text:
User:Eml4500.f08.ateam/hw 5

A-Team Teampage

Homework Report #5

This page is intended for Homework Report #5 of the A-Team.

Due date: Friday, 07 November 08, 5 pm EST = 22:00 UTC.

Justification of Reducing the Global Stiffness Matrix
The goal of eliminating rows 1,2,5 and 6 in a stiffness matrix is to find $$ {K_{2x2}}$$ in the original 2 bar truss shown below:

The Force Displacement relationship can be described as: $$ {K_{6x6}d_{6x1} = F_{6x1}}$$

This can be rearranged into the equation $$ {K_{6x6}d_{6x1} - F_{6x1} = 0_{6x1}}$$ (1)

For the Principle of Virtual Work (PVW):

$$ {W_{6x1}\cdot (K_{6x6}d_{6x1} - F_{6x1}) = 0_{1x1}}$$ (2)

This is for all $$ {W_{6x1}}$$, also known as the Weighting Matrix.

As seen above, it is trivial when going from equation 1 to equation 2, but it is very important to understand how to get from equation 2 to equation 1. The next part will be verifying the steps in going from equation 2 to equation 1.

Eq. 1 $$\Leftrightarrow $$ Eq. 2: Weighting Coefficient Choices
When proving Eq. 2 $$\Leftrightarrow $$ Eq. 1, choices have to be made for what the Weighting Coefficient will equal.

Choice 1
Let $$ {W_1 = 1}$$ and $$ {W_2 = W_3 = W_4 = W_5 = W_6 = 0}$$

This gives the Weighting Matrix $$ W = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}_{6x1}$$

$$({{W})\cdot({K} {d} - {F})=}$$

$$1\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The only equation that is used is the first one since the rest become zero.

$$ 1\cdot\left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}}\right]$$

or

$$ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}}$$

Choice 2
Next $$ W_2$$ will be equal to 1 while the rest of the weighting matrix is equal to 0.

$$ W = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}_{6x1}$$

The same process will be taken to find the next part of the equation.

$$({{W})\cdot({K} {d} - {F})=}$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

Again all of the equations are going to become zero, but this time the second equation will be the only one that matters since does not have a zero coefficient. The final equation for choice 2 is:

$$ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}}$$

Choice 3
$$ W_3$$ for this choice will be the factor equal to 1.

$$ W = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}_{6x1}$$

$$({W})\cdot({K} {d} - {F})=$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The final equation for choice 3 becomes:

$$ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}$$

Choice 4
$$ W = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}_{6x1}$$

$$({{W})\cdot({K} {d} - {F})=}$$

$$+ 0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The final equation for choice 4 becomes:

$$ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}$$

Choice 5
$$ W = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}_{6x1}$$

$$({{W})\cdot({K} {d} - {F})=}$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The final equation for choice 5 becomes:

$$ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}$$

Choice 6
Choice 6 is not any different and it ends the process of trying to prove that going from the equation with a weighting coefficient to the force displacement relationship is not trivial.

$$ W = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}_{6x1}$$

$$({{W})\cdot({K} {d} - {F})=}$$

$$+ 0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The final equation for choice 6 becomes:

$$ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}$$

Boundary Conditions
For the Principle of Virtual Work, the boundary conditions need to be accounted for. In the 2-bar truss, $$\mathbf {d_1 = d_2 = d_5 = d_6 = 0 }$$. The only two displacements the matter are $$\mathbf d_3$$ and $$\mathbf d_4$$ in the two bar truss. Weighting coefficients need to be kinematically admissible, which means they cannot violate the boundary conditions. This concludes that $$ \mathbf {W_1 = W_2 = W_5 = W_6 = 0 }$$. According to the calculus of variations, the weighting coefficients are equivalent to the virtual displacements.

The original equation $$\mathbf({W)\cdot({K} {d} - {F}) = 0}$$ can now be expressed in matrix form and can be further reduced.

$$ (W)_{6x1}K_{6x6} \begin{bmatrix} d_1=0 \\ d_2=0 \\ d_3=x  \\ d_4=y \\ d_5=0 \\ d_6=0  \end{bmatrix}_{6x1}- F_{6x1} = 0$$

Note that x and y are arbitrary values not equal to zero in this problem. The principle of virtual work allows the equation to be condensed to only what is needed and all of the zeros to be removed. This new equations will be noted as equation 3 and is shown below.

$$(W)_{6x1}\cdot \begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{bmatrix}_{6x2} \begin{bmatrix} d_3 \\ d_4 \end{bmatrix}_{2x1} - F_{2x1} = 0 $$

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

This equation can be even further reduced to

$$\begin{Bmatrix} W_3\\W_4 \end{Bmatrix}\cdot\left(\begin{bmatrix} K_{33} &K_{34} \\ K_{43}& K_{44}\end{bmatrix}_{2x2} \begin{Bmatrix} d_3\\d_4 \end{Bmatrix}_{2x1} - \begin{Bmatrix} F_3\\F_4 \end{Bmatrix}_{2x1}\right) = 0 $$

This equation can be rewritten as

$$ \mathbf {\overline{K} \overline{d} = \overline{F}}$$

Where

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

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

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

for any $$\begin{Bmatrix}W_3\\W_4 \end{Bmatrix}$$

Back to Principle of Virtual Work
An example everyone can understand for PVW is the equation for our course grade.

Course Grade = $$\alpha_{o}\,$$ x (Homework grade) + $$\sum_{i=1}^3 \alpha_{i}\,$$ x Exami

where $$\alpha_{o}\,$$ and $$\alpha_{i}\,$$ are the weighting coefficients.

For deriving $$ {k}^{(e)} = {T}^{(e)T}\quad\hat{k}^{(e)} {T^{(e)}}$$ we need to recall the force displacement relationship with the axial degrees of freedoms $$ q^{(e)} $$.

$$\quad\hat{k}_{2x2}^{(e)} {q_{2x1}^{(e)}}=  {p}_{2x1}^{(e)}$$  can be rearranged to the formula

$$\quad\hat{k}_{2x2}^{(e)} {q_{2x1}^{(e)}} - {p}_{2x1}^{(e)} = 0_{2x1} $$ (1) Note that 0 is not a scalar but the 2x1 matrix $$ 0 $$.

With the weighting coefficient added, the PVW equation becomes $$W_{2x1}\cdot \quad\hat{k}_{2x2}^{(e)} {q_{2x1}^{(e)}} - {p}_{2x1}^{(e)} = 0 $$.

Note that since $$W$$ is a 2x1 matrix, the 0 is a scalar. This works for all $$W_{2x1}$$ because equation 1 is equal to zero. Now we have again shown that equation 1 is equivalent to equation 2, like from slide 24-1.

Again recall $$ {q}_{2x1}^{(e)} = {T}_{2x4}^{(e)} {d_{4x1}^{(e)}}$$. (3) Real displacement is represented in this equation.

$$\quad\hat{W}_{2x1} = {T_{2x4}^{(e)}} {W}_{4x1}$$ (4)

The equation above shows that $$\quad\hat{W}$$ relates to the axial $$q$$, and it is also the virtual displacement.

For the next lectures, Chapter 2 sections 1,2,6 and 7 need to be read for a better understanding of what we will be taught.

As a reminder, $$\quad\hat{W}_{2x1}$$ is the virtual axial displacement corresponding to $$ {q}_{2x1}^{(e)} $$. $$ {W}_{4x1} $$ is the virtual displacement in the global coordinate system corresponding to $$ {d}_{4x1}^{(e)} $$.

Our goal is to replace equations 3 and 4 into equation 2. The result is shown below.

$$ (T^{(e)}W)\cdot [\quad\hat{k}^{(e)}(T^{(e)}d^{(e)}) - p^{(e)}] = 0 $$ for all $$W_{4x1}$$. (5)

Transposing the first matrix and multiplying it by the second matrix is equivalent to taking the dot product of two matrices.

Recall that Recall $$\left(AB \right)^{T}=B^{T}A^{T}$$ (6)

Proof
$$\left(AB \right)^{T}=B^{T}A^{T}$$ will be verified with an example.

Let $${A}_{2x3} = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}$$ and $${B}_{3x3} = \begin{bmatrix} 7 & 8 & 9\\1 & 2 & 3 \\4 & 5 & 6 \end{bmatrix}$$

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

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

$${B}^T_{3x3}=\begin{bmatrix} 7 & 1 & 4\\ 8 & 2 & 5\\ 9 & 3 & 6\end{bmatrix}$$

$${A}^T_{2x3}=\begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \\ \end{bmatrix}$$

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

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

Also recall that $${a_{nx1} \cdot b_{nx1}=a^T_{1xn}b_{nx1}=S_{1x1}}$$ (7) Now equations 7 and 6 need to be applied into equation 5.

$${\left( T^{(e)}w \right)^T} \left[ {\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)}} \right]=0_{1x1}$$

$${ w^TT^{(e)T}} \left[ {\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)}} \right]=0_{1x1}$$ Next the transposed T matrix is distributed throughout the rest of the equation.

$${ w^T}\cdot \left[ {\left(T^{(e)T}\hat{k}^{(e)}T^{(e)}\right)d^{(e)} - T^{(e)T}p^{(e)}} \right]=0_{1x1}$$

We can set $$\mathbf{(T^{(e)T}\hat{k}^{(e)}T^{(e)}) = k^{(e)}}$$ and $$\mathbf {T^{(e)T}p^{(e)} = f^{(e)}} $$ to shorten the equation to:

$${ w^T}\cdot \left[{{k}^{(e)}d^{(e)} - f^{(e)}} \right]=0_{1x1}$$

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

As shown before, it is important to find the steps going from the equation with the weighting coefficient to the one basic equation.

So far we have only done the discrete case, or the non continuous case, and now we need to learn about the continuous case, or Partial Differential Equations (PDEs). A motivational model problem is an elastic bar with varying A(x) and E(x) subject to varying axial load (distributed) and concentrated loads. The axial and the concentrated loads are both time dependent.

A(x), E(x), f(x,t), and p(x) are all given.

Example of a Composite Material
Some composite materials are examples of how the modulus of elasticity can be a function of x. One example is reinforced concrete walls. The modulus of the concrete is significantly different than that of the reinforcing steel. In order to accurately model this material the modulus would change as one moved through each material.

source: Rebar

Another material that has a modulus that changes with x is an aluminum composite panel which consists of two aluminum sheets sandwiched around a non-aluminum material.

source: (Aluminium_composite_panel)

Partial Differential Method

A very small slice of the beam shall be considered of width dx. The force balance in the x direction gives the following result.

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

$$=\frac{\partial {N}}{\partial{x}}(x,t)dx + higher\ order\ terms + f(x,t)dx-m(x)\ddot{u}dx\quad (Eqn. 1)$$

These higher order terms come from the Taylor Series expansion, shown in the general form here.

$$f(x+dx)=f(x)+\frac{df(x)}{dx}dx+\frac{1}{2}\frac{d^2f(x)}{dx^2}dx^2...$$

Where all the $$dx^2...$$ and higher terms are neglected due to their small effect on the result. After those simplifications are made, Eqn. 1 can be written in the following form.

$$\frac{\partial{N}}{\partial{x}}+f(x,t)=m(x)\ddot{u}\quad (Eqn. 2)$$

In this form, it is easy to see that this is the equation of motion. In order to get to a more usable form, N(x,t) needs to be rewritten into its constituent components.

$$N(x,t)=A(x)\ \sigma(x,t)$$

$$\sigma(x,t)=E(x)\ \epsilon(x,t)$$

$$\epsilon(x,t)=\frac{\partial{u}}{\partial{x}}(x,t)$$

Using these relationships, N(x,t) can be written as follows.

$$N(x,t)=[A(x)\ E(x)\ \frac{\partial{u}}{\partial{x}}(x,t)]$$

Now Eqn. 1 can be rewritten in the final form.

$$\frac{\partial{ }}{\partial{x}}[A(x)\ E(x)\ \frac{\partial{u}}{\partial{x}}(x,t)]+f(x,t)=m(x)\ddot{u}$$

This is the final form of the equation of motion that will be used to solve finite element problems with the partial derivative method.

Initial and Boundary Conditions
In order to solve these problems, certain aspects of the conditions must be known. For the following case, two boundary conditions are known.

 $$u(0,t)=0\ and\ u(L,t)=0$$

In another example, two initial conditions are known.

$$u(0,t)=0\ and\ N(L,t)=A(L)\sigma(L,t)=F(t)$$

Axial DOF's and Forces/Element DOF's and Forces in Global xyz-Coordinate System for 6x1 Matrices

Using the previous diagram, we can write the following relationships.

