13-Solved Problem 4-6-How to Find the Available Flexural Strength?

Last Updated on September 14, 2026 by Maged kamel

A solved problem 4- 6: How to Find the Available Flexural 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 Moment coefficient Cb = 1.0.

Determine the Beam’s available flexural 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.   

Analysis for the given section by the LRFD design.

Solved Problem 4-6 is an analysis problem; the section is given with a Beam bracing distance Lb such that Lp < Lb < Lr for LRFD design. It falls in the second Zone of inelastic buckling; see the next slide for details.

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

Determine the maximum unbraced length required for the section to reach its plastic Moment strength, Lp. The relevant equation for Lp is 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 greater than Lp. Please refer to the slide image for more details.

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

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

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

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

This is the limiting laterally unbraced length, L, from Table 3-2 for W16x48.

For the given Lb, check whether 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 LRFD: φb*Mn = 269.50 ft-kips.  

These are the values of φb*Mp and φb*Mr using Table 3-2.

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

The analysis for the given section is 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 = 6 ft, is greater than Lp but less 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).

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 flexural strength based on ASD is (1/ωb)*n = 179.00 ft-kips.

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

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

The PDF containing the data for this post is available for review and download via the button below.

This is a Link to the solved problem 4-5. Solved problem 4-5. Design a steel Beam: Lb less than Lp.

For the next post, A Solved problem 9-7: When Lb> Lr, what is flexural strength?

Here is the Link to Chapter 8, “Bending Members.” A Beginner’s Guide to the Steel Construction Manual, 14^th ed.

Here is the Link to Chapter 8, “Bending Members.” A Beginner’s Guide to the Steel Construction Manual, 15^th ed.

Here is the Link to Chapter 8, “Bending Members.” A Beginner’s Guide to the Steel Construction Manual, 16^th ed.