14- Find the Flexture Strength: Analysis Problem 9-7-lb>lr.

Find the Flexture Strength: Analysis Problem 9- 7-lb>lr.

We aim to find the flexure strength and the value of the available moment Mrx for a given beam where the distance between bracings exceeds the value of lr, where lris the limiting bracing value.

From Prof. McCormac’s book, Structural Steel Design, Example 9.7. Using AISC Equation F2-4, determine the Fcr and Mrx for a w18x97 with Fy =50 ksi and unbraced length lb =38 ft, assuming that CB =1. There are three zones for Lb and Mn for LTB in the plot.

. The first zone is Plastic, hinging between zero till Lp. The second zone is the inelastic Buckling between lp and Lr. The last zone is from lb>Lr . The following slide image includes this graph for the relation between bracing length and Mn.

Solved problem 9-7 Analysis is based on LRFD.

This is an analysis problem since the W section is given, and there is no requirement to design a section. The distance between bracing for a beam is Lb, which is bigger than Lr.
To solve the problem, we will follow the following steps:

Step 1- From Table 1-1, we can get the necessary data, such as Zx, Sx, J, h0, Cw, J, and rts. Please refer to the following slide image.

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.

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

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

The following slide image shows the detailed calculation for Phi*Mrx and Mrx/omega.

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.

Solved problem 9-7 Analysis is based on ASD.

The same procedures will be used for the ASD design as in the LRFD, except that 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 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 following slide image shows an Excel graph of lb versus Φ*Mn. I have tabulated the values of the Lb virus, the phi*Mn. The LP value is 9.36 feet, and the LR value is 38 feet. The value of Φ*Mrx equals 369 Ft. kips.

The following slide image shows an Excel graph between lb versus (1/ Ωb)*Mn. I have tabulated the values of the Lb virus, the phi*Mn. The LP value is 9.36 feet, and the Lr value is 38 feet. The value of (1/ Ωb)*Mrx equals 246 Ft. kips.

For more detailed illustrations for the CB, please follow this link–Flexural Limit State Behavior.

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