User:Uf.team.wiki/HW2

Introduction
The first step to solving a 2 truss system is to look at the picture from each element, and use the derived recipe to help solve.

Recall: Use the Derived "Recipe" to Simplify the Problem
Recalling the recipe will help us solve Truss problems with the use of FEA requirements and notation.

Statically Determinate: Problem 1.2 (continued)
The following is an example of a two member truss system which is known to be statically determinate. Given the specified information, and since it is statically determinate, this problem can be setup and solved using simple statics to determine any unknown forces, and moments. Solving this problem will also provide us with further exposure to FEAD notation in problem solving.

Figure 1: Given Two Member Truss System


 * This problem can be solved using the knowledge of two forces members. It simplifies the unknowns so we have the same number of equations and unknowns. This means that we would not have to use the Finite Element Method (FEM). This problem can be solved by simply using a statics method learned previously. The system will generate forces that act along the members themselves which will reduce unknowns.

=Step 1: Global Picture=

Figure 2: Global Forces of Two Member Truss System


 * The global forces are shown here in the global coordinate system. We see both the applied forces and the reactions forces. We see the additional unknowns (reaction forces) added to the system by removing the constraints of the ground connections.

=Step 2: Elemental Picture=  Element 1  Element 1: θ(1)=30°

Figure 3: First Element Forces

 Element 2  θ(1)=-45°
 * The element is broken down and the element forces are displayed. With this we can now determine the magnitudes and direction of the combined forces. We see here that at the pin connection we display the reaction forces created by the separation of the two elements. With the two force member method we can solve the system

Figure 4: Second Element Forces


 * The second element is broken down and the element forces are displayed. With this we can now determine the magnitudes and direction of the combined forces. We see here that at the pin connection we display the reaction forces created by the separation of the two elements. With the two force member method we can solve the system

Problem Setup
The elemental picture includes a stiffness matrix ' k ', a displacement matrix ' d ', and a force matrix ' f '. k (e) X d (e) = f (e)

For this problem there are two elements 1 and 2 (e=1,2).


 * k (e) is considered a 4 X 4 matrix
 * d (e) is considered a 4 X 1 matrix
 * f (e) is considered a 4 X 1 matrix



Figure 5: Direction Cosine Coordinate System Solve for k : $$\underset{}{k^{(e)}} \begin{bmatrix} (l^{(e)})^{2} & l^{(e)}m^{(e)}  &  -(l^{(e)})^{2}  &  -l^{(e)}m^{(e)}   \\ l^{(e)}m^{(e)} &  (m^{(e)})^{2}  &  -l^{(e)}m^{(e)}  &  -(m^{(e)})^{2}   \\ -(l^{(e)})^{2} & -l^{(e)}m^{(e)}  &  (l^{(e)})^{2}  &  l^{(e)}m^{(e)} \\ -l^{(e)}m^{(e)} &  -(m^{(e)})^{2}  &  l^{(e)}m^{(e)}  &  (m^{(e)})^{2} \\ \end{bmatrix} = \underset{--}{k^{(e)}}$$ k(e) = $$\frac{E^e A^e}{L^e }$$:

where,


 * E(e) is the Young's Modulus of each element, e
 * A(e) is the cross-sectional area of element, e
 * L(e) is the length of each element, e

$$k^1 = \frac{3(1)}{4 }=\frac{3}{4}$$, $$k^2 = \frac{5(2)}{2 } = 5$$

l(e)=cos(θe): $$l^1 = \frac{\sqrt{3}}{2}$$, $$l^2 = \frac{1}{\sqrt{2}}$$ m(e)=sin(θe): $$m^1 = \frac{1}{2}$$, $$m^2 = -\frac{1}{\sqrt{2}}$$ Here are the k (1) and k (2) matrices:

k (1) = $$\begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} &-\frac{3\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16}& -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix} $$ k (2) = $$\begin{bmatrix} \frac{5}{2}& -\frac{5}{2} &-\frac{5}{2} &\frac{5}{2}\\ -\frac{5}{2}& \frac{5}{2} & \frac{5}{2} &-\frac{5}{2} \\ -\frac{5}{2}& \frac{5}{2} &\frac{5}{2} &-\frac{5}{2}\\ \frac{5}{2}& -\frac{5}{2} & -\frac{5}{2} &\frac{5}{2} \end{bmatrix}$$

