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User:Eas4200c.f08.vqcrew.c/Homework 3

=Homework 3=

Curved Panels
Figure 1 shows a curved panel with uniform thickness $$t$$ and constant shear stress $$\tau $$. This diagram will be analyzed in order to determine the forces acting on the panel in both the y and z directions. The shear flow $$q$$ is defined as: $$q=t \tau $$

In order to find the total forces acting in the y and z direction, the panel must be divided up into small sections with differential lengths and force vectors. A side view of the panel with dimensions and the differential components is shown in Figure 2. The differential force element, $$d\bar{F}$$, is equal to the shear flow multiplied by the differential length of the element, or: $$d\bar{F}=qd\bar{L}$$

A larger view of the differential length and force element are shown in Figure 3. Figure 3 also shows the components of the differential length element. The differential force vector and length vector can be broken into y and z components and simplified as follows:

$$d\bar{F}=qd\bar{L}$$ $$d\bar{F}=q(dL_{y} \hat{j} + dL_{z} \hat{k})$$ $$d\bar{F}=q(dL cos\theta \hat{j} + dL sin \theta  \hat{k})$$ $$d\bar{F}=q(dy \hat{j} + dz  \hat{k})$$

By integrating along the path from A to B, we can find the total force acting on the curved panel.

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

The force vector can be broken into y and z components and equated to the above solution to produce equations (1.4) and (1.5): $$\bar{F}=F_{y} \hat{j} + F_{z} \hat{k}$$ $$\Rightarrow F_{y} = qa = t \tau a$$ $$\Rightarrow F_{z} = qb = t \tau b$$

The relationship between the force components in a curved panel can be expressed as: $$\frac{F_y}{F_z}=\frac{a}{b}$$

The magnitude of the force, $$\left|\left| \bar{F} \right| \right|$$, can be expressed as follows: $$\left|\left| \bar{F} \right| \right|=\sqrt{(F_y)^2+(F_z)^2} = q\sqrt{a^2+ b^2}$$

However, $$\sqrt{a^2+ b^2}$$ is equal to the straight line distance, $$d$$, from point A to B. Therefore the force vector can also be expressed as: $$\bar{F}=qd$$

Torsion in a Closed, Thin-walled Beam
Figure 4 depicts a generic, closed, thin-walled cross-section for a beam under torsion. The coordinate system may be defined from any arbitrary point in the plane of the section. The torque vector points in the positive x direction, or: $$\bar{T}=T \hat{i}$$ The differential unit of torque can be described as the cross product of the distance from the origin to the differential length segment with the differential force vector, or: $$d \bar{T}= \bar{r} \times d \bar{F}$$ The magnitude of this element is: $$dT=\rho dF$$ where $$\rho$$ is the perpendicular distance from the origin to the force vector and $$dF=qdL$$. To find the total torque, the differential torque element must be integrated about the closed path of the cross-section.

$$T=\oint{dT} = \oint{\rho dF}=q\oint{\rho dL}$$ However, $$\rho dL$$ is also equal to twice the area of the triangle swept out by the $$r$$ vector and $$dL$$. We can evaluate the integral over the average area, $$\bar{A}$$, to find the total torque as follows:

$$T=2q\int _{\bar{A}}{dA}=2q\bar{A}$$

Torsion Acting on a Uniform Bar with Circular Cross-Section
Figure 5 shows a generic case of torsion acting on a solid, circular beam.

This case does not exhibit warping. The only deformation that occurs is contained within the y-z plane. The beam behaves as a series of rigid disks rotating entirely in the plane of the cross-section. Figure 6 displays the reaction of the beam to a given torque.

The general formula for the torque is:

 * $$T = \iint_A r \tau dA $$

The shear stress is defined as the product of the shear modulus, G, and the angle of twist,  $$\gamma$$  , (Hooke's Law).

 * $$\tau = G \gamma  $$

 * $$\displaystyle \gamma = \displaystyle \frac{rd\alpha}{dx} = r\theta$$

Define $$\theta$$ as rate of twist:

 * $$\theta := \frac{d\alpha}{dx}$$

Substituting in, the torque is found to be:

 * $$\displaystyle T = \int_A rG(r\theta)dA = G\theta(\int_A r^{2} dA)$$

Where:   $$\displaystyle dA = dydz = rdrd\theta $$

The $$2^{nd}$$ polar moment of inertia for the beam is equal to the second term in the expression of the torque, $$\displaystyle J = (\int_A r^{2} dA)$$.

Thin-Walled Hollow Circular Cross-Section
Figure 7 shows two circular cross-sections, one solid and the other thin-walled, with the same area. The two cross-sections have different $$2^{nd}$$ polar moments of inertia.

For the thin-walled cross-section,

Inner radius: $$\displaystyle r_i = a$$

Outer radius: $$\displaystyle r_o = b$$

The $$2^{nd}$$ polar moment of inertia for the thin-walled cross-section can be determined through the expression:

 * $$\displaystyle J = \frac{1}{2}\pi(b^{4}-a^{4}) = \frac{1}{2}\pi(b - a)(b + a)(b^{2} + a^{2})$$

Define $$\displaystyle \bar{r}$$ as the average radius:

 * $$\displaystyle \bar{r} = \frac{a+b}{2}$$

Define $$\displaystyle t$$ as the thickness:

 * $$\displaystyle t = (a - b)$$

Substitute into the $$2^{nd}$$ polar moment of inertia equation to get:

 * $$\displaystyle J = \frac{1}{2}\pi(t)(2\bar{r})(b^{2} + a^{2}) = \pi(t)(\bar{r})(b^{2} + a^{2})$$

Approximate and substitute:

$$\displaystyle a^{2} \cong \bar{r}^{2}$$

$$\displaystyle b^{2} \cong \bar{r}^{2}$$

 * $$\displaystyle \Rightarrow J = \pi t \bar{r}(\bar{r}^{2} + \bar{r}^{2})= 2\pi t \bar{r}^{3}$$

This result can be separated to allow for a proportional factor to be found between the $$2^{nd}$$ polar moment of inertia and the the average area, $$\displaystyle \bar{A}$$.

$$\displaystyle \bar{A} = \pi\bar{r}^{2}$$

 * $$\displaystyle \Rightarrow J = (2\pi^{-\frac{1}{2}}t)(\pi\bar{r}^{2})^{\frac{3}{2}} = (2\pi^{-\frac{1}{2}}t)(\bar{A})^{\frac{3}{2}}$$

In this form, it is clear that $$\displaystyle J$$ is proprtional to $$\displaystyle \bar{A}^{\frac{3}{2}}$$ by a factor of $$\displaystyle (2\pi^{-\frac{1}{2}}t)$$.

Torsion of a Uniform Non-Circular Bar
A uniform non-circular bar leads to warping of the cross-section. Warping is defined as the axial displacement along the x-axis (i.e. along bar length) of a point on the deformed rotated cross-section.

$$U_y =$$ y component of displacement vector $$\vec{PP}'$$

$$U_z =$$ z component of displacement vector $$\vec{PP}'$$

Important Definition : $$\displaystyle \theta = \frac{\displaystyle \alpha}{x}$$       =   Rate of Twist

Referring to Figure 8, assuming that the cross-section at x=0 remains stationary, and if the rotation angle $$\displaystyle \alpha$$ is small, then the displacement components at P are given by the following:

$$u = -r \displaystyle \alpha sin\displaystyle \beta = -\alpha y$$, which can be simplified to Equation (3.11):

(1) $$U_y = - \displaystyle \theta x z$$

Now the we have the displacement in the y direction, we can solve for the displacement $$U_z$$ which yields:

$$\displaystyle U_z = + PP' cos \beta = + \alpha y_p$$, this can be simplified to Equation (3.12):

(2) $$U_z = \displaystyle \theta x y$$

The warping displacement in the x-axis direction can now be found as a function of the y and z directions.

(3) $$U_x = \displaystyle \theta \displaystyle \psi (y, z)$$

Based on these three equations come the Kinematic Assumptions.

Road Map for Torsional Analysis of Aircraft Wing:
A) Kinematic Assumption

B) Strain Displacement

C) Equilibrium Equation (stresses)

D) Prandtl Stress Function $$\displaystyle \phi$$

E) Strain Compatibility Equation

F) Equation for $$\displaystyle \phi$$

G) Boundary Conditions for $$\displaystyle \phi$$

H) $$T = 2 \iint_A \displaystyle \phi \,dA$$

$$T = G J \displaystyle \theta$$

$$ J =\frac{-4}{\nabla^2 \phi}\iint_A \displaystyle \phi \,dA$$

I) Thin-Walled Cross-Section, $$T = 2 q A $$ (3.48) This equation was also derived in Problem 1.1.

J) Twist Angle $\displaystyle \theta$ Method 1.

$$\displaystyle \theta = \frac{1}{2 G A} \oint \frac{q}{t}\,ds$$

Proving the Area of a Triangle is $$ \frac{b\ h}{2}$$
In the lecture on the 14th lecture on the 26th of September, 2008 the professor mentioned how understanding the fundamentals is important and assigned the homework problem of proving the area of a triangle.

To begin this proof, it will be assumed that the area of a parallelogram $$A_p$$ is understood and given to be

 * $$A_p = b\ h$$

To begin, if it is stated that

 * $$\bar {AB} \parallel \bar {CD}$$
 * $$\bar {BC} \parallel \bar {AD}$$

It is understood that the given shape is a parallelogram with an area given above. Since it is a parallelogram, it is understood that

 * $$\bar {AB} \cong \bar {CD}$$
 * $$\bar {BC} \cong \bar {AD}$$

And since these sides are congruent, by using the side angle side theorem of geometry (SAS) it can be understood that

 * $$\angle ADB \cong \angle CBD$$
 * $$\angle ABD \cong \angle BCD$$

Hence by the angle side angle theorem of geometry (ASA) it is understood that

 * $$\triangle ABD \cong \triangle BCD$$

Since the parallelogram is made up of these two congruent triangles, it is understood that

 * $$A_{t} = \frac{b\ h}{2}$$

It can be noted that any triangle given can be extended into one or two parallelograms, dependent on the triangle, and superimposed to have the result that was proven above.

QED

Comparison of the Polar Moment of Inertia of a Solid Circle and a Thin walled circle
The polar moment of inertia is used to quantify the ability of an object to resist torsion. It can be calculated by the definition

 * $$J_z = \iint_A \rho^2\ dA$$

Where $$\rho$$ is the radial distance from the z axis to the element $$dA$$. $$dA$$ is an infinitesimal area element. Converting $$dA$$ to polar coordinates which was done in ***homework 2*** the term $$dA$$ expands to

 * $$dA = \rho\ d\rho\ d\theta$$

Solid Circle Proof
When substituting the above expressions, the polar moment of inertia of a solid circle becomes:

 * $$J_z = \int_{0}^{2\pi} \int_{0}^{R} \rho^3\ d\rho\ d\theta$$

and after integrating results with

 * $$J_z = \frac{\pi\ R^4}{2}$$

Where $$R$$ is the radius of the circle.

Thin walled Circle Proof
As is possible with the second area moment of inertia, its possible to add and subtract the polar moment of inertia from others which result in the desired shape. Since the polar moment of inertia of a solid cross section has already been proven, it would be prudent to use this to our advantage by subtracting the polar moment of inertia of the inner "hollow" section from the polar moment of inertia of the outer shelled section.

 * $$J_{z,out} = \frac{\pi\ R_o^4}{2}$$
 * $$J_{z,in} = \frac{\pi\ R_i^4}{2}$$
 * $$J_z = J_{z,out} - J_{z,in} = \frac{\pi}{2}\ (R_o^4 - R_i^4)$$

Now if we make the assumption that $$t \ll R_o$$. Noting that

 * $$R_o \approx R_i + t$$

It makes it possible to define a new variable $$\bar r$$ which is defined as

 * $$\bar r := \frac{R_o + R_i}{2} = \frac{2\ R_o - t}{2} = \frac{2\ R_i + t}{2}$$

Now assuming that $$t \ll R_o$$ it can be approximated that

 * $$\bar r^2 = (R_o - \frac{t}{2})^2 = R_o^2 - R_o\ t + \frac{t^2}{4} \approx R_o$$
 * $$\bar r^2 = (R_i + \frac{t}{2})^2 = R_i^2 + R_i\ t + \frac{t^2}{4} \approx R_i$$

To the previous equation of the polar moment of inertia

 * $$J_z = \frac{\pi}{2}\ (R_o^4 - R_i^4) = \frac{\pi}{2}\ ((R_o^2)^2 - (R_i^2)^2)$$

can be expanded to

 * $$J_z = \frac{\pi}{2}\ (R_o - R_i)\ (R_o + R_i)\ (R_o^2 + R_i^2)$$

noting that with the approximations it can be given that

 * $$(R_o - R_i) = t$$
 * $$(R_o + R_i) = 2\ \bar r$$
 * $$(R_o^2 + R_i^2) = 2\ \bar r^2$$

and substituting these approximations

 * $$J_z = \frac{\pi}{2}\ (t)\ (2\bar r)\ (2\bar r^2) = 2\ \pi\ t\ \bar r^3$$

Comparing Solid versus thin walled
As shown above, the polar moment of inertia of a solid circle is

 * $$J_{z,solid} = \frac{\pi\ R^4}{2}$$

and for a thin walled is

 * $$J_{z,thin} = 2\ \pi\ t\ \bar r^3$$

When analyzing these equations it can be shown that most of the torsional resistance is a result of the outer material rather than the inner material. This means that the material at the center of a solid circle is essentially wasted in comparison to the outer material. If the same amount of material were used, the thin walled geometry gives a lot more bang for its buck.

In applications such as aerospace structures where weight is an extremely large factor, less material is always better. This would result in the use of thin walled circular cross-section for use in structural integrity throughout the aircraft.

Problem 1.1 Applied to General Single Cell Section
Commonly in aerospace applications it may be necessary to determine the characteristics of a hollow, thin-walled cross-section similar in shape to that of an airfoil. In this case, the cross-section can be subdivided into smaller parts, namely, one triangular and one semi-circular cross-section. Each respective $$\bar {A} $$ can be calculated and summed in order to determine the total $$\bar {A} $$ of the section as follows:

 * $$ \bar {A_{circ}} = \frac{\pi}{2}(\frac{b}{2})^2 $$ and  $$ \bar {A_{tri}} = \frac{b a}{2} $$.  Therefore,

 * $$ \bar {A_{total}} = \frac{\pi}{2}(\frac{b}{2})^2 + \frac{b a}{2} = \frac{\pi}{2}(\frac{2}{2})^2 + \frac{2(4)}{2} = \frac{\pi}{2} + 4 = 5.5708 m^2$$

The torque generated by a single cell is: $$ T_i = 2q_i \bar {A_i} $$

while the total torque is: $$ T = \sum_{i=1}^{n_c} T_i = \sum_{i=1}^{n_c} 2q_i \bar{A_i} $$ where $$ n_c $$ is the total number of cells.

