we have two classifications. There are sections with and without stringers.

Without stringers, the analysis can be summarized (and will be reviewed in more detail as we continue) that:

Shear flow is a constant for an arbitrarily shaped single-cell design. We want to know if this can resist the shear force Vx. From our analysis, the z-resultant of shear flow will equal zero.

With stringers, q (shear flow) in every section is not the same, though it is constant within each individual skin panel. In this way, it is "piecewise constant." We can superimpose the results due to their linearity, and our z-resultants will be nonzero.

Open C-section

Mean Value Theorem (MVT)

$$\int_A y dA = \bar y \int_A dA = \bar y A$$

$$\int_A z dA = \bar z \int_A dA = \bar z A$$

Where $$A=\sum_{i=1}^4 A_i$$

Equation 5.5 (from Lecture 34-3) can be particularized to: $$q(s) = (k_{yz}Q_z-k_zQ_y)V_z$$ Since all components on the RHS of the equation are independent of s. Note that for $$Q_z, Q_y$$ all areas are concentrated on the stringer.

=>shear flow q(s) is constant between 2 stringers, but q(s) would jump when crossing a stringer.

Steps to solve problems:

1) Find $$(\bar y, \bar z)$$

2) Find $$I_y, I_z, I_{yz}$$

3) Find $$k_y, k_z, k_{yz}$$

4) Follow path "s" to find $$q_{12}, q_{23}, q_{34}$$

$$q_{12}=(k_{yz}Q_z^{12}-k_zQ_y^{12})V_z$$

$$Q_z^{12} = y_1A_1$$, where y is the y-coordinate of stringer 1 with respect to the origin at C.

$$Q_y^{12} = z_1A_1$$

 * Note: $$Q_z^{23} = y_1A_1+y_2A_2$$,

Single Cells With and Without Stringers
Single Cell without Stringers where q is a constant shear flow

Q: Can the set up shown above resist shear in the z direction? A: No

$$ R^z=R_{AB}^z + R_{BA}^z$$ Where $$R^z$$ is the total resultant in the z direction, $$R_{AB}^z$$ is the resultant of q in AB, and $$R_{BA}^z$$ is the total resultant of q in BA.

$$R_{AB}^z = -qA'B' = -R_{BA}^z$$ $$R^z=0$$

Single Cells with Stringers (neglect contribution of web to bending)

$$ q_{12}, q_{23}, q_{31}$$ do not equal each other, while $$q_{ij}(s)$$ is constant within each skin panel. $$q(s)$$ is a piecewise function with respect to s. The function is assumed to be piecewise because we neglect the contribution of stringers to bending.

The presence of stringers in the cell makes the shear flow non constant.

Principle of Superpostion due to Linearity

For this you add q as a contant, like in the figure for a single cell without stringers, to all the different q's in the figure for a single cell with stringers.

$$q_{12}=q +\tilde q_{12}$$ $$q_{23}=q +\tilde q_{23}$$ $$q_{ij}=q +\tilde q_{ij}$$ - this is the general form

Analysis Algorythem Method: 1) Data: $$ V_y, V_z$$ 2) Solve problem for $$\tilde q_{12}, \tilde q_{23} (\tilde q_{31}=0)$$ 3) Moment equation: Take the moment about any point in plane (x,y) 3.1) Superpostion $$q_{ij}=q + \tilde q_{ij}$$ 3.2) Select point O in plane (x,y)

Solving Problem 2 Using Euler Cut Principle

$$\sum F_{x} = 0$$
 * $$=\int_{A_3}^{} [\sigma_{xx} (x+dx) - \sigma_{xx} (x)] dA_3$$
 * $$[-\tilde{q}_{23} - \tilde{q}_{43} + \tilde{q}_{31}] dx$$

Taking the Taylor Series of $$\displaystyle [\sigma_{xx} (x+dx) - \sigma_{xx} (x)]$$ is $$\frac{d \sigma_{xx}}{dx} + h.o.t.$$
 * $$\tilde{q}_{31}=\tilde{q}_{23} + \tilde{q}_{13} + q^{(3)}$$

$$q^{(3)} = - \int_{A_3} \frac{d \sigma_{xx}}{dx} dA_3 $$ is the contribution to shear flow by stringer 3.

Recall $$V_y=\frac{dM_z}{dx}, V_z=\frac{dM_y}{dx}$$

From Lecture 34-3: $$q^{(3)}=-(k_yV_y-k_{yz}V_z)Q_z^{(3)}-(k_zV_z-k_{yz}V_y)Q_y^{(3)}$$

where $$Q_z^{(3)}=\int_{A_3}ydA_3, Q_y^{(3)}=\int_{A_3}zdA_3$$

Stringer 2: $$\tilde q_{23}=\tilde q_{12}-\tilde q_{24}+q^{(3)} = q^{(3)}$$

$$q^{(2)}$$ is computed as in (1).

$$Q_z^{(2)}=y_2A_2$$

$$Q_y^{(2)}=z_2A_2$$

Stringer 4: $$\tilde q_{43}=\tilde q_{24}-\tilde q_{41}+q^{(4)} = q^{(4)}$$

$$Q_{y}^{(3)} = \int_{A_{3}}^{}{zdA_{3}}$$

$$Q_{y}^{(2)} = \int_{A_{2}}^{}{zdA_{2}}$$

Resuperposition of Problem 2
The basic formula for shear flow along the skin:

$$q_{ij} = \tilde{q}_{ij} + q_{k}$$

Following this formula, several equations can be found:

 * $$q_{12} = \tilde{q}_{12} + q_{1}$$, where $$\tilde{q}_{12}$$ is known as zero and $$q_{1}$$ is unknown

 * $$q_{23} = \tilde{q}_{23} + q_{1} - q_{2}$$

 * $$q_{31} = \tilde{q}_{31} + q_{1} - q_{3}$$

 * $$q_{24} = \tilde{q}_{24} + q_{2}$$

 * $$q_{43} = \tilde{q}_{43} + q_{3} - q_{4}$$

 * $$q_{41} = \tilde{q}_{41} + q_{4}$$

Looking at all these equations, there are 3 unknowns, and thus we need 3 equations.

1) The first equation would be one equation; finding the moments of Vy and Vz, and $${q_{12}, q_{23}, q_{31}, q_{24}, q_{43}, q_{41}}$$ about any convenient point.  The point about which the moment is taken should be where the lines of action of $$V_{y}$$ and $$V_{z}$$ intersect.

Because each of the three sections are combined, if one section twists by a certain angle, then the other two must also twist by the same angle. Therefore,

2) $$\theta _{1} = \theta _{2}$$ 3) $$\theta _{2} = \theta _{3}$$

Solving Problem 2 for each cell
To solve problem 2, follow each step for each cell:

 * Follow the path "$$s_{i}$$"
 * Write the equilibrium of each stringer on the path "$$s_{i}$$"

There are two ways to write the equilibrium:

1) Method shown in previous section, using FBD 2) The consequence of the first method is shown below:

The equilibrium equation is: $$\tilde{q}_{j6} = \tilde{q}_{3j} - \tilde{q}_{j5} - \tilde{q}_{j7}  q^{j} $$

The equilibrium equation is: $$-\tilde{q}_{6j} = \tilde{q}_{3j} - \tilde{q}_{j5} - \tilde{q}_{j7}  q^{j} $$

Note:

Consider what would happen if the cell walls were cut such that one stringer was isolated. Theoretically, it cannot be solved.

$$\tilde{q}_{23} = \tilde{q}_{31} = \tilde{q}_{34} = 0$$

$$\tilde{q}_{31} = \tilde{q}_{23} - \tilde{q}_{34} + q^{(3)}$$

This means that $$q^{(3)} = 0$$, but this is not true!

A second example is shown below:

$$\tilde{q}_{24} = \tilde{q}_{23} = \tilde{q}_{12} = 0$$

$$\tilde{q}_{23} = \tilde{q}_{12} - \tilde{q}_{24} + q^{(2)}$$

This shows that $$q^{(2)} = 0$$ which also is not true!

Using the peace sign, which was popular in the '60s:

 * $$\tilde{q}_{15} = \tilde{q}_{54} = \tilde{q}_{42} = \tilde{q}_{21} = 0$$

$$\tilde{q}_{15} = \tilde{q}_{31} + \tilde{q}_{21} + q^{1}$$

 * $$\Rightarrow \tilde{q}_{31} = -q^{(1)}$$

 * $$\tilde{q}_{53} = q^{(5)}$$

 * $$\tilde{q}_{43} = q^{(4)}$$

 * $$\tilde{q}_{23} = q^{(2)}$$

 * $$\tilde{q}_{15} = \tilde{q}_{31} + \tilde{q}_{21} + q^{1}$$

$$\tilde{q}_{31} = \tilde{q}_{53} + \tilde{q}_{43} + \tilde{q}_{23} + q^{(3)}$$

$$\Rightarrow \tilde{q}_{31} = q^{(5)} + q^{(4)} + q^{(2)} - q^{(1)}$$

There are four unknowns, so there will need to be four equations described in the previous section:

1) Moment equation

2) $$\theta _{1} = \theta _{2}$$

3) $$\theta _{2} = \theta _{3}$$

4) $$\theta _{3} = \theta _{4}$$

A second method of cuts for the peace sign is shown below:

$$\tilde{q}_{53} = \tilde{q}_{43} = \tilde{q}_{23} = \tilde{q}_{31} = 0$$

$$\tilde{q}_{31} = \tilde{q}_{53} + \tilde{q}_{43} + \tilde{q}_{23} + q^{(3)}$$

This leads to $$q^{(3)} = 0$$ which is not true.

The "Not Possible" zero result
Isolation of stringer 3 would tell us that

0 = q(1) + q(2) + q(4)

and this result is NOT POSSIBLE. The actual result is:

0 = q(1) + q(2) + q(3) + q(4)

0 = Σq(e) where the summation is from 1 to 4.

We then see that

q(e) = nzQz(e) + nyQy(e)

nz = -(kyVy-kyzVz)

ny = -(kzVz-kyzVy)

Shear Buckling
Plotting the buckling shape can be done by the following steps. This will also express the coefficients $$\begin{Bmatrix} C_{22} & C_{13} & C_{31} & C_{33} \end{Bmatrix}$$ in terms of $$\displaystyle C_{11}$$ with the aspect ratio $$\vartheta = 1.5$$. Equations from the wiki article by Dr. Vu-Quoc on plate buckling will be used in the following steps.

1. Find $$\displaystyle \lambda$$ for $$\vartheta = 1.5$$

$$  \displaystyle \lambda =  \left[ \frac{\vartheta^4}{81 (1 + \vartheta^2)^4} \left\{ 1	 +	 \frac{81}{625} +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{1 + 9 \vartheta^2}	 \right)^2 +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{9 + \vartheta^2}	 \right)^2 \right\} \right]^{1/2} $$ (Equation 30 for wiki article)

Substituting yields $$\displaystyle \lambda = .028753$$

2. Evaluate numerically $$\displaystyle \mathbf K _{5 \times 5}$$ using Equation 26.

$$  \displaystyle \left[ \begin{array}{lllll} \frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 \frac{4}{9} &	 0	 &	 0	 &	 0	 \\	 \frac{4}{9} &	 \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 - \frac{4}{5} &	 - \frac{4}{5} &	 \frac{36}{25} \\	 0	 &	 - \frac{4}{5} &	 \frac{\lambda (1 + 9 \vartheta^2)^2}{\vartheta^2} &	 0	 &	 0	 \\	 0	 &	 - \frac{4}{5} &	 0	 &	 \frac{\lambda (9 + \vartheta^2)^2}{\vartheta^2} &	 0	 \\	 0	 &	 \frac{36}{25} &	 0	 &	 0	 &	 \frac{\lambda (9 + 9 \vartheta^2)^2}{\vartheta^2} \end{array} \right] \left\{ \begin{array}{l} C_{11} \\	 C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} 0	 \\	 0	 \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

3. Using indicial notation, $$\displaystyle \mathbf K = [\mathbf K _{ij}]$$. Truncating $$\displaystyle \mathbf K _{5 \times 5}$$ into a $$\displaystyle \mathbf K _{4 \times 4}$$ using $$\displaystyle \mathbf K _{22}$$ through $$\displaystyle \mathbf K _{55}$$, or the lower right hand corner of $$\displaystyle \mathbf K _{5 \times 5}$$, yields

$$\displaystyle \mathbf \bar K _{4 \times 4} = \left[ \begin{array}{llll} \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 - \frac{4}{5} &	 - \frac{4}{5} &	 \frac{36}{25} \\	 - \frac{4}{5} &	 \frac{\lambda (1 + 9 \vartheta^2)^2}{\vartheta^2} &	 0	 &	 0	 \\	 - \frac{4}{5} &	 0	 &	 \frac{\lambda (9 + \vartheta^2)^2}{\vartheta^2} &	 0	 \\	 \frac{36}{25} &	 0	 &	 0	 &	 \frac{\lambda (9 + 9 \vartheta^2)^2}{\vartheta^2} \end{array} \right] \left\{ \begin{array}{l} C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} \frac{-4 C_{11}}{9} \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

4. Solving $$\begin{Bmatrix} C_{22} & C_{13} & C_{31} & C_{33} \end{Bmatrix}$$ in terms of $$\displaystyle C_{11}$$, with $$\displaystyle \mathbf K^{-1}$$ being the inverse of $$\displaystyle \mathbf K$$

$$\begin{Bmatrix} C_{22} \\ C_{13} \\ C_{31} \\ C_{33} \end{Bmatrix} = K^{-1} \begin{Bmatrix} -\frac{4}{9} C_{11} \\ 0 \\ 0 \\ 0 \end{Bmatrix}$$

$$U_{z} = C_{11} sin(\frac{\pi x}{a}) * sin (\frac{\pi y}{b}) + C_{22} sin(\frac{2 \pi x}{a}) * sin (\frac{2 \pi y}{b}) $$
 * $$ + C_{31} sin(\frac{\pi x}{a}) * sin (\frac{3 \pi y}{b}) + C_{31} sin(\frac{3\pi x}{a}) * sin (\frac{\pi y}{b}):$$
 * $$+ C_{33} sin(\frac{3\pi x}{a}) * sin (\frac{3\pi y}{b})$$

Finally, set $$\displaystyle C_{11} = 1$$, and then plot $$\displaystyle U_{z} (x,y)$$

Continuing lecture 39-2, Part 2

Stringer 1.

