Category: local buckling

All data that deals with local buckling for steel beams. How do you find out whether the beam section is compact or non-compact? How do we determine the local buckling parameters for different yield stress values? How do we find out the nominal moment strength for compact sections?

  • 9b-Practice problem 5-4-1-Check compactness for Fy=60 ksi.

    9b-Practice problem 5-4-1-Check compactness for Fy=60 ksi.

    Practice problem 5-4-1-Check compactness for Fy=60 ksi.

    Practice problem 5-4-1 For W-, M-, and S-shapes with Fy = 60 ksi: the first part a. List the noncompact shapes in Part 1 of the Manual (when used as flexural members). State whether they are noncompact because of the flange, the web, or both.
    b. List the shapes in Part 1 of the Manual that are slender. State whether they are slender because of the flange, the web, or both. Practice problem 5-4-1 is from the Steel Design Handbook.

    List the W, M, and S shapes based on compactness when Fy=60 ksi.

    The three steel sections, W, M, and S shapes, follow item 10 in Table B4.1b for the lambda value for the unstiffened flange. For the web compactness h/tw, these sections follow item 15 in the same table,

    The following slide shows the main difference between W, M, and S shapes in the profile as I section. W shape has a slope of 2:12 and a broader flange width. The S shape has a slope of 6:1 and is available in a smaller range. The M stands for Miscellaneous beams.

    Practice problem 5-4-1-The differences between W, M and S shapes.

    Determine the values of compactness ratios for Fy=60 ksi.

    Based on item 10 in Table B4.1b, the flange λFp=0.38*sqrt(E/Fy), since we have E29000 ksi and the given Fy=60 ksi, then λFp=0.38*sqrt(29000/60)=8.35. λFr=1.0*sqrt(E/Fy)=1*sqrt(29000/60)=21.964.

    for the web compactness ratio, the flange λwp=3.76*sqrt(E/Fy), λWp=3.76*sqrt(29000/60)=82.66. λwr=5.70*sqrt(E/Fy)=5.70*sqrt(29000/60)=125.31. Please refer to the next slide image for the detailed estimate of compactness ratios.

    The value of compactness ratios for fy=60 ksi.

    Table B4.1b details stiffened and unstiffened elements.

    The next two slides show the details of items 10 and 15 for compactness ratios for members subjected to flexure.

    Part 1 of The detailed B4.1B Table .

    Part 2 of The detailed B4.1B Table .

    Sort W sections based on bf/2tf>λFp but less than λfr.

    We will use an Excel sheet for Table 1-1 for W sections and sort W sections with bf/2tf bigger than 8.35, which is the value of λFp, but smaller than 21.848, the value for λFr. These are the non-compact sections for steel, with Fy=60 ksi.

    The total number of W sections is 20, starting from W30x90 and ending with W6x0.50.

    The next slide image shows the detailed dimension of these non-compact W shapes.

    The details of the 20 non-compact W sections for Fy=60 ksi.

    Find the non-compact web for W sections based on Fy=60ksi.

    There is no non-compact web for W sections based on Fy=60 ksi; the maximum h/tw for W30x90 is 57.50, smaller than the value of λwp, 82.66.

    Non compact  W section data for Fy=60 ksi.

    What are the tables for properties of M and s sections?

    The next slide shows the tables used to find the properties of W, M, and S shapes. We use Table 1-1 for W sections, and for M sections, we use Table 1-2. For S shapes, we use Table 1-3.

    The Tables used for data for W,M and S shapes.

    Sort M sections based on bf/2tf>λFp but less than λFr.

    We will use an Excel sheet for Table 1-2 for M sections and sort M sections with bf/2tf bigger than 8.35, the value of λFp, but smaller than 21.848, the value for λFr. These are the non-compact sections for steel, with Fy=60 ksi. We have only five non-compact sections, starting from M12x10 and ending with M 3×2.9.

    The list of Non compact M shapes for Fy=60 ksi.

    Find the non-compact web for M sections based on Fy=60ksi.

    There is no non-compact web for M sections based on Fy=60 ksi; the maximum h/tw for M12.5×12.40 is 74.80, smaller than the value of λwp, 82.66.

    Non compact web section data for M shape with Fy=60 ksi.

    Sort S sections based on the flange and web compactness ratio.

    e will use an Excel sheet for Table 1-3 for S sections and sort S sections with bf/2tf bigger than 8.35, which is the value of λFp but smaller than 21.848, the value for λFr. There are no non-compact S sections for steel, with Fy=60 ksi. As for the ratio h/tw, there are no non-compact S shapes for the web.