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

$$\begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix} \begin{bmatrix}q_{1}^{(e)}\\ q_{2}^{(e)}\end{bmatrix} = \begin{bmatrix}p_{1}^{(e)}\\ p_{2}^{(e)}\end{bmatrix}$$

From the diagram above, we can write the element DOF's and forces for a 3-D system.

$$\underline{f}^{(e)} = \begin{vmatrix}F_1\\F_2 \\F_3 \\ F_4\\ F_5\\ F_6\end{vmatrix}_{6x1}$$ and $$\underline{d}^{(e)} = \begin{vmatrix} d_1\\d_2\\ d_3\\ d_4\\ d_5\\ d_6\end{vmatrix}_{6x1}$$

We can now use the transformation matrix $$\underline{T}$$ to transform the 6 element DOF's in the global xyz-coordinate system to the 2 axial DOF's $$p$$ and $$q$$.

$$\underline{q}^{(e)} = \underline{T} \underline{d}^{(e)}$$

This equation expands to:

$$\begin{vmatrix}q_{1}^{(e)}\\q_{2}^{(e)} \end{vmatrix}_{2x1} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix}_{2x6} \begin{vmatrix}d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\end{vmatrix}_{6x1}$$

Similarly, this can be applied to the equation:

$$\underline{p} = \underline{T} \underline{f}$$

From either of the above equations, to solve for the element forces $$\underline{f}$$ or displacements $$\underline{d}$$ we can use the transpose of the transformation matrix $$\underline{T}^{T}$$.

$$\underline{f} = \underline{T}^{T} \underline{p}$$

$$\underline{d} = \underline{T}^{T} \underline{q}$$

Derivation of FD Relation in Global Coordinates Using the PVW
Derive the transformation matrix based on the lecture for the 2-D truss elements:

$$a_1^{(e)} = d_1^{(e)} \cdot \vec\tilde{i} = ( d_1 \tilde{i} + d_2 \tilde{j} + d_3 \tilde{k} ) \cdot \vec\tilde{i} = d_1 \cdot ( \tilde{i} \cdot \vec\tilde{i} ) + d_2 \cdot ( \tilde{j} \cdot \vec\tilde{i} ) + d_3 \cdot ( \tilde{k} \cdot \vec\tilde{i} )$$

where $$( \tilde{i} \cdot \vec\tilde{i} ) = l^{(e)}, ( \tilde{j} \cdot \vec\tilde{i} ) = m^{(e)}, and ( \tilde{k} \cdot \vec\tilde{i} ) = n^{(e)}$$

$$a_1^{(e)} = l^{(e)}d_1 + m^{(e)}d_2 + n^{(e)}d_3$$

$$a_1^{(e)} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix}\begin{vmatrix} d_1\\ d_2\\ d_3 \end{vmatrix}$$

$$a_2^{(e)} = d_2^{(e)} \cdot \vec\tilde{i} = ( d_4 \tilde{i} + d_5 \tilde{j} + d_6 \tilde{k} ) \cdot \vec\tilde{i} = d_4 \cdot ( \tilde{i} \cdot \vec\tilde{i} ) + d_5 \cdot ( \tilde{j} \cdot \vec\tilde{i} ) + d_6 \cdot ( \tilde{k} \cdot \vec\tilde{i} )$$

where $$( \tilde{i} \cdot \vec\tilde{i} ) = l^{(e)}, ( \tilde{j} \cdot \vec\tilde{i} ) = m^{(e)}, and ( \tilde{k} \cdot \vec\tilde{i} ) = n^{(e)}$$

$$a_2^{(e)} = l^{(e)}d_4 + m^{(e)}d_5 + n^{(e)}d_6$$

$$a_2^{(e)} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix}\begin{vmatrix} d_4\\ d_5\\ d_6 \end{vmatrix}$$

$$\begin{vmatrix} a_1^{(e)}\\ q_2^{(e)} \end{vmatrix} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix} \begin{vmatrix} d_1\\d_2 \\d_3 \\ d_4 \\ d_5 \\ d_6

\end{vmatrix}$$

Note: The transformation matrix is in terms of the three director cosines.

Recall the relation between axial DOF's and element DOF's in the global coordinate system:

$$\underline{q} = \underline{T} \underline{d}$$

and also the relation between axial forces and element forces in the global coordinate system:

$$\underline{P} = \underline{T} \underline{f}$$

Using the principle of virtual work (PVW) we can derive the force-displacement relationship in the global coordinate system.

$$\underline{k} \underline{d} = \underline{F}$$

therefore,

$$\underline{k} \underline{d} - \underline{F} = 0$$

With the PVW, this becomes:

$$\underline{W} \cdot ( \underline{k} \underline{d} - \underline{F} ) = 0$$

where $$\underline{W}$$ is the weighting matrix. For this system we have 6 choices for $$\underline{W}$$:

Choice 1:

$$ W = \begin{vmatrix}1 & 0 & 0 & 0 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{1,j}d_j - F_1 = 0}$$

$$\sum_{j=1}^{6}{k_{1,j}d_j = F_1}$$

Choice 2:

$$ W = \begin{vmatrix}0 & 1 & 0 & 0 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{2,j}d_j - F_2 = 0}$$

$$\sum_{j=1}^{6}{k_{2,j}d_j = F_2}$$

Choice 3:

$$ W = \begin{vmatrix}0 & 0 & 1 & 0 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{3,j}d_j - F_3 = 0}$$

$$\sum_{j=1}^{6}{k_{3,j}d_j = F_3}$$

Choice 4:

$$ W = \begin{vmatrix}0 & 0 & 0 & 1 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{4,j}d_j - F_4 = 0}$$

$$\sum_{j=1}^{6}{k_{4,j}d_j = F_4}$$

Choice 5:

$$ W = \begin{vmatrix}0 & 0 & 0 & 0 & 1 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{5,j}d_j - F_5 = 0}$$

$$\sum_{j=1}^{6}{k_{5,j}d_j = F_5}$$

Choice 6:

$$ W = \begin{vmatrix}0 & 0 & 0 & 0 & 0 & 1\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{6,j}d_j - F_6 = 0}$$

$$\sum_{j=1}^{6}{k_{6,j}d_j = F_6}$$

$$\begin{vmatrix} k_{1,1} & k_{1,2} & k_{1,3} & k_{1,4} & k_{1,5} & k_{1,6}\\ k_{2,1} & k_{2,2} & k_{2,3} & k_{2,4} & k_{2,5} & k_{2,6}\\ k_{3,1} & k_{3,2} & k_{3,3} & k_{3,4} & k_{3,5} & k_{3,6}\\ k_{4,1} & k_{4,2} & k_{4,3} & k_{4,4} & k_{4,5} & k_{4,6}\\ k_{5,1} & k_{5,2} & k_{5,3} & k_{5,4} & k_{5,5} & k_{5,6}\\ k_{6,1} & k_{6,2} & k_{6,3} & k_{6,4} & k_{6,5} & k_{6,6}\\ \end{vmatrix} \begin{vmatrix} d_1\\d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6

\end{vmatrix} = \begin{vmatrix} F_1\\F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6

\end{vmatrix}$$

Derivation of Elemental Stiffness Matrix in Global Coordinates Using PVW
The following is an expression of the element stiffness matrix $$\underline{K}_{6x6}^{(e)}$$ in terms of the transformation matrix $$\underline{T}^{(e)}$$ and the element stiffness matrix $$\underline\hat{k}^{(e)}$$ from the axial force displacement relationship. We will show that by using the priciple of virtual work, we can derive the equation for the element stiffness matrix in global coordinates.

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

Recall the force displacement relationship with axial DOF's $$ q^{(e)}$$

$$\underline\hat{k}_{2x2}^{(e)} \underline{q}_{2x1}^{(e)} = \underline{p}_{2x1}^{(e)}$$

Rearranging this equation yields:

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

We can now apply the principle of virtual work (PVW) to get:

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

We now can recall the equation:

$$\underline{q}_{2x1}^{(e)} = \underline{T}_{2x6}^{(e)} \underline{d}_{6x1}^{(e)}$$    (3)

Similarly,

$$\hat{W}_{2x1} = \underline{T}^{(e)}_{2x6} \underline{W}_{6x1}$$ (4)

Now, substitute equations (3) & (4) into equation (2) which yields:

$$( \underline{T}^{(e)} \underline{W} ) \cdot [ \underline\hat{k}^{(e)} ( \underline{T}^{(e)} \underline{d}^{(e)} ) - \underline{p}^{(e)} ] = 0$$  for all $$ \underline{W}_{6x1}$$   (5)

Recall that:

$$ ( \underline{A}_{pxq} \underline{B}_{qxr} )^{T} = \underline{B}^{T} \underline{A}^{T}$$  (6)

And also that:

$$\underline{a}_{nx1} \cdot \underline{b}_{nx1} = \underline{a}^{T}_{1xn} \underline{b}_{nx1}$$   (7)

Now we can apply equations (6) and (7) into (5) to get:

$$ ( \underline{T}^{(e)} \underline{W} )^{T} [ \underline\hat{k}^{(e)} ( \underline{T}^{(e)} \underline{d}^{(e)} ) - ( \underline{p}^{(e)} ] = 0 $$ for all $$\underline{W}_{6x1}$$

In this equation, $$\underline\hat{W}_{2x1}$$ represents the virtual axial displacement and from eq. (4) and the matrix $$\underline{W}_{6x1}$$ represents the virtual displacement in the global coordinate system corresponding to $$\underline{d}^{(e)}$$

The previous equation now becomes:

$$\underline{W}^{T} \underline{T}^{(e)T} [ \underline\hat{k}^{(e)} ( \underline{T}^{(e)} \underline{d}^{(e)} ) - \underline{p}^{(e)} ] = 0$$

$$\underline{W} \cdot [ ( \underline{T}^{(e)T} \underline\hat{k}^{(e)} \underline{T}^{(e)} ) \underline{d}^{(e)} - ( \underline{T}^{(e)T} \underline{p}^{(e)}) ] = 0$$ for all $$\underline{W}_{6x1}$$

In the equation above, the element stiffness matrix $$\underline{K}_{6x6}^{(e)}$$ is represented by $$( \underline{T}^{(e)T} \underline\hat{k}^{(e)} \underline{T}^{(e)} )$$and the force matrix $$\underline{f}_{6x1}^{(e)}$$ is represented by $$\underline{T}^{(e)T} \underline{p}^{(e)}$$

The equation now becomes:

$$\underline{W} \cdot [ \underline{K}^{(e)} \underline{d}^{(e)} - \underline{f}^{(e)} ] = 0$$ for all $$\underline{W}$$

We have the original expression for the elemental stiffness matrix in terms of the axial force displacement relationship.

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

Expression for element forces in terms of transformation matrix and element axial forces
From the previous derivations, it was determined that the element axial force was related to the element forces in global coordinates by

$$p^{(e)}_{2x1}=T^{(e)}_{2x6}f^{(e)}_{6x1}$$

Therefore, the expression for the element forces in global coordinates in terms of the transformation matrix and the element axial forces is

$$f^{(e)}_{6x1}=T^{(e)T}_{6x2}p^{(e)}_{2x1}$$

Element Stiffness Matrix in Global Coordinates
The equation for the elemental stiffness matrix in global coordinates is

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

This equation can be combined with the director cosines to yield

$$ \mathbf{k}_{6\times 6}^{(e)} = k^{(e)}\begin{bmatrix} l^{(e)^{2}} & m^{(e)}l^{(e)} & n^{(e)}l^{(e)} & -l^{(e)^{2}} & -m^{(e)}l^{(e)} & -n^{(e)}l^{(e)} \\ m^{(e)}l^{(e)}& m^{(e)^{2}} & m^{(e)}n^{(e)} & -m^{(e)}l^{(e)} & -m^{(e)^{2}} & -m^{(e)}n^{(e)} \\ n^{(e)}l^{(e)}& m^{(e)}n^{(e)} & n^{(e)^{2}} & -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -n^{(e)^{2}} \\ -l^{(e)^{2}}& -m^{(e)}l^{(e)} & -n^{(e)}l^{(e)} & l^{(e)^{2}} & m^{(e)}l^{(e)} & n^{(e)}l^{(e)} \\ -m^{(e)}l^{(e)}& -m^{(e)^{2}} & -m^{(e)}n^{(e)} & m^{(e)}l^{(e)} & m^{(e)^{2}} & m^{(e)}n^{(e)} \\ -n^{(e)}l^{(e)}& -m^{(e)}n^{(e)} & -n^{(e)^{2}} & n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & n^{(e)^{2}} \end{bmatrix} $$

where $$l^{(e)}=\frac{x^{(e)}_2-x^{(e)}_1}{L}$$, $$m^{(e)}=\frac{y^{(e)}_2-y^{(e)}_1}{L}$$ , $$n^{(e)}=\frac{z^{(e)}_2-z^{(e)}_1}{L}$$.

Essential MATLAB Functions for Finite Element Analysis
Similar to the five-bar truss system, all of the following MATLAB programs require additional functions to run. These functions are found at the course textbook website, Fundamental Finite Element Analysis and Applications. They include,  , and. The details of these functions are outlined below.

This function generates the element stiffness matrix for each element. When the function is called it calculates the X and Y coordinate positions of the element ends from the coordinates passed in. It produces the element length from this information. Next it calculates the direction cosines of the element. Using the Young's modulus and cross-sectional area of the element, the element stiffness matrix, k, is assembled. The stiffness matrix is passed back to the.

This function solves the force-displacement relationship for each node, resulting in the displacements and reactions at each node. This information is passed back to the.

This function outputs the axial strain, axial stress, and axial force of each element.

Two-Bar Truss System: MATLAB Code Debugging
In Homework Report #2, the team analyzed the MATLAB code for a two-bar truss system. This code was provided by Dr. Vu-Quoc and can be found here. After the homework report submission deadline, a bug was found in the provided code. This bug occurs during the building of the  array. Instead of passing the proper Young's modulus,, and cross-sectional area,  , indicies into the   function, the program passes in the entire   and   arrays. This is sufficient for cases in which all truss elements have an identical cross-sectional area and Young's modulus, but fails when these properties vary among elements.

Note: Applied load P = 7

To fix the bug, a simple modification was made to the following program segment:

Where the  array is assembled the variables   and   are replaced by the array indices   and   respectively.

was modified to be

The resulting MATLAB program code is as follows:

Running
Running this code produces the following output: >> twobar_debug k_local = 0.75       -0.75         -0.75         0.75 k = 0.5625     0.32476      -0.5625     -0.32476       0.32476       0.1875     -0.32476      -0.1875        -0.5625     -0.32476       0.5625      0.32476      -0.32476      -0.1875      0.32476       0.1875 k_local = 5   -5     -5     5 k = 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 K = 0.5625     0.32476      -0.5625     -0.32476            0            0       0.32476       0.1875     -0.32476      -0.1875            0            0       -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5      -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5             0            0         -2.5          2.5          2.5         -2.5             0            0          2.5         -2.5         -2.5          2.5 R = 0     0      0      7      0      0 d = 0            0         4.352        6.1271             0             0 reactions = -4.4378      -2.5622        4.4378       -4.4378 results = 1.7081      5.1244       5.1244        0.6276        3.138        6.276

Compared to the previously obtained  array shown below, this new   array makes much more sense. The previous array contained additional meaningless values along with the desired values.