The equilibrium equation, or the FBD method, yields the equation for stringer 1 to be

$$\tilde{q}_{12} = \tilde{q}_{31} - \tilde{q}_{41} - q^{(1)} $$

With $$\tilde{q}_{12}$$ and $$\tilde{q}_{41}$$ being unknown, and $$\tilde{q}_{31} = 0$$, due to a cut in the panel between stringer 1 and stringer 3.

Canceling out $$\tilde{q}_{31}$$,

$$\tilde{q}_{12} = -\tilde{q}_{41} - q^{(1)} $$ (Eq. 1)

Stringer 2

Again the equilibrium equation about stringer 2

$$\tilde{q}_{24} = \tilde{q}_{12} - \tilde{q}_{23} - q^{(2)} $$

This time with $$\tilde{q}_{23} = 0$$ yielding

$$\tilde{q}_{24} = \tilde{q}_{12} - q^{(2)} $$ (Eq. 2.)

With $$\tilde{q}_{24} $$ and $$ \tilde{q}_{12} $$ being unknown.

Stringer 4

The equilibrium equation about stringer 4 gives

$$\tilde{q}_{41} = \tilde{q}_{24} - \tilde{q}_{34} - q^{(4)} $$ (Eq. 3)

Substituting Eq. 2 into Eq. 3 and canceling out $$\tilde{q}_{34} = 0$$ due to a cut.

$$\tilde{q}_{41} = (\tilde{q}_{41} + q^{(1)}) + q^{(2)} +  q^{(4)} $$

Subtracting $$\tilde{q}_{41}$$ from both sides

$$\displaystyle q^{(1)} + q^{(2)} +   q^{(4)} = 0 $$

Bidirectional Bending
The revisiting of this topic has already been covered in Team Carbon's previous HW.

Wiki Comparison
In concluding this course, our group, Team Carbon, has unanimously concluded that the Wikiversity/Mediawiki work we have done far outweighs a system like e-Learning.

By best summary, e-learning is a tool to interface student and teacher. Mediawiki is a tool to interface student to student, team to team, instructor to class, and the world can view and benefit from our work in "the spirit of copyleft."

Mediawiki excels in the following sense: It is a teamwork based cooperative communication and learning tool. The capability of logging all changes in history prevents vandalism of others' work. Account privacy is maintained, yet work can be shared! Furthermore, any individual can challenge and question and correct the mistakes of another!

E-learning loses work year after year, semester after semester, and Mediawiki retains this work for the future and for future students. Mediawiki can be useful in our employment and in industry communication. Other media can be inserted to our work. And there are many more benefits out there for students to yet discover.

Team Carbon has chosen the following 5 best features of the Wiki-system for this class.

1) Saving work for future reference.

2) Help for employment/resume building.

3) Individuals can pace their work and contribute in multiple manners.

4) Anti-vandalism history log.

5) Example set for future students to build on.


text:
User:Eas4200c.f08.carbon.orear/HW7/Contributors

 * Jeff Wright Eas4200c.f08.carbon.w 03:56, 8 December 2008 (UTC)
 * Jeff O'Rear Eas4200c.f08.carbon.orear 17:57, 8 December 2008 (UTC)
 * Aaron Clausen Eas4200c.f08.carbon.clausen 17:58, 8 December 2008 (UTC)
 * Brian Taylor Eas4200c.f08.carbon.taylor 06:55, 9 December 2008 (UTC)
 * Beau Guidry Eas4200c.f08.carbon.guidry 15:27, 9 December 2008 (UTC)
 * Abby Booth Eas4200c.f08.carbon.booth 18:36, 9 December 2008 (UTC)

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


text:
User:Eas4200c.f08.carbon.orear/HW7/MATLAB

=MATLAB Code=

Mon, 24 Nov 2008, 07:21:38 EST Dear class (and TAs):

Again, the goal of HW7 is the NACA 2415 airfoil. This time, you will analyze the NACA airfoil to find the shear flows in its panels (skin and spar webs) when the airfoil is subjected to transverse shear forces in both y and z directions, as a result of the drag and the lift forces on the airfoil.

See for example a figure of the 3-cell NACA airfoil with all defined points by Team Aero_Eelman at:

http://en.wikiversity.org/w/index.php?title=User:Eas4200c.f08.aero.e/Week_8&oldid=356604

Now, consider the following shear forces:

 * horizontal shear force $$V_y = 10000 N$$, applied at point L (leading edge)

 * transverse shear force $$V_z = 5000 N$$, applied at point B

=Part I: Single-cell NACA airfoil=

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

Stringer Centroid = (0, 0.2083 , 0.0081) m

 * Area_B = 0.0002 m2
 * Area_E = 0.0002 m2
 * Area_F = 0.0001 m2
 * Area_H = 0.0001 m2

Stringer locations
 * B =    0    0.1250    0.0458
 * F =    0    0.3750    0.0263
 * H =    0    0.3750   -0.0132
 * E =    0    0.1250   -0.0285

In your shear flow analysis, using the method explained in class(see also the examples in the book), solve problem P2 by using the following single cut:

 * cut off the skin at the trailing edge point T

The path "s" to use in your solution of problem P2 is the same path of the shear flow as shown in red in the above figure, starting from the trailing edge point T, following the upper skin to the leading edge point L, then wrap around the airfoil to the lower skin to end at the trailing edge.

For the moment equation, take the moment of all shear forces and shear flows about point D on the spar web BDE.

Find the final true shear flow in each skin panel.

$$q'_{TF} = 0$$

$$q'_{FB} = q'_{TF} + q'_{F} = 0 + (V_z * Q_fy)/(I_y) + (V_y * Q_fz)/(I_z) = -(10000N * 0.0001 m2 *(0.2083-0.1250))/(6.3913e-007) - (5000N *  0.0001 m2 * (0.0458 - 0.0081))/( 8.3333e-006) = -1.3260e+005$$

$$q'_{FB} = -1.3260e+005$$

$$q'_{BL} = q'_{FB} + q'_{B} = -1.3260e+005 + (V_z * Q_by)/(I_y) + (V_y * Q_bz)/(I_z) = -(10000N * 0.0002 m2 *(0.3750-0.2083))/(6.3913e-007) - (5000N *  0.0002 m2 * (0.0263 - 0.0081))/( 8.3333e-006) = -1.3260e+005 +  -5.2383e+005 =  -6.5643e+005$$

$$q'_{LE} = q'_{BL} = -6.5643e+005$$

$$q'_{EH} = q'_{LE} + q'_{E} = -6.5643e+005 + (V_z * Q_by)/(I_y) + (V_y * Q_bz)/(I_z) = -(10000N * 0.0002 m2 *(0.3750-0.2083))/(6.3913e-007) - (5000N *  0.0002 m2 * (0.0081 + 0.0132))/( 8.3333e-006) = -6.5643e+005 + + 5.2420e+005 = -1.3223e+005$$

$$q'_{HT} = q'_{EH} + q'_{H} = -1.3223e+005 + (V_z * Q_by)/(I_y) + (V_y * Q_bz)/(I_z) = -(10000N * 0.0001 m2 *(0.2083-0.1250))/(6.3913e-007) - (5000N *  0.0001 m2 * (0.0081 + 0.0285))/( 8.3333e-006) = -1.3223e+005 + 1.3253e+005 = 0.0030+005$$

We see that this result will be the same if we insert more cells but make cuts in those cells! The only change will occur when those hypothetical cuts are replaced by the complete skin.

=Part II: 3-cell NACA airfoil=

In your shear flow analysis, using the method explained in class (see also the examples in the book), solve problem P2 by using the following triple cuts:

 * cut off the skin at the trailing edge point T

 * cut off the spar web FH

 * cut off the spar web BE

The path "s" to use in your solution of problem P2 is the same path of the shear flow as shown in red in the above figure, starting from the trailing edge point T, following the upper skin to the leading edge point L, then wrap around the airfoil to the lower skin to end at the trailing edge.

For the moment equation, take the moment of all shear forces and shear flows about point D on the spar web BDE.

Find the final true shear flow in each skin panel and spar web.

$$q'_{TF} = 0$$

$$q'_{FB} = q'_{TF} + q'_{F} = 0 + (V_z * Q_fy)/(I_y) + (V_y * Q_fz)/(I_z) = -(10000N * 0.0001 m2 *(0.2083-0.1250))/(6.3913e-007) - (5000N *  0.0001 m2 * (0.0458 - 0.0081))/( 8.3333e-006) = -1.3260e+005$$

$$q'_{FB} = -1.3260e+005$$

$$q'_{BL} = q'_{FB} + q'_{B} = -1.3260e+005 + (V_z * Q_by)/(I_y) + (V_y * Q_bz)/(I_z) = -(10000N * 0.0002 m2 *(0.3750-0.2083))/(6.3913e-007) - (5000N *  0.0002 m2 * (0.0263 - 0.0081))/( 8.3333e-006) = -1.3260e+005 +  -5.2383e+005 =  -6.5643e+005$$

$$q'_{LE} = q'_{BL} = -6.5643e+005$$

$$q'_{EH} = q'_{LE} + q'_{E} = -6.5643e+005 + (V_z * Q_by)/(I_y) + (V_y * Q_bz)/(I_z) = -(10000N * 0.0002 m2 *(0.3750-0.2083))/(6.3913e-007) - (5000N *  0.0002 m2 * (0.0081 + 0.0132))/( 8.3333e-006) = -6.5643e+005 + + 5.2420e+005 = -1.3223e+005$$

$$q'_{HT} = q'_{EH} + q'_{H} = -1.3223e+005 + (V_z * Q_by)/(I_y) + (V_y * Q_bz)/(I_z) = -(10000N * 0.0001 m2 *(0.2083-0.1250))/(6.3913e-007) - (5000N *  0.0001 m2 * (0.0081 + 0.0285))/( 8.3333e-006) = -1.3223e+005 + 1.3253e+005 = 0.0030+005$$

Plot the perspective figure of the shear buckling shape u_z using 5 buckling mode shapes for the plate aspect ratio a/b = 1.5 as explained in class on Fri, 5 Dec 08. The derivation of the transverse buckling shape u_z and the perspective plot form an important part of HW7.

=Buckling analysis= Buckling analysis of skin panels FB and EH due to shear flows.

 * refer to my plate buckling wiki page at http://en.wikiversity.org/wiki/User:Eas4200c.f08/Plate_buckling, particularly the section on shear loading. Do the following:

 * consider Eq.(27); take the determinant of matrix K (2x2), and verify Eq.(28) for lambda

 * compute the lowest critical buckling stress $$(\sigma_{xy})_{cr}^{2,\star}$$ shown in Eq.(29) for a range of aspect ratio a/b from 0.5 to 2.

 * compute the more accurate critical buckling stress $$(\sigma_{xy})_{cr}^{5,\star}$$ shown in Eq.(31) for the same range of aspect ratio a/b from 0.5 to 2.

 * Plot the values of $$(\sigma_{xy})_{cr}^{2,\star}$$ and $$(\sigma_{xy})_{cr}^{5,\star}$$ versus the aspect ratio a/b from 0.5 to 2.

 * the above is for simply-supported plate; for clamped plate, report on the same plot above the 3 points corresponding to K = 1, 2, infinity. For K=infinity, also report the experimental data point with K=4.1.

For each skin panel (FB or EH), do the following:

 * find the curvilinear length of the skin panel by adding the infinitesimal length as you integrate along the skin panel (you may have done this computation in HW6, but verify to weed out possible theoretical or coding bugs)

 * compute the shear stress from the shear flow obtained for the skin panel, and compare the skin panel shear stress to the theoretical buckling shear stress $$(\sigma_{xy})_{cr}^{5,\star}$$ and draw some conclusion.


