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User:EML4500.f08.JAMAMA/LectureNotesHW2

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9/8/08
The Global Forces are determined by the following matrix

$$\begin{Bmatrix} f1\\ f2\\ f3\\ f4\\ f5\\ f6\\ \end{Bmatrix} = \begin{bmatrix} & &  &  &  & \\ &  &  &  &  & \\ &  &  K&  &  & \\ & &  &  &  & \\ &  &  &  &  & \\ &  &  &  &  & \end{bmatrix} \times \begin{Bmatrix} d1\\ d2\\ d3\\ d4\\ d5\\ d6\\ \end{Bmatrix} $$

6X1      6X6           6X1

Global Force      Global Stiffness           Global Displacement

Column Matrix      Matrix           Column Matrix

Before the global matrix can be constructed, you first need to find the displacements in the elements themselves. To do this for each element, the following equation may be used:

$$K^{\left(e \right) } \times D^{\left(e \right)} = F^{\left(e \right)} $$

4X4      4X1           4X1

where: $$K^{\left(e \right) }$$=Element stiffness matrix for element e

$$D^{\left(e \right)}$$=Element displacement matrix of element e

$$F^{\left(e \right)}$$=Element force matrix

Element Stiffness Matrix:

The element stiffness matrix $$K^{\left(e \right)}$$ for the equation above is calculated with the matrix shown (below).

$$k^{\left(e \right)}\times \begin{bmatrix} \left(l^{\left(e \right)} \right)^2 & l^{\left(e \right)}m^{\left(e \right)} & -\left(l^{\left(e \right)} \right)^2 & -l^{\left(e \right)}m^{\left(e \right)}\\ l^{\left(e \right)}m^{\left(e \right)} & \left(m^{\left(e \right)} \right)^2 & -l^{\left(e \right)}m^{\left(e \right)} & -\left(m^{\left(e \right)} \right)^2 \\ -\left(l^{\left(e \right)} \right)^2& -l^{\left(e \right)}m^{\left(e \right)} & \left(l^{\left(e \right)} \right)^2 & l^{\left(e \right)}m^{\left(e \right)} \\ -l^{\left(e \right)}m^{\left(e \right)}& -\left(m^{\left(e \right)} \right)^2 & l^{\left(e \right)}m^{\left(e \right)} & \left(m^{\left(e \right)} \right)^2 \end{bmatrix} $$

where: $$k^{\left(e \right)}= \frac{E^{\left(e \right)}A^{\left(e \right)}}{L^{\left(e \right)}}$$ Axial stiffness of bar element "e"

and $$l^{\left(e \right)}, m^{\left(e \right)}$$=Director cosines of the $$\tilde{x}$$ axis with respect to the (x,y) coordinates for element e.

Director Cosines

Director cosines relate the angle at which the element is on to the traditional x,y axis. These values are needed to calculate the stiffness matrix k.

$$l^{\left(e \right)}=\vec{\tilde{i}}\bullet \vec{i}=\cos \theta^{\left(e \right)}$$

$$m^{\left(e \right)}=\vec{\tilde{i}}\bullet \vec{j}=\cos \left(\frac{\pi }{2} - \theta^{\left(e \right)}\right)=\sin \theta ^{\left(e \right)}$$

$$\vec{\tilde{i}}=\cos \theta ^{\left(e \right)} \vec{i} + \sin \theta ^{\left(e \right)}\vec{j}$$

$$\vec{\tilde{i}}\bullet \vec{i}=\cos \theta ^{\left(e \right)} \vec{i}\bullet \vec{i} + \sin \theta ^{\left(e \right)}\vec{j}\bullet \vec{i}$$

$$\vec{\tilde{i}}\bullet \vec{j}=\cos \theta ^{\left(e \right)} \vec{i}\bullet \vec{j} + \sin \theta ^{\left(e \right)}\vec{j}\bullet \vec{j}$$

9/10/08
Model of the bar turss system (continued)

Elem. 1
θ(1) = 30°

l(1) = cos(1)(θ) = cos(1)(30°) = $$ \frac{\sqrt{3}}{2} $$

m(1) = sin(1)(θ) = sin(1)(30°) = $$ \frac{1}{2} $$

k(1) = $$ \frac{ E^{(1)} * A^{(1)}}{L^{(1)}} = \frac {(3)(1)}{4} = \frac {3}{4} $$

k(1)= $$ \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} \\

k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} \\

k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)} & k_{34}^{(1)} \\

k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)} & k_{44}^{(1)}

\end{bmatrix}$$ = [k$$_{ij}^{(e)}]$$4 × 4, where $$_{j = 1,2,3,4\  \Rightarrow \ column\ index}^{i = 1,2,3,4\ \Rightarrow \ row\ index}$$  and $$e $$ = number of element

$$k_{11}^{(1)} = k^{(1)} [l^{(1)}]^2 = \frac {3}{4} * [\frac{\sqrt{3}}{2}]^2 = \frac {9}{16}$$

$$k_{12}^{(1)} = k^{(1)} [l^{(1)}m^{(1)}] = \frac {3}{4} * \frac{\sqrt{3}}{2}]*\frac{1}{2} = \frac {3\sqrt{3}}{16}$$

$$k_{42}^{(1)} = k^{(1)} (-[m^{(1)}]^2) = \frac {3}{4} * (-\frac{1}{4}) = -\frac {3}{16}$$

Note: this matrix is symetircal with respect to the diagnol $$k_{xx}^{(1)}, where\; x= 1,2,3... \to k_{13}^{(1)}=k_{31}^{(1)}. $$

In genral, $$ k_{ij}^{(e)}= k_{ji}^{(e)}\ or\ [{\mathbf{k}^{(e)}}]^T = \mathbf{k}^{(e)}$$

Elem. 2

k(2) = $$ \frac{ E^{(2)} * A^{(2)}}{L^{(2)}} = \frac {(5)(2)}{2} = 5 $$

θ(2) $$= -45^\circ \Rightarrow _{m^{(2)} =\ sin(-45^\circ)\ =-\frac{\sqrt{2}}{2}}^{l^{(2)} =\ cos(-45^\circ)\ =\frac{\sqrt{2}}{2}} $$

$$\mathbf{k}^{(e)}$$ = [k$$_{ij}^{(e)}]$$4 × 4$$,\qquad \begin{matrix} k_{11}^{(2)} = k^{(2)} [l^{(2)}]^2 = 5 * [\frac{\sqrt{2}}{2}]^2 = \frac {5}{2}\\ k_{12}^{(2)} = k^{(2)} l^{(2)}m^{(2)} = 5 * \frac{\sqrt{2}}{2}* (-\frac{\sqrt{2}}{2}) = -\frac {5}{2}\\ k_{42}^{(2)} = k^{(2)} (-[m^{(2)}]^2) = 5 * (-[\frac{\sqrt{2}}{2}]^2) = -\frac {5}{2} \end{matrix} $$

Observe: Absolute values for all the coeficients are the same [k$$_{ij}^{(e)}]$$4 × 4, where $$ e =\ 2,and\ _{j = 1,2,3,4}^{i = 1,2,3,4}$$. All that differs are the positive/negative signs in front of the magnitudes of the coeficients.

Element F-D relation: $$ \mathbf{k}_{4 \times 4}^{e}*\mathbf{d}_{4\times1}^{e} = \mathbf{f}_{4\times1}^{e} $$

$$ \mathbf{d}^{e}\ ={ \begin{Bmatrix} d_1^{(e)}\\ \vdots\\ d_4^{(e)} \end{Bmatrix}}_{4 \times 1};\quad

\mathbf{f}^{e}\ = {\begin{Bmatrix} f_1^{(e)}\\ \vdots\\ f_4^{(e)} \end{Bmatrix}}_{4 \times 1} $$

Global F-D Relation("free-free" structure)

$$ \mathbf{K}_{n \times n}*\mathbf{d}_{n\times1} = \mathbf{F}_{n\times1} \Longrightarrow $$ there are $$\mathbf{6} $$ degrees of freedom (6 forces are acting upon the sturcture) $$\quad \Rightarrow \mathbf{n=6} $$

9/12/08
Global Force Displacement Relationship

 * Note: The following matrix formulas show that there is a difference between the notation for the stiffness and force matrices $$k$$, $$K$$, and $$f$$, $$F$$.  For example, $$K$$ is the global stiffness matrix, and $$k$$ is the element stiffness matrix.

The matrix form for the example problem in progress is as follows:

$$ \begin{bmatrix} k_{11} & k_{12} & k_{13} & k_{14} & k_{15} & k_{16}\\ k_{21} & k_{22} & k_{23} & k_{24} & k_{25} & k_{26}\\ k_{31} & k_{32} & k_{33} & k_{34} & k_{35} & k_{36}\\ k_{41} & k_{42} & k_{43} & k_{44} & k_{45} & k_{46}\\ k_{51} & k_{52} & k_{53} & k_{54} & k_{55} & k_{56}\\ k_{61} & k_{62} & k_{63} & k_{64} & k_{65} & k_{66} \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix} = \begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix} $$

The matrix formula for the element matrices is written as follows:

In order to express the element matrices in global matrices to solve for the displacements, there is an assembly process.

You must identify the correspondence between the element displacement degrees of freedom and the global displacement degrees of freedom.

For the current example, the global displacement degrees of freedom are: $$\{d_1,d_2,d_3,d_4,d_4,d_5,d_6\}$$

At the element level:

Element 1: $$\left\{d_1^{(1)},d_2^{(1)},d_3^{(1)},d_4^{(1)},d_5^{(1)},d_6^{(1)}\right\}$$

Element 2: $$\left\{d_1^{(2)},d_2^{(2)},d_3^{(2)},d_4^{(2)},d_5^{(2)},d_6^{(2)}\right\}$$

Then it is necessary to identify the global - element degrees of freedom. In other words, determine the element displacement degrees of freedom that are actually equivalent to one another if the separate elements were to be put back together, as well as which are in turn actually describing a global displacement degree of freedom:

$$\left. \begin{array}{c} d_1 = d^{(1)}_1 \\ d_2 = d^{(1)}_2 \end{array} \right\}$$ at Global Node 1

$$\left. \begin{array}{c} d_3 = d^{(1)}_3 = d^{(2)}_1\\ d_4 = d^{(1)}_4 = d^{(2)}_2 \end{array} \right\}$$ at Global Node 2

$$\left. \begin{array}{c} d_5 = d^{(2)}_3\\ d_6 = d^{(2)}_4 \end{array} \right\}$$ at Global Node 3

Conceptual Step of Assembly: Topology of K

After calculation of the local stiffness matrices K1 and K2, they can be combined following the conceptual step of assembly method called: The Topology of K as follows:

Within the intersection of the two matrices, the terms that would overlap there are simply added together. (See 9/15/08 for the matrix completed for the current example.)