    Are there Non compact sections for web for S section with fy=60 ksi?

    Part b-Sort W, M, and S sections are based on slender sections.

    Based on Fy=60 ksi, the W, M, and S sections do not have slender sections. We have reached the end of our post. Thanks a lot.

    What are the slender sections for W,M and S section for Fy=60 ksi?

    Here is the link for Chapter 8 – Bending Members.
    This links to the next post, 10-lateral-torsional buckling for steel beams.

    There is a newly added post, which is a practice problem 5-5-6-Compute Lp and Lr and φb*Mn.

  • 9A-Solved problem-7-4-1-part 2.

    9A-Solved problem-7-4-1-part 2.

    Solved problem-7-4-1-part 2. 

    This is the second part of the solved problem 7-4-1, where it is required to design a W section for a given beam with a span of 20 Feet. The steel is A572, where Fy=65 ksi.

    For Case-3, design a steel beam with Fy=65 ksi. Solve the problem: 7-4-1-part 2.

    Find the Ultimate Moment for the given beam.

    The following slide includes the value of the ultimate load without considering the beam’s weight, which will be added after selecting the section. The ultimate load is 1.52 kips/ft, and the ultimate moment is 76 ft. kips. The beam’s flange is fully supported against lateral movement.

    Solved problem-7-4-1-part 2, The ultimate load for steel beam grade 65.

    We will equate the ultimate moment with the φb*Mn or φb*Fy*Zx.the Fy=65 ksi, φb=0.90, Mu=76 Ft.kips.

    Determine the required Zx-select a section.

    The required Zx value is 15.589 inch3. Using Table 3-2 for W sections sorted based on Zx, we will select a W12x14 with a Zx value of 17.40 inch3, Zx selected > Zx required. Please refer to the next slide image for more details.

    Determine the plastic section Zx for Fy=65 ksi.

    We need to check whether W12x14 meets the requirement of local buckling parameters. The next slide shows the controlling parameters for the flange and web.

    What are the local buckling parameters based on Fy=65 ksi?

    Based on Fy=65 ksi, the lambda for flange at plastic stage λFp=8.03. The lambda for flange at elastc stage λFr=21.12. The lambda for web at plastic stage λwp=79.42. The lambda for flange at elastc stage λwr=120.40. The next slide shows the detailed estimate and the corresponding values for Flange λ and Web λ based on Fy=65 ksi.

    The values for parameters λFp and λr for flange, λWp and λwr for web-Fy=65 ksi.

    Check flange λ and web λ against local buckling parameters.

    In the second part, we use Table 1-1 for W12x14 to find the value bf/2tf, which equals 8.82, while h/tw equals 54.30. if we check, we will discover that bf/2tf is bigger than λFp but smaller than λFr, which means that the flange is non-compact. For λw is smaller than λwp, meaning the web is compact. So the entire W section will be considered as non-compact.

    The Nominal Moment for W12x14 is between the value of (Fy*Zx) and (0.7*Fy*Sx). The elastic section modulus Sx=14.90 inch3, while the plastic section modulus Zx equals 17.40 inch3.

    check whether W12x14 is compact or non compact-Fy=65 ksi.

    Find the Nominal moment of W12x14.

    For the next slide, the flange lambda value λF =8.82,Fy=65 ksi.

    Our λf=8.80, in between λf-p and λf-r. It is required to get the Mn value.From Equation F3-1, Mn=(Mp-(Mp-0.70Fy*Sx)/(λfr-λfp)). This is the last step of designing a steel beam under Fy=65 ksi.

    Mn is the same as the equation for a straight line y=m*x. We will estimate the Mn for the upper  Point, which equals Mp=Fy*Zx=65*17.40=1131 inch. Kips. The second point Mn=0.70*Fy*Sx, Sx for the section=14.90 inch2.
    0.70*Fy*Sx=0.70*65*14.90= 677.95 inch. Kips.

    Find the slope of the Linear portion.

    We can find the slope of the linear portion, which equals (Fy*Zx-0.70Fy*Sx)/λfr-λfp). The next slide shows the slope’s value as 34.61-inch. Kips.

    Find the slope of the linear part of M n and lambda for flange.

    Now, we can find the Nominal moment equal to Mp-S*(λf-λp)=1103.66 inch. Kips. We can find the value in Feet Kips as 92.0 ft. kips. The Φb value is 0.90. the factored Mn=82.80 Ft.kips.