Also, if the value of shear flow is desired, the above equations could be rearranged to yield:

 * $$ shear flow = q = \frac{T}{2 \bar {A}} $$

Additionally, the twist angle could be calculated as:

 * $$ twist angle = \theta = \frac{1}{2 G \bar {A}} \sum_{j=1}^{n_c} \frac{q_j l_j}{t_j} $$

where G is the shear modulus and $$ n_c $$ is the number of cells.

Quadrature
The concept of quadrature (also known as squaring the circle) has its roots in ancient geometry. The problem consists of constructing a square from a circle of equal area using only a compass and straightedge and accomplishing this task in only a finite number of steps. The problem dates back to the 19th century BCE and has been attempted by numerous widely-known mathematicians. Although many approximations of the solution exist, it has been proven that a solution to the problem is impossible. It was proven in the early 19th century CE that if $$ \pi $$ were a transcendental, then a solution to the problem does not exist. In 1882, Ferdinand von Lindemann proved in his Lindemann-Weierstrass Theorem that $$ \pi $$ is, in fact, a transcendental, therefore, proving that squaring the circle is impossible.

MATLAB Question
The task for the MATLAB portion of the HW report involves plotting an arbitrary NACA airfoil, finding the average area of the airfoil, and locating the airfoil's area centroid.

Using the following input (in accordance with the syntax presented in the code header), the code generates Figure 17, a graph of the NACA 2415 airfoil with a cross-hair at the centroid. The output references "chord-units", as the code was purposely generated to handle any unit system for chord length. For this particular case, the chord length is set at .5 m.

 abar(1,.5,2,4,15,1000,[0;0])

The average area of the airfoil is 0.025 square chord units

The centroid is located at (0.210,0.008) chord units

= Contributing Team Members = The following students contributed to this report:

John Saxon   Eas4200c.f08.vqcrew.c 20:40, 8 October 2008 (UTC) Javier Stober Eas4200c.f08.vqcrew.b 14:25, 8 October 2008 (UTC) Darin Toscano Eas4200c.f08.vqcrew.d 17:58, 8 October 2008 (UTC) Kevin Klauk  Eas4200c.f08.vqcrew.E 19:34, 8 October 2008 (UTC) Steven Hepsworth Eas4200c.f08.vqcrew.f 17:57, 8 October 2008 (UTC) Adam Edstrand Eas4200c.f08.vqcrew.a 20:20, 8 October 2008 (UTC)

=References=

title: User:Eas4200c.f08.vqcrew.c/Homework 3


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User:Eml 4500.f08.delta 6.specht/HW3

Determination of Element FD (wrt global coordinates)
We have the FD for springs:

k$$^{(e)}$$ d$$^{(e)}$$ = f$$^{(e)}$$   (1) Where each represents a 4x4, 4x1, and 4x1 matrix, respectively.

This is represented by the following diagram.

Also, we can treat this element as a spring element turned 45° with the following analysis.

The reaction forces have been taken along the direction of the element so that we can analyze the stretching or shrinkage of the bar element as a force is applied on it. The p vectors translate to be the force on the node and the q vectors are the displacement of the node as the bar stretches or shrinks. These pertain to the FD relationship mentioned above as:

$$\hat{\mathbf{k}}^{(e)} \mathbf{\hat{q}}^{(e)} = \hat{\mathbf{q}}^{(e)}$$ (2)

In long-hand form:

$$k^{(e)} \begin{bmatrix} 1 & -1 \\ -1 & 1\\ \end{bmatrix} \begin{pmatrix} q^{(e)}_{1}\\ q^{(e)}_{2} \end{pmatrix} = \begin{pmatrix} p^{(e)}_{1}\\ p^{(e)}_{2} \end{pmatrix}$$

Where $$q^{(e)}_i$$ is the axial displacement of element e at local node i, and $$p^{(e)}_i$$ is the axial force of element e at local node i. Also, these matrices are sized as 2x2, 2x1, and 2x1, respectively.

Derivation of eqn (1) from eqn (2)
We want to find the relationship between $$q^{(e)}_{2x1}$$ and $$d^{(e)}_{4x1}$$, and $$p^{(e)}_{2x1}$$ and $$f^{(e)}_{4x1}$$. It can be expressed as follows:

 * $$q^{(e)}_{2x1}$$ = $$T^{(e)}_{2x4}$$ $$d^{(e)}_{4x1}$$

Consider the displacement of node 1, denoted as $$\vec{d}^{(e)}$$.

 * $$\vec{d}^{(e)} = d^{e}_{1} \vec{i} + d^{e}_{2} \vec{j}$$

We say that $$q^{(e)}_{1}$$, the axial displacement of local node 1, is the orthogonal projection of the displacement. It is defined as:

$$q^{(e)}_{1}$$ = $$\vec{d}^{(e)} \vec{\tilde{i}}$$
 * =($$d^{e}_{1} \vec{i}$$ + $$d^{e}_{2} \vec{j}$$)•$$\vec{\tilde{i}}$$
 * =$$d^{e}_{1}$$ ($$\vec{i}$$•$$\vec{\tilde{i}}$$) + $$d^{e}_{2}$$ ($$\vec{j}$$•$$\vec{\tilde{i}}$$)

where
 * $$\vec{i}$$•$$\vec{\tilde{i}}$$ = cosθ^{(e)} = l$$^{(e)}$$
 * $$\vec{j}$$•$$\vec{\tilde{i}}$$ = sinθ^{(e)} = m$$^{(e)}$$

Which leads to the construction of the matrix equation that relates $$d$$ and $$q$$:

$$q^{(e)}_{1}$$ = l$$^{(e)}$$ $$d^{e}_{1}$$ + m$$^{(e)}$$ $$d^{e}_{2}$$
 * = $$\begin{bmatrix}

l^{(e)} & m^{(e)}\\ \end{bmatrix}_{1x2}$$ $$\begin{pmatrix} d^{e}_{1}\\ d^{e}_{2}\\ \end{pmatrix}_{2x1}$$

Similarily for node 2:

$$q^{(e)}_{2}$$ = l$$^{(e)}$$ $$d^{e}_{2}$$ + m$$^{(e)}$$ $$d^{e}_{2}$$
 * = $$\begin{bmatrix}

l^{(e)} & m^{(e)}\\ \end{bmatrix}_{1x2}$$ $$\begin{pmatrix} d^{e}_{3}\\ d^{e}_{4}\\ \end{pmatrix}_{2x1}$$

These two matrix equations combine to yield:

 * $$\begin{pmatrix}

q^{e}_{1}\\ q^{e}_{2}\\ \end{pmatrix}_{2x1}$$ = $$\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ \end{bmatrix}_{2x4} \begin{pmatrix} d^{e}_{1}\\ d^{e}_{2}\\ d^{e}_{3}\\ d^{e}_{4}\\ \end{pmatrix}_{4x1}$$

and in short-hand form:

 * $$q^{(e)}$$ = $$T^{(e)}$$ $$d^{(e)}$$

Similarly: $$\begin{Bmatrix}P_{1}^{(e)}\\P_{2}^{(e)}\end{Bmatrix}_{2x1}= T^{(e)}\begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \\ f_{3}^{(e)} \\ f_{4}^{(e)} \end{Bmatrix}_{4x4} => P^{(e)}=T^{(e)}*F^{(e)}$$ $$\hat{K}^{(e)}_{2x2}*q^{(e)}_{2x1} = P^{(e)}_{2x1}$$ where $$ q^{(e)}_{2x1}=T^{(e)}*d^{(e)}$$ and $$ P^{(e)}_{2x1}= T^{(e)}*F^{(e)} $$ therefore $$\Rightarrow \hat{K}^{(e)}_{2x2}*T^{(e)}*d^{(e)}= T^{(e)}*F^{(e)}$$

The goal with these matrices is to obtain $$K^{(e)}*d^{(e)}=F^{(e)}$$ in order to do that we need to move $$T^{(e)}$$ from the right hand side to the left hand side and this usually can be done by multiplying by the inverse $$T^{(e)-1}$$ but since $$T^{(e)}$$ is a rectangular matrix with (2x4) dimensions it can not be inverted therefore what needs to be done is the transpose $$T^{(e)^(T)}$$

$$[T^{(e)^(T)}_{4x2}*\hat{K}^{(e)}_{2x2}*T^{(e)}_{2x4}]$$$$ * d^{(e)}_{4x1}$$ = $$F^{(e)}_{4x1}$$ where $$K^{(e)}_{4x4}=[T^{(e)^(T)}_{4x2}*\hat{K}^{(e)}_{2x2}*T^{(e)}_{2x4}]\Rightarrow $$ $$K^{(e)}*d^{(e)}=F^{(e)}\Rightarrow K^{(e)}= T^{(e)^(T)} * \hat{K}^{(e)} * T^{(e)}$$

By the principle of virtual work (PVW) the reduction of the global force displacement relation is $$\Rightarrow$$

$$K_{6x6}*d_{6x1}=F_{6x1}\Rightarrow K_{2x2}*d_{2x1}=F_{2x1}$$

Then why not solve these matrices at this stage as follows $$d=K^{-1}*F$$ ? The reason why this can not be done is due to the matrix singularity of $$ K $$ Since the determinant of matrix $$ K $$ is equal to zero its inverse will be $$ \frac{1}{det [K]} $$ which will be undefined since the denominator is equal to zero.

The Eigenvalue and Stiffness Matrix K Relationship
$$K= \ \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}_{6X6}\,$$ $$\ \ = \begin{bmatrix}  \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & (\frac{9}{16} + \frac{5}{2}) & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & (\frac{3}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} \\ 0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$

For an unconstrained structure system there are 3 possible rigid body motions in 2-D (2 in translation plus 1 in rotation) The eigenvalues of the stiffness matrix are really important for design because they are related to the vibrations of the structure. A zero eigenvalue is related to a rigid body movement since it lacks the strain energy according to the equation $$K*v=\lambda *M*v $$ where $$K$$ is the stiffness matrix, λ is the eigenvalue related to vibrational frequency, and $$M$$ is the mass matrix. A zero eigenvalue means there is no elastic energy therefore there is no energy stored in the structure.

Closing the Loop Between FEA and Statics
Use global force displacement relation. $$ \begin{bmatrix} K_{13} & K_{14}\\ K_{23} & K_{24}\\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \textbf{F} $$ This matrix is a 6x2 multiplied by a 2x1 matrix and equals a 6x1 matrix. The forces $$F_{3}$$ and $$F_{4}$$are the known applied loads. Therefore, there is no need to compute them. The computations that need to be done are for rows 1,2,5 and 6. This will give The reactions $$F_{1}$$, $$F_{2}$$, $$F_{5}$$ and $$F_{6}$$. Closing the loop between The finite element method (FEM) and statics using Virtual Displacement Solving the two bar truss system uses a process shown in the figure below. Where side 1 is the finite element method and side two is the statics approach. The statics approach is used to confirm the finite element method and is known as closing the loop. The process of determinig the displacements using statics was discussed above in "Determination of Element FD (wrt global Coordinates). In the statics process the reactions are known. Therefore, the member forces $$P^{(1)}_{1}$$ and $$P^{(2)}_{2}$$ are known. To compute the axial displacement dof's ($$q$$) or the amount of extension of the bars the member forces and the element stiffiness's need to be used. The following formulas show the relationship between the member forces, element stiffness's and the axial displacements. $$ P=Kq$$ so, $$ q=P/K$$ $$q^{(1)}_{2}= \frac{P^{(1)}_{1}}{K^{(1)}}= \frac{P^{(1)}_{2}}{K^{(1)}}= AC$$ $$q^{(2)}_{1}= \frac{-P^{(2)}_{2}}{K^{(2)}}= \frac{-P^{(2)}_{2}}{K^{(2)}}= -AB$$ The axial displacements dof's at node 1 and 3 are equal to zero.  $$q^{(1)}_{2}=q^{(2)}_{2}= 0$$  Question: How does one back out from the displacement dof's of node 2?  Where: A = node 2, AC is the extension of 1A and AB is the extension of 3A. First an extensions must be drawn perpendicular to AC and AB. Then a line can be drawn between the point D, where the two extensions meet, and node 1. A line is also drawn between node 3 and point D. These two lines are the dotted line and denote the deformed shape of the two-bar truss system. The vector R is the displacement of A or node 2. It is found by taking the square root of the sum of the squared magnitudes of AC and AB. $$R= (AC^{2} + AB^{2})^{\frac{1}{2}}$$ Review of the small angle approximation:

Where A is the angle between the two bars. $$U_{y}= Rsin(A)$$ $$U_{x}= R(1-cos(A))$$ If the angle is small then $$sin(A)$$ is approximately equal to A and $$cos(A)$$ is approximately equal to 1. The above equations become: $$U_{y}= R*A$$ $$U_{x}= 0$$ The image above presents the two bar truss system that is being analyzed. The vector AD is the displacement vector of the whole system, which has an unknown magnitude. The length of the vectors AB and AC can be determined using the method shown below. The computation shows the length of the component AC. The magnitude of AB is determined in the homework section of this report. $$ AC = \begin{vmatrix} \frac{P^{(1)}_{2}}{k^{(1)}} \end{vmatrix}=\frac{5.1243}{\frac{3}{4}}=6.8234 $$ $$ AB=\begin{vmatrix} \frac{P_{2}^{1}}{k^{2}} \end{vmatrix} $$ The x and y coordinates of points B, C, and D are denoted as: $$ \begin{pmatrix} x_{B}, y_{B} \end{pmatrix}    , \begin{pmatrix} x_{C}, y_{C} \end{pmatrix}    and \begin{pmatrix} x_{D}, y_{D} \end{pmatrix} $$ The values for these points are computed in the homework. The slopes of the lines BD and CD are given below. $$ Slope of CD = \theta ^{(1)}+ 90^{\circ} $$ $$ Slope of BC = \theta ^{(2)}- 90^{\circ} $$ If the magnitudes of the vectors BD and CD are known, the point D can be found using the equations given below. Points B and C must first be found.

$$ \vec{PQ}=(PQ)\vec{\tilde{i}}=PQ(cos\vec{i}+sin\vec{j}) $$ $$ \vec{PQ}=(x-x_{P})\vec{i}+(y-y_{P})\vec{j} $$ $$ x-x_{P}=(PQ)cos\theta $$ $$ y-y_{P}=(PQ)sin\theta $$ $$ \frac{y-y_{P}}{x-x_{P}}=tan\theta $$ $$ y-y_{P}=(tan\theta )(x-x_{P}) $$ Once the points B and C have been found, the equation shown below can be used to find the point D. The equation below is the equation of a line, passing through a point P.  If the values for B and C are input into these equations, the point D can be found geometrically. $$ y-y_{P}=(tan\theta + \frac{\pi }{2})(x-x_{P}) $$ The vector AD can be expressed as shown below, where the x and y values of the point A are zero, since the origin is defined at A.  $$ \vec{AD}= (x_{D}-x_{A})\vec{i}+(y_{D}-y_{A})\vec{j} $$ The vector AD can also be expressed in terms of the global displacement vectors, as shown in the diagram earlier (d3 and d4). $$ \vec{AD}=d_{3}\vec{i}+d_{4}\vec{j} $$

Three Bar Truss System
The three bar truss system is shown in the image below. The characteristics of each element is shown in the table that follows the image.