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User:Eml4500.f08.wiki1.aguilar/TeamHW7

=Team Homework 7=

Wikiversity vs. E-Learning - Wiki1 Group Verdict
While both the Wikiversity and the E-Learning systems are very useful, each was designed for a specific purpose in mind. While E-Learning was designed to facilitate the communication between instructors and students, wikis are more designed to allow groups and organizations to easily post, edit, and share information. Because they are used with different purposes in mind, they are difficult to compare.

While E-Learning is in no way perfect, it does facilitate communication between professors and students when used correctly. By using E-learning as the standard and trying to familiarize all the professors with it, it bypasses the need for teachers to create a website to post information that needs to be accessible to all the students. We are not familiar with the back end of E-Learning, but we are sure that using E-learning to post files and information is easier than teaching professors how to build their own websites with attached files and documents. E-Learning does allows easy methods for professors to do a variety of different things. For example, E-Learning allows professors and TA's to upload grades and then for the various assignments and exams, students can view their grades at any time throughout the semester. Professors can also allow it to provide statistics on the assignment or exam like the average, the high and low scores, as well as histograms showing the grade distributions. E-Learning allows the professors to easily post announcements and documents as well, like homework solution PDFs and course syllabi. There is a feature for a discussion board which could easily be used to allow students to ask other students and TAs various questions, and there is also a calendar to allow various important dates to be posted. This are just some of the features provided for by the E-Learning system, many of which cannot be adequately provided for through Wikiversity.

It becomes more of a question what is it that we are trying to do? If, for example, the end goal is to provide a large, multimedia report that is to be worked on by an entire group, then Wikiversity does indeed provide a very good solution for this. But E-Learning is currently not meant to serve this function. For the purposes of this class, "E-Learning" is more of an indirect method for group collaboration. If you had to use E-Learning to do this, the process would vary group by group. Groups could come up with various ways to create these reports. Some groups might use sharing features of Google documents to write up the report and allow all members of the group to contribute text, images and various other elements to the report. This could then be saved as either a PDF or a Word document which could then be submitted on E-Learning. The advantage to this method would be that there isn't any "coding" to learn so that would minimize some of the learning curve, and it would also protect reports from being viewed by others outside the group. The downside to this would be that reports from group to group would be much less streamlined, and having to save equations as images could become quite a hassle. Another alternative would be to use Word to allow groups to create the report. Word 2007 does contain a much improved equation editor and almost everyone uses Microsoft Word already. E-Learning could be set up to allow groups members to upload their parts, and then someone at the end would combine into a final report. This method would have similar pros and cons to the Google Shared documents method.

Clearly, for a group to put together an entire report like what was required of us, Wikiversity does seem to provide a very good solution, keeping in mind that no solution will be perfect. Wikiversity allows for students to be introduced to wikis, something that is becoming much more widespread throughout the "real world." These skills are much more likely to becomes valuable later on down the road. Wikiversity is major project and it's performance will not be affected by the activity of our groups even under peak conditions. The downside is that the Wikiversity approach does, however, lend itself to the possibility of being vandalized or copied by members from other groups.

As a group, we think that for our class purposes, the Wikiversity method could be improved. This could be done by cutting out much of the tedious busy work that tends to be more "filler" within our reports. While each homework report usually looked impressive when it was turned in, by essentially having each of the groups rewrite all the notes from the course lectures was of little added benefit to each student. Much time was spent tediously formatting the lengthy equations that time was taken away from more beneficial areas of the homework. Each student should be responsible for taking his or her own notes thus emphasizing the importance of class attendance. The reports should focus on actual homework with each group's solutions to various problems and MATLAB assignments with included images, videos, and coding from all the members of the group. In our opinion, this approach to the Wikiversity homework reports would be more beneficial to the students, and should be used in conjunction with many of the features of E-Learning, such as grade statistics and feedback, as well as others.

Course Notes
This consists of course notes from 11/17 - 12/8.

Frame Analysis
A frame element is a combination of a truss (bar element) with a beam element.

The differences between a truss and a beam is that because of it's pin connections, a truss only incurs only axial deformation while the fixed nature of the nodes of a beam experience transverse deformation.

2-D Frame Local DOFs

In general, $$d_{(i)}^{(e)}\Rightarrow f_{(i)}^{(e)}$$, where $$f_{(i)}^{(e)}$$ are the generalized forces. Here, e=1,2 and i=1,...,6

$$\left.\begin{matrix} d_{(3)}^{(e)}\\d_{(6)}^{(e)} \end{matrix}\right\}$$ rotational degrees of freedom

$$\left.\begin{matrix} f_{(3)}^{(e)}\\f_{(6)}^{(e)} \end{matrix}\right\}$$ bending moments

2-D Frame Global DOFs

Therefore, there are two element stiffness matrices, $$k_{6\times6}^{(e)}$$ where e=1,2.

This means the global stiffness matrix is $$K_{9\times9}^{(e)}=Ak_{6\times6}^{(e)}$$.

This is assembled by combining the two element stiffness matrices as shown below.

Transformed Coordinate System

$$\mathbf{\tilde{k}}_{(6\times 6)}^{(e)} \mathbf{\tilde{d}}_{(6\times 1)}^{(e)}=\mathbf{\tilde{f}}_{(6\times 1)}^{(e)}$$, where each is defined below.

$$\mathbf{\tilde{d}}_{(6\times 1)}^{(e)}=\begin{Bmatrix} {\tilde{d}}_{1}^{(e)}\\ {\tilde{d}}_{2}^{(e)}\\ {\tilde{d}}_{3}^{(e)}\\ {\tilde{d}}_{4}^{(e)}\\ {\tilde{d}}_{5}^{(e)}\\ {\tilde{d}}_{6}^{(e)} \end{Bmatrix}$$     and $$\mathbf{\tilde{f}}_{(6\times 1)}^{(e)}=\begin{Bmatrix} {\tilde{f}}_{1}^{(e)}\\ {\tilde{f}}_{2}^{(e)}\\ {\tilde{f}}_{3}^{(e)}\\ {\tilde{f}}_{4}^{(e)}\\ {\tilde{f}}_{5}^{(e)}\\ {\tilde{f}}_{6}^{(e)} \end{Bmatrix}$$

Note: The rotational displacements and moments remain unchanged. $$\tilde{d}_{3}^{(e)}={d}_{3}^{(e)}$$ $$\tilde{d}_{6}^{(e)}={d}_{6}^{(e)}$$ $$\tilde{f}_{3}^{(e)}={f}_{3}^{(e)}$$ $$\tilde{f}_{6}^{(e)}={f}_{6}^{(e)}$$

and

$$ \mathbf{\tilde{k}}_{(6\times 6)}^{(e)}= \overbrace{

\begin{bmatrix} \frac{EI}{L} & 0 & 0 & -\frac{EI}{L} & 0 & 0\\ 0 & \frac{12EI}{L^3} & \frac{6EI}{L^2} & 0 & -\frac{12EI}{L^3} & \frac{6EI}{L^2}\\ 0 & \frac{6EI}{L^2} & \frac{4EI}{L} & 0 & -\frac{6EI}{L^2} & \frac{2EI}{L}\\ -\frac{EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0\\ 0 & -\frac{12EI}{L^3} & -\frac{6EI}{L^2} & 0 & \frac{12EI}{L^3} & -\frac{6EI}{L^2}\\ 0 & \frac{6EI}{L^2} & \frac{4EI}{L} & 0 & -\frac{6EI}{L^2} & \frac{4EI}{L} \end{bmatrix}}

^{\begin{matrix} \;\tilde{d}_{1}^{(e)} \;\;\;& \tilde{d}_{2}^{(e)} \;\;\;& \tilde{d}_{3}^{(e)} \;\;\;& \tilde{d}_{4}^{(e)} \;\;\;\;& \tilde{d}_{5}^{(e)} \;\;\;\;& \tilde{d}_{6}^{(e)} \end{matrix}}

\left.\begin{matrix} \\ \\ \\ \\ \\ \end{matrix}\right\}

\begin{matrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)}\\ \tilde{d}_{3}^{(e)}\\ \tilde{d}_{4}^{(e)}\\ \tilde{d}_{5}^{(e)}\\ \tilde{d}_{6}^{(e)} \end{matrix} $$

Dimensional Analysis
The square bracket notation around a variable means "dimension of".

$$ \begin{bmatrix} \tilde{d}_1 \end{bmatrix}=L = \begin{bmatrix} \tilde{d}_i \end{bmatrix} $$, where i=1,2,4,5 and L signifies that it has a dimension of length.

Note: There is a difference between 'units' and 'dimensions'. Displacement has a dimension of length Mass, length, time, charge and temperature are all fundamental dimensions. They cannot be broken down any further, while a dimension such as force can be broken down even further as shown below.

$$Force=mass \times acceleration = \frac {mass \times distance}{{time}^2}=\frac{m\times L}{t^2}$$

$$\begin{bmatrix} \tilde{d}_3 \end{bmatrix}=1$$, meaning it has no dimension.

$$\sigma = E\varepsilon \Rightarrow \begin{bmatrix}\sigma \end{bmatrix}= \begin{bmatrix}E\end{bmatrix} \begin{bmatrix}\varepsilon   \end{bmatrix} =1$$

$$\left[\varepsilon \right]={[du] \over [dx]}={L \over L}=1$$

$$[ \sigma ]=[E]={F \over L^2} \,$$

$$[A]=L^2, [I]=L^4 \,$$

$$\left[ {EA \over L} \right]=[\tilde{k_{11}}]={{F \over L^2}L^2 \over L}={F \over L}$$

$$[\tilde{k}_{11} \tilde{d}_1]=[\tilde{k}_{11}][\tilde{d}_1]=F$$

$$[\tilde{k}_{14} \tilde{d}_4]=[\tilde{k}_{14}][\tilde{d}_4]=\left[ {-EA \over L} \right][L]={-{F \over L^2}L^2 \over L}L=-F$$

$$[\tilde{k}_{22} \tilde{d}_2]=[\tilde{k}_{22}][\tilde{d}_2]={12EI \over L^3}L={{F \over L^2}L^4 \over L^3}L=F$$

$$[\tilde{k}_{23} \tilde{d}_3]=[\tilde{k}_{23}][\tilde{d}_3]={[6][E][I] \over [L^2]}={1 {F \over L^2}L^4 \over L^2}=F$$

$$[\tilde{k}_{25} \tilde{d}_5]=[\tilde{k}_{25}][\tilde{d}_5]=-[\tilde{k}_{22} \tilde{d}_2]=-F$$

$$[\tilde{k}_{26} \tilde{d}_6]=[\tilde{k}_{26}][\tilde{d}_6]=[\tilde{k}_{23} \tilde{d}_3]=F$$

$$[\tilde{k}_{33} \tilde{d}_3]=[\tilde{k}_{33}][\tilde{d}_3]={[4][E][I] \over [L]}={1{F \over L^2}L^4 \over L}=F*L$$

$$[\tilde{k}_{35} \tilde{d}_5]=[\tilde{k}_{35}][\tilde{d}_5]=-[\tilde{k}_{23} \tilde{d}_3]=-F$$

$$[\tilde{k}_{36} \tilde{d}_6]=[\tilde{k}_{36}][\tilde{d}_6]={[2][E][I] \over [L]}={1{F \over L^2}L^4 \over L}=F*L$$

$$[\tilde{k}_{44} \tilde{d}_4]=[\tilde{k}_{44}][\tilde{d}_4]={[E][A] \over [L]}[L]={{F \over L^2}L^2 \over L}L=F$$

$$[\tilde{k}_{55} \tilde{d}_5]=[\tilde{k}_{55}][\tilde{d}_5]=[\tilde{k}_{22} \tilde{d}_2]=F$$

$$[\tilde{k}_{56} \tilde{d}_6]=[\tilde{k}_{56}][\tilde{d}_6]=-[\tilde{k}_{23} \tilde{d}_3]=-F$$

$$[\tilde{k}_{66} \tilde{d}_6]=[\tilde{k}_{66}][\tilde{d}_6]=[\tilde{k}_{33} \tilde{d}_3]=F*L$$

Element Force-Displacement Relationship in Global Coordinates from the Force-Displacement Relationship in Local Coordinates
$$\mathbf{k}^{(e)}_{6x6} \mathbf{d}^{(e)}_{6x1}=\mathbf{f}^{(e)}_{6x1}$$

