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Stability bracing for flexural members
2

Stability bracing for flexural members

Stability bracing for flexural members

(OP)
I am designing a freestanding scoreboard with wideflange columns. The axial load on these columns is small compared to the flexural demand. I would like to design a bracing system (at 10 ft above grade) to justify an unbraced length of the compression flange of 10 ft. In order to design the bracing members (HSS braces and stringer), should I assume that the bracing is relative or nodal? I have read AISC 360 Appendix 6 and commentary as well as several other related questions on this forum but I am still unsure. See attached sketch. Thanks in advance for your recommendations.

RE: Stability bracing for flexural members

Sekwahrovert:
You don’t know which direction the lateral force may come from, or which col. flg. might be the buckling flg. And, you don’t want these cols. to be like drunken sailors leaning against each other for support, and hoping one of them doesn’t start to move. Why not make two identical, light symmetrical trusses, which bolt inside the flgs. on the cols. Light, but still fairly rigid in their own plane, so the cols. can’t twist. These would have two fairly strong chords and diags. as needed and bolt to the col. flgs. They could also be used in the bracing and col. setting process to help space and hold the cols. during erection.

RE: Stability bracing for flexural members

(OP)
dhengr: thanks for the reply and for the concern. I have always felt comfortable saying that the HSS brace and stringer can resist lateral-torsional buckling of either flange, making the wind direction irrelevant. Do you disagree? If so, can you explain?

I understand your truss idea, but the concept I sketched requires very few members and connections in comparison, and as long as the braces and stringer are sized appropriately, accomplishes the same thing.

RE: Stability bracing for flexural members

It is nodal bracing. Let me see if I can explain this properly. To be relative bracing, the column/beam would need at least two brace points on the beam/column and if one brace point began to rotate or displace the other brace point on the same beam/column would also need to rotate/displace.

I am not 100% clear on the theory of bracing cantilevers. But the code requires the tension flange to be laterally braced at the free end of the cantilever. See AISC 360-10 app 6.3.1(1). I don't know if lean-on or rotational bracing at the top of the scoreboard columns will accomplish this.

RE: Stability bracing for flexural members

(OP)
wannabeSE: Thanks for your response. I think I agree with you that this is nodal bracing, but I'm not sure I follow your explanation. Which member are you referring to as the column/beam and which member are you referring to as the beam/column? In my case, I think I do have two brace points on the wideflange column: the 45 degree brace is attached to one flange and the stringer is attached to the other. Regardless of the direction of the wind load, the 45 degree brace will experience an axial load and will transfer a component of that load into the stringer in flexure. For the purposes of this discussion I am only interested in the flexural (LTB) bracing, not the axial bracing. I am not using my stringer and brace to reduce the axial unbraced length. However, I feel that the definition of nodal vs. relative bracing is more straightforward for columns (axial members) than for beams (flexural members).

RE: Stability bracing for flexural members

I am sorry that I can not explain this any better. I should have just said beam not beam/column or column/beam. The bracing scheme shown only braces the beam compression flange in one location (and/or it only braces the beam from rotating in one location).

RE: Stability bracing for flexural members

Are you sure that would not be relative bracing?
Here is an explanation from someone who is smarter than I:

Frome thread507-194560: Relative vs. Nodal bracing

Quote (WillsV)


WillisV (Structural)
9 Aug 07 14:52
Fig. C-A-6.2 on p16.1-422 does a pretty good job of illustrating this concept for both relative and nodal bracing on columns and beams.
A relative brace is a brace between two objects that moves along with the objects - they are objects being braced are being braced RELATIVE to one another - but the entire system can still be swaying. For instance the diagonals in a braced frame are relative - they brace the columns however the entire frame can still sway. Think of it as INTERNAL bracing to the system. These are the most common types of braces in building construction.
A nodal brace braces an object from an exterior "rigid" point. Think of a big concrete shear wall with beams over to a moment frame. The beams brace the columns at descrete points, the columns are not braced relative to one another.

Another link:
https://www.aisc.org/store/p-2296-notes-on-the-nod...

EIT
www.HowToEngineer.com

RE: Stability bracing for flexural members

(OP)
RFreund: You provided some great background info for me. I'm thrown off by the following statement from the Commentary to Appendix 6 (page 16.1-421 in the 13th Edition AISC manual):

"For beams a cross frame between two adjacent beams at midspan is a nodal brace because it prevents twist of the beams only at the particular cross frame location."

When I first read this, it sounded like it was talking about beam lateral bracing (6.3.1), but perhaps it's talking about torsional bracing (6.3.2). Unless someone can convince me otherwise, my current position is that the bracing system shown in my sketch is a relative lateral brace and a nodal torsional brace. I wish there were a simpler explanation than provided by AISC that could make the concept "click" in my brain.

RE: Stability bracing for flexural members

Yes, they are referring to a nodal torsional brace. Refer to the Fig. C-A-6.2 on the next page. They show a crossframe and label this as a Torsional Brace (nodal). Because this prevents twist of the section the unbraced length for LTB is limited to the distance between these points. or in the case of the figure L/2

EIT
www.HowToEngineer.com

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