= Step 3: Global Force Displacement Relationships = $$\begin{bmatrix} K_{11} &  K_{12}  &  K_{13}  &  K_{14}  &  K_{15}  &  K_{16} \\ K_{21} &  K_{22}  &  K_{23}  &  K_{24}  &  K_{25}  &  K_{26} \\ K_{31} &  K_{32}  &  K_{33}  &  K_{34}  &  K_{35}  &  K_{36} \\ K_{41} &  K_{42}  &  K_{43}  &  K_{44}  &  K_{45}  &  K_{46} \\ K_{51} &  K_{52}  &  K_{53}  &  K_{54}  &  K_{55}  &  K_{56} \\ K_{61} &  K_{62}  &  K_{63}  &  K_{64}  &  K_{65}  &  K_{66} \\ \end{bmatrix} \begin{Bmatrix} d_{1}\\ d_{2}\\ d_{3}\\ d_{4}\\ d_{5}\\ d_{6}\\ \end{Bmatrix} = \begin{Bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}\\ \end{Bmatrix} $$

In Compact Notation:

$$ \left[K_\underset{6 \times 6}{ij} \right]\left\{d_\underset{6 \times 1}{j} \right\}=\left\{F_\underset{6 \times 1}{i} \right\} $$

More General Form of nxn Matrix:

$$ \sum_{j=1}^{6}{K_{ij}d_{j}} = F_{i} $$ ,where i = 1,...,6

$$ \mathbf{\underset{n \times n}K}=\left[K_\underset{n \times n}{ij} \right] $$ = Global Stifness Matrix

$$ \mathbf{\underset{n \times 1}d}=\left\{d_\underset{n \times 1}{j} \right\} $$ = Global Displacement Matrix

$$ \mathbf{\underset{n \times 1}F}=\left\{F_\underset{n \times 1}{i} \right\} $$ = Global Force Matrix

Recall the Element Force Displacement Relation:

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

$$ \mathbf{\underset{4 \times 4}k^{(e)}}=\left[k^{(e)}_{ij} \right] $$ = Element Stifness Matrix

$$ \mathbf{\underset{4 \times 1}d^{(e)}}=\left\{d_{i}^{(e)} \right\} $$ = Element Displacement Matrix

$$ \mathbf{\underset{4 \times 1}f^{(e)}}=\left\{f_{i}^{(e)} \right\} $$ = Element Force Matrix

How to go from element matricies (stiffness, displacement, force) to global matrix?

Through an assembly process:
 * Identify the correspondence between element displacement definitions and global displacement definitions.

Global Level:

$$ \left\{d_{1},d_{2},d_{3},d_{4},d_{5},d_{6} \right\} $$

Element Level: $$ \left\{d_{1}^{(1)},d_{2}^{(1)},d_{3}^{(1)},d_{4}^{(1)},d_{5}^{(1)},d_{6}^{(1)} \right\} $$ $$ \left\{d_{1}^{(2)},d_{2}^{(2)},d_{3}^{(2)},d_{4}^{(2)},d_{5}^{(2)},d_{6}^{(2)} \right\} $$
 * Element 1
 * Element 2

Identification of Global to Local Degrees of Freedoms:

$$ \left.\begin{matrix} d_{1} = d_{1}^{(1)}\\ d_{2} = d_{2}^{(1)}\\ \end{matrix}\right\} $$ = Node (1)

$$ \left.\begin{matrix} d_{3} = d_{3}^{(1)} = d_{1}^{(2)}\\ d_{4} = d_{4}^{(1)} = d_{2}^{(2)}\\ \end{matrix}\right\} $$ = Node (2)

$$ \left.\begin{matrix} d_{5} = d_{3}^{(2)}\\ d_{6} = d_{4}^{(2)}\\ \end{matrix}\right\} $$ = Node (3)

Conceptual Step of Assembly:

(Topology of K)

$$\begin{Bmatrix} d_{1}\\ d_{2}\\ d_{3}\\ d_{4}\\ d_{5}\\ d_{6}\\ \end{Bmatrix} = \begin{Bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}\\ \end{Bmatrix} $$

