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

Last Updated on September 14, 2026 by Maged kamel

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

We will discuss why we should use Table 3-10 to design the section instead of Table 3-2 when Lb is large, as we will see.

A solved problem 4-7 from Prof. Alan Williams’s book. A 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=1.0. I have included the steps to be followed if we start the design by selecting AISC Table 3-2.

Solving problem 4-7 using LRFD design.

If we proceed using AISC Table 3-2 for the solved problem 4-7, where tZx sorts the W sections, we preliminarily estimate Zx from the relation (Φb* Mn) = (Φb* Fy*Zx), so Zx = (Φb* M_n)/Φb* Fy.

Based on the given data, Fy = 50 ksi, Φb = 0.90, and (Φb* Mn) = 190.0 ft-kips, the estimated Zx = 50.66 in^4, as shown in the next slide image.

Check our section against FLB, WLB, and LTB. If the section does not have an F designation, it means that there is no local buckling for the Flange or web.

Sorting by Zx in Table 3-2, we have the following sections: W10x45, W14x34, W16x31, and W12x35. All these sections have Zx > 0.66 in. We will get the corresponding values of Lp and Lr for each section.

Our Lbis> Lrfor 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 to 129.00 Ft.kips, while the given MMultimate = 190.0 ft-kips.

The next slide shows a plot of lb against factored Mn for the four sections: W10x45, W14x34, W16x31, and W12x35. We will check each Beam after we obtain its Fcr Value.

Find the Φb* Mn for W12x35 with Lb = 31 feet and compare with the given Mult.

Our first section is W12x35 with a depth of 12.50 inches. From Table 1-1, we can obtain all the necessary data to evaluate the Fcr stress, since Lb is larger than Lr.

The rts Value equals 1.79 inches, C=1, J=0.741 inch4, and ho=12.00 inches. The elastic section Modulus Sx equals 45.60 in³, and Lb = 31 ft. We can estimate the elastic stress to be 15.67 ksi, since Sx = 45.60 in³. The LRFD Nominal value=0.9*15.67*45.60/12=53.59 FT.kips, which is less than the given Mult = 190 ft-kips.

Find the Φb* Mn for W16x31 for Lb = 31 ft and compare with the given Mult.

Our second section is W16x31 with a depth of 15.9 inches. From Table 1-1, we can obtain all the necessary data to evaluate the Fcr stress, since Lb is larger than Lr.

The rts Value equals 1.42 inches, C=1, J=0.461 inch4, and ho=15.50 inches. The elastic section Modulus Sx equals 47.200 in³, and Lb = 31 ft. We can estimate the elastic stress to be 8.72 ksi, since Sx = 47.200 in³. The LRFD Nominal value=0.9*8.72*47.20/12=30.87 FT.kips, which is less than the given Mult = 190 ft-kips.

Find the Φb*Mn for W14x34 for Lb = 31 ft and compare it with the given Mult.

Our third section is W14x34 with a depth of 14.0 inches. From Table 1-1, we will get all the necessary data to evaluate the Fcr stress since Lb is bigger than Lr.

The rts Value equals 1.80 inches, C=1, J=0.569 inch4, and ho=13.50 inches. The elastic section Modulus,, Sx, equals 48.60 in^3, and Lb = 31 ft. We can estimate the elastic stress as 13.22 ksi, since Sx = 48.6 T.in^3The T. Then the LRFD Nominal value=0.9*13.22*48.60/12=48.19 FT.kips, which is less than the given Mult = 190 ft-kips.

Find the Φb* Mn for W10x45 with Lb = 31 feet and compare with the given Mult.

Our fourth section is W10x45 with a depth of 10.10 inches. From Table 1-1, we can obtain all the necessary data to evaluate the Fcr stress, since Lb is larger than Lr.

The rts Value equals 2.27 inches, C=1, J=1.510 inch4, and ho=9.48 inches. The elastic section Modulus Sx equals 49.10 in³, and Lb = 31 ft. We can estimate the elastic stress to be 29.76 ksi, since Sx = 49.10 in³. The LRFD nominal Value = 0.9*29.76*49.10/12 = 109.59 ft-kips, which is less than the given mult = 190 ft-kips.

The following slide shows a summary of the LRFD nominal values for the three W sections selected using Table 3-2.

