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

A-Team Teampage

EML 4500 Notes: September 5, 2008
Read: Chapter 4 – Trusses, beams, frames

Chapter 1: Big Picture

Section 1.1: Discretization

Section 1.1.1: Plane truss elements

Section 1.2 Assembly of element equations
 * Example: Five-bar truss

Section 1.4 Element solution and model validity

Section 1.4.1 Plane truss elements

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_{11}&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&K_{66} \\ \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_{11}&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&K_{66} \\ \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix}$$