STEEL MANUAL - TABLE 3-10

[EBAT75]

Lb absolutely is zero if the top flange is continuously braced

The definition of Lb in section F2-3, page 16.1-47 of the 14th edition: 'Lb = length between points that are either braced against lateral displacement of the compression flange OR braced against twist of the cross section'

E720 is quicker to the draw!

(c) When Lb > Lr
Mn = FcrSx ≤ Mp (F2-3)
where
Lb = length between points that are either braced against lateral displacement of
the compression flange or braced against twist of the cross section, in.

This definition of Lb in ONLY WHEN Lb>Lr. Please remember what I said earlier -Up to only Lr, fully braced, you can use 3-6.

The danger is when we do not match the definitions to the context in hand. When the context or boundary conditions change, the definition, in this case the definition of Lb changes. It is not static.

 

[thedaywa1ker]

This definition of Lb in ONLY WHEN Lb>Lr. Please remember what I said earlier -Up to only Lr, fully braced, you can use 3-6.

The danger is when we do not match the definitions to the context in hand. When the context or boundary conditions change, the definition, in this case the definition of Lb changes. It is not static.

I remember what you said earlier, I just don't think that is true.  Can you give me a code reference for this alternate definition of Lb?

'Up to Lr, fully braced'

I am confused. If a beam is unbraced up to Lr, then it is not fully braced.

 

[EBAT75]

'Up to Lr, fully braced'

I had the code reference in the post itself - Eq. F2-3.

As for 'Up to Lr, fully braced', beyond Lr, braced or not does not matter. Why? Because, Fcr in Eq F2-4 has Lb squared in the denominator. The effect on Mn is not linear now. The (1/Lb squared) factor makes it curvilinear and as Lb approaches say infinity becomes asymptotic to zero i.e. even a gentle breathing can cause buckling. AISC cuts the 3-10 table values at 30 times beam depth as a practical limit. 

Reading the full sections in F2.2 1 through 3 will answer the questions on this topic. On the definitions of bracing, lateral restrained, segments, points of inflection, twist, bracing of which flange etc, I will get to them later.

I am glad we started with the beam weight and got to the meat of LTB.

 

[thedaywa1ker]

I had the code reference in the post itself - Eq. F2-3.

As for 'Up to Lr, fully braced', beyond Lr, braced or not does not matter. Why? Because, Fcr in Eq F2-4 has Lb squared in the denominator. The effect on Mn is not linear now. The (1/Lb squared) factor makes it curvilinear and as Lb approaches say infinity becomes asymptotic to zero i.e. even a gentle breathing can cause buckling. AISC cuts the 3-10 table values at 30 times beam depth as a practical limit. 

Reading the full sections in F2.2 1 through 3 will answer the questions on this topic. On the definitions of bracing, lateral restrained, segments, points of inflection, twist, bracing of which flange etc, I will get to them later.

I am glad we started with the beam weight and got to the meat of LTB.

F2-3 has the definition of Lb that I am familiar with.  I am asking why you think that the definition of Lb changes, and where that is in the code. 

Your changing definition of Lb is why your interpretations of 3-10 and 3-6 are off.  3-6 assumes Lb of zero (for all intents and purposes), aka fully laterally braced. Lb/Lr/Lp aren't relevant to that table at all.

This definition of Lb in ONLY WHEN Lb>Lr. Please remember what I said earlier -Up to only Lr, fully braced, you can use 3-6.

the definition of Lb changes. It is not static.

I don't think these statements are correct.

 

[thedaywa1ker]

Looking at the capacities in 3-6

Say you have a beam spanning 40 feet.

You are saying that you can use this table for unbraced lengths up to Lr.

How do you tell the table what your unbraced length is?

A particular beam that spans 40 feet only has one capacity per table 3-6.  As the beams laterally unbraced length increases, the capacity will decrease.  How is this reflected in that table?  It isnt.

 

[EBAT75]

I am asking why you think that the definition of Lb changes, and where that is in the code. 

Fig.C-J-10.2 might help.

There are many other instances where this is borne out. Why I mentioned On the definitions of bracing, lateral restrained, segments, points of inflection, twist, bracing of which flange etc, I will get to them later.

 

[thedaywa1ker]

Fig.C-J-10.2 might help.

