 # 14- A guide to solved problem 9-7-Lb is bigger than Lr.

## Solved problem 9-7-When Lb is bigger than Lr, What Is Flexure Strength?

### Brief description of the video.

We will talk about the third case, where Lb, buckling length or distance between bracing, Lb is bigger than Lr. This is through a solved problem 9-7.

We are talking about the curved portion, the MCr follows the equation F2-4 from AISC, M value, which is called Mcr, Mcr=Fcr*Sx.

Mcr can be evaluated from Fcr equation that includes, the coefficient cb where Cb is the coefficient for the bending moment also includes E the modulus of elasticity  E =29000 ksi,  but E will be taken as 29000 ksi in this equation.

Lb is the bracing distance rts, the effective radius of gyration, in inch, the first term=Cb*Pi^2*E/(Lb/rts)^2 all to be *(sqrt(1+0.078*J*C/(Sx*ho))*(Lb/rts)^2)), where J =torsional constant in inch4.

C=1 for doubly symmetric I shapes. Sx is the elastic section modulus. The ho distance between the Cg of flanges( Lb/rts), bracing length/ effective radius of gyration. This is a part of the video, which has a subtitle and closed caption in English.

You can click on any picture to enlarge it and then press the small arrow to review all the other images as a slide show.

From Prof. Mccormac’s book Structural steel design Example 9.7. Using AISC Equation F2-4, determine the values of and for a w18x 97 with fy =50 ksi and unbraced length lb =38 ft,  assume that CB =1.

As a  reminder, this is the graph for the relation between bracing length and Mn.

### Solved problem 9-7 Analysis is based on LRFD.

This an analysis problem, since the W section is given, for which, the distance between bracing for a beam is Lb is bigger  Lr, and also Lb is bigger than LP.
To solve the problem, we will follow the following steps:

This is the equation used to get the value of fcr for a beam when we have the relevant values of Sx,Jc,CB, h0.

3-Since the given Lb is bigger than Lp and also Lb is bigger than Lr, in our solved problem 4-7,  the section is in the elastic slender zone, we need to estimate the stress Fcr from the equation F2-4 as shown in the next slide.

We need to find the data expressed in that equation, we can get these data from table 1-1 for the W- section our given section is W18x97, our data can be obtained from table 3-7.

The next slide shows how to get the other values and how we find out that our given bracing distance is >lr. the table includes the value of Lr which is similar to the value estimated by the equation.

5-Estimate  the strength of the section for the LRFD by using the formula Φb*Mn.= Φb* Fcr *Sx=0.90*26.156*(188)=369 ft.kips
This is the detailed calculation for The LRFD design shown in the next slide image for the solved problem 9-7.

### Solved problem 9-7 Analysis is based on ASD.

For the ASD design:  the same procedures that have been done in the LRFd, except step no.5 will be modified.
1-From table 1-1 get both Lp, Lr values for the given section W18x97.
2- since the given Lb>Lp and >Lr, in our example, the section is in the elastic slender zone.

3-We need to get Sx, J, ho =d-2tf, rts from table 1-1 for section W18x97, and also we need Lr value for the same section but from table 3-2, then Estimate Fcr from equation F2-4.

4-after evaluating Fcr, we get the value of Mn of the section, where Mn= Fcr *Sx.  cr=25.156 ksi.

Mn=fcr*Sx=25.16*188/12=410.0 Ft.kips.

5-Estimate  the strength of the section for the ASD by using the formula Φb*Mn.= (1/ Ωb)* Fcr*Sx=.(1/1.67)*26.156*(188)=246 Ft.kips
This is the detailed calculation for The LRFD design shown in the next slide image for the solved problem 9-7.

This is the pdf file used for the illustration of this post.