User:Eml4500.f08.group.w

=Finite Element Analysis=

Course Information

Website - www.mae.ufl.edu/~vql → "teaching" → "my course website" → "EML 4500 Finite Element Anslysis and Design" EML 4500 Course website Password: fall08.wiki

MWF Period 8 (3:00-3:50PM), FLG 270 Office: NEB135

Course Content

Finite Element Analysis (FEM) makes solving partial differential equations (PDEs) convenient. This is integral to the study of Aerospace Structures, or its structural components.

The emphasis of the course will be to understand the mechanics, formulate problems, judge solution correctness, and avoiding ad-hoc structural analysis methods.

The plan of the course is to maintain confidentiality while using large-scale peer-to-peer domains including Wikiversity and MIT's OpenCourseWare in addition to the textbook. The course will examine the method of work instead of other approaches and such as e-learning and the old approach to EML4500 (10% HW and 30% x 3 for each exam). This format allows students to be introduced to Media Wiki and stimulate collaboration between students with other students, teams, the professor, and the class as a whole.

There are several relevant time zones important to us (UTC, EDT, and EST). HW will be turned in according to the UTC time to maintain consistency with the Wikipedia time stamp format.

HW shortcuts should be created by attaching the URL from the archived version by going to "History" → click on the date to open archived version → copy this address to submit

=Chapter 4: Trusses, Beams, and Frames=

Trusses
The image below is a two-truss system with elastic (deformable) bars.



Global node numbers are denoted by circles while element numbers are denoted by triangles. The elements are separable members of a system which can be individually analyzed. Nodes are points on a system on which forces appear. The x-y coordinate system can be located in any rotation about the z-axis, however in this case the y-axis is chosen to be parallel to the applied force P. Furthermore, the triangles beneath nodes 1 and 2 denote that the node displacement is fixed to zero in both the x- and y-direction.

Forces acting on the system are shown below with a systematic example of how to name the forces on the member. The names given to each of forces are arbitrary, but must be systematic and consistent.



The R1 and R3 forces can be replaced with two forces instead of four that act along the line of the trusses. Taking the moment about node 2 equal to zero, gives no conclusion because both R1 and R3 forces run through the node. The figure then has two unknown reactions, and only two equations of equilibrium (the sum of forces about the x- and y-axes. This means that the system is statically indeterminate (equations of statics cannot be used to solve R1 and R3.

To examine the forces, the system can be split into two independent bar elements.

Bar element 1:



This picture is another example of how to name the forces, nodes and elements in a systematic manner, rather than a rigid, consistent manner. The element number is denoted by a triangle again. The local nodes (nodes relative only to the specific FBD section) are denoted by squares as opposed to the global nodes (nodes relative to the entire structure) which are circled. The subscript of the forces are the degrees of freedom (dof) or internal force of the element. Each force is a separate degree of freedom and should have their own number. Also they are numbered in a consistent order: x-, y-, x-, y-axis. The superscript represent the element number in parentheses. For this global system, the dof's are 1, 2, 3, and 4 and the element numbers are 1, 2, which come to eight total numbers in both elements.

Bar element 2:



Force Displacement (FD) Relation


The diagram above shows the relationship between force and displacement in a spring with one end fixed. It both ends are free, the case becomes as follows.



Because there are now two independent variables, the spring must be analyzed with a matrix form of F = kd. Matrices are names by "row x column"



\begin{pmatrix} f_1 \\ f_2 \end{pmatrix}_{2x1} = \begin{bmatrix} k & -k \\ -k & k \end{bmatrix}_{2x2} \begin{pmatrix} d_1 \\ d_2 \end{pmatrix}_{2x1} $$

There are 2 cases which can be examined:

Case 1 is the result with respect to node 1:

$$f_2 = k (d_2 - d_1)$$

Case 2 is the result with respect to node 2:

$$f_1 + f_2 = 0$$

→ $$f_1 = -f_2 = -k (d_2 - d_1) = k (d_1 - d_2)$$


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!Contribution title, from team group
 * Text
 * }
 * }

=MATLAB Primer=

Hard copy
Command:  filename

This command pulls up the hardcopy and copies the content excluding graphics to that file, and will go to a default file if no filename is given. It continues to copy to the file until you command it to  or.

