User:Eml4500.f08.ateam/hw 1/trusses

EML 4500 Notes:
Trusses are structural systems consisting of long slender members. The members are arranged so that they are loaded axially and do not experience bending or shear forces. For more information on trusses, please refer to the Truss article on Wikipedia.

Global Picture
Before trusses are analyzed with the Matrix Method or any other method, it is important to develop a standardized system for labeling and drawing the systems. First, the truss system is drawn and the components are labeled. Global nodes are labeled with a number circumscribed in a circle and elements are labeled with a number circumscribed in a triangle.



Global Free Body Diagram
The next step in the process is to draw the global free body diagram (FBD). The global FBD is a diagram of the whole structure that shows the applied forces, as well as the reaction forces. The FBD includes known and reactionary forces represented by arrows. The reaction forces are labeled Rna, where a denotes the direction of the force in reference to the global coordinate system and n represents the global node number. For example, R1X represents the reaction force R applied along the X axis at node 1.



Element Free Body Diagrams
After the global FBD is created, an element FBD must be created. The element FBD is a diagram of each element of the whole structure that includes known, reaction, and internal forces applied to that element. The elemental diagram is labeled with the element number as stated above. Each node of the element is labeled with its global node number as stated above, as well as a local node number that is a number circumscribed in a square. Each internal force of the element is labeled fi(e), where i is the force number and e is the element number. Force numbers are assigned first by local node number and then orientation related to the coordinate system. For example, a force f1(1) is the first degree of freedom (DOF) in element one.



Force-Displacement Relation
The next "big step" in applying the matrix method to a truss problem is to develop a relation between the applied force and the resulting displacement. The deformation of an elastic truss member can be modeled as the deformation of a spring. Therefore, mass-spring relations will be examined.

Spring with One End Constrained
First, consider a one dimensional spring with one end fixed. The force-displacement relation for this system is represented by the equation $$f=kd$$, where f is the force, d is the displacement, and k is the spring constant that it a property of the spring. This relation shows that there is linear relationship between the applied load and the resulting displacement.



Spring with Both Ends Free
Another spring system to consider is a one dimensional spring with both ends free. This system has two degrees of freedom so there is a system of two equations that govern the relationship between the applied force and the resulting displacement. Consider the spring system shown in the figure. Force 1, F1, acts at node 1, causing a displacement, d1 and force 2 acts at node 2, resulting in a displacement d2. The equations of motion for this system are developed for two cases. For the first case, consider an observer sitting on node 1. The resulting equation of motion is $$f_2 = k(d_2-d_1)$$. The second case is when an observer sits on node 2, resulting in the equation $$f_1 = k(d_1-d_2)$$. Therefore, the resulting force-displacement relation for this system is represented by the system of equations below.

$$\begin{bmatrix} F_1 \\ F_2 \\ \end{bmatrix} $$ = $$\begin{bmatrix} k & -k \\ -k & k \\ \end{bmatrix} $$ $$\begin{bmatrix} d_1 \\ d_2 \\ \end{bmatrix} $$



Steps to solve simple truss system described previously
1. Global picture

The global picture consists of a diagram of the entire structure detailing all of the nodes and nodal forces relevant to the problem. A global coordinate system is selected. At the structure level there are two components to be concerned about. These are the global degrees of freedom, or displacement degrees of freedom, and the global forces. Both are divided into known parts and unknown parts. In the previous truss example, the known parts consist of the applied force and the fixed degrees of freedom for nodes 1 and 3. These two nodes are fixed, therefore the degrees of freedom for each is constrained to zero. The unknown part is solved using the Finite Element Method (FEM). FEM is a numerical method used to obtain solutions to ordinary and partial differential equations. In the previous truss system, the reactions forces are unknown.

