From Wikiversity

1 Introduction
2 Recall: Use the Derived "Recipe" to Simplify the Problem
3 Statically Determinate: Problem 1.2 (continued)
4 Step 1: Global Picture
5 Step 2: Elemental Picture
- 5.1 Problem Setup
6 Step 3: Global Force Displacement Relationships
7 Step 4:Elimination of Known Degrees of Freedom
8 Step 5: Computation of Element Forces
9 Step 6: Computation of Reaction Forces
10 Further Analysis
11 Method 2: Statics and Euler Cut Principle
12 Example 1.2 Solved in MathCAD
13 Solution for example 1.2 using two force member principle In MathCAD
14 Help with MathCAD
15 References
16 Contributing Members

[edit] 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.

[edit] 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.

Step 1: Develop a Global Picture

This Global Picture will give us a description of the problem at the structure level as well a schematic diagram, complete with the following:

Global Degrees Of Freedom (DOFs)
Known Displacement DOFs (ex.) Fixed DOFs, Constraints
Unknown Displacement DOFs (ex.) Found on lecture from 9/3/05 slide #1
Global Forces
Known forces (ex.) Applied forces
Unknown forces (ex.) Reaction forces

Step 2: Develop a Picture For Each Element

It is important to draw element pictures for all elements in the system, so that one can use information already know in static mechanics to continue to breakdown the problem, and finally solve it. The element pictures should include:

Element Degrees Of Freedom (DOFs)
Known Displacement DOFs (ex.) Fixed DOFs, Constraints
Unknown Displacement DOFs (ex.) Found on lecture from 9/3/05 slide #1
Element Forces
Known forces (ex.) Applied forces
Unknown forces (ex.) Reaction forces

Note: The Element picture may be place in the global coordinate system or in a local coordinate system

Step 3: Define Global Force Displacement Relationships

By defining global force displacement relationships, will allow for the following:

Compute the element stiffness matrices in the global coordinate system
Compute the element force matrices in the global coordinate system
Assemble the element stiffness matrices and element force matrices to form the global FD relations.

K_nxn x d_nx1 = F_nx1

K is the global stiffness matrix
- K is a n x n matrix
d is the global displacement matrix
- d is a n x n matrix
F is the global force matrix
- F is a n x n matrix
  - n is the # of displacement degrees of freedom

Step 4: Elimination of known Degrees of Freedom

This can serve to simplify the problem so that unknown forces may be found, and solved for.

Elimination of know DOFs to reduce the global Force Displacement relations

K_mxm x d_mx1 = F_mx1

K is the global stiffness matrix
- K is a m x m matrix
- K is a non-singular(invertible)
d is the global displacement matrix
- d is a m x m matrix
F is the global force matrix
- F is a m x m matri
  - m is the # of unknown displacement degrees of freedom

With the preceding information, we will use the recipe to solve for the required displacement degree of freedom vectors as well as the global forces.

Step 5: Computation of Element Forces

The matrix K is non-singular, which allows inversion of the matrix With this known we can now say

d_mx1 = K^-1_mxm x F_mx1

Know we may solve for the matrix d which can be used to find the element forces/element stresses.

Step 6: Computation of Reaction Forces
We know have all the necessary tools to compute the unknown reaction forces.

[edit] 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.

[edit] 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.

[edit] Step 2: Elemental Picture

Element 1

Element 1: θ⁽¹⁾=30^°

Figure 3: First Element Forces

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

Element 2
θ⁽¹⁾=-45^°

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

[edit] 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

Contribution: Directional Proofs

The following two diagrams will serve as proofs for the "dot products" for the directional coordinate system

Figure 6: Proof #1

Figure 7: Proof #2

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) ^{= :}}

^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(θ^e): $l^1 = \frac{\sqrt{3}}{2}$ , $l^2 = \frac{1}{\sqrt{2}}$

m^(e)=sin(θ^e): $m^1 = \frac{1}{2}$ , $m^2 = -\frac{1}{\sqrt{2}}$

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⁾ = $\begin{bmatrix} \frac{5}{2}& -\frac{5}{2} &-\frac{5}{2} &\frac{5}{2}\\ -\frac{5}{2}& \frac{5}{2} & \frac{5}{2} &-\frac{5}{2} \\ -\frac{5}{2}& \frac{5}{2} &\frac{5}{2} &-\frac{5}{2}\\ \frac{5}{2}& -\frac{5}{2} & -\frac{5}{2} &\frac{5}{2} \end{bmatrix}$

[edit] 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:

Element 1

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

Element 2

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

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.

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

[edit] 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: $\mathbf K \mathbf d = \mathbf F$

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

$\begin{pmatrix} d_3 \\ d_4 \end{pmatrix} = \mathbf K^{-1}$ $\begin{pmatrix} 0 \\ P \end{pmatrix}$

What is $\mathbf K^{-1}$ ?

To find $\mathbf K^{-1}$ (the inverse of $\mathbf K$ ) without a calculator:

det $\mathbf K = K_{33}K_{44} - K_{34}K_{43}$

$\mathbf K^{-1} = \tfrac{1}{det K} \begin{bmatrix} K_{44} & -K_{34} \\ -K_{43} & K_{33} \\ \end{bmatrix}$

Verification:

$\mathbf K \mathbf K^{-1} = \mathbf K^{-1} \mathbf K = \mathbf I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} = \tfrac{1}{det K} \begin{bmatrix} det K & 0 \\ 0 & det K \\ \end{bmatrix}$

NOTE:

$\mathbf K^T = \begin{bmatrix} K_{33} & K_{43} \\ K_{34} & K_{44} \\ \end{bmatrix}; \mathbf K^{-1}$ ≠ $\tfrac{1}{det K} \mathbf K^T$

For this example:

$\mathbf K = \begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \\ \end{bmatrix} = \begin{bmatrix} 3.0625 & -2.1752 \\ -2.1752 & 2.6875 \\ \end{bmatrix}$

$det \mathbf K = K_{33}K_{44} - K_{34}K_{43} = 3.4990$

$\Rightarrow \mathbf K^{-1} = \tfrac{1}{3.4990} \begin{bmatrix} 2.6875 & 2.1752 \\ 2.1752 & 3.0625 \\ \end{bmatrix} = \begin{bmatrix} 0.7681 & 0.6217 \\ 0.6217 & 0.8753 \\ \end{bmatrix}$

$\Rightarrow \mathbf K^{-1} = \begin{bmatrix} 0.7681 & 0.6217 \\ 0.6217 & 0.8753 \\ \end{bmatrix}$

$\Rightarrow P = 7$

$\begin{pmatrix} d_3 \\ d_4 \end{pmatrix} =$ $\begin{bmatrix} 0.7681 & 0.6217 \\ 0.6217 & 0.8753 \\ \end{bmatrix}$ $\begin{pmatrix} 0 \\ 7 \end{pmatrix} =$ $\begin{pmatrix} 4.3519 \\ 6.1271 \end{pmatrix}$

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

$\Rightarrow \mathbf f^{(2)} = \begin{pmatrix} f^{(2)}_1 \\ f^{(2)}_2 \\ f^{(2)}_3 \\ f^{(2)}_4 \end{pmatrix} = \begin{pmatrix} -4.442 \\ 4.442 \\ 4.442 \\ -4.442 \end{pmatrix}$

[edit] 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

[edit] 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:

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

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

[edit] Method 2: Statics and Euler Cut Principle

Second method for solving the two-bar truss example using [Statics] 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]

[edit] 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.

[edit] 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

[edit] Help with MathCAD

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

[edit] 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]

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

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

[edit] 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)

User:Uf.team.wiki/HW2