Previous  array:

results = 1.7081      5.1244       8.5406       5.1244       17.081        0.6276       1.8828        3.138       1.8828        6.276

Verifying Array Accuracy
The new  array values correspond to the table shown below.

The  array for finite element analysis of the truss system is verified using principles of statics by simply summing the forces in the X and Y directions and setting these equations equal to zero. This produces two equations with two unknown values from which the axial forces of each element are derived. From here, using the element properties, the element stresses and strains can be determined. This is shown below.

$$\sum{F_{y}} = 7 - F_1\cdot sin(30^{\circ} ) - f_2\cdot sin(45^{\circ}) = 0$$

$$\sum{F_{x}} = F_1\cdot cos(30^{\circ} ) - F_2\cdot cos(45^{\circ}) = 0$$

Where $$F_1$$ is the axial force of Element 1 and $$F_2$$ is the axial force of Element 2.

Solving these equations simultaneously yields the element axial forces.

$$F_1 = 5.1244$$

$$F_2 = 6.2760$$

Now the axial stress for each element can be computed using the axial force and cross-sectional area for each element.

$$\sigma_1 = \frac{F_1}{A_1} = \frac{5.1244}{1.00} = 5.1244$$

$$\sigma_2 = \frac{F_2}{A_2} = \frac{6.2760}{2.00} = 3.1380$$

The values for axial stress can now be used along with the Young's modulus to determine the axial strain for each element.

$$\epsilon_1 = \frac{\sigma_1}{E_1} = \frac{5.1244}{3.00} = 1.7081$$

$$\epsilon_2 = \frac{\sigma_2}{E_2} = \frac{3.1380}{5.00} = 0.6276$$

These values are almost identical to the values obtained through finite element analysis. The MATLAB results have been verified to be accurate. For this case, principles of statics proved to be a much more efficient method to solve the problem. This is not the case as mesh complexity increases.

Six-Bar Truss System: MATLAB Finite Element Analysis
The problem to be evaluated in this section is found as Example 4.1 on pg.226 of the course textbook. The MATLAB code used to evaluate the problem is from the  file found at the Student Companion Site for the class textbook.

Additionally, a modified version of the problem is analyzed accordingly in this section.

The load $$P = 20 kN$$ acts at an angle $$\alpha = 30^\circ$$ for each version of the problem.

Original Problem: Example 4.1
Analysis of the original six-bar truss system is important before comparing to the modified six-bar truss system. The following code is from the  file found at the website for the class textbook.

Running
Running this code produces the following results.

>> SixBarTrussEx K = Columns 1 through 6 85355       35355       -50000            0            0            0         35355        35355            0            0            0            0        -50000            0        85355       -35355            0            0             0            0       -35355  1.0202e+005            0            0 0           0            0            0        71554       -35777             0            0            0            0       -35777        17889             0            0            0            0            0            0             0            0            0       -66667            0            0        -35355       -35355       -35355        35355       -71554        35777        -35355       -35355        35355       -35355        35777       -17889   Columns 7 through 10 0           0       -35355       -35355             0            0       -35355       -35355             0            0       -35355        35355             0       -66667        35355       -35355             0            0       -71554        35777             0            0        35777       -17889         71554        35777       -71554       -35777         35777        84555       -35777       -17889        -71554       -35777  2.1382e+005            0 -35777      -17889            0  1.0649e+005 R = 0            0         10000         17321             0             0             0             0             0             0 d  = 0            0       0.21311       0.24998             0             0             0             0    -0.0060971      0.012242 reactions = -10873      -217.27        874.27       -437.13       -1.7279        -16666 results = 5.3276e-005      10.655        10655 -4.6334e-006    -0.92669      -926.69 -4.8873e-006    -0.97746      -977.46 -8.3326e-005     -16.665       -16665 1.5363e-006     0.30727       307.27 -9.659e-009  -0.0019318      -1.9318

The  array values correspond to the table shown below.

The original displacements are shown in the following matrix. It is important to note that the nodal positions were magnified by a factor of 1000 in the original  file.

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \\ d_7 \\ d_8 \\ d_9 \\ d_{10} \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 0 \\           0 \\   0.21311 \\   0.24998 \\            0 \\            0 \\            0 \\            0 \\   -0.0060971 \\     0.012242 \\ \end{Bmatrix}$$

Plotting Example 4.1
To plot the undeformed and deformed shapes, the following code was appended to the original  file. For viewing, the nodal magnification factor in the original file was removed and the resulting nodal displacements were multiplied by a factor of 2500.

Modified Problem: Varying Young's Modulus
This modification of Example 4.1 involves varying the Young's modulus, E, of the elements as shown in the table below. This requires re-writing the  code to allow for an array of E values. The subsequent MATLAB file is called.

To allow for the passing of array indices, the MATLAB code was modified as shown. Much like fixing the bug in the two-bar truss system code,  replaces   inside the   loop that builds the stiffness matrices and the results array.

Running
The results obtained by running the code are as follows. Again, it is important to note that the nodal positions are multiplied by a factor of 1000. >> sixbarmod K = Columns 1 through 6 76391       38891       -37500            0            0            0         38891        38891            0            0            0            0        -37500            0        69320       -31820            0            0             0            0       -31820        98486            0            0             0            0            0            0        71554       -35777             0            0            0            0       -35777        17889             0            0            0            0            0            0             0            0            0       -66667            0            0        -38891       -38891       -31820        31820       -71554        35777        -38891       -38891        31820       -31820        35777       -17889   Columns 7 through 10 0           0       -38891       -38891             0            0       -38891       -38891             0            0       -31820        31820             0       -66667        31820       -31820             0            0       -71554        35777             0            0        35777       -17889         89443        44721       -89443       -44721         44721        89027       -44721       -22361        -89443       -44721  2.3171e+005        16015 -44721      -22361        16015  1.1096e+005 R = 0            0         10000         17321             0             0             0             0             0             0 d = 0            0       0.26485       0.26083             0             0             0             0    0.00063864     -0.001246 reactions = -9908.3       23.619       -90.274        45.137       -1.4008        -17389 results = 6.6213e-005      9.9319       9931.9 5.347e-007    0.096245       96.245 5.0465e-007     0.10093       100.93 -8.6943e-005     -17.389       -17389 -1.5183e-007   -0.033402      -33.402 -6.2645e-009  -0.0015661      -1.5661

The  array values correspond to the table shown below.

The original displacements are shown in the following matrix. It is important to note that the nodal positions were magnified by a factor of 1000 in the  file.

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \\ d_7 \\ d_8 \\ d_9 \\ d_{10} \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 0 \\           0 \\      0.26485 \\      0.26083 \\            0 \\            0 \\            0 \\            0 \\   0.00063864 \\    -0.001246 \\ \end{Bmatrix}$$

Plotting Example 4.1 and the Modified Six-Bar Truss System
To plot the undeformed and deformed shapes, the following code was appended to the original  file. The subsequent file was plotted over the existing plot for Example 4.1 to show how varying Young's modulus throughout the truss system can affect deformation. Subsequent deformation clearly differed from that of elements with identical Young's modulus. However, deformation did not vary greatly.

The undeformed truss system is represented by the blue dashed line. The deformed truss system of Example 4.1 is represented by the solid red line. The deformed modified truss system is represented by the green dotted line. For viewing, the nodal magnification factor in the original files was removed and the resulting nodal displacements were multiplied by a factor of 2500.

Analysis of Three-Bar Space Truss System: Example 4.2
This problem is found on page 230 of the course textbook.

The truss system in Example 4.2 is a pin-jointed structure with three elements. Much like problems solved earlier, the element properties vary in this truss system and are displayed below. This system adds another dimension to the FEA and results in a total of 12 degrees of freedom. Nodal displacements for this problem are denoted by,    and   in the  ,    and   directions, respectively.

Before implementing FEA to solve the problem, it is important to properly set up the  and   arrays. $$= \begin{bmatrix} 1&4\\ 2&4\\ 3&4 \end{bmatrix}$$

$$= \begin{bmatrix} 1 & 2 & 3 & 10 & 11 & 12\\ 4 & 5 & 6 & 10 & 11 & 12\\ 7 & 8 & 9 & 10 & 11 & 12 \end{bmatrix} $$

The  file for this problem was found on the Student Companion Site for the course textbook. It is as follows.

Additional MATLAB Functions
These functions were found on the Student Companion Site for the course textbook. They behave similarly to their corresponding functions for plane trusses detailed in the previous section.

Note: The program also uses, detailed in the previous section.

Running
Running this code produces the following results. Note that the nodal positions are multiplied by a factor of 1000 for simplicity. >> ThreeBarSpaceTrussEx K = Columns 1 through 6 1459.7      2919.3      -3040.9            0            0            0        2919.3       5838.6      -6081.9            0            0            0       -3040.9      -6081.9       6335.3            0            0            0             0            0            0       3566.7      -3566.7       4953.8             0            0            0      -3566.7       3566.7      -4953.8             0            0            0       4953.8      -4953.8       6880.3             0            0            0            0            0            0             0            0            0            0            0            0             0            0            0            0            0            0       -1459.7      -2919.3       3040.9      -3566.7       3566.7      -4953.8       -2919.3      -5838.6       6081.9       3566.7      -3566.7       4953.8        3040.9       6081.9      -6335.3      -4953.8       4953.8      -6880.3   Columns 7 through 12 0           0            0      -1459.7      -2919.3       3040.9             0            0            0      -2919.3      -5838.6       6081.9             0            0            0       3040.9       6081.9      -6335.3             0            0            0      -3566.7       3566.7      -4953.8             0            0            0       3566.7      -3566.7       4953.8             0            0            0      -4953.8       4953.8      -6880.3             0            0            0            0            0            0             0            0            0            0            0            0             0            0        60000            0            0       -60000             0            0            0       5026.4      -647.45       1912.9             0            0            0      -647.45       9405.4       -11036             0            0       -60000       1912.9       -11036        73216 d  = 0            0             0             0             0             0             0             0             0      -0.18705        -2.592       -0.3858 reactions = 6666.7        13333        -13889       -6666.7        6666.7       -9259.3             0             0         23148 results = 0.00050936      101.87        20375    0.00033036       66.072        13214    -0.0001929       -38.58       -23148

The  array values correspond to the table shown below.

The displacement matrix is as follows.

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \\ d_7 \\ d_8 \\ d_9 \\ d_{10} \\ d_{11} \\ d_{12} \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 0 \\        0 \\         0 \\         0 \\         0 \\         0 \\         0 \\         0 \\         0 \\   -0.1871 \\   -2.5920 \\   -0.3858 \\ \end{Bmatrix}$$

The reactions are as follows.

$$\begin{Bmatrix} R_1 \\ R_2 \\ R_3 \\ R_4 \\ R_5 \\ R_6 \\ R_7 \\ R_8 \\ R_9 \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 6667 \\  13333 \\  -13889 \\   -6667 \\    6667 \\   -9259 \\       0 \\       0 \\   23148 \\ \end{Bmatrix}$$

Static Determinacy
The three bar space truss was found to be statically determinate. There are three unknowns and three equations to use, summing the forces in x, y, and z directions. By using these three equations, the following relationships are made.

$$\sum{F_{x}}= F_{1}l^{(1)} + F_{2}l^{(2)} = 0$$

$$\sum{F_{y}}= F_{1}m^{(1)} + F_{2}m^{(2)} = -P$$

$$\sum{F_{z}}= F_{1}n^{(1)} + F_{2}n^{(2)} + F_{3}n^{(3)} = 0$$

Plugging in P = 20,000 N, and solving the set of linear equations gives the following forces along each element.

$$F_{1} = 20375\ N$$

$$F_{2} = 13124\ N$$

$$F_{3} = -23149\ N$$

In order to solve for the axial stress and strain, the following equations will be used.

$$\sigma_{n} = \frac{F_{n}}{A_{n}}$$

$$\epsilon_{n} = \frac{\sigma_{n}}{E_{n}}$$

Plugging in the given values for the Young's Modulus and area for each element yields these values. (Sigma is expressed in MPa and epsilon is expressed in mm.)

$$\sigma_{1} = \frac{20375 N}{200\ mm^2}= 101.875\qquad \qquad \epsilon_{1} = \frac{\sigma_{1}}{200\ GPa}= 0.5094$$

$$\sigma_{2} = \frac{13214 N}{200\ mm^2}= 66.07\qquad \qquad \ \ \ \epsilon_{2} = \frac{\sigma_{2}}{200\ GPa}= 0.33035$$

$$\sigma_{3} = \frac{-23149 N}{600\ mm^2}= -38.58\qquad \qquad \epsilon_{3} = \frac{\sigma_{3}}{200\ GPa}= -0.1929$$

These values for the displacement of each element matches the results from the MATLAB confirming the method.

MATLAB Plots of Example 4.2
To plot the undeformed and deformed states of the three-bar space truss system the following code was appended to the original  file. Each plotting view was manipulated within the MATLAB plot editor.

Perspective View: (1, 1, 1)
To make the deformation more clear the following perspective view was included.

