Statics: Trusses Method of Joints

 We haven't used any tools we haven't acquired we're just going to apply them to different things different ways, and we're going to look at the general category of what's called structures. There are 3 different types of structures:

1. Trusses

We try to figure out what reaction forces are looking at certain external forces. Now, we're going to look at forces that are interior to these structures, and trusses are structures made out of exclusively 2-force members. 

2-force members are many members with only the 2 forces in them meaning they must be pinned at each and and no more than pinned. So, if we're looking at some sort of bridge structure, each of the joined intersections are just simple pins, which means we're not taking into account the weight of any of these which generally means they're massless. Most of the stuff are guesses which you see a big pin with rivets all over the place. 

The deal with the pin joints are they are free to rotate as needed. Most bridge/truss has something or plates attached to each one of the joints, meaning we're trying to look for the forces of each of the joints, and as a result, the forces of each of the members themselves. It's going to have a huge factor of safety in it, and we're going to go through mostly planar 2D trusses, and we're also going to look at frames, trusses with at least one member with 3 or more forces. These are FRAMES. 

The last type of structure is the class of machines, frames with movable parts in which some of the members actually move. 

So trusses, frames and machines. 

Let's look at the analysis of a simple truss, planar with pinned joints. There's 2 different ways to analyze these. 

1. Method of Joints

2. Method of Sections.

We're looking at the force of each of the joints and as a result, the forces of each of the members of the truss.

We look at the free body diagram generally of the entire structure. What this will do for us is to find the reaction forces. The best way to find that is via free-body diagram of the entire structure and the tools apply, summing the forces and the moments, and making sure that they are zero (0). 

We'll then do a free body diagram of each joint itself. We find a forces on the joint which are forces of the members that connect to that joint. One of the rules of the 2 forces on the joint are equal and opposite, so that means to us that the forces always lie along the line connecting the 2 adjacent joints. 

I know that AB will go up because we need an upward force. then I know CA goes to the left. Once we know what the forces are on a joint, we know what the forces are on the member that connects to that joint. If the weight is pulling on joint a, then joint a must be pulling up as an equal and opposite direction and this is tension.  and the other member AC has equal and opposite forces and these forces are in compression. You can do very different things designing something in tension and other things in something that is compression. 

Time for a problem.

To keep things simple, just draw a straight line to represent a member. So, we have a very simple 3-triangular structure. Why triangles? That's not just a coincidence. All the trusses are nothing but triangles because a triangle is much more structurally sound. In a triangle, when there is no material failure, will not collapse. If we have a square truss, the joints are freely pinned, and can collapse. So we can't have any polygons above a triangle. Let's say the structure is pinned in support at one corner, has a roller support at another corner. 

There's also 2 loads in pounds with all the measurements in feet. Label the pieces for reference such that we are all talking about the some things. From the diagram, we want to analyze the forces in each of the members. 

If we're wrong in determining the sense, we'll get a negative number. The roller gives a bit of transverse support. Here's the result of the trapezoid, to be double checked in the future:

We can either start at joint C or joint A (2 lines coming out of it, therefore 2 unknowns). Let's try to start with joint A. There is no use going to some other point because that tends to overcomplicate things. The rest is algebra.  This just comes by solving the 2 force balances in the 2 different directions. The angle between AB and AD is 53.1 degrees. Remember: every member exerts a force! We see a member and draw the forces equal and opposite to those forces and in this way I can determine whether a member is in tension or in compression. 

Draw the joints in the center where they lay, at least in the same spatial arrangement. Try to assess the direction if you can, but don't make a huge deal about it. Most of the forces we can mirror and we already know their magnitude, the rest is just trigonometry. Here's the diagram (unit is in lbs):