    Check whether Mult is bigger than Φb*Mn or not.

    We will finalize the ultimate moment value by adding the W section’s own weight moment. The final Mu value is 77.0 feet. KIPS. The Φb*Mn=83.00 feet. Kips. The section is safe since Φb*Mn>Mult. This is the end of the post. Thanks a lot.

    Check whether section is adequate.

    For the first part of this post, please refer to post 8: How to design a steel beam? Solved problem-7-4-1.

    Here is the link for Chapter 8 – Bending Members.
    This links to the next post, 9b-Practice problem 5-4-1-Check compactness for Fy=60 ksi.

  • 9-How to design a steel beam? Solved problem-7-4-1.

    9-How to design a steel beam? Solved problem-7-4-1.

    How to design a steel beam? Solved problem-7-4-1.

    This is the first part of the post on designing a steel beam using various steel types. In this first part, we will consider Fy=36 ksi and Fy=50 ksi. The second part of this post will consider A572 steel.

    How to design a steel beam? Solved problem-7-4-1.

    A steel beam with the lightest W or M section must be designed under a uniformly distributed load and a given live load. The beam’s compression flange is fully supported against lateral movement.

    For Case#1- Design a steel beam with Fy=36 ksi.

    For the first case of The yield stress of Fy=36 ksi, it is required to design a steel beam of W—section or M section for a simply supported beam of span =20′ for the three cases of Fy values.

    Find the Ultimate Moment-Fy=36 ksi.

    Starting with ASTM A36, where the Fy=36 ksi. The first step of designing a steel beam is to estimate the Ultimate load value of 1.20D+1.6L of the steel beam. The ultimate uniform load, ultimate uniform load, Wult=1.2*0.20+1.60*0.80=1.52kips/Ft. The ultimate moment for a supported beam, Mult= wult*L^2/8. Mult=1.52*20^2/8=76 Ft.kips. This is the value of Multimate.

    Solved problem 7-4-1 for the design of a steel beam.

    Determine the required Zx-select a section.

    Our first trial is to equate Mult/(0.90*Fy)=Zx, get the plastic section modulus Zx, and then check Table 3-2. The next slide shows the limiting slenderness parameters lambda λF and lambda λw for ASTM A36, where Fy=36 ksi,  lambda λfp=0.38*sqrt*(E/Fy)=0.38*sqrt(29000/36) =10.79, while flange lambda λfr=1*sqrt*(E/Fy)=28.38. For web lambda λwp=3.76*sqrt(E/Fy)=106.72, whie for lambda λwr=5.70*sqrt*(E/Fy)=922.141.

    The plastic section modulus equals 28.15 inch3.

    The required plastic section modulus value Zx.

    We will use Table 3-2, where Zx sorts W sections; we find that W12x22 has a Zx value equal to 29.30 inch3; the Zxc value is bigger than the required Zx.

    Check flange and web compactness.

    Flange λ and web λ against local buckling parameters.

    We will estimate the slenderness ratios for both the flange and web and the flange lambda λF and lambda λw for the selected W12x22 beam and compare these values with the limiting values of lambda λp; based on the comparison, we will determine if lambda λF or lambda λw< lambda λp – plastic. From Table 1-1, in the second part, the flange λf=4.74 and web λ=41.80.

    The next slide shows that section W12x22 is a compact section.

    Check whether section is compact or non-compact.

    We will readjust the value of the ultimate moment by adding its weight—the new Mult=1.555*20^2/8=77.32 ft. kips.

    Since our selected section is compact, Mn=Mp. Zx=Fy*Zx. For the LRFD, the value of Φb*Mn=Φb*Fy*Zx, which is Φb*Fy*Zx=79.11 Ft.kips.

    Since the acting Multimate is only 77.32 ft. kips, the section is safe since 79.11 Ft. kips is >77.32 ft. kips.

    Check the final ultimate moment against the fatcored moment.

    For Case#2- design a steel beam with Fy=50 ksi.

    Determine the required Zx-Fy=50ksi.

    The second case for ASTM A 992, where Fy=50 ksi. Mult without the superimposed load was =76 ft. kips.
    We need Zx to start with Table 3-2. Zx=Mult/(0.90*50)=20.22 inch3.
     We will proceed to table 3-2, but select Zx>20.22 inch3, which will be W10x19 with Zx=21.60 inch3. Zx selected=21.60 inch3. We proceed to get the properties of our section. W10x19.