Contributing Team Members
The following students contributed to this report:

Mark Barry Eml4500.f08.Ateam.barry 03:17, 7 November 2008 (UTC)

Daniel DiDomenico Eml4500.f08.ateam.didomenico 00:37, 3 November 2008 (UTC)

Sean Miller EML4500.f08.Ateam.Miller 05:22, 7 November 2008 (UTC)

Jil Paquette Eml4500.f08.Ateam.paquette 00:31, 7 November 2008 (UTC)

Matthew Philbin Eml4500.f08.Ateam.philbin 14:13, 3 November 2008 (UTC)

Ian Slevinski Eml4500.f08.Ateam.Slevinski 01:11, 7 November 2008 (UTC)

title: User:Eml4500.f08.ateam/hw 5


text:
User:Eas4200c.f08.spars.lee/Homework 5

27 October
Bidirectional bending recipe

Diagram

Equation 4.26

Equation 4.27

The moment of inertia tensors, $$ \ I_y ,I_z, I_{yz}, $$ (also written as $$ \ I_y = I_{yy} , I_z = I_{zz} $$) are determined by

Equation 4.28a

Equation 4.28b

Equation 4.28c

Which is related to the bending normal stress, $$ \ \sigma_{xx} $$, according to the equation

Equation 4.29

The denominator of the above equation is also the determinant of a submatrix of the inertial tensor matrix

matrix

proof that $$ \ I_{yz} = I_{zy} $$

Along the neutral axis the bending normal stress, $$ \ \sigma_{xx} $$, is equal to zero

Equation 4.30

Diagram

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

title: User:Eas4200c.f08.spars.lee/Homework 5


text:
User:Uf.team.wiki/HW5

 Comment: This is a resubmission of HW 5 and fixes one slight mistake made in the original submission, specifically located in the contributing member section. The new revision is listed and is indicated with (NEW) as part of the link name.

A comparison between the original submission and the new submission can be seen here

Eml4500.f08.wiki.barcia 21:43, 27 November 2008 (UTC)

Principal of Virtual Work I
Looking back to our discussion to virtual work,

Global Force-Displacement Relationship vs. Principal of Virtual Work
Justification of eliminating rows 1,2,5,6 to obtain $$\overline {\underset{2 \times 2}\mathbf K}\quad$$ in the original 2-bar truss problem:

Global Force-Displacement Relationship:

$${\underset{6 \times 6}\mathbf K} {\underset{6 \times 1}\mathbf d}= {\underset{6 \times 1}\mathbf F} \quad \Rightarrow \quad \mathbf {Kd}-\mathbf F = {\underset{6 \times 6}\mathbf 0} \qquad (1)$$

Principal of Virtual Work:

$${\underset{6 \times 1}\mathbf W} \cdot (\mathbf {Kd}- \mathbf F) = \underbrace_{scalar}\hookrightarrow for all \quad{\underset{6 \times 1}\mathbf W} \qquad (2)$$

Proof: $$(1) \iff (2)$$
$$Eq. (1) \iff Eq. (2)$$

Proof:
 * A) $$\quad (1) \iff (2)\quad$$ : trivial
 * B) Want to show $$\quad (2) \Rightarrow (1)\quad$$:
 * $$(2) \Rightarrow \mathbf W \cdot (\mathbf {Kd}- \mathbf F) = 0 \quad$$ for\quad all $$\mathbf W $$

CHOICE 1: Select W such that W1=1, W2=...=W6=0
 * $${\underset{1 \times 6}\mathbf W^T}= \begin{bmatrix}

1 & 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix}_{1 \times 6}$$

Thus,

$$ \mathbf W \cdot (\mathbf{Kd}-\mathbf F) = 1 \cdot [\sum_{j=1}^6 K_{1j}d_j - F_1] + 0 \cdot [\sum_{j=1}^6 K_{2j}d_j - F_2] + ... + 0 \cdot [\sum_{j=1}^6 K_{6j}d_j - F_6] = 0 \Rightarrow$$

$$\sum_{j=1}^6 K_{1j}d_j = F_1 \qquad \qquad (1st \quad Equation)$$

CHOICE 2: Select W such that W1=0, W2=0, W3=...=W6=0
 * $${\underset{1 \times 6}\mathbf W^T}= \begin{bmatrix}

0 & 1 & 0 & 0 & 0 & 0 \\ \end{bmatrix}_{1 \times 6}$$

Thus,

$$ \mathbf W \cdot (\mathbf{Kd}-\mathbf F) = 0 \cdot [\sum_{j=1}^6 K_{1j}d_j - F_1] + 1 \cdot [\sum_{j=1}^6 K_{2j}d_j - F_2] + 0 \cdot [\sum_{j=1}^6 K_{3j}d_j - F_3] + ... + 0 \cdot [\sum_{j=1}^6 K_{6j}d_j - F_6] = 0 \Rightarrow$$

$$\sum_{j=1}^6 K_{2j}d_j = F_2 \qquad \qquad (2nd \quad Equation)$$

CHOICE 3: Select W such that W1=W2=0, W3=1, W4=...=W6=0
 * $${\underset{1 \times 6}\mathbf W^T}= \begin{bmatrix}

0 & 0 & 1 & 0 & 0 & 0 \\ \end{bmatrix}_{1 \times 6}$$

Thus,

$$ \mathbf W \cdot (\mathbf{Kd}-\mathbf F) = 0 \cdot [\sum_{j=1}^6 K_{1j}d_j - F_1] + 0 \cdot [\sum_{j=1}^6 K_{2j}d_j - F_2] + 1 \cdot [\sum_{j=1}^6 K_{3j}d_j - F_3] + 0 \cdot [\sum_{j=1}^6 K_{4j}d_j - F_4] + ... + 0 \cdot [\sum_{j=1}^6 K_{6j}d_j - F_6] = 0 \Rightarrow$$

$$\sum_{j=1}^6 K_{3j}d_j = F_3 \qquad \qquad (3rd \quad Equation)$$

CHOICE 4: Select W such that W1=...=W3=0, W4=1, W5=W6=0
 * $${\underset{1 \times 6}\mathbf W^T}= \begin{bmatrix}

0 & 0 & 0 & 1 & 0 & 0 \\ \end{bmatrix}_{1 \times 6}$$

Thus,

$$ \mathbf W \cdot (\mathbf{Kd}-\mathbf F) = 0 \cdot [\sum_{j=1}^6 K_{1j}d_j - F_1] + ... + 0 \cdot [\sum_{j=1}^6 K_{3j}d_j - F_3] + 1 \cdot [\sum_{j=1}^6 K_{4j}d_j - F_4] + 0 \cdot [\sum_{j=1}^6 K_{5j}d_j - F_5] + 0 \cdot [\sum_{j=1}^6 K_{6j}d_j - F_6] = 0 \Rightarrow$$

$$\sum_{j=1}^6 K_{4j}d_j = F_4 \qquad \qquad (4th \quad Equation)$$

CHOICE 5: Select W such that W1=...=W4=0, W5=1, W6=0
 * $${\underset{1 \times 6}\mathbf W^T}= \begin{bmatrix}

0 & 0 & 0 & 0 & 1 & 0 \\ \end{bmatrix}_{1 \times 6}$$

Thus,

$$ \mathbf W \cdot (\mathbf{Kd}-\mathbf F) = 0 \cdot [\sum_{j=1}^6 K_{1j}d_j - F_1] + ... + 0 \cdot [\sum_{j=1}^6 K_{4j}d_j - F_4] + 1 \cdot [\sum_{j=1}^6 K_{5j}d_j - F_5] + 0 \cdot [\sum_{j=1}^6 K_{6j}d_j - F_6] = 0 \Rightarrow$$

$$\sum_{j=1}^6 K_{5j}d_j = F_5 \qquad \qquad (5th \quad Equation)$$

CHOICE 6: Select W such that W1=...=W5=0, W6=1
 * $${\underset{1 \times 6}\mathbf W^T}= \begin{bmatrix}

0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}_{1 \times 6}$$

Thus,

$$ \mathbf W \cdot (\mathbf{Kd}-\mathbf F) = 0 \cdot [\sum_{j=1}^6 K_{1j}d_j - F_1] + ... + 0 \cdot [\sum_{j=1}^6 K_{5j}d_j - F_5] + 1 \cdot [\sum_{j=1}^6 K_{6j}d_j - F_6] = 0 \Rightarrow$$

$$\sum_{j=1}^6 K_{6j}d_j = F_6 \qquad \qquad (6th \quad Equation)$$

Hence, $$ \quad \mathbf{Kd}=\mathbf F \quad$$ [or Eqn. (1)] is verified.

Principal of Virtual Work: Revisited
Accounting for boundary conditions:

2 bar truss example:

d1 = d2 = d5 = d6 = 0  →   W1 = W2 = W5 = W6 = 0

 * Weighting Coefficient must be "kinematically admissible", ie: cannot violate the boundary conditions.

 * Weighting Coefficients = Virtual Displacements (Calculus of Variations)

$$\mathbf W \cdot (\mathbf {Kd}- \mathbf F) = \begin{pmatrix} W_3 \\ W_4 \end{pmatrix} \cdot (\overline{\underset{2 \times 2}\mathbf K} \overline{\underset{2 \times 1}\mathbf d} - \overline{\underset{2 \times 1}\mathbf F}) = 0 \qquad \qquad  for \quad all \quad \begin{pmatrix} W_3 \\ W_4 \end{pmatrix} \qquad (3)$$

Recall:

\begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \\ \end{bmatrix}$$
 * $$ \quad \overline \mathbf K =

d_3 \\ d_4 \end{pmatrix}$$
 * $$ \quad \overline \mathbf d =\begin{pmatrix}

F_3 \\ F_4 \end{pmatrix}$$
 * $$ \quad \overline \mathbf F =\begin{pmatrix}

$$\overline \mathbf K \overline \mathbf d = \overline \mathbf F \qquad (4)$$

Proof: $$ (3) \iff (4) $$
$$Eq. (3) \iff Eq. (4)$$

Proof:
 * A) $$\quad (4) \iff (3)\quad$$ : trivial
 * B) Want to show $$\quad (3) \Rightarrow (4)\quad$$:
 * $$(3) \Rightarrow \mathbf W \cdot (\overline \mathbf {Kd}- \overline \mathbf F) = 0 \qquad for \quad all \quad \begin{pmatrix}

W_3 \\ W_4 \end{pmatrix}$$

CHOICE 1: Select W such that W3=1, W4=0
 * $${\underset{2 \times 1}\mathbf W}= \begin{pmatrix}

1 \\ 0 \\ \end{pmatrix}_{2 \times 1}$$

Thus,

$$ \mathbf W \cdot (\mathbf{Kd}-\mathbf F) = 1 \cdot [\sum_{j=3}^4 K_{3j}d_j - F_3] + 0 \cdot [\sum_{j=3}^4 K_{4j}d_j - F_4] = 0 \Rightarrow$$

$$\sum_{j=1}^6 K_{3j}d_j = F_3 \qquad \qquad (1st \quad Equation)$$

CHOICE 2: Select W such that W3=0, W4=1
 * $${\underset{2 \times 1}\mathbf W}= \begin{pmatrix}

0 \\ 1 \\ \end{pmatrix}_{2 \times 1}$$

Thus,

$$ \mathbf W \cdot (\mathbf{Kd}-\mathbf F) = 0 \cdot [\sum_{j=3}^4 K_{3j}d_j - F_3] + 1 \cdot [\sum_{j=3}^4 K_{4j}d_j - F_4] = 0 \Rightarrow$$

$$\sum_{j=1}^6 K_{4j}d_j = F_4 \qquad \qquad (2nd \quad Equation)$$

Hence, $$ \quad \overline \mathbf K \overline \mathbf d= \overline \mathbf F \quad$$ [or Eqn. (4)] is verified.

Review
$$d^{i}_{4x1}\Rightarrow\mathbf{d(lmm(i,:))}$$

 * Element DOF'S in the global (x,y) coordinate system for element (i)

$$d_{6x1}$$

 * 6 bar truss DOF's in global (x,y) coordinate system

 * Note: On a side note, the bug is explained in component element forces. Correct the code and rerun 2 bar truss problem and interpret "results"

$$ \mathbf{\underset{2 \times 1}q^{(e)}}=\mathbf{\underset{2 \times 4}T^{(e)}}\mathbf{\underset{4 \times 1}d^{(e)}} $$
 * Remember

Principle of Virtual Work II

 * The course grade is an example of a linear combination:
 * $$\alpha_o$$ x hw grade + $$sum_i=1)^3 \alpha_i$$, Exam i

 * Dereving:

$$\mathbf k^{e} = \mathbf {T^{(e)^T} \hat k^{(e)} T^{(e)}}$$ which is also found on p14-3 and p32-3

 * Recall FD relation with axial DOF'S  $$\mathbf{q^{e}}$$

$$ \mathbf{\underset{2 \times 2} \hat k^{(e)}}\mathbf{\underset{2 \times 1}q^{(e)}}=\mathbf{\underset{2 \times 1}p^{(e)}} $$

$$\Rightarrow\mathbf{\underset{2 \times 2} \hat k^{(e)}}\mathbf{\underset{2 \times 1}q^{(e)}}-\mathbf{\underset{2 \times 1}p^{(e)}}=0_{2x1}\;\;(1)$$

 * PVW: $$\mathbf{w}\cdot\mathbf{\underset{2 \times 2} \hat k^{(e)}}\mathbf{\underset{2 \times 1}q^{(e)}}-\mathbf{\underset{2 \times 1}p^{(e)}}=0_{2x1}\;\;(2)$$  for all $$\mathbf{\hat w_{2x1}}$$

 * Proof:

 * We showed that (1)   $$\Leftrightarrow$$   (2)     p24-1

$$ \mathbf{\underset{2 \times 1}q^{(e)}}=\mathbf{\underset{2 \times 4}T^{(e)}}\mathbf{\underset{4 \times 1}d^{(e)}} $$

 * Where $$\mathbf{\underset{2 \times 1}q^{(e)}}$$ is the real displacement

$$ \mathbf{\underset{2 \times 1}\hat w}=\mathbf{\underset{2 \times 4}T^{(e)}}\mathbf{\underset{4 \times 1}w} $$
 * Similarly
 * Where $$\mathbf{\underset{2 \times 1}\hat w}$$ is the virtual displacement

Conversion of Force Displacement

 * It is important to take note and go back into the book and review Chapter 2 sections 2.1,2.2,2.6,2.7

$$\mathbf{\underset{2 \times 1}\hat w}=$$ virtual axial displacement

$$\mathbf{\underset{4 \times 1}w}=$$ virtual displacement in global coordinate system, corresponding to $$\mathbf{\underset{4 \times 1}d^{(e)}}$$

Now we replace Equations (3) and (4) into equation (2) (p 26-2)

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

Recall: $$(\mathbf{AB^T}=\mathbf{B^T}\mathbf{A^T}\;\;(6)$$

For example: $$\begin{bmatrix} 1 & 2 & 3     \\  3 & 4 & 5 \end{bmatrix} $$

Recall:

$$\mathbf{a_{1}b}=\mathbf{a^{T}b}\;\;(7)$$ is a scalar

now applying (7) and (6) to (5)

$$(\mathbf{T^{(e)}}\mathbf{w})^{T}\cdot\mathbf{ \hat k^{(e)}}(\mathbf{T^{(e)}}\mathbf{d^{(e)}})-\mathbf{p^{(e)}}=0_{1x1}$$ for all $$\mathbf{w_{4x1}}$$

$$\rightarrow$$  $$(\mathbf{T^{(e)T}}\mathbf{w^{T}})^{T}\cdot\mathbf{ \hat k^{(e)}}(\mathbf{T^{(e)}}\mathbf{d^{(e)}})-\mathbf{p^{(e)}}=0_{1x1}$$  for all $$\mathbf{w_{4x1}}$$

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

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

$$\rightarrow$$  $$(\mathbf{T^{(e)}}\mathbf{p^{(e)}})=\mathbf{f^{(e)}}$$

 * Now we can say and prove

$$\rightarrow$$   $${\mathbf W} \cdot [(\mathbf {K^{(e)}d^{(e)}}- \mathbf {f^{(e)}}) = 0$$ for all $$\mathbf{w}$$

Review:

 * So far we have been dealing with a discrete case (not continuous) [matrices]

Setup for a Continous Case Using Partial Differential Equations (PDE's)
Using a sample motivational problem, we look at the following two conditions:

 * The elastic bar with varying cross sectional area $$\mathbf{A(x)}$$,$$\mathbf{E(x)}$$, also subject to a varying axial load(distributed and time independent) + concentrated load + inertia force(dyn)

Working with Partial Differential Equations and Boundary Conditions
Looking back on the infinitesimal element, dx it is shown below in Figure 3.