In the Global Stiffness Matrix, The first Element Stiffness Matrix occupies the top left, while the second Element Stiffness Matrix occupies the bottom left. The two Element Stiffness Matrices overlap in the middle, where the values are summed. The resulting empty locations in the Global Stiffness Matrix are zero.

$$ \begin{bmatrix} k_{11}^{(1)} &  k_{12}^{(1)}  &  k_{13}^{(1)}  &  k_{14}^{(1)}  &  0  &  0 \\ k_{21}^{(1)} &  k_{22}^{(1)}  &  k_{23}^{(1)}  &  k_{24}^{(1)}  &  0  &  0 \\ k_{31}^{(1)} &  k_{32}^{(1)}  &  k_{33}^{(1)}+k_{11}^{(2)}  &  k_{34}^{(1)}+k_{12}^{(2)}  &  k_{13}^{(2)}  &  k_{14}^{(2)} \\ k_{41}^{(1)} &  k_{42}^{(1)}  &  k_{43}^{(1)}+k_{21}^{(2)}  &  k_{44}^{(1)}+k_{22}^{(2)}  &  k_{23}^{(2)}  &  k_{24}^{(2)} \\ 0 &  0  &  k_{31}^{(2)}  &  k_{32}^{(2)}  &  k_{33}^{(2)}  &  k_{34}^{(2)} \\ 0 &  0  &  k_{41}^{(2)}  &  k_{42}^{(2)}  &  k_{43}^{(2)}  &  k_{44}^{(2)} \\ \end{bmatrix} = \begin{bmatrix} \frac{9}{16} &  \frac{3\sqrt{3}}{16}  &  -\frac{9}{16}  &  -\frac{3\sqrt{3}}{16}  &  0  &  0 \\ \frac{3\sqrt{3}}{16} &  \frac{3}{16}  &  -\frac{3\sqrt{3}}{16}  &  -\frac{3}{16}  &  0  &  0 \\ -\frac{9}{16} &  -\frac{3\sqrt{3}}{16}  &  \frac{49}{16}  &  \frac{3\sqrt{3}-40}{16}  &  -\frac{5}{2}  &  \frac{5}{2} \\ -\frac{3\sqrt{3}}{16} &  -\frac{3}{16}  &  \frac{3\sqrt{3}-40}{16}  &  \frac{43}{16}  &  \frac{5}{2}  &  -\frac{5}{2} \\ 0 &  0  &  -\frac{5}{2}  &  \frac{5}{2}  &  \frac{5}{2}  &  -\frac{5}{2} \\ 0 &  0  &  \frac{5}{2}  &  -\frac{5}{2}  &  -\frac{5}{2}  &  \frac{5}{2} \\ \end{bmatrix} = \begin{bmatrix} 0.5625 &  0.3248  &  -0.5625  &  -0.3248  &  0  &  0 \\ 0.3248  &  0.1875  &  -0.3248  &  -0.1875  &  0  &  0 \\ -0.5625  &  -0.3248  &  3.0625  &  -2.1752  &  -2.5  &  2.5 \\ -0.3248  &  -0.1875  &  -2.1752  &  2.6875  &  -2.5  &  2.5 \\ 0  &  0  &  -2.5  &  2.5  &  2.5  &  -2.5 \\ 0  &  0  &  2.5  &  -2.5  &  -2.5  &  2.5 \\ \end{bmatrix} $$

= Step 4:Elimination of Known Degrees of Freedom =

Elimination of Known Degrees of Freedom=>

Reduce the Global Force Displacement Relation (See Lecture 5-3)

$$ d_{1} = d_{2} = d_{5} = d_{6} = 0 $$

$$\underset{6 \times 6}{K} \begin{pmatrix}

d_1= 0 \\

d_2 = 0 \\

d_3 = ? \\

d_4 = ? \\

d_5 = 0 \\

d_6 = 0 \end{pmatrix} \Rightarrow \mathbf \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{pmatrix}

d_3 \\

d_4 \end{pmatrix} = \underset{6 \times 1}{F}$$


 * Apply fixed boundary conditions $$ \Rightarrow $$ delete corresponding columns in the global stiffness matrix, K


