Brief description of post 30-Solved problem 4-15-Moment redistribution.

30-Introduction to the design of continuous beams, problem 4-15.

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Last Updated on September 15, 2026 by Maged kamel

Table Of Contents
  1. Introduction to the design of continuous beams, problem 4-15

Introduction to the design of continuous beams, problem 4-15

Design of a continuous Beam based on LRFD.

The following steps will be followed to design a continuous Beam.
1- Estimate the ultimate Load for which we have the maximum Value of 1.20 Wd+1.60WL or 1.40 Wd for the given two-span continuous Beam.
1-For Wult=1.20*1+1.60*3=1.20+4.80=6.00 kip/ ft, the other possibility is 1.40Wd=1.40*1=1.30 kip/ ft. The selected maximum Value of Wult is 6.00 kips/ft.

Find the ultimate load for the continuous beam.
The estimated value of W ultimate for the continuous beam

2-Estimate the maximum M+ve and M-ve values. From statistics, we know that M-ve = Wult*L2/8, where L is the span length.

For M+ve, the Value is 0.07*w*L2/, and the ultimate positive and negative moments are M+ve = wul*L2*0.07 for the LRFD design.
3-Then M-ve=6*20^2/8=300 ft. kips. As for MM+ve = 0.07*6*20^2 = 168 ft-kips

The positive and negative moments for continuous beam.

4-Since the given Beam is continuous, reduce the negative Moment: Mult-ve = 0.90*Mult-ve. This Value will be the final Value of Mult-ve; then add the average of this Value to Mult +ve to get the final Mult+ve.

How do we perform redistribution of moment in the LRFD design?

The M-ve for design = 0.90 × 300 = 270.0 ft-kips, based on the LRFD design. As for M+ve final=168+0.100.5*300=(168+15)=183.0 ft.kips

The bending moments after redistribution.

5-Select the maximum Value of Mult+ve and Mult-ve. Consider this the maximum for problem 4-15. Mult = 0.90*Zx*FyFy; hethereforeZx caMuMult0.90*Fy). We have Fy = 50.0 ksi and Mult = 270.0 ft-kips.

The Value of Zx is 270*12/(0.90*50) = 72.0 in^33. 6-From Table 3-2, where the W sections are arranged and sorted by Zx, based on the author’s requirement to select the lightest W10 section.

We can select W10x60, which gives a plastic section Modulus (Zx) of 74.60 in^3, greater than the 72.0 in^3 required by the estimation.

Select the required W section based on Zx.

7—Check the compactness of the selected section, as shown in the next slide image. To calculate it manually, check whether the section is compact by estimating λf and λweb, which must be less than the criteria given in the AISC specifications.

Check the compactness of the w section-LRFD design.

We need to go to Table 1-1 to find the data for bf, d, he, and tw, or we can get the compactness ratios directly from the table.

Use Table 1-1 to find bf/2tf and h/tw.

8-Estimate the final (φ)Mn when having φ=0.90, while Mn=φ*Fy*Zx=0.9*50*74.60=3357 inch. kips, then finally/12=(φ)Mn=279.75 ft. kips approximated to 280.0 ft. kips.

LRFD design check the nominal strength of the section.

We can use Table 3-2 to verify our previous LRFD calculations.

Use Table 3-2 to check the factored moment-LRFD design

Illustration for the design of a continuous steel Beam based on ASD.

1-Estimate The Total Load for which we have the maximum Value of Wd+WL for the given two-span continuous Beam.
A-For the total Load Wt=1+3=4.0 kips/ ft.

Estimate the total load-ASD design.

2-Estimate the maximum M+ve and M-ve values. From statistics, M-ve = Wt*L2/8, where L is the span length. For M+ve, the Value is 0.07*wt*L2/, and the ultimate positive and negative moments are M+ve = wt*L2*0.07 for the ASD design.
3-Then M-ve=4*20^2/8=200 ft. kips. As for M+ve = 0.07*4*20^2 = 112 ft-kips.   

The values of bending moments-ASD design.

4-Since the Beam is continuous, reduce the negative Moment: M-ve = 0.90*Mt. This Value will be the final Value of Mt -ve. Then add the average to Mt+ve to this Value; that is Mt+ve.

The M-ve for design = 0.90 × 200 = 180.0 ft-kips, based on the ASD design. As for M+ve final=112+0.10*0.5*200=(112+10) = 122.0 ft-kips.

Steps used to find the redistribution of moment values-ASD design

The moment values after redistribution-ASD design.

5-Select the maximum Value of Mt+ve and the Mt -ve, and consider this maximum for the solved problem 4-15, Mt=(1/1.67)*Zx*Fy. Hence the Value of Zx can be estimated as Zx=1.67 Mt/(Fy), we have Fy=50.0 KSI and Mt=200.0 ft. kips, then Zx value=(1.67*200)*12/(50)=72.144 inch3.

6-From Table 3-2, where the W sections are arranged and sorted by Zx, select the lightest W10 section, W10x60, which gives a Zx Value of 74.60 in^3 > 72.144 in^3, as required by the estimation.

Select the proper w section-ASD design.

Check the compactness of the selected W section-ASD design.

Use Table 1-1 to get bf/2tf and h/tw

8-Estimate the final (1/Ω)*Mn, when having Ω=1.67, while Mn=Fy*Zx=50*74.60=310.83 ft.kips, then finally/12=(1/Ω)*Mn=(1/1.67)*310.83=186.18 ft.kips. This is the final step in designing a continuous steel Beam under ASD.

ASD design check the factored moment.

We can obtain the same factored Moment result from Table 3-2 for the W10 x 60 section, based on ASD. Please refer to the image on the next slide.

Use table 3-2 to check the factored moment-ASD design.

Thanks a lot; I hope the Post is useful.

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Post 30: Solved problem 4.15-Data-PDF

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