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

$$= \frac{\partial N}{\partial x}(x,t)dx + h.o.t. + f(x,t)dx + m(x)\ddot{u}dx\;\;(1)$$

$$f(x +dx) = F(x) + \frac{df}{dx}(x) dx + \frac{1}{2}\frac{{d^2}f}{dx^2} + ..... $$

Equation (1) $$\rightarrow$$ $$\frac{\partial N}{\partial x}+ f = m\ddot{u}\;\;(2)$$ which is the equation of motion (EDM)

$$N(x,t)=A(x)\sigma(x)\;\;(3)$$ is the constructive relation where

$$\sigma(x) = E(x)\epsilon(x,t)$$ where

$$\epsilon(x,t)=\frac{\partial U}{\partial x}(x,t)$$

PDE of motion

(3) in (2) yields: $$\frac{\partial}{\partial x}[A(x)E(x)\frac{\partial u}{\partial x}] + f(x,t) = m(x) \ddot{u}\;\;(4)$$ where $$\ddot{u} = \frac{{\partial^2}u}{\partial t^2}$$

 * Need:
 * 2 BC'S (2nd order derivation with respect to x)
 * 2 IC'S (initial displacement, initial velocity)

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

(2nd derivative with respect to t)

1) $$u(o,t) = 0$$

$$N(L,t) = F$$

$$N(L,t)=A(L)\sigma(L,t)$$ where $$(\sigma(L,t) = E(L)\epsilon(L,t)$$ where $$\epsilon(x,t)=\frac{\partial U}{\partial x}(x,t)$$

3-Bar Space Truss System
This section solves a 3-Bar Space Truss System found in p230 of the textbook. It is similar to the 3-Bar Truss System in HW3, but is in the x-y-z plane.

Statics Method
The 3-Bar Truss System was proved to be Statically Indeterminate. (See the previous link for the proof.) This 3-Bar Space Truss System is Statically Determinate. The addition of the z-axis gives another equation to solve for the reaction forces. Therefore, three equations are given to solve for three unknowns.

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

The following MATLAB code helps solve the reaction forces for each element.

The results are shown below:

Therefore, according to the Statics Method the reaction forces for each element are:

$$\R_{1}=20375 N$$ $$\R_{2}=13214 N$$ $$\R_{3}=-23148 N$$

FEA Method
Another method to solve the 3-Bar Space Truss System is using the FEA Method. This will implement the cross-sectional area and modulus of elasticity of each element to solve for the displacement.

Required MATLAB Functions
Before running the 3-Bar Space Truss System program, the following three MATLAB program functions must be in the same directory. They are obtained from the textbook website (See HW1 for textbook information). Matlab Truss Files

SpaceTrussElement.m
This function takes the Modulus of Elasticity (e), the Cross-Sectional Area (A) and the Nodal Coordinates (coord) of the element and generates the local stiffness matrix.

NodalSoln.m
This function takes the Global Stiffness Matrix (K), the Global Force vector (R), and two inputs that are the boundary conditions. The output is an array of the displacement and reaction forces.

SpaceTrussResults.m
This function inputs the Modulus of Elasticity (e), Cross-Sectional Area (A), Nodes of Each Element (coord), and the Displacement calculated in the previous function (disps). The output are the Strain, Stress, and Axial Force of each element.

Main MATLAB Function
The following MATLAB program (ThreeBarSpaceTrussEx.m) was also taken from the ZIP file provided by the textbook. This uses the FEA Method to calculate the displacement and forces. Note: "% ***" comments are modifications to the function. They decrease the amount of code lines, but still performs the same way.

Results
The following are the results of the previous program. The Connectivity Array (conn), Location Matrix Master Array (lmm), Displacement (d), Reactions (reactions), and Strain/Stress/Axial Forces (results)

The Reactions calculated using the FEA Method are equal to the Statics Method.

$$\R_{1}=20375 N$$ $$\R_{2}=13214 N$$ $$\R_{3}=-23148 N$$

Plot
The following MATLAB function is used to plot the 3-Bar Space Truss System. There are comments embedded in the function.

The Deformed Displacement is magnified by a factor of 500 for better viewing.

Plane View Down the X-Axis

Plane View Down the Y-Axis

Plane View Down the Z-Axis

Overall View

2-Bar Truss System
The 2-Bar Truss System of HW2 must be revisited due to a bug found in the MATLAB program to calculate the Strain, Stress, and Axial Forces of the elements.

Required MATLAB Functions
Before running the 2-Bar Truss System program, the following three MATLAB program functions must be in the same directory. They are obtained from the textbook website (See HW1 for textbook information). Matlab Truss Files

PlaneTrussElement.m
This function takes the Modulus of Elasticity (e), the Cross-Sectional Area (A) and the Nodal Coordinates (coord) of the element and generates the local stiffness matrix.

NodalSoln.m
NodalSoln.m

PlaneTrussResults.m
This function inputs the Modulus of Elasticity (e), Cross-Sectional Area (A), Nodes of Each Element (coord), and the Displacement calculated in the previous function (disps). The output are the Strain, Stress, and Axial Force of each element.

Bugged Original MATLAB Function
The source of the bug is in this line of code:

The array of e and A are passed though, where it should only be the e and A specifically to the element being passed though. The whole bugged function can be found below.

Fixed MATLAB Function
To fix this bug, "(i)" must be added to the e and A inputs in order for the specific elemental values to be passed into the function. The fixed function can be found below.

Results Array
The following results array is from the bugged function. The values have no meaning to the problem.

The following results array is from the fixed function. These values are right at showing the Strain, Stress, and Axial Force of each element.

These values can be compared to the results of the Statics Solution

6-Bar Space Truss System
This section solves the 6-Bar Truss System found in p226 of the textbook.

Required MATLAB Functions
Before running the 6-Bar Truss System program, the following three MATLAB program functions must be in the same directory. They are obtained from the textbook website (See HW1 for textbook information). Matlab Truss Files

PlaneTrussElement.m
PlaneTrussElement.m

NodalSoln.m
NodalSoln.m

PlaneTrussResults.m
PlaneTrussResults.m

Main MATLAB Function
The following MATLAB program (SixBarTrussEx.m) was also taken from the ZIP file provided by the textbook. It is the main function to break down the 6-Bar Truss System.

Results
The following are the results of the 6-Bar Truss System

Plot
The following MATLAB function is used to plot the 6-Bar Truss System. There are comments embedded in the function.

The Deformed Plot displacement is magnified by 1000 for better viewing.

Modified Main MATLAB Function
The following function is a modification to SixBarTrussEx.m when the elements do not have the same Modulus of Elasticity (e).

Results
The following are the results to the modified 6-Bar Truss System

Plot
The following MATLAB function is used to plot the two cases of the 6-Bar Truss System. Case 1 is where the Modulus of Elasticity are equal for all the elements. Case 2 is where the Modulus of Elasticity are not equal for all the elements. There are comments embedded in the function.

In both Deformed Plot displacements is magnified by 2000 for better viewing.

Contributing Members
The following Team Wiki members contributed to this report.

Rodney Dagulo --Eml4500.f08.wiki.dagulo.r 6:44, 6 November 2008 (UTC)

Renee Hood --Eml4500.f08.wiki.hood 6:44, 7 November 2008 (UTC)

Thomas Kehoe --Eml4500.f08.wiki.Kehoe 8:49, 7 November 2008 (UTC)

Cortland Glowacki --Eml4500.f08.wiki.cort 11:13, 7 November 2008 (UTC)

Ricardo Lopez --Eml4500.f08.wiki.Lopez 4:.08, 7 November 2008 (UTC)

Gonzalo Barcia --Eml4500.f08.wiki.barcia 12:12, 12 November 2008 (UTC)

title: User:Uf.team.wiki/HW5


text:
International Building Code/Who is responsible for funding special inspections?

The owner of the project is responsible for funding the expense for special inspections.


text:
User:Eas4200c.f08.blue.a/Lecture 23

Reading page 70

$$\left\{\epsilon_{ij} \right\}_{6x1} = \begin{bmatrix} \bar{A}_{3x3} & o\\ 0 & \bar{B}_{3x3} \end{bmatrix}_{6x6}\left\{\sigma_{ij} \right\}_{6x1}$$

The $$\sigma$$ - $$\epsilon$$ relationship: Assuming diagonal coefficient is not zero

$$\left\{\sigma_{ij} \right\}_{6x1} = \begin{bmatrix} \bar{A}^{-1}_{3x3} & 0\\ 0 & \bar{B}^{-1}_{3x3} \end{bmatrix}_{6x6}\left\{\epsilon_{ij} \right\}_{6x1}$$

If $$\begin{bmatrix} \bar{A}^{-1}_{3x3} & 0\\ 0 & \bar{B}^{-1}_{3x3} \end{bmatrix}_{6x6}$$ we can varify that the relationship by $$\bar{c}^{-1}\bar{c} = \bar{I}$$ then $$\bar{c}^{-1}\bar{c} = \begin{bmatrix} \bar{A}^{-1}\bar{A} & 0 \\ 0 & \bar{B}^{-1}\bar{B} \end{bmatrix} = \bar{I}$$ therefore,

$$\left\{\sigma_{ij} \right\} = \begin{bmatrix} \bar{A}^{-1} & 0 \\ 0 & \bar{B}^{-1} \end{bmatrix}\begin{Bmatrix} 0\\ 0\\ 0\\ 0\\ \epsilon_{31}\\ \epsilon_{12} \end{Bmatrix}$$

Road map

Equilibrium Equation for Stress in a nonuniform stress field (section 2.4 in the book)

Consider a one dimensional case as a model.For a nonuniform field, the distribution is not constant, therefore $$f\left(x \right)\neq constant$$

and showing the shear forces acting on this

title: User:Eas4200c.f08.blue.a/Lecture 23


text: File:Fkdtorture.JPG

title: File:Fkdtorture.JPG


text:
File talk:021 PM Project Proposal for Approvement.png

So what is the license of this file?--Juan 23:56, 27 October 2008 (UTC)


text:
User:Eas4200c.f08.aeris.guan/hw5

Strain Displacement Relationship and Equilibrium Equations (Stresses)
With the theory of elasticity, we first considered the kinematic assumptions (21.3) which resulted in realizing that the four strain components in the xx, yy, zz, and yz directions were equal to zero (21.3)and consequently the stresses in these directions were zero (22.4). Comparing this to the equations in page 70 of the book, we find that since the book has used the convention of an (x,y) coordinate plane in order to analyze the strain and stress, however we used a (y,z) coordinate system. These expressions are easily equated by using a cyclic permutations of the variables. Then, the stress strain relationship (tensorial or engineering) were established (23.2). Now, we will rewrite the stress strain relationship as shown below. $$ [\varepsilon_{ij}]_{6x1}= \begin{bmatrix} A_{3x3} & 0_{3x3}\\ 0_{3x3} & B_{3x3}\\

\end{bmatrix}_{6x6} [\sigma_{ij}]_{6x1}= C[\sigma_{ij}]_{6x1} $$ Where A and B are matrices with the values shown below and B is a diagonal matrix. The 6x6 block diagonal matrix will be referred to as the C matrix. $$A= \begin{bmatrix} \frac{1}{E} &\frac{-\nu }{E} &\frac{-\nu}{E} \\ \frac{-\nu }{E} &\frac{1}{E} &\frac{-\nu }{E}  \\ \frac{-\nu }{E} &\frac{-\nu }{E} &\frac{1}{E} \\ \end{bmatrix}, B= \begin{bmatrix} \frac{1}{2G} &0  &0 \\ 0 &\frac{1}{2G}  &0 \\ 0 &0  &\frac{1}{2G} \end{bmatrix} $$ However, we will keep the C in the block diagonal form. To solve for the stress, we multiply both sides by the inverse of C. To invert a diagonal matrix, one must take the inverses of each value of the diagonal. It is the same for a block diagonal matrix as seen below. Invert C by inverting the diagonal matrices. $$ [\sigma_{ij}]_{6x1}= \begin{bmatrix} A_{3x3}^{-1} & 0_{3x3}\\ 0_{3x3} & B_{3x3}^{-1}\\

