brief illustration for the content of post 26 for steel beam posts

26-Part-2-4 of the Solved problem 9-9-6, How To Find LL?

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Last Updated on September 14, 2026 by Maged kamel

Table Of Contents
  1. Part-2-4 for the Solved Problem 9-9-6: How To Find LL? 

Part-2-4 for the Solved Problem 9-9-6: How To Find LL? 

The following slide image summarizes the content of Post 26-steel Beam.

Summary of the content of post 26-steel beam.

Sx for the built-up section.

This is part-2-4 of the solved problem 9.9.6 from Prof. Charles Salmon’s book. We will estimate the elastic section Modulus Sx for the built-up section.
We can estimate Ix for this section using the full rectangular section. For the Ix, where the x-axis is the major axis, the easiest method to evaluate the Ix for the section is to consider the whole rectangular section (16″x27.25″) and then deduct the two void rectangles of dimensions (7.843″x26″).
The Moment of Inertia can then be estimated as Ix = (16*27.25^3/12) – (2*7.843*(26)^3/12) = 4002.8116 in^4.

The distance from the N>A to the top of the upper Flange Y Value =27.25/2=13.625″. the section Modulus Ix/y=Sx=(4002.8116/13.625)=293.80 inch3.

Zx should be > Sx. We have estimated Zx=319.06 inch3, which is > 293.80 inch3. We can proceed to estimate the Nominal moments.

Part-2-4 for the solved problem 9-9-6-The value of Sx.

Mn-based on the local buckling of the Flange.

In part-2-4 of the solved problem 9.9.6. Two graphs are drawn: the left-side graph is for the local buckling of the Flange, and the values are as follows: the x-axis represents lambda, while the vertical axis represents the nominal Moment Value.

Both the Flange and the Web are noncompact sections; the Mn vValuewill be >0.70 Fy*Sx but less than Fy*Zx.
λp=8.026  For which the Mn=Fy*Zx=65*319.06/12=1728 Ft- kips. λr=15.90  For which the nominal Moment Mn=0.70FySx=0.765*293.806/12=1114 Ft- kips. The nominal Moment of the Flange will be between 1728 and 1114 ft-kips.

M nominal value for built-up section flange.

When the factor λflange is equal to 12.80, for which the nominal Moment Value Mn=1728-(1728-1114)*(12.80-8.026)/(15.90-8.026)=1355.70 Ft-kips, this represents the Mn based on the limit state for the local buckling of the Flange.

Mn for the flange of the built-up section.

The following slide shows a table of Mn for the Flange, generated in Excel.

This is a table for the value of Moment for flange.

The following slide shows a graph of Mn for the Flange generated in Excel.

Graph for Mn for flange.

The nominal Moment Mn is based on the local buckling of the Web.

In part-2-4 of the solved problem 9.9.6. Two graphs are drawn from the right-side graph that represents local buckling of the Web; the values are as follows: λp = 79.46, for which the nominal Moment Mn = Fy*Zx = 65*319.06/12 = 1728 ft-kips.

λr=120.40  For which the Mn=0.70Fy*Sx=0.7*65*293.806/12=1114 Ft- kips.

When the factor λweb=83.20, the nominal Moment can be estimated as Mn=1728-(1728-1114)*(183.20-79.46)/(120.40-79.46)=1627.0 Ft-kips; this is the Mn based on the limit state for the local buckling of the Web. The next slide image shows the detailed calculations for the nominal Moment.

Mn for the web of the built-up section.

The following slide image shows a table of Mn for the Web, generated from an Excel sheet.

This is a table for the value of Moment for web.

The following slide image shows a graph of Mn for the Web generated from an Excel sheet.

Graph for Mn for web.

You can view or download the PDF for this Post from the following document.

Post 26: Problem 9-9-6-Part 2-4- PDF

The nexPostst will be Part 3/4 for Solved Problem 9-9-6. How do you find LL?

Here is thLinknk to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 14th ed.

Here is thLinknk to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 15th ed.

Here is thLinknk to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 16th ed.