12- Easy introduction to plastic bracing length Lp, Lr, Fcr.

Last Updated on February 20, 2026 by Maged kamel

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

How to estimate Lp?

Lp is the maximum unbraced length that will allow the shape to 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 the limiting laterally unbraced length Lr?

Lr is the limiting laterally unbraced length for the limit state. The following 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 in 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 lr is shown in the slide image.

Limiting bracing length Lb=Lr.

Bracing length conditions.

The next slide shows the different values of Mn, or the nominal moment of a section, depending on the bracing length: 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 by AISC to determine the nominal moment of 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 it is larger or smaller than Lp or Lr.

The Different zones for bracing length for beams

Practice problem for Lp & Lr, and Fcr-W24x176-Fy=36 ksi.

The bracing length at the plastic stage, Lp, the limiting laterally un-braced length, and the values 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 the limiting lateral 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 obtain the values ry, rts, Sx, ho, and J, assuming 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 slide image below. The value for limiting laterally unbraced length for the limit state = 49.01.’

The  detailed 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 into itself, we obtain the plastic moment, Mp. The z value is 511 in³, 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 sectional modulus for W24x176 and is 450 in^3.

The value of Mp and Mr.

We can get the slope S between Mp and 0.70*Fy*Sx. For more information, please refer to the image on the next slide.

The graph between lb and Mn.

Let us check the value of fcr for lb = lr. Using the equation for far, the value must be equal to 0.7*Fy, which in our case equals 25.20 ksi. Since we have fy = 36 ksi, please refer to the next slide for more information.

How can critical stress Fcr be estimated for lb=50 feet?

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, 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 Mrx = 922.5 ft-kips.

Use an Excel graph to plot Lb virus Mn.

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

Use excel graph to plot Lb and Mn.

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

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

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

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

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