Category: lateral torsional buckling

Introduction to Lateral torsional buckling for steel beams. How do we find the bracing length Lp and Lr? How do you see the nominal moment value for a steel section for a given bracing length Lb?

  • Practice problem 5-5-8-Compute Mn.

    Practice problem 5-5-8-Compute Mn.

    Practice problem 5-5-8-Compute Mn for W18x71 with lb=9 feet.

    Practice problem 5-5-8 A W18 x 71 is used as a beam with an unbraced length of 9 feet. Use Fy=65 ksi and Cb = 1 and compute the nominal flexural strength. Compute everything with the equations in Chapter F of the AISC Specification. Practice problem 5-5-8 is from the Steel Design Handbook.

    How do we estimate Lp for W18x71 with Fy=65 ksi?

    In the absence of design aids, Table 3-3, where W sections are sorted by their plastic section Zx, and with different Fy, which equals Fy=65 ksi, we need to estimate Lp and lr values by using the equations given by the specifications.

    The following slide shows The detailed estimate of the Lp value; Lp is the value of the un-braced length that the plastic hinge should be developed, which is given by equation F2-5 as equal to 1.76*rysqrt(E/Fy).

    From Table 1-1, we will get the radius of gyration ry for W18x71. The ry value is 1.70 inches from the data that Fy=65 ksi. The E value is 29000 ksi. The value of Lp will be 63.20 inches and can be rounded to 5.30 feet.

    What is the value of lp for  W18x71-Fy=65 ksi?

    How do we estimate Lr for W18x71 with Fy=65 ksi?

    We use table 1-1 for W section W18x71 to get the necessary data for estimating the Lr. Lr represents the unbraced length at which lateral-torsional buckling transitions from the inelastic range to the elastic range. These items are ry, rts, Zx,SX, Cw,C, J, H0. The Sx value equals 127 inch3 while Zx=146 inch3.

    We can get the value of bf/2tf and h/tw. Please refer to the next slide image for details.

    Use Table 1-1 part2-to get the data for lr value.

    What is the value of lr?

    We use the equation F2-6 provided by the AISC specification. The value of lr for W18x71 equals 196 inches and is rounded to 16.33 feet. Please refer to the next slide image for more details.

    What is the value of lr?

    Check buckling parameters for W18x71 where Fy=65 ksi.

    We will find the local buckling parameters for the flange and web-based Fy=65 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=65 ksi, then λFp=0.38*sqrt(29000/65)=8.03. λFr=1.0*sqrt(E/Fy)=1*sqrt(29000/65)=21.12.

    For the web compactness ratio, the flange λwp=3.76*sqrt(E/Fy), λWp=3.76*sqrt(29000/65)=79.42. λwr=5.70*sqrt(E/Fy)=5.70*sqrt(29000/65)=120.4. Please refer to the next slide image for the detailed estimate of compactness ratios.

    Local buckling parameters based of Fy=65 ksi.

    Check whether W18x71 is compact or non-compact.

    In the next slide image, we have the bf/2tf and h/tw values of W18x71 against the required local buckling parameters based on Fy=65 ksi. We will find out that the flange and web are compact, so the whole section of W18x71 is compact.

    Check  whether W18x71 is compact section or not.

    Find the equation of Mn-based on Fy=65 ksi.

    The Plastic moment value equals Zx*Fy, which is equal to 65×146=9490 inch. Kips. The term (0.7*Fy*Sx) value equals 0.7*65*127=5778.50 Inch. Kips. The first term corresponds to bracing length Lp=5.30 ft; the second corresponds to lr=16.33 feet. We have a given bracing lengthlb=9 ft.

    We get the values of Mp and 0.70*Fy Sx in Ft.kips by using the conversion factor of 1Ft=12 inches. Mp value equals 791 Ft. kips, while 0.70*Fy Sx=482, we could round these values to the nearest ones.

    There is a linear relation for the required Mn with Mp and(0.7*Fy*Sx) together with Lp and Lr. The BF factor is the slope value, which equals (Mp-0.7*Fy*Sx)/(Lr-Lp)=28.01.

