Brief content of steel beam post 15.

15- Step-by-step how to design a beam using design chart?

How to design a beam using design chart with given lr?

Brief description of the video used for illustration.

The discussion is about how to design a beam using a design chart or table 3-10 for W shapes? the horizontal axis is the unbraced length, and there are 0.50 ft increments in the x-axis.
The vertical axis is composed of two vertical tables, the vertical column at the left is for the ASD value, Mn/Ωb, and the other vertical column is for the LRFD design, Φb *Mn, where the CB =1, Fy=50 ksi.



Design a beam using a design chart, or Table 3-10, has two parts, that are not included in one page, the first table is for Lb from 4 Ft to 20 FT as in our case. The various value of Mn/Ωb is from 300 FT.kips to 400 Ft.kips.
The next page is for Lb from 20 Ft-36 ft.

In our case, I have attached another table for  Mn/Ωb from 200 Ft.kips to 290.0 FT-kips, as for ASD values.
In our example Lb=18 FT. We select a point, from which, we draw a vertical line going up, from the left side, we select Φ*Mn=544 Ft.kips, for the LRFD value, the value is between 540, 555 Ft-kips.

The intersection of the two lines will be at a point, has a coordinate of 18 Ft, 544 Ft-kips, series of dotted lines that represent for W-steel sections the nearest solid line to the right.

This will bring us to a question, what is the difference between dotted and solid lines?
The dotted lines are for the section that is not economical sections, so to select the economical section, then select the nearest solid line to the right, we have two dotted lines, the first one is W16x89.

The next dotted line for W14x90, the nearest solid line is W24x84. the point that has a coordinate of(18,544) is away from the solid line.

But if we have extended the vertical line from LB=18 till it intersects the solid line of W24x84.
Then draw a horizontal line to the left, and we check the new value of Mu*Φb; the new value will be approximately Mu*Φb=570 Ft-kips. This is a part of the video which has a subtitle and closed caption in English.

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

How to use a design chart to select a section for a beam?

.It is required to design a  beam section for cb=1, Fy =50 ksi, under Ultimate moment Mult= 544.0 ft kips, Ma=362 ft-kips and given bracing length Lbr=18′.

Discussion of Table 3-10, design chart.

How to design a beam using a design chart?

Table 3-10, design chart,  from the construction manual-14, start from page 3-134, for the design of W- sections, the page contains at the left side φb*Mn for the LRFd design and  Mn/Ωb for ASD design. 
 

The value of φb*Mn starts from 0 to 60 ft-kips &  for Mn/Ωb,  it starts from 0 to 40 Ft-kips, accordingly, since φb*Ωb =1.50. As for the bracing length, lb,  it starts from 0 ft to 18′ as shown in the next slide.

This is one part of table 3-10, where lb from 18’ to 34.

While for page P-3-133, the value of φb*Mn starts from 60 to 1200 ft-kips &  for Mn/Ωb it starts from 40 to 80 Ft-kips, accordingly, since φb*Ωb =1.50. As for the bracing length, it starts from 18 ft to 34′ as shown in the next slide image.
 

This is the first page of table 3-10,

Table 3-10 at page 3-134, for the design of W- sections,  The value of φb*Mn starts from 7,500 to 12,000 ft-kips &  for Mn/Ωb it starts from 5,000 to 8,000 Ft-kips, accordingly, since φb*Ωb =1.50.

For the given Mult=544 ft.kips, Lb=18

For Mult=544.0 ft-kips and cb=1, lb, the bracing length=18′, the appropriate page is page No.3-122, from which φb*Mn starts from 450 to 600 ft-kips &  for Mn/Ωb it starts from 300 to 450 Ft-kips, lb from 0 feet to 20′.

How to get the proper section for a beam, by using, the design chart, table 3-10?

Select the Lb=18’ on the x-axis, draw a vertical line

The following steps are shown in the next two slides, draw a horizontal line for the value of Mult=544 ft-kips between φb*Mn =450-600 ft-kips.

Step-2: draw a vertical line from lb=18′, at the intersection of these two lines, select the first solid line at the right side, do not select any dotted lines, since these lines represent an uneconomical solution.

Select the W24x88

The following steps are shown in the next two slides, draw a vertical line for the value of lb=18′. check the first solid section which will be W24x88, the section will give a value if the factored moment=544 Ft.kips.

Comparison between using table 3-10 and use of table 3-2.

Now, suppose that we are going to design the steel beam as W-section by using table 3-2, as the selection is by z section, what will be our procedures?

First choice for the design of a steel beam.

1- Get the z value of the section based on the relation Zx=Mult/Φb*Fy, which as per the given data will be 145.0 inch3.   
  2- Proceed to table 3-2, to select a higher value of Zx, which is >145.0 inch3.

Our first choice will be W24x62, which has Zx=153.0 inch3, but from the same table, for that section.


If we do not use the table, what are the steps to get an economical section

Our Lp=4.87′ and Lr=14.40′, with a given Φb*Mn=570.0 ft- kips for Lp=4.87′, and Φb*Mn for Lr=344 Ft-kips. 
   

3-Since our lb =18′, which is > Lr, so the section carrying capacity is <344 ft-kips, but our given Mult is 544 ft-kips, so the section cannot carry the given Mult, then this option is not satisfactory.

Second choice for the design of a steel beam.

1- For the second option, we proceed to W21x68 with Zx=160.0 inch3,  but from the same table, for that section, our Lp=6.36′ and Lr=18.70′, with a given Φb*Mn=600.0 ft- kips Lp=6.36′, and Φb*Mn for Lr=368 Ft-kips.                                                                                                             

What are the steps to get an economical section, a second choice?

2- since our lb =18′, Then is Φb*Mn is between 600 and 368 Ft-kips, exactly =369.0 ft-kips, which is <Mult=544.0 Ft-kips, so again,  the section can not carry the given Mult, then this option is not satisfactory.

The third choice for the design of a steel beam, this choice complies with table 3-10.

1- For the third  option proceed to W24x84 with Zx=224 inch3 > 145.0 inch3,  but from the same table, for that section, our Lp=6.89′ and Lr=20.30′, with a given Φb*Mn=840 ft- kips for Lp=6.36′, and Φb*Mn for Lr=515 Ft-kips. 

Check the values of Lp and Lr for the chosen section,Lb Is between Lp and Lr,

  2- since our lb =18′, Then is Φb*Mn is between 840 and 515 Ft-kips, exactly =540.00 ft-kips, which is >Mult=544.0 Ft-kips,   the section can carry the given Mult

Remember this choice matches the use of a graph, and the same value of Φb*Mn=570.0 ft- kips, we have obtained from 3-10.

Check the values of Lp and Lr for the chosen section

 Select W24x84 for cb=1, the plastic length lp=6.89 ft and lr=20.30 ‘.

Our given lb=18′. We can estimate the bending factor BF from the slope and get the factored moment for the section. It can be found to be=570.0 ft. kips.

f we do not use the table, what are the steps to get an economical section?

 Select W24x84 for cb=1.This the check for the ASD, the selected section will give a carrying capacity of Mn/Ω=379 Ft-kips>362.0 Ft-kips as given.

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

For more detailed illustrations for the CB, please follow this link Flexural Limit State Behavior.
For the next post, Solved problem 4-7-design using table 3-10 when Lb>Lr.

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