How to identify members in a truss as compression or tension?

[Vicinity24]

Homework Statement

Is this correct? The sum of the forces in the Y have to be zero and there is only two y-components so it's pretty simple to determine this. There are also only two x-components so those will also cancel each other out as they have to go in opposite directions to balance as it is in static equilibrium = 0

I don't know what to do from here. If the diagonal '?' is tension because it has to balance the compression in the y direction then what would the diagonal force be in the horizontal '?'.

I don't know how to balance this because of the 5th member being there. Since 2 is compressing in the horizontal then 7 has to compressing too but after I've founds this, how would I know what 5 is?

This one has really got me troubled. 5 is what I believe a zero force member.

Homework Equations

ΣFx=0, ΣFy=0, this is meant to be a qualitative question but even when I do it quantitatively I cannot figure some out.

The Attempt at a Solution

Pointing to the joint is compression, pointing away from the join is tension.

 

[JBA]

In the top figure, you are correct except for the reaction at the roller. Since the force vector is exactly aligned with the left triangle leg there is no moment about the pin and therefore forces applied to the roller (assuming a weightless triangle).

In the second figure, the top horizontal member is in compression, so what does that tell you about the two interior diagonals? You arrows you have displayed for the right triangle are correct and mirrored on the left triangle.

In the third figure, 4 is in compression, so, what does that tell you about 5 and 7?

In the fourth figure the the force arrow should only be vertical upward (the roller, by definition cannot generate any horizontal forces, and 5 cannot be 0. So, see if you can resolve the other forces based upon those two facts.

 

[Vicinity24]

In the top figure, you are correct except for the reaction at the roller. Since the force vector is exactly aligned with the left triangle leg there is no moment about the pin and therefore forces applied to the roller (assuming a weightless triangle).

In the second figure, the top horizontal member is in compression, so what does that tell you about the two interior diagonals? You arrows you have displayed for the right triangle are correct and mirrored on the left triangle.

In the third figure, 4 is in compression, so, what does that tell you about 5 and 7?

In the fourth figure the the force arrow should only be vertical upward (the roller, by definition cannot generate any horizontal forces, and 5 cannot be 0. So, see if you can resolve the other forces based upon those two facts.

For the second one if the top horizontal member is in compression then the interior diagonals are in tension, I was about to ask how you determined the top to be horizontal but then I figured out that to balance the y forces from A, the diagonal has to be in tension.

In the third one, the top horizontal is in compression, 6 is also 0. You say member 4 is in compression which I understand, so member 9 is in compression to balance the x force from 4 and member 5 is in tension to balance the y force from 9. I think I got it. 7 would therefore be compression.

Alright I applied what you said and I started working from the roller but I get stuck after finding 9 to be compression and 8 to be tension. So I move to the pin and find out 1 is compression and 2 is tension but again I am stuck now at figuring out member 3 and 4. All of these 3 members have both vertical and horizontal components so I flip it to make 1 and 4 horizontal:

But at this point I need to balance the x components. 3 and 4 could both be tension or compression when I look at it. So instead should I try to find force 5 and work for 3 from that direction?

thanks btw!

 

[JBA]

In the third one, the top horizontal is in compression, 6 is also 0. You say member 4 is in compression which I understand, so member 9 is in compression to balance the x force from 4 and member 5 is in tension to balance the y force from 9. I think I got it. 7 would therefore be compression.

With 4 and 9 in compression, both 7 and 8 are going to be in tension.just same as 2.On the fourth figure , first of all I failed to correct your mistake about the vertical reaction force on the pin, that force should be directed down to resist the tension force of 1, because the vertical offset between the force at the top and the roller is trying to lift the structure off of the pin. As a result, both 1 and 4 are in tension; and, likewise, 6 and 9 are in compression to resist the upward reaction at the roller. Also 8 just like 2 is in tension.
With all of the above results in place. To resolve 5, 3 and 7. First visualize 5 with 3 and 7 removed; and, next visualize 3 and 7 with 5 removed. That should help you see the answers for those three members.

I hope this helps you complete your analyses.

 

[Vicinity24]

With 4 and 9 in compression, both 7 and 8 are going to be in tension.just same as 2.On the fourth figure , first of all I failed to correct your mistake about the vertical reaction force on the pin, that force should be directed down to resist the tension force of 1, because the vertical offset between the force at the top and the roller is trying to lift the structure off of the pin. As a result, both 1 and 4 are in tension; and, likewise, 6 and 9 are in compression to resist the upward reaction at the roller. Also 8 just like 2 is in tension.
With all of the above results in place. To resolve 5, 3 and 7. First visualize 5 with 3 and 7 removed; and, next visualize 3 and 7 with 5 removed. That should help you see the answers for those three members.

I hope this helps you complete your analyses.

Aha At first I was still confused then I realized that member 3 would be in tension since the arrows of both 1 and 4 point to the left. Would 7 be compression then? Since it has one y component over the axis pointing down (6) and two (9 and 7) below, the ones below have to point to the joint to balance it? Is it a logical way to think about it? Would 5 then be in compression too?

Edit: Or they're just zero... I thought that but you said five isn't zero so...

 

[JBA]

From my view as best as I can see it:

5, is in compression;

3 cannot be in tension and my best judgement is that it is in compression but it could be 0

7 remains a bit of a mystery for me and appears that it could also be 0 as well

I am sorry I can't be more specific on figure 4, but I don't want to risk giving you some answer that I am not fully convinced is correct.

 

[insightful]

If you can take the truss out, and nothing moves, it is a "0."

If only one truss, say x, attaches at a mid-point of a straight truss, x is a "0."

 

[Vicinity24]

Alright I got it, thanks JBA and insightful!

 

[JBA]

Vicinity 24, Since I could not resolve the issue of trusses 3 and 7 in figure 4, I decided to do what I would have done during 20 years of doing product development and testing, make a model, load it and see what happens. Being able to do this type of thing is why I enjoyed almost every day of my 45 year engineering career.

As you can see in the below photo sequence the final answer is: Truss 3 is in compression (this can be implied by the slight displacement to the right of the top of truss 5) ; and, Truss 7 is (had it been permanently attached it) in tension. (The truss assembly is permanently attached to the left end block and free to slide on the right end block.)

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[insightful]

The bowing of your model truss means that truss has a bending moment. Truss analysis assumes only tension or compression in a truss; no moments. I believe 3, 5, and 7 are "0" trusses. Unlike real truss members, ideal trusses cannot fail in a buckling (bending) mode, as your model did.

 

[JBA]

If it is assumed that the truss members are ideally rigid under both tension and compression loads, then all triangles are perfectly rigid; and, I agree with your conclusion that 3,5 & 7 would be "0" trusses.

 

[Vicinity24]

Vicinity 24, Since I could not resolve the issue of trusses 3 and 7 in figure 4, I decided to do what I would have done during 20 years of doing product development and testing, make a model, load it and see what happens. Being able to do this type of thing is why I enjoyed almost every day of my 45 year engineering career.

As you can see in the below photo sequence the final answer is: Truss 3 is in compression (this can be implied by the slight displacement to the right of the top of truss 5) ; and, Truss 7 is (had it been permanently attached it) in tension. (The truss assembly is permanently attached to the left end block and free to slide on the right end block.)

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I forgot to reply when I saw it as I was busy but that really helped. It's really cool how you just made that and tested it. Thanks again!

I've gotten better at trusses since then too :P