\end{bmatrix}_{6x6} [\varepsilon_{ij}]_{6x1}= C^{-1}[\varepsilon_{ij}]_{6x1}

$$ $$ C^{-1}C= \begin{bmatrix} AA^{-1} & 0\\ 0 & BB^{-1}\\

\end{bmatrix} =I

$$ Where I is the identity matrix, a diagonal matrix with the diagonal values all equaling one. Then, we will continue on proving $$\displaystyle \sigma_{xx}=\sigma_{yy}=\sigma_{zz}=\tau_{yz}=0$$. First, we will apply the zero strain relations in which the strains in the xx, yy, zz, and yz directions are zero. Knowing that there are essentially six strain components since the strain tensor is a 3 by 3 symmetrical matrix, a 6 by 1 column matrix can be constructed where the first four rows are zero and the last two are non zero. Subscripts x, y, and z will be replaced with indices 1, 2, and 3 respectively. This gives the following expression. $$ [\sigma_{ij}]= \begin{bmatrix} A^{-1} & 0\\ 0 & B^{-1}\\ \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ \varepsilon _{31}\\ \varepsilon _{12}\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ \sigma _{31}\\ \sigma _{12}\\ \end{bmatrix}

$$ Multiplying the top three rows of the 6 x 6 matrix to the top three rows of the strain tensor results in $$\displaystyle \sigma_{11}=\sigma_{22}=\sigma_{33}$$. The resulting last three rows are derived in the collapsible box below. (16.2) Recall the road map step: equilibrium equation for stresses (section 4.2) in a nonuniform stress field. To understand the nonuniform stress state, let's consider a 1 dimensional case (a cantilever beam) as a model:

It is necessary to point out that the axial stress in the uniform case is the same for every value of x (i.e. $$\displaystyle \sigma(x)=\sigma(x+dx)$$) and F(x)=0. For the nonuniform case, axial stress DOES vary with x (i.e. $$\sigma(x)\ne\sigma(x+dx)$$) and F(x) is either a nonuniform function or constant. We will use this concept to derive the equilibrium equations for the stresses.

title: User:Eas4200c.f08.aeris.guan/hw5


text: File:Cyclicperm.JPG

title: File:Cyclicperm.JPG


text:
MediaWiki:Ipbreason-dropdown

 * Common block reasons
 * Vandalism
 * Advertising on Wikiversity is not appropriate
 * External link spam or spambot
 * Repeat use of incapable copyright information
 * Intimidation, harassment, or vulgar language
 * Repeat disruptive editing of pages
 * Inappropriate username
 * Wikiversity is not your personal podium
 * Abusing multiple accounts
 * Open proxy or zombie (more info)
 * Self-block or block of additional account, for testing the administrative tools
 * Cross-wiki issues
 * Cross-wiki vandalism
 * Cross-wiki advertising or link spam
 * Cross-wiki intimidation or harassment
 * Cross-wiki abuse of multiple accounts
 * Cross-wiki legal threats
 * Cross-wiki use of open proxy or zombie (more info)
 * Cross-wiki disruptive editing
 * Cross-wiki impersonation

title: MediaWiki:Ipbreason-dropdown


text: File:Nonunistressfield.JPG

title: File:Nonunistressfield.JPG


text:
User:EML4500.f08.echo.mott/HW5

=HW Report 5= The contents of this page were moved to User:Eml4500.f08.echo.mott/HW5

Axial FD Relation
The axial force-displacement relation for a bar element remains the same whether 2-D or 3-D:

$$k^{(e)}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix}=\begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix}$$

Element dofs & Element Forces in Global x-y-z Coordinates
The transformation matrix in 3-D can be shown by

$$\mathbf{T}= \begin{bmatrix}l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)}\end{bmatrix}$$

The axial F-D relation can be written as:

$$\begin{bmatrix}F_{1} \\ F_{2}\end{bmatrix}=\begin{bmatrix}k & -k \\ -k & k\end{bmatrix}\begin{bmatrix}d_{1} \\ d_{2}\end{bmatrix}=\begin{bmatrix}P_{1} \\ P_{2}\end{bmatrix}$$

The conversion from local to global axial coordinates is:

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

$$\begin{bmatrix}d_{1} \\ d_{2}\end{bmatrix}=\begin{bmatrix}l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)}\end{bmatrix}\begin{bmatrix}{d_{1}}^{(e)} \\ {d_{2}}^{(e)} \\ {d_{3}}^{(e)} \\ {d_{4}}^{(e)} \\ {d_{5}}^{(e)} \\ {d_{6}}^{(e)}\end{bmatrix}$$

Where l, m, and n are the director cosines of each axis.

$$l^{(e)}=\frac{x_{2}-x_{1}}{L}$$

$$m^{(e)}=\frac{y_{2}-y_{1}}{L}$$

$$l^{(e)}=\frac{z_{2}-z_{1}}{L}$$

$$L^2={(x_{2}-x_{1})}^{2} + {(y_{2}-y_{1})}^{2} + {(z_{2}-z_{1})}^{2}$$

Beginning with the element F-D relation and substituting the global transformation matrix,

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

$$\mathbf{k^{(e)}Td=f^{(e)}}$$

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

The global 3-D F-D relation can be written as:

$$ \frac{EA}{L}\begin{bmatrix}l^{2(e)} & m^{(e)}l^{(e)} & n^{(e)}l^{(e)} & -l^{2(e)} & -m^{(e)}l^{(e)} & -n^{(e}l^{(e)} \\ m^{(e)}l^{(e)} & m^{2(e)} & m^{(e)}n^{(e)} & -m^{(e)}l^{(e)} & -m^{2(e)} & -m^{(e)}n^{(e)} \\ n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & n^{2(e)} & -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -n^{2(e)} \\ -l^{2(e)} & -m^{(e)}l^{(e)} & -n^{(e)}l^{(e)} & l^{2(e)} & m^{(e)}l^{(e)} & n^{(e)}l^{(e)} \\ -m^{(e)}l^{(e)} & -m^{2(e)} & -m^{(e)}n^{(e)} & m^{(e)}l^{(e)} & m^{2(e)} & m^{(e)}n^{(e)} \\ -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -n^{2(e)} & n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & n^{2(e)} \\\end{bmatrix}\begin{bmatrix}u_{1} \\ v_{1} \\ w_{1} \\ u_{2} \\ v_{2} \\ w_{2} \\\end{bmatrix}=\begin{bmatrix}F_{1x} \\ F_{1y} \\ F_{1z} \\ F_{2x} \\ F_{2y} \\ F_{2z} \\ \end{bmatrix} $$

Matlab Solutions

 * Please refer to this page.

Monday Oct. 20, 2008
Justification of eliminating rows 1, 2, 5, and 6 to obtain $$\mathbf{\bar{k}}_{2x2}$$ in 2-bar truss: (p 10-1 & 23-2)

FD Relation:

$$\mathbf{K}_{6x6}\mathbf{d}_{6x1}=\mathbf{F}_{6x1} \Rightarrow \mathbf{K}\mathbf{d}-\mathbf{F}=\underline{0}_{6x1} $$ . . . (1)

PVW:

$$\mathbf{W}_{6x1}\cdot (\mathbf{K}\mathbf{d}-\mathbf{F})=0_{1x1}$$ for all $$\mathbf{W}_{6x1}$$. . . (2)

Where the weighting matrix is $$\mathbf{W}_{6x1}$$

$$Eq (1) \Leftrightarrow Eq (2)$$

Proof:

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

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

(2) $$\Rightarrow \mathbf{W}\cdot(\mathbf{Kd}-\mathbf{F})=0$$

for all W

Choice 1: Select W at $$w_1=1, w_2=w_3=w_4=w_5=w_6=0$$

$$\mathbf{W}_{1x6}^T=\begin{vmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{vmatrix}$$

$$\mathbf{W}\cdot (\mathbf{kd}-\mathbf{F})=1.\left[\sum_{j=1}^6 k_{1j}d_j-F_1\right]+0\left[\sum_{j=1}^6 k_{2j}d_j-F_2\right]+...+0\left[\sum_{j=1}^6 k_{6j}d_j-F_6\right]=0$$

This leads to:

$$\sum_{j=1}^6 k_{1j}d_j=F_1$$ (1st Equation)

Choice 2: Select W at $$w_1=0, w_2=1, w_3=w_4=w_5=w_6=0$$

$$\mathbf{W}_{1x6}^T=\begin{vmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{vmatrix}$$

$$\mathbf{W}\cdot (\mathbf{kd}-\mathbf{F})=0\left[\sum_{j=1}^6 k_{1j}d_j-F_1\right]+1\left[\sum_{j=1}^6 k_{2j}d_j-F_2\right]+0\left[\sum_{j=1}^6 k_{3j}d_j-F_3\right]...+0\left[\sum_{j=1}^6 k_{6j}d_j-F_6\right]=0$$

This leads to:

$$\sum_{j=1}^6 k_{2j}d_j=F_2$$ (2nd Equation)

Hence kd=F (or Eq (1))

PVW: Accounting for boundary conditions:

2 bar truss: $$d_1=d_2=d_5=d_6=0$$

Weight coefficient must be "kinematically admissible," i.e. cannot violate the boundary conditions.

Weighting coefficients $$\equiv$$ Virtual Displacements as on p24-4 (Calculus of Variations)

Previously, we wrote:

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

Where k was a 6x2 matrix and d was a 2x1 matrix.

Simplifying results in:

$$=\begin{Bmatrix} w_3 \\ w_4 \end{Bmatrix} \cdot (\mathbf{\bar K}_{2x2}\mathbf{\bar d}_{2x1}-\mathbf{\bar F})=0$$ for all $$\begin{Bmatrix} w_3 \\ w_4 \end{Bmatrix}$$. . . (3)

Specifically:

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

$$\mathbf{\bar d}=\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix}$$

$$\mathbf{\bar F}=\begin{Bmatrix} F_3 \\ F_4 \end{Bmatrix}$$

HW Begin

Show $$(3) \Leftrightarrow \mathbf{\bar K \bar d}=\mathbf{\bar F} (4)$$

Take w3=w4=1

$$\mathbf{1}\cdot\mathbf{\bar K}\mathbf{\bar d}-\mathbf{\bar F}=\mathbf{0}$$

Therefore, $$\mathbf{\bar K}\mathbf{\bar d}=\mathbf{\bar F}$$

Monday Oct. 27, 2008
HW: 6 bar truss, page 226

 * Same E, A for all elements

Bug: for two bar truss used different E's and A's for the elements

$$ \mathbf {d}_{4x1}^{(i)} \harr d(lmm(i,:))$$
 * right hand side is Matlab code

Where: element degree of freedoms in global (x,y) coordinate system for element (i)

$$\mathbf {d}_{6x1}$$ 2-bar truss degree of freedoms in global coordinate (x,y) system.

Bug Explained

In computation of element forces $$\rarr$$ change (e,A) to e(i) and A(i). Correct the code, and rerun the 2-bar truss problem. Interpret the "results".

Back to PVW

$$ \mathbf{q}_{2x1}^{(i)} = \mathbf{T}_{2x4} \mathbf{d}_{4x1}^{(i)}$$

 * At this point Dr. Vu-Quoc referred to slides 24-1 and 21-2

Course grade $$\alpha_o \cdot$$ Hw grade +$$\sum \alpha_i \cdot Exam_i$$

 * Examine pages 23-3 and 14-3

Deriving $$\mathbf{k}_{4x4}^{(e)}= \mathbf{T}_{4x2}^{(e)T} \mathbf{\hat{k}}_{2x2}^{(e)}\mathbf{I}_{2x4}^{(e)}$$

Recall FD relation with axial degree's of freedom $$ \mathbf{q}^{(e)}$$

$$ \mathbf{\hat{k}}_{2x2}^{(e)} \mathbf{q}_{2x1}^{e} = \mathbf{p}_{2x1}^{e}$$

$$\rightarrow \mathbf{\tilde{k}}^{(e)} \mathbf{q}^{(e)} - \mathbf{p}^{(e)}= \mathbf{0}_{2x1}$$. . . (1)

PVW: $$ \mathbf{W}_{2x1} \cdot (\mathbf{\hat{k}}^{(e)} \mathbf{q}^{(e)} - \mathbf{p}^{(e)}) = 0$$. . . (2)

we showed $$ (1) \leftrightarrow (2)$$

Recall: $$ \mathbf{q}_{2x1}^{(e)} = \mathbf{T}_{2x1}^{(e)} \mathbf{d}_{4x1}^{(e)}$$. . . (3)

Similarly $$\mathbf{\hat{W}}_{2x1} = \mathbf{T}_{2x4}^{(e)} \mathbf {W}_{4x1}$$. . . (4)

Wednesday Oct. 29, 2008
$$\mathbf{\hat{w}}_{2x1}$$ = virtual displacement corresponding to $$\mathbf{q}_{2x1}^{(e)}$$

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

Replacing Equations 3 and 4 into Equation 2:

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

Recall: $$\left ( \mathbf{A}_{pxq} \mathbf{B}_{qxr} \right )^T = \mathbf{B}^T \mathbf{A}^T $$ (6)

Recall: $$\mathbf{a}_{nx1} \cdot \mathbf{b}_{nx1}=\mathbf{a}_{1xn}^T \mathbf{b}_{nx1}$$ (7)

Apply Equation 7 & 6 in 5:

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

Which results in:

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

Further simplification results in:

$$\mathbf{w} \cdot \left [ \left (\mathbf{T}^{(e)T} \mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)} \right ) \mathbf{d}^{(e)} - \mathbf{T}^{(e)} \mathbf{p}^{(e)} \right ] = 0 $$ for all $$\mathbf{w}_{4x1}$$

Generalizing:

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

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

So for discrete case (matrices) (non-continuous)

Now, continuous case (PDEs) Motivational model problem: Elastic bar with varying A(x), E(x), subjected to varying axial (time dependent) load (distributed) + concentrated (time dependent) load + inertial force (dynamic) This problem is illustrated in the figure below.