9/15/08
The resulting global stiffness matrix is:

$$\begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\[1ex] k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\[1ex] k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)}+k_{11}^{(2)}) & (k_{34}^{(1)}+k_{12}^{(2)}) & k_{13}^{(2)} & k_{14}^{(2)}\\[1ex] k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)}+k_{21}^{(2)}) & (k_{44}^{(1)}+k_{22}^{(2)}) & k_{23}^{(2)} & k_{24}^{(2)}\\[1ex] 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)}\\[1ex] 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)}\\ \end{bmatrix}$$

A sample calculation of the values of K:

 * $$K_{11} = k_{11}^{(1)}=\dfrac{9}{16}$$
 * $$K_{12} = k_{12}^{(1)}=\dfrac{3\sqrt 3}{16}$$
 * $$K_{33} = k_{33}^{(1)}+k_{11}^{(2)}=\dfrac{9}{16}+\dfrac{5}{2}=3.0625$$
 * $$K_{34} = k_{34}^{(1)}+k_{12}^{(2)}=\dfrac{3\sqrt 3}{16}+\left(-\dfrac{5}{2}\right)=-2.175$$
 * $$K_{43} = K_{34}=k_{43}^{(1)}+k_{21}^{(2)}=k_{34}^{(1)}+k_{12}^{(2)}=-2.175$$
 * $$K_{44} = k_{44}^{(1)}+k_{22}^{(2)}=\dfrac{3}{16}+\dfrac{5}{2}=2.6875$$

9/17/08
Step 4: Introduction of Boundary Equations

In this step eliminating known degrees of freedom will reduce the global Force Displacement Relationship. By knowing the displacements for global nodes 1 and 3 are zero because of the boundary conditions expressed in the problem statement. The boundary conditions have been applied to the global stiffness matrix below.

$$ \begin{bmatrix} K_{11} & K_{12} & K_{13} & K_{14} & K_{15} & K_{16}\\ K_{21} & K_{22} & K_{23} & K_{24} & K_{25} & K_{26}\\ K_{31} & K_{32} & K_{33} & K_{34} & K_{35} & K_{36}\\ K_{41} & K_{42} & K_{43} & K_{44} & K_{45} & K_{46}\\ K_{51} & K_{52} & K_{53} & K_{54} & K_{55} & K_{56}\\ K_{61} & K_{62} & K_{63} & K_{64} & K_{65} & K_{66} \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ d_3 \\ d_4 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} K_{13} & K_{14}\\ K_{23} & K_{24}\\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix} \begin{bmatrix} d_3 \\ d_4 \end{bmatrix} = \begin{bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{bmatrix} $$

The global stiffness matrix reduces to a 6X2 matrix when applying the boundary equations. The top and bottom two rows are also eliminated due to the principle of virtual work. The Force Displacement relation then becomes the equation shown below.

$$\begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44} \end{bmatrix} \begin{bmatrix} d_3 \\ d_4 \end{bmatrix} = \begin{bmatrix} F_3 \\ F_4 \end{bmatrix}$$

To solve for the displacement of global node 2. The equation above needs to be multiplied on each side by K-1, and the values for the stiffness matrix and the force matrix inserted. This solves d3 and d4 to be 4.352 and 6.1271 respectively.

9/19/08
In order to solve for the displacements of the example, each element is considered:

The displacements are found by solving the 4x4 matrix, after the eliminations discussed previously. This method provides the solutions:

$$\underline{k}^{(1)} \underline{d}^{(1)} = \underline{f}^{(1)}$$

$$\underline{k}^{(2)} \underline{d}^{(2)} = \underline{f}^{(2)}$$

The internal forces for nodes 1 and 3 will be zero because they are fixed points. (See "Boundary Conditions") Therefore, the matrix will reduce to:

$$ \begin{bmatrix} 3.0625 & 2.7152\\ -2.7152 & 2.6875 \end{bmatrix} \begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix} = \begin{Bmatrix} F_3 \\ F_4 \end{Bmatrix} $$

Where the known external forces will be plugged in.

To continue, for example, using specifically element 1, the now known displacements from the previous step, can be plugged back in for the variable labels, and the reactions at the supports can be found. This is the last step.

$$ \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0\\ 0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0\\  0      & 0      & -2.5    & -2.5    &2.5&2.5\\  0      & 0      & -2.5    & -2.5    &2.5&2.5 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix} = \begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2 \\ f^{(1)}_3 \\ f^{(1)}_4 \end{Bmatrix} $$

Where the subscripts 1 and 2 denote the reactions at the supports, and the subscripts 3 and 4 denote the internal forces. (As labeled on the free body diagrams for element 1 above.)

The solutions of the reactions and internal forces for element 1 are as follows:

$$ \begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2 \\ f^{(1)}_3 \\ f^{(1)}_4 \end{Bmatrix} = \begin{Bmatrix} -4.4378 \\ -2.5622 \\ 4.4378 \\ -2.5622 \end{Bmatrix} $$

Take note of the fact that each of the internal forces and reactions for the x and y directions will sum to zero, providing an algebraic proof that the element is in equilibrium.

$$\begin{array}{l} \sum f_1^{(1)}+f_2^{(1)}=0\\ \sum f_3^{(1)}+f_4^{(1)}=0 \end{array}$$

Solving a Two-Bar Truss System using MatLab
The same system can be solved using MatLab.

The following code solves for the parameters: "Stiffness matrices, displacements, and reactions."

Here are the calculaled results from MatLab:

It can be observed that the values that were calculated previously are similar to the ones calculated using Matlab.

The two "k_local" that MatLab calculated are our k(1) and k(2) respectively.

The two lowercase k's (k) from MatLab are our local stiffness matrices k(1) and k(2).

The Uppercase k (K) found in the Matlab Results is equivalent to our global stiffness matrix K.

The "R" matrix is the known forces matrix.

The "d" matrix is the displacement matrix.

The "reaction" matrix is for the reaction forces.

Contributing Team Members
--Eml4500.f08.jamama.justin 18:26, 26 September 2008 (UTC)

Cedric Adam--Eml4500.f08.jamama.adam 18:29, 26 September 2008 (UTC)

William Mueller Eml4500.f08.jamama.mueller 19:07, 26 September 2008 (UTC)

Chris Alford--Eml4500.f08.jamama.chris 20:28, 26 September 2008 (UTC)

Megan Alvarez - --EML4500.f08.jamama.megan 21:00, 26 September 2008 (UTC)

Arunas Janulevicius --Eml4500.f08.jamama.jan 22:42, 26 September 2008 (UTC)

title: User:EML4500.f08.JAMAMA/LectureNotesHW2


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User:Eml4500.f08.team.allen.oca/hw2-1

P. 8-1 In compact notation, the Global Force Displacement Equation is $$\left[K_{ij} \right]_{6x6} \left\{d_{j} \right\}_{6x1}=\left\{F_{i} \right\}_{6x1}$$ Here the $$\mathbf{K}$$ matrix is generally n by n. Note that a capital K is a global stiffness matrix and a lowercase k is a element stiffness matrix. $$\sum_{j=1}^{6}{k_{ij}d_{j}}=F_{i} $$ where i = 1,...,6 The Global Stiffness Matrix is $$\mathbf{K}_{nxn} = \left[k_{ij} \right]_{nxn}$$ The Global Displacement Matrix is $$\mathbf{d}_{nxn}=\left\{d_{j} \right\}_{nxn}$$ P. 8-2 Recall the Element Force Displacement rel. from P. 7-5 $$\mathbf{k}_{4x4}^{e}\mathbf{d}_{4x1}^{e}=\mathbf{f}_{4x1}^{e}$$ How would one go from element matrices (stiffness, displacement and force) to global matrices? The answer is through an assembly process: $$\bullet$$ Identify the correspondence between elements displacement dofs and global displacement dofs.

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

title: User:Eml4500.f08.team.allen.oca/hw2-1


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User:Eml4500.f08.team.allen.oca/hw2-2

P. 10-1 Previously, the global matrices were found to be

$$\mathbf{K}_{6x6} \; \begin{Bmatrix} d_{1}=0\\ d_{2}=0\\ d_{3}\\ d_{4}\\ d_{5}=0\\ d_{6}=0\\ \end{Bmatrix}_{6x6} = \begin{bmatrix} K_{13} &K_{14} \\ K_{23} &K_{24} \\ K_{33} &K_{34} \\ K_{43} &K_{44} \\ K_{53} &K_{54} \\ K_{63} &K_{64} \end{bmatrix}_{6x6}

\begin{Bmatrix} d_{3}\\ d_{4}\\ \end{Bmatrix}_{2x1} =\mathbf{F}_{6x1} $$ More rows in the K matrix can be removed by using the principle of virtual work. Also, incorporating the fixed boundary conditions leds to the deletion of the corresponding columns in the global stiffness matrix K. By principle of virtual work (PVW) the corresponding rows are also deleted, here rows 1, 2, 5, and 6. The corresponding rows in F are deleted as well. P. 10-2 The resulting Force Displacement relationship becomes $$ \begin{bmatrix} K_{3x3} &K_{3x4} \\ K_{4x3} &K_{4x4} \end{bmatrix}_{2x2} \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix}_{2x1} = \begin{Bmatrix} F_{3}\\ F_{4} \end{Bmatrix}_{2x1} $$ which can alternately be written $$ \mathbf{\bar{K}}_{2x2} \mathbf{\bar{d}}_{2x1} = \mathbf{\bar{F}}_{2x1} $$ $$ \mathbf{\bar{F}}= \begin{Bmatrix} F_{3}\\ F_{4} \end{Bmatrix} = \begin{Bmatrix} 0\\ P \end{Bmatrix} $$ where F3 is known to be 0 and P is 7. Recall that for $$\mathbf{\bar{K}}_{2x2}$$ above, the determinate is $$det\; \mathbf{\bar{K}}=k_{33}k_{44}-k_{34}k_{43}$$ and also that the inverse of $$\mathbf{\bar{K}}_{2x2}$$ is $$ \mathbf{k^{-1}}=\frac{1}{det \; \mathbf{k}} \begin{bmatrix} k_{44} &-k_{34} \\ -k_{43} & k_{33} \end{bmatrix} $$ P. 10-3

Verifying the above two properties $$ \mathbf{\bar{K}}\; \mathbf{\bar{K}}^{-1} =\mathbf{\bar{K}}^{-1}\; \mathbf{\bar{K}} = \mathbf{I}=\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix} = \frac{1}{det \boldsymbol{\bar{k}}} \begin{bmatrix} det \boldsymbol{\bar{k}} &0 \\ 0 & det \boldsymbol{\bar{k}} \end{bmatrix} $$

For finite element method, $$ k_{43} = k_{34}$$? Note: $$ \mathbf{\bar{k}}^{T}=\begin{bmatrix} k_{33} &k_{43} \\ k_{34} & k_{44} \end{bmatrix} $$ $$ \mathbf{\bar{k}}^{-1}\neq \frac{1}{det\mathbf{\bar{k}}}\mathbf{\bar{k}}^{T} $$ Row echelon method

Back to 2D bar truss system: $$ \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix} = \mathbf{\bar{k}}^{-1} \begin{Bmatrix} 0\\ \mathbf{P} \end{Bmatrix} = \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix} = \mathbf{\bar{k}}^{-1} \begin{Bmatrix} 4.352\\ 6.1271 \end{Bmatrix} $$ which is the displacement of node 2.

title: User:Eml4500.f08.team.allen.oca/hw2-2


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User:Waldyrious

Waldir@meta.wikimedia

title: User:Waldyrious


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User:Eml4500.f08.Ateam.philbin/HW1

September 8, 2008 - Lecture 6

Statically Determinate versus Statically Indeterminate
A statically determinate problem is one that can be solved using simple statics relationships. There are three main equations, summing forces in the x and the y directions and solving for the moment about a point. With these three equations, any problem with up to three unknowns can be solved. Problems with multiple beams or more than three reaction forces are statically indeterminate since there are insufficient equations to solve for all the unknowns.