Convenient methods of numbering the local nodes are shown in the images below for each element. Element 1

Element 2

Element 3

$$ \sum{F_{x}}=0 $$ $$ \sum{F_{y}}=0 $$ $$ \sum{M_{A}}=0 $$

Question: What if we take the moment about a point B?

$$ \sum{M_{B}} = \vec{BA}\; X\; \vec{ F} = \vec{BA'}\; X\; \vec{ F} $$ The above equation stands only when A' lies on the line of action. We can derive this moment by defining the BA' vector, then substituting into the moment equation.

$$ \vec{BA'} = \vec{BA}+\vec{AA'} $$

$$ \vec{M_B} = {(\vec{BA}+\vec{AA'})}\; X\; \vec{F}= \vec{BA}\; X\; \vec{F} + \vec{AA'}\; X\; \vec{F} $$

where $$ \vec{AA'}\; X\; \vec{F} = 0 $$

The derivation of the moment about point B is quite simple:
 * NOTE

$$ \sum{\vec{F}_{i}} = 0 $$ $$ \sum{\vec{M}_{B_{i}}} = \sum{\vec{BA'}_{i}}\; X\; \vec{F}_{i} = \sum{\vec{BA}_{i}}\; X\; \sum{\vec{F}_{i}} $$ where $$ \sum{\vec{F}_{i}} = 0 $$ Therefore, the moment about point B is also equal to zero.

The 8x8 global stiffness matrix is found below, where the local stiffness matrices for each element correspond to the color-coded areas. The k values for the local stiffness matrix for element 3 can be found as shown below. Only two examples are shown, yet the values can be found under the homework section of this page. $$ k_{33} = k^{(1)}_{33} + k^{(2)}_{11} + k^{(3)}_{11} $$ $$ k_{34} = k^{(1)}_{34} + k^{(2)}_{12} + k^{(3)}_{12} $$

Determination of Point Coordinates
The following are the unknown variables that will be determined:

(XD, YD) (XB, YB) (XC, YC)

We will begin by finding the coordinates of point D. Note: Point A has been designated as the origin.

Using the definition of line AD, (XD, YD) can be found.

$$\vec{AD} = (X_{D} - X_{A}) \vec{i} + (Y_{D} - Y_{A}) \vec{j}$$

Since point A is referenced as the origin, XA and YA are equal to zero.

By definition of the displacement vector of A,

$$\vec{AD} = d_{3}\vec{i} + d_{4}\vec{j}$$

Thus,

(XD, YD) = (d3, d4) = (4.352, 6.1271)

Next, using the relationship of two points on the same line gives,

$$y - y _{C} = tan(\theta) (x - x_{C})$$

where θ = 30°, (XC, YC) = (5.92, 3.42)

The relationship between point B and point D gives,

$$y - y _{B} = tan(\theta + \frac{\pi }{2}) (x - x_{B})$$

where θ = 135°,(XB, YB) = (-0.88, 0.88)

Results:

Point D: (4.35, 6.13) Point C: (5.92, 3.42) Point B: (-0.88, 0.88)

Global K Matrix for Three Bar System
The following variables are known: E(1) = 2 A(1) = 3 L(1) = 5 θ(1) = 30° E(2) = 4 A(2) = 1 L(2) = 5 θ(2) = -30° E(3) = 3 A(3) = 2 L(3) = 10 θ(3) = -135° The following variables can be determined:

$$k_{(1)} = \frac{E^{(1)}A^{(1)}}{L^{(1)}} = \frac{6}{5} = 1.2$$ $$k_{(2)} = \frac{E^{(2)}A^{(2)}}{L^{(2)}} = \frac{4}{5} = 0.8$$ $$k_{(3)} = \frac{E^{(3)}A^{(3)}}{L^{(3)}} = \frac{6}{10} = 0.6$$

$$l^{(1)} = cos\theta ^{(1)} = 0.866$$ $$m^{(1)} = sin\theta ^{(1)} = 0.5$$ $$l^{(2)} = cos\theta ^{(2)} = 0.866$$ $$m^{(2)} = sin\theta ^{(2)} = -0.5$$ $$l^{(3)} = cos\theta ^{(3)} = -0.707$$ $$m^{(3)} = sin\theta ^{(3)} = -0.707$$ Note: l(3) = m(3) The following are the element stiffness matrices, which are needed to compute the global stiffness matrix.

$$k^{(e)} = k_{(e)}\begin{bmatrix} l^{(e)2} & l^{(e)}m^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & m^{(e)2} & -l^{(e)}m^{(e)} & -m^{(e)2} \\ -l^{(e)2} & -l^{(e)}m^{(e)} & l^{(e)2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -m^{(e)2} & l^{(e)}m^{(e)} & m^{(e)2} \end{bmatrix}$$ Thus,

$$k^{(1)} = \begin{bmatrix} 0.9 & 0.52 & -0.9 & -0.52\\ 0.52 & 0.3 & -0.52 & -0.3 \\ -0.9 & -0.52 & 0.9 & 0.52\\ -0.52 & -0.3 & 0.52 & 0.3 \end{bmatrix}$$

$$k^{(2)} = \begin{bmatrix} 0.6 & -0.346 & -0.6 & 0.346\\ -0.346 & 0.2 & 0.346 & -0.2 \\ -0.6 & 0.346 & 0.6 & -0.346\\ 0.346 & -0.2 & -0.346 & 0.2 \end{bmatrix}$$

$$k^{(3)} = \begin{bmatrix} 0.3 & 0.3 & -0.3 & -0.3\\ 0.3 & 0.3 & -0.3 & -0.3\\ -0.3 & -0.3 & 0.3 & 0.3\\ -0.3 & -0.3 & 0.3 & 0.3 \end{bmatrix}$$

Now the global stiffness matrix can be computed.

$$K = \begin{bmatrix} 0.9 & 0.52 & -0.9 & -0.52 & 0 & 0 & 0 & 0\\ 0.52 & 0.3 & -0.52 & -0.3 & 0 & 0 & 0 & 0\\ -0.9 & -0.52 & 1.8 & 0.474 & -0.6 & 0.346 & -0.3 & -0.3\\ -0.52 & -0.3 & 0.474 & 0.8 & 0.346 & -0.2 & -0.3 & -0.3\\ 0 & 0 & -0.6 & 0.346 & 0.6 & -0.346 & 0 & 0\\ 0 & 0 & 0.346 & -0.2 & -0.346 & 0.2 & 0 & 0\\ 0 & 0 & -0.3 & -0.3 & 0 & 0 & 0.3 & 0.3\\ 0 & 0 & -0.3 & -0.3 & 0 & 0 & 0.3 & 0.3 \end{bmatrix}$$

Matrix Proof
Note: Since the inverse of T(e) cannot be used the transpose is utilized. The following demonstrates that both sides are equal.

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

$$k^{(e)} = k_{(e)}\begin{bmatrix} l^{(e)2} & l^{(e)}m^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & m^{(e)2} & -l^{(e)}m^{(e)} & -m^{(e)2} \\ -l^{(e)2} & -l^{(e)}m^{(e)} & l^{(e)2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -m^{(e)2} & l^{(e)}m^{(e)} & m^{(e)2} \end{bmatrix}$$ where,

$$k_{(e)} = \frac{E^{(e)}A^{(e)}}{L^{(e)}}$$

thus,

$$k_{(e)}\begin{bmatrix} l^{(e)2} & l^{(e)}m^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & m^{(e)2} & -l^{(e)}m^{(e)} & -m^{(e)2} \\ -l^{(e)2} & -l^{(e)}m^{(e)} & l^{(e)2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -m^{(e)2} & l^{(e)}m^{(e)} & m^{(e)2} \end{bmatrix} = T^{(e)T}\hat{k}^{(e)} T^{(e)}$$

To prove that both sides are equal, the matrices must be multiplied in the correct order. The matrices are given as:

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

$$\hat{k}^{(e)} = k_{(e)}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

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

Multiplying out the first two matrices gives:

$$T^{(e)T}\hat{k}^{(e)} = \begin{bmatrix} l^{(e)} & 0 \\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)} \end{bmatrix} k_{(e)}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} = k_{(e)} \begin{bmatrix} l^{(e)} & -l^{(e)}\\ m^{(e)} & -m^{(e)}\\ -l^{(e)} & l^{(e)}\\ -m^{(e)} & m^{(e)} \end{bmatrix}$$

Taking the result from this and now multiplying it by the third matrix T(e) gives:

$$k_{(e)} \begin{bmatrix} l^{(e)} & -l^{(e)}\\ m^{(e)} & -m^{(e)}\\ -l^{(e)} & l^{(e)}\\ -m^{(e)} & m^{(e)} \end{bmatrix} T^{(e)} = k_{(e)} \begin{bmatrix} l^{(e)} & -l^{(e)}\\ m^{(e)} & -m^{(e)}\\ -l^{(e)} & l^{(e)}\\ -m^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix} = k_{(e)}\begin{bmatrix} l^{(e)2} & l^{(e)}m^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & m^{(e)2} & -l^{(e)}m^{(e)} & -m^{(e)2} \\ -l^{(e)2} & -l^{(e)}m^{(e)} & l^{(e)2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -m^{(e)2} & l^{(e)}m^{(e)} & m^{(e)2} \end{bmatrix} $$

Observation: both sides are equal.

Determination of Reaction Forces using 3 Different Methods
Three methods of solving the same problem are shown below. Each method returns the same result upon careful consideration of the geometry of the situation, and the properties of the elements. The only difference between each method of solving this problem is the amount of effort put into the calculations. It can be seen that the Global Reduced Matrix Method can take significantly less time to find a solution than for each of the other methods.

Global Reduced Matrix Method
This method uses the global K matrix (which has been reduced to only the 3rd and 4th columns of the 6 x 6 matrix), and the global displacement vector to calculate the resultant forces. These forces are given as F1, F2, F5, and F6 in the global force matrix. $$ F=\begin{bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}\\ \end{bmatrix}=\begin{bmatrix} K_{13} &K_{14} \\ K_{23} &K_{24} \\ K_{33} &K_{34} \\ K_{43} &K_{44} \\ K_{53} &K_{54} \\ K_{63} &K_{64} \end{bmatrix}*\begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} $$ $$ F=\begin{bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}\\ \end{bmatrix}=\begin{bmatrix} -0.5625 &-0.3248 \\ -0.3248 &-0.1875  \\  (2.5+0.5625) &(-2.5+0.3248)  \\  (-2.5 +0.3248) &(2.5+0.1875)  \\  -2.5 &2.5  \\  2.5 &-2.5  \end{bmatrix}*\begin{Bmatrix} 4.352\\ 6.1271 \end{Bmatrix}=\begin{Bmatrix} -4.44\\ -2.56\\ 0.00\\ 7.00\\ 4.44\\ -4.44\\ \end{Bmatrix} $$

Element Method (Section 5.1)
The second method computes the reaction forces at the fixed nodes by considering each element indivudually. First, the reaction forces for element 1 are computed using the local k matrix and the local displacement vector. Second, the reaction forces for element 2 are computed using the same method. The general equation is presented first, followed by the specific equation for each element, with values for the k matrices and the displacement vectors inserted. $$ f^{(e)}=k^{(e)}*d^{(e)} $$ $$ f^{(1)}=\begin{bmatrix} 0.5625 &0.3248 &-0.5625  &-0.3248 \\ 0.3248 &0.1875  &-0.3248  &-0.1875 \\ -0.5625 &-0.3248  &0.5625  &0.3248 \\ -0.3248 &-0.1875  &0.3248  &0.1875 \end{bmatrix}*\begin{Bmatrix} 0\\ 0\\ 4.352\\ 6.1271 \end{Bmatrix}=\begin{Bmatrix} -4.44\\ -2.56\\ 4.44\\ 2.56 \end{Bmatrix} $$ $$ f^{(2)}=\begin{bmatrix} 2.5 &-2.5 &-2.5  &2.5 \\ -2.5 &2.5  &2.5  &-2.5 \\ -2.5 &2.5  &2.5  &-2.5 \\ 2.5 &-2.5  &-2.5  &2.5 \end{bmatrix}*\begin{Bmatrix} 4.352\\ 6.1271\\ 0\\ 0 \end{Bmatrix}=\begin{Bmatrix} -4.44\\ 4.44\\ 4.44\\ -4.44 \end{Bmatrix} $$

Statics Method
The last method computes the reaction forces by using equations from statics considerations. The equations used are presented below, along with the values obtained for the reaction forces.