Graphics
MATLAB plots curves, 3D plots of curves , #D mesh surface plots , and 3D faceted surface plots. allows you to preview these features.

Planar plots
The command  creates an x-y plot as shown below the format for the range is 'lower rang:meshsize: upper range'.

Figures can be suppressed or exposed as the "current" figure by the command according to its figure number, and command   will give you the number of the current figure.

To make graphs, first put a function in an M-file then command it open with a domain:

Parametric graphs can also be defined as shown:

To enter text in the graph, follow these commands:

title of graph x-axis label y-axis label text on graph (placement defined by mouse) text position at specific coordinates places grid on the graph axis limits makes standard axis for all graphs auto-scaling shows current scaling for vector v puts same scale on both axes puts same scale and tic marks on both axes turns off both axis scaling and tic marks turns on both axis scaling and tic marks

Put the  commands after the   command.

To make multiple plots on a graph the format may be one of the two as follows

You can also command  to freeze the current plot and from there superimpose another plot on the same graph. This can be turned off with. To make line types use the codes: solid(-), dashed (--), dotted, dashdot (-.). To make marktypes, use the codes: point(.), plus (+), star (*), circle (o), x-mark (x). To make colors, use the codes: yellow(y), magent(m), cyan(c), red(r), green(g), blue(b), white(w), black(k). An example is a green dotted line:.

Several other features are:

Graphics hardcopy
To send the graphics to the printer, command, while   filename saves the figure to the specified format or to a default format if not specified.

3-D line plots
An example format of a 3-D line plot is as follows:

A 3-D graph has all the same commands plus.

3-D mesh and surface plots
3-D mesh plots shows a plots of matrix z with the command. A surface plot is controlled by. To obtain a mesh graph of a function over a rectangle, an example is as follows:

The shading can be  and are entered after the   command. Color profiles may be:.

Handle Graphics
To see a more complete list of the options to control a graph enter  and.

Sparse matrix computations
=HW3=

If you use the same argument, $$\begin{Bmatrix} p^{(e)}_{1}\\ p^{(e)}_{2} \end{Bmatrix}_{2x1} = \bar{T}^{(e)}_{2x4} \begin{Bmatrix} f^{(e)}_{1}\\ f^{(e)}_{2}\\ f^{(e)}_{3}\\ f^{(e)}_{4} \end{Bmatrix}_{4x1}$$

This yields the symbolic simplification:

P (e) = T (e) f (e)

For each node, the relationship can be expressed as:

q (e)2x1 = T (e)2x4 d (e)4x1

Substituting variables into the element axial Force-Distance Relationship:

$$\hat{k}^{(e)}_{2x2} q^{(e)}_{2x1}= p^{(e)}_{2x1}$$ → $$\hat{k}^{(e)} (T^{(e)}d^{(e)})= (T^{(e)}f^{(e)})$$

The goal is to have k (e) d (e) = f (e). Therefore, we want to move the T (e) from the right hand side to the left hand side by multiplying both sides by T (e)-1 (or the inverse of T). But because T is not a square matrix, it cannot be inverted.

k (e) d (e)= f (e) → $$[T^{(e)T}_{4x2} (transpose) \hat{k}^{(e)}_{2x2}T^{(e)}_{2x4}]_{(4x4)}d^{(e)}_{4x1} = f^{(e)}_{4x1}$$

HW: Verify that $$k^{(e)} = T^{(e)T}\hat{k}^{(e)}T^{(e)}$$ $$\bar{T}^{(e)T}k^{(e)}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)} \end{bmatrix}k^{(e)}\begin{bmatrix} l^{(e)} & m^{(e)} & -l^{(e)} & -m^{(e)}\\ -l^{(e)} & -m^{(e)} & l^{(e)} & m^{(e)} \end{bmatrix} = k^{(e)}\begin{bmatrix} (l^{(e)})^{2} & l^{(e)}m^{(e)} & -(l^{(e)})^{2} & -(l^{(e)}m^{(e)})\\ l^{(e)}m^{(e)} & (m^{(2)})^2 & -(l^{(e)}m^{(e)}) & -(m^{(2)})^2\\ -(l^{(e)})^{2} & -(l^{(e)}m^{(e)}) & (l^{(e)})^{2} & l^{(e)}m^{(e)}\\ -(l^{(e)}m^{(e)}) & -(m^{(2)})^2 & l^{(e)}m^{(e)} & (m^{(2)})^2 \end{bmatrix} = \bar{k}^{(e)}$$

