Last Updated on June 18, 2024 by Maged kamel

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 to be followed if we start the design by selecting table AISC 3-2.

Solving problem 4-7 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* M_n )/Φb* Fy.

Based on the given data Fy=50 ksi, Φb=0.90,(Φb* M_n )=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, W16x31, and W12x35 sections. 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) are sketched for all the chosen sections, yet their capacity (Φb* M_rx) will be from 120.0 ft. kips and 129.00 Ft.kips, while the given Multimate=190.0 ft. kips.

The next slide shows The plot of lb against factored Mn for the four sections of W10x45, W14x34, W16x31, and W12x35.

This means all these sections cannot carry the given moment since Lb>lr. Again, we have to check other selections for W steel shapes. But this time, we will focus on w sections with Φb* M_rx bigger than 190 Ft.kips.

Continue using Table 3-2 will not be our best choice because of the many trials we will have to do Since we have 7 sections with Φb* M_rx bigger than 190 Ft.kips.

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

Instead, we will use Table 3-10. For solved problem 4-7, we have a given factored moment of 190 ft. kips (LRFD) and bracing length Lb=31.’
We will use Table 3-10. The appropriate page is P-3-129 for CM#14 and page P3-122 for CM-15, Where lb is from 18′ to 34′ and φb*Mn is between 180 and 240 Ft.kips.

Table 3-10 is used for the nominal strength value.

Considering the bracing length Lb=31′, we draw a vertical line. We are interested in the first solid graph, which will give a w-section, which is W12x58. This will be our choice for 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 value of 127 Ft. Kips.

Please find the calculations in the next slides if we want to check the value of φb*Mn for the new section using the equation AISC F2-4. 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 inch3 from Table 1-1. The other related data are also shown in the next slide image.

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

Fcr value by the equation.

We will substitute the equation for Fcr with the different parameters shown, 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.

Solving problem 4-7 by using ASD design in terms of FCr.

The factored(1/Ω_b* M_r) value can be estimated as =(1/Ω_b)*F_cr*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 solved problems 4-7.

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

For the next post, Cb-The coefficient of bending-1/3.