P = 7 L(1) = 4 L(2) = 2 θ(1) = 30° θ(2) = 45°

Observation:

By using trigonometry and the angles given the unknown lengths can be determined. These lengths will be needed when analyzing the summation of the moments of each node.

$$\sin {30^{\circ} } = \frac{L_c}{4} \Rightarrow L_c = 4 \sin {30^{\circ} } = 2$$

$$\cos {30^{\circ} } = \frac{L_x}{4} \Rightarrow L_x = 4 \cos {30^{\circ} } = 3.46$$

$$\sin {45^{\circ} } = \frac{L_b}{2} \Rightarrow L_b = 2 \sin {45^{\circ} } = 1.4$$

$$\cos {45^{\circ} } = \frac{L_y}{2} \Rightarrow L_y = 2 \cos {45^{\circ} } = 1.4$$

Lz = Lx + Ly = 3.46 + 1.4 = 4.86

La = Lc - Lb = 2 - 1.4 = 0.6

Statics equations can now be analyzed since all lengths are known. In order to do this, two moment equations and the equations for the sum of forces about the x and y axes will be used, and are presented below.

$$ \sum{M_{2}^{(1)}}=(-F_{1}*L_{C})-(F_{2}*L_{X})=0 $$

$$ \sum{M_{2}^{(2)}}=(F_{3}*L_{B})-(F_{4}*L_{Y})=0 $$

$$ \sum{F_{X}}=F_{1}+F_{3}=0 $$

$$ \sum{F_{Y}}=P+F_{2}+F_{4}=0 $$

Solving these equations simultaneously, with the value of P=7 will yield these results:

$$ F_{1}=-4.44 $$

$$ F_{2}=-2.56 $$ $$ F_{3}=4.44 $$ $$ F_{4}=-4.44 $$

Determination of q for Node 2
The displacement vector for node 2 can be found by breaking it down into vector components. $$q^{(e)}_2= (d^{(e)}_3\vec{i}+d^{(e)}_4\vec{j}) \cdot \vec{\tilde{i}}$$ This is equivalent to $$q^{(e)}_2= d^{(e)}_{3}(\vec{i}\cdot \vec{\tilde{i}})+d^{(e)}_4(\vec{j} \cdot \vec{\tilde{i}})$$

q2(e) is shown below, after substituting in the director cosines.

$$q_2^{(e)}=l^{(e)}d_3^{(e)}+m^{(e)}d_4^{(e)}$$ where $$ l^{(e)} = cos\theta $$ $$ m^{(e)} = sin\theta $$

$$q_2^{(e)}=\begin{bmatrix}l^{(e)} & m^{(e)} \end{bmatrix}\begin{bmatrix}d_3^{(e)}\\ d_4^{(e)}\end{bmatrix}$$

Determining p values
We know that $$p^{(e)}$$ = $$T^{(e)}$$ $$f^{(e)}$$, where

 * $$T^{(e)}$$ = $$\begin{bmatrix}

l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ \end{bmatrix}$$

and, from past exercises,

 * $$f^{(e)}$$ = $$\begin{bmatrix}

-4.4378\\ -2.5622\\ 4.4378\\ 2.5622\end{bmatrix}$$

Therefore, we can say that the p values become:

 * $$p^{(1)}_{2}$$ = |-4.4378cos(30°) + (-2.5622)sin(30°)|= 5.1243
 * $$p^{(2)}_{1}$$ = |4.4378cos(30°) + 2.5622sin(30°)|= 5.1243

Determining k values
From past exercises it is known that the axial stiffness for both bar elements are as follow:

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

Therefore,

 * $$k^{(1)}$$ = $$\frac{(3)(1)}{4}$$ = 0.75

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

Determining the displacements
Following the same idea of $$d = \frac{f}{k}$$, we have

 * AC = $$\frac{p^{(1)}_{2}}{k^{(1)}}$$ = $$\frac{5.1243}{0.75}$$ = 6.8324

 * AB = $$\frac{p^{(2)}_{1}}{k^{(2)}}$$ = $$\frac{5.1243}{5}$$ = 1.0249

MATLAB Plot of a Two-bar Truss Structure (undeformed and deformed)
The following code plots the analysis of an undeformed and deformed two-bar truss system using MATLAB MATLAB Code

Note: this MATLAB code was created using the principles obtained from:

 * The MATLAB script for the analysis of the Two Bar Truss System obtained from Dr. L. Vu-Quoc website http://clesm.mae.ufl.edu/~vql/courses/fead/2008.fall/codes/twoBarTrussEx.txt

 * The MATLAB script for the plot of a crab-leg structure created by X.G. Tan (2003) and obtained from Dr. L. Vu-Quoc website http://apollo.mae.ufl.edu/~vql/courses/fead/2006.fall/codes/matlab_plot_example2.m

 * The MATLAB script for the plot of an electric pylon (undeformed five-bar truss structure) created by Team MeshWorks (2003) and obtained from Dr. L. Vu-Quoc website http://apollo.mae.ufl.edu/~vql/courses/fead/2006.fall/code.web.pages/electric_pylon.m

Results

Contributing Team Members
Iain Specht--Eml 4500.f08.delta 6.specht 18:13, 5 October 2008 (UTC) Nicolas Castrillon --Eml4500.f08.delta 6.castrillon 02:37, 6 October 2008 (UTC) Diana Guzman--Eml4500.f08.delta 6.guzman 21:34, 6 October 2008 (UTC) Sean Lopez--Eml4500.f08.delta 6.lopez 16:02, 7 October 2008 (UTC) Micheal Krueger--Eml4500.f08.delta 6.krueger 20:29, 7 October 2008 (UTC) Eric Ramirez--Eml4500.f08.delta 6.ramirez 01:49, 8 October 2008 (UTC)

title: User:Eml 4500.f08.delta 6.specht/HW3


title: File:Diagram2hw3.jpg


title: File:Diagram3hw3.jpg


text:
User:Uf.team.wiki/HW3

= Recall: Element Force-Displacement Relationship with respect to the Global Coordinate System =

The Traditional Element Force-Displacement Relationship:

$$ \mathbf{\underset{4 \times 4}k^{(e)}}\mathbf{\underset{4 \times 1}d^{(e)}}=\mathbf{\underset{4 \times 1}f^{(e)}} $$ (Eqn. 1)

Bar Element 1:

= Element Force-Displacement Relationship: Axial Forces =

Element Force-Displacement Relationship: Axial Forces

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

$$ \Rightarrow 	\underbrace{k^{(e)} \begin{bmatrix}

1 &-1 \\

-1 & 1 \\ \end{bmatrix}}_{\hat k^{(e)}} \underbrace{\begin{pmatrix}

q^{(e)}_1 \\

q^{(e)}_2 \end{pmatrix}}_{\underset{2 \times 1}q^{(e)}} = \underbrace{\begin{pmatrix}

p^{(e)}_1 \\

p^{(e)}_2 \end{pmatrix}}_{\underset{2 \times 1}p^{(e)}} $$

Where:

 * q(e)i is the axial displacement of element "e" at local node "i"
 * p(e)i is the axial force of element "e" at local node "i"

GOAL:  Derive equation (1) from equation (2) in order to find:
 * Relationship between q (e) and d (e)
 * Relationship between p (e) and f (e)

These relationships can be expressed as:

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

Consider the displacement vector of local node [1], denoted by $$\mathbf d_{[1]}^{(e)}$$:

Where:

 * $$\overrightarrow{d}^{(e)}_{[1]} = d^{(e)}_1\overrightarrow{i}+{d}^{(e)}_2\overrightarrow{j}$$

 * $$\overrightarrow{q}^{(e)}_{[1]}$$ = axial displacement of node [1] is the orthogonal projection of the displacement vector d(e)[1] of node [1] on the axis $$\overrightarrow{\tilde{x}} $$ of element "e".

$$ \mathbf q_1^{(e)} = \overrightarrow{d}^{(e)}_{[1]}\overrightarrow{i} $$
 * NOTE: Very important relationship***

$$ = (d^{(e)}_1 \overrightarrow{i}+d^{(e)}_2 \overrightarrow{j}) \overrightarrow{\tilde{i}} $$

$$ = d^{(e)}_1 (\overrightarrow{i} \cdot \overrightarrow{\tilde{i}}) + d^{(e)}_2 (\overrightarrow{j} \cdot \overrightarrow{\tilde{i}})$$

Recall:
 * $$(\overrightarrow{i} \cdot \overrightarrow{\tilde{i}}) = \cos \Theta\,\! = l^{(e)}$$
 * $$(\overrightarrow{j} \cdot \overrightarrow{\tilde{i}}) = \sin \Theta\,\! = m^{(e)}$$

$$ q_1^{(e)}= l^{(e)} d_1^{(e)} + m^{(e)} d_2^{(e)} \Rightarrow \begin{bmatrix}

l^{(e)} & m^{(e)} \\ \end{bmatrix} \begin{pmatrix}

d^{(e)}_1 \\

d^{(e)}_2 \end{pmatrix} $$

Similarly for node [2]:

$$ q_2^{(e)}= \begin{bmatrix}

l^{(e)} & m^{(e)} \\ \end{bmatrix} \begin{pmatrix}

d^{(e)}_3 \\

d^{(e)}_4 \end{pmatrix} $$

This leads to the following relationship:

$$	\underbrace{\begin{pmatrix}

q^{(e)}_1 \\

q^{(e)}_2 \end{pmatrix}}_{\underset{2 \times 1}q^{(e)}} = \underbrace{\begin{bmatrix}

l^{(e)} & m^{(e)} & 0 & 0 \\

0 & 0 & l^{(e)} & m^{(e)} \\ \end{bmatrix}}_{\underset{2 \times 4}T^{(e)}}\underbrace{\begin{pmatrix}

d^{(e)}_1 \\ d^{(e)}_2 \\ d^{(e)}_3 \\ d^{(e)}_4 \end{pmatrix}}_{\underset{4 \times 1}d^{(e)}}$$

Similarly, the following represents the same argument:

$$	\underbrace{\begin{pmatrix}

p^{(e)}_1 \\

p^{(e)}_2 \end{pmatrix}}_{\underset{2 \times 1}p^{(e)}} = \mathbf\underset{2 \times 4}T^{(e)}\underbrace{\begin{pmatrix}

f^{(e)}_1 \\ f^{(e)}_2 \\ f^{(e)}_3 \\ f^{(e)}_4 \end{pmatrix}}_{\underset{4 \times 1}f^{(e)}}$$

Recall the element axial force-displacement relationship:

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

$$ \Rightarrow \mathbf \hat k^{(e)} (\underbrace{\mathbf T^{(e)} \mathbf d^{(e)}}_{\mathbf q^{(e)}}) = (\underbrace{\mathbf T^{(e)} \mathbf f^{(e)}}_{\mathbf p^{(e)}}) $$

GOAL: To get into the form $$\mathbf{k^{(e)}d^{(e)}=f^{(e)}}$$.

 * Must "move" $$ \mathbf T^{(e)} $$ from left-hand side to right-hand side by premultiplying the equation by $$ \mathbf T^{(e)^{-1}} $$ {the inverse of T}.
 * However, $$ \mathbf T^{(e)} $$ is a rectangular matrix, it cannot be inverted!

Solution: Divide left-hand side by the transpose of T:

$$ \underbrace{\mathbf{ [T^{(e)^T}\hat k^{(e)} T^{(e)} }]}_{4 \times 4} \mathbf {\underset{4 \times 1}d^{(e)}}\mathbf {\underset{4 \times 1}f^{(e)}} \Rightarrow \mathbf{k^{(e)}d^{(e)}=f^{(e)}}$$

$$\mathbf k^{e} = \mathbf {T^{(e)^T} \hat k^{(e)} T^{(e)}}$$ Equation (3)

Further justification for (3) : Principal of Virtual Work

Example:

$$ \underset{6 \times 6}\mathbf K \underset{6 \times 1}\mathbf d = \underset{6 \times 1}\mathbf F \longrightarrow \bar \underset{2 \times 2}\mathbf K \bar \underset{2 \times 1}\mathbf d = \bar \underset{2 \times 1}\mathbf F$$

For an unconstrained structural system, there are 3 possible rigid body motions in two-dimensional space (2 translational and 1 rotational).

Example: Dynamics evaluation problem:

$$ \mathbf {Kv} = \lambda \mathbf {Mv} $$

Where:
 * K : stiffness matrix
 * λ : eigenvalue (related to the vibrational frequency)
 * M : mass matrix

Use example from HW2 to compute eigenvalues of $$ \underset{6 \times 6}\mathbf K$$:

Recall:

$$ \underset{6 \times 6}\mathbf K = \begin{bmatrix} k_{11}^{(1)} &  k_{12}^{(1)}  &  k_{13}^{(1)}  &  k_{14}^{(1)}  &  0  &  0 \\ k_{21}^{(1)} &  k_{22}^{(1)}  &  k_{23}^{(1)}  &  k_{24}^{(1)}  &  0  &  0 \\ k_{31}^{(1)} &  k_{32}^{(1)}  &  k_{33}^{(1)}+k_{11}^{(2)}  &  k_{34}^{(1)}+k_{12}^{(2)}  &  k_{13}^{(2)}  &  k_{14}^{(2)} \\ k_{41}^{(1)} &  k_{42}^{(1)}  &  k_{43}^{(1)}+k_{21}^{(2)}  &  k_{44}^{(1)}+k_{22}^{(2)}  &  k_{23}^{(2)}  &  k_{24}^{(2)} \\ 0 &  0  &  k_{31}^{(2)}  &  k_{32}^{(2)}  &  k_{33}^{(2)}  &  k_{34}^{(2)} \\ 0 &  0  &  k_{41}^{(2)}  &  k_{42}^{(2)}  &  k_{43}^{(2)}  &  k_{44}^{(2)} \\ \end{bmatrix} = \begin{bmatrix} \frac{9}{16} &  \frac{3\sqrt{3}}{16}  &  -\frac{9}{16}  &  -\frac{3\sqrt{3}}{16}  &  0  &  0 \\ \frac{3\sqrt{3}}{16} &  \frac{3}{16}  &  -\frac{3\sqrt{3}}{16}  &  -\frac{3}{16}  &  0  &  0 \\ -\frac{9}{16} &  -\frac{3\sqrt{3}}{16}  &  \frac{49}{16}  &  \frac{3\sqrt{3}-40}{16}  &  -\frac{5}{2}  &  \frac{5}{2} \\ -\frac{3\sqrt{3}}{16} &  -\frac{3}{16}  &  \frac{3\sqrt{3}-40}{16}  &  \frac{43}{16}  &  \frac{5}{2}  &  -\frac{5}{2} \\ 0 &  0  &  -\frac{5}{2}  &  \frac{5}{2}  &  \frac{5}{2}  &  -\frac{5}{2} \\ 0 &  0  &  \frac{5}{2}  &  -\frac{5}{2}  &  -\frac{5}{2}  &  \frac{5}{2} \\ \end{bmatrix}$$

$$ \lambda \Rightarrow \begin{pmatrix} 10.03 &  1.47  &  0  &  0  &  0  &  -1.75 \\ \end{pmatrix} $$

NOTE: THREE of the eigenvalues are 0 → zero eigenvalue corresponds to zero stored elastic energy (in rigid body modes).

= Two-Bar Truss System MATLAB Plot = The following MATLAB Code plots the two-bar truss system solved in HW2 in the initial and deformed state.

Output:

= Closing the loop between FEM and Statics =

Use Force Displacment relation on (p10-1)

 * (6x2 matrix) x (2x1 matrix) = (6x1 matrix)
 * Closing the Loop between the Finite Element Method (FEM) and Statics is virtual displacement

 * How do we find the displacement to close the loop?
 * Statics tells us the reactions known and therefore the member forces P1(1), P2(2) are equivalent and in opposing directions.

 * Computing axial displacement DOF's, the amount of extension of the bars themselves:

$${\mathbf{q}_2^{(1)}}= \frac{\mathbf{K}^{(1)}}= \frac{|{\underset{2}\mathbf{P}^{(1)}}|}{\mathbf{K}^{(1)}} $$ $${\mathbf{q}_1^{(2)}}= \frac{-{\underset{2}\mathbf{P}^{(2)}}}{\mathbf{K}^{(2)}} $$
 * where q1(1) = 0 is the fixed node 1
 * where q2(2) = 0 is the fixed node 3

= Infinitesimal Displacement =

Figure 5 to the right is a display of the 2 bar truss system we solved previously.

