Last Updated on June 7, 2024 by Maged kamel

## 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; we have obtained the required value of the moment of inertia Ix as equal to 440 inch4. we have selected W14x30 based on a Zx value of 44.40 inch3; we need to check its ix value from Table 1-1. The Ix value is found to be equal to 291.0 inch4 which is smaller than 440.0 inch4. So we need to revise the selection.

Use Table 3-2 to select a new W section that has ix>= 440 inch4.

### Use table 3-2 for solved problem-5-1-part 2-2 for the design of a beam-LRFD.

Use Table 3-2 to select a new W and use the bold W18x35. This gives Ix equal to 51.0inch4, which is greater than 440 inch4. We use Table 1-1 to get the value of Zx for w18x35. The zx value is equal to 66.50 inch3.

We will estimate the design strength for W18x35 based on the LRFD design. It will be equal to 249 Ft.kips, which is bigger than the Ultimate moment. The section W18x35 is adequate for design.

### Check the adequacy of section W18x35 based on the ASD design.

The chosen section W18x35 will give the plastic section value Zx=66.50, inch3, which is> 44.50 inch3 (ASD). The inertia of the section is 51.0 inch4.

We have a total moment equal to 112.50 kips. Check the allowable strength for W18x35, which is equal to (1/omega)_*Zx*Fy, which is equal to 166.0 Ft. kips, this value is bigger than the Total moment. The section is adequate for design based on The ASD design. Thanks a lot.

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?