41-Practice problem 5-6-1-What is the available strength for W12x65?

Last Updated on September 12, 2026 by Maged kamel

Practice problem 5-6-1: What is the available strength for W12x65?

Practice problem 5-6-1. A W12x65 is used as a supported, uniformly loaded Beam with a 50-foot span and continuous lateral support. The yield stress, Fy, is 50 ksi. If the ratio of live load to dead load is 3, what is the available strength for W12x65, and what is the maximum total service load, in kips/ft, that can be supported? a. Use LRFD. b. Use ASD.

I will include both the LRFD and ASD designs in this post. Practice problem 5-6-1 is from the Steel Design Handbook.

Timestamps for the video.

• 00:00 Introduction to practice problem 5-6-1.
• 03:01 Check the information on W12x65 from Table 1-1 in Part 1.
• 04:06 What are the values of λ Flange and λweb for W12x65?
• 05:01 Values of λp and λr for the Flange, and λp and λr for the web, based on Fy = 50 ksi
• 08:34 Sketch the relation between λ and Mn and show Mp and Mr.
• 10:00 How can we determine the nominal Moment value of W12x65?
• 11:50 Introduction to the list of noncompact sections for Fy=50 ksi.
• 13:30 Excel sheet list of noncompact sections for Fy=50 ksi.
• 15:17 How can we determine lp-dash length?
• 18:50 Confirm our Estimate with Table 3-2 for lp-dash and factored nominal Moment.
• 19:30 Estimate service load -LRFD design.
• 21:05 Estimate service load -ASD design.

Check the local buckling parameters for W12x65.

For the given W12x65 section with Fy = 50 ksi and continuous lateral support, we need to determine the local buckling parameters bf/2tf and h/tw to confirm whether the section is compact or non-compact.

We checked Table 1-1 for W12x65 and found that the section has a symbol f’, indicating that W12x65 doesn’t conform to the local buckling parameters.

We obtain the data needed to solve practice problem 5-6-1: bf/2tf = 9.52 and h/tw = 24.90. The plastic section modulus Zx = 96.80 inch^3, and the elastic section modulus = 87.90 inch^3.

As a reminder, please find the local buckling parameters for W sections with 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, 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

Since bf/2tf is bigger than λFp, the Flange is noncompact; hence, W12x65 is a noncompact section.

Find the values for Mpx and 0.7*Fy*Sx for W12x65.

Practice problem 5-6-1.The Plastic Moment value equals Fy*Zx, which equals 50*96.8 = 4840. Kips. The term (0.7*Fy*Sx) value equals 0.7*50*87.90=3076.50 Inch. Kips. The first term corresponds to λFp=9.15; the second corresponds to λFr=24.08.

Practice problem 5-6-1: What is the available strength for W12x65 based on local buckling?

To find the available strength Mn for W12x65 based on local buckling, we plot the relationship between λF and Mn; Mn can equal 4749 in. Kips can be approximated to 396 Ft. For the available strength φ*Mn, multiply by 0.9; the final answer is 356 ft. kips. This is the answer for part a.

What are the noncompact W sections based on Fy=50 ksi?

The following slide, from the companion to The AISC Steel Construction Manual, Volume 1- Design Examples, lists nine noncompact W sections, starting with W21x48 and ending with W6x8.50. The %centage reduction of the nominal Moment is included for each W section.

Create an Excel sheet for noncompact W sections with Fy = 50 ksi.

In the Excel sheet for noncompact W sections based on Fnoncompactsi, I sorted the W sections with bf/2>> 9.15 using an Excel sheet quoted from AISC W sections V15; the result is 10 W sections. Please refer to the next slide for more details.

Adjustment in the Lp distance based on Mn.

The bracing Length Lp, which is equal to 10.70 feet, corresponds to Mpx equal to Fy*Zx, which is equal to 403 Ft.kips, and the Bracing length lr, 35.10 ft, corresponds to 0.7*Fy*Sx, which equals 256 Ft.kips. But our W12x65 has an Mn of only 396 ft-kips, so we need to add an extra length to lp.

This distance equals the difference between (403-396)/B.F value. The B.F.F. value is tBF slope of the line joining Lp and Lr and equals 6.02.

The final L’p = 11.86 ft, approximated to 11.90 Ft.

Check the values for Lp’,lr, and φ*Mn from Table 3-2.

To check the Lp’, lr values, and φ*Mn for W12x65, we find the New Lp value equals 11.90 ft, lr = 35.10 ft, and φ*Mn=356 Ft.kips.

Check the maximum Total service load.

For Practice problem 5-6-1, what is the available strength for W12x65? We will equate Mult to φ*Mn. We have l/D=3, the uniform ultimate Load Wul=1.2*D+1.6*3D=6D, the Mu=6D*(50)^2/8=356, then D=0.19 K/Ft, the value of L=3*D=3*0.17=0.51 k/F.

We will add D+L to achieve a Maximum Total service load of 0.76 K/Ft. Thanks a lot.

You can view or download the PDF used to solve this problem from the following Link.

Please refer to post 41m for the metric version of this post.

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

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

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

For more information about LP and Lr, please see the post: Step-by-step guide to Lateral-torsional buckling.