It also displays the displaced element bar system and the location ff points $$\mathbf{B}$$, $$\mathbf{C}$$, and $$\mathbf{D}$$

 * Line $$\mathbf{A}\mathbf{B}$$ is $$\perp$$ to $$\mathbf{B}\mathbf{D}$$

 * Line $$\mathbf{A}\mathbf{C}$$ is $$\perp$$ to $$\mathbf{C}\mathbf{D}$$

$$\mathbf{A}\mathbf{C}= \frac{|{\underset{2}\mathbf{P}^{(1)}}|}{\mathbf{K}^{(1)}}= \frac{|5.124|}{.75}=6.832 $$

$$\mathbf{A}\mathbf{B}= \frac{|{\underset{1}\mathbf{P}^{(2)}}|}{\mathbf{K}^{(2)}}= \frac{|-6.276|}{5}=1.255 $$

The equations for $${\mathbf(U)_y}$$ and $${\mathbf(U)_x}$$ are given as follows

$${\mathbf(U)_y} = \mathbf(R)\sin\alpha$$

 * for small $$\alpha$$

$${\mathbf(U)_x} = \mathbf(R)({1} - \cos\alpha)\cong {0}$$

 * for first order

 * note that $$({1} - \cos\alpha)\simeq {1}$$

Figure 7 displays the trigonometric ratios given by the components above.

We take note that for very small $$\boldsymbol{\alpha}$$,

 * $$\sin\boldsymbol{\alpha}\simeq{\boldsymbol{\alpha}}$$

We use these following equations to fin the x and y coordinates of point B and C

$$\mathbf{X} - {\mathbf{X}_p} = (\mathbf{P}\mathbf{Q})\cos\theta$$

$$\mathbf{Y} - {\mathbf{Y}_p} = (\mathbf{P}\mathbf{Q})\sin\theta$$

We also know that $$\frac{\mathbf{Y} - {\mathbf{Y}_p}}{\mathbf{X} - {\mathbf{X}_p}} = \tan\theta$$

We also deduce that $$\mathbf{Y} - {\mathbf{Y}_p} = \tan\theta(\mathbf{X} - {\mathbf{X}_p}) $$

For a line that is perpendicular to the line above the equation is

$${\mathbf{Y} - {\mathbf{Y}_p}} = (\tan{\theta + \frac{\pi}{2}}){\mathbf{X} - {\mathbf{X}_p}}$$

To determine $$\mathbf{A}\mathbf{D}$$ and point $$\mathbf{D}$$'s components.

$$\overrightarrow{\mathbf{A} \mathbf{D}}= ({\mathbf{X}_d} - {\mathbf{X}_a})\tilde{i} + ({\mathbf{Y}_d} - {\mathbf{Y}_a})\tilde{j} $$

From the definition of the displacement vectors we know

$$\overrightarrow{\mathbf{A} \mathbf{D}} = {\mathbf{d}_3}\tilde{i} + {\mathbf{d}_4}\tilde{j}$$

$${\mathbf{X}_d} = {\mathbf{d}_3} = {4.35}$$

$${\mathbf{Y}_d} = {\mathbf{d}_4} = {6.13}$$

We can now solve for the x and y coordinates of point B and C

For point B we used the following formula to find thew coordinates

$${\mathbf{Y} - {\mathbf{Y}_b}} = (\tan{\theta + \frac{\pi}{2}}){\mathbf{X} - {\mathbf{X}_b}}$$

and our results were

$${\mathbf{X}_b} = {-0.88}$$

$${\mathbf{Y}_b} = {0.88}$$

For point C we used the following formula to find thew coordinates

$${\mathbf{Y} - {\mathbf{Y}_c}} = (\tan{\theta + \frac{\pi}{2}}){\mathbf{X} - {\mathbf{X}_c}}$$

and our results were

$${\mathbf{X}_c} = {5.92}$$

$${\mathbf{Y}_c} = {3.42}$$

= 3 Bar Truss System =

As shown in our previous Wikipedia page, HW2, the 3 bar truss system below is statically indeterminate. The moment equation for this system is proven useless in the section below.

We simply have to many unknowns and not enough equations.

To solve this system we must turn to FEM (Finite Element Method) to compute the reactions and displacement vectors.

Figure 10 clearly shows the 3 bar truss system configuration.

Clearly defined in the figure for each element is:

 * The Young's Modulus, $$\mathbf{E}$$

 * The cross sectional area, $$\mathbf{A}$$

 * The Length of the element, $$\mathbf{L}$$

 * The angle associated with each element, $$\boldsymbol{\theta}$$

We now have what we need to begin forming elemnt stiffness matrices for each element.

Figure 10 shows the 3 bar truss system in its element form. It displays the corresponding $$\theta$$ and the numbering system used.

All of the nodes are labeled accordingly and the elements are numbered. It shows the proper numbering system to allow for the most overlap. This simplifies the compilation of the global stiffness matrix.

This assembly process is shown below in the following section.

 * $$\theta^{(1)} = {30^0}$$

 * $$\theta^{(1)} = {-120^0}$$

 * $$\theta^{(1)} = {-30^0}$$

Going back to the 3 bar truss system: We can begin to start solving the problem by writing the moment equation, with sigma notation and for moment compenents, i. Therefore we take the sum, $$\sum_{i}{\overrightarrow{\mathbf{M_{B,i}}}}=\sum_{i}{(\overrightarrow{\mathbf{BA'_{i}}}\times\overrightarrow{\mathbf{F_{i}}})}$$ $$\mathbf{A'_{i}}$$= any point on line of action of $$\overrightarrow{\mathbf{F_{i}}}$$ $$\sum_{i}{\overrightarrow{\mathbf{M_{B,i}}}}=\sum_{i}{\overrightarrow{\mathbf{BA_{i}}}\times\overrightarrow{\mathbf{F_{i}}}}$$ $$=\overrightarrow{\mathbf{BA}}\times\overrightarrow{\mathbf{F_{i}}}=\overrightarrow{\mathbf{0}}$$ $$\sum_{i=0}^{3}{} = \overrightarrow{\mathbf{0}}$$
 * Note: A is in equilibrium

Next, we continue the problem by looking to begin to setup the entire global stiffness matrix from all 3 element stiffness matricies. Below is the assembly block diagram of the global stiffness matrix. It displays the proper method to combine the 3 element stiffness matrices to allow for proper calculation.

This is the global stiffness matrix fully labeled. It displays the addition used from all of the element stiffness matrices.

$$\begin{bmatrix} k_{11}^{(1)} &  k_{12}^{(1)}  &  k_{13}^{(1)}  &  k_{14}^{(1)}  &  0  &  0  &  0  &  0\\ k_{21}^{(1)} &  k_{22}^{(1)}  &  k_{23}^{(1)}  &  k_{24}^{(1)}  &  0  &  0  &  0  &  0\\ k_{31}^{(1)} &  k_{32}^{(1)}  &  [k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)}]  &  [k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)}]  &  k_{13}^{(2)}  &  k_{14}^{(2)}&k_{13}^{(3)} &k_{14}^{(3)}\\ k_{41}^{(1)} &  k_{42}^{(1)}  &  [k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)}]  &  [k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)}]  &  k_{23}^{(2)}  &  k_{24}^{(2)} &k_{23}^{(3)}&k_{24}^{(3)}\\ 0 &  0  &  k_{31}^{(2)}  &  k_{32}^{(2)}  &  k_{33}^{(2)}  &  k_{34}^{(2)}     &  0  &  0\\ 0 &  0  &  k_{41}^{(2)}  &  k_{42}^{(2)}  &  k_{43}^{(2)}  &  k_{44}^{(2)}     &  0  &  0\\ 0 & 0 &k_{31}^{(3)} & k_{32}^{(3)} & 0 & 0 &k_{33}^{(3)} & k_{34}^{(3)}\\ 0 & 0 &k_{41}^{(3)} & k_{42}^{(3)} & 0 & 0 &k_{43}^{(3)} & k_{44}^{(3)}\\ \end{bmatrix}$$

= References =

In this section we will list some of our references and sources to help readers improve and further there understanding of the material covered in this assignment.

For matrices, the transpose was used to help evaluate an expression. [Source: Wikipedia]

The eigenvalues of a matrix were also mentioned. [Source: Wikipedia]

It is helpful for readers to review methods in ][Source: Wikipedia] to successfully navigate harder problems.

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

Rodney Dagulo --Eml4500.f08.wiki.dagulo.r 16:52, 7 October 2008 (UTC)

Thomas Kehoe --EML4500.f08.Wiki.Kehoe 22:36, 7 October 2008 (UTC)

Ricardo Lopez --EML4500.f08.Wiki.Lopez 15:13, 8 October 2008 (UTC)

Cortland Glowacki --EML4500.f08.Wiki.cort 09:46, 8 October 2008 (UTC)

Renee Hood --EML4500.f08.Wiki.hood 16:19, 8 October 2008 (UTC)

Gonzalo Barcia --EML4500.f08.Wiki.barcia 14:21, 8 October 2008 (UTC)

title: User:Uf.team.wiki/HW3


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text:
User:Kaihsu/Bylaws database

= Bylaws database for British local authorities = Presentation at the Free Legal Web Barcamp, 2008-10-18.

American databases
In the United States of America, where the government structure is more decentralized, largish cities usually have their ordinances codified and sometimes even online. For example, the San Diego Municipal Code.

Monrovia, California, Code of Ordinances § 1.04.010(K): Or, and. “Or” may be read “and,” and “and” may be read “or” if the sense requires it.

British idea
Here in the United Kingdom, things are more centralized, but local authorities can still make bylaws and orders. I have recently put in Freedom of Information requests for the local authorities in my area:
 * Oxfordshire County Council: Constitution; Oxfordshire list of bylaws; Oxfordshire text of bylaws; Oxfordshire County Council Byelaw on the Employment of Children 1998.
 * Oxford City Council: Constitution; Oxford list of bylaws; Oxford text of bylaws; Oxford City Council Bylaws.

In addition, there are orders under other statutory provisions, for example ‘controlled drinking zones’ under section 13(2) of the Criminal Justice and Police Act 2001: City of Oxford Alcohol in Public Places Order 2006; text of City of Oxford Alcohol in Public Places Order 2006.

For the UK central government, there is the Statute Law Database containing statutes in force. I wonder whether there is scope for a web-based database of local authority bylaws in force.

Section 27 of Copyright Act 1994 (New Zealand)
Section 27 of Copyright Act 1994 (New Zealand) (No copyright in certain works).

Update: Communication with my Member of Parliament and relay of ministerial correspondence revealed that the Open Government Licence now exists in the United Kingdom. – Kaihsu 18:46, 21 November 2011 (UTC)

title: User:Kaihsu/Bylaws database


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text:
User:Eas4200c.f08.aeris.krammer/HW3

Shear flow and shear force in a thin-walled member
Open thin-walled sections are very common members in aircraft structures. Shear flow travels through these members, as can be seen by the directional arrows in the figure below.

A shear force is induced through this member due to the shear flow through the member.

 * $$\ dF=q*dl=q(dl_{x}\hat{i}+dl_{z}\hat{k}) $$

 * $$\ dF=q*dl=q(dl*cos(\theta)*\hat{i}+dl*sin(\theta)*\hat{k}) $$

We now integrate, from A to B as given in the figure above, to find the resultant force.

 * $$\ F=\int_{A}^{B}{dF}=q\left[\left( \int_{A}^{B}{dx}*\hat{i}\right) + \left( \int_{A}^{B}{dz}*\hat{k}\right) \right]$$

 * $$\ F=q\left(a*\hat{i}+b*\hat{k} \right)$$

NOTE: q is a constant with respect to x & z.

Now that we have the resulting sheer force vector we can calculate the resultant magnitude of this vector. Noting that this sheer force is broken down into an x and z component, we use Pythagorean Theorem to find the resulting magnitude.

 * $$\ \left| \left| F\right|\right|=\sqrt{F_{x}^2+F_{z}^2} $$

 * $$\ \left| \left| F\right|\right|=\sqrt{(q*a*\hat{i})^2+(q*b*\hat{k})^2}$$

 * $$\ \left| \left| F\right|\right|=q*\sqrt{a^{2}+b^{2}} $$

Torque, Shear Flow, and Polar Moment of Inertia
In order to calculate torque on a member induced by shear flow, it is necessary to analyze the average area in which that shear flow acts. In the collapsible box below, it is proved that a triangle's area is half the product of its base length and its height even if the height is measured outside of the triangle.

The torque on an arbitrarily shaped shear panel can be expressed in terms of shear flow, and the average area. The shear flow is the force per unit length the shear panel experiences and the average area is the area bounded by the contour of the shear panel relative to the designated origin of the coordinate system in which the shear panel is described. The math process of obtaining this equation is shown below. $$\displaystyle T= \oint{\rho qds}= \int\int_{ \bar{A}}{2qdA}=2q \bar{A}$$ The below picture describes the forces, shear flow, average area and coordinate system of an open thin-walled cross section.  PQ is the shear panel contour, A is the average area enclosed by the panel relative to the origin, e is the perpendicular vector arm from the origin to the force F (which is calculated by multiplying the shear flow with the length in with the shear flow acts). F is the essentially the equivalent force of the shear flow and is used to calculate the torque. Another equation can be written to accommodate for F and the vector e. $$\displaystyle Fe=T=2Q\bar{A}$$ Now, let's consider a uniform bar with circular cross section subject to torque and resulting torsion. This is not a case in which warping of the member is a problem and thus the cross section behaves like rigid disks with only angular displacement rather than lateral displacement (along the length of the bar). The picture below represents an infinitesimally small part of the circlar bar where dA is the infinitesimally small area. This area will be displaced in the (y,z) plane an angel gamma when the object is torqued.  The derivation of an equation to find the torque in terms of the bar's shear modulus, the polar moment of inertia of the cross section, and the rate of twist (gamma). $$\displaystyle T=\int\int_{A}{r \tau dA}=G \gamma \int\int_{A}{rdA}$$ where $$\tau =G \gamma $$ is obtained from Hooke's Law The below picture describes the deformation after torque when the vector r (vector from the origin to dA) is displaced by a small angle alpha to a new vector (r') in the (y,z) plane. $$\displaystyle \gamma= \frac{rd \alpha}{dx}$$ where $$\displaystyle \frac{d \alpha}{dx}=: \theta$$ (rate of twist) so $$\displaystyle T= \int_A{rG(r \theta)dA}$$ where dA= dydz, and G and $$\theta$$ are independent of y and z. $$\displaystyle = G \theta \int_A{r^2dA}=G \theta J$$ G is the shear modulus, theta is the rate of twist, and J is the polar moment of inertia. If a is the radius of the circular cross section then $$ J=\frac{1}{2} \pi a^4$$.

