User:Eml4500.f08.group/Homework1/LectureNotes

=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.



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:

$$f_1 + f_2 = 0$$

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

Steps to solve simple truss system
Steps to solve simple truss system described on Wed, 3 Sep 08. (Recipe)

Global picture
(description) at structure level
 * Global degrees of freedom (disp coeffs)
 * Global Forces

Actually the displacement degrees of freedom are partitioned into Simularly for the global forces
 * A known part, eq, fixed defs, constraints
 * an unknown part, solved using FEA
 * a know part: applied force
 * an unknown part: reactions

Element picture
either in global coord system on in local coord system
 * element degrees of freedom
 * element forces

Global Force Displacement relationship

 * Elemenet Stiffness matrix in global coord
 * element force matrices in global coord.
 * Assembly of element stiffness matrix and element force matrix into a global Force Displacement relationship

Elimination of known degrees of freedom
Elimination of known degrees of freedom to reduce the global Force Displacement Relationship (stiffness matrix non-singular => invertible)
 * m = number of unknown displacement degrees of freedom
 * n = number of known and unknown and displacement degrees of freedom

Kbar non singular => Kbar^-1   (Kbar invertible)

Compute Element forces
Compute Element forces from now known dbar => element stresses

Compute Reactions
Compute reactions (unknown forces)

Example
Take a specific example to see how the recipe works (no justification yet)

Element Length - L(1) = 4, L(2) = 2

Youngs Modulus - E(1) = 3, E(2) = 5

Cross Section Area - A(1) = 1, A(2) = 2

Inclination Angle - θ(1) = 30°, θ(2) = -45°

=Mon, 8 Sep 08= 1.) Global Picture 2.) Element Picture

3.) Global Force Displacement at element level



k is the element stiffness matrix for element e

d is the the element displacement matrix

f is the element force matrix

Matrix p 225 book, 2nd equation form the bottom





axial stiffness of bar elem "e". e = 1,2

l^(e), m^(e) = director cosines of ~x axis (goes from [1] to [2]) with respect to global (x,y) coords