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

from $$\tilde{\mathbf{k}}^{(e)} \tilde{\mathbf{d}}^{(e)}=\tilde{\mathbf{f}}^{(e)}$$

$$\begin{Bmatrix} \tilde{d}_1\\ \tilde{d}_2\\ \tilde{d}_3\\ \tilde{d}_4\\ \tilde{d}_5\\ \tilde{d}_6 \end{Bmatrix}= \begin{bmatrix} l^{(e)}&m^{(e)}&0 &0& 0& 0\\ -m^{(e)} &l^{(e)} &0 &0 &0 &0\\ 0 &0 &1 &0 &0 &0\\ 0 &0 &0 &l^{(e)}& m^{(e)} &0\\ 0 &0 &0 &-m^{(e)} &l^{(e)} &0\\ 0 &0 &0 &0 &0 &1 \end{bmatrix} \begin{Bmatrix} d_1\\d_2\\d_3\\d_4\\d_5\\d_6\end{Bmatrix}$$

$$\tilde{d}_3=d_3$$

$$\tilde{d}_6=d_6$$

Derivation of $$\tilde{\mathbf{k}}^{(e)}$$ from the Principle of Virtual Work, focusing only on the bending effect.

$$-{\partial^2 \over \partial x^2}((EI){\partial^2 v \over \partial x^2}) + f_t(x)=m(x)\ddot v$$

$${\partial \over \partial x}((EA){\partial u \over \partial x}) + f_a(x,t)=m(x) \ddot u$$

Motivation: deformed shape of truss element interpolation of transverse displacement $$v(s)=v(\tilde{x})$$

The Principle of Virtual Work for Beams
$$\int_0^L W(\tilde{x})[-{\partial^2 \over \partial x^2}(EI){\partial^2 v \over \partial x^2}+f_t-m \ddot v] dx=0$$ for all W(x)Equation (1)

Integration by Parts of the 1st term:

$$\alpha =\int_0^L W(x) {\partial^2 \over \partial x^2}((EI){\partial^2 v \over \partial x^2})dx$$

$$W(x){\partial \over \partial x}({\partial \over \partial x}((EI){\partial^2 v \over \partial x^2}))$$

$$=[W{\partial \over \partial x}((EI){\partial^2 v \over \partial x^2})]_0^L-\int_0^L {dW \over dx}{\partial \over \partial x}((EI){\partial^2 v \over \partial x^2})dx$$

$$=[W{\partial \over \partial x}((EI){\partial^2 v \over \partial x^2})]_0^L-[{dW \over dx}(EI){\partial^2 v \over \partial x^2}]_0^L+\int_0^L{d^2W \over dx^2}(EI){\partial^2 v \over \partial x^2}dx$$

Thus, Equation (1) becomes:

$$[W{\partial \over \partial x}((EI){\partial^2 v \over \partial x^2})]_0^L-[{dW \over dx}(EI){\partial^2 v \over \partial x^2}]_0^L+\int_0^L{d^2W \over dx^2}(EI){\partial^2 v \over \partial x^2}dx+\int_0^L W f_t dx-\int_0^L W m \ddot v dx =0 for all possible W(x)$$

Derivation of the Stiffness Matrix and Shape Functions for Beams
Focus on stiffness term, $$\int_0^L{d^2W \over dx^2}(EI){\partial^2 v \over \partial x^2}dx$$, to derive the beam stiffness matrix and identify shape functions for beams

$$v(\tilde{x})=N_2(\tilde{x})\tilde{d}_2+N_3(\tilde{x})\tilde{d}_3+N_5(\tilde{x})\tilde{d}_5+N_6(\tilde{x})\tilde{d}_6$$Equation (2)

Recall:$$u(\tilde{x})=N_1(\tilde{x})\tilde{d}_1+N_4(\tilde{x})\tilde{d}_4$$Equation (1)

$$N_2(\tilde{x})=1-{3\tilde{x}^2 \over L^2}+{2\tilde{x}^3 \over L^3}; N_3(\tilde{x})=\tilde{x}-{2\tilde{x}^2 \over L} + {\tilde{x}^3 \over L^2}; N_5(\tilde{x})={3 \tilde{x}^2 \over L^2} - {2\tilde{x}^3 \over L^3}; N_6(\tilde{x})=-{\tilde{x}^2 \over L} + {\tilde{x}^3 \over L^2}$$

$$\mathbf{d}^{(e)}_{6x1}=\tilde{T}^{(e)}_{6x6}\tilde{d}^{(e)}_{6x1}$$

Computation of $$u(\tilde{x})$$ and $$v(\tilde{x})$$:

$$u(\tilde{x})=u(\tilde{x})\tilde{\vec{i}}+v(\tilde{x})\tilde{\vec{j}}$$

$$=u_x(\tilde{x})\vec{i}+u_y(\tilde{x})\vec{j}$$

Computation of $$u(\tilde{x})$$ and $$v(\tilde{x})$$ using Equations (1) and (2):

$$\begin{Bmatrix}u_x(\tilde{x})\\u_y(\tilde{x})\end{Bmatrix}=\mathbf{R}^T \begin{Bmatrix}u(\tilde{x})\\v(\tilde{x})\end{Bmatrix}$$

$$\begin{Bmatrix}u_x(\tilde{x})\\u_y(\tilde{x})\end{Bmatrix}=\begin{bmatrix}N_1 &0&0&N_4&0&0\\0&N_2&N_3&0&N_5&N_6\end{bmatrix}\begin{Bmatrix}{\tilde{d}_1^{(e)}}\\{\tilde{d}_2^{(e)}}\\{\tilde{d}_3^{(e)}}\\{\tilde{d}_4^{(e)}}\\{\tilde{d}_5^{(e)}}\\{\tilde{d}_6^{(e)}}\end{Bmatrix}$$

$$\mathbb{N}(\tilde{x})=\begin{bmatrix}N_1 &0&0&N_4&0&0\\0&N_2&N_3&0&N_5&N_6\end{bmatrix}$$

$$\begin{Bmatrix}{u_x(\tilde{x})}\\{u_y(\tilde{x})}\end{Bmatrix}=\mathbf{R}^T\mathbb{N}(\tilde{x})\tilde{\mathbf{T}}^{(e)}\mathbf{d}^{(e)}$$

 Note:  Eq. (2) (p. 39-2, Dimensional Analysis)

[u] = L p. 31-4: [N1] = [N4] = 1

Eq. (2) p. 39-2: $$\left[N_{1} \right]\left[\tilde{d}_{1} \right]+\left[N_{4} \right]\left[\tilde{d}_{4} \right]$$

$$\left[N_{1} \right]=1$$ $$\left[\tilde{d}_{1} \right]=L$$ $$\left[N_{4} \right]=1$$ $$\left[\tilde{d}_{4} \right]=L$$

[v] = L $$\left[N_{2} \right]\left[\tilde{d}_{2} \right]=L$$ $$\left[N_{2} \right]=1$$ $$\left[\tilde{d}_{2} \right]=L$$

$$\left[N_{3} \right]\left[\tilde{d}_{3} \right]=L$$ $$\left[N_{3} \right]=L$$ $$\left[\tilde{d}_{3} \right]=1$$

$$\left[N_{5} \right]\left[\tilde{d}_{5} \right]=L$$ $$\left[N_{5} \right]=1$$ $$\left[\tilde{d}_{5} \right]=L$$

$$\left[N_{6} \right]\left[\tilde{d}_{6} \right]=L$$ $$\left[N_{6} \right]=L$$ $$\left[\tilde{d}_{6} \right]=1$$

$$N_{2}, N_{3}, N_{5}, N_{6}$$ (p. 38-4) p. 38-3 plots Recall: Governing PDE for beams p. 37-4, Eq. (1), without f2 (dist. transverse load) and without inertia force $$m\ddot{v}$$ (static case):

$$\frac{\partial }{\partial x^2}\left\{(EI)\frac{\partial^2v }{\partial x^2} \right\}=0$$

Further, consider constant EI : $$\frac{\partial^4 }{\partial x^4}v = 0$$

Integrate 4 times 4 constants (c0, c1, c2, c3) $$\Rightarrow v(x)=c_{0}+c_{1}x^{1}+c_{2}x^{2}+c_{3}x^{3}$$

To obtain N2(x) ($$\tilde{x}=x$$for simplicity)

$$v(0)=1$$ $$v(L)=0$$ $$v'(0)=v'(L)=0$$

Use above boundary conditions to solve for c0, c1, c2, c3

v(0)=1=c0 $$(1) \; \; v(L)=1+c_{1}L+c_{2}L^2+c_{3}L^3=0$$ $$v'(x)=c_{1}+2c_{2x}+3c_{3}x^2$$ v'(0)=c_(1)=0 $$(2)\; \; v'(L)=2c_{2}L+3c_{3}L^2=0$$

$$c_{0}=1$$ $$c_{1}=0$$ $$c_{2}=\frac{-3}{L^2}$$ $$c_{3}=\frac{-2}{3}\frac{1}{L}\frac{-3}{L^2}=\frac{2}{L^3}$$ Compute with expression for N2 on p. 38-4

 For N3:  (Boundary Conditions) v(0)=v(L)=0 v'(0)=1 v'(L)=0

 For N5:  ($$\tilde{d_{5}}$$ rotation) v(0)=0 v(L)=1 v'(0)=0 v'(L)=0

 For N6:  ($$\tilde{d_{6}}$$ rotation) v(0)=0 v(L)=0 v'(0)=0 v'(L)=1

See p. 39-1 for plots of N5, N6

From p. 36-2: $$\tilde{k}_{22}=\frac{12EI}{L^3}$$

From p. 38-2: $$\tilde{k}_{22}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)}\frac{d^2N_2}{dx^2}dx$$

In general:

$$\tilde{k}_{ij}=\int_{0}^{L}{\frac{d^2N_i}{dx^2}(EI)}\frac{d^2N_j}{dx^2}dx$$ (i,j)=2,3,5,6

Therefore, knowing that the stiffness matrix $$\tilde{K}_{6x6}$$ is symmetric about the diagonal:

From p. 36-2: $$\tilde{k}_{23}=\tilde{k}_{32}=\frac{6EI}{L^2}$$

From p. 38-2: $$\tilde{k}_{23}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)}\frac{d^2N_3}{dx^2}dx=\tilde{k}_{32}=\int_{0}^{L}{\frac{d^2N_3}{dx^2}(EI)}\frac{d^2N_2}{dx^2}dx$$

Similarly: $$\tilde{k}_{25}=\tilde{k}_{52}=\frac{-12EI}{L^3}$$

$$\tilde{k}_{25}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)}\frac{d^2N_5}{dx^2}dx=\tilde{k}_{52}=\int_{0}^{L}{\frac{d^2N_5}{dx^2}(EI)}\frac{d^2N_2}{dx^2}dx$$

$$\tilde{k}_{26}=\tilde{k}_{62}=\frac{6EI}{L^2}$$

$$\tilde{k}_{26}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)}\frac{d^2N_6}{dx^2}dx=\tilde{k}_{62}=\int_{0}^{L}{\frac{d^2N_6}{dx^2}(EI)}\frac{d^2N_2}{dx^2}dx$$

$$\tilde{k}_{33}=\frac{4EI}{L}$$

$$\tilde{k}_{33}=\int_{0}^{L}{\frac{d^2N_3}{dx^2}(EI)}\frac{d^2N_3}{dx^2}dx$$

$$\tilde{k}_{35}=\tilde{k}_{53}=\frac{-6EI}{L^2}$$

$$\tilde{k}_{35}=\int_{0}^{L}{\frac{d^2N_3}{dx^2}(EI)}\frac{d^2N_5}{dx^2}dx=\tilde{k}_{53}=\int_{0}^{L}{\frac{d^2N_5}{dx^2}(EI)}\frac{d^2N_3}{dx^2}dx$$

$$\tilde{k}_{36}=\tilde{k}_{63}=\frac{2EI}{L}$$

$$\tilde{k}_{36}=\int_{0}^{L}{\frac{d^2N_3}{dx^2}(EI)}\frac{d^2N_6}{dx^2}dx=\tilde{k}_{63}=\int_{0}^{L}{\frac{d^2N_6}{dx^2}(EI)}\frac{d^2N_3}{dx^2}dx$$

$$\tilde{k}_{55}=\frac{12EI}{L^3}$$

$$\tilde{k}_{55}=\int_{0}^{L}{\frac{d^2N_5}{dx^2}(EI)}\frac{d^2N_5}{dx^2}dx$$

$$\tilde{k}_{56}=\tilde{k}_{65}=\frac{-6EI}{L^2}$$

$$\tilde{k}_{56}=\int_{0}^{L}{\frac{d^2N_5}{dx^2}(EI)}\frac{d^2N_6}{dx^2}dx=\tilde{k}_{65}=\int_{0}^{L}{\frac{d^2N_6}{dx^2}(EI)}\frac{d^2N_5}{dx^2}dx$$

$$\tilde{k}_{66}=\frac{4EI}{L}$$

$$\tilde{k}_{66}=\int_{0}^{L}{\frac{d^2N_6}{dx^2}(EI)}\frac{d^2N_6}{dx^2}dx$$

Elastodynamics
(used with trusses, frames, 2-D, 3-D, elasticity)

p. 31-1: Model problem

Discrete PVW
(Boundary Conditions already applied) $$\mathbf{\bar{w}\bullet \left[\bar{M}\ddot{\bar{d}}+\bar{K}\bar{d}-\bar{F} \right]}=0$$ for all $$\mathbf{\bar{w}}$$ where $$\mathbf{\bar{K}}$$ is the reduced stiffness matrix