If it is a hollow, thin-walled (t<$$J=\frac{1}{2}\pi(b^4-a^4)=\frac{1}{2}\pi(b-a)(b+a)(b^2+a^2)=\frac{1}{2}\pi t(2 \bar{r})(b^2+a^2)$$ where $$\bar{r}$$ is the average radius ((a+b)/2) $$\displaystyle b^2 \cong \bar{r}^2$$ and $$a^2 \cong \bar{r}^2$$ so $$J=2\pi t \bar{r}^3$$

$$\displaystyle J=(2\pi^{1/2}t)(\pi \bar{r}^2)^{3/2}$$ where $$\displaystyle \bar{a}^=\pi \bar{r}^2$$ This means that J is proportional to the average area to the 1.5 power with a proportional factor of $$2\pi^{1/2}t$$

NACA Airfoil and Torsion of Uniform, Non-circular Bars
In homework 3, we are examining the NACA 4 digit airfoil series by using Matlab to do the analysis. Below are some pictures of some examples of the NACA airfoil. The one to the left describes fluid flow around the airfoil cross section (source: rit.edu) and the one to the right is a simple and crude cross sectional representation of the NACA airfoil with a (y,z) coordinate system where the y axis passes through the trailing edge. The y-axis is discretized in "ns" segments. The red circle highlights the infinitesimally small segment that will be analyzed.

Here is the picture of the enlarged segment. As a crucial goal while analyzing the NACA airfoil is to determine the "average" area, we must devise a method to do this. It is possible to use a summation of infinitesimally small trapezoids, however, this method is not very elegant. In turn, we will use a method that utilizes triangles to determine the quadrature of the airfoil. The triangle will be a cross product a vector from an observation point (can be positioned anywhere) to a point on the airfoil and a vector of infinitesimally small length along the airfoil. These two vectors enclose a triangle of infinitesimally small area dA. This is shown in the picture below.  Torsion of uniform, non-circular bars are also examined in homework 3. This is the case in which warping will occur as opposed to the case of uniform, circular bars. Warping is the axial displacement along the x-axis of the bar (along the length of the bar) of a point on the deformed (rotated) cross section. The below pictures describe the bar and the bar's cross section when torsion is applied. Point P is a point on the (y,z) plane which moves to P' as the torque is applied. Beta describes the initial angular displacement measured from the y-axis and alpha is the angular displacement due to the torsion. Alpha is also measured from the y-axis.  Below is the magnified version of point P and its movement and displacement on the (y,z) plane of the cross section. $$u_y$$ and $$u_z$$ are the components of the displacement vector $$\vec {PP'}$$.  Why is the vector PP' perpendicular to OP? Isn't it more intuitive that one would draw an arc with radius |OP| and plot point P' at an angle alpha away from vector OP? The perpendicular nature of PP' to OP is because the angle alpha is very small. The proof is below. Let there be two vectors OP and OP' of equal length (|OP|=|OP'|=R). The displacement differences in the points P and P' are denoted by $$u_z$$and $$u_y$$ where z denotes vertical displacement and y denotes horizontal displacement.

$$\displaystyle Rsin\alpha \cong R\alpha (\alpha small)$$ $$\displaystyle R(1-cos\alpha) \cong 0 (first order)$$ Thus, it is easy to see that since the horizontal displacement of point P' from point P is zero, point P' directly above point P in this case. That makes vectors OP and PP' perpendicular.

What is the displacement component $$\displaystyle u_{y}$$ and $$\displaystyle u_{z}$$? $$\displaystyle u_{y}=-(PP')sin\beta = -\alpha (OP)sin\beta$$

Where $$\displaystyle (PP')$$ is $$\displaystyle (OP)\alpha$$, $$\displaystyle - \alpha (OP)$$ is $$\displaystyle \alpha x$$, and $$\displaystyle sin\beta$$ is $$\displaystyle z_{P}$$. The rate of twist is denoted by theta, which is the twist angle per unit length $$\theta =\frac{\alpha }{x}$$ Therefore, the horizontal displacement $$\displaystyle u_{y}$$ is $$\displaystyle u_{y}=-\theta xz$$ and the vertical displacement $$\displaystyle u_{z}$$ is $$\displaystyle u_{z}=+(PP')cos\beta =+\alpha y_{z}$$ Where $$\displaystyle (PP')$$ is $$\displaystyle (OP)\alpha$$, $$\displaystyle + \alpha$$ is $$\displaystyle \theta y_{z}$$,. $$\displaystyle u_{y}=+ \theta xy$$. The warping displacement along the x-axis is
$$\displaystyle u_{x}=\theta \psi (y,z)$$ Equations (1), (2) and (3) are kinematic assumptions.

Roadmap for Torsional Analysis of an Aircraft Wing
Wing skin on the wing section is stiffened by stringers, spar webs and ribs. The supporting thin vertical webs form  a  structure that comprises of two or more cells.

A. Kinematic assumptions (C.T. Sun: Mechanics of Aircraft Structures. Section 3.2.)

B. Strain-displacement relationship (C.T. Sun: Mechanics of Aircraft Structures. Section 3.2.)

C. Equilibrium equation for stresses (C.T. Sun: Mechanics of Aircraft Structures. Chapter 2; Section 3.6.)

D. Prandtl stress function $$\phi$$ (C.T. Sun: Mechanics of Aircraft Structures. Section 3.2, Equation 3.15.)

E. Strain compatibility equation (C.T. Sun: Mechanics of Aircraft Structures. Equation 3.17)

F. Equation for $$\phi$$ (C.T. Sun: Mechanics of Aircraft Structures. Equation 3.19.)

G. Boundary conditions for $$\phi$$ (C.T. Sun: Mechanics of Aircraft Structures. Equation 3.24.)

H.
   $$\displaystyle T=2\int\int_{A}^{}{}\phi dA$$ $$\displaystyle T = GJ\theta $$ $$\displaystyle J=\frac{-4}{\nabla^{2}\phi }\int \int _{A}\phi dA$$

I. Thin-walled cross-section Ad-hoc assumption on shear flow. Formal derivation can be found in C.T. Sun: Mechanics of Aircraft Structures. Section 3.5.
   $$\displaystyle T = 2q\bar{A}$$ (C.T. Sun: Mechanics of Aircraft Structures. Equation 3.48.)

J. Twist angle $$\theta$$: Method 1
   $$\displaystyle \theta = \frac{1}{2G\bar{A}}\oint{\frac{q}{t}ds}$$ (C.T. Sun: Mechanics of Aircraft Structures. Equation 3.56.)

K. Section 3.6 on multicell thin-walled cross-section cell $$\displaystyle i=1, ...,n_{c}$$ for $$\displaystyle n_{c}=3$$

K1.
     $$\displaystyle T=2\sum_{i = 1}^{n_{c}}{q_{i}\bar{A}_{i}}$$ Where, $$\displaystyle q_{i}$$ = shear flow in cell i      $$\displaystyle \bar{A}_{i}$$ = "average" area in cell i
$$\displaystyle T_{i}=2q_{i}\bar{A}_{i}\qquad$$ thus the torque generated by one cell:
$$\displaystyle T=\sum_{i=1}^{n_{c}}{T_{i}}$$

K2. Shape of airfoil is "rigid" in the (y,z) plane (but it can warp out of plane).
     $$\displaystyle \theta = \theta _{1} = \ldots = \theta _{n_{c}}$$ $$\displaystyle \theta _{i}=\frac{1}{2G_{i}\bar{A}_{i}}\oint\frac{q_{i}}{t_{i}}ds$$ Where, $$\displaystyle G_{i}$$= shear modulus of cell i.      $$\displaystyle t_{i}$$= thickness of cell i.       $$\displaystyle t_{i}(s)$$= curvilinear coordinate along the wall of the cell.

Single Cell Section
Problem 1.1 was an example for a rectangular single cell section. Now, let's consider a more general single cell section:

Where $$\displaystyle t_{1} = 0.008 m$$ $$\displaystyle t_{2} = t_{3} = 0.001 m$$ $$\displaystyle a = 4 m $$ $$\displaystyle b = 2m $$ The area of the shape is the area of the semicircle plus the area of the triangle
$$\frac{}{A}= \frac{1}{2} \Pi ({\frac{b}{2}})^2 + \frac{1}{2}ba=5.57m$$

The shear flow of the shape is defined as the torque divided by two times the area:
$$\displaystyle q=\frac{T}{2\bar{A}}$$ The twist angle is calculated by breaking up an integral into a summation of three parts:
$$\displaystyle \theta=\frac{1}{2G\bar{A}}\sum_{j=1}^{3}\frac{q_{j}l_{j}}{t_{j}}$$ where $$\displaystyle j=1,2,3$$ is the index for the part number.

Airfoil Matlab Program
For the third homework assignment we were asked to develop a Matlab program to plot an airfoil profile and compute the average $$\bar{A}$$.

To begin we needed to start with a simple profile, a circle. The equation for a circle with the origin at the "leading edge" is $$z=\sqrt{R^2-(y-R)^2}$$. The circle is then divided up into tiny segments of length $$dy$$ along the y-axis. The circular foil is then integrated along the length of the chord (2R) while computing the small area, $$dA$$, for each segment. The area was calculated by taking half of the cross product of the first position vector (y) with the next (y+dy). $$S = \frac{1}{2}|{AB}\times{AC}|$$ The plot of the foil is shown below. The profile is displayed in the lower half while the segmented nodes are shown in the top half. As expected, the centroid of the circular airfoil is located in the center.

To determine how many steps were needed to accurately calculate the airfoils $$\bar{A}$$, the code was ran for different number of segments. The results are displayed on the graph below. To receive an error of 1% or less, the value of $$ns_{max}$$ is estimated to be 17 steps.

The same technique for finding $$\bar{A}$$ and the centroid location was applied to an NACA 2415 airfoil. Using the four digit number the equation for the mean camber line can be expressed using 2nd order polynomials. The thickness of the airfoil is calculated using another polynomial. Using trigonometry and the slope of the camber line ($$\frac{dz_c}{dy}$$), the y and z values for the nodes of each segment can be determined. The result of the NACA 2415 computation is shown below. The y-value of the centroid is 0.25 meaning the airfoil is a quarter-chord. The average $$\bar{A}$$ for the airfoil was calculated to be 0.0990.

$$\bar{A}_{airfoil}=0.0990$$

$$\bar{A}_{airfoil}=0.0247$$

= Contributions = Jin Yu Guan Eas4200c.f08.aeris.guan 13:21, 08 October 2008 (UTC) Anett Krammer Eas4200c.f08.aeris.krammer 13:22, 08 October 2008 (UTC) Josh Holladay Eas4200c.f08.aeris.holladay 17:33, 8 October 2008 (UTC) Nelson Caceres Eas4200c.f08.aeris.caceres 13:24, 08 October 2008 (UTC) Jesse W. McRae Eas4200c.fo8.aeris.mcrae 13:25, 08 October 2008 (UTC)

title: User:Eas4200c.f08.aeris.krammer/HW3


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User:Eas4200c.f08.gator.edwards/Week Three Class Notes

 Class Notes (current) archived

 Homework Problems (current) archived

 Matlab Program (current) archived

Contributing Team Members:

The following members contributed to this report:

Greg Edwards Eas4200c.f08.gator.edwards 13:41, 6 October 2008 (UTC)

Taylor Lusk Eas4200c.f08.gator.lusk 23:31, 7 October 2008 (UTC)

William Mueller Eas4200c.f08.gator.mueller 04:15, 8 October 2008 (UTC)

Justin Dickinson Eas4200c.f08.gator.dickinson 05:33, 8 October 2008 (UTC)

Jacob Papp Eas4200c.f08.team12.papp 13:46, 8 October 2008 (UTC)


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User:Eas4200c.f08.gator.edwards/Wk3 Class Notes

Curved Panels
Looking at the shear flow in a curved panel: $$d\vec{F} = qd\vec{l} = q(dl_{y}\hat{j} + dl_{z}\hat{k})$$
 * $$= q(dlcos\theta\ \hat{j} \ +\ dlsin\theta \ \hat{k})$$

Therefore the resultant shear force vector is:

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

Now, the magnitude of the resultant shear force vector is:
 * $$\parallel \vec{F} \parallel = \sqrt{F_{y}^{2}+F_{z}^{2}}$$
 * $$=q\sqrt{a^{2}+b^{2}}$$

 * $$=qd$$

 * $$\parallel \vec{F} \parallel =qd$$

Now we will relate the magnitude of the resultant shear force vector (R) to the torque $$\ T=2q\bar{A}$$
 * $$\vec{T}= T\hat{i}$$

 * $$dF = qdl$$

 * $$d\vec{T}=\vec{r}$$ x $$d\vec{F}$$

 * $$ dT = rdFsin\theta= \rho dF$$

 * $$ = \rho q dl$$

 * $$dA = \frac{1}{2}\ \rho dl$$

 * $$T=\oint dT = q\oint \rho dl = 2q\int_{\bar{A}}dA$$

 * $$\Rightarrow T=2q\bar{A}$$

Circular Cross-Section
Uniform bar with circular cross section non warping.

$$T = \iint_A{r\tau \; \mathrm{d}A} = \int_A{rG(r\theta) \; \mathrm{d}A} = G\theta \int_A{r^2 \; \mathrm{d}A} = G\theta J = \frac{\pi G\theta}{2}r^4$$

Hollow thin walled cross section.

$$r_i = a \; r_o = b$$

$$J = \frac{\pi}{2} \left( b^4 - a^4 \right) = \frac{\pi}{2}(b-a)(b+a)(b^2+a^2) = \frac{\pi}{2}(t)(2\bar{r})(2\bar{r}^2) = 2\pi t\bar{r}^3$$

So the torque equation becomes

$$T=2 \pi G \theta t {\bar r}^3$$

NACA Airfoils
For the Matlab portion of the homework assignment, a portion is dedicated to finding the area enclosed by the airfoil. To do this, the airfoil will be broken up into many small segments, each of area dA. All of these segment areas will then be added up to come up with a total area. The second image on the right shows an enlarged region of the first image of the airfoil.

Looking at the image of the enlarged section, to find the area:
 * $$ d\vec{A} = \frac{1}{2} \ \vec{r}\ $$ x $$\ \vec{PQ}$$
 * $$=||d\vec{A}|| \ \ \hat{i}$$
 * $$ d\vec{A} = \frac{1}{2} \ \vec{r}\ $$ x $$\ \vec{PQ}$$
 * $$=||d\vec{A}|| \ \ \hat{i}$$
 * $$=||d\vec{A}|| \ \ \hat{i}$$

Torsion of Non-Circular Bars
warping = axial displacement along x-axis (i.e. along bar length) of a point on the deformed (rotated) cross-section.

Note: $$\vec{PP'}\quad$$ is the displacement vector
 * $$u_{z}\quad$$ is the z-component of the displacement vector
 * $$u_{y}\quad$$ is the y-component of the displacement vector

From the figure, we can see that the displacement is perpendicular to OP, but why? We can see that $$OP=OP'=R\qquad$$ and that $$u_{z} = Rsin(\alpha )\quad$$ and with $$\alpha\quad$$ being small, $$u_{z} = R\alpha\quad$$ By the same logic: $$u_{y}=R(1-cos(\alpha ))=R(1-1) = 0\quad$$. Therefore we can approximate the displacement as being perpendicular to OP.