    Design a steel beam with Fy=50 ksi.
    Design a steel beam with Fy=50 ksi.

    Local buckling parameters-Fy=50 ksi.

    In the next slide, we find the local buckling parameters for Flange and Web-based on Fy=50 ksi.

    The local buckling parameters for For Fy=50 ksi.

    From Table 1-1 part 2, the λ for flange equals 5.09, which is <λfp, which λ for web is smaller than λwp, and the section is compact.

    Check whether section is compact or non compact.

    For the Mult, we will adjust due to the superimposed load, the weight of the beam (19 lb/ft), and the final Mult=77.14 ft. kips.

    The Φb*Mn=81.00 Ft.kips >Mult. The compact section is the last step in designing a steel beam under Fy=50 ksi. This is the end of part 1, which will be followed by part 2.

    Check the LRFD moment against Mult

    Here is the link for Chapter 8 – Bending Members.
    This links to the next post, 10-lateral-torsional buckling for steel beams.

How do we 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 provided, and it is required to find the section.

Analysis of steel beam at zone-1.

The solved example is 5-3 from Prof. Segui’s handbook. The beam shown in Figure 5.11 is W16x31 of A992 steel, for which 16″is 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. 
Here, the compression flange is mentioned as being supported continuously, which means we are dealing with a plastic range or zone -1.

Our lambda λ between 0 and λp, and Mn=Mp=Fy*Zx.

The beam shown in Figure 5.11 is W16x31 of A992 steel. W16x31, 16″ 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

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 in 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 λF, which is 6.28   for the flange, is<λp, which is 9.15, then the section will be in the compact zone for the flange, while λw, which is  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  Mn=Mp=50*54=2700 inch kips. To convert into ft. kips, we will divide by 12.

LRFD Design-Beam Moment capacity.

Mp=225.0 kips ft. For the LRFD, Our phi is Φb=0.90, then Φb*Mn=0.90*225.0=202.50 ft. Kips.
The Ultimate moment, acting on the section, should be <= Φb*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 Φb*Mn for the steel beam section is bigger than M-ultimate.
Check the Φb*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 Ωb, 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 λF was< lambda λp flange also, λw is<λp for the web.

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

A second problem is to check the compactness of a given section.

This is an example from Lindeburg. Establish whether a W21x55  beam of A992 steel is compact or non-compact. An analysis of steel beams is required.
There are four options. First is the A992 steel,  with Fy=50 ksi.
For W21x55, the overall depth is 20.80″, which is highlighted. The web has a Thickness of 3/8″ or 0.375″ t-web. I draw the section, Bf=8.22″, and its thickness =0.522″.
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, λp for flange =64.70/sqrt(50)=9.15. Then, λf<λp 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″, 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″/0.375 = 52.68.

Estimate h/w compactness for W21x55.

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

For bending members, please refer to this link from Prof  T. Bart Quimby, P.E., Ph.D. F.ASCE site.


For the next post, A Solved problem-4-7-1, how to design a steel beam.

  • 7- A guide to Local buckling parameters for steel beams.

    7- A guide to Local buckling parameters for steel beams.

    local buckling parameters for steel beams.

    The relation between Nominal Moment Mn, the λ Values, and the three regions.

    The topics included in our discussion are shown in the next slide.

    The content of the lecture for post 7.

    The relation between λ and Mn can be viewed in three regions. The first region is the compact region, which we call F2-1, chapter F provision 2 equation 1-which states that Mn=Mp=Zx*Fy.
    To determine the plastic modulus Zx, we must evaluate the lambda and its value to reach this first region.

    To get a compact doubly symmetric W section, for instance, bf/2tf<λpF or 0.38*sqrt(E/Fy) and h/tw<λpw or less than 3.76*sqrt(E/Fy).

    To get a non-compact W section, bf/2tf >λpF or 0.38*sqrt(E/Fy) and smaller than λrF or 1.0*sqrt(E/Fy) and h/tw>λpw or 3.76*sqrt(E/FY) and less than 5.70*sqrt(E/Fy).

    To get a slender W section, bf/2tf >λpr or 1.0*sqrt(E/Fy) and h/tw>λrw or 5.70*sqrt(E/Fy).

    Please refer to the next slide image for more details.

    The values of λ plastic for both flange and web and λr.

    We must evaluate the lambda and its value to determine Zx to reach this first region. The lambda value λ should be from 0 to λp for the first zone.