Principle of Virtual Work: K 6x6 d 6x1 = F 6x1 → K' 2x2 d' 2x1 = F' 2x1

We cannot solve for d as d = K -1 F because of the singularity of K -1 where the det[ K ]=0 and therefore we cannot invert K.

In 2D motion, there are three possible rigid body motions: two in translation and one rotation

HW: To find the eigenvalues: $$det\begin{bmatrix} k_{11}^{(1)} -\lambda & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\ k_{21}^{(1)} & k_{22}^{(1)} - \lambda & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\ k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)}  k_{11}^{(2)}) - \lambda & (k_{34}^{(1)}   k_{12}^{(2)}) & k_{13}^{(2)} & k_{14}^{(2)}\\ k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)}  k_{21}^{(2)}) & (k_{44}^{(1)}   k_{22}^{(2)}) - \lambda  & k_{23}^{(2)} & k_{24}^{(2)}\\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} - \lambda & k_{34}^{(2)}\\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} - \lambda \end{bmatrix} = 0$$ This can be reduced to (because of the Principle of Virtual Work) $$det\begin{bmatrix} (k_{33}^{(1)} + k_{11}^{(2)}) - \lambda & (k_{34}^{(1)} + k_{12}^{(2)})\\ (k_{43}^{(1)} + k_{21}^{(2)}) & (k_{44}^{(1)} + k_{22}^{(2)}) - \lambda \end{bmatrix} = 0$$ Solving becomes $$[(k_{33}^{(1)} + k_{11}^{(2)}) - \lambda]*[(k_{44}^{(1)} + k_{22}^{(2)}) - \lambda] - [(k_{34}^{(1)} + k_{12}^{(2)})(k_{43}^{(1)} + k_{21}^{(2)})] = 0$$ But because K43=K34, this can be written as $$\lambda ^{2}- \lambda(k_{33}^{(1)} + k_{11}^{(2)} + k_{44}^{(1)} + k_{22}^{(2)}) + [(k_{33}^{(1)} + k_{11}^{(2)})(k_{44}^{(1)} + k_{22}^{(2)}) - (k_{34}^{(1)} + k_{12}^{(2)})^{2}] = 0 $$ Therefore, the eigenvalues are λ = 0,$$(k_{33}^{(1)} + k_{11}^{(2)})(k_{44}^{(1)} + k_{22}^{(2)}) - (k_{34}^{(1)} + k_{12}^{(2)})^{2}$$, and $$k_{33}^{(1)} + k_{11}^{(2)} + k_{44}^{(1)} + k_{22}^{(2)}$$ There is a zero eigenvalue because there is no rotational component of the system. This means that there is no elastic energy stored in the structure in the rotational mode. → Dynamic eigenvalue as realted to vibration frequency: K v =λ M  v, where K is the stiffness matrix, λ is the eigenvalue, and M is the mass matrix.

=HW 4=

Lecture 19
$$\mathbf{\tilde{d}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{d}^{(e)}_{4x1} \Rightarrow \begin{Bmatrix} \tilde{d^{(e)}_{1}}\\ \tilde{d^{(e)}_{2}}\\ \tilde{d^{(e)}_{3}}\\ \tilde{d^{(e)}_{4}} \end{Bmatrix} = \begin{bmatrix} \textbf{R}^{(e)}_{2x2} & \textbf{0}_{2x2}\\ \textbf{0}_{2x2} & \textbf{R}^{(e)}_{2x2} \end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\\ d^{(e)}_{3}\\ d^{(e)}_{4} \end{Bmatrix}$$



$$\mathbf{\tilde{f}}^{(e)} = k^{(e)}\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\mathbf{\tilde{d}}^{(e)}$$

$$\mathbf{\tilde{f}}^{(e)}_{4x1} = \mathbf{\tilde{k}}^{(e)}_{4x4}\mathbf{\tilde{d}}^{(e)}_{4x1}$$

