Brief data for post 16 -steel beam

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

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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.

Solved problem 4-7-it is required to design a w section

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* 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.

Estimate the plastic section modulus Zx for Solved problem 4-7.

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

Check the requirements to have a compact section.

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.

The various selections of beams.

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* Mrx) 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.

The corresponding Lp and Lr & factored Mp

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. but this time we will focus on w sections that have Φb* Mrx bigger than 190 Ft.kips.

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

A new selection of W- section from table 3-2

A lot of efforts have 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 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, for CM#14 and page P3-122 for CM-15.where lb is from 18′ to 34′ and φb*Mn is between 180 to 240 Ft.kips.

A new selection of W- section from table 3-2-check by table 3-10

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 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.

From Lb=31’, draw a vertical line

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.

Get the values for Lb, Lr from table 3-2, also the values for φb*MP, and φb*Mr for W12x58.

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.

Get the values for cw and J from table 1-1.

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.

Sketch the relation between bracing length Lb and the values of Lp, Lr, φb*MP, and φb*Mr,

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.

Detailed estimation for fcr and φb*Mn

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

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.

The estimation for (1/ Ω)*Mn for section W12x58.

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

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.

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