AISC Appendix 6 - Beam Bracing 5

[JAE]

Through past work by Yura the AISC specification has added appendix 6 which provides requirements for bracing.

Section 6.3 is for beams and includes both relative and nodal bracing requirements. The general concept I understand but we've been trying to understand the explicit application of this section to brace designs.

The main issue is that the brace strength equations ((A-6-5 and A-6-7) provide values for required brace strength, Pbr. This is given as a force in pounds as a function of bending moment and beam depth (h_o).

But if we provide braces at 4 feet on center, we get a Pbr value. If we put braces at 8 feet on center we still get a similar Pbr value since the moment is the same. The beam might be a bit deeper since Lb would be larger, but that means that the brace force actually gets smaller since h_o is in the denominator....and that seems counter-intuitive...fewer braces means less brace strength required.

Anyone have any views on this?

 

[haynewp]

Suppose, though, that you design a beam spanning 30 feet taking a total distributed load w, with a single brace at midspan. The moment is wL^2/8 and you get a beam size based upon Lb = 15 feet.

You use the AISC forumlae to get a Pbr capacity for your brace at that point. So far so good.

But now if you look at using 3 brace points, the Lb now equals 10 feet, and you get a new beam that is perhaps a bit smaller in weight but the same depth.

With the second design, AISC implies that you need almost 3 times the brace strength of the first design since the Pbr's for each brace will be approximately the same as for the single brace.

Mr (required flexural strength) changes for the braces at the first and third points for a UDL, SS beam. The brace force required at these locations will be less than the center brace and you now have the advantage of being able to use a lighter beam. Though you have to install 3 braces instead of 1.

 

[slickdeals]

Any luck trying to contact the AISC Solutions Center?

 

[JAE]

haynewp - that is intuitive (I think)

Here's an example design:

Span = 30 feet
Factored uniform load = 1.5 k/ft.
C_d = 1.0

[blue]Case 1 - Brace only at midspan
Factored moment at midspan = 2025 in-kips
Lb = 15 feet
Required beam size = W16x40
h_o = 15.49 inches

Per AISC Equation A-6-7,
P_br = 0.02M_rC_d / h_o

Required brace strength P_br= 2.61 kips

Required brace stiffness [β]_br = (1/[φ])(10M_rC_d)/(L_bh_o)

Required brace stiffness [β]_br = 8.07 kips/in.

Case 2 - Braces at 1/4 points
Factored moment at midspan = 2025 in-kips
Factored moment at 1/4 point = 1519 in-kips
Lb = 7.5 feet
Required beam size = W16x36
h_o = 15.47 inches

Required midspan brace strength P_br= 2.62 kips
Required 1/4 point brace strength P_br= 1.96 kips (this is based on the 1/4 point moment)

Required brace stiffness at midspan [β]_br = 16.16 kips/in.

Required brace stiffness at 1/4 points [β]_br = 12.12 kips/in.[/blue]


So for Case 1 we need a brace with strength = 2.61 kips and a stiffness = 8.07 k/in.

For Case 2 we need just a bit smaller beam, but the brace strengths required are 2.6 and 1.96 kips - almost the same individually but in total almost 3 times that required of the single brace.

For the stiffness in Case 2 - we need a lot more stiffness due to Lb being in the denominator and being 1/2 as much as case 1.

I guess I can go with this - but again, it seems strange that for simply adding braces, we need twice as much stiffness and in total three times as much brace strength. But I guess the moment in the beam is the moment in the beam and at any point it requires 2% or so strength capacity and the key is to optimize your braces through trial and error.

 

[kslee1000]

By inspection, it makes sense to me, though the word "required" does not seem fit.

The brace stiffness is derived by Pbr/d (d=lateral displacement), let's exam the center brace, as Pcr remains constant, however, "d" becomes much smaller from effect of adding brace at 1/4 point, thus the brace stiffness increases. Work out?

 

[BAretired]

A hefty beam may be adequate without lateral bracing, but for very long beams, the beam would need to be sized to resist lateral buckling. A more economical beam may be chosen by providing lateral bracing to the top chord at discrete intervals.

Building codes recognize that no beam can be erected perfectly straight, so a tolerance is permitted. The permissible deviation from a straight line may be expressed as k*L where L is the clear distance between braced points. The value of k may vary amongst codes but is likely to be in the order of 1/500. So D₀ = k*L is simply the permissible deviation from a straight line for any beam.

If single point bracing is chosen at midspan, the brace must resist a moment of Cf*D₀ where Cf is the compressive force in the flange and D₀ is the permissible deviation from a straight line. But, in resisting that force, the brace must deflect a bit more. Let us assume that, at equilibrium, the additional deflection is D_b. Let us say the force applied at the brace point is Pb.

It is a simple beam. The reaction at each end is Pb/2. The moment at midspan is Pb*L/2. But the moment is equal to Cf*(D₀+D_b). So Pb*L.2 = Cf(D₀+D_b) and Pb is found to be 2Cf(D₀+D_b)
The beam size may be further reduced by bracing at more frequent intervals. If the beam is braced at an infinite number of brace points, a point of inflection will occur at the midpoint of a typical interior span. The shear at the inflection point will be Pb/2 and the moment at the typical node will be Pb/2*(L/2) = Pb*L/4. But the eccentric moment from Cf is still Cf(D₀+D_b). So Pb = 4*Cf(D₀+D_b)/L.

In all cases, the spring stiffness is given by Pb/(D₀+D_b).

That is why the brace force is larger with multiple brace points.