An exception to the indeterminate rules is shown in the following example.

Here there are four reaction forces, two at each support and a known applied force at the top. It can be noted that the beams in this problem are both two-force members. By looking at each element separately and summing the forces in any direction it is found that they must be both equal and opposite. Summing the moment about either end shows that both forces' line of action must go through each end. Otherwise the moments would not sum to zero.

Therefore the two forces are constrained in one direction. With these simple straight beams, the line of action is along the beam itself. If the beam were curved or had an angle in it, the same logic would apply and the line of action of the two forces will be along the invisible line going through the two attachment points.

This simplification of the problem lowers the degrees of freedom (unknowns) from four to three. Since the direction of the reaction forces are now known, only the magnitude is left to be found.

Lecture Notes
2) Element picture

Here the element degrees of freedom and element forces are shown in either the global system or in a local coordinate system. The element number is in the parentheses and the forces (and displacements) are numbered from node 1 to 3, each starting with the x and then y axes.

3) Global force / displacement relationship

The next step is to create a relationship between the forces and displacements in the global coordinate system. The matrices will be in the following form.

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

Where k (e) is a 4x4 matrix and d (e) and f (e) are 4x1 matrices. In order to find the values of k (e), the formula found on page 225 of Fundamental Finite Element Analysys and Applications must be used. It states that the scalar, $$K = \frac{EA}{L}$$, is multiplied by a matrix composed of functions of the director of cosines.

The director of cosines is a function of the angle of the element and given as, $$l^{(e)}= \vec \tilde{i} \centerdot \vec i = \cos \ \Theta^{(e)}$$ and  $$m^{(e)}= \vec \tilde{i} \centerdot \vec j = \sin \ \Theta^{(e)}$$. The completed matrix for K (e) is as follows.

$$ \underline K^{(e)}= K*\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)} \\ l^{(e)}m^{(e)} & (m^{(e)})^2 & -l^{(e)}m^{(e)} & -(m^{(e)})^2 \\ -(l^{(e)})^2 & -l^{(e)}m^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)} \\ -l^{(e)}m^{(e)} & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2 \end{bmatrix} $$

The k (e) matrices for elements 1 and 2 were calculated based on the given data and are as follows.

$$ \underline K^{(1)}= \begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & \frac{-9}{16} & \frac{-3\sqrt{3}}{16} \\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & \frac{-3\sqrt{3}}{16} & \frac{-3}{16} \\ \frac{-9}{16} & \frac{-3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16} \\ \frac{-3\sqrt{3}}{16} & \frac{-3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix}=\begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 \\ -0.5625 & -0.3248 & 0.5625 & 0.3248 \\ -0.3248 & -0.1875 & 0.3248 & 0.1875 \end{bmatrix} $$

$$ \underline K^{(2)}= \begin{bmatrix} \frac{5}{2} & \frac{-5}{2} & \frac{-5}{2} & \frac{5}{2} \\ \frac{-5}{2} & \frac{5}{2} & \frac{5}{2} & \frac{-5}{2} \\ \frac{-5}{2} & \frac{5}{2} & \frac{5}{2} & \frac{-5}{2} \\ \frac{5}{2} & \frac{-5}{2} & \frac{-5}{2} & \frac{5}{2} \end{bmatrix}=\begin{bmatrix} 2.5 & -2.5 & -2.5 & 2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ 2.5 & -2.5 & -2.5 & 2.5 \end{bmatrix} $$

title: User:Eml4500.f08.Ateam.philbin/HW1


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User:Eml4500.f08.ateam.carr/HW2

Since displacements of Element 1 are now known, the reaction forces can be determined.

Recall Element 1:

$$ k^{(1)} \cdot d^{(1)} = f^{(1)} $$

Where k and d have previously been defined:

k^{(1)} = \begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix} $$

d^{(1)} =

\begin{bmatrix} 0\\     0\\      4.352\\      6.1271\\     \end{bmatrix} $$

To compute the reaction forces in Element 1, simply multiply the k and d matrix of Element 1. Note that the first two columns of matrix k will be multiplied to zero and can be ignored if a manual calculation is performed.

\begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix}\cdot

\begin{bmatrix} 0\\     0\\      4.352\\      6.1271\\     \end{bmatrix}

= F^{(1)}

$$

Executing matrix multiplication to solve for the reaction forces, F:

\begin{bmatrix} -4.4378\\ -2.5622\\ 4.4378\\ 2.5622\\ \end{bmatrix}= \begin{bmatrix} f_{1}^{(1)}\\ f_{2}^{(1)}\\ f_{3}^{(1)}\\ f_{4}^{(1)} \end{bmatrix}

$$

Observation: Note Element 1 is in equilibrium. If the sum of the forces in the x and y direction are computed, one will note that they sum to zero.

 * $$\sum f_{1}^{(1)} + f_{3}^{(1)}=0$$
 * $$\sum f_{2}^{(1)} + f_{4}^{(1)}=0$$

HW: Verify also that:

To prove that the magnitude and direction of the forces acting on Element 1 are equal and opposite, first compute the magnitude of the reaction. This is done by the root-sum-square method:

 * $$P_1^{(1)}=[(f_1^{(1)})^2 + (f_2^{(1)})^2]^{1/2}$$

 * $$P_1^{(1)}=[(-4.4374)^{2} + (-2.5622)^{2}]^{1/2} $$

 * $$P_1^{(1)}=5.124 $$

The magnitude of P2 is computed the same way:

 * $$P_2^{(1)}=[(f_3^{(1)})^2 + (f_4^{(1)})^2]^{1/2}$$

 * $$P_2^{(1)}=[(4.4374)^{2} + (2.5622)^{2}]^{1/2} $$

 * $$P_2^{(1)}=5.124 $$

Thus, the forces are equal in magnitude but opposite in direction. The element is in equilibrium.

Element 2 is also in equilibrium and this can be proven the same way as above.

Method 2: To solve for 2 bar truss system using statics method:

Two different cut methods can be used as demonstrated above. The red outlines the different applications of the Euler cut principle. Individual free body diagrams can be drawn for each element circled in red and the problem can be solved as statically determinate.

HW: Verify the equilibrium of node 2 statically:

 * $$\sum F_{x}=0=-P_{1}^{(1)}cos(30) + P_{2}^{(1)}cos(45)$$
 * $$\sum F_{y}=0=P - P_{1}^{(1)}sin(30) - P_{2}^{(1)}sin(45)$$

P, P1, and P2 are known from previous steps:

 * $$P_{applied}^{} = 7 $$
 * $$P_{1}^{(1)}=5.124$$
 * $$P_{2}^{(2)}=6.276$$

Plugging these known values into the force equations:

 * $$\sum F_{x}=-5.124 \cdot cos(30) + 6.276 \cdot cos(45)$$
 * $$\sum F_{y}=7 - 5.124 \cdot sin(30) - 6.276 \cdot sin(45)$$

 * $$\sum F_{x}=0$$
 * $$\sum F_{y}=0$$

Global node 2 is in equilibrium.

title: User:Eml4500.f08.ateam.carr/HW2


text:
User:Eas4200c.f08.vqcrew.d/hw2

Problem 1.1
Given Figure 1, find the optimum ratio $$\frac{b}{a}$$  to maximize the load bearing capability of the thin-walled beam.

The assumptions for this problem are as follows:
 * 1) The cross-sectional perimeter is constant, i.e. $$L=2(a+b)=Constant$$
 * 2) The torque $$T$$ is equal to the bending moment $$M$$, i.e. $$T = M$$
 * 3) The maximum allowable bending normal stress is twice the maximum allowable shear stress, i.e. $$\sigma_{allow} = 2\tau_{allow}$$

The assumption that $$t$$ is very small compared to $$a$$ and $$b$$ is also valid.

Since the cross-section walls are very thin, the shear stress can be assumed to be uniform along the walls, as shown in the shear flow diagram in Figure 2. The shear distribution and resultant shear vector $$V$$ for a small piece of the wall is also shown. This distribution is also mechanically equivalent to the linear distribution shown in Figure 3, where $$\tau =\frac{V}{t}$$.

The total torque ($$T$$) can be described as the sum of the torques on each wall of the box beam as follows: $$T = T_{AB}+T_{BC}+T_{CD}+T_{DA}$$

$$T_{AB}=(\frac{b}{2})Va=\frac{1}{2}\tau tab$$

$$T_{BC}=(\frac{a}{2})Vb=\frac{1}{2}\tau tab$$

$$T_{CD}=(\frac{b}{2})Va=\frac{1}{2}\tau tab$$

$$T_{DA}=(\frac{a}{2})Vb=\frac{1}{2}\tau tab$$

Therefore, $$T=2t\tau ab$$, and $$\tau =\frac{T}{2abt}$$.

There are two cases that must be analyzed in order to solve the problem:
 * 1) Assume the bending normal stress (σ) reaches $$\sigma_{allow}$$ first, then verify that $$\tau \le \tau_{allow}$$.
 * 2) Assume the shear stress (τ) reaches $$\tau_{allow}$$ first, then verify that $$\sigma \le \sigma_{allow}$$.

Case 1

Case 1 begins by recalling the formula for bending normal stress: $$\sigma =\frac{Mz}{I}$$, where $$M$$ is the bending moment, $$z$$ is the ordinate of a point on the axis perpendicular to the bending neutral axis, and I is the second area moment of inertia defined by $$I=\iint_A z^2 \ dy\, dz$$. In this case, $$z=\frac{b}{2}$$. Figure 4 shows each item pictorially.

Rewriting this equation yields $$M=\frac{2I\sigma _{max}}{b}$$, where $$\sigma _{max}=\sigma _{allow}$$ from the Case 1 assumption. We proceed to find $$M_{max}$$ by maximizing the ratio of $$\frac{I}{b}$$.

The expression for finding the moment of inertia $$I$$ of the box beam is: $$I=\sum_{i=1}^{4}{\left[\frac{b_{i}h_{i}^3}{12} + A_{i}d_i^2\right]}$$, where $$i$$ represents each wall of the box beam, varying from one to four. This simplifies to $$I=2 \frac{tb^3}{12}+2\left[\frac{at^3}{12}+\frac{atb^2}{4} \right]$$.

The first sum is the result of the vertical bars, and the second sum is the result of the horizontal bars. Because $$t\ll b$$, the $$t^3$$ term will be very small compared to $$b$$ and can be neglected. After some algebra, $$I$$ can be simplified to: $$I=\frac{tb^2}{6}\left(3a+b \right) $$.

