Last Updated on October 5, 2024 by Maged kamel

Practice problem 5-6-1-find the total service load for W12x65.

Practice problem 5-6-1. A W12x65 is used as a supported, uniformly loaded beam with a span length of 50 feet and continuous lateral support. The yield stress, Fy, is 50 ksi. If the ratio of live load to dead load is 3, compute the available strength and determine the maximum total service load, in kips/ft, that can be supported. a. Use LRFD. b. Use ASD. I will include only the LRFD design in this post. Practice problem 5-6-1 is from the Steel Design Handbook.

Check the local buckling parameter for W12x65.

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

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

We get the data required to solve the practice problem 5-6-1: the values of bf/2tf=9.52 and h/tw=24.90. The plastic section modulus Zx=96.80 inch23, and the elastic section moduls=87.90 inch3.

As a reminder, please find the local buckling parameters for W sections for 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, the flange 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 non-compact section.

Find the values for Mpx and 0.7*FySx for W12x65.

The Plastic moment value equals Fy*Zx, which equals 50×96.8=4840 inch. 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 lλFr=24.08.

Find the values for φ*Mn for W12x65 based on local buckling.

To find the values for Mn for W12x65 based on local buckling, we plot the relation between λF and Mn; the Mn can equal 4749 inch. 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.

Press on any image, and you will see a slide show of pictures from 1 to 6, which are included above.

Practice problem 5-6-1-find the total service load for W12x65.

Table 1-1 for W12x65.

Table 1-1 part 2 find bf/2tf nd h/tw for W12x65.

Part 2-table 1-3, h/tw=25.90.

Find Mpx and 0.70*Fy*Sx for W12x65.

Find the values for φ*Mn for W12x65.

What are the non-compact W sections based on Fy=50 ksi?

The following slide, from the companion of The AISC steel construction Manual volume 1-design examples, lists Nine non-compact sections starting from 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 sections based on Fy=50 ksi.

I have sorted the W sections with bf/2tf bigger than 9.15 using an Excel sheet quoted from The 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 equals 256 Ft.kips. But our W12x65 has an Mn equal to only 396 ft. kips, an extra length to add to lp.

This distance equals the difference between (403-396)/B.F value. The BF value is the slope of the line joining Lp and Lr and equals to 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 Lp’, lr values, and φ*Mn for W12x65, we find the New Lp value equals 11.90 ft, lr=35.10Ft, and φ*Mn=356 Ft.kips.

Check the values for the maximum total service load.

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 get the Maximum Total service load equal to 0.76 K/Ft. Thanks a lot.

Press on any image, and you will get a slide show for pictures from 7-13, which are included above.

List of Non compact W sections via AISC design examples data.

Local buckling parameters based of Fy=50 ksi.

An excel sheet for non compact W sections.

Adjustement of Lp value based on Mn .

Check Lp and lr and φ*Mn.

The value of D and L after estimation.

The total service load value.

Here is the link for Chapter 8 – Bending Members.
For more information about LP and Lr, please see the post: Step-by-step guide to Lateral-torsional buckling.