    The relation between reuired Mn and lb, BF values.

    What is the value of Mn for W18x71 at lb=9 feet?

    The relation of 791-28.01*(Lb-Lp) represents the nominal moment value. The Lb value equals 9 feet, and the Lp value equals 5.30 feet. The Final value for Mn equals 687 Ft. kips.

    Mn value for W12x30 for a bracing length of 10 ft.

    We could use an Excel sheet to graph the relation between Lb and the Flexural design strength-Mn in ft. kips for W18x71 with Fy=65 ksi. We can see the linear portion and the Value of Mn at 9 feet equals 687 ft. kips. Thanks a lot.

    Graph betwen lb and Nominal moment for W18x71.

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

    A new post is added 41-Practice problem 5-6-1-find the total service load for W12x65. The section is non-compact.

  • Practice problem 5-5-6-Compute Lp and Lr and φb*Mn.

    Practice problem 5-5-6-Compute Lp and Lr and φb*Mn.

    Practice problem 5-5-6-Compute Lp and Lr, φb*Mn and Mn/Ωb for lb=10 feet.

    Practice problem 5-5-6 A W12 x 30 of A992 steel has an unbraced length of 10 feet. Using Cb = 1.0, a. Compute Lp and Lr. Use the equations in Chapter F of the AISC Specification. Do not use any of the design aids in the Manual.
    b. Compute the flexural design strength, φb*Mn.
    c. Compute the allowable flexural strength Mn/Ωb. Practice problem 5-5-6 is from the Steel Design Handbook.

    How do we estimate Lp for W12x30?

    In the absence of design aids, Table 3-3, where W sections are sorted by their plastic section Zx, in which we can easily find the values of lp and lr and other values of flexure design moments, We need to estimate Lp and lr values by using the equations given by the specifications.

    The following slide shows The detailed estimate of the Lp value; Lp is the value of the un-braced length that the plastic hinge should be developed, which is given by equation F2-5 as equal to 1.76*rysqrt(E/Fy).

    From Table 1-1, we will get the radius of gyration ry for W12x30. The ry value is 1.52 inches. From the data on A992 steel, we have the yield stress Fy=50 ksi. The E value is 29000 ksi. The value of Lp will be 64.43 inches and can be rounded to 5.40 feet.

    Practice problem 5-5-6-Compute Lp and Lr, φb*Mn.
And Mn/Ωb for lb=10 feet.

    How do we estimate Lr for W12x30?

    We use table 1-1 for W section W12x30 to get the necessary data for estimating the Lr. Lr represents the unbraced length at which lateral-torsional buckling transitions from the inelastic range to the elastic range. These items are ry, rts, Zx,SX, Cw,C, J, H0.

    We can get the value of bf/2tf and h/tw. Please refer to the next slide image for details.

    Use Table 1-1 part2-to get the data for lr value.

    What is the value of lr?

    We use the equation F2-6 provided by the AISC specification. The value of lr equals 187.24 inches and is rounded to 15.60 feet. Please refer to the next slide image for more details. Thus, we have completed the required part a) in the practice problem 5-5-6.

    What is teh value of lr?

    Check buckling parameters for W12x30 where Fy=50 ksi.

    We will find the local buckling parameters for the flange and web-based Fy=50 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=50 ksi, then λFp=0.38*sqrt(29000/50)=9.15. λFr=1.0*sqrt(E/Fy)=1*sqrt(29000/50)=24.08.

    for the web compactness ratio, the flange λwp=3.76*sqrt(E/Fy), λWp=3.76*sqrt(29000/50)=90.55. λwr=5.70*sqrt(E/Fy)=5.70*sqrt(29000/50)=137.27. Please refer to the next slide image for the detailed estimate of compactness ratios.

    Local buckling parameters based of Fy=50 ksi.

    Check whether W12x30 is compact or non-compact.

    In the next slide image, we have the values of bf/2tf and h/tw of W12x30 against the required local buckling parameters based on Fy=50 ksi. We will find out that the flange and web are compact, so the whole section of W12x30 is compact.