Friday Oct. 31, 2008
FBD:

$$\sum F_x=0=-N(x,t)+N(x+dx,t)+f(x,t)dx-m(x)\ddot u dx=\frac{\partial N}{\partial x}(x,t)du+h.o.t.+f(x,t)dx-m(x)\ddot u dx $$  eqn(1)

Where h.o.t. = higher order terms

Recall Taylor series expansion:

$$f(x+dx)=f(x)+\frac{df(x)}{dx}dx+\frac{1}{2}\frac{d^2f(x)}{dx^2}dx^2...$$

Where the final part of the equation above represents the beginning of the higher order terms.

Eq.(1) yields the Equation of Motion (EoM): $$\frac{\partial N}{\partial x}+f=m\ddot u $$  eqn(2)

$$N_{ }^{ }(x,t)=A(x)\sigma (x,t)$$  eqn(3)

Where $$\sigma (x,t)=E(x)\varepsilon (x,t)$$

and $$\varepsilon (x,t)=\frac{\partial u}{\partial x}(x,t) $$

Where equation 3 is a constitutive relation.

Plugging equation 3 into 2 yields the PDE of Motion:

$$\frac{\partial}{\partial x} \left [ A(x)E(x)\frac{\partial u}{\partial x}\right ] + f(x,t)=m(x)\ddot u$$  eqn(4)

Where $$\ddot u = \frac{d^2 u}{dt^2}$$

In order to proceed further, we need two boundary conditions (2nd order derivative with respect to x)

We also have two initial conditions which are the initial displacement and initial velocity (2nd derive with respect to t)

Applying these to the two possible options shown in the figures to the right:

For scenario 1,

$$u_{ }^{ }(0,t)=u(L,t)=0$$

For scenario 2,

$$u_{ }^{ }(0,t)=0$$ and $$N_{ }^{ }(L,t)=F(t)$$

where $$N_{ }^{ }(L,t)=A(L)\sigma(L,t)$$

where $$\sigma(L,t)=E(L)\varepsilon(L,t)$$

where $$\varepsilon(L,t)=\frac{\partial u}{\partial x}(L,t)$$

Contributing Team Members
Chris Mott 20:14, 7 November 2008 (UTC)

Bradley LaCroix 20:08, 7 November 2008 (UTC)

Brandon Sell 20:10, 7 November 2008 (UTC)

Charles Marshall 20:15, 7 November 2008 (UTC)

--Chris Sell 20:27, 7 November 2008 (UTC)

Past Homeworks
[Homework 4]

[Homework 3]

[Homework 2]

[Homework 1]

title: User:EML4500.f08.echo.mott/HW5


text:
User:Eas4200c.f08.aero6.behnke/HW5

Continuing the Roadmap
The kinematic assumption, the 4 zero strain components, and the 4 zero stress components can be found on page 70 of C.T. Sun's "Mechanics of Aircraft Structures."

$$\varepsilon -\sigma $$ Relationship

You can rewrite $$\varepsilon -\sigma $$ Relationship as follows

$$\begin{Bmatrix} \varepsilon_{ij}\end{Bmatrix}_{6x1} = \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}\begin{Bmatrix} \sigma _{ij} \end{Bmatrix}$$

You can also rewrite $$\sigma -\varepsilon $$ Relationship as follows

$$\begin{Bmatrix} \sigma_{ij}\end{Bmatrix}_{6x1} = \begin{bmatrix} A^{-1} & 0 \\ 0 & B^{-1} \end{bmatrix}\begin{Bmatrix} \varepsilon _{ij} \end{Bmatrix}$$

In order to prove that the above matrices are inverses of one another, multiply them together. If an identity matrix is produced, than the matrices are inverses of one another.

$$\sigma_{11} = 0 ;\sigma_{23} = 2G\epsilon_{23} = 0 ;\sigma_{31} = 2G\epsilon_{31} ;\sigma_{12} = 2G\epsilon_{12}\,$$
 * \sigma_{22} = 0
 * \sigma_{33} = 0

Stresses on a unit cube

Figure 1 only has stresses that have x-components acting on the block. Therefore, we can use this figure to sum all the forces in the x-direction. This procedure will lead to the following equation:

 * $$0 = \frac{d\sigma_{xx}(x)}{dx}+\frac{d\sigma_{yx}(y)}{dy}+\frac{d\sigma_{zx}(z)}{dz}\!$$

The same can be done for the stresses acting in the y- and z-directions. However, it is the same procedure as the stresses in the x-direction. Therefore, in the calculations below only the x-component stresses have been used to obtain the above formula.

The procedure to obtain the above formula follows. The formulas produced from the y- and z-component stresses are also recorded and are obtained in a similar fashion.

 * $$\Sigma F_{x} = 0 = \sigma_{xx}(x+dx)(dydz)-\sigma_{xx}(x)(dydz)+\sigma_{yx}(y+dy)(dxdz)-\sigma_{yx}(y)(dxdz)+\sigma_{zx}(z+dz)(dxdy)-\sigma_{zx}(z)(dxdy)\!$$

 * $$0 = dydz[\sigma_{xx}(x+dx)=\sigma_{xx}(x)]+dxdz[\sigma_{yx}(y+dy)-\sigma_{yx}(y)]+dxdy[\sigma_{zx}(z+dz)-\sigma_{zx}(z)]\!$$

Using the following theorem and neglecting the squared terms

 * $$g(x+dx) = g(x) + \frac{dg(x)}{dx}dx + \frac{d^{2}g(x)}{dx^{2}}dx^{2} + ...\!$$

 * $$0 = dydz[\sigma_{xx}(x)+\frac{d\sigma_{xx}(x)}{dx}dx-\sigma_{xx}(x)]+dxdz[\sigma_{yx}(y)+\frac{d\sigma_{yx}(y)}{dy}dy-\sigma_{yx}(y)]+dxdy[\sigma_{zx}(z)+\frac{d\sigma_{zx}(z)}{dz}dz-\sigma_{zx}(z)]\!$$

 * $$0 = dydz[\frac{d\sigma_{xx}(x)}{dx}dx]+dxdz[\frac{d\sigma_{yx}(y)}{dy}dy]+dxdy[\frac{d\sigma_{zx}(z)}{dz}dz]\!$$

We can then divide this equation by $$dxdydz\!$$ to obtain:

 * $$0 = \frac{d\sigma_{xx}(x)}{dx}+\frac{d\sigma_{yx}(y)}{dy}+\frac{d\sigma_{zx}(z)}{dz}\!$$

The same thing could be done for the stresses in the y- and z-direction to obtain the following equations:

 * $$0 = \frac{d\sigma_{yy}(y)}{dy}+\frac{d\sigma_{xy}(x)}{dx}+\frac{d\sigma_{zy}(z)}{dz}\!$$

 * $$0 = \frac{d\sigma_{zz}(z)}{dz}+\frac{d\sigma_{xz}(x)}{dx}+\frac{d\sigma_{yz}(y)}{dy}\!$$

Bidirectional Bending
Read section 4.1 and 4.2

$$ M_y = \int_A z\sigma_{xx} dA = \int_A x\sigma_{xx} dy dz \, $$ $$ M_z = \int_A y\sigma_{xx} dA = \int_A y\sigma_{xx} dy dz \, $$

Moment of Inertia tensor: $$ I_{y}, I_{z}, I_{yz} \,$$

$$ I_y = \int_A z^2 dA \, $$
 * $$ I_z = \int_A y^2 dA \, $$

$$ I_{yz} = \int_A y z dA \, $$

$$ \sigma_{xx} = E \epsilon_{xx} = \frac {I_y M_z - I_{yz}M_y}{I_yI_z - I_{yz}^2 }y + \frac {I_zM_y - I_{yz} M_z}{I_yI_z - I_{yz}^2 }z \, $$

Recall

$$\mathbf{\bar{I}} = \begin{bmatrix} I_{11} & I_{12} & I_{13} \\ I_{21} & I_{22} & I_{23} \\ I_{31} & I_{32} & I_{33} \end{bmatrix}$$

Note the denominator the determinant of the matrix $$\, \begin{bmatrix} I _{y} & I _{yz}\\ I _{zy} & I _{z}\\ \end{bmatrix}$$

$$ D = I_y I_z - I_{yz}^2 \, $$

$$ D = I_{22} I_{33} - I_{23}^2 \, $$

NACA Airfoil Problem
Part 1 Results:

Part 2 Results:

As expected, the inclusion of the skin and walls in the calculation resulted in an increase in $$I_{22}$$, $$I_{33}$$, and $$I_{23}$$. The largest increase can be seen in $$I_{33}$$, while there is very little increase in $$I_{22}$$. This is because the additional elements are a greater distance from the z-axis than the y-axis.

The addition of the skin and partitions also results in slightly lower maximum stresses. This change is due to the increased moment of inertia.

Team Members
Felipe OrtegaEas4200c.f08.aero6.ortega 11:55, 6 November 2008 (UTC)

Kris Loper Eas4200c.f08.aero6.loper 00:32, 7 November 2008 (UTC)

Eduardo Villalba Eas4200c.f08.aero6.villalba 19:58, 7 November 2008 (UTC)

Matt Inman Eas4200c.f08.aero6.inman 20:05, 7 November 2008 (UTC)

Marlana Behnke Eas4200c.f08.aero6.behnke 20:06, 7 November 2008 (UTC)

Scott Chastain Eas4200c.f08.aero6.chastain 21:00, 7 November 2008 (UTC)

title: User:Eas4200c.f08.aero6.behnke/HW5


text:
User:Shanel/monobook.js

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

title: User:Shanel/monobook.js


text:
Oscillators

Theoretical oscillators
Def.

"a tuned electronic circuit used to generate a continuous output waveform", or

a "pattern that returns to its original state, in the same orientation and position, after a finite number of generations"

is called an oscillator.

title: Oscillators


text:
User:Muhraz (old)

Hello
Hello! I am from Wikipedia Malay. Look at my page in Wikipedia Malay.

title: User:Muhraz (old)


text:
Internet and Project Competence/groups

open education

group 1: open education

 * Earnhart, Lee W.
 * Nowlan, Mary K.
 * St.Charles, Angela R.
 * Marlowe, Ashley N.
 * Gilmore, Ashley L.
 * Stewart, Michelle R.
 * Maultsby, Sarah A.
 * Trotter, Allison L
 * Daniela Schweigl
 * Veronica Belcheva

discussion board
elementary education program requirements

group 2: elementary education program requirements

 * Christiane Rohr
 * Martin, Amanda R.
 * Mclain, Laura K.
 * Reese, Ashley D.
 * Novak, Ashley B.
 * Boris Aleksiev
 * Ledford, Jessica L.
 * Sos, Ashley L.

discussion board
types of instruction and learning theories

group 3: types of instruction and learning theories

 * Ott, Sherry L.
 * Capps, Taylor K.
 * Hinson, Jennifer S.
 * Ryan, Ashton B.
 * Travis, Amie J.
 * Reynolds, Danielle R
 * Reyes, Valerie
 * Verena Kroeger
 * Veronika Pfaller => Wikipage

discussion board
areas of specialization for elementary teachers

group 4: areas of specialization

 * Truesdale, Sidney L.
 * Rhoden, Mark A.
 * Williams, Marissa J.
 * Watson, Chelsea M.
 * Coberly, Ashley F.
 * Barnhardt, Regina L.
 * McGuirt, Meredith N.
 * Kathrin Bogenrieder -> Kathrin B.
 * Katharina Band -> Katharina B

discussion board
Interaction between children and nature

group 5:Interaction between children and nature

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

discussion board
The teacher's (emotional) influence on first graders

group 6a: The teacher's (emotional) influence on first graders

 * Graham, Ashley M.
 * Valentine, Casey D.
 * Murrell, Genevieve N.
 * Sain, Jennifer L.
 * Goodson, Jordan C.
 * Daniela Widdmann  =>   Wikipage
 * Lilia Popova

discussion board
The teacher's (emotional) influence on first graders

group 6b: The teacher's (emotional) influence on first graders

 * Endaya, Verna F.
 * Register, Cayton P.
 * Grimm, Erica R. => http://en.wikiversity.org/wiki/User:Egrimm
 * Surette, Megan E. => http://en.wikiversity.org/wiki/User:MeganSurette
 * Mariana Igova
 * Kathrin Lehnert =>  Wikipage

title: Internet and Project Competence/groups


text:
User talk:Jenny O/Ghosts of Rwanda

Feedback
This is an exceptionally good page which could (down the track) be migrated/merged to Ghosts of Rwanda if you're willing. -- Jtneill - Talk - c 11:56, 28 October 2008 (UTC)
 * Thanks James for your postive feedback. Yes, I would be willing to have it merged. Jenny O 21:40, 28 October 2008 (UTC)

title: User talk:Jenny O/Ghosts of Rwanda


text:
User:Eas4200c.f08.aeris.krammer/HW5

 Please do not remove this comment box; in case you amend your work, just add some comments to this box.