Complete Ordinary Differential Equations (ODE's)
(2nd order in time) and initial conditions governing the elastodynamics of discretized continuity problem (MDOF) $$(1)\Rightarrow \mathbf{\bar{M}\ddot{\bar{d}}+\bar{K}\bar{d}=\bar{F}(t)}$$ $$\; \; \; \; \mathbf{\bar{d}(0)=\bar{d_{0}}}$$ $$\; \; \; \; \mathbf{\dot{\bar{d}}(0)=\bar{v}_{0}}$$

'''Solving Eq. (1):'''

Method 1: Considering the unforced vibration problem
$$\mathbf{\bar{M}_{nxn}\ddot{v}_{nx1}+\bar{K}_{nxn}v_{nx1}=0_{nx1}}$$ Assume $$\mathbf{v(t)_{nx1}=(sin\omega t)\phi _{nx1}}$$ $$\mathbf{\ddot{v}=-\omega ^2sin\omega t\phi}$$ $$\mathbf{-\omega ^2sin\omega t\bar{M}\phi+sin\omega t\bar{K}\phi=0}$$ Generalized eigenvalue problem $$\Rightarrow \mathbf{\bar{K}\phi =\omega ^2\bar{M}\phi}\; \; \; (1)$$ General form: $$\mathbf{Ax=\lambda Bx}$$, where $$\lambda$$ is an eigenvalue Standard eigenvalue problem: $$\mathbf{Ax=\lambda x}$$,where $$\mathbf{(B=I)}$$ The identity matrix $$\mathbf{I}=\begin{bmatrix}1 & & 0\\ & ... & \\ 0 & & 1\end{bmatrix}$$ $$\lambda =\omega ^2$$ $$(\lambda _i,\mathbf{\phi _i)}$$ are eigenpairs and i = 1,...,n Mode i $$\Rightarrow \mathbf{v_i(t)=(sin\omega _it)\phi _i}$$ where i = 1,...,n

Method 2: Model Superposition method
Orthogonal properties of eigenpairs:

$$\mathbf{\phi _{i}^T\bar{M}\phi _j=\delta _{ij}=\begin{cases}1 & \text{ if } i=j \\0 & \text{ if } i\neq j\end{cases}}$$ where $$\delta _{ij}$$ is the Kronecker delta

Mass Orthogonality of eigenvectors: Eq. (2) p. 41-2: Eq. (1) p. 41-3:

$$\mathbf{\bar{M}\phi _j=\lambda _j^{-1}\bar{K}\phi _j}$$

$$\mathbf{\phi _i^T\bar{M}\phi _j=\lambda _j^{-1}\phi _i^T\bar{K}\phi _j}$$ where $$\mathbf{\phi _i^T\bar{M}\phi _j=\delta _{ij}}$$

$$\Rightarrow \mathbf{\phi _i^T\bar{K}\phi _j=\lambda _j\delta _{ij}}$$

Eq. (1) p. 41-2: $$\mathbf{\bar{d}(t)_{nx1}=\sum_{i=1}^{n}{\xi _i(t)_{1x1}}\phi _{nx1}}$$

$$\mathbf{M\left(\sum_{j}^{}{}\ddot{\xi _j}\phi _j\right)+\bar{K}\left(\sum_{j}^{}{}\xi _j\phi _j \right)=F}$$ where $$\sum_{j}^{}{}\ddot{\xi _j}\phi _j=\mathbf{\ddot{\bar{d}}}$$ and $$\sum_{j}^{}{}\xi _j\phi _j=\mathbf{\bar{d}}$$

$$\mathbf{\sum_{j}^{}{}\ddot{\xi _j}\left(\phi _i^T\bar{M\phi _j}\right)+\sum_{j}^{}{}\xi _j\left(\phi _i^T\bar{K}\phi _j \right)=\phi _i^TF}$$ <p style="text-align:center;">where $$\phi _i^T\bar{K}\phi _j=\delta _{ij}$$ and $$\phi _i^T\bar{K}\phi _j=\lambda _j\delta _{ij}$$

<p style="text-align:center;">$$\Rightarrow \mathbf{\ddot{\xi _i+\lambda _i\xi _i=\phi _i^TF}}$$ where i = 1,...,n

Correction to HW 6
After HW6 was submitted, an important error was discovered in the pylon code. The -1000 lb force was intended to be applied at node 33 in the y direction. This was mistakenly entered into the R matrix as row 33. The problem is that row 33 is DOF 33, not node 33. As a result, the code has been changed so the 1000 lbs is in row 66, which is the vertical DOF on node 33.

With these new results, it is determined that the highest compressive stress is -8.6957e+006 Pa, which is located in element 55. The highest tensile stress is 9.0511e+006 Pa in element 81.

Electric Pylon as Frame Elements
The highest bending moment is 6.287 N-m which is located on element 81. The highest shear stress is -5.48269 Pa, which is also located on element 81.

Problem Statement and Given Information
The following depicts the two bar system to be analyzed using MATLAB. It is very similar to the 2-bar system analyzed in Homework 3 except for this system is a frame system rather than a simple truss system. Here, Element 1 is a frame element with a square cross-sectional area and Element 2 is a truss system.

Since the frame element is assumed to be sqaure, the values for the moment of inertia were calculated by using the equation $$ I=\frac{s_{1}s_{2}^{3}}{12} $$ where the s-values are the sides of the square and thus $$s_{1}=s_{2}$$. These values along with the values given in the problem are stated in the following table. P is known to be 7.

Matlab Code
Below is the matlab code used to obtain the comparison between the undeformed 2 bar truss, the 2 bar truss system (determined in homework 3), and the combined frame/truss system explained above.

<p style="text-align:center;"> "2 Bar Truss/Frame Comparison Matlab Code"

Below is the matlab code for the m-file PlaneFrameElement.m as given in the companion for the textbook. This file obtains the K matrix and R matrix. The one given in the book is for an element with a distributed load. Plugging in zeros for these values simplifies the distributed load to the one point load we have in this problem. These lines of code could have been removed but were left in to keep it simple. <p style="text-align:center;">PlaneFrameElement.m Matlab Code

Below is the matlab code used to obtain the results for shear, bending moment, and axial forces. It is important to note that while this file is similar to the books companion site, some changes were made. One of the changes was to increase the number of increments along the frame element from 3 points in the book (i = 0:L/2:) to 16 points in our code (i=0:L/15:L). This gives us more points along the element to analyze so that the curve is more accurate. Other changes were just simple definition changes. Instead of using the word 'inertia', it is defined as I. Similarly, the word 'modulus' was replaced with E.  This serves just to simplify the code. <p style="text-align:center;"> Modified PlaneFrameResults.m Matlab Code

Results Obtained
Below are the results obtained. Note too that we also have the reaction forces as defined by rf. The shear, bending moment, and axial forces are much larger matricies because they are the results at each of the points along element 1 as defined by (i = 0:L/15:L). <p style="text-align:center;"> Results

Images
Below is the image obtained by the code. The undeformed system is the dotted blue line. The deformed truss system is the green line. These are the same as obtained in HW3. The red line shows the deformations for the frame/truss system given in this homework. As you can see, for the frame system, the angle at node 1 does not change significantly. This is due to the change in the type of connection at that node. Instead, there is a sort of wavy motion going up to node 2. This is due to the added degree of freedom and moments at each node. There is also a close up view to see these differences.

Contributing Members
Eml4500.f08.wiki1.aguilar 20:06, 9 December 2008 (UTC) Eml4500.f08.wiki1.ambrosio 19:46, 9 December 2008 (UTC) Eml4500.f08.wiki1.oatley 16:26, 9 December 2008 (UTC) Eml4500.f08.wiki1.schaet 19:34, 9 December 2008 (UTC) Eml4500.f08.wiki1.brannon 17:42, 9 December 2008 (UTC)

title: User:Eml4500.f08.wiki1.aguilar/TeamHW7


text: File:Model frame.png

title: File:Model frame.png


text: File:Free body diagram.png

title: File:Free body diagram.png


text: File:Frame global.png

title: File:Frame global.png


text: File:Frame element.png

title: File:Frame element.png


text:
User:Eml4500.f08.ateam.didomenico/hw 7/animation

Generating Video Using MATLAB
To create video using MATLAB image files must be stored for assembly into movie format. The  command is useful for this. It stores images as movie frames to be accessed and assembled later. The user can manipulate it to define the size and other properties of the frame to be stored. Detailed information on this command can be found here.

MATLAB can create video from these stored frames using the  command. More information on this command can be found here. This command can be used to determine the speed at which the movie is shown as well as how many frames per second are shown and the size and location of the movie. It can also be used to play a movie in reverse. While useful within MATLAB, this type of video cannot be played by other programs. For widespread application, it is advantageous to generate more compatible video files.

Using the command  allows generation of MATLAB movie as an Audio Video Interleave or AVI file. This format is more widely compatible than MATLAB movies. Using this command allows the specification of a filename, video quality and other parameters. More information on this command can be found here.

Example of Video Creation
This example depicts the graph of a function changing over time. The animation helps visualize the change. It creates an array,, of movie frames which are played three times by the   command. An AVI file called  is also created. This AVI file is converted for Wikimedia and shown in the next section.

Conversion for Wikimedia
Wikimedia Commons currently allows video files only in Theora format. All other formats must be converted prior to upload to avoid distribution of patent-encumbered, and therefore not freely distributable, file formats.

A useful and simple tool for converting videos to Theora format is. The Windows version used in this example is a command-line program and can be obtained here. Information pertaining to releases and other versions can be found at the ffmpeg2theora page.

To use this program, open the command prompt window from the Start window. Next click and drag the  file in to the command window. The directory location of the file can also be entered instead of dragging. This should result in the command window seen below.

Next, input a space and then click and drag the file you want to convert into the command window as seen below. Again, the file's location can also be entered manually.

Pressing the  key begins conversion. Depending on the size of the file being converted, the conversion time will vary. The converted video will have the same file name but will be Theora format with a  extension. It will be located in the same directory as the original video file. This format can be uploaded to Wikimedia Commons similarly to other multimedia types. A completed video conversion can be seen below.


text: File:Beam shape.png

title: File:Beam shape.png


text: File:Isolatedstringercarbon.JPG

title: File:Isolatedstringercarbon.JPG


text: File:Deflection points.png

title: File:Deflection points.png


title: File:Isostringer1carbon.JPG


title: File:Isolatedstringercarbon2.JPG


text: File:Global stiffness matrix.png

title: File:Global stiffness matrix.png


title: File:Peacesign1carbon.JPG


title: File:Peacesign2carbon.JPG


text:
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User talk:Eml4500.f08.bottle.hipps

== clear instead of lot of ==

Better way : Use one clear instead a lot of .Crochet.david 12:20, 8 December 2008 (UTC)

title: User talk:Eml4500.f08.bottle.hipps


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Wikiversity talk:Policy review

Reviewing policies
We've had a number of tensions lately about policies, and after looking through them there seem to be very few of them (seven!) that have actually been approved after discussion. We should really revisit all of the old proposals as well as the the approved policies... most of these are from the earliest days of Wikiversity when we hadn't actually seen what sort of content would be developed.

At least 6 policy proposals have been in contention lately, so perhaps it's best to start with those so that we can avoid further tensions between those on either side of the issues. The naming policies in particular are a slow-burning issue that's been going on in the background for years, and the deletion policy (and in particular how it should relate to "stubs") has been a subject of major contention for at least the past 4 months.

I suggest we go through these one by one, perhaps giving a quick up-or-down straw poll to see if the policies can be at least temporarily adopted as-is, and then move on to adding, amending, or subtracting sections from the policy. A lot of policies might also be good for merging together, though it might be easier to come to consensus on "small" policies, and then merge them into larger policies later (even 1-sentence "fragment policies" might be a good idea since we've seen lately how easily a policy discussion gets bogged down). Come to think of it, we might want to have a standard approach to amending policies, including using the sitenotice to announce a discussion at least a week before something is "officialized" so that we don't revisit the situation where we find policies moved to "enforced" long afterward and without there having been any public notification.--SB_Johnny talk 14:03, 8 December 2008 (UTC)

title: Wikiversity talk:Policy review


text: File:Truss 7.jpg

title: File:Truss 7.jpg


title: File:1.GIF


text:
User:Eml4500.f08.group/HW7

 * Lecture Notes (current version).Lecture Notes (archived version)

 * MATLAB Code (current version).MATLAB Code (archived version)

 * HW7 Contributors (current version). HW7 Contributors (archived version)

title: User:Eml4500.f08.group/HW7


text:
User:Eml4500.f08.group/HW7/Lecture Notes

Lecture Notes

=Rotational Displacement in 2-bar Frames=

Frame element = truss(bar) element + beam element

A truss has axial deformation

A beam has transverse deformation (bends)

Model Frame Problem with 2 Elements:

With the free body diagram of each element with 6 degrees of freedom:

Where $$ \left.\begin{matrix} d_{3}^{(e)}\\ d_{6}^{(e)}

\end{matrix}\right\} $$ rotational dofs and $$ \left.\begin{matrix} f_{3}^{(e)}\\ f_{6}^{(e)}

\end{matrix}\right\} $$ bending moments

Frame Global dofs (9):

2 element stiffness matrices k(e)6x6, where e=1,2

Then the global stiffness matrix K9x9 is the assembly of the two element stiffness matrices shown below.