Let us now define a function $$f(b)\equiv \frac{I}{b}$$ and maximize it to find the maximum bending moment, $$M_{max}$$. By using Assumption 1, we can substitute $$a$$ with $$a=\frac{L}{2}-b$$. Replacing this in the expression for $$I$$ results in the concave-downward quadratic shown in Figure 5: $$f(b)=\frac{tb}{12}\left(3L-4b \right)$$. By setting $$f(b)=0$$, the roots are shown to be $$b=0$$ and $$b=\frac{3L}{4}$$. Since the function is a quadratic, the maximum value of $$f(b)$$ is located halfway between the two roots at $$b^{(1)}=\frac{3L}{8}$$. By Assumption 1, we find $$a^{(1)}=\frac{L}{8}$$.

The maximum bending moment $$M_{max}^{(1)}$$ can then be written as:

$$M_{max}^{(1)} = \left(2\sigma _{allow} \right)\left(\frac{I^{(1)}}{b^{(1)}} \right) = \left(2\sigma _{allow} \right)\left(\frac{tb^{(1)}}{6} \right)\left(3a^{(1)}+b^{(1)} \right)$$

Substituting in for $$a^{(1)}$$ and $$b^{(1)}$$ yields: $$M_{max}^{(1)} = \frac{3tL^2}{32}\sigma _{allow}$$

This is also equal to $$T_{max}^{(1)}$$ based on Assumption 2. We will now use this assumption and the shear stress equation to determine the maximum shear stress for Case 1.

$$\tau _{max}=\frac{T_{max}^{(1)}}{2a^{(1)}b^{(1)}t} = \frac{\frac{3tL^2}{32}\sigma _{allow}}{2\frac{L}{8}\frac{3L}{8}t}$$

This reduces to:

$$\tau _{max}= \sigma _{allow}$$

Substituting Assumption 3 for $$\tau _{max}$$ gives:

$$\tau _{max}= 2 \tau _{allow}$$.

Therefore, Case 1 is an unacceptable solution because $$\tau _{max} > \tau _{allow}$$.

title: User:Eas4200c.f08.vqcrew.d/hw2


text: User:Eml4500.f08.qwiki.nobles/monobook.js


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text:
User talk:Takei~enwikiversity

 Hello Takei, 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

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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! --Emesee mobi 02:08, 25 September 2008 (UTC)

Hello, I'm curious what the pages that you have created refer to... Emesee mobi 02:08, 25 September 2008 (UTC)

Japanese Wikiversity
Hi, I was wondering if you know about 日本語? --mikeu talk 14:38, 29 September 2008 (UTC)
 * Konichiwa, Takei. Do you think ja: could be a better place for your contributions? -- Jtneill - Talk - c 13:35, 4 October 2008 (UTC)

Copyright problem with Image:01.gif

 * And other files the same:-(--Juan 00:13, 28 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 Takei. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Takei~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 23:44, 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) 06:23, 19 April 2015 (UTC)

title: User talk:Takei~enwikiversity


text: File:Director cosines.JPG

title: File:Director cosines.JPG


text:
User:Eas4200c.f08.vqcrew.E/Homework1

Introduction to Aerospace Structures
The main consideration during the aircraft design process is to reduce weight while maintaining strength in the structure. Strength, stiffness, and toughness all play important roles in the overall performance of an aircraft. The role of a design engineer is to create a balance of these properties that best satisfies all of the requirements of the aircraft.

 * Strength - the resistance of a material to deformation and rupture.
 * The yield stress, $$\displaystyle \sigma_Y$$, of a material is the point at which the material will begin to plastically and permanently deform.
 * The ultimate stress, $$\displaystyle \sigma_u$$, of a material is the critical point at which the material fails catastrophically.
 * Stiffness - the ability of a material to resist deformation caused by forces acting on the object.
 * The strain, $$\displaystyle \epsilon$$, of a material is the measure of the deformation caused by an applied force.
 * Young's modulus, $$\displaystyle E$$, is the measure of a material's stiffness. Young's modulus is the ratio of the stress applied over the strain caused in response, $$\displaystyle E = \sigma/\epsilon$$.  Figure 1 displays the stress-strain curve for steel.  Young's modulus represents the slope of the curve in the linear, or "elastic", region.  In this region, no permanent or "plastic" deformation has taken place and the material will return to its original dimensions.
 * Toughness - the overall resilience of a material and resistance to fracture. Toughness is measure in units of energy per unit volume a material can withstand before failure.

Examles of Materials and Their Properties
 Glass  - high stiffness, low toughness. Unlikely to bend or deform, yet vulnerable to fracture.

 Plastics  - low stiffness, high toughness. Easily deformed, but resistant to rupture.

 Metals  - Aluminum has high toughness, but is weak in terms of stiffness. Aluminum will not resistant deformation under high loads. Steel alloys are better suited for a situation requiring high stress handling capabilities.

All of these materials have their strengths and weaknesses. All of these materials can be used safely and effectively as long as these constraints are recognized. In the search for highly stiff, highly tough materials, the strengths of different materials were combined and their weaknesses diminished. Materials known as composites are formed from a network, or "matrix", of strong, flexible fibers held in place by stiff epoxy. The union of these materials creates a new, strong and lightweight material ideal for the aerospace industry.

In order to optimize performance, aerodynamic considerations must be made during the design process. Even though a material is structurally robust, there many be other factors that make that material detrimental to the overall function of the aircraft. For example, a material with vulnerability to cold or corrosion could not be used on the outer skin of an aircraft. The shape of major features of the aircraft must function both structurally and aerodynamically. The fuselage must be strong enough to withstand years of stress associated with pressurizing and depressurizing the aircraft while also maintaining a small cross-sectional area to keep aerodynamic drag to a minimum. The wings must be shaped to create lift to allow the aircraft to fly, but they must also act to support the weight of the fuselage.

Fall 2008 EAS4200C Introduction

title: User:Eas4200c.f08.vqcrew.E/Homework1


text:
User:Gien-Debbido

Hi.

 * Well I'm a spanish native speaker, I can speak English fluently, I can understand a little French, I can read Russian and a few words of Japanese and Arabic, altough I'm a beginner on the last three.

 * If I can help you with anything do not hesitate to ask me.

title: User:Gien-Debbido


text:
User:Eas4200c.f08.vqcrew.E/Homework2

Case 2:

For this case, the assumption is made that $$ \tau_{max} = \tau_{allow}$$  (i.e.  $$ \tau_{max}$$  reaches  $$ \tau_{allow}$$  first).

$$ \tau = \frac{T}{2abt} = \tau_{max} = \tau_{allow} $$

Rearrange to solve for the torque, T.

 $$ T = (2t\tau_{allow})(ab) $$

 $$ T_{max} = (2t\tau_{allow})(ab)_{max} $$

From Assumption 1, the perimeter of the beam is constant, L = 2(a + b). This allows for a to be solved in terms of only L and b.

$$a = \frac{L}{2} - b$$

Now, substituting in for a the maximum torque becomes

 $$ T_{max} = (2t\tau_{allow})(\frac{bL}{2} - b^{2})_{max} $$

The first term in the maximum torque equation $$(2t\tau_{allow}) $$  is a constant. In order to maximize the torque, the second term $$(\frac{bL}{2} - b^{2})$$ must be maximized. A function f(b) is defined as the second term to allow for the maximum to be found.

 $$ f(b) = (\frac{bL}{2} - b^{2}) $$

The derivative of f(b) is taken and set to zero to find the maximum of the function.

 $$ f'(b) = (\frac{L}{2} - 2b) = 0 \Rightarrow b^{(2)} = \frac{L}{4} $$

Next, $$ a^{(2)}$$ is found through the perimeter relation.

 $$ a^{(2)} = (\frac{L}{2} - \frac{L}{4}) = \frac{L}{4}$$

The value of $$ T^{(2)}$$ can now be determined and equated to the maximum moment by Assumption 2, T = M.

 $$ T^{(2)}_{max} = (2t\tau_{allow})(\frac{L}{4})^{2} = \frac{1}{8}tL^{2}\tau_{allow} = M_{max}^{(2)}$$

The allowable normal stress can now be substituted in to the equation per Assumption 3, $$\sigma_{allow} = 2\tau_{allow}$$

 $$M_{max}^{(2)} = \frac{1}{8}tL^{2}\tau_{allow} = \frac{1}{16}tL^{2}\sigma_{allow}$$

Solving for $$\sigma_{allow}$$  ,

 $$\sigma_{allow} = \frac{16M_{max}^{(2)}}{tL^{2}}$$ Recall:  $$I=\frac{tb^2}{6}(3a+b)$$

For $$ a=b=\frac{L}{4} $$

 $$I^{(2)} = \frac{t(\frac{L}{4})^2}{6}(3(\frac{L}{4})+(\frac{L}{4})) = \frac{tL^3}{96}$$

Solve for $$\sigma_{max}^{(2)}$$  and substitute in values for $$I^{(2)}$$ and $$b^{(2)}$$ ,

 $$\sigma_{max}^{(2)} = \frac{M_{max}^{(2)}b^{(2)}}{2I^{(2)}} = \frac{M_{max}^{(2)}\frac{L}{4}}{2\frac{tL^3}{96}} = \frac{12M_{max}^{(2)}}{tL^2}$$

Substituting in the previously found expression for $$M_{max}^{(2)}$$ ,

 $$\sigma_{max}^{(2)} = \frac{12(\frac{1}{16}tL^{2}\sigma_{allow})}{tL^2} = \frac{12}{16}\sigma_{allow}$$

Since $$ \frac{12}{16}\sigma_{allow}$$ is less than $$\sigma_{allow}$$ , this configuration of dimensions for the thin-walled beam are acceptable.

title: User:Eas4200c.f08.vqcrew.E/Homework2


text:
User:Eas4200c.f08.blue.b/Lecture 5

 Problem 1.1

The problem is to find the optimal $$b/a$$ in a thin walled beam to maximize the load-bearing capacity of a beam while assuming that the moment (M) is equal to the torsion (T).This is done by optimizing the cross section that carries the maximum bending moment and the maximum torsion.

Finding the Solution begins with the following assumptions: 1. The wall thickness is much less than the width and height of the beam (t<<a, t<<b)  2. The Moment on the beam is equal to the Torsion on the beam (M=T)  3. The allowable stress is equal to twice the allowable shear ($$\sigma _{allowable}= 2 \tau _{allowable}$$)  4. The beam is rectangular in shape, ie. the perimeter of the beam face (L) is equal to twice the sum of length a and b (L=2(a+b)).

Because of the assumption that t is very small, it is also assumed that the shear stress is uniform about the cross sectional walls.

Next, the shear stress ($$\tau$$) due to Torsion (T) is found. First, the total Torsion is found by separating the beam into four different elements, finding the Torsion in each, and adding them together.

$$T=T_{AB}+T_{BC}+T_{CD}+T_{DA}$$

Knowing that $$\tau =\frac{\nu }{t}$$, we are able to solve for Torsion in each element of the area of the beam.

$$T_{AB}=\frac{b}{2}\times (\nu \times a)=\frac{1}{2}\tau abt$$

$$T_{BC}=\frac{a}{2}\times (\nu \times a)=\frac{1}{2}\tau abt$$

$$T_{CD}=T_{BC}$$

$$T_{DA}=T_{CD}$$

$$T=2 \tau abt$$

$$\tau =\frac{T}{2abt}$$

We consider two cases for the solution.