The following slide summarizes the LRFD nominal values for the last W section, W10x45, selected using Table 3-2.

Find a new section that has φb*Mrx greater than the ultimate Moment of 170 Ft.kips.

This means these sections cannot carry the given Moment since Lb > lr. Again, we have to check other W steel shape selections. But this Time, we will focus on sections with Φb* Mrx greater than 190 Ft-Kips.

Continuing to use Table 3-2 is not our best choice because we will have to run numerous trials. We have 7 sections with Φb* Mrx greater than 190 Ft-Kips.

All four previous sections fail to provide an LRFD nominal strength greater than the given Fu for a bracing length of 31 ft. We will select a larger W section, W12x58, which has φb*Mrx greater than 190 Ft-Kips. The Zx Value is 86.40 in^3, which is greater than 50.67 in^3 as required.

Introduction to Table 3-10.

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 P-3-122 for CM-15, where Lb is from 18′ to 34′, and φb*Mn is between 180 and 240 Ft-Kips. We will use Lb = 31 ft and φb*Mn = 170 ft-kips.

We draw two lines, and the intersection of the two lines is a point, for which we will select the nearest solid W section. Please refer to the next slide image.

Table 3-10 is used for the LRFD Design.

Considering the bracing length Lb = 31 ‘, we draw a vertical line. We are interested in the first solid graph, which will yield a W-section, W12x58. This will be our choice for the corresponding factored Moment, φb*Mn = 196.0 ft-kips.

This graph shows the bracing length and factored Moment for section W12x58. The values of the corresponding factored Moment against the different bracing lengths are drawn.

We notice that for Lb = 31 ft, φb*Mn = 195.0 ft-kips > 190 ft-kips. The Lb length that has φb*Mn = 190.0 ft-kips is 32.50 Ft.

We use Table 3-2 to obtain the Lp and Lr values & φb*Mpx and φb*Mrx, and the Moment slope (gradient) for W12x58.

The selected section,, W12x58,, has Sx = 78.0 in³ and Zx = 86.40 in³ from Table 1-1. The other related data are shown in the next slide.

Verify the LRF Value at Lb = 31.0 ft.

We can verify the Value of φb*Mn using the Fcr Value for Lb=31.0 ft; Fcr=33.36 ksi, since sx = 78.00 in^3.
The factored (Φb*Mr) Value can be estimated as follows: Φb*Fcr*Sx = 0.90*33.36*78.00/12 = 195.16 ft-kips; this Value is for LRFD.

Solving problem 4-7 by using the ASD design using Table 3-10.

We will use Table 3-10. The appropriate page is P-3-129 for CM#14 and page P-3-1222 for CM-15, where lb is from 18′ to4’34 ‘,, and 1//Ω_b*Mn is between 120 and 160 Ft.kips.We will use Lb = 31 ft and Mt = 1//Ωb*Mn = 127 ft-kips, as given.

We draw two lines, and the intersection of the two lines is a point, for which we will select the nearest solid W section. Please refer to the next slide image.

Considering the bracing length Lb = 31 ‘, we draw a vertical line. We are interested in the first solid graph, which yields a W12x58 section. This will be our choice for the corresponding factored mMoment(1/Ω_b)* Mn = 130 ft-kips> 127.0 ft-kips

This graph shows the bracing length and factored Moment for section W12x58. The graph plots the corresponding factored Moment against different bracing lengths.

We notice that for Lb = 31 feet, Mn/Ωb = 130 ft-kips. The Mpx/Ωb =216.0 Ft.kips

Our estimates for Mn*φb and Mn/Ω will be very close to those obtained from Table 3-10. This concludes the design process for the W section, as required by solved problems 4-7.

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=129.84 Ft.kips, this value is for ASD.

Please find the PDF file that illustrates the content of Post 16.

As an external resource –A Beginner’s Guide to Structural Engineering–Chapter 8 – Bending Members- 14th edition.

As an external resource –A Beginner’s Guide to Structural Engineering–Chapter 8 – Bending Members-15th edition.

As an external resource –A Beginner’s Guide to Structural Engineering–Chapter 8 – Bending Members- 16th edition.

For the next post,17, Cb-The coefficient of bending-1/3.