Steel truss bottom chord in compression 5

[LJ_]

Hello,

I have this truss that has the bottom chord in compression. The length between the truss supports is 90ft and the section is HSS. The joints in the truss are every 8 ft. The trusses are spaced 20ft.

When I check for compression in the bottom chord, the length is 90ft so my HSS fails. Can I use gusset plates in the joints so I can provide for out of plane rigidity and thus reduce the length of analysis?. If the length was 8ft my member wouldn’t fail. Or is it necessary to provide bracing by connecting the trusses in the joints.

What do you think?

 

[dold]

It sounds like OP may have these trusses bearing on concrete columns/pilasters - there was talk of embeds... Depending on the top restraint of the columns/pilasters (i.e., concrete slab (for some reason...) tying in to the top of the column/wall, just above the truss bearing), I could see that leaning towards a pin-pin support system. Any other situation (bearing on steel columns/beams/other trusses/etc) that's a pin-roller - all day long. For what it's worth, I've done my fair share of piddling around in RISA, etc., modeling trusses and other 'exerimental' setups. It does matter, even though many programs only run linear analysis. Non-linear analysis would only amplify this effect. See pic below. Even though we're not seeing bottom chord compression (which I suspect might have been caused by either: 1) fully fixed supports and fixed members or, 2) some sort of restraint at bottom chord in plane of the truss), we do see significant differences in top chord axial forces. So, food for thought. This does not address any buckling, etc. Just a discussion about fixity.

OP, it would be very helpful (even just for our curiosity) if you provided more information: sketch, bearing condition detail, screenshot of your model, etc.

 

[r13]

Josh,

Here is a simple verification model (using RISA2D). The first two models are Pin-Roller support trusses with external gravity load, and lateral load respectively. The third is the superposition of results obtained above. The fourth is Pin-Pin support truss with the identical external gravity load as in model 1. (Red - applied load; D: deflection)

 

[JoshPlumSE]

Retired13 -

1) Your gravity load of 2 kips vertical causes a 1 kip lateral reaction at the pinned-pinned supports. Does feel right to you? And, a no force in the bottom chord. Still feel right?
2) Whereas if you make it pinned-roller, you get a deflection of 0.003 inches. 3 mils. If I were to apply a 1 kip force to the top of a 12ft tall W18x97 column (which should be pretty stiff in the strong axis) then I get a deflection of 0.021 inches. 7 times the deflection you'd get from the pinned roller model. 14 times when you consider that the 1 kip reactions happen on both sides of the truss.

Does that make it more clear why the pinned-roller model should be much much closer to the real results than the pinned-pinned? If not, then create a model that includes the supporting columns and compare the forces in the truss compared to what you get with your pinned-pinned and pinned-roller models. You'll see that the results are much, much closer to the pinned-roller model.

I hope that clarifies things.

 

[Agent666]

Well said JoshPlum, hopefully that puts to bed all the misinformation and pursuit of incorrect fantasy theories regarding infinitely rigid support fixity being a real thing!

Retired13, you're trying to prove a solution/theory for a problem that just doesn't exist in reality. Please stop, you've tried multiple times to convince people of your views but no one is in agreement as you can see.

 

[r13]

My first three models are on pin-roller support. A horizontal load (1 in this case) was introduced on model 2 to close the gap, then model 1 is imposed on model 2 to yield model 3, which in turn agrees with model 4 that was supported by pin at both end. The thing I want to point out is the lateral deflection play an important role in truss analysis, as it will alter the result.

 

[r13]

Agent,

This is simple engineering mechanics.

 

[JoshPlumSE]

This is simple engineering mechanics.

And yet, you (retired13) are having a really, really tough time understanding it. Let's put it another way:

a) In your Pinned-pinned model you have an INFINITELY stiff lateral support. Right?
b) We know that's not correct, we know that it has to be somewhere between a pinned and a roller.
c) If the truss were allowed to deflect 0.003 inches, then the lateral reaction at the support would be zero. But, your pinned-pinned model shows a lateral reaction of 1 kip.
d) Therefore, in order for your pinned-pinned model to be correct, the actual lateral stiffness at each end would have to be significantly greater than 2* 1kip / 0.003 inches. That's a lateral stiffness significantly greater than 667 kips per inch. What support is going to give you a stiffness like that? If the column support were a 12 ft tall W44x335, then that lateral stiffness of each support would be about 714 k/in. Probably still not stiff enough. But, at least it's in the ball park. However, I don't think the lateral stiffness of your column support for this truss will (in reality) be anywhere near that stiff.

e) By all means, I accept the argument that modeling in the actual lateral stiffness of your column supports is going to be the most accurate solution. All I'm saying is that if you actually do this, you will find out that the truss forces will be almost identical to the Pinned-Roller case. I'll follow this up with a similar type of truss modeled this way to show you how good the pinned-roller model really is.

 

[r13]

Josh,

This confirms your point. Thanks.

See revised diagram below.

 

[r13]

[Correction] I found a glitch on my previous model. The forces remain the same, but deflection differs, which is anticipated.

 

[r13]

Josh,

I think my original model is more useful in truss sitting on rigid abutment, the type of support will alter the results, so the physical support must in agreement with the analysis. However, your thinking is correct for truss supported by typical frames, or walls. Very good point, I appreciate your effort in clearing up the air.

 

[JoshPlumSE]

Retired13 -

No problem.... I appreciate that you didn't take my attempts at correcting you as a personal attack.

For trusses on a rigid abutment, you may be correct. However, my tendency would be to envelope the design for both methods. I just tend to think that in the field there is often some give in the connections or bolts or such. That's where engineering gets tough. When the real world doesn't want to cooperate with our design / modeling assumptions.

 

[r13]

Josh,

Envelop is a good/sound practice, no doubt there. I accept and appreciate pointed comments, nobody knows every tricks until enlightened by others. I think this is the most important feature/purpose of this forum. Thanks again, as I really didn't think about it until your comments.

 

[BAretired]

Looking back at dold's truss, it is top chord bearing. With a pin at each end, the total strain in the top chord must be zero. Green lines represent compression, blue lines tension. The thickness of the lines indicates magnitude to some scale. Interior panels are in compression but the end panels are in tension, a necessary outcome to satisfy the boundary conditions.

If the truss had been bottom bearing, there would be a similar stress reversal in the bottom chord with the end panels in compression and interior panels in tension.

BA