 # 16-Solved problem 4-7- design using table 3-10 when Lb>Lr.

## Solved problem 4-7-design using table 3-10 when Lb>Lr.

We will discuss why we should use Tabel 3-10, for the design of the section, instead of table 3-2, since Lb>Lr as we will find out.
A solved problem 4-7 from Prof. Alan Williams’s book. A simply supported beam of Grade 50 has an unbraced length of 31′. Determine the lightest adequate w-shape if the beam is subjected to a uniform factored bending moment of 190 ft-kips(LRFD) or 127.0 Ft-kips (ASD) with cb=1. I have included the steps which are to be followed if we are going to start the design by selecting table AISC 3-2.

### Solving by using LRFD design.

If we will proceed by using AISC table 3-2, for the solved problem 4-7, where the W sections are sorted based on Zx. We have Zx estimated preliminary from the relation (Φb* Mn ) =(Φb* Fy*Zx ) then Zx=(Φb* Mn )/Φb* Fy.

Based on the given data Fy=50 ksi, Φb=0.90,(Φb* Mn )=190.0 ft. kips, the estimated Zx=50.66 inch4 as we can see from the next slide image.

Our section should be checked against FLB, WLB, and LTB.

Sorting by Zx , we have various W10x45, W14x34, W 16×31, and W12x35 all these sections have Zx >50.66 inch4.
We will get the corresponding values of Lp and Lr for each section.

Our Lb is>Lr, for all of the different sections, the factored(Φb* MP) and the factored (Φb* Mr) is sketched for all the chosen sections, yet their capacity (Φb* Mr) will be from 120.0 ft. kips and 129.00 Ft.kips, while the given Multimate=190.0 ft. kips.

This means that all these sections cannot carry the given moment since Lb>lr. Again we have to check other selections of W steel shape..

Continue using table 3-2, will not be our best choice, because of the many trials that we will have to do, instead, we will use table 3-10.

For the solved problem 4-7, in which we have a given factored moment of190 ft. kips(LRFD) and bracing length Lb=31′
We will use table 3-10, the appropriate page is P-3-129, where lb is from 18′ to 34′ and φb*Mn is between 180 to 240 Ft.kips.

Again we proceed with a new selection of W-section that can carry more than 190.0 Ft.kips, factored Moment at lb=lr. We will check these sections in the next slide image.

We have a variety of sections that can give factored LRFD moment > 190.0 ft. kips we can see these sections as included in the next slide image.

A lot of efforts have been made for selection through table 3-2, but using the graph will give in one step to the economic section as we will see next.

### Using table 3-10 for nominal strength value.

By considering the bracing length Lb=31′, we draw a vertical line, and we are interested in the first solid graph, which will give a w-section, which is W12x58. which it will be our choice for which the corresponding, factored moment φb*Mn=196.0 ft. kips.

The factored moment based on the ASD design will be Mn/Ω =130.0 ft. kips, these values are higher than the given values.

If we want to check the value of φb*Mn for the new section, by using the equation AISC F2-4, please find the calculations in the next slides. The steps show how to derive the same values as obtained from table 3-10 can be obtained.

The selected section W12 x58 for the solved problem 4-7 has Sx=78.0 inch3 and Zx=86.40 inch 3 from table 1-1. the other related data are also shown in the next slide image.

This is a graph between the bracing length and the factored moment value for section W12x58. the values of the corresponding factored moment against the different values of bracing lengths are drawn.

### Fcr value by the equation.

We will substitute in the equation for Fcr by the shown different parameters, Fcr=33.30 KSI.
The factored(Φb* Mr) value can be estimated as =Φb*Fcr*Sx=0.90*33.30*78.00/12=196.0 Ft.kips, this value is for LRFD.

The factored(1/Ωb* Mr) value can be estimated as =(1/Ωb)*Fcr*Sx=(1/1.66)*33.30*78.00/12=130.0.0 Ft.kips, this value is for ASD. Our estimation for Mn*φb and Mn/Ω will be very close to that obtained by table 3-10. this is the end of the design process for the W section as required by the solved problem 4-7.

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

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

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