Case1

In Case 1 we assume that the bending normal stress ($$\sigma $$) reaches the allowable stress ($$\sigma _{allowable}$$). The goal is to verify that the shear stress in the beam ($$\tau $$) is less than or equal to the allowable shear stress ($$\tau _{allowable}$$). The first step in this case is to define the bending moment as follows:

$$\sigma =\frac{Mz}{I}$$

Where M is the bending moment, z is the ordinate of a point adn I is the second area moment of inertia defined as: $$I=\int \int_{A}^{}{z^{2}dydz}$$

title: User:Eas4200c.f08.blue.b/Lecture 5


text: File:HW2 oppvector.jpg

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User:Eas4200c.f08.blue.b/Lecture 12

Assigned Reading from Mechanics of Aircraft Structures (C. T. Sun): pg. 451 - Shear Panel pg 85-86 - Shear Flow (Eq. 3.49, 3.50) pg. 20 - Fig. 2.2 pg. 64 - Fig, 3.2 pg 115 - Fig. 4.1

The lecture began with a discussion of shear in non-uniform objects such as the one in Figure 1 below.

Figure 1

Then, curved panels were discussed and reference was made to equations 1.4 and 1.5 (Sun, 5). To illustrate the concept of shear in a curved panel, problem 1.5 from the text book (Sun, 17) was discussed.

Problem 1.5
Problem statement: Derive the relations given by equations 1.4 $$\left( V_{x}=\tau ta\right)$$ and 1.5 $$\left( V_{y}=\tau tb\right)$$ in a curved panel as in Figure 2 below.

Figure 2

A Closer look is taken at the z,y plane and the shear flow on the edge of the curve as seen below:

Figure 3

Now let q be equal to the shear flow traveling from point A to point B as shown by the direction of the arrows above in Figure 3. $$q= \left( \tau \right) \left( t \right)$$

The infinitesimal change of shear force $$\left( d\vec{F}\right)$$ is shown below as

$$d\vec{F}=qd\vec{l}=q(dl_{y}\hat{j}+dl_{z}\hat{k})=q(dlcos\theta \hat{j}+dlsin\theta \hat{k})=qdydz$$

A closer look at $$d\vec{F}$$is shown in Figure 4 below.

Figure 4

The Resultant shear force vector is found as

$$\vec{F}=\int_{B}^{A}{d\vec{F}}=q\left(\left(\int_{B}^{A}{dy}\right)\hat{j}+\left(\int_{A}^{B}{dz} \right) \hat{k}\right)$$

Substituting equations 1.4 and 1.5,

$$\vec{F}=q\left(a\hat{j} +b\hat{k}\right)=F_{y}\hat{j}+F_{z}\hat{k}$$

where

$$F_{y}=\left( q\right)\left(a\right)$$ as in equation 3.49b

and

$$F_{z}=\left( q\right)\left(b\right)$$ as in equation 3.49c

yielding

$$\frac{F_{y}}{F_{z}}=\frac{a}{b}$$

The Resultant magnitude is

$$\left|\left|\vec{F} \right| \right|=\sqrt{\left[\left(F_{y} \right)^{2} +\left(F_{z} \right)^{2}\right]}=q\left[a^{2} +b^{2}\right]^{1/2}$$

$$R=\left|\left|\vec{F} \right| \right|$$

$$T=2\left( q\right)\left( A\right)$$

Now, a thin walled cross section can be analyzed in the same way as the curved panel. The thin walled cross section is seen in the z,y plane below.

Figure 5

a close up of the section in question:

Figure 6

It is assumed that the Torque (T) is in the x direction ie. $$\vec{T}=T\hat{i}$$

$$\vec{T}=\vec{r}\bowtie d\vec{F}$$

It follows that

$$dT = \rho dF=\rho \left(qdl \right)$$

and therefore

$$T=\oint_{}^{}{dT}=q\oint_{}^{}{\rho dl}$$

From Figure 6 above, it can be seen that $$ dA=\frac{1}{2}\rho dl$$

Therefore

$$T=\oint_{}^{}{dT}=q\oint_{}^{}{2dA}=2q\oint_{}^{}{dA}=2q\bar{A}$$

Where $$\bar{A}$$ is equal to the average area as measured from the middle of the thickness dl.

title: User:Eas4200c.f08.blue.b/Lecture 12


text:
Youth delinquency

Essay question: What socio-psychological factors contribute to youth delinquency?

Abstract: This paper explores primarily the social factors but also psychological factors such as personality and intelligence that impact on youth delinquency. It has been found that the media increases aggression in youths and that socio-economic status along with peer relations, education and parenting neglect can increase the probability of juveniles engaging in delinquent behaviour. Two theories of personality and intelligence are looked at which show that psychological factors may also contribute to juvenile delinquency.

Youth Delinquency: Youth delinquency is essentially a criminal act committed by a juvenile (usually defined as being between seven and eighteen) (Schafer &Knudten 1970). It is usually a by-product of antisocial behaviour. Antisocial behaviour refers to behaviour that either damages interpersonal relationships or is culturally unacceptable (Baumeister & Bushman 2008), or in some cases both. Commonly antisocial behaviour is identified in self harming youths or in youths that are truant with school and engage in theft or drug taking activity (Luncheon, Bae, Gonzalez, Lurie and Singh 2008). A large factor in youth delinquency can also be attributed to a form of antisocial behaviour known as aggression.

Aggression is a major factor in youth delinquency as acts of aggression are usually carried out on other people and as such are a crime. Two types of aggression identified by Baumeister and Bushman (2008) are “hostile aggression” and “instrumental aggression”. Hostile aggressions constitute crimes or acts with impulsive or emotive motivations whereas instrumental aggression is more calculated and motivated by goal driven behaviour. The difference in motivation behind aggressive behaviour has led researchers to explore whether aggression in youths and subsequently adults, is a result of the increasing violence shown in the media, the situational circumstances one is in or if there are physiological factors such as personality that determine how aggressive one is.

Media, video games and the effects on aggression.

In the early 1950s horror comics were criticised and linked to juvenile delinquency. Since then television as well as video and computer games have been accused of undermining moral values and cultivating a more violent and criminally oriented social climate (Gunter, 1994). Clint Eastwood’s movie “Dirty Harry” has been linked to copy cat serial killings and more recently the school shootings at Columbine (1999) have been linked to violent video games (Carnagey, Anderson and Bartholow). Numerous studies have been undertaken to see what effects video game playing has on feelings of aggression and subsequent acts of aggression.

Video Games: There is a large number of studies that look at different factors of video games that increase frustration and feelings of hostility in youths and adolescents who engage in video playing activity. The past research has led to the application of the General Aggression Model (GAM) in violent video game studies (Bartlett, Harris and Baldassaro 2007). The GAM encompasses all past theories on aggression and relies on short term affect, arousal and cognition components (Anderson & Bushman 2002) According to Anderson (2002), the GAM can account for the wide variety of effects seen in the media violence literature which lead to the child exposed becoming more desensitized to violence and habitually more aggressive. The GAM suggests that individual factors interact with the situational factors, which may lead to a person’s feeling impacting on real world actions (Bartlett et al 2007). Therefore if a youth has been playing a violent video game and has experienced an increase in frustration or hostility it is more likely that they are to act out. Bartlett et al. sought to demonstrate this in an experiment where participants played a video game with measures being taken at a timed interval. Participants were given story stems to respond to where participants had to say how they would react to certain scenarios and their heart rate was taken.

The results supported the theory that aggression increased in the individual while engaging in violent video games as there was an increase in physiological arousal (heart rate) and the responses to the story stems increased in aggression significantly between baseline to the time interval. The study found that the GAM offered an adequate explanation of the short-term effects of video games (Bartlett et al.) Although this particular study was conducted on participants with an average age of 19 (1 year older than what is recognised as being juvenile), the GAM suggests the desensitisation would have occurred in previous exposures to stimuli, while they were juveniles and the high internal validity does not suggest that findings would significantly differ if participants were 18.

Television and movies: The concern surrounding movies, television and their effects on the youth watching them stem from the social learning theory of imitation (Leyens, Herman and Dunand 1984). In 1961 Albert Bandura conducted an experiment measuring levels of aggression in children (1961). The experiment consisted of an adult exhibiting physical and verbal aggression towards the doll. Afterwards Bandura would place the child in a room with the doll and see what would happen. It was found that it was much more likely for those who had witnessed acts of aggression to act out such acts when placed in the same situation than those who did not (Bandura 1961). The experiment was run with live models as well as a video taped model with no difference in the results. Thus it can be reasonably applied to youths that witness violent acts in movies and television would be more likely to repeat those acts than those who don’t. Criticism came about the contrived nature of the experiment and the use of artificial films however, was quietened by numerous field experiments that yielded the same results (Leyens et al.). Numerous studies have also found the proclivity to act out aggressively strengthen upon watching violent acts carried out and that the movie or show will act as a primer for an individual to act out (Berkowitz, 2008). An experiment demonstrating this effect was carried out by Jopenson (1987). In her experiment a group of school boys were first frustrated by some act, after which they watched either a violent or non-violent television show. After watching the show the participants were observed playing a game of hockey. The groups displaying more aggression were the ones who displayed more aggressive acts throughout the game were those that had seen the violent show.

Social Factors:

Socio-economic factors- Class is considered an important social marker that plays an undeniable role in deviance (Wahrman, 1972). Studies have been done and replicated using a range of measures of socioeconomic factors including income, poverty and status (Ferguson, Campbell & Horwood, 2004). Each study has led to the same conclusion; youth in lower socioeconomic standing are more likely to be delinquent. This idea was furthered explored by the strain theory (Merton, 1938). According to the strain theory, individuals in a lower socioeconomic status are more likely to engage in delinquent behaviour to try and alleviate the imbalance and strain of the social situation (Ferguson et al.). The dilemma faced is able to be adapted to in five ways according to Merton, they are: 1.	 Innovation: individuals who accept socially approved goals, but not necessarily the socially approved means. Individuals who may adapt using innovation may aim to achieve socially approved goals but in order for them to attain them may be more likely to engage in delinquent behaviour as they do not have the same opportunities as the higher classes may have. 2.	Retreatism: those who reject socially approved goals and the means for acquiring them. These individuals may entirely shun societal norms and follow just what they want without regard for societal laws thus engaging in delinquent activity. 3.	Ritualism: those who buy into a system of socially approved means, but lose sight of the goals. Merton believed that drug users are in this category. 4.	Conformity: those who conform to the system's means and goals. 5.	Rebellion: people who negate socially approved goals and means by creating a new system of acceptable goals and means (Wikipedia 2008) Peer relationships Ferguson, Campbell and Horwood (2004) further suggest a differential association theory that may act as an influencing social factor in youth delinquency. The differential association states that an increase in youth delinquency and its relation to socioeconomic status are due to the fact that youths in the lower socioeconomic class have a larger exposure to criminal peers and environments. Sutherland’s (1947) original finding that personal networking leads to either a favourable or unfavourable view of delinquency, which supports Haynie’s (2002) finding that adolescents who report that their friends are delinquent tend to report higher levels of delinquency than adolescents with fewer or no delinquent friends. If the strain theory is applied, then it is not unreasonable to deduce that there would be more individuals in the lower socio-economic class engaging in delinquent activities, more delinquents associating and forming peer relationships with other delinquents or nondelinquents increasing the likelihood of non offenders engaging in delinquent behaviour and repeat offending in already delinquent individuals and therefore perpetuate the cycle of youth crime.

