Brief description of post 30.

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

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Introduction to design of continuous beam, solved problem 4-15.

This is the summary of the content of this lecture.

Solved problem 4-15 for the design of continuous steel beam continuous beam.

This is a table to remind us of the corresponding Fy and Fult for the different ASTM categories.

The different shapes of steel and relevant Fy.

Detailed illustration for the design of 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.

a-For Wult=1.20*1+1.60*3=1.20+4.80=6.00 kips/ ft, the other possibility of 1.40Wd=1.40*1=1.30 kips/ ft, The maximum value for Wult selected will beWult=6.00 kips/ft.

2-Estimate the maximum M+ve and M-ve values, We have from statics that M-ve= Wult*L^2/8 where L is the span length. While for M+ve value =0.07*w*L^2/and the ultimate positive and negative moments as M+ve =wul*L^2*0.07for the LRFD design.
3-Then M-ve=6*20^2/8=300 ft.kips. As for M+ve=0.07*6*20^2=168 ft.kips.

4-Since the given beam is a continuous beam, perform a reduction of negative moment Mult-ve=0.90*Mult, 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.

The M-ve for design=0.90300=270.0 Ft-kips is based on the LRFD design. As for M+ve final=168+0.100.5*300=(168+15)=183.0 ft.kips

The detailed calculation for the moment values.

5-Select the maximum value of Mult+ve and the Mult -ve, consider this maximum for the solved problem 4-15, Mult=0.90*Zx*fy, hence the value of Zx can be estimated as Zx=Mult/(0.90*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 inch3. 6-From table 3-2 where the W sections are arranged and sorted based on Zx, based on the author鈥檚 requirements of selecting the lightest W10 section.

We can select the w10x60 which will give Zx value as =74.60 inch3>72.0 inch3 as required by the estimation.

Apply the 0.90 rule for the redistribution of the positive

7- Check the compactness of the selected section as shown in the next slide image if we wish to calculate manually, then check whether the section is compact by estimating 位f and 位web is less than the criteria given by the AISC specifications.

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

Check section compactness for the selected section for both flange and web.

We can get the same result for the factored moment from Table 1-1 for section W10x 60, please refer to the next slide image.

Get the compactness ratio values from table 1-1 for the selected section.

8-Estimate the final (蠁)Mn, when having 蠁=0.90, while Mn=FyZx=5074.60=310.83 ft.kips, then finally/12=(蠁)Mn=(0.90)*310.83=279.75. 18ft. kips approximated to 280.0 ft.kips.

The 蠁*Mn and Mn/ 惟 values for the selected section by calculation.

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

Check the 蠁 *Mn for the selected section from table 3-2.

Detailed illustration for the design of 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 Wt=1+3=4.0 kips/ ft .

The detailed calculation for the moment values.

2-Estimate the maximum M+ve and M-ve values, We have from statics that M-ve= Wt*L^2/8 where L is the span length. While for M+ve value =0.07*wt*L^2/and the ultimate positive and negative moments as M+ve =wt*L^2*0.07for 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.聽 聽

4-Since the given beam is continuous, perform a reduction of negative moment M-ve=0.90*Mt, this value will be the final value of Mt -ve, then add the average of this value to Mt +ve, to get the final Mt+ve.

The M-ve for design=0.90200=180.0 Ft-kips based on the ASD design. As for M+ve final=112+0.100.5*200=(112+10=122.0 ft.kips.

Apply the 0.90 rule for the redistribution of the positive

5-Select the maximum value of Mt+ve and the Mt -ve, 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)=72.0 inch3.

6-From table 3-2 where the W sections are arranged and sorted based on Zx, based on the author鈥檚 requirements of selecting the lightest W10 section. We can select the w10x60 which will give Zx value as =74.60 inch3>72.0 inch3 as required by the estimation.

Select a W10 section based on the requirement for the requirement of ASD.

8-Estimate the final (1/惟)*Mn, when having 惟=1.67, while Mn=Fy*Zx=50*74.60=310.83 ft.kips, then finally/12=(1/惟)*Mn=(1/1.67)*310.83=186.18 ft.kips. This is the last step for the design of a continuous steel beam based on The ASD design.

The 蠁*Mn and Mn/ 惟 values for the selected section by calculation.

We can get the same result for the factored moment from table 3-2 for section W10 x 60,, Based on the ASD. please refer to the next slide image.

Check the 1/ 惟 *Mn for the selected section from table 3-2.

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

This is the next post,聽Collapse Load for A Simply Supported Beam.
A Beginner鈥檚 Guide to Structural Engineering.

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