There are many other instances where this is borne out. Why I mentioned On the definitions of bracing, lateral restrained, segments, points of inflection, twist, bracing of which flange etc, I will get to them later.

I'm on the edge of my seat

 

[thedaywa1ker]

Are you saying that you back-calculate phi*Mpx from 3-6, to then use with equation 3-4a?

 

[EBAT75]

Are you saying that you back-calculate phi*Mpx from 3-6, to then use with equation 3-4a?

I am a bit lost in the woods. 

F3. Doubly Symmetric I-Shaped Members with Compact Webs and Noncompact
or Slender Flanges Bent About Their Major Axis

The tables have footnotes to cover compact, non-compact, slender..........all F sub-sections. Looking for 3-4a, I cannot find a 3-4a.

TDW, I will pick up after work. Appreciate your spirit.

 

[thedaywa1ker]

I am a bit lost in the woods. 

F3. Doubly Symmetric I-Shaped Members with Compact Webs and Noncompact
or Slender Flanges Bent About Their Major Axis

The tables have footnotes to cover compact, non-compact, slender..........all F sub-sections. Looking for 3-4a, I cannot find a 3-4a.

TDW, I will pick up after work. Appreciate your spirit.

It is on page 3-9

I am most curious for your answer to this post:

Looking at the capacities in 3-6

Say you have a beam spanning 40 feet.

You are saying that you can use this table for unbraced lengths up to Lr.

How do you tell the table what your unbraced length is?

A particular beam that spans 40 feet only has one capacity per table 3-6.  As the beams laterally unbraced length increases, the capacity will decrease.  How is this reflected in that table?  It isnt.

 

[dauwerda]

I agree with thedaywa1ker on this one. Lb is not only defined in F2-3, it is also defined in the index of the Manual (under "General Nomenclature") the definition given here is the exact same as mentioned above.

The intent of F2.2(c) is not to change the definition of Lb, it is saying the definition of Mn given by that equation is only applicable when Lb is greater than Lr.

 

[EBAT75]

I agree with thedaywa1ker on this one. Lb is not only defined in F2-3, it is also defined in the index of the Manual (under "General Nomenclature") the definition given here is the exact same as mentioned above.

The intent of F2.2(c) is not to change the definition of Lb, it is saying the definition of Mn given by that equation is only applicable when Lb is greater than Lr.

Have either of you  (you and daywa1ker) looked at Fig.C-J10.2 and also Fig. C-F1.4. ?

Here is the text that goes with the latter, Fig. C-F1.4:  "In this case, the Cb = 5.67 would be used with the lateral-torsional buckling strength
for the beam using an unbraced length of 20 ft which is defined by locations where twist or lateral movement of both flanges is restrained."

I have also seen the General Nomenclature thing, but it is "General" or generic. I also have seen the following definitions for Lb in the same General Nomenclature. They are :

Lb Length between points that are either braced against lateral
displacement of compression flange or braced against twist
of the cross section, in. (mm) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . F2.2 (ALREADY HACKED TO DEATH)

Lb Distance between braces, in. (mm) . . . . . . . . . . . . . . . . . . . . . . . . . . . App. 6.2
Lb Largest laterally unbraced length along either flange at the point
of load, in. (mm) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . J10.4
 

The eng-tips thread had one gem from a contributor after a lot of posts from him. I have no time to look at it again, but it ran something like  he doesn't care about all of this, what really matters is going back to the fundamentals.

 

[EBAT75]

It is on page 3-9

Page 3-9 of what?

 

[thedaywa1ker]

Page 3-9 of the steel manual...immediately prior to tables 3-6 and 3-10...

I don't see anything in what you've posted that would lead me to believe that the writers of that code mean anything by Lb other than what they have explicitly defined.

The question of specifically what kind of restraint is sufficient in real life to prevent LTB is a can of worms.  But, AISC simplifies it for us dumb designers with their definition of Lb, so that we can do a quick check and move on 

 

[EBAT75]

Looking at the capacities in 3-6

Say you have a beam spanning 40 feet.

You are saying that you can use this table for unbraced lengths up to Lr.

How do you tell the table what your unbraced length is?

A particular beam that spans 40 feet only has one capacity per table 3-6.  As the beams laterally unbraced length increases, the capacity will decrease.  How is this reflected in that table?  It isnt.

