Statics - Trusses Method of Sections

 We are trying to see how to find the forces in each of the members. This is a slightly different pointer. 

What we did was doing the method of joints where we take any one of the connection points and figure out the forces onto the connection point. 

For instance, let's look at point C. Remember each force member means that the forces in them and the forces exerted on the pin are aligned with the members themselves. 

Here's the force diagram for Point C:

Thus, by observation of Joint C, there is a 0 force member and we can remove such member. 

These members of obviously in Tension, so If AC is only going to be in tension, let's replace things with a table. 

We still need a roller, but we can heavily simplify the problem in that manner. Watch for 0-force members, and watch for members only in tension and you can simplify things in that way. The roller can be the normal force.

The roller is just boing to be a normal force from the wall, which in hindsight, can result in equilibrium. 

The next thing we're going to do is the method of sections. In the general sense, when you need only a few of the forces and members, we might want to consider the method of sections. 

There might be hundreds of sections, but sometimes we only need to look at the force of one member. For method of sections we try to imagine a cut through a specific point. Anything across the imaginary cut we have to supply the force that each member gets applied. Here is an example.

Here, the green members have some type of force applied to the member. I save a little bit of trouble by not having to solve for all of the reaction forces anymore. We'll get exactly what we received before. 

Let's draw a pinion and a crane boom with a mass at the end.We have gigantic boom thing hanging at the end, and each side of these squares is 1 meter and we need some cross members in there since squares are not stable, but triangles are. 

 For whatever reason we want to know the force in member CJ Either way, we're going to have to do at least 5 different free-body diagrams, 5 different joins, and 5 different solutions where the sum of the forces equal to 0. Remember you must put out a force in the right direction to place a member for any member you cut through and don't forget to put the reaction force down. Now, you can solve that section. 

We then sum the forces in the y direction, given that y is our typical up and down and given this you can solve for CJ. 

Cj is 141.421356239kN as a result. 

Let's try another problem. 

There's some type of train with some squares so as a result we want triangles so we put in a nice piece like that. 

We can cut through a member of the force in order to see the inclination maybe, just maybe, figure out the precise direction. 

Also don't forget moment equations apply here, since this is a full section, in contrast to just a pin.

 

What we can do is sum the moments about point C. We can apply method of joins. We can also change our cut and come up ith a different cut to use as a set of independent equations. We can either sum the moments in the D direction or sum the forces in the X direction.