    Check  whether W12x30 is compact section or not.

    Find the equation of Mn-based on Fy=50ksi.

    The Plastic moment value equals Zx*Fy, which is equal to 50×43.10=2155 inch. kips. The term (0.7*Fy*Sx) value equals 0.7*50*38.60=1352 inch. kips. The first term corresponds to bracing length Lp=5.40 ft; the second corresponds to lr=15.60 feet. We have a given lb=10 ft.

    We get the values of Mp and 0.70*Fy Sx in Ft.kips by using the conversion factor of 1Ft=12 inches. Mp value equals 179.583 Ft. kips, while 0.70*Fy Sx=112.583, we could round these values to the nearest ones.

    There is a linear relation for the required Mn with Mp and(0.7*Fy*Sx) together with Lp and Lr. The BF factor is the slope value which equals (Mp-0.7*Fy*Sx)/(Lr-Lp)=6.5686.

    The relation between reuired Mn and lb, BF values.

    What is the value of Mn at lb=10 feet?

    The relation of 179.583-6.53*(Lb-Lp) represents the nominal moment value. The Lb value equals 10 feet, and the Lp value equals 5.40 feet. The Final value for Mn equals 149.367 Ft. kips.

    Part b) Compute the Flexural design strength-φb*Mn.

    For part b of the practice problem 5-5-6, for Flexural design strength-φb*Mn, we multiply the Nominal value by the factor φb, which equals 0.90. We get the value of 134 ft.kips.

    Part c. Compute the allowable flexural strength Mn/Ωb.

    For part b of practice problem 5-5-6, for allowable flexural strength Mn/Ωb, we multiply the Nominal value by the factor (1/ Ωb), which equals (1/1.67). We get the value of 89 Ft.kips.

    Mn value for W12x30 for a bracing length of 10 ft.

    We could use an Excel sheet to graph the relation between Lb and the Flexural design strength-φb*Mn for W12x30 Fy=50 ksi. We can see the linear portion and the Value of φb*Mn at 10 feet equals 134 ft. kips.

    Graph betwen lb and design strength flexture strength for W12x30.

    We could use an Excel sheet to graph the relation between Lb and allowable flexural strength Mn/Ωb for W12x30 and Fy=50 ksi. We can see the linear portion and the Value of Mn/Ωb at bracing length 10 feet equals 89 ft. kips. Thanks a lot.

    Graph betwen lb and  allowable flexture strength for W12x30.

    Here is the link for Chapter 8 – Bending Members.
    This links to the next post, Practice problem 5-5-8-Compute Mn.

  • 14- A guide to solved problem 9-7-Lb is bigger than Lr.

    14- A guide to solved problem 9-7-Lb is bigger than Lr.

    Solved problem: 9-7-When Lb is bigger than Lr, What Is Flexure Strength?

    From Prof. Mccormac’s book Structural Steel Design Example 9.7. Using AISC Equation F2-4, determine the values of and for a w18x 97 with fy =50 ksi and unbraced length lb =38 ft, assuming that CB =1. As a reminder, this is the graph for the relation between bracing length and Mn.

    The different zones for the bracing length Lb

    Solved problem 9-7 Analysis is based on LRFD.

    This is an analysis problem since the W section is given, for which the distance between bracing for a beam is Lb is bigger than Lr, and also Lb is bigger than LP.
    To solve the problem, we will follow the following steps:

    Step-1- From Table 1-1, we can get the necessary data, such as Sx, J, h0, and rts.

    Solved problem 9-7

    This is the equation used to get the value of Fcr for a beam when we have the relevant values of Sx, Jc, CB, and h0.

    The formula for the value of Fcr

    3-Since the given Lb is bigger than Lp and also Lb is bigger than Lr, in our solved problem 4-7, the section is in the elastic slender zone. We need to estimate the stress Fcr from the equation F2-4, as shown in the next slide.

    The calculation for lr using formula.

    This is a continuation of the estimate of Lr and the value of lr from Table 3-2, which matches the Lr value.