 Kinematic Assumptions
 * Starting with the definition of the rate of twist $$\theta = \frac{\alpha}{x}\quad \rightarrow \alpha = \theta x$$

$$u_{y}=-\theta x z \qquad \qquad (3.11)$$ $$u_{z}=PP'cos(\beta ) =OP\alpha cos(\beta ) = \alpha y $$ $$u_{z}=\theta x y \qquad \qquad (3.12)$$ $$u_{z}\quad$$ is the z displacement of a point on the cross-section. $$u_{x}=\theta \psi (y,z) \qquad \qquad (3.13)$$ The three equations above are the Kinematic Assumptions
 * Warping displacement along x-axis

A. Kinematic Assumptions
(Section 3.2 in textbook) See above three equations.

B. Strain-Displacement Relationship
(Section 3.2 in textbook) Strain is a non-dimensional value representing the ratio of the amount the material has deformed in one direction versus the original length of the material in that direction. $$\epsilon = \frac{\delta\ell}{\ell_{0}}$$

C. Equilibrium Equation for Stresses
(Chapter 2, Section 3.6) If a body is in equilibrium the stress field must satisfy these equations everywhere:

$$\frac{\partial\sigma_{xx}}{\partial x} + \frac{\partial\tau_{yx}}{\partial y} + \frac{\partial\tau_{zx}}{\partial z} = 0$$

$$\frac{\partial\tau_{xy}}{\partial x} + \frac{\partial\sigma_{yy}}{\partial y} + \frac{\partial\tau_{zy}}{\partial z} = 0$$

$$\frac{\partial\tau_{xz}}{\partial x} + \frac{\partial\tau_{yz}}{\partial y} + \frac{\partial\sigma_{zz}}{\partial z} = 0$$

$$\tau_{xy}=\tau_{yx}\quad\tau_{yz}=\tau_{zy}\quad\tau_{xz}=\tau_{zx}$$

D. Prandtl Stress Function $$\phi$$
(3.2, Eq. 3.15) The Prandtl stress function is defined such that its derivative in the y-direction equals the shear stress in the plane perpendicular to the y-direction (the x-z plane), and that its derivative in the x-direction equals the negative shear stress in the plane perpendicular to the x-direction (the y-z plane). $$\tau_{xy} = \frac{\partial \phi}{\partial y}$$ $$\tau_{yz} = -\frac{\partial \phi}{\partial x}$$

E. Strain Compatibility Equation
(Eq. 3.17) $$\frac{\partial\gamma_{yz}}{\partial x} - \frac{\partial\gamma_{xz}}{\partial y} = 2\Theta $$

F. Equation for $$\phi$$
(Eq. 3.19) Combining the compatibility equation for torsion and Prandtl's stress function we get: $$\frac{\partial^{2}\phi}{\partial x^{2}} + \frac{\partial^{2}\phi}{\partial y^{2}} = -2G\Theta$$

G. Boundary Conditions for $$\phi$$
(Eq. 3.24) The traction free boundary condition tz = 0 is now: $$\frac{\partial\phi}{\partial s} = 0$$ Meaning that the stress function is constant at the surface. For a solid regular body the constant is arbitrary and is often made to be 0 for simplicity. $$\phi = 0$$

H. Torque
$$T=2\int\int_{A}^{}{}\phi dA$$ (Eq. 3.25)

$$T=GJ\theta $$

$$J=\frac{-4}{\nabla^{2}\phi }\int \int _{A}\phi dA$$

(pg. 74)

I. Thin-walled Cross-section
Ad-hoc assumption on shear flow. Formal derivative (Section 3.5). We previously did the end of this section, which is easier than the beginning.

$$T = 2q\bar{A}$$

J. Twist Angle $$\theta$$: Method 1
$$\theta = \frac{1}{2G\bar{A}}\oint{\frac{q}{t}ds}$$

K. Section 3.6 on multicell thin-walled cross-section
Looking at the picture of the multicell wing we find that the number of cells, $$n_{c}$$ is equal to 3.

K1.
$$T=2\sum_{i = 1}^{n_{c}}{q_{i}\bar{A}_{i}}$$

 * Where $$q_{i}$$ = shear flow in cell i
 * and $$\bar{A}_{i}$$ = the "average" area of cell i

Define: $$T_{i}=2q_{i}\bar{A}_{i}\qquad$$ torque generated by one cell
 * $$\Rightarrow T=\sum_{i=1}^{n_{c}}{T_{i}}$$

K2.
Shape of airfoil is "rigid" in the yz plane (can warp out of plane). $$\theta = \theta _{1} = \ldots = \theta _{n_{c}}$$

$$\theta _{i}=\frac{1}{2G_{i}\bar{A}_{i}}\oint\frac{q_{i}}{t_{i}}ds$$

 * $$G_{i}$$= shear modulus of cell i.
 * $$t_{i}$$= thickness of cell i.
 * $$t_{i}(s)$$= curvilinear coordinate along cell wall.

Shear Flow: $$T = 2q \bar{A}$$
 * $$q=\frac{T}{2\bar{A}} \qquad$$ express q as a function of T

Twist Angle:
 * $$\theta = \frac{1}{2G\bar{A}}\sum_{j=1}^{3}{\frac{q_{j}l_{j}}{t_{j}}}\qquad$$ with l as the length of the segment and j as the index number.

NOTE:

 * $$\int \qquad$$: elongated S, standing for summation (continuous)
 * $$\sum \qquad$$: (discrete) sum


text:
User:Eas4200c.f08.gator.edwards/Homework Problems

Area of a Triangle
Prove $$A = \frac{1}{2}bh$$

First begin by considering a rectangle with base of BD and height of AD. The area of this rectangle would be (BD)(AD). Now, if we just consider half of this rectangle (triangle ABD) we would have an area of $$\frac{1}{2}$$(BD)(AD). This is the area of the triangle pictured including the area enclosed by the dotted lines. If we subtract the area enclosed by the dotted lines from the area we just found, we will get the area of the triangle we are interested in. The area of the dotted line would be $$\frac{1}{2}$$(AD)(CD) by the same logic as before. Performing this, we find that $$A = \frac{1}{2}\left(AD \right)\left(BD \right)- \frac{1}{2}\left(AD \right)\left(CD \right) = \frac{1}{2}\left(AD \right)\left[ (BD)-(CD)\right]$$

Looking at the picture, we can see that by definition BD - CD = b and AD = h. Plugging these relationships into the above equation yields $$A = \frac{1}{2}bh$$

Polar Moment of Inertia of Circular Cross-Section
For a circle of radius a, the polar moment of inertia ($$J_{0}$$) can be calculated as follows:

$$J_{0} = \int_{A}\rho^{2}dA = \int_{0}^{a}\int_{0}^{2\pi}\rho^{2}(\rho d\theta)d\rho $$
 * $$= \int_{0}^{a} \rho^{2}\left(2\pi\rho d\rho \right) $$
 * $$= 2\pi\int_{0}^{a}\rho^{3}d\rho = \frac{1}{2}\pi a^{4}$$
 * $$\Rightarrow J_{0} = \frac{1}{2}\pi a^{4}$$

Solid vs. Hollow Cross-Section
Compare a solid circular cross section to a hollow thin walled circular cross section.

$$r_{o}^{(a)}=1cm$$

$$r_{i}^{(b)}=5cm$$

$$t^{(b)}=0.1cm$$

$$A^{(a)}=\pi r^{2}=\pi (1cm)^{2}=3.1415cm^{2}$$

$$A^{(b)}=\pi (r_{o}^{2}-(r_{o}-t_{b})^{2})=\pi (5cm^{2}-(5cm-0.1cm)^{2})=3.1102cm^{2}$$

$$J^{(a)}=\frac{\pi }{2}r^{4}=\frac{\pi }{2}(1cm)^{4}=1.5708cm^{4}$$

$$J^{(b)}=2\pi t\bar{r}^{3}=2\pi t(\frac{2r_{o}-t^{(b)}}{2})^{3}=2\pi (0.1cm)(\frac{2(5cm)-(0.1cm)}{2})^{3}=76.207cm^{4}$$

$$\frac{J^{(a)}}{J^{(b)}}=\frac{1.5708cm^{4}}{76.207cm^{4}}=0.0206$$

Find $$r_{i}^{(c)}$$ if $$t^{(c)}=0.02r_{o}^{(c)}$$ and $$J^{(b)}=J^{(c)}$$.

$$J^{(b)}=76.207cm^{4}=2\pi (0.02r_{i}^{(c)})(\frac{2(r_{i}^{(c)}+(0.02r_{i}^{(c)}))}{2})^{3}=1.06684r_{o}^{(c)}=J^{(c)}$$

$$r_{i}^{(c)}=2.907$$

Single Cell Section Area

To calculate the area of this figure, we break it down into two smaller parts: a semicircle and a triangle. The area of a semicircle is $$\frac{1}{2}\pi R^{2}\quad$$ with R being $$\frac{b}{2}$$ = 1 m. The area of a triangle is $$\frac{1}{2}bh\quad$$ with the base being a (4 m) and the height being b (2 m).

Therefore the overall area is:

$$\bar{A} = \frac{1}{2}\pi \left( \frac{b}{2}\right)^{2} + \frac{1}{2}ab $$
 * $$= \frac{1}{2}\pi \left( \frac{1}{2}\right)^{2} + \frac{1}{2}\left(4 \right)\left(2 \right)$$
 * $$=\frac{\pi}{8} + 4 \quad m.$$
 * $$= \frac{1}{2}\pi \left( \frac{1}{2}\right)^{2} + \frac{1}{2}\left(4 \right)\left(2 \right)$$
 * $$=\frac{\pi}{8} + 4 \quad m.$$
 * $$=\frac{\pi}{8} + 4 \quad m.$$
 * $$=\frac{\pi}{8} + 4 \quad m.$$


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User:Eml4500.f08.ramrod.D/HW3

MATLAB Code to Plot Undeformed and Deformed Two Bar Truss System
The code below plots the two bar truss system that was solved in class with the deformed system superimposed on top. The plot of the system is below the code.

% Matlab Script %********************************************************************* % filename: two_bar_truss.m  % % PURPOSE: %  Plot the two bar truss system discussed in class and the deformed %  system with matlab % % AUTHOR: %   Daniel Olivero %   Team Ramrod %   EML 4500 FEAD, Fall 2008. % % Modified on: vql, Fri, 03 Oct 2003, 08:37:46 EDT % Created on : Sun, 05 Oct 2008, 16:22:00 EDT % % DEPENDENCIES: %  call: % % REMARKS: % %*********************************************************************  %Plot Two beam truss system % clear; close;

%model with 2-D beam elements dof = 2; %dof per node: axial disp x, 2= disp y %obtain the coordinatates of all nodes % n_node = 3;             %number of nodes n_elem = 2;           %number of elements total_dof = 2 * n_node; %total dof of system

position(:, 1) = [0; 0]; position(:, 2) = [3.46; 2]; position(:, 3) = [4.9; .59];

% print the node coord. for i = 1 : n_node; x(i) = position (1,i); y(i) = position (2,i); end

node_connect (1, 1) = 1;   %element 1 node_connect (2, 1) = 2;

node_connect (1, 2) = 2;   %element 2 node_connect (2, 2) = 3;

% connect all beam elements by connectivity array for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; axis([ -2 5 -2 5]) plot(xx,yy,'--') hold on end

text(x(node_connect(1,1)),y(node_connect(1,1)),'Global Node 1', 'HorizontalAlignment', 'center') text(x(node_connect(2,1)),y(node_connect(2,1)),'Global Node 2', 'HorizontalAlignment', 'center') text(x(node_connect(2,2)),y(node_connect(2,2)),'Global Node 3', 'HorizontalAlignment', 'center') text(x(node_connect(2,1))/2,y(node_connect(2,1))/2,'Element 1', 'HorizontalAlignment', 'center') text( x(node_connect(2,1)) + (x(node_connect(2,2))-x(node_connect(2,1)))/2, y(node_connect(2,1)) +(y(node_connect(2,2))-y(node_connect(2,1)))/2,'Element 2', 'HorizontalAlignment', 'center')

hold on

position_disp(:,1) = [0;0]; position_disp(:,2) = [7.812; 8.1271]; position_disp(:,3) = [4.9; .59];

% print the node coord. for i = 1 : n_node; x(i) = position_disp (1,i); y(i) = position_disp (2,i); end

node_connect_disp (1, 1) = 1;   %element 1 node_connect_disp (2, 1) = 2;

node_connect_disp (1, 2) = 2;   %element 2 node_connect_disp (2, 2) = 3;

% connect all beam elements by connectivity array for i = 1 : n_elem; node_1 = node_connect_disp(1,i); node_2 = node_connect_disp(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; axis([ -2 10 -2 10]) plot(xx,yy,'-') hold on end

text(x(node_connect_disp(2,1)),y(node_connect_disp(2,1)),'Global Node 2', 'HorizontalAlignment', 'center') text(x(node_connect_disp(2,1))/2,y(node_connect_disp(2,1))/2,'Element 1', 'HorizontalAlignment', 'center') text( x(node_connect_disp(2,1)) + (x(node_connect_disp(2,2))-x(node_connect_disp(2,1)))/2, y(node_connect_disp(2,1)) +(y(node_connect_disp(2,2))-y(node_connect_disp(2,1)))/2,'Element 2', 'HorizontalAlignment', 'center')

title ('Two Barr Truss System') xlabel('x') ylabel('y')

title: User:Eml4500.f08.ramrod.D/HW3


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User:EML4500.f08.FEABBQ.Jayma/HW3

Class Notes: Friday, October 3, 2008
There is a clarification/correction to the angle for element 3, $$\theta^{(3)}$$, as shown by the following image and equation...

$$ \theta^{(3)}=-90^{\circ}-45^{\circ}=-135^{\circ} $$

Now that the angle for element 3 is clear, we can examine the three-bar truss system with respect to the reaction forces and the external force P...

We know the following equations to be true for this three-bar truss...

$$\Sigma{F}_{x}=0\, $$

$$\Sigma{F}_{y}=0\, $$

$$\Sigma{M}_{A}=0\, $$

where $$\Sigma{M}_{A}=0$$ is trivial, due to the fact that all force vectors pass through this point.

But what about point B ~ $$\Sigma{M}_{B}\, $$?

We will use the following image and subsequent equations to explain the moment about point B. Note: Points A and B are unrelated to the original three-bar truss figure...