2. Element picture

An element picture is made of each individual part of the structure. These elements are considered "free" and a free-body diagram with a local coordinate system is made for each. Included in the element picture are both the element degrees of freedom and the element forces. These can be described either in the global coordinate system or the local coordinate system.
 * Element degrees of freedom: $$d_1,\ d_2,\ d_3,\ d_4,$$
 * Element forces: $$f_1,\ f_2,\ f_3,\ f_4$$

3. Global force-displacement relationship

The global force-displacement relationship is shown through matrices. These matrices consist of:
 * Element stiffness matrix in global coordinates ($$k$$) . The dimensions of this matrix are $$n$$ x $$n$$ where $$n$$ is the number of both known and unknown displacement degrees of freedom.
 * Element displacement matrix in global coordinates ( $$d$$) . The dimensions of this matrix are $$n$$ x $$1$$.
 * Element force matrix in global coordinates ($$f$$) . The dimensions of this matrix are $$n$$ x $$1$$.

To assemble the element stiffness matrices, the element displacement matrices, and the element force matrices into global force-displacement relationship matrix multiplication is applied in the form of:


 * $$kd= f$$

This is considered a free-free system where $$k$$ is singular, meaning $$k$$ is not invertible, or $$k^{-1}$$ does not exist.

4. Elimination of known degrees of freedom

In order to make the method more manageable, the global force-displacement relationship is reduced by eliminating the known degrees of freedom. This makes the stiffness matrix non-singular, or invertible. The form is:


 * $$KD= F$$

$$K$$ is the new stiffness matrix with dimensions $$m$$ x $$m$$, where $$m$$ is the number of unknown displacement degrees of freedom. Therefore, $$m$$ < $$n$$. Similarly, $$D$$ is the new displacement matrix with dimensions $$m$$ x $$1$$ and $$F$$ is the new force matrix with dimensions $$m$$ x $$1$$.

To compute the displacement, the equation is rearranged:


 * $$D = K^{-1}F$$

5. Compute element forces

Solving for $$D$$ above, the element stresses can now be determined.

6. Compute reactions

From the element stresses, the reaction forces can be determined.

Example: Truss problem


 Data 

Element length:

L(1) = 4

L(2) = 2

Young's Modulus:

E(1) = 3

E(2) = 5

Cross section area:

A(1) = 1

A(2) = 2

Inclination angle:

θ(1) = 30°

θ(2) = -45°

1. Global picture

Global degrees of freedom: $$d_1,\ d_2,\ d_3,\ d_4,\ d_5,\ d_6$$



n = global node number

Note: When numbering the displacement degrees of freedom, it is useful to follow the order of the global node number. For each individual node, number the displacement degrees of freedom in the order of the global coordinate axes.

Global forces: $$f_1,\ f_2,\ f_3,\ f_4,\ f_5,\ f_6$$



Global force matrix


 * $$\begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix}$$

Global stiffness matrix


 * $$\begin{bmatrix}

\; \; K & -K & \; \; K & -K & \; \; K & -K \\ -K & \; \; K & -K & \; \; K & -K & \; \; K \\ \; \; K & -K & \; \; K & -K & \; \; K & -K \\ -K & \; \; K & -K & \; \; K & -K & \; \; K \\ \; \; K & -K & \; \; K & -K & \; \; K & -K \\ -K & \; \; K & -K & \; \; K & -K & \; \; K \end{bmatrix} $$

Global displacement matrix


 * $$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix}$$

To solve for the unknowns, the matrices are arranged according to the equation above.


 * $$\begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix} = \begin{bmatrix}

\; \; K & -K & \; \; K & -K & \; \; K & -K \\ -K & \; \; K & -K & \; \; K & -K & \; \; K \\ \; \; K & -K & \; \; K & -K & \; \; K & -K \\ -K & \; \; K & -K & \; \; K & -K & \; \; K \\ \; \; K & -K & \; \; K & -K & \; \; K & -K \\ -K & \; \; K & -K & \; \; K & -K & \; \; K \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix}$$