=Stiffness Matrix with Rotational Displacement Included=

Below is a deformable truss with horizontal, vertical and rotational displacements:

The matrix formulas are as follows:

$$\mathbf{\tilde{k}}^{(e)}_{6x6}\mathbf{\tilde{d}}^{(e)}_{6x6} = \mathbf{\tilde{f}}^{(e)}_{6x6}$$, where

$$\mathbf{\tilde{d}}^{(e)} = \begin{Bmatrix} \tilde{d}^{(e)}_{1}\\ ...\\ \tilde{d}^{(e)}_{6} \end{Bmatrix}$$ and $$\mathbf{\tilde{f}}^{(e)} = \begin{Bmatrix} \tilde{f}^{(e)}_{1}\\ ...\\ \tilde{f}^{(e)}_{6} \end{Bmatrix}$$

Note: $$\tilde{f}_{3}^{(e)} = f_{3}^{(e)}$$ and $$\tilde{f}_{6}^{(e)} = f_{6}^{(e)}$$

Taking the moments about $$\tilde{z}$$ (where $$\tilde{z} = z$$), the displacement matrix is as follows:

.......$$\tilde{d}_{1}$$  .........   $$\tilde{d}_{2}$$   .........  $$\tilde{d}_{3}$$   .........   $$\tilde{d}_{4}$$  .........   $$\tilde{d}_{5}$$    ......... $$\tilde{d}_{6}$$

$$\begin{Bmatrix} \frac{EA}{L} & 0 & 0 & -\frac{EA}{L} & 0 & 0\\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}}\\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{2EI}{L}\\ \frac{-EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0\\ 0 & \frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}}\\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{4EI}{L} \end{Bmatrix}\begin{matrix} \tilde{d}_{1}\\ \tilde{d}_{2}\\ \tilde{d}_{3}\\ \tilde{d}_{4}\\ \tilde{d}_{5}\\ \tilde{d}_{6} \end{matrix}$$

The dimensional analysis is as follows:

$$[\tilde{d}_{1}] =L=[\tilde{d}_{i}]$$, where i = 1,2,3,4,5

dimensions of rotation = $$[\tilde{d}_{3} = 1 \Rightarrow$$ (no dimension, because the unit is degree)

The length of the arc AB is equal to $$R\theta \Rightarrow \theta = \frac{arcAB}{R}$$

$$[\theta ] = \frac{[arcAB]}{[R]} = \frac{1}{1} = 1$$

To note, the unit is not the same as the dimension. $$\theta$$ of course does not have a unit of 1!

$$\sigma = E\varepsilon \Rightarrow [\sigma ] = [E][\varepsilon ]$$

=Dimensional Analysis=

The validity of the stiffness matrix can be verified by using dimensional analysis. Since this matrix approximated the stiffness of an element as a spring, the resulting dimension of this stiffness should be the same as the dimensions for a spring's spring constant or Force per unit Length.

As an example, take stress ($$\sigma$$)which has units of force per unit area. In the case the dimensional analysis would yield:

$$ [\sigma ] = \frac{F}{L^2} $$

We also know that $$ [E] = \frac{F}{L^2},[A] = L^2,[L] = L,[I]=L^4 $$

With this, we can now check and see if all the terms of the element stiffness matrix come out to be Force per unit Length. There are 5 terms that must be checked, these terms are:

$$ \frac{EA}{L},\frac{12EI}{L^3},\frac{6EI}{L^2},\frac{4EI}{L},and \frac{2EI}{L} $$

$$\frac{[E][A]}{[L]}=\frac{F}{L^2}*L^2*\frac{1}{L}=\frac{F}{L}$$

$$\frac{12[E][I]}{[L^3]}=\frac{F}{L^2}*L^4*\frac{1}{L^3}=\frac{F}{L}$$

$$\frac{6[E][I]}{[L^2]}=\frac{F}{L^2}*L^4*\frac{1}{L^2}=F$$

$$\frac{4[E][I]}{[L]}=\frac{F}{L^2}*L^4*\frac{1}{L^2}=F*L$$

$$\frac{2[E][I]}{[L]}=\frac{F}{L^2}*L^4*\frac{1}{L^2}=F*L$$

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

!MATLAB 2-bar frame with one fixed node, from team Carbon
 * A two bar frame shown below is examine with the left-most node fixed (i.e. no rotational component):
 * A two bar frame shown below is examine with the left-most node fixed (i.e. no rotational component):

This makes the leftmost truss deformation parabolic instead of linear. The MATLAB code is shown below:

close; clear; % model with 3-D beam elements with 6 dofs per node dof = 9;        %  dof per node % 1 : axial disp x,  2 : disp y  , 3 : disp z                    %  4 : torsion x    ,  5 : bend y  , 6 : bend z  % % input the nodal coordinates %  n_node = 3;             % total number of nodes n_elem = 2;            % total number of elements total_dof = 6 * n_node; % total dofs of system Le(1) = 4; % Element 1 beam length Le(2) = 2; % Element 2 beam length position(:, 1) = [     -Le(1) *cosd(30) ; -Le(1) *sind(30); 0]; position(:, 2) = [     0; 0; 0]; position(:, 3) = [      Le(2) *cosd(45) ; -Le(2)*sind(45); 0]; % inital postion of the nodes in the truss system % set up the nodal coordinate arrays x, y, x  x = zeros(1,3); y = zeros(1,3); z = zeros(1,3); for i = 1 : n_node x(i) = position(1,i); y(i) = position(2,i); z(i) = position(3,i); end % set up the element connectivity array node_connect, defined as % follows: % node_connect(local node number, element number) = global node number node_connect(1, 1) = 1;  % element 1 node_connect(2, 1) = 2; node_connect(1, 2) = 2;  % element 2 node_connect(2, 2) = 3; % plot the whole truss system % loop over the elements for i = 1 : n_elem % for element i, do the following: % node_1 = global node number corresponding to the local node 1 node_1 = node_connect(1,i); % node_2 = global node number corresponding to the local node 1 node_2 = node_connect(2,i); % xx : 1x2 array containing x coordinates of node_1 and node_2 xx = [x(node_1),x(node_2)]; % yy : 1x2 array containing y coordinates of node_1 and node_2 yy = [y(node_1),y(node_2)]; % zz : 1x2 array containing z coordinates of node_1 and node_2 zz = [z(node_1),z(node_2)]; % plot the element i in 3-D using command plot3 % use axis command to avoid having the plot window too tight % axis([]) % dashed line plot3(xx,yy,zz,'--') % hold the current figure for the next element hold on  end text(x(1) + .25, y(1)+ .25, z(1),'Node 1'); text(x(2) + .25, y(2)+ .25, z(2),'Node 2'); text(x(3) + .25, y(3)+ .25, z(3),'Node 3'); text((x(2) + x(1))/2, (y(1) + y(2))/2, (z(1) + z(2))/2,'Element 1'); text((x(3) + x(2))/2, (y(3) + y(2))/2, (z(3) + z(2))/2,'Element 2'); view([0 0 1]); x(2) A(1) = 1; A(2) = 2; %   Cross Sectional Area of Elements 1 and 2 E(1) = 3; E(2) = 5; %   Youngs Modulus of Elements 1 and 2 Te(1) = 30; Te(2) = -45; I(1) = (1/8) + 16; I(2) = (2/3) + 8; Ke = zeros(6,6); for i = 1 : n_elem Ke(:,:,i) = [ ((E(i)*A(i))/Le(i)), 0                         , 0                         , -((E(i)*A(i))/Le(i)) , 0                        , 0                          ; 0                , ((12*E(i)*I(i))/(Le(i)^3)) , ((6*E(i)*I(i))/(Le(i)^2))  , 0                   , ((12*E(i)*I(i))/(Le(i)^3)) , ((6*E(i)*I(i))/(Le(i)^2))  ; 0                , ((6*E(i)*I(i))/(Le(i)^2))  , (4*E(i)*I(i))/Le(i)        , 0                   , -((6*E(i)*I(i))/(Le(i)^2)) , (2*E(i)*I(i))/Le(i)        ; -((E(i)*A(i))/Le(i)), 0                        , 0                         , (E(i)*I(i))/Le(i)    , 0                         , 0                         ; 0                 , ((12*E(i)*I(i))/(Le(i)^3)) , -((6*E(i)*I(i))/(Le(i)^2)) , 0                   , ((12*E(i)*I(i))/(Le(i)^3)) , -((6*E(i)*I(i))/(Le(i)^2)) ; 0                 , ((6*E(i)*I(i))/(Le(i)^2))  , (2*E(i)*I(i))/Le(i)        , 0                   , -((6*E(i)*I(i))/(Le(i)^2)) , (4*E(i)*I(i))/Le(i)       ]; %   Element stiffness matrix for Element i     end Kg = zeros(9,9); Kg(4:9,4:9) = Ke(:,:,2); Kg(1:6,1:6) = Ke(1:6,1:6,1); Kg(4:6,4:6) = Ke(4:6,4:6,1) + Ke(1:3,1:3,2) %   Creation of the Global Matrix    % Fg = zeros(9,1); Fg(4,:) = 0; Fg(5,:) = 7; %   Known Forces %Fg(9,:) = 0; %   Set Matrix Size Dg = zeros(9,1); Dg(1,:) = 0; Dg(2,:) = 0; Dg(3,:) = 0; % no rotational component on node 1 Dg(7,:) = 0; Dg(8,:) = 0; %   Known Displacements Dg(4:6,:) = inv(Kg(4:6,4:6)) * Fg(4:6,:) %   d1 = d2 = d7 = d8 = 0, reducing matrix and sloving for d3 and d4   Fe(:,:,1) = Ke(:,:,1) * Dg(1:6,:); Fg(1:3,:) = Fe(1:3,1); %   sloving for f1 and d2   Fe(:,:,2) = Ke(:,:,2) * Dg(4:9,:); Fg(7:9,:) = Fe(4:6,2) %   sloving for f5 and d6   position(:, 1) = [      -Le(1) *cosd(30) ; -Le(1) *sind(30); 0]; position(:, 2) = [      Dg(4)           ;  Dg(5)          ; 0]; position(:, 3) = [      Le(2) *cosd(45) ; -Le(2) *sind(45); 0]; % displaced postion of the nodes in the truss system (only node 2 moves) % set up the nodal coordinate arrays x, y, x  for i = 1 : n_node x(i) = position(1,i); y(i) = position(2,i); z(i) = position(3,i); end % plot the whole truss system % loop over the elements for i = 1 : n_elem % for element i, do the following: % node_1 = global node number corresponding to the local node 1 node_1 = node_connect(1,i); % node_2 = global node number corresponding to the local node 1 node_2 = node_connect(2,i); % xx : 1x2 array containing x coordinates of node_1 and node_2 xx = [x(node_1),x(node_2)]; % yy : 1x2 array containing y coordinates of node_1 and node_2 yy = [y(node_1),y(node_2)]; % zz : 1x2 array containing z coordinates of node_1 and node_2 zz = [z(node_1),z(node_2)]; % plot the element i in 3-D using command plot3 % use axis command to avoid having the plot window too tight %axis([]) % solid line plot3(xx,yy,zz,'-') % hold the current figure for the next element hold on  end text(x(1) + .25, y(1)+ .25, z(1),'Node 1'); text(x(2) + .25, y(2)+ .25, z(2),'Node 2'); text(x(3) + .25, y(3)+ .25, z(3),'Node 3'); text((x(2) + x(1))/2, (y(1) + y(2))/2, (z(1) + z(2))/2,'Element 1'); text((x(3) + x(2))/2, (y(3) + y(2))/2, (z(3) + z(2))/2,'Element 2'); % put the title on the figure title('2-Bar Truss structure w/ 3-D beam elements') % label x axis xlabel('x') % label y axis ylabel('y') % label z axis zlabel('z') % attempt to view the plotted 3-D truss structure from different % viewpoint. % plot viewed from z axis, i.e., projection on xy plane view([0 0 1])     % xy plane view % plot viewed from y axis, i.e., projection on xz plane % view([0 1 0])    % xz plane view % plot viewed from x axis, i.e., projection on yz plane % view([1 0 0])    % yz plane view The results of the code is shown below.