Considering the case where $$\mathbf{\tilde{d}}^{(e)}_{4} \neq 0 \Rightarrow \mathbf{\tilde{d}}^{(e)}_{1} = \mathbf{\tilde{d}}^{(e)}_{2} = \mathbf{\tilde{d}}^{(e)}_{3} = 0$$

$$\mathbf{\tilde{f}}^{(e)}_{4x1} = \mathbf{\tilde{k}}^{(e)}_{4x4}\mathbf{\tilde{d}}^{(e)}_{4x1} = \textbf{0}_{4x1}$$, the 4th column of the stiffness matrix.

Interpretation of Transverse Degrees of Freedom

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!HW Force relationship, from team group
 * Using the relationship between distance and T, we can find the relationship between force and T: $$\mathbf{\tilde{d}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{d}^{(e)}_{4x1}$$:
 * Using the relationship between distance and T, we can find the relationship between force and T: $$\mathbf{\tilde{d}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{d}^{(e)}_{4x1}$$:

$$\mathbf{\tilde{f}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{f}^{(e)}_{4x1} \Rightarrow \begin{Bmatrix} \tilde{f^{(e)}_{1}}\\ \tilde{f^{(e)}_{2}}\\ \tilde{f^{(e)}_{3}}\\ \tilde{f^{(e)}_{4}} \end{Bmatrix} = \begin{bmatrix} \textbf{R}^{(e)}_{2x2} & \textbf{0}_{2x2}\\ \textbf{0}_{2x2} & \textbf{R}^{(e)}_{2x2} \end{bmatrix}\begin{Bmatrix} f^{(e)}_{1}\\ f^{(e)}_{2}\\ f^{(e)}_{3}\\ f^{(e)}_{4} \end{Bmatrix}$$

contribution by carbon.w
 * }

In addition, using the force-distance relationship:

$$\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)} \Rightarrow \mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}_{(e)}\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{T}}_{(e)}\mathbf{\tilde{f}}^{(e)}$$

If $$\mathbf{\tilde{T}}_{(e)}$$ is invertible, then the relationship becomes:

$$[\mathbf{\tilde{T}}^{(e)-1}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)}]\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)}$$

$$\mathbf{\tilde{T}}_{(e)}$$ is a block diagonal matrix. The general form of a block diagonal matrix is as follows:

$$\textbf{A} = \begin{bmatrix} \textbf{D}_{1} & & \textbf{0}\\ & ... & \\ \textbf{0} & & \mathbf{D_{s}} \end{bmatrix}$$

A simpler example of the diagonal matrix to display more clearly the format:

$$\textbf{B} = \begin{bmatrix} d_{11} & & & \textbf{0}\\ & d_{22} & & \\ & & ... & \\ \textbf{0} & & & d_{nn} \end{bmatrix} = Diag[d_{11}, d_{22},...,d_{nn}]$$

$$\mathbf{B^{-1}} = Diag[\frac{1}{d_{11}}, \frac{1}{d_{22}},...,\frac{1}{d_{nn}}]$$

If we assume that dii does not equal zero for i=1,...,n and a block diagonal matrix would be:

$$\mathbf{A} = Diag[\mathbf{D_{1}},...,\mathbf{D_{s}}]$$

$$\mathbf{A^{-1}} = Diag[\mathbf{D^{-1}_{1}},...,\mathbf{D^{-1}_{s}}]$$

$$\mathbf{T^{(e)-1}} = Diag[\mathbf{R^{(e)-1}},\mathbf{R^{(e)-1}}]$$

$$\mathbf{R^{(e)T}} = \begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)} & l^{(e)} \end{bmatrix} * \begin{bmatrix} l^{(e)} & m(e)\\ -m^{(e)} & l^{(e)} \end{bmatrix} = [(l^{(e)})^{2}  (m^{(e)})^{2}][(-l^{(e)}m^{(e)})   (m^{(e)}l^{(e)}]$$

