Brief description of post 28.

28- Part 4/4 of the Solved problem 9-9-6 How To Find LL?

Part 4/4 of the Solved Problem 9-9-6, How To Find LL? 

Part 4 of the solved problem video.

This is the fourth video, we are concerned about the lateral-torsional buckling. We continue discussing the solved problem from Prof. Salmon’s book, which includes how to compare the bracing length lb as given by Part 4/4 of the Solved Problem 9-9-6, as =15′. The CB and how to estimate, what are the final values of Mn and the safe LL that the section can carry?

How to estimate the Cb for the beam-Part 4/4 of the Solved Problem 9-9-6.

Cb for a simply supported beam under uniformly distributed, this is included in part 4/4 of the Solved Problem 9-9-6.

We have one roller support at one end and hinged support at the other end.

The bracings spacing is every L/3 distance where L is the span length. We will take the first part of L/3 and magnify it.

The reaction for the beam at the left hinge=W*L/2=Reaction at the right support. We need to estimate the moment values at A, B, C, and D.

The L/3 length is to be divided into four sections. So we have L/12 intervals between these points. Then we can estimate the CB equation. CB=(12.50*M-max)/(2.50 M-max+3*Ma+4*Mb+3*Mc)

First, Ma=(w*L/2)*(L/12)-Moment from the free span=L/12, which is (w*L/12)*(L/24).
While for Mb, distance becomes larger, distance=2*(L/12)=L/6, the resisting moment=(W*L/6)*(L/12).

Mc for point c that has a distance of 3*L/12, this will lead to span =L/4, the resisting moment=(W*L/4)*(L/8).
The last point D  is at a distance of L/3 from the left support it will be the second bracing point. Md=W*(L/2)*(L/3)- W*L/3*L/6.

MA=11*WL^2/288, Mb=5*W*L^2/72, Mc=3*W*l^2/32, and  Md=W*L^2/9.

From the next slide. Use the equation of CB=(12.50*M-max)/(2.50 M-max+3*Ma+4*Mb+3*Mc).

The maximum moment is at point D Mmax=W*L^2/9, for the CB equation the factor W*L^2 will cancel each other Cb=12.50*1/9/(2.50*(1/9)+3*11/288+4*5/72+3*3/32). Cb=(12.50/9)/take 288 as a common factor. Cb=12.50(1/9)/((80+33+80+81)/288). Cb=12.50288/(274)*(9). Cb=1.46. Due to symmetry for the next bracing point, the Cb =1.46.

Part 4/4 of the Solved Problem 9-9-6

The middle L/3 portion can be easily calculated and can be divided into four portions each is of L/12 span.
M max=W*L^2/8. We need to estimate MA which is =Mc.

The bending moment value for point A, at the middle strip, can be estimated after knowing the distance from the left support to that point.

The distance=L/3+(L/12)=5*L/12. MA= W*L/2*(5L/12)-W*(5*L/12)*(5/24MA=35*W*L^2/(388). Mc value=35*W*L^2/(388). While Mb =W*L^2/8, Mb is the maximum moment. Cb can be estimated for the middle portion, WL^2 will cancel each other, then Cb= 12.50 (1/8)/2.50(1/8)+3(35/288)2+4(1/8). The last item is for Mb to take 288 as a common factor.

How to estimate CB for the beam for Part 4/4 for the solved problem 9-9-6?

The value of Cb=12.50/8*(234+210)/288=1.01, the beam has three values of Cb as (1.46,1.01,1.46), the CB can be used to increase the Mn

But in the end, Mn must not be >MP. where Mp is the maximum moment the section can carry.

How to estimate the Mn for the beam?

If we take CB as 1.46 and the check Mn value. Cb=1.46 for the first and last third and cb=1.01 for the middle third. The next step is to estimate the Mn value based on the limit state of the lateral-torsional buckling.

We have Lp =12.10′ while Lr=30.41′, actual bracing length=15’as given in the solved problem. We have Mp=1728 at Lb=12.10 and Mr=0.70*Fy*Sx=1114 at Lb=Lr.

At the 15′ distance Mn=Cb* (1728-1728* (1728-1114)/(30.41-12.10)*(15-12.10), if we take cb=1.46, Mn will be more than1728 Ft-kips, so we can use cb as=1.01, taken as =1.0 by prof. Salmon.

In the next slide, there is a graph between Lb and Mn values. At Lp=12′ we have Mn=1728 ft, while for Lr=30.41′, Mn=1114 ft.kips.

The nominal moment value for Part 4/4 for the solved problem 9-9-6.

The Mn value for cb=1.00 will be equal to 1630 ft. kips for the case of lateral torsional buckling. The three values are shown in the slide. This is a reminder of the Mn in the case of flange local buckling.

Mn for the flange of the built-up section.

This is a reminder of the Mn from the case of web local buckling.

Mn for the web of the built-up section.

We have three Mn values, one from the local buckling of the flange which is equal to 1355.7 ft-kips. The second value for the Mn, from the local buckling of the web=1672 Ft-kips. The third value for Mn is 1630 ft.kips.

The maximum selected value of Mn is 1355.70 ft-kips.
The LRFD value multiplied by φb will be φb*Mn=0.90*1355.70=1220 ft-kips.

How to estimate the LL for the beam- part 4/4 of the Solved problem 9-9-6?

The ultimate load for the simple beam Mult must be =< 1220 ft.kips.
From the next slide, the Dead load included the own weight=(0.15*1.2+1.60*LL)*(45^2)/8<=1220 ft-kips. Then (0.18+1.60*LL)*253.125<=1220.  1.60*LL=(1220/253.125)-0.18=4.639. LL=4.639/1.60=2.899 Kips/ft. Taken as LL=2.90 kip/ft, which matches the final solution by Prof. Salmon.

How to estimate the LL for the beamPart 4/4 of the Solved problem 9-9-6?

This is the pdf file used in the illustration of this post.

This is the next post,  Moment Redistribution For Continuous Steel Beams. The post is an introduction to the method for moment redistribution set the AISc code for the reduction of the negative moments and add 0.10 to the average value between supports to the positive moment.

For useful external resources for steel beams. Chapter 8 – Bending Members

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