 * By Principal of [Virtual work] $$ \Rightarrow $$ delete corresponding rows (here, rows 1,2,5,6) of K; therefore, corresponding rows of F may also be deleted

Resulting force-displacement relation:

$$ \begin{bmatrix}

K_{33} & K_{34} \\

K_{43} & K_{44} \\ \end{bmatrix} \begin{pmatrix}

d_3 \\

d_4 \end{pmatrix} = \begin{pmatrix}

F_3 \\

F_4 \end{pmatrix} $$

$$ \mathbf F = \begin{pmatrix}

F_3 \\

F_4 \end{pmatrix} = \begin{pmatrix}

0 \\

P \end{pmatrix} $$  {Known$$\Rightarrow$$ Problem Statement}

Recall: $$ \mathbf K  \mathbf d = \mathbf F $$

$$ \Rightarrow \mathbf d = \mathbf F \mathbf K^{-1}$$

$$ \begin{pmatrix}

d_3 \\

d_4 \end{pmatrix} = \mathbf K^{-1} $$  $$ \begin{pmatrix}

0 \\

P \end{pmatrix}$$

$$ \Rightarrow \mathbf K^{-1} = \begin{bmatrix}

0.7681 & 0.6217 \\

0.6217 & 0.8753 \\ \end{bmatrix}$$

$$ \Rightarrow P = 7 $$

$$ \begin{pmatrix}

d_3 \\

d_4 \end{pmatrix} = $$   $$ \begin{bmatrix}

0.7681 & 0.6217 \\

0.6217 & 0.8753 \\ \end{bmatrix}$$ $$ \begin{pmatrix}

0 \\

7 \end{pmatrix} = $$ $$ \begin{pmatrix}

4.3519 \\

6.1271 \end{pmatrix} $$

= Step 5: Computation of Element Forces =

Use the global force-displacement relationship to compute element forces:

$$ \Rightarrow \begin{pmatrix}

d_3 \\

d_4 \end{pmatrix} = \begin{pmatrix}

d^{(1)}_3 \\

d^{(1)}_4 \end{pmatrix} = \begin{pmatrix}

d^{(2)}_1 \\

d^{(2)}_2 \end{pmatrix} = \begin{pmatrix}

4.3519 \\

6.1271 \end{pmatrix} $$

Recall:

$$\underset{4 \times 4}{k^{(e)}}\underset{4 \times 1}{d^{(e)}} = \underset{4 \times 1}{f^{(e)}} $$     {e = 1, 2}

For Element 1: (e=1)

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

$$ \mathbf d^{(1)} = \begin{pmatrix}

0 \\ 0 \\ 4.3519 \\ 6.1271 \end{pmatrix} $$

Recall, $$

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

$$ \Rightarrow \mathbf f^{(1)} = \begin{pmatrix}

f^{(1)}_1 \\ f^{(1)}_2 \\ f^{(1)}_3 \\ f^{(1)}_4 \end{pmatrix} = \begin{pmatrix}

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

Similarly, For Element 2: (e=2)

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

$$ \mathbf d^{(2)} = \begin{pmatrix}

4.3519 \\ 6.1271 \\ 0 \\ 0 \end{pmatrix} $$

Recall, $$ \mathbf k^{(2)} = \begin{bmatrix} \frac{5}{2}& -\frac{5}{2} &-\frac{5}{2} &\frac{5}{2}\\ -\frac{5}{2}& \frac{5}{2} & \frac{5}{2} &-\frac{5}{2} \\ -\frac{5}{2}& \frac{5}{2} &\frac{5}{2} &-\frac{5}{2}\\ \frac{5}{2}& -\frac{5}{2} & -\frac{5}{2} &\frac{5}{2} \end{bmatrix}$$

$$ \Rightarrow \mathbf f^{(2)} = \begin{pmatrix}

f^{(2)}_1 \\ f^{(2)}_2 \\ f^{(2)}_3 \\ f^{(2)}_4 \end{pmatrix} = \begin{pmatrix}

-4.442 \\ 4.442 \\ 4.442 \\ -4.442 \end{pmatrix}$$

= Step 6: Computation of Reaction Forces =



Figure 8: Axial Forces in Member 1

$$ P^{(1)}_1 = [(f^{(1)}_1)^2 + (f^{(1)}_2)^2]^{1/2} $$

and

$$ P^{(1)}_2 = [(f^{(1)}_3)^2 + (f^{(1)}_4)^2]^{1/2} $$

Element 1 is a two-force member, thus:

$$ P^{(1)}_1 = P^{(1)}_2 $$

Similarly, in element 2, the global nodes are translated into the local nodal coordinate system and are shown in the figure below.