    Use Table 3-2 to find lr and Lrfd values of Mp and Mr.

    Estimate the strength of the section for the LRFD by using the formula ΦbMn.= Φb Fcr Sx=0.9026.156*(188)=369 ft.kips
    This is the detailed calculation for The LRFD design shown in the next slide image for the solved problem 9-7. Please refer to the next slide image for a detailed estimate of Fcr.

    The detailed estimation of Fcr and Mn value

    Solved problem 9-7 Analysis is based on ASD.

    For the ASD design, the same procedures will be used as in the LRFd, except step 5 will be modified.
     1-From table 1-1, get both Lp and Lr values for section W18x97.
    2- since the given Lb>Lp and >Lr, the section is in the elastic slender zone in our example.
    3-We need to get Sx, J, ho =d-2tf, and rts from Table 1-1 for section W18x97. We also need the Lr value for the same section but from Table 3-2. Then, we need to Estimate Fcr from equation F2-4.

    4-After evaluating Fcr, we get the section’s Mn value, where Mn = Fcr *Sx. cr=25.156 ksi.
    Mn=fcr*Sx=25.16*188/12=410.0 Ft.kips.

     5-Estimate  the strength of the section for ASD using the formula (1/ Ωb)* Fcr*Sx=.(1/1.67)*26.156*(188)=246 Ft.kips
    This is the detailed calculation for The ASD design shown in the next slide image for the solved problem 9-7.   
     

    The detailed estimation of Fcr and Mnx/ Ωb

    The following slide image shows an Excel graph between lb versus Φ*Mn.

    Excel plot Lb virsus Phi*Mn.

    The following slide image shows an Excel graph between lb versus (1/ Ωb)*Mn.

    Excel plot Lb virsus 1/omega *Mn.

    For more detailed illustrations for the CB, please follow this linkFlexural Limit State Behavior.

    For the next post, How to design a beam with a design chart?

                              

                                                                                      

  • 13-Solved problem 4-6-how to find the available flexure strength?

    13-Solved problem 4-6-how to find the available flexure strength?

    A Solved Problem 4-6-How to Find The Available Flexure  Strength?

    A solved problem 4-6 for lateral-torsional buckling when lb>Lp but<Lr.

    From Prof. Alan Williams’s book Structure -Reference manual,  solved problem 4.6 A W16x40 beam of grade 50 steel is laterally braced at 6 ft intervals. It is subjected to a uniform bending moment with the coefficient of moment Cb =1.0.

    Determine the beam’s available flexure strength. This graph represents the relation between Lb and the nominal moment Mn; it has three zones based on the value of bracing length Lb and its relation with Lp and Lr.   

    Solved problem 4-6, for which the flexure strength is required.

    Analysis for the given section by the LRFD design.

    The solved problem 4-6 is an analysis problem, the section is given for which, the distance between bracing for a beam is Lb > Lp, but Lb <Lr, for the LRFD design. it will fall in the second zone of inelastic buckling; please refer to the next slide for more details.

    Solved problem 4-6, for which the flexure strength is required.

    We need to find both bracing lengths lp and Lr for the given W section, which can be obtained from Tables 1-1 and Table 3-2. From Table 1-1, we need the value of ry to estimate the Lp value. For lr we need plenty of data to find lr as we will see next.

    Get the maximum un-braced length to let the shape reach its plastic moment strength Lp. The relevant equation for Lp is introduced from Lp=300*ry/sqrt(Fy). We need to get these data from AISC table 1-1. The LP value is 5.55 feet. Our given bracing length is 6 feet, and the condition is that the bracing length is bigger than Lp. Please refer to the slide image for more details.

    Estimate the lp value from the formula

    We need the following values from the Torsional properties: J, CW, and other properties, tf, Sx, rts, ho, selected from Table 1-1.

    The different parameters for estimating Lr

    This is the equation for the Lr formula using equation F2-6.

    The formula used to estimate the value of Lr.

    This is the detailed reference equation number as presented in the AISC code.

    The equation of Lr.

    This is the detailed estimation of the value of Lr using the equation.Lr=15.9′.