Education Youth delinquency has many interactive and causal effects which is the case when it comes to education. Blackorby and Wagner (1996) found that many juvenile delinquents are unable to attain skills and knowledge that would help them in employment opportunities or the chance to further their academic career due to expulsion or dropping out of school. This inability to attain the relevant skill can be linked back to the strain theory previously discussed as the disadvantage of lack of education and employment may create an imbalance in the social situation which may lead to delinquent activity to achieve the goal.

Parenting: The family circumstances have a considerable impact on the risk of engaging in some form of delinquency. Studies show that children who are provided with adequate parental supervision are less likely to engage in criminal activity, while children from dysfunctional family settings such as conflict, inadequate parental control and premature autonomy are more closely associated with juvenile delinquency (World Youth Report 2003) hostility and rejection as well as low child involvement are the most salient predictors of behavioural problems and delinquency (Simons, Simons, Chen, Brody, & Lin 2007). These lines of study are important as Gerstien and Briggs (1993) found 30 percent of violent offenders in their study were reared in the absence of a father. These types of studies has also allowed for intervention programs to be brought in to try and control increasing delinquent behaviour in youths. One such program outlined by Connel, Dishion, Yasui and Kananagh (2007) focuses on preventing substance abuse in youths. It does this by targeting problems in the family arena, primarily in parental monitoring and management of children engaging in delinquent activities. The research demonstrates that the motivation of parents to manage and monitor their children results in less delinquent behaviour exhibited by the youth. Further research should look into developing similar programs of intervention for the other aspects of parenting that contribute to an increased likelihood of delinquent activity.

Psychological Factors:

Personality: The three major personality factors according to Eysenck (1977) are: Psychoticism, extraversion and neuroticism. According to Eysenck’s criminal theory, juvenile delinquents would score highly on all three of the personality dimensions (Van Dam, Coleta, De Bryun & Janssens 2007). To test his theory Eysenck surveyed a sample of males in juvenile detention to assess their levels in the personality dimensions compared to a control group of college participants. The study found that of statistically high significance were the high extraversion levels in juvenile offenders, suggesting that highly extraverted juveniles that score low on neuroticism and psychotocism are more at risk of becoming delinquent.

While Eysenck described personality in three dimensions, Block and Block (1980) looked at personality in two area’s: Ego control and ego resiliency, later used to determine three types of personalities: Over-controllers, under-controllers and resilients (Akse, Hale, Engels, Raaijmakers & Meeus 2007). Over controllers tend to be internalising problems, under controllers tend to externalise problems while resilients strike a healthy balance (Akse et al.) In internalising their behaviour, over controllers tend to reject help from others, isolate themselves and have increased anxiety and depression, whereas under controllers who externalise their problems are more likely to act out in a deviant manner (Akse et al.)

Intelligence: Delinquency is found to be more prevalent and more frequent among young males with a low IQ (Koolhof, Loeber, Wei, Pardini & D’Escury 2007). An experiment run by Koolhof et al. compared impulsivity, psychopathy and empathy between high and low IQ individuals. The results found a significant difference in the impulsivity of individuals with a lower IQ as well as finding those with a lower IQ less empathetic and with less reported feelings of guilt. This is an important finding as those factors are related to delinquency, and would seem to suggest that due to those factors individuals with a low IQ are more prone to juvenile delinquent behaviour (Koolhof et al.).

Attachment: Bowlby (1969), theorised that as children we create internal working models which are based on the responsiveness of our primary caregiver. He states that these internal working models would allow for us to predict the future and how to react to our environment and the people in it. He predicted that children who formed secure attachment would feel free to explore their environment and interact freely with it as they would feel comfortable having their mother as a secure base should anything happen. This idea is based on previous experience and the mother or primary caregiver responding to needs. This is likely to continue on through life and set them up to be able to maintain strong social connections (Sigelman & Rider 2006). Conversely, if a child had not had all needs consistently met as a child they may form an insecure attachment. Children who develop an insecure form of attachment may develop a penchant to avoid social situations or have trouble regulating mood, emotion and behaviour.

A study by Elger, Knight, Sherman and Worrall (2003) found support for Bowlby’s insecure attachment theory. In surveys completed by youth delinquents reporting on attachment characteristics, substance abuse and behavioural problems, it was found that insecure attachment was related to the internalising and externalising of behaviours. As previously discussed those who have problems in doing this often act out in a deviant manner. It also showed a relation to antisocial and aggressive behaviour, which is a precursor to delinquent behaviour.

Conclusion: Delinquency in children can be affected by a myriad of factors either in the social realm or the psychological realm. The media increases hostility and aggression in youth, socio-economic status may effect how an individual acts to try and reap the same benefits those who are more fortunate already have and may also in turn effect who the individual forms peer relationships with, which may serve to increase delinquent behaviour. Parenting neglect results in higher instances of juvenile delinquency, as do certain psychological traits such as personality and intelligence. Attachment theory in infants effecting adolescents has also shown to have an effect in studies. The best method of reducing the risk of delinquent behaviour appears to be intervention programs such as motivating parents to monitor and manage their children and their behaviours, and as such further research might look at other area’s where similar programs may be introduced to alleviate some of the problems contributing to youth delinquency.

title: Youth delinquency


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Culture shock

Culture shock refers to the anxiety and feelings (of surprise, disorientation, uncertainty, confusion, etc.) felt when people have to operate within an entirely different cultural or social environment, such as a foreign country. It grows out of the difficulties in assimilating the new culture, causing difficulty in knowing what is appropriate and what is not. This is often combined a dislike for or even disgust (moral or aesthetical) with certain aspects of the new or different culture. The term was introduced for the first time in 1954 by Kalvero Oberg.

title: Culture shock


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Emotion

Emotion refers to short-lasting affective states which occur in response to significant life events.

Emotions are complex - they involve physiology, cognition, behaviour, and social communications, as well as feelings. The main reason for the occurrence of different emotions is that they each serve a motivational function.

What is emotion?
Emotion is not easily defined. It involves the complex psychophysiological experience of an individual's state of mind as interacting with biochemical and environmental influences. In humans, emotion fundamentally involves "physiological arousal, expressive behaviours, and conscious experience".

More simply, it's how we feel about things. Many of the chapters which follow will delve into conceptualising emotion and particular aspects of emotion - both what affects emotion and how emotion affects behaviour and thinking.

 * "short-lived, feeling-purposive-expressive-bodily responses that help us adapt to the opportunities and challenges we face during important life events" (Reeve, 2018, p. 288)
 * "synchronised brain-based systems that coordinate feeling, bodily response, purpose, and expression so to ready the individual to adapt successfully to life circumstances" (Reeve, 2018, p. 289)
 * "short-lived psychological-physiological phenomena that represent efficient modes of adaptation to changing environmental demands" (Levenson, 1994, p. 123)

Questions about emotion

 * What is an emotion?
 * What causes an emotion?
 * What type of emotions are there?
 * What functions do emotions serve?
 * How can we regulate emotions?
 * What is the difference between emotion and mood?
 * How can emotion be measured?
 * What are the consequences of emotion?
 * How can emotion be changed?
 * How and why did emotions evolve?
 * How are the emotions of animals and humans similar and how do they vary?

Types of emotion
Some of the main types of emotion are:
 * Fear
 * Anger
 * Sadness
 * Disgust
 * Contempt
 * Guilt
 * Surprise
 * Joy

Emotion causes

The causes of emotion are debated; they could be caused primarily by:
 * physiological responses first, followed by cognitive interpretation
 * cognitive responses first, followed by physiological responses
 * a combination of physiological and cognitive responses

Emotion images
What emotions do you think are depicted in each of these images?

To find more emotion pictures, visit the Motivation and Emotion gallery or go to the emotions category on Wikimedia Commons.

title: Emotion


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Social environment

The social environment (context), also known as the milieu (a French word for environment), refers to the culture that a person lives in, and the people and institutions with whom the person interacts. A given social environment is likely to create a feeling of solidarity amongst its members, who are more likely to keep together, trust, and help one another. Members of the same social environment will often think in similar styles and patterns even when their conclusions differ.

title: Social environment


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Social capital

 is a term used to refer to the extent of positive, collective, socially-stored potential which can be drawn upon within a society. It is analogous to notions of economic capital.

Social capital is poorly understood, loosely measured, and chronically undervalued. Current economic activities do not properly account for social capital, resulting in its unnecessary diminishment. - Conservation Economy

Lin (2001) defines social capital as resources embedded in a social structure that are accessed and/or mobilised in purposive actions.

title: Social capital


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User:Eas4200c.f08.blue.a/Homework2

Homework Report #2

Continuation of Problem 1.1
Because the shear stress is greatest at the top and bottom of the beam section (where z = b/2), the equation of shear stress can be rewritten as follows: $$\sigma = \frac{Mz}{I}  \Rightarrow   M=\frac{2I\sigma\ _{allowable}}{b} $$

Thus, it is necessary to find the value of I in terms of b (recalling that a can be written as a function of b: a=L/2-b). From the problem parameters we know that for this instance, Mmax = Tmax. Because Mmax has been determined, it must only be put into the relation:

$$\tau _{max}=\frac{T^{(1)}_{max}}{2a^{(1)}b^{(1)}t}=\frac{M^{(1)}_{max}}{2a^{(1)}b^{(1)}t}$$

Now check the values of $$\tau _{max}$$ with the value of $$\tau _{allowed}$$.If $$\tau _{max}$$ is larger than $$\tau _{allowable}$$ the optimal ratio of $$\frac{b^{(1)}}{a^{(1)}}$$would be acceptable. But $$\tau _{max}$$ is smaller than $$\tau _{allowable}$$, then the optimal ratio of $$\frac{b^{(1)}}{a^{(1)}}$$ would be unacceptable

Problem 1.1 Case 2
For this case we assume that the shear stress is eqaul to the maximum allowable shear stress in the section. Then we will test to see if the normal stress falls within acceptable tolerances (notably, less than $$\sigma _{max}$$)

By the equation given earlier in the problem the value of T becomes:

$$T=(2t\tau _{allowable})(ab)$$

Where the dimensions of the rectangle ab are equal to one another, hence a square. Therefore if the total perimeter of the square is L and all sides are equal, both a and b are equal to L/4. Plugging this quantity into the previous equation and using the given relation that M=T, we have:

$$M^{(2)}_{max}=\frac{1}{8}tL^{2}\tau _{allowable}$$

Using this relation and the previously defined equation for normal stress:

$$\sigma ^{(2)}_{max}=\frac{M^{(2)}_{max}b^{(2)}}{2I^{(2)}}$$

We can find $$\sigma ^{(2)}_{max}$$ and compare it with the allowable normal stress. If the maximum normal stress is less than the allowable shear stress, then the ratio of b to a for case 2 is acceptable. But, if it is greater, then that ratio is not correct.

