Last Updated on August 31, 2024 by Maged kamel

Practice problem 5-5-8-Compute Mn for W18x71 with lb=9 feet.

Practice problem 5-5-8 A W18 x 71 is used as a beam with an unbraced length of 9 feet. Use Fy=65 ksi and Cb = 1 and compute the nominal flexural strength. Compute everything with the equations in Chapter F of the AISC Specification. Practice problem 5-5-8 is from the Steel Design Handbook.

How do we estimate Lp for W18x71 with Fy=65 ksi?

In the absence of design aids, Table 3-3, where W sections are sorted by their plastic section Zx, and with different Fy, which equals Fy=65 ksi, 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 value of the un-braced length that the plastic hinge should be developed, which is given by equation F2-5 as equal to 1.76*rysqrt(E/Fy).

From Table 1-1, we will get the radius of gyration ry for W18x71. The ry value is 1.70 inches from the data that Fy=65 ksi. The E value is 29000 ksi. The value of Lp will be 63.20 inches and can be rounded to 5.30 feet.

How do we estimate Lr for W18x71 with Fy=65 ksi?

We use table 1-1 for W section W18x71 to get the necessary data for estimating the 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. The Sx value equals 127 inch3 while Zx=146 inch3.

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 for W18x71 equals 196 inches and is rounded to 16.33 feet. Please refer to the next slide image for more details.

Check buckling parameters for W18x71 where Fy=65 ksi.

We will find the local buckling parameters for the flange and web-based Fy=65 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=65 ksi, then λFp=0.38*sqrt(29000/65)=8.03. λFr=1.0*sqrt(E/Fy)=1*sqrt(29000/65)=21.12.

For the web compactness ratio, the flange λwp=3.76*sqrt(E/Fy), λWp=3.76*sqrt(29000/65)=79.42. λwr=5.70*sqrt(E/Fy)=5.70*sqrt(29000/65)=120.4. Please refer to the next slide image for the detailed estimate of compactness ratios.

Check whether W18x71 is compact or non-compact.

In the next slide image, we have the bf/2tf and h/tw values of W18x71 against the required local buckling parameters based on Fy=65 ksi. We will find out that the flange and web are compact, so the whole section of W18x71 is compact.

Find the equation of Mn-based on Fy=65 ksi.

The Plastic moment value equals Zx*Fy, which is equal to 65×146=9490 inch. Kips. The term (0.7*Fy*Sx) value equals 0.7*65*127=5778.50 Inch. Kips. The first term corresponds to bracing length Lp=5.30 ft; the second corresponds to lr=16.33 feet. We have a given bracing lengthlb=9 ft.

We get the values of Mp and 0.70*Fy Sx in Ft.kips by using the conversion factor of 1Ft=12 inches. Mp value equals 791 Ft. kips, while 0.70*Fy Sx=482, we could round these values to the nearest ones.

There is a linear relation for the required Mn with Mp and(0.7*Fy*Sx) together with Lp and Lr. The BF factor is the slope value, which equals (Mp-0.7*Fy*Sx)/(Lr-Lp)=28.01.

What is the value of Mn for W18x71 at lb=9 feet?

The relation of 791-28.01*(Lb-Lp) represents the nominal moment value. The Lb value equals 9 feet, and the Lp value equals 5.30 feet. The Final value for Mn equals 687 Ft. kips.

We could use an Excel sheet to graph the relation between Lb and the Flexural design strength-Mn in ft. kips for W18x71 with Fy=65 ksi. We can see the linear portion and the Value of Mn at 9 feet equals 687 ft. kips. Thanks a lot.

Here is the link for Chapter 8 – Bending Members.
This links to the next post, 10-lateral-torsional buckling for steel beams.

A new post is added 41-Practice problem 5-6-1-find the total service load for W12x65. The section is non-compact.