Stiffness Matrix:

Kg = 0.7500        0         0   -0.7500         0         0         0         0         0         0    9.0703   18.1406         0    9.0703   18.1406         0         0         0         0   18.1406   48.3750         0  -18.1406   24.1875         0         0         0   -0.7500         0         0   17.0938         0         0   -5.0000         0         0         0    9.0703  -18.1406         0   74.0703   46.8594         0   65.0000   65.0000         0   18.1406   24.1875         0   46.8594  135.0417         0  -65.0000   43.3333         0         0         0   -5.0000         0         0   21.6667         0         0         0         0         0         0   65.0000  -65.0000         0   65.0000  -65.0000         0         0         0         0   65.0000   43.3333         0  -65.0000   86.6667

Displacement Matrix:

Dg = 0        0         0    0.1211   -0.0420         0         0         0         0

Force Matrix:

Fg = 0.3361  -3.2129         0    7.0000         0         0   10.6017    6.0499         0

Contribution by group.w (Jeff Wright)
 * }

Obtain Element Force-Displacement Relationship in Global Coordinates
$$\begin{Bmatrix} \tilde{d_1}\\ \tilde{d_2}\\ \tilde{d_3}\\ \tilde{d_4}\\ \tilde{d_5}\\ \tilde{d_6} \end{Bmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 & 0 & 0\\ -m^{(e)}& l^{(e)} & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & l^{(e)} & m^{(e)}\\ 0 & 0 & 0 & 0 & -m^{(e)} & l^{(e)}\\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}*\begin{Bmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6 \end{Bmatrix}$$

=PVW for Beams=

$$\displaystyle \int_0^L w(x) [ -\frac{\partial^2}{\partial x^2}((EI)\frac{\partial^2 v}{\partial x^2}) + f_t - m \dot \dot v] dx = 0 $$ for all possible w(x) → Eq. 1

Integration by parts of first term: (p.30-1)

$$\displaystyle \alpha := \int_0^L w(x) \frac{\partial^2}{\partial x^2}[(EI)\frac{\partial^2 v}{\partial x^2}] dx$$

Where: $$\displaystyle s(x) = w(x)$$ $$\displaystyle r'(x) = \frac{\partial^2}{\partial x^2}[(EI)\frac{\partial^2 v}{\partial x^2}]$$ $$\displaystyle r(x) = \frac{\partial}{\partial x}[(EI)\frac{\partial^2 v}{\partial x^2}]$$

$$\displaystyle \alpha = \beta_1 - \int_0^L \frac{dw}{dx} \frac{\partial}{\partial x}[(EI)\frac{\partial^2 v}{\partial x^2}] dx$$

where: $$\displaystyle \beta_1 = [w \frac{\partial}{\partial x} [(EI) \frac{\partial^2 v}{\partial x^2}] ]_0^L $$

Repeat intergration by parts

Therefore $$\displaystyle \alpha = -(\beta_1 - \beta_2 + \gamma)$$

Where: $$\displaystyle \beta_2 = [\frac{dw}{dx} (EI) \frac{\partial^2 v}{\partial x^2}]_0^L $$ $$\displaystyle \gamma = \int_0^L \frac{d^2 w}{dx^2}(EI)\frac{\partial^2 v}{\partial x^2} dx$$

Equation 1 becomes: $$\displaystyle -\beta_1 + \beta_2 - \gamma) + \int_0^L w f_t dx - \int_0^L w m \dot \dot v dx = 0$$ for all possible w(x)

Lets focus on the stiffness term $$\displaystyle \gamma$$ for now to derive the beam stiffness matrix and to identify the shape functions of a beam.

$$ \displaystyle V(\tilde x) = N_2(x) \tilde d_2 + N_3(x) \tilde d_3 + N_5(x) \tilde d_5 + N_6(x) \tilde d_6$$ → Eq. 2

Recall:

$$\displaystyle u(\tilde x) = N_1(x) \tilde d_1 + N_4(x) \tilde d_4$$

pg. 246 in book: $$\displaystyle N_2(\tilde x) = 1 -\frac{3 \tilde x^2}{L^2} + \frac{2 \tilde x^3}{L^3}$$

$$\displaystyle N_3(\tilde x) = \tilde x -\frac{2 \tilde x^2}{L} + \frac{\tilde x^3}{L^2}$$

$$\displaystyle N_5(\tilde x) = \frac{3 \tilde x^2}{L^2} - \frac{2 \tilde x^3}{L^3}$$

$$\displaystyle N_6(\tilde x) = -\frac{\tilde x^2}{L} + \frac{\tilde x^3}{L^2}$$

=Deformation Analyzed Graphically=

From Lecture 37-3: $$\boldsymbol{\tilde d^{(e)}_{6x1} = \tilde T^{(e)}_{6x6}d^{(e)}_{6x1}}$$

Compute $$u(\tilde x), v(\tilde x)$$ using Equations (1) and (2)from Lecture 38-3.

$$\boldsymbol u(\tilde x) = u(\tilde x)\hat{\tilde i}+v(\tilde x)\hat{\tilde j} = u_x(\tilde x)\hat i + u_y(\tilde x)\hat j$$

$$\begin{Bmatrix} u_x(\tilde x) \\ u_y(\tilde x)\end{Bmatrix}= R^T\begin{Bmatrix} u(\tilde x) \\ v(\tilde x)\end{Bmatrix} $$

$$\begin{Bmatrix} u(\tilde x) \\ v(\tilde x)\end{Bmatrix}=\begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & n_3 & 0 & N_5 & N_6 \end{bmatrix} \begin{Bmatrix} \tilde d_1^{(e)} \\\tilde d_2^{(e)} \\\tilde d_3^{(e)} \\\tilde d_4^{(e)} \\\tilde d_5^{(e)} \\ \tilde d_6^{(e)}\end{Bmatrix}$$

where, $$\begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & n_3 & 0 & N_5 & N_6 \end{bmatrix} = \mathbb N$$

$$\begin{Bmatrix} u_x(\tilde x) \\ u_y(\tilde x)\end{Bmatrix}=\boldsymbol R^T\mathbb N \boldsymbol T^{(e)}\boldsymbol d^{(e)}$$

=Displacements in Deformed Trusses=

$$[u]=L$$

$$[N_{1}]=[N_{4}]=1$$ and [d]=L

$$[N_{1}][\tilde{d_{1}}] + [N_{4}][\tilde{d_{4}}] = L$$

[v]=L

$$[N_{2}][\tilde{d_{2}}] = L$$

$$[N_{3}][\tilde{d_{3}}] = L$$

$$[N_{5}][\tilde{d_{5}}] = L$$

$$[N_{6}][\tilde{d_{6}}] = L$$

Derivation of beam shape functions

N2, N3, N5, N6

Recall the governing partial differential equation for beams, but without $$f$$ (distributed frames load) and without the inertia force $$m\ddot{v}$$

$$\frac{\partial ^{2}}{\partial x^{2}}\left\{(EI)\frac{\partial ^{2}V}{\partial x^{2}} \right\}=0$$

Then when (EI) is considered constant the equation reduces to $$\frac{\partial ^{4}}{\partial x^{4}}V=0$$ Integrate 4 times and you get:

$$V(x)=C_{o}+C_{1}x +C_{2}x^{2}+C_{3}x^{3}$$

The boundary conditions are: $$V(0)=1, V(L)=0, V^{'}(0)=V^{'}(L)=0$$

Coefficient Analysis
$$ v(x) = C_0+C_1x^1+C_2x^2+C_3x^3 $$ $$ v'(x) = C_1+2C_2x+3C_3x^2 $$

N_2 Coefficients
Boundary Conditions

$$ v(0) = 1,v(L) = 0,v'(0) = 0, v'(L) = 0 $$

$$ v(0) = 1 = C_0 $$ $$ v(L) = 1 + C_1L+C_2L^2 + C_3L^3 = 0 $$ $$ v'(L) = 2C_2L + 3C_3L^2$$

$$ C_2L^2 + C_3L^3 = -1 $$ $$ 2C_2l + C_3L^2 = 0 $$

This yields:

$$ C_2 = \frac{-3}{L^2},C_3 = \frac{2}{L^3} $$

Therefore:

$$ v(x) = 1 - \frac{-3}{L^2}x^2 + \frac{2}{L^3}x^3 $$

HW: Find the Coefficients for Shape Functions N_3, N_5, and N_6

N_3 Coefficients
Boundary Conditions

$$ v(0) = 0,v(L) = 0,v'(0) = 1, v'(L) = 0 $$

Using $$ v(0) = 0 $$, $$ 0 = C_0 + 0 + 0 + 0 $$ $$ C_0 = 0 $$

Using $$ v'(0) = 1 $$, $$ 1 = C_1 + 0 + 0 $$ $$ C_1 = 1 $$

Using $$ v(L) = 0 $$, $$ 0 = C_0 + C_1L + C_2L^2 + C_3L^3 $$ $$ 0 = L + C_2L^2 + C_3L^3 $$ $$ C_2L^2 + C_3L^2 = -L $$

Using $$ v'(L) = 0 $$, $$ 0 = C_1 + 2C_2L + 3C_3L^2 $$ $$ 0 = 1 + 2C_2L + 2C_3L^2 $$ $$ C_22L + C_33L^2 = -1 $$

Solving the 2 above equations simultaneously yields the last 2 coefficients. $$ C_2 = \frac{-2}{L}, C_3 = \frac{1}{L^2} $$

N_5 Coefficients
Boundary Conditions

$$ v(0) = 0,v(L) = 1,v'(0) = 1, v'(L) = 0 $$

Using $$ v(0) = 0 $$, $$ 0 = C_0 + 0 + 0 + 0 $$ $$ C_0 = 0 $$

Using $$ v'(0) = 1 $$, $$ 1 = C_1 + 0 + 0 $$ $$ C_1 = 1 $$

Using $$ v(L) = 1 $$, $$ 1 = C_0 + C_1L + C_2L^2 + C_3L^3 $$ $$ 1 = L + C_2L^2 + C_3L^3 $$ $$ C_2L^2 + C_3L^2 = 1-L $$

Using $$ v'(L) = 0 $$, $$ 0 = C_1 + 2C_2L + 3C_3L^2 $$ $$ 0 = 1 + 2C_2L + 2C_3L^2 $$ $$ C_22L + C_33L^2 = -1 $$

Solving the 2 above equations simultaneously yields the last 2 coefficients. $$ C_2 = \frac{-2L+3}{L^3}, C_3 = \frac{L-2}{L^3} $$

N_6 Coefficients
Boundary Conditions

$$ v(0) = 0,v(L) = 0,v'(0) = 0, v'(L) = 1 $$

Using $$ v(0) = 0 $$, $$ 0 = C_0 + 0 + 0 + 0 $$ $$ C_0 = 0 $$

Using $$ v'(0) = 0 $$, $$ 0 = C_1 + 0 + 0 $$ $$ C_1 = 0 $$

Using $$ v(L) = 0 $$, $$ 0 = C_0 + C_1L + C_2L^2 + C_3L^3 $$ $$ 0 = L + C_2L^2 + C_3L^3 $$ $$ C_2L^2 + C_3L^2 = 0 $$

Using $$ v'(L) = 1 $$, $$ 1 = C_1 + 2C_2L + 3C_3L^2 $$ $$ 1 = 2C_2L + 2C_3L^2 $$ $$ C_22L + C_33L^2 = 1 $$

Solving the 2 above equations simultaneously yields the last 2 coefficients. $$ C_2 = \frac{-1}{L}, C_3 = \frac{1}{L^2} $$

Derive Stiffness Coefficient $$\displaystyle \tilde k$$
$$\displaystyle \tilde k_{23} = \frac{6EI}{L^3} = \int_0^L \frac{d^2 N_2}{dx^2} (EI) \frac{d^2 N_3}{dx^2} dx$$

In general:

$$\displaystyle \tilde k_{ij} = \int_0^L \frac{d^2 N_i}{dx^2} (EI) \frac{d^2 N_j}{dx^2} dx$$ for $$\displaystyle i,j = 2,3,5,6$$

=Elastodynamics=

Discrete PVW:

$$\displaystyle w * [M \dot \dot d + Kd - F] = 0$$ for all w

=Wikiversity Vs. E-Learning=

wikiversity pros and cons
Pros:

 * Can collaborate with team members even if you can not meet with them in person.

 * Many people can be adding to a page simultaneously.

 * The wiki network was never down, so could always be accessed.

 * The class can see the best team's homework and notes which each student can then use when studying for the test.

Cons:

 * Have to use basic code to type notes.

 * Have to use outside sources to make decent formulas.

 * The pages can be changed by other people possibly resulting in wrong or incomplete sections of notes.

 * Cheating is easy to do with no restricted access to each teams pages.

 * studying other teams notes can lead to misleading information.

E-Learning pros and cons
Pros:

 * Everyone in the group was previously familiar with the system and how it works.

 * A chat feature where you can communicate with any students also in the class without actually changing any posted notes.

 * Cheating is very difficult because homework is submitted directly to the teacher and other students do not have access to it.

Cons:

 * The organization of group pages would not be as efficient.

 * Usually only instructors have access to adding or changing folders and links on a class page.

 * It is a tool meant for easy communication between a teacher and students, and between students via discussion topics, not as a tool for students to type up homework and class notes for everyone to see.