Thx for the info where 3-9 is. My edition of 14th is page 3-10. Anyway, let us look at the 40 ft span. Say Lp is 10 ft. 

from manual  for table 3-6: Maximum total uniform loads on braced (Lb ≤ Lp) simple-span beams bent about the strong
axis are given for W-shapes with Fy = 50 ksi (ASTM A992).

So, say you have that beam with a span of only 10 ft, Lb=10ft, the max load corresponds to that which generates Mp (yield limit state ie no LTB). 

Increase the span to 20 ft. Isn't there going to be LTB? In an earlier post, you said "Table 3-6 is based on braced (Lb always < Lp, therefore not considering LTB) beams". 

Why are spans longer than Lp listed? Because they are braced at intervals of Lp or less, but beyond the first Lp, LTB kicks in, the Mn is not Mp anymore but the table 3-6 is based on beam being braced at least every Lp. "..............braced (Lb ≤ Lp) simple-span beams". Keep going to say Lr for that beam section, because of inelastic LTB the Mn (consequently load carrying capacity goes down linearly. Beyond Lr, the limit state is non-linear. 

From your earlier thread -Lb absolutely is zero if the top flange is continuously braced

Does this mean Mn = Mp regardless of the span? 

Anyway, it was an interesting discussion and thank you for it. Software programs are there for those smart engineers. Good luck.

 

[thedaywa1ker]

So, say you have that beam with a span of only 10 ft, Lb=10ft, the max load corresponds to that which generates Mp (yield limit state ie no LTB). 

Increase the span to 20 ft. Isn't there going to be LTB? In an earlier post, you said "Table 3-6 is based on braced (Lb always < Lp, therefore not considering LTB) beams".

Yes there will be LTB if Lb>Lp, but then you can't use table 3-6 anymore, because the table values assume Lb<Lp.

Why are spans longer than Lp listed? Because they are braced at intervals of Lp or less, but beyond the first Lp, LTB kicks in, the Mn is not Mp anymore but the table 3-6 is based on beam being braced at least every Lp. "..............braced (Lb ≤ Lp) simple-span beams". Keep going to say Lr for that beam section, because of inelastic LTB the Mn (consequently load carrying capacity goes down linearly. Beyond Lr, the limit state is non-linear. 

From your earlier thread -Lb absolutely is zero if the top flange is continuously braced

Does this mean Mn = Mp regardless of the span?

If you are looking at table 3-6, then yes, the moments from those loads equal Mp (unless it is above the bold line where shear controls)

 

[dauwerda]

Have either of you  (you and daywa1ker) looked at Fig.C-J10.2 and also Fig. C-F1.4. ?

Here is the text that goes with the latter, Fig. C-F1.4:  "In this case, the Cb = 5.67 would be used with the lateral-torsional buckling strength
for the beam using an unbraced length of 20 ft which is defined by locations where twist or lateral movement of both flanges is restrained."

Fig C-F1.4 and the associated text is speaking to the situation where reverse curvature is possible (the example for this used in the text is wind uplift on a roof beam). So, yes, in this case the the bottom flange sees compression so the unbraced length is 20 ft.  

Chapter J.10 is dealing with concentrated forces, not flexure. It specifically states that Lb is the unbraced length along either flange at the point of the concentrated load for equations J10-6 and J10-7 exclusively. The figure C-F1.4 is referring to this specific case, not the Lb used when determining the moment capacity of a beam in flexure. 

Why are spans longer than Lp listed? Because they are braced at intervals of Lp or less, but beyond the first Lp, LTB kicks in, the Mn is not Mp anymore but the table 3-6 is based on beam being braced at least every Lp. "..............braced (Lb ≤ Lp) simple-span beams". Keep going to say Lr for that beam section, because of inelastic LTB the Mn (consequently load carrying capacity goes down linearly. Beyond Lr, the limit state is non-linear. 

I'm not completely following this - which questions are yours, what you are agreeing or disagreeing with, etc.

From your earlier thread -Lb absolutely is zero if the top flange is continuously braced

Does this mean Mn = Mp regardless of the span? 

Yes, for a simply supported beam that does not undergo uplift (meaning only the top flange ever sees compression) and the top flange is continuously braced Lb = 0 and Mn = Mp regardless of the overall span.

 

[EBAT75]

From your earlier thread -Lb absolutely is zero if the top flange is continuously braced

Does this mean Mn = Mp regardless of the span?