    The value of lr after estimation.

    This is the value of limiting laterally unbraced length Lr using Table 3-2 for W16x48.

    The value of Lr, from table 3-2

    For the given Lb check if  Lb>Lp and Lb  <Lr, then the section is not compact, the value of φb*Mn is < φb*(Mpx), but φb*Mn > φb*(Mrx),  where Mpx=Fy*Zx,  while  Mrx= (0.70*Fy*Sx).

    Estimate φb*Zx*Fy for Lb and φb*Fy*Sx for  Lr. Estimate the value of φb *BF.

    The next picture explains the final φb*Mn= φb (Zx*Fy)—φb *BF*(Lr-Lb).
    The available flexure strength is based on the LRFD, φb *Mn=269.50 Ft.kips.  

    The value of Mn from BF.

    This is the value of φb*Mp and φb*Mr by using Table 3-2.

    using table 3-2 to get fatored moments.

    I have used an Excel plot to show the relation between Lb and φb*Mn.

    Using excel graph to represent lb virsus phi Mn

    The analysis for the given section by ASD.

    We will use Table 3-2 to get (1/Ωb)*Mp and (1/Ωb)*Mr. We also get Lp and Lr values for W16x40.
      The given bracing length lb of 6 feet is bigger than Lp but smaller than Lr.

    The section is not compact, the value of Mn/Ω is < (Mpx)/ Ωb, > (1/Ωb)*(Mrx)
    Mpx=Fy*Zx, Mrx= (0.70*Fy*Sx).

    The values of Mpx/ Ωb and (1/ Ωb )*Mrx from table 3-2.

    Estimate (1/Ωb)*Zx*Fy for Lb and (1/Ωb)* 0.70*Fy*Sx for Lr.   
    The final (1/Ωb)*Mn= (1/Ωb) (Zx*Fy)- (1/Ωb) *Bf *(Lr-Lb).
    The available flexure strength based on the ASD, (1/Ωb)*Mn=179.00 Ft.kips.

    These are the detailed calculations for the ASD moment value using BF, as shown in the next slide image.

    This is the case with the ASD design.

    I have used an Excel plot to show the relation between Lb and (1/Ωb)*Mn.

    Excel graph between Lb and Mn/omega for the ASD design.

    Please follow the Lateral Torsional Buckling Limit State for a valuable external source.
    This is the solved problem 4-5.
    For the next post, A Solved problem 9-7, When Lb>Lr, what is flexure strength?

Review of the information on plastic bracing length Lp, Lr, Fcr.

How to estimate Lp?

Lp is the maximum un-braced length that will let the shape reach its plastic moment strength. The relevant equation for lp is introduced from Lp=300*ry/sqrt(Fy).

This equation gives Lp the bracing length at the plastic stage for any given section. We need to find the radius of gyration at the y-direction and the corresponding Fy or yield stress for the section in question.

Review of the bracing length when lb=LP, how to estimate Lp value?

Lp can be obtained if Fy=50 ksi from table 3-2 for the sorted w section by Zx.

How do we estimate limiting laterally unbraced length Lr?

Lr is the limiting laterally unbraced length for the limit state. The next slide image shows the relevant equation for limiting the laterally unbraced length for the limit state.

All requirements for this equation can be obtained from Table 1-1 for the W sections. Sx is the elastic section modulus at the x-direction, Fy is the yield stress, and E is the Modulus of elasticity.

Lr is the Limiting laterally unbraced length for the limit state. The general equation for the value of lr is included in the slide image.

Bracing length Lb=Lr.

Bracing length conditions.

The next slide shows the different values of Mn, or the nominal moment of a section, based on the bracing length, whether Lp, between Lp and Lr, or >Lr. Based on these conditions, we have three zones.

The different zones for the bracing length Lb, based on the values of Lp and Lr.

The relation between Nominal moment Mn and bracing length Lb.

The next slide shows the equations used in the AISC to determine the value of a nominal moment for a steel beam. What does lb stand for?

The equations for Mn for different cases of Lb and equation of Fcr.