The following is true for all $$A'\,$$ on the line of action of $$\vec{F}$$... $$ \Sigma{\vec{M}}_{B}=\vec{BA}\times\vec{F}=\vec{BA'}\times\vec{F}$$ $$ \vec{BA'}=\vec{BA}+\vec{AA'}$$ $$ \Sigma{\vec{M}}_{B}=(\vec{BA}+\vec{AA'})\times\vec{F}=\vec{BA}\times\vec{F}+\vec{AA'}\times\vec{F}$$

where we know that $$\vec{AA'}\times\vec{F}=\vec{0}\, $$. Now we go back to the three-bar truss: Node $$A$$ is in equilibrium: $$ \sum_{i=0}^3\vec{F}=\vec{0}\, $$

And we know that $$ \sum_{i}\vec{M}_{B,i}=\sum_{i}\vec{BA'_{i}}\times\vec{F_{i}}\, $$

$$A'_{i}\, $$ is any point on the line of action of $$\vec{F_{i}}\, $$

$$ \sum_{i}\vec{M}_{B,i}=\sum_{i}\vec{BA'_{i}}\times\vec{F_{i}}=\vec{BA}\times\sum_{i}\vec{F_{i}}=\vec{0}\, $$, therefore proving $$ \Sigma{M}_{B}=\vec{0}\, $$.

Next, we move on to the global stiffness matrix for the three-bar truss. A general outline of the overlapping nature of the three individual element stiffness matrices is shown below...

Next, we fill this matrix...

title: User:EML4500.f08.FEABBQ.Jayma/HW3


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User talk:Pathophysiology

Hey just wanted to see how this worked..kinda cool

title: User talk:Pathophysiology


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User:Eml4500.f08.delta 6.castrillon/HW 3

=Determination of Element FD (wrt global coordinates)=

We have the FD for springs:

k $$^{(e)}$$ d $$^{(e)}$$ = f $$^{(e)}$$   (1) Where each represents a 4x4, 4x1, and 4x1 matrix, respectively.

This is represented by the following diagram.

Also, we can treat this element as a spring element turned 45° with the following analysis.

The reaction forces have been taken along the direction of the element so that we can analyze the stretching or shrinkage of the bar element as a force is applied on it. The p vectors translate to be the force on the node and the q vectors are the displacement of the node as the bar stretches or shrinks. These pertain to the FD relationship mentioned above as:

$$\hat{\mathbf{k}}^{(e)} \mathbf{\hat{q}}^{(e)} = \hat{\mathbf{q}}^{(e)}$$ (2)

In long-hand form:

$$k^{(e)} \begin{bmatrix} 1 & -1 \\ -1 & 1\\ \end{bmatrix} \begin{pmatrix} q^{(e)}_{1}\\ q^{(e)}_{2} \end{pmatrix} = \begin{pmatrix} p^{(e)}_{1}\\ p^{(e)}_{2} \end{pmatrix}$$

Where $$q^{(e)}_i$$ is the axial displacement of element e at local node i, and $$p^{(e)}_i$$ is the axial force of element e at local node i. Also, these matrices are sized as 2x2, 2x1, and 2x1, respectively.

=Derivation of eqn (1) from eqn (2)=

We want to find the relationship between $$q^{(e)}_{2x1}$$ and $$d^{(e)}_{4x1}$$, and $$p^{(e)}_{2x1}$$ and $$f^{(e)}_{4x1}$$. It can be expressed as follows:

 * $$q^{(e)}_{2x1}$$ = $$T^{(e)}_{2x4}$$ $$d^{(e)}_{4x1}$$

Consider the displacement of node 1, denoted as \vec{d}^{(e)}.

 * $$\vec{d}^{(e)} = d^{e}_{1} \vec{i} + d^{e}_{2} \vec{j}$$

We say that $$q^{(e)}_{1}$$, the axial displacement of local node 1, is the orthogonal projection of the displacement. It is defined as:

$$q^{(e)}_{1}$$ = $$\vec{d}^{(e)} \vec{\tilde{i}}$$
 * =($$d^{e}_{1} \vec{i}$$ + $$d^{e}_{2} \vec{j}$$)•$$\vec{\tilde{i}}$$
 * =$$d^{e}_{1}$$ ($$\vec{i}$$•$$\vec{\tilde{i}}$$) + $$d^{e}_{2}$$ ($$\vec{j}$$•$$\vec{\tilde{i}}$$)

where
 * $$\vec{i}$$•$$\vec{\tilde{i}}$$ = cosθ^{(e)} = l $$^{(e)}$$
 * $$\vec{j}$$•$$\vec{\tilde{i}}$$ = sinθ^{(e)} = m $$^{(e)}$$

Which leads to the construction of the matrix equation that relates $$d$$ and $$q$$:

$$q^{(e)}_{1}$$ = l$$^{(e)}$$ $$d^{e}_{1}$$ + m$$^{(e)}$$ $$d^{e}_{2}$$
 * = $$\begin{bmatrix}

l^{(e)} & m^{(e)}\\ \end{bmatrix}_{1x2}$$ $$\begin{pmatrix} d^{e}_{1}\\ d^{e}_{2}\\ \end{pmatrix}_{2x1}$$

Similarily for node 2:

$$q^{(e)}_{2}$$ = l$$^{(e)}$$ $$d^{e}_{2}$$ + m$$^{(e)}$$ $$d^{e}_{2}$$
 * = $$\begin{bmatrix}

l^{(e)} & m^{(e)}\\ \end{bmatrix}_{1x2}$$ $$\begin{pmatrix} d^{e}_{3}\\ d^{e}_{4}\\ \end{pmatrix}_{2x1}$$

These two matrix equations combine to yield:

 * $$\begin{pmatrix}

q^{e}_{1}\\ q^{e}_{2}\\ \end{pmatrix}_{2x1}$$ = $$\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ \end{bmatrix}_{2x4} \begin{pmatrix} d^{e}_{1}\\ d^{e}_{2}\\ d^{e}_{3}\\ d^{e}_{4}\\ \end{pmatrix}_{4x1}$$

and in short-hand form:

 * $$q^{(e)}$$ = $$T^{(e)}$$ $$d^{(e)}$$


text: File:Bar - spring similarity.JPG

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User:Eml4500.f08.ateam.boggs.t/week 5

Lecture 12

title: User:Eml4500.f08.ateam.boggs.t/week 5


text:
User:Eml4500.f08.ateam.boggs.t/Lecture 12

The equation for a two-force element,

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

can also be expressed with axial forces and displacements where,

$$

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

This can also be expressed as,

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

where,

$$ K^{(e)} = \frac{EA}{L} $$

The relationship between q(e) and d(e) can be expressed in the form of the equation,

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

where q(e) is a 2X1 matrix, T(e) is a 2X4 matrix, and d(e) is a 4X1 matrix.

Consider the displacement vector of local node one, it is denoted by d[1](e) and can be seen at right where,

$$ d_{[1]}^{(e)} = d_{1}^{(e)}\overrightarrow{i} + d_{2}^{(e)}\overrightarrow{j} $$

relating q to d,

$$ q_{1}^{(e)} = d_{[1]}^{(e)}\overrightarrow{\tilde{i}} $$

$$ q_{1}^{(e)} = (d_{1}^{(e)}\overrightarrow{i} + d_{2}^{(e)}\overrightarrow{j})\cdot \overrightarrow{\tilde{i}} $$

$$ q_{1}^{(e)} = d_{1}^{(e)}(\overrightarrow{i}\cdot \overrightarrow{\tilde{i}}) + d_{2}^{(e)}(\overrightarrow{j}\cdot \overrightarrow{\tilde{i}}) $$

Where,

$$ \overrightarrow{i}\cdot \overrightarrow{\tilde{i}} = \cos\theta = l^{(e)} $$

$$ \overrightarrow{j}\cdot \overrightarrow{\tilde{i}} = \sin\theta = m^{(e)} $$

$$ q_{1}^{(e)} = l^{(e)}d_{1}^{(e)} + m^{(e)}d_{2}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{bmatrix} $$

For node 2, denoted by d[2](e),

$$ d_{[2]}^{(e)} = d_{3}^{(e)}\overrightarrow{i} + d_{4}^{(e)}\overrightarrow{j} $$

relating q to d,

$$ q_{2}^{(e)} = d_{[2]}^{(e)}\overrightarrow{\tilde{i}} $$

$$ q_{2}^{(e)} = (d_{3}^{(e)}\overrightarrow{i} + d_{4}^{(e)}\overrightarrow{j})\cdot \overrightarrow{\tilde{i}} $$

$$ q_{2}^{(e)} = d_{3}^{(e)}(\overrightarrow{i}\cdot \overrightarrow{\tilde{i}}) + d_{4}^{(e)}(\overrightarrow{j}\cdot \overrightarrow{\tilde{i}}) $$

$$ q_{2}^{(e)} = l^{(e)}d_{3}^{(e)} + m^{(e)}d_{4}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} d_{3}^{(e)}\\ d_{4}^{(e)} \end{bmatrix} $$

Therefore,

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

Here q(e) is a 2X1 matrix, T(e) is a 2X4 matrix, and d(e) is a 4X1 matrix.


text: File:Loa pic.JPG

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title: File:Uniform Circular2.png


text:
User:Eas4200c.f08.vqcrew.E/Homework3

Torsion Acting on a Uniform Bar with Circular Cross-Section
This case does not exhibit warping. The only deformation that occurs is contained within the y-z plane. The beam behaves as a series of rigid disks rotating entirely in the plane of the cross-section. Figure ## displays the reaction of the beam to a given torque.

The general formula for the torque is:

 * $$T = \iint_A r \tau dA $$

The shear stress is defined as the product of the shear modulus, G, and the angle of twist,  $$\gamma$$  , (Hooke's Law).

 * $$\tau = G \gamma  $$

 * $$\displaystyle \gamma = \displaystyle \frac{rd\alpha}{dx} = r\theta$$

Define $$\theta$$ as rate of twist:

 * $$\theta := \frac{d\alpha}{dx}$$

Substituting in, the torque is found to be:

 * $$\displaystyle T = \int_A rG(r\theta)dA = G\theta(\int_A r^{2} dA)$$

Where:   $$\displaystyle dA = dydz = rdrd\theta $$

The $$2^{nd}$$ polar moment of inertia for the beam is equal to the second term in the expression of the torque, $$\displaystyle J = (\int_A r^{2} dA)$$.

Thin-Walled Hollow Circular Cross-Section
Figure ## shows two circular cross-sections, one solid and the other thin-walled, with the same area. The two cross-sections have different $$2^{nd}$$ polar moments of inertia.

For the thin-walled cross-section,

Inner radius: $$\displaystyle r_i = a$$

Outer radius: $$\displaystyle r_o = b$$

The $$2^{nd}$$ polar moment of inertia for the thin-walled cross-section can be determined through the expression:

 * $$\displaystyle J = \frac{1}{2}\pi(b^{4}-a^{4}) = \frac{1}{2}\pi(b - a)(b + a)(b^{2} + a^{2})$$

Define $$\displaystyle \bar{r}$$ as the average radius:

 * $$\displaystyle \bar{r} = \frac{a+b}{2}$$

Define $$\displaystyle t$$ as the thickness:

 * $$\displaystyle t = (a - b)$$

Substitute into the $$2^{nd}$$ polar moment of inertia equation to get:

 * $$\displaystyle J = \frac{1}{2}\pi(t)(2\bar{r})(b^{2} + a^{2}) = \pi(t)(\bar{r})(b^{2} + a^{2})$$

Approximate and substitute:

$$\displaystyle a^{2} \cong \bar{r}^{2}$$

$$\displaystyle b^{2} \cong \bar{r}^{2}$$

 * $$\displaystyle \Rightarrow J = \pi t \bar{r}(\bar{r}^{2} + \bar{r}^{2})= 2\pi t \bar{r}^{3}$$

This result can be separated to allow for a proportional factor to be found between the $$2^{nd}$$ polar moment of inertia and the the average area, $$\displaystyle \bar{A}$$.

$$\displaystyle \bar{A} = \pi\bar{r}^{2}$$

 * $$\displaystyle \Rightarrow J = (2\pi^{-\frac{1}{2}}t)(\pi\bar{r}^{2})^{\frac{3}{2}} = (2\pi^{-\frac{1}{2}}t)(\bar{A})^{\frac{3}{2}}$$

In this form, it is clear that $$\displaystyle J$$ is proprtional to $$\displaystyle \bar{A}^{\frac{3}{2}}$$ by a factor of $$\displaystyle (2\pi^{-\frac{1}{2}}t)$$.

title: User:Eas4200c.f08.vqcrew.E/Homework3


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User:Eml4500.f08.Ateam.philbin/HW3

Homework 3

Mtg. 17 10-05-08

Q: How to sum the moment about point B?

A: The moment is the same as any point along the line of action of the force. In this example, the point A' is used to simplify the cross product since the vector $$ \vec {BA'}$$ > is perpendicular to $$\vec F$$ >.

$$\sum M_B = \vec {AB} x  \vec F$$

$$\sum M_B = \vec {BA'} x  \vec F $$

$$ \vec {BA'} = \vec {BA} + \vec {AA'}$$

$$ M_B = (\vec {BA} + \vec AA') x  \vec F $$

$$ M_B = \vec {BA} x \vec F + \vec {AA'} x  \vec F  = 0$$

3-bar Problem

In this problem, there are three bars instead of two in the last example. The numbering of the elements and nodes is important to make the subsequent matrix manipulations as simple as possible. As a rule of thumb, one should use the overlapping numbering system. Put local node 1 on local node 2 of the previous element, if possible. This is shown in the following image.

The angle theta is determine by the vector going from local node 1 to 2 in each element. Therefore the angle from local node 1 to 2 in element three is pointing down and the corresponding angle is negative. By following these conventions when the global stiffness matrix is made, the three elements add into the same place in order to solve for the unknown forces. In this case, they all add into the columns and rows 3 and 4.

The finished matrix without numerical values is given in the following figure.

 \underline K= \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)}&0&0&0&0\\ k_{21}^{(1)} &k_{22}^{(1)} &k_{23}^{(1)} &k_{24}^{(1)} &0 &0 &0 &0 \\ k_{31}^{(1)} &k_{32}^{(1)} &k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)} &k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)} &k_{13}^{(2)} &k_{14}^{(2)} &k_{13}^{(3)} &k_{14}^{(3)}\\ k_{41}^{(1)} &k_{42}^{(1)} &k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)} &k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)} &k_{23}^{(2)} &k_{24}^{(2)} &k_{23}^{(3)} &k_{24}^{(3)}\\ 0 &0 &k_{31}^{(2)} &k_{32}^{(2)} &k_{33}^{(2)} &k_{34}^{(1)} &0 &0\\ 0 &0 &k_{41}^{(2)} &k_{42}^{(2)} &k_{43}^{(2)} &k_{44}^{(1)} &0 &0\\ 0 &0 &k_{31}^{(3)} &k_{32}^{(3)} &0 &0 &k_{33}^{(3)} &k_{34}^{(3)}\\ 0 &0 &k_{41}^{(3)} &k_{42}^{(3)} &0 &0 &k_{43}^{(3)} &k_{44}^{(3)}\\ \end{bmatrix}

title: User:Eml4500.f08.Ateam.philbin/HW3


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