# Analyzing Truss w/ Counters: Find Force in Green & Pink Members

• stinlin
In summary: Well, wait, be careful. The only reason why we can do this problem as described, is because it is given that the pink diagonals can only take tensile forces. If that's what you mean by 'boundary condition', that's fine. But in the general case where the diagonals can take both tension AND compression, you can't ignore them, and you would then have to do some sort of indeterminate analysis (as suggested by Radou in an earlier response) to determine the forces (the upward "v" diagonals...).In summary, if you are given a truss with a "boundary condition" where one of the members can only take tension, you can solve
stinlin
How would you begin to analyze a truss of this style? The picture I posted is the situation given. Say you want the force in the green member and the pink members given loads 1,2,3. If the pink members can only act in tension, where you begin?

For those who may not know, the support on the left is a pin (2 force reactions, and on the right is a roller, one force in the y direction). Any thoughts??

http://img300.imageshack.us/img300/1917/trusscn0.png

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stinlin said:
How would you begin to analyze a truss of this style? The picture I posted is the situation given. Say you want the force in the green member and the pink members given loads 1,2,3. If the pink members can only act in tension, where you begin?

For those who may not know, the support on the left is a pin (2 force reactions, and on the right is a roller, one force in the y direction). Any thoughts??

http://img300.imageshack.us/img300/1917/trusscn0.png
I'd take out 2 of the pink members...the one that goes from load point 1 and the one that goes from load point 3. They would go into compression, and thus buckle out or go slack. So pretend they are not there. Then you should note by isolating that top joint that the green member is a zero force member. Solve in the usual manner.

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So that definitely would solve the problem...

Now my question is this - how do you know those 2 would go into compression and not the other 2 pink ones?

stinlin said:
So that definitely would solve the problem...

Now my question is this - how do you know those 2 would go into compression and not the other 2 pink ones?

This is a statically undeterminate system, unless I'm missing something. You may want to investigate the force method, which seems appropriate to solve this type of system.

stinlin said:
So that definitely would solve the problem...

Now my question is this - how do you know those 2 would go into compression and not the other 2 pink ones?
Just look at the loading at joint 2. When you take out the 2 pink ones I noted, you will see that the green member is a zero force menmber, and that the remaining pink diagonals must be in tension. Now relook at the problem and take out the the other pink members instead. Now at joint 2, you should see that the green member will be in tension, and when you look at the top joint of that green member, you should now see that the pink diagonals will be in compression, which is not allowed.

So these members and the green member may be removed for the analysis, however, it doesn't mean you don't need them, due to other loading considerations not shown (lateral forces, seismic, etc.), or in the case of the green member, l/r reduction in the top chord.

PhanthomJay said:
Just look at the loading at joint 2. When you take out the 2 pink ones I noted, you will see that the green member is a zero force menmber, and that the remaining pink diagonals must be in tension. Now relook at the problem and take out the the other pink members instead. Now at joint 2, you should see that the green member will be in tension, and when you look at the top joint of that green member, you should now see that the pink diagonals will be in compression, which is not allowed.

So these members and the green member may be removed for the analysis, however, it doesn't mean you don't need them, due to other loading considerations not shown (lateral forces, seismic, etc.), or in the case of the green member, l/r reduction in the top chord.

That makes SO much sense! Thanks! So I'm guessing in the future, if I'm ever given a problem that is overly constrained with some sort of "boundary" condition, I can analyze it like this, right? Thanks for the clear explanation! :)

stinlin said:
That makes SO much sense! Thanks! So I'm guessing in the future, if I'm ever given a problem that is overly constrained with some sort of "boundary" condition, I can analyze it like this, right? Thanks for the clear explanation! :)
Well, wait, be careful. The only reason why we can do this problem as described, is because it is given that the the pink diagonals can only take tensile forces. If that's what you mean by 'boundary condition', that's fine. But in the general case where the diagonals can take both tension AND compression, you can't ignore them, and you would then have to do some sort of indeterminate analysis (as suggested by Radou in an earlier response) to determine the forces (the upward "v" diagonals from joint 2 would be in tension, and the downward "^" diagonals would be in compresion). What would happen, for example, is that when the diagionals are 'tension only", you get say 10 kips tension, whereas if they can take both Tension and Compression, you might get 5 kips in the tension diagonals and 5 kips inh the compression diagonals.

PhanthomJay said:
Well, wait, be careful. The only reason why we can do this problem as described, is because it is given that the the pink diagonals can only take tensile forces. If that's what you mean by 'boundary condition', that's fine. But in the general case where the diagonals can take both tension AND compression, you can't ignore them, and you would then have to do some sort of indeterminate analysis (as suggested by Radou in an earlier response) to determine the forces (the upward "v" diagonals from joint 2 would be in tension, and the downward "^" diagonals would be in compresion). What would happen, for example, is that when the diagionals are 'tension only", you get say 10 kips tension, whereas if they can take both Tension and Compression, you might get 5 kips in the tension diagonals and 5 kips inh the compression diagonals.

Yeah - what I meant by "boundary condition" was the fact that counters are like cables in that they must be in tension or they go slack. We discussed it in lecture today, the idea behind counters. =) Thanks again.

## 1. How do I determine the force in the green and pink members of a truss with counters?

To find the force in the green and pink members, you will need to use the method of joints or method of sections. First, draw a free body diagram of the truss and label all known forces and angles. Then, use the equations of equilibrium (sum of forces in the x and y directions and sum of moments) to solve for the unknown forces.

## 2. What is the purpose of counters in a truss?

Counters are used in truss design to provide stability and prevent buckling. They are diagonal members placed in the truss to resist lateral forces and keep the structure from collapsing.

## 3. How do I determine if the force in a green or pink member is compressive or tensile?

If the force in the member is pulling away from the joint, it is tensile. If the force is pushing towards the joint, it is compressive. You can also use the sign convention of positive forces being tensile and negative forces being compressive.

## 4. Can I use the same method to find the force in all members of a truss?

Yes, the method of joints or method of sections can be used to find the force in any member of a truss, as long as there are enough equations available to solve for the unknown forces. However, it is important to note that the method of joints is more suitable for trusses with fewer members, while the method of sections is better for more complex truss structures.

## 5. What are some common assumptions made when analyzing trusses with counters?

Some common assumptions include: all joints are pinned and can only support axial forces, all members are 2-force members (meaning they only have axial forces), and the truss is in static equilibrium. These assumptions simplify the analysis and allow for easier calculation of forces in the members.

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