The stress values are based on the lb, whether bigger or smaller than Lp, Lr.

Information of  diffeerent cases of Lb.

Practice problem for Lp & Lr and Fcr.

The bracing length at the plastic stage Lp, the limiting laterally un-braced length, and the value of Mp and Mr for a given steel W24x176 at Fy=36 ksi are required. The steps will be as follows:
1-From AISC table 1-1, we get the values ry to get the Lp from the equation Lp=300*ry/sqrt Fy. The ry value can be found in AISC table 1-1 for the given section.
2-Substitute for Fy=36 ksi, ry=3.04 inch in the previous formula.

Lb, Lr, Mp, and Mr for a given steel section with known Fy.

3- Apply in the formula of Lp=300*(3.04)/sqrt(36)=152″, we convert into feet by dividing /12.The value for Lp=12.70′ approximately.

The value of Lp, the unbraced length up to which yielding will control.

How do we get limiting laterally unbraced length for the limit state value?

1-If we refer to the next slide image, we know to estimate the limiting laterally unbraced length for the limit state value. From AISC 1-1, we get the values ry, rts, Sx, ho, and J, considering C =1. These are the parameters for estimating the limiting laterally unbraced length Lr for the limit state from formula F2-6.

The radius of gyration about y, ry=3.04 inch2,rts=3.57″ and the section modulus=450 inch3, ho=23.90″ n j polar=23.90 inch4, cw=68400.

The  data required to find lr.

2-Substitute for Fy=36 ksi and all the needed items. The relevant calculation is shown in the coming slide image. The value for limiting laterally unbraced length for the limit state=49.01.’

The calculation for Lr.

How to estimate MP and Mr.

3—The calculations are shown in the next slide images. When we substitute the equation Mp=Fy*Zx with the equation Mp=Fy*Zx, we can get the Mp, the plastic moment value. The z value is 511 inch3, while Fy=36 ksi.

The value of Mp is 1533.0 Ft.kips. We can get the Mr, which is the inelastic moment value, from the equation Mr=0.70Fy*Sx, where Sx is the statical section modulus for W24x176 and can be found to be 450 inch3.

The value of Mp and Mr.

We can get the slope S between Mp and 0.70Fy*Sx. Please refer to the next slide image for more information.

The graph between lb and Mn.

How to estimate critical stress Fcr?

For a given bracing length Lb=50′, which is >Limiting laterally unbraced length for the limit state, we can find the value of Fcr by substituting it in the shown equation in the next slide image. The critical stress Fcr value is 24.60 KSI.

Fcr Flexural Buckling stress and the elastic critical moment (Mcr).

Using the following equation, Mrx =Fcr*Sx. When we substitute, we will get the value of Mrx=922.55 Ft.kips.

Use Excel graph to plot Lb virus Mn.

We can use an Excel graph to plot the bracing length Lb versus the nominal moment; please refer to the next slide image.

Use excel graph to plot Lb and Mn.

For a valuable external source, please follow this link—lateral Torsional Buckling Limit State.

For the next post, A Solved problem 4-6 how to get the available flexure beam strength?

  • 11-Solved problem 4-5-How to design a steel beam?

    11-Solved problem 4-5-How to design a steel beam?

    Solved Problem 4-5-How To Design A Steel Beam?

    A Solved problem 4-5.

    A solved problem 4-5 from Prof. Alan Williams‘s Structural Engineering Reference Manual.

    Design of beam according to LRFD for part a.

    Part A includes the lightest adequate W section for design. We must identify which region That w section is located according to bracing. This is a design problem for which the distance between bracing for a beam is Lb < Lp. After the design, we will get the Lb value from the next step 3.

    For the LRFD design:
    1-Estimate the preliminary Zx value by considering that φbMn=Mult, since Mn=ZxFy.

    Solved problem 4-5, it is required to determine the lightest W section.

    We can get the plastic section modulus Zx= Mult /(φb*Fy). We go to Table 3-2, where sections are sorted by Zx, and select the first bold section with Zx selected > Zx estimated.

