Brief content of steel beam post 14.

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.

The different zones for the bracing length Lb

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:

Solved problem 9-7

Step-1- from table 1-1 we can get the necessary data, as Sx, J, h0, rts.

The formula for the value of Fcr

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.

Solved problem 9-7

 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.

The different parameters for estimating Fcr value.

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 bracing length lb is> Lp

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.

The detailed estimation of Fcr

 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.
 

The detailed estimation of Fcr and Mnx/ Ωb

 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.

For more detailed illustrations for the CB, please follow this linkFlexural Limit State Behavior.

For the next post, How to design a beam with a design chart?

                          

                                                                                  

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