Solved problems for local buckling parameters-post 8 steel.

8- How to make an analysis of steel beams? Solved problems.

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How to analyze steel beams? Solved problems.

The difference between analysis and design of steel beam.

The difference between analysis and design problems is that for the analysis, the section is given, and a check of stress is needed, while for the design, the loads are given, and it is required to find the section.

Analysis of steel beam at zone-1.

The solved example is 5-3 from Prof. Segui鈥檚 handbook. The beam shown in Figure 5.11 is W16x31 of A992 steel, for which 16鈥砳s the overall height or nominal depth, while 31 is the weight in lbs per linear ft of A992 steel, where Fy=50 ksi. It supports a reinforced concrete floor slab that provides continuous lateral support of the compression flange.聽

Here, the compression flange is mentioned as being supported continuously, which means we are dealing with a plastic range or zone -1.

Our lambda 位 between 0 and 位p, and Mn=Mp=Fy*Zx.

The beam shown in Figure 5.11 is W16x31 of A992 steel. W16x31, 16鈥 is the overall height, while 31 is the weight in lbs per linear ft of A992 steel, where Fy=50 ksi.

Solved problem 5-3 for analysis of steel beam, the W section for a steel beam
Solved problem 5-3 for analysis of steel beam, the W section for a steel beam

How do we estimate the Ultimate load and Ultimate Moment?

The weight of the beam, which equals 31 lb/ft, is added to the given uniform dead load; The ultimate load equals 1.2*Wd+1.6*wl=1.2*(0.45+0.031)+1.6*(0.55)=1.4572 k/Ft. We estimate the ultimate moment as equal to Wu*L^2/8.The span length equals 30 feet. The final value of the Ultimate moment equals 164 Ft.kips.Please refer to the Next slide image.

Estimate the ultimate load and Ultimate Moment.

How to check the compactness of the flange and the web of a beam?

We use Table 1 as the first step of analyzing the steel beam. For W16x31 A=9.13 inch2, the overall depth=15.90 inch, the web thickness =0.275 inches, bf =5.53 inch.
The thickness of the t flange is 7/16 inches.

The controlling factor is 3.76*sqrt(E/Fy) for the web=90.55. The 位F, which is 6.28聽聽 for the flange, is<位p, which is 9.15, then the section will be in the compact zone for the flange, while 位w, which is聽 51.60, is also <90.55. The section will be in the compact zone for the web, the first zone.

check the compactness of W16x31.

The first step of the steel beam analysis is to get the nominal moment value Mn=Mp=Fy*Zx. From table 1-1, we can get the Zx value, Zx=54.0 inch3.
To get the聽 Mn=Mp=50*54=2700 inch kips. To convert into ft. kips, we will divide by 12.

LRFD Design.

Mp=225.0 kips ft. For the LRFD, Our phi is 桅b=0.90, then 桅b*Mn=0.90*225.0=202.50聽ft. Kips.
The Ultimate moment, acting on the section should be <= 桅b*Mn.

As estimated earlier, the section can carry 202.50 Ft.kips. This section is adequate for analyzing the steel beam for the LRFD design. This is the end of the analysis of the steel beam.

Check the 桅b*Mn for the steel beam section is bigger than M-ultimate.
Check the 桅b*Mn for the steel beam section is bigger than the M-ultimate.

ASD design.

While for the ASD Design, Wd= uniform load+own weight=450+31, Wd=481 lb/ft, and WL=550 lb/ft, adding together for Wt, the Mt=(480+550)*30^2/8/1000=116.0 ft. kips.

This is the total Moment. For Mn =225.0 ft. kips, divide by the omega 惟b, which is 1.67. M all =225/1.67=1350.0 ft.kips. The section can carry 135.0 ft. kips and is only subjected to 116.0 ft. kips; the section is safe.

This section is adequate for analyzing steel beams for the ASD design. The idea of the example is that the section of the beam carries a slab with studs to provide the continuous bracing, then a section was selected Bf/2tf, lambda 位F was< lambda 位p flange also, 位w is<位p for the web.

Check the compactness of the flange for a given W section for a steel beam.

A second problem was solved by checking the compactness of a given section.

This is an example from Lindeburg. Establish whether a W21x55聽 beam of A992 steel is compact. An analysis of steel beams is required.

There are four options鈥擣irst the A992 steel,聽 with Fy=50 ksi.
For W21x55, the overall depth is 20.80鈥, which is highlighted. The web has a Thickness of 3/8鈥 or 0.375鈥 t-web.
I draw the section, Bf=8.22鈥, and its thickness =0.522鈥.
To estimate the controlling lambda, first, we need to find the value of Tf/2Tf for flange, which is 8.22/2*0.22=7.87, 位p for flange =64.70/sqrt(50)=9.15. Then 位f<位p for the flange.

Solved problem to check whether a given section W21x55 of the steel beam is compact or not.

For the second part, we will correct the web thickness as 0.375鈥, h, as estimate =(20.80-2*0.522)/0.375, where tweb=0.375. If we divide 20.80-2*0.522 by the calculator, we get hw, hw=19.756鈥/0.375 = 52.68.

Estimate h/w compactness for W21x55.

Check against 位p, which is聽 640/sqrt(50)=90.55, then 52.55 is <90.55; since bf/2tf< 位p f and hw/tw is< 位p t, then option A is correct. This is the end of the analysis of the steel beam. Option A is the correct selection.

Option A is the correct option for the given steel beam.


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

For bending members, please refer to this link from Prof聽 T. Bart Quimby, P.E., Ph.D. F.ASCE site.
For the next post, Solved problem-4-7-1 how to design a steel beam.

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