Last Updated on February 16, 2026 by Maged kamel
Practice problem 5-5-6-Compute Lp and Lr, φb*Mn and Mn/Ωb for lb=10 feet.
Practice problem 5-5-6 A W12 x 30 of A992 steel has an unbraced length of 10 feet. Using Cb = 1.0, a. Compute Lp and Lr. Use the equations in Chapter F of the AISC Specification. Do not use any of the design aids in the Manual.
b. Compute the flexural design strength, φb*Mn.
c. Compute the allowable flexural strength Mn/Ωb. Practice problem 5-5-6 is from the Steel Design Handbook.
How do we estimate Lp for W12x30?
In the absence of design aids, Table 3-3, where W sections are sorted by their plastic section Zx, in which we can easily find the values of lp and lr and other values of flexure design moments, we need to estimate Lp and lr values by using the equations given by the specifications.
The following slide shows the detailed estimate of the Lp value. Lp is the unbraced length at which the plastic hinge should develop, given by equation F2-5 as 1.76*rysqrt(E/Fy).
From Table 1-1, we will get the radius of gyration, ry, for W12x30. The ry value is 1.52 inches. From the data on A992 steel, we have the yield stress Fy=50 ksi. The E value is 29000 ksi. The value of Lp is 64.43 inches, which can be rounded to 5.40 feet.
How do we estimate Lr for W12x30?
We use Table 1-1 for the W12x30 W section to obtain the necessary data for estimating Lr. Lr represents the unbraced length at which lateral-torsional buckling transitions from the inelastic range to the elastic range. These items are: ry, rts, Zx, SX, Cw, C, J, H0.
We can get the value of bf/2tf and h/tw. Please refer to the next slide image for details.
What is the value of lr?
We use the equation F2-6 provided by the AISC specification. The value of lr equals 187.24 inches and is rounded to 15.60 feet. Please refer to the next slide image for more details. Thus, we have completed the required part a) in the practice problem 5-5-6.
Check the buckling parameters for W12x30 with Fy = 50 ksi.
We will determine the local buckling parameters for the flange and web, 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, 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. Please refer to the next slide image for the detailed estimate of compactness ratios.
Check whether W12x30 is compact or non-compact.
On the next slide, we show the values of bf/2tf and h/tw for W12x30 against the required local buckling parameters for Fy=50 ksi. We will find that the flange and web are compact, so the entire W12x30 section is compact.
Find the equation of Mn-based on Fy=50ksi.
The Plastic moment value equals Zx*Fy, which is 50×43.10 = 2155 in.- kips. The term (0.7*Fy*Sx) value equals 0.7*50*38.60=1352 inches. kips. The first term corresponds to bracing length Lp=5.40 ft; the second corresponds to lr=15.60 feet. We have a given lb=10 ft.
We get the values of Mp and 0.70*Fy Sx in ft · kip using the conversion factor of 1 ft = 12 in. Mp value equals 179.583 ft ·kips, while 0.70*Fy Sx = 112.583. We could round these values to the nearest.
There is a linear relationship between the required Mn and Mp, and between (0.7*Fy*Sx) and Lp and Lr. The BF factor is the slope value, which equals (Mp-0.7*Fy*Sx)/(Lr-Lp)=6.5686.
What is the value of Mn at lb=10 feet?
The relation of 179.583-6.53*(Lb-Lp) represents the nominal moment value. The Lb value equals 10 feet, and the Lp value equals 5.40 feet. The Final value for Mn is 149.367 ft-kips.
Part b) Compute the Flexural design strength-φb*Mn.
For part b of practice problem 5-5-6, for the flexural design strength φb*Mn, we multiply the Nominal value by the factor φb, which equals 0.90. We get a value of 134 ft-kips.
Part c. Compute the allowable flexural strength Mn/Ωb.
For part b of practice problem 5-5-6, for allowable flexural strength Mn/Ωb, we multiply the Nominal value by the factor (1/ Ωb), which equals (1/1.67). We get the value of 89 Ft.kips.
We could use an Excel sheet to graph the relationship between Lb and the Flexural design strength, φb*Mn, for W12x30, Fy=50 ksi. We can see the linear portion, and the Value of φb*Mn at 10 feet is 134 ft-kips.
We could use an Excel spreadsheet to plot the relationship between Lb and the allowable flexural strength, Mn/Ωb, for W12x30 with Fy=50 ksi. We can see the linear portion, and the Value of Mn/Ωb at a bracing length of 10 ft is 89 kips. Thanks a lot.
The PDF file for this post can be viewed or downloaded from the following link.
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.
This links to the next post, post 9d: Practice problem 5-5-8-Compute Mn.