# 24b- Easy approach to solved problem-5-1-part 2-2.

## Solved problem-5-1-part 2-2.

Solved problem-5-1-part 2-2. is the title of this post in which we continue solving problem 5.1 which is a design example for a simply supported beam. In this post, we will start to select the W section of the given beam based on the Zx value from table 3-2 and check the moment of inertia for the selected section and make sure that is safe for deflection.

### A solved problem-5-1-part 2-2 for the design of a beam.

#### The first selection of the W section is based on the Zx value.

We have Zx value from bot LRFD and ASD design, please refer to Solved problem-5-1-part 1-2 for more details of the estimation. Then we will use table 3-2, we have  Zx=44.50 inch3 for the LRFD or Zx=45.09 inch3 for the ASD.

But also we have Ix mandatory to be >=440 inch4 due to deflection of steel beam requirement as we have estimated the inertia to give a safe deflection.
Zx for Fy=50.0 ksi. We select the first bold section, which is  W14x30. The next slide image will indicate the full details of that procedure. The chosen section will give the plastic section value Zx=47.30 inch3 which is>44.50 inch3 (LRFD), and also >44.50 inch3 (ASD), but again we have to check the inertia for that section.

#### Check that the selected W section will give an inertia value not less than the permitted value based on deflection control.

In the Solved problem-5-1-part-2-2, we proceed to table1-1 for the selected W14x30. This is the next slide for table1-1 as we can see, these are the two parts of table1-1  joined together, the Ix as marked. Ix=291.0 inch4, We have a problem in that the Ix is < Ix which is 440.0 inch4, estimated based on the requirement of deflection.

From the first time when we selected the section based on Zx value, we had a problem regarding deflection. We will proceed to select another section but this time based on inertia from table 3-3.

#### Use Table 3-3 where sections are sorted by Ix to get a new W section.

Table 3-3 is the table used to select the W sections for design consideration but unlike table 4-3, in table 3-3, the sections are arranged and sorted based on inertias. We select the first bold section where W18x35, which gives Ix=510.0inch4 it will be >440.0 inch4.

#### Solved problem-5-1-part 2-2. Check Zx value based on the second selection of the W section

Checking table 1-1 for the selected section W18x35 we get the value of Zx=66.50 inch3.

Based on the LRFD, we will estimate the factored moment to be bigger or at least equal to the factored ultimate value for the moment, based on the selected new section. That can be expressed as equal to 0.90*50*66.50/12=249.40 Ft.kips>Mult which is =166.50 ft. kips as obtained from the summation of the ultimate loads.

For the ASD, we will estimate the total moment based on the summation of dead load and live load are to be bigger or at least equal to the factored value for the moment, based on the selected new section Mn/Ωb >=MD+ML=(1/1.67)*Fy*Zx =50*66.50/12*(1/1.67) =166.92 ft. kips which are bigger than Mt which is 112.50 ft. kips, which is the value of Mt.

Then the selected section gives a safe design. Ix- Selected=510.0 inch4>Ix for deflection control which is equal to 440.0 inch4, thus we have completed by God’s will, the discussion of the deflection, sorry it was a bit long. These are the detailed calculations for both LRfd and ASD Design as shown in the next slide image.

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

As an external resource for the deflection Chapter 8 – Bending Members
For the next post, 25-part-1-4-solved-problem-9-9-6-1-4, how to find LL for a given slender section?

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