BA

 

[WillisV]

JAE - I guess I'm not quite understanding your questioning of the required brace strength and stiffness going up with more braces. Thinking of it in terms of column-type bracing, it takes more and more force/stiffness from each brace point to get a column to buckle into higher and higher modes (read the first paragraph of Commentary App 6.2 - the most severe case is an infinite number of braces).

 

[miecz]

Sticking with JAE's example from above, let's say we provided a W16x40 with 3 braces, each with a stiffness of 10 kips/in and a strength of 3 kips. It seems like that wouldn't meet the code unless we removed (or ignored?) the two outer braces. Is that right?

 

[kslee1000]

I think all the confusion is centered on the phrase "required brace stiffness". Without look into the code, judged by concept, shouldn't it be "achieved stiffness (by providing brace with strength XXX, in a spacing XX)"?

 

[BAretired]

JAE said:

For Case 2 we need just a bit smaller beam, but the brace strengths required are 2.6 and 1.96 kips - almost the same individually but in total almost 3 times that required of the single brace.

The forces in Case 2 are alternating in direction, so the total brace force acting on the whole beam is 2(1.96) - 2.6 = 1.32 kips which is only half the force required in Case 1.

If a beam requires more than three braces in order to satisfy the moment, individual forces will be larger but the sum of all brace forces acting on the beam will be close to zero.

BA

 

[JAE]

I guess I was intuitively expecting that if I take a W16 beam and provide 3 braces instead of one, I'd need overall about the same total strength and stiffness, not exponentially more.

WillisV - I can see that with additional braces on a column, that the resulting column "design" would allow more axial load...perhaps a lot more, and that would kick up the brace requirements...so the light is beginning to shine a bit on my befuddled brain based on that.

But if you have a column with the SAME axial load, but add more braces, the formulae present much higher brace requirements that don't seem necessary and this has led me to suggest that the designer can use these brace formulae to economize the design by using the minimum number of braces to offer the lowest cost.

miecz - yes, that is the sort of conundrum that I see in these specifications. In my Case 1 above I have one brace with Pbr = 2.61 k and [β]br = 8.07 k/in. It works with a W16x40.

Now if I still use a W16 x 40 beam and add two additional braces at the 1/4 points, both with the same strength and stiffness, the equations say that I need a strength Pbr = 2.64 kips at the center (which now doesn't work) and 1.96 kips at the 1/4 points (which works).

For stiffness, I need 16.33 k/in and 12.1 k/in at the center and 1/4 points respectively, neither of which work now....before, with one brace, it worked. This makes NO sense to me.

 

[WillisV]

JAE - regarding the column example, once you add enough braces to basically develop the squash load, you can still add more braces, and you will still get the same strength, but the required brace loads and stiffness will continue to creep upwards because you must force the column to buckle in higher and higher modes which takes more energy for lack of a better word.

 

[kslee1000]

JAE:

One suggestion. Make a simple beam model with parameters identical to those used in the cal you provided, and set the stiffness as spring support with schemes as in the cal. Now apply an unit load in the center, then observe the change in reactions and deflections. Take a long walk and think it again. Hope this would help.

 

[BAretired]

JAE,

I agree with you to a considerable extent. If you have a 30' beam, say W16x40 which works with one brace in the middle, it cannot be made more critical by adding quarter point braces. In fact, the quarter point braces should help alleviate the stress on the midspan brace.

In that situation, the quarter point braces are not "required" braces. You can add them if you wish but the code does not recognize that they help the midspan brace. In fact, there is no code requirement to be satisfied for the two added braces.

If, on the other hand, the beam needs quarter point braces to carry the moment, then all three braces are "required" braces. In that case, AISC dictates that each brace carry 2% of the flange force at the brace point.

So where is the problem?

BA

 

[JAE]

BAretired, kslee1000, WillisV, miecz - thanks for the posts - the last question BA asked "where is the problem" is really just me trying to presume that everything in the AISC specification makes perfect sense, which I know no spec will.

I know the Appendix 6 is an appendix and so maybe needs some maturing before making it into the full body of the specification - I was just struck by the fact that the section doesn't attempt to deal with multiple braces along a span directly but forces you to play around with it a bit to get a feel for it. In this instance, the "feel" I was getting didn't feel right at all.

Thanks again all of you.

 

[JAE]

...and thank you haynewp as well

 

[kslee1000]

haynewp:

Sorry, you are one step late. I have already stole the star

 

[haynewp]

Thanks for not leaving me out JAE

 

[miecz]

JAE

Thanks for bringing this up. A beam with three braces cannot be weaker than the same beam with one brace. I think the solution is to ignore some of your braces if the beam works with less braces than you provide. So, if a beam/bracing meets the code with one brace and not with more, ignore all the braces but one.

 

[ARLORD]

My two cents: According to the commentary, page 16.1-421, 4th paragraph last sentence: " Member inelasticity has no significant effect on the brace requirement." Therefore, for beams the size and bending unbraced length is not a function to the brace load or stiffness.

The required brace force, Pbr, is only a function of the axial load in the member, Pr for columns and Pr=Mr/ho for beams. For beams, Mr does vary with the length of the beam according to the moment diagram.

The stiffness of the brace has the units k/in. The "inch" part of the units is the lateral movement perpendicular to the member. See commentary Figure C-A-6.3(a), it shows a little diagram relating the axial force P=Pr, and the braced length L=Lb, the brace force Pbr and delta. It looks to me, for the same axial load P, as the length of the "member" increases it becomes less rigid and therefore requires a lower force, Pbr, less stiff member, to resist that lateral movement. Again, no relation to the bending unbraced length, the member size, etc.

 

[ARLORD]

The required brace force, Pbr, can also be modified per commentary equation C-A-6-1. If the provided brace is stiffer than that required, this equation reduces the required brace force.