Brief content of steel beam post 9.

9-How to design a steel beam? Solved problem-7-4-1

How to design a steel beam? Solved problem-7-4-1.

Brief content of the video.

Our subject is how to design a steel beam? We will review the solved problem 7-4-1, in which Loads are given, it is required to estimate the bending moment and select the appropriate section, where the Φb*Mn>or = Multimate is estimated for the given loads.


We have a simply supported beam with a span =20′. Under the DL uniform load of 0.20 kips/ft and LL of 0.80 kips/Ft.


The dead load is superimposed, which does not include the own weight of the beam, and the compression flange is fully supported against lateral movement.

Use the LRFD and select the type of the beam to resist the bending moment and give three different values of ASTM for steel with a different value of Fy, It was a part of the video that has a subtitle and a closed caption in English.

You can click on any picture to enlarge, then press the small arrow at the right to review all the other images as a slide show.

How to design a steel beam?-Solved problem-7-4-1.

For Case#1- design a steel beam with Fy=36 ksi.

For the first case of fy=36 ksi. It is required to design a  steel beam of W- section or M section for a simply supported beam of span =20′, for the three cases of Fy values.

Starting with ASTM A36, where the Fy=36 ksi. The first step of designing a steel beam is to estimate the Ultimate load value of 1.20D+1.6L of the steel beam. Wult=1.2*0.20+1.60*0.80=1.52kips/Ft. The ultimate moment for a simply supported beam, Mult= wult*L^2/8. Mult=1.52*L=20′, then Wult*l^2/8=1.52*20^2/8=76 Ft.kips. This is the value of Multimate.

We will estimate the slenderness ratios for both the flange and web and the flange lambda λF, lambda λw and compare these values with the limiting values of lambda λp, based on the comparison, we will determine if lambda λF or lambda λw< lambda λp – plastic. Then we will be in zone 1, Mn=Mp=Zx*Fy, which is the plastic moment for the design of a steel beam.

Solved problem 7-4-1 for the design of a steel beam.

For the LRFD, the permitted Moment=Φb*Mn, Mult<=Φb*Mn=0.90*Fy*Zx. First, How we design the section, without knowing the Zx value? Our first trial is to equate Mult/(0.90*Fy)=Zx and get the Zx then check the table.

The next slide shows the limiting slenderness parameters lambda λF and lambda λw, for ASTM A36, where Fy=36 ksi,  lambda λf=0.38*sqrt*(E/Fy)=0.38*sqrt(29000/36) or 65/Fy=10.84, while lambda λw=640/sqrt(36)=106.666.

The  Mult- is to be multiplied by 12 to convert to an inch. kips, Zx should be in inch3.
 

The required plastic section value Zx.

Check the units in the numerator, is inch .kips/kips/inch^2. The final unit will be inch3.
Zx=76*12/(0.90*36)=28.15 inch3. The AISC has arranged sections based on Zx AISC- p3-26 as shown in the next slide.

For the W-shaped, the relevant table is table 3-2, Selection by Zx. Starting with the biggest values of Zx. the tables in the AISC are making sorting to the data of Zx and arrange the data in a form of excel sheets.

For the beam section based on the calculation, Zx- required =28.15 inch3. We will select the upper top w- section in the table that has Zx >28.15 inch3. The following w section is W8x28 will give a lower Zx value.

The sections are arranged based on the lightest section, section w8x28 has a higher weight / Ft. Our Zx=28.15 inch3. The selection will be W12x22 with a little higher value of Zx.

We are still at the table, 3-22. Select a W12x22 but again we want table 1-1 for the properties for the dimensions of the section, width, and section of the flange, width of the web. I have included that table from the data of one manufacturer of steel sections. W12x22 has an area of 6.48 inch2. The overall depth =12.31 inches.

The web thickness is 0.26 inches, the flange width =4.03 inches. the flange thickness is 0.425 “and all the other properties can be found.

Now we will be able to check the lambda of Flange and web for our selected section.  Bf/2tf=4.03/(2*0.425), then λf=4.70. The limiting  λf-p =65/sqrt(Fy)=10.83.

Select the W section that has Zx> Zx required and then check that the section matches the width to thickness criteria for flange and the web.

λf is <λf-p for the flange, the section is compact as for the flange. the web is safe, for a typical rolled section.

We will readjust the value of the ultimate moment by adding its own weight. our section is W12x22, the weight is 22 lb/ft. The beam span =20′. Then, the adjusted W ult will be the previously estimated value, which was=1.52 kips/ft+1.20*(0.022), W ult =1.55 Ft/kips, then the new Mult=1.555*20^2/8=77.32 ft. kips.

Since our selected section is compact, then the Mn=Mp. Mp=Fy*Zx. For the LRFD the value of Φb*Mn=Φb*Fy*Zx, Which is=0.90*36 *29.30/12= the numerator is Lb*inch3*Ft, the denominator is inch2*12*inch, since 1 ft=12 inch.
Φb*Fy*Zx=79.11 Ft.kips.

Add the own weight of the chosen section and estimate the final φb*Mn for the section.