    2- From table 3-2, we get the section W16x 40, which has Zx =73.0 inch3 >72.0 inch3, the preliminary value for Zx. 
    The bracing length required can be obtained from Table 3-2 at the plastic stage Lp and lr value.

    The steel section W16x40 has no f symbol, meaning there is no problem with local buckling.

    Select W16x30 from Table 3-2 based on Zx value-LRFD design.

    This is a reminder about the Mn graph, the bracing distance, and the different zones—the value of Lp according to Fy and the radius of gyration ry.

    Lp realtion between ry and Fy.

               

    3- From Table 1-1, we can check the Lp value and find the φb*Mn.

    Estimate the value of Lp for W16x40.

    4—The section is compact since the given bracing length Lb is smaller than Lp. The value of φb*Mn= φb*Zx*Fy is to be divided by 12 to get the value in Ft-kips-LRFD. We get the φb*Mn=274 ft. kips. The exact value φb*Mn can be obtained from Table 3-2, as seen in the next slide.

    As we can see that φb*Mn is bigger than the given ultimate moment of 270 Ft.kips.

    Estimate the value of φ*Mn of the selected section using Table 3-2.

    Design of beam according to ASD for part a.

    The ASD calculation is shown in the next slide; here are the following steps to implement:
    1-Get a preliminary Zx value by considering that (1//Ω)*Mn=Mtotal, since Mn=Zx*Fy.

    We can get Zx= Mtotal /(1/Ω)*Fy). 2-From table 3-2, select the lightest w section that gives Zx>Zx preliminary.
    The selected W section is W16x40, Zx of the selected section=73.00 inch3, which is >72.144 inch3 as per requirement.

    Estimate the Zx value-ASD design.

    2—From Table 3-2, we get the section W16x 40 with Zx =73.0 inch3 >72.0 inch3, the preliminary value for Zx. 

    The bracing length required can be obtained from Tables 3-2 at the plastic stage Lp. Lp can be estimated from the relevant formula  Lp=ry* (300/sqrt(Fy)), but we need the ry value. 

    Select W16x40 based on Table 3-2-ASD design.

      3- From Table 1-1, get the Sx value, ry, for the selected section. Find the value of lp.

    4- Since the given bracing length Lb is smaller than Lp, the section is compact,  (1/ Ω)*Mn = (1/ Ω)*Zx*Fy, to be divided by 12 to get the value in Ft-kips-ASD.
    5- Check that the estimate (1/ Ω)*Mn is > the total moment Mt. The exact value (1/ Ω)*Mn can be obtained from Table 3-2, as seen in the next slide.

    Estimate the value of (1/ Ω)*Mn of the selected section

    I have added a graph from an Excel sheet to show the values of φb*Mn and (1/ Ω)*Mn for bracing length=4 feet; the section is W16x40 part a.

    use an excel sheet to find factored moment.

    Design of beam according to LRFD for part b.

    Why do we select W10x60?

    Part b determines the W shape with minimum allowable depth,  as per LRFD.

    The selection is based on the minimum depth. We will select W10x60 since the depth is smaller < depth of W16x40 as shown in the next slide, then check that the φb*Mn> Mult.

    Select W section of the minimum depth.

    The φb*Mn of the selected section is bigger than 270 ft.kips.

    Why do we select W10x60?

    Design of beam according to ASD for part b.

    This is part b, W shape with minimum allowable depth,  as per ASD,  for the selection based on the minimum depth.
    We will select W10x60 since the depth is smaller < depth of W 16×40 as shown in the next slide, then check that the (1/ Ω)*Mn > Mt.

    Estimate the value of 1/ Ω)*Mn of the selected section.

    I have added a graph from an Excel sheet to show the values of φb*Mn and (1/ Ω)*Mn for bracing length=4 feet; the section is W10x60 for part b. Thanks a lot. part a.

    Use an excel sheet to find factored moment for part b.

    For a valuable external source, please follow this link– lateral Torsional Buckling Limit State.

    Review the information for Lp and Lr for the next post. This post introduces the different terms for Lp and Lr for a steel beam.