Figure 9: Axial Forces in Member 2

=Further Analysis=

Looking at the axial forces shown in section below in the two figures, we may use statics to relate the relationship between all the forces. Both elements are in equilibrium.

For Element 1:

$$ \sum F_x = f^{(1)}_1 + f^{(1)}_3 = 0$$

$$ \sum F_y = f^{(1)}_2 + f^{(1)}_4 = 0 $$

For Element 2:

$$ \sum F_x = f^{(2)}_1 + f^{(2)}_3 = 0 $$

$$ \sum F_y = f^{(2)}_2 + f^{(2)}_4 = 0 $$

= Method 2: Statics and Euler Cut Principle=

Second method for solving the two-bar truss example using ] and the Euler cut principle. It is useful to use Euler's method rather than statics or the FEM "recipe" method to solve since it allows us to break away specific sections of the system, and recreate them specifically into distinct sections.



Figure : One example of Euler Cut Principle with Two Bar System



Figure : Another example of Euler Cut Principle with same Two Bar System

More information on Euler's Law and Cut Principle can be found by clicking here: [Euler's Laws and Cut Principles]

= Example 1.2 Solved in MathCAD =

In the following pictures we solve example problem 1.2 using PTC's MathCAD software.



Figure 11: Element 1's Stiffness Matrix


 * Here we are simply defining all of the variables for Element 1's Stiffness Matrix.Defined here is The Young's Modulus, Cross sectional area, and Length. The Matrix coefficient is defined here. We also and solve for Element 1's Stiffness Matrix.



Figure 12: Element 2's Stiffness Matrix


 * Here we are simply defining all of the variables for Element 2's Stiffness Matrix.Defined here is The Young's Modulus, Cross sectional area, and Length. The Matrix coefficient is defined here. We also and solve for Element 2's Stiffness



Figure 13: Combining Element Stiffness Matrices



Figure 14: Combined Global Stiffness Matrix


 * Above we combine the Element Stiffness Matrices as stated in step 3 of example 1.2 solution above.



Figure 15: What is left of the Global Matrices after Virtual Work


 * By Principal of Virtual Work we delete corresponding rows (here, rows 1,2,5,6) of K; therefore, corresponding rows of F may also be deleted. We end up with the following variables left. We also take the inverse of the remaining Global Stiffness Matrix.

Figure 16: Global Displacement Matrix values


 * Here we simply solve for the Global Displacement Matrix values.



Figure 17: Reaction forces


 * We now take the Global Displacement values and return them respectively to their corresponding Element Stiffness Matrix. The we compute the reaction forces at each pin.

= Solution for example 1.2 using two force member principle In MathCAD =

In the following picture we solve the axial load picture from the notes using PTC's MathCAD software.



Figure 18: Solution of Example 1.2 with two force member.


 * You can refer to Figure 7 to see the system solved using the principle of two force members

= Help with MathCAD = For help with ], we found some practical help on LSU's webpages. [Source: LSU]

= 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.

Principles in [Virtual work] were touched upon to help facilitate learning in the subject. [Source: Wikipedia]

] method of work was also used to help solve some problems. [Source: Wikipedia]

For help with ], we found some practical help on LSU's webpages. [Source: LSU]

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

Rodney Dagulo --Eml4500.f08.wiki.dagulo.r 11:51, 25 September 2008 (UTC)

Renee Hood -- Eml4500.f08.wiki.hood 22:52, 25 September 2008 (UTC)

Cortland Glowacki -- Eml4500.f08.wiki.cort 8:34, 25 September 2008 (UTC)

Ricardo Lopez --Eml4500.f08.wiki.Lopez 1:23, 26 September 2008 (UTC)

Gonzalo Barcia --Eml4500.f08.wiki.Barcia 3:14, 26 September 2008 (UTC)