Solving Problem 1.1 Case 1
First the moment of inertia, I, must be computed. This is done by summing the moments of inertia of each of the four parts of the section. In order to do this, the Parallel axis theorem must be applied. Solving for the horizontal axis yields:

$$I=\frac{tb^{2}}{6}(3a+b)$$

And with the relation a=L/2-b, the plot of f(b)=I/b is shown to the right.

Taking the derivative of the function f(b) yields the maximum value of the curve to be b(1)=3L/8. Thus the ratio of $$\frac{b^{(1)}}{a^{(1)}}$$ is equal to 3 and the value of a(1) is found to be L/8. Solving for $$\tau _{max}$$ with the values of a(1) and b(1) yield:

$$\tau _{max}=\frac{M^{(1)}_{max}}{2a^{(1)}b^{(1)}t}=\frac{32M^{(1)}_{max}}{3tL^{2}}$$

but because this larger than half of $$\tau _{allowable}$$ this case is not possible

Motivation For Problem 1.1
The motivation for this problem is that we are trying to study a box beam with a rectangular cross-section as a model for aerospace structures. These structures have a variety of applications in the aerospace industry such as in the fuselage and the wings, for example.

In continuation of problem 1.1, we left off on lecture 7 discussing the results of the Case 1 example. We saw that the strongest structure ended up having a height that was 3 times longer than it was wide.

Lecture 8 presented the idea of looking into the shear stress caused by Tmax(1). Where:

Recalling that we allow this equivalences due to Assumption 2. Furthermore:

Recalling that this is due to Assumption 3. As you can see, this is not acceptable.

The Mohr's Circle defines the region by which sigma and tau interact on this beam. Notice that tau is exactly half of sigma.

Solving Problem 1.1 Case 2
Now, let's assume the maximum shear stress reaches the allowable shear stress first.

This is valid due to the assumption that there is a uniform stress distribution throughout the member.

It is clear to see that the only way to maximize T is by maximizing (ab), since thickness and allowable shear stress are fixed.

where (ab)max is the maximized value of (ab), which coincidentally is only when a=b=L/4

Problem Statement The problem is to find the optimal ratio of b/a in a thin walled beam to maximize the load-bearing capacity of a beam while assuming that the moment (M) is equal to the torsion (T).This is done by optimizing the cross section that carries the maximum bending moment and the maximum torsion.

Assumptions

In this problem, the first assumption$$^{(1)}$$ for the first case is that the bending normal stress reaches the allowable stress. In the second case$$^{(2)}$$, the shear stress is assumed to be equal to the maximum allowable shear stress. Lastly, the third assumption$$^{(3)}$$ is that the allowable stress is equal to the two times the allowable sheer stress.

Solution

We last left Problem 1.1 stating that $$ T_{max} $$ occurred when $$ a=b $$ and by recalling that $$ L=2(a+b) $$ the resulting value of $$ ab $$ is $$ \frac{L}{4}^{4}$$. This leads to

By Case 2: $$T_{max} = (2t\tau _{allowable})(\frac{L}{4})^{2} = \frac{1}{8}tL^{2}\tau _{allowable} = M_{max}^{(2)}$$

From these and by way of assumption 3, the allowable stress is

By Case 3: $$M_{max}^{(2)} = \frac{tL^{2}}{16}\sigma _{allowable}$$

$$\Rightarrow \sigma_{allowable}=\frac{16M_{max}^{(2)}}{tL^{2}}$$

Now, recall that $$f\left(b \right)=\frac{I}{b}$$

$$\Rightarrow f\left(b^{(2)} \right)=\frac{I^{(2)}}{b^{(2)}}=\frac{tb^{(2)}(3L-4b^{(2)})}{12}=\frac{tL^{2}}{24}$$ with $$b^{(2)} = \frac{L}{4}$$

$$\Rightarrow \sigma_{max}^{(2)} = \frac{M_{max}^{(2)}b^{(2)}}{2I^{(2)}}$$

remember that $$\sigma = \frac{My}{I}$$ with $$y = \frac{b}{2}$$

so $$M_{max}^{(2)} = \frac{tL^{2}}{16}\sigma_{allowable}$$

$$\Rightarrow \sigma_{max}^{(2)}=\frac{12}{16}\sigma_{allowable}\leq \sigma_{allowable}$$

This result concludes that Case 2 is acceptable because the maximum stress is less than the allowable stress.

Aircraft Structure
Aircraft structures are assembled from various basic structural elements to help withstand loading conditions like axial, bending, and torsion. Some basic elements of an aircraft are the axial members(truss and bar elements), sheer panels, bending members(like beams), and torsional members(bar and beam torsion).

Homework Assignments - Moments of Inertia and Stringers
The first homework assigned in class consists of four individual problems.

The first Problem is to define the moment of inertia $$I_y$$ for a solid circular cross section through integration using polar coordinates. The integral being $$I_y=\int\int_A(Z^2dA)$$.

Also the derivation of $$I_y$$ for a square cross section is required which would be derived by the use of the integral $$I_y=\int\int_A(Z^2dYdZ)$$

The second problem assigned required comparing the moment of inertia defined for the circular cross section (case:1) in the previous problem to a redistributed equivalent area (case:2)

Where $$a=b$$, $$ A_2=3(a)(t)=\pi(r^2)$$,  and $$ t=a/10$$

The moment of inertia will be calculated assuming the y-axis passes through the centroid of case:1 and case:2 for fair comparison. The moment of inertia for case:2 will be calculated by using the fact that the moment of inertia for a rectagular cross section is $$I_y=1/12(b)(h^3)$$ along with the parallel axis theorem which is $$I_y=I+Ad^2$$

The third problem assigned was a modified version of problem 1.7. The modification was that all rectangular surfaces be changed to circular surfaces. The problem is to compare the moment of inertia $$I_y$$ of a circular cross section (case:1) to a modified cricular cross section (case:2) calculating the moment of inertia $$I_y$$ for both cases through the centroid of the cross section.

Where $$r_0=10cm$$, $$t=r_0/10$$, and $$A_1=A_2$$

The moment of inertia will be calculated for the circle from the answer to problem number one since it will be the same the moment of inertia for case 2 will be from a combination of rectangle moment of inertia and circular moment of inertia with use of the parallel axis theorem.

The fourth problem assigned was to show why a open rectangular cross section would be preferable to a hollow recangular cross section by comparing moment of inertia, manufacturing, and assembly techniques.

How Ahead We Have Been
While the pace of the course seems to be slow, we have actually been covering material several chapters ahead of where we are currently at. An example is found when comparing problems 1.1 and 3.5. As it turns out, problem 1.1 is just a specific case of problem 3.5.

For the first case, the box, $$tau = T/2abt$$. In the second case, T = 2qA, where A is the average area and $$q = tau*t$$. This is a case where the first example, the box is a specific case of the second figure. The box problem is solved using the ad-hoc method and the generic shape is solved using a method based on the elasticity of the object.

Getting back to the airplane stringer problem, the question was posed last time as to why the walls of the stringer are sloped outward rather than being parallel as shown below. Also the question of why an open cross section is used rather than a closed cross section.

First off, there are 2 reasons why an open cross section is used. 1) Manufacturing - sheet metal is used to make the stringer. It is much easier to stamp an open cross section than to extrude a closed cross section. 2) Airplane construction - stringers are rivited to the outside skin of the plane and it is next to impossible to rivit a closed cross section. To use a closed cross section, spot wels would be needed.

Next, why are the walls of the stringer not parallel. It was suggested that the reason why they are not parallel is for storage purposes. Stringers can be stacked vertically when being stored prior to construction which saves a lot of space in a factory.

Shear Panels, Shear Stress/Strain
Engineering Shear Strain (not torsional shear strain)

The angle between the original and deformed box is shear strain which is defined as dv/dx. Using small angle approximations, the angle is equal to the tangent of the angle(i.e deg = tan(deg) based on small angle approximations)

shear strain = (du/dy)+(dv/dx)

Tensorial shear strain is defined as (1/2)(shear strain)xy.

Contributing Team Members
Mark Barry Eas4200c.f08.blue.a 13:42, 25 September 2008 (UTC)

Clay Robertson eas4200c.f08.blue.f 18:44, 25 September 2008 (UTC)

Thomas Sexton EAS4200c.f08.blue.c 22:27, 25 September 2008 (UTC)

Marc Pelletier EAS4200c.f08.blue.b 05:18, 26 September 2008 (UTC)

Christian Garabaya Eas4200c.f08.blue.d 08:25, 26 September 2008 (UTC)

Jeffrey Shea EAS4200c.f08.Blue.E 05:21, 26 September 2008 (UTC)

title: User:Eas4200c.f08.blue.a/Homework2


text: File:HW2 node2.jpg

title: File:HW2 node2.jpg


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User:MandaG/Lecture week10

Relationships, be they friendships or more intimate are tricky. As social beings we naturally seek out contact and belonging with others and strive to maintain lasting relationships. This need to belong may explain why when relationsihps end we are left upset and troubled leading to the occasional crying session with a tub of ice cream watching a sad movie.

The need to Belong: The notes suggest that belongingness consists of regular social contact with others that is close, stable and mutual. That people typically seek 4-6 close relationships and generally do not continue to form relationships after reaching that point. I know its not scientifically based, but in my personal experience I find it very hard to believe this theory. My uni friendship group consists of 11 people that satisfy the ideas of belonging and then I have work friends and sporting friends who also meet the criteria set out for belonging and I know many people in similar social situations to myself.

Marriage The idea of Happily married people living longer is kinda sweet. However, sweet as it may be (unless of course you're not married and afraid of dying alone, earlier and not as happy), I would be interested in seeing research on whether other social connections also play a major role.

Similarity, Reinforcement Theory and Reciprosity The idea of opposites attract, is quite well attractive to me. I like the idea of opposing views and constant debating (note different to arguements), however it is suggested that a prolonged period of confliction may result in cognitive disonance. Self-monitoring suggests that we may change ourselves to be more similar with those that we interact with, thus I suppose if the partner acts in a way that opposes our own beliefs it is plausible to see how the dissonance would arise. I'm not sure how much of a part culture plays in our choices of interaction, although research suggests that due to culture there is a greater need for "complementarity" and I guess that makes sense as there would not be cultural barriers, I wonder whether in Australia given our multiculturalistic (not sure if thats a word...)society it is less important than it would be in say China? Reinforcement theory I guess basically operates the smae way in a relationship as in other types of behaviour and is very logical, we are likely to repeat behaviour that has been positively reinforced (operant) and associate things that are attractive to positive outcomes(classical). Now according to our notes reciprocity plays a huge part in relationships, this whole liking begets liking thing. Well while true in many cases I think this theory is flawed in that it doesn't account for abusive relationships, stalkers and of course my inevitable marriage to Johnny Depp who I'm pretty sure does not know me or that he loves me yet.

Playing Hard To Get? Oh yeah this has a special heading purely because I think it's hilarious in our notes James has put that attractiveness increases as bar closing time approaches. I find it hilarious because I have noticed this happening to myself and friends frequently and have often wondered if everyone else notices to...apparently they have. I also note that we are more likely to be attracted to those that are "moderately selective." Not to offend anyone but I think this whole area of research is ridiculous as I fail to see how individuals perceptions of other individuals can be made into a theory as surely it is hugely varied and dependent on other variables to?