Eml4500.f08 15:48, 11 November 2008 (UTC)

- Finished the matlab problem

Eas4200c.f08.aeris.holladay 03:03, 21 November 2008 (UTC)

Strain Displacement Relationship and Equilibrium Equations (Stresses)
With the theory of elasticity, we first considered the kinematic assumptions (21.3) which resulted in realizing that the four strain components in the xx, yy, zz, and yz directions were equal to zero (21.3)and consequently the stresses in these directions were zero (22.4). Comparing this to the equations in page 70 of the book, we find that since the book has used the convention of an (x,y) coordinate plane in order to analyze the strain and stress, however we used a (y,z) coordinate system. These expressions are easily equated by using a cyclic permutations of the variables. Then, the stress strain relationship (tensorial or engineering) were established (23.2). Now, we will rewrite the stress strain relationship as shown below. $$ [\varepsilon_{ij}]_{6x1}= \begin{bmatrix} A_{3x3} & 0_{3x3}\\ 0_{3x3} & B_{3x3}\\

\end{bmatrix}_{6x6} [\sigma_{ij}]_{6x1}= C[\sigma_{ij}]_{6x1} $$ Where A and B are matrices with the values shown below and B is a diagonal matrix. The 6x6 block diagonal matrix will be referred to as the C matrix. $$A= \begin{bmatrix} \frac{1}{E} &\frac{-\nu }{E} &\frac{-\nu}{E} \\ \frac{-\nu }{E} &\frac{1}{E} &\frac{-\nu }{E}  \\ \frac{-\nu }{E} &\frac{-\nu }{E} &\frac{1}{E} \\ \end{bmatrix}, B= \begin{bmatrix} \frac{1}{2G} &0  &0 \\ 0 &\frac{1}{2G}  &0 \\ 0 &0  &\frac{1}{2G} \end{bmatrix} $$ However, we will keep the C in the block diagonal form. To solve for the stress, we multiply both sides by the inverse of C. To invert a diagonal matrix, one must take the inverses of each value of the diagonal. It is the same for a block diagonal matrix as seen below. Invert C by inverting the diagonal matrices. $$ [\sigma_{ij}]_{6x1}= \begin{bmatrix} A_{3x3}^{-1} & 0_{3x3}\\ 0_{3x3} & B_{3x3}^{-1}\\

\end{bmatrix}_{6x6} [\varepsilon_{ij}]_{6x1}= C^{-1}[\varepsilon_{ij}]_{6x1}

$$ $$ C^{-1}C= \begin{bmatrix} AA^{-1} & 0\\ 0 & BB^{-1}\\

\end{bmatrix} =I

$$ Where I is the identity matrix, a diagonal matrix with the diagonal values all equaling one. Then, we will continue on proving $$\displaystyle \sigma_{xx}=\sigma_{yy}=\sigma_{zz}=\tau_{yz}=0$$. First, we will apply the zero strain relations in which the strains in the xx, yy, zz, and yz directions are zero. Knowing that there are essentially six strain components since the strain tensor is a 3 by 3 symmetrical matrix, a 6 by 1 column matrix can be constructed where the first four rows are zero and the last two are non zero. Subscripts x, y, and z will be replaced with indices 1, 2, and 3 respectively. This gives the following expression. $$ [\sigma_{ij}]= \begin{bmatrix} A^{-1} & 0\\ 0 & B^{-1}\\ \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ \varepsilon _{31}\\ \varepsilon _{12}\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ \sigma _{31}\\ \sigma _{12}\\ \end{bmatrix}

$$ Multiplying the top three rows of the 6 x 6 matrix to the top three rows of the strain tensor results in $$\displaystyle \sigma_{11}=\sigma_{22}=\sigma_{33}$$. The resulting last three rows are derived in the collapsible box below. (16.2) Recall the road map step: equilibrium equation for stresses (section 4.2) in a nonuniform stress field. To understand the nonuniform stress state, let's consider a 1 dimensional case (a cantilever beam) as a model:

It is necessary to point out that the axial stress in the uniform case is the same for every value of x (i.e. $$\displaystyle \sigma(x)=\sigma(x+dx)$$) and F(x)=0. For the nonuniform case, axial stress DOES vary with x (i.e. $$\sigma(x)\ne\sigma(x+dx)$$) and F(x) is either a nonuniform function or constant. We will use this concept to derive the equilibrium equations for the stresses.

Bidirectional Bending
Let us consider a bar with a noncircular cross section. $$\displaystyle Px(x)$$ and $$\displaystyle Py(x)$$ are distributed loadings on the bar, $$\displaystyle dA=dyzdz$$ is an infinitely small area and $$\displaystyle \sigma _{xx}$$ is the normal bending stress. The bending moments about the y and z axes are $$M_{y}=\int_{A}^{}{z\sigma_{xx}dA}$$ $$M_{z}=\int_{A}^{}{y\sigma_{xx}dA}$$

The moment of inertia tensors are shown also in indicial notations $$\displaystyle I_{y}=I_{yy}=I_{22}$$ $$\displaystyle I_{z}=I_{zz}=I_{33}$$ $$\displaystyle I_{yz}=I_{23}$$ And they are defined as $$I_{y}=\int_{A}^{}{}\int z^{2}dA $$ $$I_{z}=\int_{A}^{}{}\int y^{2}dA $$ $$I_{yz}=\int_{A}^{}{}\int y zdA$$

Examining Hooke's Law we can see that $$\sigma _{xx}=E\varepsilon _{xx}= m_{y}y+m_{z}z=\frac{I_{y}M_{z}-I_{yz}M_{y}}{I_{y}I_{z}-(I_{yz})^{2}}y+\frac{I_{z}M_{y}-I_{yz}M_{z}}{I_{y}I_{z}-(I_{yz})^{2}}z$$

Where the denominator of the equation that is the determinant of the matrix
$$\displaystyle D:= {I_{y}I_{z}-(I_{yz})^{2}} = {I_{22}{I_33}-(I_{23})^{2}}$$ Is the determinant of
$$\begin{bmatrix} I_{22} & I_{23}\\ I_{32} & I_{33} \end{bmatrix}$$

At the neutral axis  the bending moment  $$\displaystyle \sigma _{xx}=0$$
$$\displaystyle\sigma _{xx}= m_{y}y+m_{z}z=0$$ Solving the above equation, we get
$$\displaystyle z=(-\frac{m_{y}}{m_{z}})y$$ The above equation is the equation of a straight line. Where $$\displaystyle (-\frac{m_{y}}{m_{z}})= tan\beta$$  is the slope of the equation.

Because the distribution of bending stresses is linear over this cross-section, it can be said that it is also linear with respect to the neutral axis. This means that the further away from the neutral axis we go, greater bending stress we experience.

Zero Stress Components
Using indicial notation for $$i = 1,2,3$$ $$\sum_{j=1}^{3}{\frac{\delta \sigma _{ij}}{\delta x_{j}}}=0$$, the equations expanded out look like the following:
 $$\frac{\delta \sigma _{11}}{\delta x_{1}}+\frac{\delta \sigma _{12}}{\delta x_{2}}+\frac{\delta \sigma _{13}}{\delta x_{3}}=0 $$ $$\frac{\delta \sigma _{21}}{\delta x_{1}}+\frac{\delta \sigma _{22}}{\delta x_{2}}+\frac{\delta \sigma _{23}}{\delta x_{3}}=0 $$ $$\frac{\delta \sigma _{31}}{\delta x_{1}}+\frac{\delta \sigma _{32}}{\delta x_{2}}+\frac{\delta \sigma _{33}}{\delta x_{3}}=0 $$ Analyzing the forces for an infinitesimally small square, like the following:
 the sum of the forces, $$\sum{F}=0=-\sigma _{x}A+\sigma _{x+\delta x}A+f(x)\delta x$$. This equation can also be written as follows:
 $$\frac{\delta \sigma }{\delta x}+\frac{f(x)}{A}=0$$ after neglecting the higher order terms of its Taylor series expansion. Now, summing all the forces in the x-direction:
 $$\sum Fx=[-\sigma_{xx}(x,y,z)+\sigma_{xx}(x+dx),y,z)]dydz+[-\sigma_{xy}(x,y,z)+\sigma_{xy}(x,y+dy,z)]dxdz+[\sigma_{xz}(x,y,z)+\sigma_{xz}(x,y,z+dz)]dxdy=0$$ Summimg all the forces in the y-direction:
 $$\sum Fy=[-\sigma_{xy}(x,y,z)+\sigma_{xy}(x+dx),y,z)]dydz+[-\sigma_{yy}(x,y,z)+\sigma_{yy}(x,y+dy,z)]dxdz+[\sigma_{yz}(x,y,z)+\sigma_{yz}(x,y,z+dz)]dxdy=0$$ Summimg all the forces in the z-direction:
 $$\sum Fz=[-\sigma_{xz}(x,y,z)+\sigma_{xz}(x+dx),y,z)]dydz+[-\sigma_{yz}(x,y,z)+\sigma_{yz}(x,y+dy,z)]dxdz+[\sigma_{zz}(x,y,z)+\sigma_{zz}(x,y,z+dz)]dxdy=0$$ The sum of the forces in each direction are also written as:
 $$(\frac{\delta \sigma_{xx}}{\delta_x}+\frac{\delta \sigma_{xy}}{\delta_y}+\frac{\delta \sigma_{xz}}{\delta_z})dxdydz=0$$
 $$(\frac{\delta \sigma_{xy}}{\delta_x}+\frac{\delta \sigma_{yy}}{\delta_y}+\frac{\delta \sigma_{yz}}{\delta_z})dxdydz=0$$
 $$(\frac{\delta \sigma_{xz}}{\delta_x}+\frac{\delta \sigma_{yz}}{\delta_y}+\frac{\delta \sigma_{zz}}{\delta_z})dxdydz=0$$ respectively.

Equilibrium Equations for a non-symmetrical beamshape

The equilibrium equations are broken down into the x, y, and z directions of forces (i.e. ΣFx, ΣFy, and ΣFz respectively). In the x-direction, the ultimate goal equilibrium equation is:
$$\frac{\delta \sigma_{yx}}{\delta_y}+\frac{\delta \sigma_{zx}}{\delta_z}=0$$

Changing these terms into their indicial notation brings about $$\frac{\delta \sigma_{21}}{\delta x_{2}}+\frac{\delta \sigma_{31}}{\delta x_{3}}=0$$

Recalling that $$\frac{\delta \sigma_{xx}}{\delta_x}+\frac{\delta \sigma_{yx}}{\delta_y}+\frac{\delta \sigma_{zx}}{\delta_z}=0$$ we can also write the similarity equations for ΣFy, and ΣFz as
$$\frac{\delta \sigma_{xy}}{\delta_x}+\frac{\delta \sigma_{yy}}{\delta_y}+\frac{\delta \sigma_{zy}}{\delta_z}=0$$ and $$\frac{\delta \sigma_{xz}}{\delta_x}+\frac{\delta \sigma_{yz}}{\delta_y}+\frac{\delta \sigma_{zz}}{\delta_z}=0$$ respectively. The conversion to indicial notation for each would be as follows:

$$\frac{\delta \sigma_{12}}{\delta y_{2}}+\frac{\delta \sigma_{22}}{\delta y_{2}}+\frac{\delta \sigma_{32}}{\delta y_{2}} =0$$ and $$\frac{\delta \sigma_{13}}{\delta z_{2}}+\frac{\delta \sigma_{23}}{\delta z_{2}}+\frac{\delta \sigma_{33}}{\delta z_{2}} =0$$ respectively.

HW5 MatLAB Problem Statement
In this assignment we will modify the MatLAB code from the previous homework to include stringers attaching the spars to the flange, as can be seen in the figure below.

Given an input NACA airfoil (NACA 2415 shown in the above figure), we are to compute the centroid of the stringers (as well as the airfoil), the components of the moment of inertia tensor, the highest normal bending stress and location of this stress, and the ultimate bending moments of the airfoil.

We first compute this data neglecting the bending effects of the skin and the partition walls (spar webs). We are then to recompute the data, only this time NOT neglecting the bending effects of the skin or partition walls (or the stringers themselves). A comparison of the difference will then be made so we can access the contribution of these bending effects.

In order to make these calculations, we need to know some initial data on the stringers to be used. We first make an assumption on the bending moments, shown below.

$$\ M_{y}=-125,000 N*cm $$

$$\ M_{x}=50,000 N*cm $$

We also need to know the cross sectional area of these stringers, which is given below.

$$\ A_{B}=2 cm^{2} $$

$$\ A_{E}=2 cm^{2} $$

$$\ A_{H}=1 cm^{2} $$

$$\ A_{F}=1 cm^{2} $$

For the second part of our calculations, we are to take into account the bending affects of the skin, partition walls, and stringers themselves. We can note that the stringers are essentially stamped out sheet metal which is then bent into the L-shape as seen in the first figure. Because of this, thickness of the stringers are uniform, and we let this thickness equal 5 mm, or 0.005 m. Knowing this, we can compute the width of the metal sheet used to manufacture the stringers, as seen below.

$$\ \frac{\frac{0.0002}{2}}{0.5}=0.02m=2cm $$

Because the thin skin would buckle under high normal bending stress acting on the plane of the skin, we will next find the highest normal bending stress in any of the stringers, and take a closer look at that stringer. Assuming the ratio of My and Mz are constant, we will then set this maximum normal bending stress to the ultimate stress for our material, steel 300M. Upon doing this, we are now able to compute the ultimate bending moments My and Mz for our full blown airfoil (which we will be imputing as the NACA 2415 airfoil).

The output from the program can be seen below. The dashed line is the neutral axis. $$\bar{M_{y,ult}}=2.7368*10^4 Nm$$ $$\bar{M_{z,ult}}=-1.0947*10^4 Nm$$ For part 2 we take into account the spars and the skin in the stress analysis. Inherently we would assume a larger moments of inertia, slightly different centroid, and a larger maximum bending moment for steel. $$\bar{M_{y,ult}}=9.7364*10^4 Nm$$ $$\bar{M_{z,ult}}=-3.8946*10^4 Nm$$ As expected, the moments of inertia increased. The point of maximum stress remained the same and the ultimate bending moments more than tripled.

Contributions
Jin Yu Guan Eas4200c.f08.aeris.guan 05:30, 7 November 2008 (UTC) Nelson Caceres Eas4200c.f08.aeris.caceres 05:40, 7 November 2008 (UTC) Anett Krammer Eas4200c.f08.aeris.krammer 19:27, 7 November 2008 (UTC) Josh Holladay Eas4200c.f08.aeris.holladay 21:33, 7 November 2008 (UTC) Jesse McRae Eas4200c.f08.aeris.mcrae 21:28, 7 November 2008 (UTC)

title: User:Eas4200c.f08.aeris.krammer/HW5