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′.
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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 3-2.
If we will proceed by using 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, 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- sections.
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 LRFDmoment > 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 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.