Brief data for post 14 -steel beam

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

Last Updated on June 17, 2024 by Maged kamel

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

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, assuming that CB =1. As a reminder, this is the graph for the relation between bracing length and Mn.

The different zones for the bracing length Lb

Solved problem 9-7 Analysis is based on LRFD.

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

Step-1- From Table 1-1, we can get the necessary data, such as Sx, J, h0, and rts.

Solved problem 9-7

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

The formula for the value of Fcr

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 calculation for lr using formula.

This is a continuation of the estimate of Lr and the value of lr from Table 3-2, which matches the Lr value.

Use Table 3-2 to find lr and Lrfd values of Mp and Mr.

Estimate the strength of the section for the LRFD by using the formula ΦbMn.= Φb Fcr Sx=0.9026.156*(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. Please refer to the next slide image for a detailed estimate of Fcr.

The detailed estimation of Fcr and Mn value

Solved problem 9-7 Analysis is based on ASD.

For the ASD design, the same procedures will be used as in the LRFd, except step 5 will be modified.
 1-From table 1-1, get both Lp and Lr values for section W18x97.
2- since the given Lb>Lp and >Lr, the section is in the elastic slender zone in our example.
3-We need to get Sx, J, ho =d-2tf, and rts from Table 1-1 for section W18x97. We also need the Lr value for the same section but from Table 3-2. Then, we need to Estimate Fcr from equation F2-4.

4-After evaluating Fcr, we get the section’s Mn value, 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 ASD using the formula (1/ Ωb)* Fcr*Sx=.(1/1.67)*26.156*(188)=246 Ft.kips
This is the detailed calculation for The ASD design shown in the next slide image for the solved problem 9-7.   
 

The detailed estimation of Fcr and Mnx/ Ωb

The following slide image shows an Excel graph between lb versus Φ*Mn.

Excel plot Lb virsus Phi*Mn.

The following slide image shows an Excel graph between lb versus (1/ Ωb)*Mn.

Excel plot Lb virsus 1/omega *Mn.

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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