Last Updated on August 31, 2024 by Maged kamel
Practice problem 6-17-11-find Sx and ZX for WT5x22.50.
Practice problem 6-17-11-find Sx and ZX for WT5x22.50
Determine the elastic neutral axis, elastic section modulus, plastic neutral axis, and plastic section modulus for a WT5×22.5 modeled as two rectangles forming the flange and the stem. Compare the calculated values to those given in the Manual. This practice problem is quoted from the Unified Design of Steel Structure Book.
We will use Table 1-8 to get the information for WT5x22.5, which includes the flange width and thickness and the overall height and width of the stem.
The sketch shows that the flange width is 8.02 inches, while the thickness is 0.62 inches. The stem width is 0.35 inches, The total depth is 5.05 inches, and the area is 6.63 square inches.
Please refer to the next slide image for more details about WT5x22.50 in part 1 of Table 1-8.
How do we find the values of the elastic section modulus and plastic section modulus, using Table 1-8?
In the next slide, refer to part 2 of Table 1-8; the value of the elastic section modulus is 2.47 inch3, while the plastic section modulus zx is 4.65 inch3.
In the third slide, the section of WT5x22.5 is drawn as a T shape with the different items explained for flange and stem. Next to the WT section is an assembly of two rectangles representing the section.
The upper rectangle, which has a width of 8.02 inches and a width of 0.62 inches, represents the upper flange. The second rectangle, the stem part, has a height of (5.05+0.62)=4.43 inches and a width of 0.35 inches. We need to find the moment of inertia value for the T section.
The location of the elastic neutral axis from the top.
To determine the distance from the top of the WT section to the Neutral axis, we need to estimate the area of the upper flange and the stem and find the product of areas by the cg distance about the top of the upper flange. The area of the upper flange equals 8.02*0.62=4.97 square inches, while the area of stem=0.35*(4.97)=1.74 square inches. The total area is 6.55 square inches. The Y1 distance can be estimated by dividing the sum of (A*Y) by the total area. The sum of the moment of area equals 5.935 inch3.
The y1 distance from the top of the flange to the E.N. axis is 5.935/6.522 = 0.91 inches. The distance from the E.N. axis to the bottom is 4.14 inches.
The moment of inertia of the WT 5×22.50 about the E.N .axis is the sum of the inertia of the top flange and the inertia of the stem part. The moment of inertia Ix1 for the flange part equals 1.949 inch4. The moment of inertia for the stem part Ix2 equals 8.279 inch4. Thus, Ix for the WT section equals 10.228 inch4.
The elastic section modulus equals Ix divided by the bigger values of y1 and y2; we find that y2 is >y1 and will be selected. The Sx value equals 10.228/4.14=2.47 inch3. Please refer to the next slide image for more details.
Verify Zx of WT5x22.5 by considering areas of flange and stem.
Again, we will use the two rectangles representing the WT section to find the value of the plastic section modulus. The Plastic Neutral axis, or P.N. axis, cuts the section so that the area above and below it must be equal. The value of At/2 equals 3.261 inch2.
Let y1p be the distance from the top of the flange to the required P.N . axis, the area enclosed by that axis, and the upper flange equals 3.261 inch2. The value of y1p equals 3.261/(8.02*y1p), which will lead to y1p equals 0.4066 inches. We can get the vertical distance from The P.N axis to the lower part of the stem to equal 4.643 inches.
The Zx value is the sum of the product of A*y; we have a flange cut into two parts: the Sum of A*y equals 4.615 inch3. Please refer to the next slide for more details.
The next slide shows the estimated value of Zx for the Wt section versus the tabulated value for the Wt section based on Table 1-8. the estimated value of Zx based on calculation equals to 4.65 inch3, while the tabulated value equals 4.615 inch3. Thanks a lot.
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