Attractiveness and Beauty I am very hesitant to tackle this area as I think there are many factors that are not considered. Superficially we can base attractiveness and beauty at a glance and I think it is unique to the individual as to what is thought to be attractive and admittedly culture does certainly play a part. However, yes a cliche, I honestly believe it is inner beauty that is most attractive in a person. Some people it is impossible to see as beautiful due to their exceptionally ugly nature and personality.

Social Rejection and Loneliness Bad news for all the recluses out there, apparently not satisfying the need to belong leads to a higher death rate (I assume that means more premature...) and more physical and mental problems. I think the problem in ideas such as these is that they are very generalised. I am and have always been more about the individual in psychology rather than the groups or context, and I don't think everyone has the same "need" to belong, I suggest that perhaps the level varies with the individual. With that in mind I struggle to understand how more mental and physical problems in those without "good" social networks is viewed. Perhaps one who had significant problems in life only had a social network of 2, but perhaps that is all they needed and the problems relate to some unknown cause, how have the researchers assesed enough people and their relationships and needs to enable such an accepted idea? To me the idea of not belonging differs from rejection primarily in the way that rejection is an active rebuff of an attempt to belong. Rejection can lead to rejection sensitivity, the expectation of rejection and hypersensitivity to possible rejection. If an individual is sensitive to all the possible scenario's in which they may be rejected ostracism and isolation is almost a guarantee. It is also suggested that rejection of an individual initiates feelings of emotional numbness, interferes with psychological and cognitive functioning, a decline in intelligent thought, skepticism, impulsiveness and typically the individual tends to be less co-operative and helpful while more willing to break the rules. I suppose if this is indeed the case then the poor people feeling this through hypersensitivity to rejection are struggling with those emotions a great deal of the time? In saying that as a law student with a relatively black and white view on crime I still rejct this idea as a defence to any illegal activity. Loneliness affects everyone at times and it is particularly evident through tranisition stages. Loneliness occurs generally when the atcual level of contact is below the desired level, and just to throw my own thoughts out there I think that it doesn't just have to do with the number of people or time of contact playing into the addage of being surrounded by people yet being entirely alone, but also and I'm not quite sure how to word this, but I'm sure you are all familiar with or have seen friends fight with other another friend or break up with their boy/girlfriend and no matter how much you're there for them they are still lonely? well that kind of scenario.

Love's me, Love's me not, Love's me as a companion? Ever been told I love you but I'm not IN love with you? I've always though this was a lame man way of escaping a relationship, however Sternbergs "Triangular Model of Love" suggests that there may be some validity to this statement. Passion- Physiological arousal, longing, sexual attraction Intimacy- Closeness and sharing, support, understanding, concern Commitment- Concious decision, willing to define as love, long term. Having all these three factors is more likely to lead to succesful relationship while having a missing component may burst that relationship bubble. My history with relationships has allowed me to be in or observe a relationship with a missing element. Loving someone but not being in love with someone suggests the passion is laggaing. Intimacy, in my view is closely related to trust and the freedom to be yourself and without it...well I'd be giving up fairly quickly. And lastly the for some people deal breaker. If both parties in a relationship are not willing to define what they have as love and be in it for the long haul the relationship lifespan is cut down significantly.

title: User:MandaG/Lecture week10


text:
User:Eml4500.f08.qwiki.bishop/Lecture 9

=Lecture 9 - Monday, September 15, 2008= $$\begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)}+k_{11}^{(2)} & k_{34}^{(1)}+k_{12}^{(2)} & k_{13}^{(2)} & k_{14}^{(2)}\\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)}+k_{21}^{(2)} & k_{44}^{(1)}+k_{22}^{(2)} & k_{23}^{(2)} & k_{24}^{(2)}\\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)}\\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)}\\ \end{bmatrix}\begin{pmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\\ \end{pmatrix}=\begin{pmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6\\ \end{pmatrix}$$
 * Global Force/Displacement Relationship
 * (Global stiffness matrix, K) X (global displacement, d) = (global force, F)
 * $$\mathbf{K_{(6x6)}d_{(6x1)} = F_{(6x1)}}$$
 * The global stiffness matrix, K, is found by combining the element stiffness matrices $$k^{(1)}$$ & $$k^{(2)}$$ as seen below.
 * NOTE: There is an overlap for displacements $$d_1$$ & $$d_1$$, which results from Element 1 and Element 2 sharing a node; and their k values are then summed together.
 * The values for the global stiffness matrix are as follows:
 * $$K_{11} = k_{11}^{(1)} = \tfrac{9}{16} = 0.5625$$
 * $$K_{12} = k_{12}^{(1)} = \tfrac{3\sqrt{3}}{16} = 0.3248$$
 * $$K_{13} = k_{13}^{(1)} = -\tfrac{9}{16} = -0.5625$$
 * $$K_{14} = k_{14}^{(1)} = -\tfrac{3\sqrt{3}}{16} = -0.3248$$
 * $$K_{15} = 0$$
 * $$K_{16} = 0$$
 * $$K_{21} = k_{21}^{(1)} = \tfrac{3\sqrt{3}}{16} = 0.3248$$
 * $$K_{22} = k_{22}^{(1)} = \tfrac{3}{16} = 0.1875$$
 * $$K_{23} = k_{23}^{(1)} = -\tfrac{3\sqrt{3}}{16} = -0.3248$$
 * $$K_{24} = k_{24}^{(1)} = -\tfrac{3}{16} = 0.1875$$
 * $$K_{25} = 0$$
 * $$K_{26} = 0$$
 * $$K_{31} = k_{31}^{(1)} = -\tfrac{9}{16} = -0.5625$$
 * $$K_{32} = k_{32}^{(1)} = -\tfrac{3\sqrt{3}}{16} = -0.3248$$
 * $$K_{33} = k_{33}^{(1)}+k_{11}^{(2)} = \tfrac{9}{16}+\tfrac{5}{2} = 3.0625$$
 * $$K_{34} = k_{34}^{(1)}+k_{12}^{(2)} = \tfrac{3\sqrt{3}}{16}-\tfrac{5}{2} = -2.1752$$
 * $$K_{35} = k_{13}^{(2)} = -\tfrac{5}{2} = -2.5$$
 * $$K_{36} = k_{14}^{(2)} = \tfrac{5}{2} = 2.5$$
 * $$K_{41} = k_{41}^{(1)} = -\tfrac{3\sqrt{3}}{16} = -0.3248$$
 * $$K_{42} = k_{42}^{(1)} = -\tfrac{3}{16} = 0.1875$$
 * $$K_{43} = k_{43}^{(1)}+k_{21}^{(2)} = \tfrac{3\sqrt{3}}{16}-\tfrac{5}{2} = -2.1752$$
 * $$K_{44} = k_{44}^{(1)}+k_{22}^{(2)} = \tfrac{3}{16}+\tfrac{5}{2} = 2.6875$$
 * $$K_{45} = k_{23}^{(2)} = \tfrac{5}{2} = 2.5$$
 * $$K_{46} = k_{24}^{(2)} = -\tfrac{5}{2} = -2.5$$
 * $$K_{51} = 0$$
 * $$K_{52} = 0$$
 * $$K_{53} = k_{31}^{(2)} = -\tfrac{5}{2} = -2.5$$
 * $$K_{54} = k_{32}^{(2)} = \tfrac{5}{2} = -2.5$$
 * $$K_{55} = k_{33}^{(2)} = \tfrac{5}{2} = 2.5$$
 * $$K_{56} = k_{34}^{(2)} = -\tfrac{5}{2} = -2.5$$
 * $$K_{61} = 0$$
 * $$K_{62} = 0$$
 * $$K_{63} = k_{41}^{(2)} = \tfrac{5}{2} = 2.5$$
 * $$K_{64} = k_{42}^{(2)} = -\tfrac{5}{2} = -2.5$$
 * $$K_{65} = k_{43}^{(2)} = -\tfrac{5}{2} = -2.5$$
 * $$K_{66} = k_{44}^{(2)} = \tfrac{5}{2} = 2.5$$
 * Step 4: Eliminate known degrees of freedom
 * $$d_1 = d_2 = d_5 = d_6 = 0$$
 * This reduces the global stiffness matrix from a (6x6) matrix to a (2x2) matrix

title: User:Eml4500.f08.qwiki.bishop/Lecture 9


text:
User:Lizzi~enwikiversity

I am studying to become a Yoga instructor in my spare time through the British School for Yoga and so far it seems to be going fairly well but it is always good to have a safety net. I had no idea that there would be so my psychology.'''

'I am also well on my way through a classical education, including Greek, Latin and the esential classical literature. I work in an art gallery and although i am passionate about art i could be more fluent in the ins-and-outs of the art world so i hope for all the above reasons that Wikiversity will help.''

title: User:Lizzi~enwikiversity


text:
User:EML4500.f08.JAMAMA/Homework2

Lecture Notes (September 8 - September 19)

title: User:EML4500.f08.JAMAMA/Homework2


text:
User:EML4500.f08.JAMAMA/Homework1

Lecture Notes 1

Matlab Tutorial

Click to go back to Team JAMAMA's Home Page ->

title: User:EML4500.f08.JAMAMA/Homework1


text:
User talk:Lizzi~enwikiversity

 Hello Lizzi, 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

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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! Jtneill - Talk - c 14:44, 25 September 2008 (UTC)

 Hi , I am Percy. Nice to meet you. I love the beach, swimming, fishing, and I am currently doing some research at Wikiversity on oceanography. I am also learning to play piano. What do you like? Can I be your mascot?

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 Lizzi. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Lizzi~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.

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Sorry for the inconvenience.

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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) 06:14, 19 April 2015 (UTC)

title: User talk:Lizzi~enwikiversity


text:
Combinatorics

"'There is no problem in all mathematics that cannot be solved by direct counting.'-Ernst Mach"

Combinatorics is a branch of pure mathematics concerning the study of discrete (and usually finite) objects. It is related to many other areas of mathematics, such as algebra, probability theory, ergodic theory and geometry, as well as to applied subjects in computer science and statistical physics. Aspects of combinatorics include "counting" the objects satisfying certain criteria (enumerative combinatorics), deciding when the criteria can be met, and constructing and analyzing objects meeting the criteria, finding "largest", "smallest", or "optimal" objects (as in combinatorial designs, extremal combinatorics and combinatorial optimization), and finding algebraic structures these objects may have (algebraic combinatorics).

Developed resources

 * Combinatorics/Rule of product counts what might be called "decisions". It is still under construction, but quite developed.

 * Combinatorics/Binomial coefficients introduces binomial coefficients using a tree diagram that recovers Pascal's triangle. A number of advanced topics are then discussed, including Catalan numbers, and Lattice paths

Undeveloped resources
The following resources are undeveloped; some are little more than stubs:
 * 1) Combinatorics/About this resource is currently a stub that contains comments that compare the various wikis.
 * 2) Combinatorics/Graph & Ramsey Theory
 * 3) Combinatorics/Structural Algebra (truly a stub)

title: Combinatorics