If you are looking at table 3-6, then yes, the moments from those loads equal Mp (unless it is above the bold line where shear controls)

Feeding Lb=0   Mn becomes > Mp because the second term will become positive. How is Lb absolutely is zero if the top flange is continuously braced satisfying Mn= Mp? or, Eq F2-2?

Ex: W16X100;  Zx = 198 in^3 ; span = 20 ft ;  W = 297 K; 

Lp = 8.87 ft  ;  phi Mp'= 743 k-ft ; Note the ', this is not Mp as you infer in the posts.

Lr = 32.8 ft ; phi Mr' = 459 k-ft ; Note '

w = 297/20 = 14.85 k/ft  ;  Span L = 20 ft; Mu = 14.85x20^2/8 = 742.5, rounded to 743 k-ft. 

Quote:  " If you are looking at table 3-6, then yes, the moments from those loads equal Mp". 

From Manual:  Mp′ Maximum available flexural strength for noncompact shapes, when Lb ≤ Lp′, kip-in. or kip-ft, as indicated

 

[thedaywa1ker]

View attachment 19143

Feeding Lb=0   Mn becomes > Mp because the second term will become positive. How is Lb absolutely is zero if the top flange is continuously braced satisfying Mn= Mp? or, Eq F2-2?

Ex: W16X100;  Zx = 198 in^3 ; span = 20 ft ;  W = 297 K; 

Lp = 8.87 ft  ;  phi Mp'= 743 k-ft ; Note the ', this is not Mp as you infer in the posts.

Lr = 32.8 ft ; phi Mr' = 459 k-ft ; Note '

w = 297/20 = 14.85 k/ft  ;  Span L = 20 ft; Mu = 14.85x20^2/8 = 742.5, rounded to 743 k-ft. 

Quote:  " If you are looking at table 3-6, then yes, the moments from those loads equal Mp". 

From Manual:  Mp′ Maximum available flexural strength for noncompact shapes, when Lb ≤ Lp′, kip-in. or kip-ft, as indicated

I don't understand why you are feeding Lb=0 into F2-2 when (a) says that LTB doesn't apply when Lb<Lp.  F2-2 only applies when Lp<Lb<Lr

Regardless, the end of F2-2 says <=Mp, so if the equation does equal more than Mp, then you still use Mp.  This is meant for the case of Cb > 1, because you shouldn't even be going down this rabbit hole if Lb=zero, per (a).

I'm not sure what you're saying with the W16x100.  The moment from the loads in table 3-6 equal Mp, like I've been saying.  Am I missing something?

 

[organix]

I'm getting fairly lost in this topic or maybe I'm missing something.  You only seem to be proving their point now, EBA75.

View attachment 19143

Feeding Lb=0   Mn becomes > Mp because the second term will become positive. How is Lb absolutely is zero if the top flange is continuously braced satisfying Mn= Mp? or, Eq F2-2?

Again, not sure if I'm missing the point, but are you trying to show Mn would exceed Mp?  For one, F2-2 simply doesn't apply if you're continuously braced.  But even if it did, it says less than or equal to Mp.  So Lb is zero, the second term is positive and you get Mp plus some value less than or equal to Mp = Mp.  So yes, Mn would exceed Mp if you use the left side of the equation, but you can't ever exceed Mp anyway.

Ex: W16X100;  Zx = 198 in^3 ; span = 20 ft ;  W = 297 K; 

Lp = 8.87 ft  ;  phi Mp'= 743 k-ft ; Note the ', this is not Mp as you infer in the posts.

Lr = 32.8 ft ; phi Mr' = 459 k-ft ; Note '

w = 297/20 = 14.85 k/ft  ;  Span L = 20 ft; Mu = 14.85x20^2/8 = 742.5, rounded to 743 k-ft. 

Quote:  " If you are looking at table 3-6, then yes, the moments from those loads equal Mp". 

From Manual:  Mp′ Maximum available flexural strength for noncompact shapes, when Lb ≤ Lp′, kip-in. or kip-ft, as indicated

I don't get this example either.  You've proven their point that the distributed load is equal to Mp.  If you're point is based on the -`- next to the terms, that's simply for cross sections with noncompact flanges.  You have to adjust Lp and Mp accordingly.