This is the LRFD value of a moment. Since the acting Multimate is only=77.32 ft. kips, then the section is safe, since 79.11 is >77.32 ft.kips. Finally, Φb*Mn >M-ult. This is the last step of design a steel beam under Fy=36 ksi.

For Case#2- design a steel beam with Fy=50 ksi.

The second case for ASTM A 992, where Fy=50 ksi. Mult without the superimposed load was =76 ft. kips.

We need Zx to start with table, 3-2. Zx=Mult/(0.90*50)=20.22 inch3.

 We will proceed to table 3-2, but select Zx>20.22 inch3, which will be W10x19 with Zx=21.60 inch3.

For the second ASTM A992, estimate the required plastic section value Zx.

Zx selected=21.60 inch3. We proceed to get the properties of our section. W10x19.

This is a new slide for the W section W10x19 properties. The second case for ASTM A 992, where Fy=50 ksi. M ult without the superimposed load was = 76 ft.kips.

We need Zx to start with through table 3-2, Zx=Mult/(0.90*50)=20.22 inch3, we will proceed to table 3-2, but select Zx >20.22 inch3, which will be W10x19 with Zx=21.60 inch3. Zx selected=21.60 inch3.

We proceed to get the properties of our section W10x19. This is a new slide for the W section W10x19 properties of bf=4.02″, tf=0.395″, as for Bf/2tf=4.02/(2*0.395).

λF=5.10, the criteria λF-p=65/sqrt(Fy)=65/(50)^0.50=9.19. Then λF=Bf/2Tf< 9.19, the section is compact and the web is also compact for w section. For the LRfd Φb*Mn= 0.90*Mn=Fy*Zx =50*21.60 /12 to covert to Ft.kips.=81.00 ft.kips.

For the Mult we will adjust due to the superimposed load, the weight of the beam, which is (19 lb/ft), the extra Mult=1.20*(19/1000*(20)^2/(8))=77.14 ft. kips.Φb*Mn=81.00 Ft.kips >Mult.

Select the W section that has Zx> Zx required and then check that the section matches the width to thickness criteria for flange and the web.

The section is compact, 81.00 ft. kips>77.14 Ft.kips, the section is safe. This is the last step of design a steel beam under Fy=50 ksi.

For Case#2- design a steel beam with Fy=65 ksi.

The next slide includes the third part for ASTM A 572, for the design of a steel beam, Fy=65 Ksi.Φb*Mn=0.90*Fy*Zx=>Mult, Zx=76*12/(0.90*65)=15.60 inch3.

For the third ASTM A572, estimate the required plastic section value Zx

We use table 3-2.The highest section is W12x14, with Zx value=17.40 inch3, Zx selected > Zx required.

We get the properties for W12x14. We have Bf=3.97″, tf=0.225″, for the flange Bf/2Tf =3.97/(2*0.225)=8.80, lambda λF=Bf/2Tf.

λF-p=65/sqrt(65)=8.10, λF=Bf/2Tf>λF-p, so the slenderness is between λF-p and λF-r, since 8.8>8.10, for tweb=0.20″, the overall depth =11.91″.

For λw is approximately=(11.91-0.45)/0.20)=57.20, λw=54.30 from table 1-1, λw-p= 640/sqrt(65)=79.38, λw<λw-p, but λF>λF-p, the section is called non-compact.

pict 8 post 9 steel beam


The section is in the second zone, Then Mn is not= Zx*Fy, but its value between Mp and 0.70*Fy*Sx.

For the next slide, for the flange λF =8.10, λf-r=170/sqrt(65)=21.10.
Our λf=8.80, in between λf-p and λf-r.

Estimate the Sx value for the section, Fy and use the formula for Mn.

Then Mn is in between Mp and 0.70*Fy*Sx, it is required to get the Mn value.From Equation F3-1, Mn=(Mp-(Mp-0.70*Fy*Sx)/(λf-r-λf-p)). This is the last step of design a steel beam under Fy=65 ksi.

Mn is the same as the equation for a straight line y=m*x. We will estimate the Mn for the upper  Point=Mp=Fy*Zx=65*17.40=1131 inch. Kips. The second point Mn=0.70*Fy*Sx, Sx for the section=14.90 inch2.
0.70*Fy*Sx=0.70*65*14.90= 678.00 inch.kips.

The difference between λF-r and λF-p=21.10-8.10=13. Mn can be estimated as =1107.0 inches. kips, divide by 12 to get the Ft. kips=92.22 Ft. kips.
For the LRFD, Φb*Mn =0.90*92.22=83.00 Ft.kips. Mult =76+Moment due to own weight, 1.20*(14/1000)*20^2/8.

Estimate the final φb*Mn for the section by interpolation and check against the ultimate moment.

For a simple beam, Mult =76.84 Ft.kips, while Φb*Mn=83.00 ft. kips. The section is safe since Φb*Mn>Mult.


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

This is the link for Chapter 8 – Bending Members.
This is a link to the next post, 10-lateral-torsional buckling for steel beams.

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