    For the second region represented by the equation, the Lateral-Torsional buckling region, F2-2, the value of the line for the second region starts from Mn=Mp and ends at Mn=0.70 Fy*Sx.

    The values of λ plastic for both flange and web and λr.

    The value of 0.70 Fy at the other end is due to the residual stresses during fabrication. The residual stress will cause Internal stresses in opposite directions. So 0.30 Fy*Sx is reduced from the Mn.
    We have a linear equation for the second region.

    If your lambda value λ is between λp and λr, then the line equation Y=mx will be Mn=Mp- (the difference between the x’s)(λ-λp).


    The final form will be Mn=Mp-(Mp-0.70FySx)*((λf-λpf) /(λrf-λpf), where λrf for limiting slenderness for Flange as an upper limit on non-compact.

    What are the values for λp?

    For the Flange, our lambda is λF= (Bf/2Tf) used to estimate λF.

    While in the case of the web, λw= (h/tw), is used for the estimation of λw.

    The first local buckling parameter for the flange is λpf, which is = 0.38*sqrt (Et/Fy).

    For instance, if Fy=36 ksi, put the value of 36 inside the sqrt. λpf =0.38*sqrt(29000/36)=10.785.

    The first local parameter λpf for Fy=50 ksi,λpf=0.38*sqrt(29000/50)=9.15.

    For the web, the first local parameter λpW for web =3.76*sqrt(E/Fy). Again, the numerator will be set =3.76*sqrt(29000/36)=106.717 for Fy=36 ksi.

    We can use the sketch of the three regions in the two cases of Flange and Web, but consider the different value λp for each case for Flange and Web.

    The slenderness ratio values for the non-compact sections.

    The non-compact section slenderness ratio starts from λp till λr.

    λr will be our second local parameter.
    For the value of λr for flange λrf, for A 992 steel with Fy=50 ksi, for the case of flange λrf=1.0*sqrt(E/Fy)=1*sqrt(29000/50)=24.08.

    The values of λ plastic for both flange and web and λr.

    While the second local parameter. for Fy=50 ksi, λr equation for the web equals 5.70*sqrt(E/Fy)=5.7*sqrt(29000/50)=137.27.

    What are the unstiffened -elements?

    Unstiffened – elements are elements that are supported only at one edge, for example, the flange of the W section and the flange of C -the channel.

    How do you estimate the bf for the un-stiffened elements?

    The code defined the value of b, which is used in estimating slenderness for four items. The first item is for an I-shaped member, where b is 1/2 the full flange width of bf.
    For the second case, where the section for legs of angles and flanges of channels and zees, b, is the full leg width, refer to item 12, Table B4.10.

    For the third case, where the section is a plate, b is the distance from the free edge to the first row of fasteners.
    For the fourth case w, here the section is the stem of tees, d is the full depth of thction item no.14.

    What are unstiffened elements?

    If we check Table B4.1b, which gives the ratio between width to thickness for members subject to flexure, item 10 is for flanges or rolled I beams, and item 11 is for flanges of doubly or singly Symmetric I-shaped built-up section.

    What are stiffened elements definition?

    The list from No.10 to No. 14 for the un-stiffened elements and the λp, the first local buckling parameter for members subject to flexure. The limiting slenderness of the compact flange. The second local buckling parameter, λr, limits the non-compact flange’s slenderness.

    What are the stiffened elements?

    Stiffened elements are those supported at two edges, such as the web of the W section, the web of the C channel, and the tube sections.

    How do you estimate h for stiffened elements?

    The B4.1 Table shows the different elements and how to find the h value.

    The different items of stiffened elements.

    There is also a graphical table for the unstiffened elements, so check items from 10 to 14. Items 1 to 9 are for the elements subjected to axial compression.

    Local buckling parameters, width to thickness ratios for flexure members-stiffened elements

    Stiffend items list inAISC-360-10

    The list from No.15 to No. 20 for the stiffened elements in Aisc-360-10.

    The stiffened item list in AISC-360-10 has only 20 items; please refer to the next slide image.

    Stiffend items list in AISC-360-10

    These tables show the values of λp, the limiting slenderness for a compact flange, and λr, the limiting slenderness for a non-compact flange. The AISC360-16 has 21 items, the last three of which are shown in the next slide image.

    Aisc 360-16 -table for width to thickness.

    For the buckling concept, please refer to this link from Prof  T. Bart Quimby, P.E., Ph.D., F.ASCE site.
    This is a link for the next post, 8- How to make an analysis of steel beams? Solved problems.