$$\mathbf{R^{(e)T}}*\mathbf{R^{(e)}} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}_{2x2} = \mathbf{\bar{I}_{2x2}}$$ (the identity matrix)

$$\Rightarrow \mathbf{R^{(e)-1}} = \mathbf{R^{(e)T}}$$

$$\mathbf{\tilde{T}^{(e)-1}} = Diag[\mathbf{R^{(e)T}}, \mathbf{R^{(e)T}}] = (Diag[\mathbf{R^{(e)}}, \mathbf{R^{(e)}}])^{T}$$

$$\Rightarrow \mathbf{\tilde{T}^{(e)-1}} = \mathbf{\tilde{T}^{(e)T}}$$


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!HW Force relationship, from team group
 * Using the relationship shown above,
 * Using the relationship shown above,

$$[\mathbf{\tilde{T}}^{(e)-1}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)}]\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{T}}_{(e)}\mathbf{\tilde{f}}^{(e)}$$,

and plugging in $$\mathbf{\tilde{T}^{(e)-1}} = \mathbf{\tilde{T}^{(e)T}}$$, we arrive at:

$$[\mathbf{\tilde{T}}^{(e)T}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)}]\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)}$$

$$\mathbf{k}^{(e)} = \mathbf{\tilde{T}}^{(e)-1}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)} $$

contribution by carbon.w
 * }

=HW6=

Stiffness term:





In the figure above, assuming the displacement u(x) for $$x_{i} \leq x \leq x_{i+1}$$ (for example, $$x\in [x_{i}, x_{i+1}]$$, where the symbol $$\in $$ means "belongs to". The figure shows the interpolation of u(x) where there is only axial displacement (or, no transverse displacement)



The motivation for linear interpolation of u(x)
In a 2-bar truss the deformation can be shown as:



The deformed shape is a straight line, meaning that there is an implicit asumption of linear interpolation of displacement between two nodes.

u(x) can be expressed in terms of $$d_{i} = u(x_{i})$$ and $$d_{i+1} = u(x_{i+1})$$ as a linear function in x (for example, linear interpolation) as follows:

$$u(x) = N_{i}(x)d_{i} + N_{i+1}(x)d_{i+1}$$

where $$N_{i}(x)$$ and $$N_{i+1}$$ are linear functions in x.

$$N_{i} = \frac {x-x_{i+1}}{x_{i} - x_{i+1}}$$

$$N_{i+1} = \frac {x - x_{i}}{x_{x+1} - x_{i}}$$

Therefore, $$u(x) = \frac {x-x_{i+1}}{x_{i} - x_{i+1}}d_{i} + \frac {x - x_{i}}{x_{x+1} - x_{i}}d_{i+1}$$



$$\mathbf{k^{(i)}} = \int_{x=0}^{\tilde{x} = L^{(i)}}{\mathbf{B^{T}}(\tilde{x})(EA)(\tilde{x})\mathbf{B}(\tilde{x})d\tilde{x}}$$


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!HW Expression for $$\mathbf{k^{(i)}}$$, from team group
 * The expression for $$\mathbf{k^{(i)}}$$ can be solved as such:
 * The expression for $$\mathbf{k^{(i)}}$$ can be solved as such:

$$\mathbf{k^{(i)}} = \int_{x=0}^{\tilde{x} = L^{(i)}}{\mathbf{B^{T}}(\tilde{x})(EA)(\tilde{x})\mathbf{B}(\tilde{x})d\tilde{x}} = \frac {EA}{\tilde{x}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}^{L^{(i)}}_{0} = \frac {EA}{L^{(i)}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

The equation of the area is: $$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

The equation of the modulus of elasticity is: $$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$



Substituting the equations above into the stiffness matrix, is as follows:

$$\mathbf{k^{(i)}} = \frac {[N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2][N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2]}{L^{(i)}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

contribution by carbon.w
 * }

=HW6 MATLAB=

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

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

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



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

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

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

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

Consider the node at (25, 30):



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

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

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

Lecture 36
Below is a deformable truss with horizontal, vertical and rotational displacements:



The matrix formulas are as follows:

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

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

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

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

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

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

The dimensional analysis is as follows:

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

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



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

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

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

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