Last Updated on August 11, 2024 by Maged kamel
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.
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.
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.
3- From Table 1-1, we can check the Lp value and find the φb*Mn.
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.
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.
2—From Table 3-2, we get the section W16x 40 with Zx =73.0 inch3 >72.0 inch3, the preliminary value for Zx.Â
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.Â
 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.
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.
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.
The φb*Mn of the selected section is bigger than 270 ft.kips.
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.
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.
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.