SimpleTrussExample

SimpleTrussExample - CE 3101 Fall 2011 A Simple Truss...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CE 3101 Fall 2011 A Simple Truss Exposed Problem Consider the simple, statically determinant truss shown in Figure 1. This figure also presents the externally applied forces. [1] [2] [3] 1 2 3 2 2 1 2 3 1 1 1 Figure 1: A simple, planar, pin-jointed, statically determinant truss. The joint numbers are in the circles, the member numbers are in brackets. The specified external loads at the joints are shown with arrows. Let M be the number of members (in this case M = 3), and N be the number of joints (in this case N = 3). Note that this truss satisfies the necessary condition for a statically determinant truss 1 M = 2 N - 3 (1) Using the notation and nomenclature that we have developed in class, this truss problem is fully described by three matrices. The ( N × 2 ) joint geometry matrix G = 0 0 1 1 2 0 (2) The ( M × 2 ) member connection matrix C = 1 2 2 3 3 1 (3) The ( N × 2 ) external force matrix F = 0 1 - 3 2 2 2 (4) 1 See, for example, http://en.wikipedia.org/wiki/Truss#Statics_of_trusses . Simple Truss 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
CE 3101 Fall 2011 Some Geometric Calculations It will be convenient to have some basic geometric quantities pre-computed. Define the ( M × 2 ) matrix D , where D ( m , 1 ) is the “delta X” for member m , and where D ( m , 2 ) is the “delta Y” for member m . The code fragment for computing D would look like: D = G(C(:,2),:) - G(C(:,1),:); This is very compact code (a speciality of MATLAB ); i.e., there is a lot going on in this one line. To make sense out of this line of code, we break it into pieces and see what we get for our small example. Think of every member as having a “FROM” joint and a “TO” joint, with the “FROM” given by the first column of C and the “TO” given by the second column of C . C (: , 2 ) = 2 3 1 (5) This is nothing more than the second column of C ; that is the ( M × 1 ) matrix of “TO” joint indices. Then G ( C (: , 2 ) , :) = 1 1 2 0 0 0 (6) So, this is the ( M × 1 ) matrix of coordinates of the “TO” joints for each member. Similarly, G ( C (: , 1 ) , :) = 0 0 1 1 2 0 (7) is the ( M × 1 ) matrix of coordinates of the “FROM” joints for each member. Taking the difference between these two gives us the ( M × 2 ) matrix of Δx and Δy for each member (we are denoting this D ): D = 1 1 1 - 1 - 2
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 5

SimpleTrussExample - CE 3101 Fall 2011 A Simple Truss...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online