 * The web page is sometimes down or slow due to the number of people trying to access it at once.

=Useful software for typing notes and homework on Wikiversity=

For writing wiki articles the members of our group just typed directly into the text area, using basic html code. We did not use any of the mentioned software in class such as WikED, AutoWikiBrowser, MWiki-Browser so we cannot provide a comparison between them.

Equations
For the Equations most of us found it easiest to use the latex equation editor, without it many of the equations would have been very difficult to type. Although sometimes the latex site was down and then our team members used Microsoft Word Equation Editor and saved the equation as a picture to put on the wiki page.

Pictures and Figures
Our team members preferred to use paint or Microsoft Powerpoint to create our drawings. We used which ever programs we were already familiar with to create the figures in our notes. No one used Inkscape so a comparison cannot be made.


text:
User:Eml4500.f08.group/HW7/Matlab

=MATLAB Code=

Mon, 24 Nov 2008, 13:17:17 EST Dear class:

In all of the HW7 problems involving the 2-element frame system under different boundary conditions, the deformation inside each frame element due to bending and extension/compression is not a straight line as for a truss element. As a result, the *complete* deformed shape of the frame system is more complex than the deformed shape of the truss system.

The deformed shape of a truss system only involves the displacement at the two nodes for each element.

To obtain the *complete* deformed shape of a frame system, you need to subdivide each frame element into segments (say 10, 15, etc.), then for each point at the end of each segment, you need to figure out the displacement vector for that point based on the nodal displacements and rotations of the frame element and the frame shape functions (which involve both 2 shape functions for a bar element, and 4 shape functions for a beam element). More on these details should be done in my lectures (today).

=Frame Model=

Next, go back to the problem:

Electric pylon: http://apollo.mae.ufl.edu/~vql/courses/fead/2006.fall/code.web.pages/electric_pylon.m

and convert the finite element model for the electric pylon from a truss system (HW6) to a frame system, i.e., all elements now are frame elements.

Use the clamped boundary conditions at the base of the frame model, instead of the simply supported boundary conditions (pin connections) at the base of the truss model.

Redo the static analysis using the same geometric, material, and loading data as for HW6, assuming in addition that the cross section for each frame element is a square.

Compare the nodal displacements of the node under the applied force for the frame model to those of the truss model (which you already obtained in HW6).

Identify the frame element(s) with the highest nodal bending moment and *transverse* shear force. To compute the transverse shear force, you need to work in the tilde coordinate system, i.e., all internal force components in the global coordinate system should be transformed back to the element tilde coordinate system.

Max Nodal Bending Moment is 60 N m at Node 176 (element 58, green)

Max Transverse Stress is 37749 N/m^2 at Node 149 (element 49, red)

Plot the following on the same figure:

 * underformed shape of the electric pylon in dotted line (as in HW6)

 * deformed shape of the truss model (as in HW6)

 * the "incomplete" deformed shape of the frame model in which only a straight line connecting two displaced nodes of a frame element is plotted (not the complete deformed shape inside each element as done above).

You should use the same magnification factor for both the truss model and the frame model for comparison.

 * is the frame model statically determinate?

 * If yes, provide an argument to support your answer, and compute the reactions. Extend your matlab code to compute the reactions, and compare the computed reactions to the reactions obtained from statics.

 * If no, provide an argument to support your answer. You can still compute the reactions using the truss model and the frame model.

The truss system is statically indeterminate. It requires the use of the Principle of Virtual Work or Mechanics of Materials to solve.

Consider isolating an arbitrary node located within the truss system. Also, determine each reaction at a node as a single transverse reaction along the line of action of a beam rather than two reactions along the vertical and horizontal 2-D coordinate system.

Now begin with the node on the far right tip of the arm of the pylon as shown below:

As you can see there are two unknowns and three possible statics equations. Taking a moment however at the node in consideration reveals no conclusion and cannot be considered a possible statics equation in this example.

Using $$\sum{F_{y}} = 0$$, it is easily found that the y-component of R2 = 1000N.

Using $$\sum{F_{x}} = 0$$, it is also easily found that $$R_{1} = R_{2}cos\theta $$

From here, the final reactions cannot be computed and equations of Mechanics must be introduced (if not using FEM). The node shown above is the simplest FBD in the system.

Consider the node at (25, 30):

From this node, it is clear to see why six reactions cannot be computed with only two equations of statics.

With each node there are two, three, four, five or six unknown reactions. There are six nodes with two unknown reactions at the extreme locations of the truss system which cannot be solved as shown above. The nodes with four, five or six unknown reactions can't be computed using only two equations of statics.

The reason we cannot use the moment equation as a third equation of statics is because if the moment is taken at any point the moment equals zero. At each node, each isolated truss intersects. If the moment is taken at the node, it is clear why the moment equals zero, however at any other point it does as well (as shown in previous homework assignments). Each member is a two-force body and there is no resulting moment arm and no moment. This shows that the nodes with three unknown reactions cannot be solve as well.

 * Compare the reactions obtained from the truss model to those from the frame model (remembering that the boundary conditions are different from one model to the other).

The deformation of both the frame and truss structure are nearly identical. This is likely due to the small moment of inertia of the beams, 1.3 x 10^-8 meters.

=Lumped Mass Matrix=

Tthe mass density of the 300 M steel is rho = 7.8 g/cm^3. Consider a lumped mass model in which the mass of each bar element is divided equally by two and is lumped at each of the two nodes of the bar.

Construct the lumped mass matrix for the frame model of the electric pylon. Note that this lumped mass matrix is "almost" the same as the lumped mass matrix for the truss model, except that the dimension of the matrix is larger (due to the rotation dofs); there are 3 dofs per node in the frame model, instead of just 2 dofs per node in the truss model. Just put zero mass at the rotation dofs, i.e., we neglect the effects of rotational inertia.

 * Use matlab to solve the generalized eigenvalue problem $$K_{bar} v = \lambda M_{bar} v$$ where $$K_{bar}$$ is the reduced stiffness matrix, $$M_{bar}$$ the associated mass matrix, $$v$$ the eigenvector and lambda the eigenvalue.

 * Find the lowest 3 eigenpairs ($$\lambda_i, v_i$$), for i=1,2,3.

>> lam(121:123,121:123)

ans =

11.83           0            0            0        25.81            0            0            0       30.927

 * In 3 separate plots, plot the eigenvectors $$v_1, v_2, v_3$$ as deformed shapes (solid line) superposed onto the underdeformed shape (dotted line). Again, use a magnification factor to clearly distinguish each eigenvector from the undeformed shape.

 * Also, you only need to plot an "incomplete" eigenmode for the frame model in the sense that there is only a straightline between two displaced nodes for each element, and not the *complete* deformed shape for each frame element.

 * given that $$\lambda = \omega^2$$ where omega is the circular frequency of vibration. The vibrational frequency is given by $$f = \omega / (2 \pi)$$ and the vibrational period is given by $$T = 1 / f = 2 \pi / \omega$$ Find the 3 lowest vibrational periods in the frame model of the electric pylon.

ans =

3.4395           0            0            0       5.0804            0            0            0       5.5612

 * Compare the 3 lowest vibrational periods for the frame model to those of the truss model.

=Animation= We want to create an animation of the 3 lowest eigenmodes of the electric pylon. To this end, we choose to animate the simpler truss model.

An animation is nothing but a graphic file in gif format in which you just include a sequence of many picture frames together just like in a movie.

So what sequence of picture frames would you create? Let $$v_i$$ be an eigenvector of the truss model (i=1,2,3), and let omega_i the corresponding circular vibrational frequency. According to the theory of vibration (to be done in class if time permits), the following time-dependent displacement matrix

$$d_i (t) = v_i sin (\omega_i t)$$

for t = 0 to t = T_i (the vibrational period of mode i) can be used to create a sequence of deformed shapes to create an animation. All you need to do now is to discretize the time domain into small intervals so you can produce the picture frames. For example, you can have a time sequence of

$$t_0 = 0, t_1, t_2, t_3, ..., t_n$$

with $$\delta_t = T / n$$

being the time increment. The value of n should be at least 8, i.e., each period should be subdivided into 8 intervals for a minimum animation. The larger n is the smoother the animation, but also the larger file size and the longer the download time. So a reasonable value of n would be 8 or 16.

=Code=

title: User:Eml4500.f08.group/HW7/Matlab


text:
User:Eml4500.f08.group/HW7/Contributors

Contributors

 * Tom Cady Eml4500.f08.group.cady 15:05, 8 December 2008 (UTC)

 * Aaron Clausen Eas4200c.f08.carbon.clausen 17:58, 8 December 2008 (UTC)

 * Julia Setlak Eml4500.f08.group.setlak 22:33, 8 December 2008 (UTC)

 * Jeff Wright Eml4500.f08.group.w 02:00, 9 December 2008 (UTC)

 * Brian Taylor Eml4500.f08.group.taylor 07:24, 9 December 2008 (UTC)

 * Melisa Gaar Eml4500.f08.group.gaar 21:55, 9 December 2008 (UTC)

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User:Eas4200c.f08.blue.b/Lecture 36

A plan for attacking the problem of solving for the shear flow in a multi-cell section with stringers was introduced.

This miniPLAN is shown below. S) Singe-cell sections                   S.1)  without stringers S.2) with stringers               M)  Multi-cell sections M.1) without stringers                   M.2)  with stringers

First, a single-cell section with no stringers(as in Figure 1) is analyzed to find the shear flow q through the entire section (S.1 from the miniPLAN). The section can be viewed as a thin walled cross section. As is shown below, a thin walled cross section cannot resist transverse shear (Vz).

Figure 1: Shear flow (q) in a single-cell section

Next, a single-cell section with stringers is analyzed as is shown in. The solution to this problem (P) can be analyzed by decomposing the problem into two parts: A section with stringers must be considered without stringers (as in problem P1 in Figure 3) and an open cross section with stringers (as in problem P2 in Figure 4).

Figure 2: Single-cell section with stringers

Figure 3: Single-cell section without stringers

Figure 4: Open single-cell section with stringers

P=P1+P2

It is known that the sum of shear forces in P is not equal to zero and that the shear flows between stringers in P are not equal but the shear flow is constant in the within each parcel. Therefore, the presence of stringers results in a non-constant shear flow.

$$R^{z}=\sum{V_{z}}\neq 0$$

$$q_{12}\neq q_{23}\neq q_{31}$$

$$q_{ij}(s)=const$$

Problem P1 is solved just as in S.1 of the miniPLAN and the resulting shear flows between stringers can be solved with the superposition of P1 and P2.

$$q_{ij}=q+\tilde{q}_{ij} $$

The method for solving the problem consists of the following steps:

1) Solve P2 for $$\tilde{q}_{12},\tilde{q}_{23},\tilde{q}_{31} $$

2) Take the moment about any point in the plane of the section

2.1) Superposition $$q_{ij}=q+\tilde{q}_{ij} $$

2.2) Selection point in plane and find moment about that point

title: User:Eas4200c.f08.blue.b/Lecture 36


text:
User:Eas4200c.f08.blue.a/Lecture 35

A plan for attacking the problem of solving for the shear flow in a multi-cell section with stringers was introduced.

This miniPLAN is shown below. S) Singe-cell sections                   S.1)  without stringers S.2) with stringers               M)  Multi-cell sections M.1) without stringers                   M.2)  with stringers

First, a single-cell section with no stringers(as in Figure 1) is analyzed to find the shear flow q through the entire section (S.1 from the miniPLAN). The section can be viewed as a thin walled cross section. As is shown below, a thin walled cross section cannot resist transverse shear (Vz).

Figure 1: Shear flow (q) in a single-cell section

Next, a single-cell section with stringers is analyzed as is shown in. The solution to this problem (P) can be analyzed by decomposing the problem into two parts: A section with stringers must be considered without stringers (as in problem P1 in Figure 3) and an open cross section with stringers (as in problem P2 in Figure 4).

Figure 2: Single-cell section with stringers

Figure 3: Single-cell section without stringers

Figure 4: Open single-cell section with stringers

P=P1+P2

It is known that the sum of shear forces in P is not equal to zero and that the shear flows between stringers in P are not equal but the shear flow is constant in the within each parcel. Therefore, the presence of stringers results in a non-constant shear flow.

$$R^{z}=\sum{V_{z}}\neq 0$$

$$q_{12}\neq q_{23}\neq q_{31}$$

$$q_{ij}(s)=const$$

Problem P1 is solved just as in S.1 of the miniPLAN and the resulting shear flows between stringers can be solved with the superposition of P1 and P2.

$$q_{ij}=q+\tilde{q}_{ij} $$

The method for solving the problem consists of the following steps:

1) Solve P2 for $$\tilde{q}_{12},\tilde{q}_{23},\tilde{q}_{31} $$

2) Take the moment about any point in the plane of the section

2.1) Superposition $$q_{ij}=q+\tilde{q}_{ij} $$

2.2) Selection point in plane and find moment about that point

title: User:Eas4200c.f08.blue.a/Lecture 35


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