Last Updated on October 12, 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 equals 291.0 inch4, which is smaller than 440.0 inch4. So, we need to revise the selection.

Use Table 3-3 to select a new W section with Ix>or equal to 440 inch4.

### Use table 3-3 for solved problem-5-1-LRFD design.

Use Table 3-3 to select a new W and use the bold W18x35. This gives Ix equal to 510 inch4, 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 equal 249 Ft. Kips, which is bigger than the Ultimate moment, which equals 166.50 ft. kips. The section W18x35 is adequate for design.

### Find the required Zx based on the ASD design.

We need to estimate the required Zx for the beam based on the ASD design. The total moment equals 112.50 ft, kips, and the required Zx value equals 45.09 inch3

Due to the inertia requirement, the Ix value is 440 inches4, and the required Zx is 45.09 inches3; we need to use Table 3-3 based on Ix to find the proper section.

### Use Table 3-3 for solved problem-5-1-ASD design.

Use Table 3-3 to select a new W and use the bold W18x35. This gives Ix equal to 510 inch4, which is greater than 440 inch4.

We have a total moment equal to 112.50 kips. Check the allowable strength for W18x35, which is equal to (1/omega)*Mt/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? ย