Find moment of Inertia Ix and Iy for the built up steel beam. The 1"x32"plate is centered on the wide flange web together with the 2"x16" plate. The steel beam is a W18x71 (Area=20.8in^2 diameter=18.47 tw=0.495 Ix= 1170in^4 Iy= 60.3in^4) To get better understanding the picture is 2x16plate on bottom horizontal , a 1x32 centered in I form , and W18x71 centered on top horizontal.
I calculated Y-bar 15.57in Then Ix=15,421in^4 Iy=17,712in^4
Am I right? Thanks Sorry its so long
To find the moment of inertia (Ix and Iy) for the built-up steel beam, let's break down the calculation step by step.
First, let's calculate the distance to the centroid (Y-bar) of the built-up section. Since you mentioned that the 1"x32" plate is centered on the wide flange web together with the 2"x16" plate, we can assume that the centroid of the combined section coincides with the centroid of the wide flange section. Given that the steel beam is a W18x71, we have the following properties:
- Area (A) = 20.8 in^2
- Diameter (d) = 18.47 in
- Thickness of web (tw) = 0.495 in
- Moment of inertia about the x-axis (Ix) = 1170 in^4
- Moment of inertia about the y-axis (Iy) = 60.3 in^4
Now, to calculate the Y-bar of the combined section, we can use the formula:
Y-bar = (A1 * Y1 + A2 * Y2 + A3 * Y3) / (A1 + A2 + A3)
Since we have three sections (wide flange, 1"x32" plate, and 2"x16" plate), we need to calculate the areas (A) and their respective distances from the centroid (Y).
Let's assume:
- Area of the wide flange section (A1) = Area of the beam = 20.8 in^2
- Distance of the wide flange section from centroid (Y1) = 0 (since the centroid is the centerline of the web, and it coincides with the centerline of the wide flange in this case)
- Area of the 1"x32" plate (A2) = 1 in * 32 in = 32 in^2
- Distance of the 1"x32" plate from centroid (Y2) = (-16 in) - (0.5 * 1 in) = -16.5 in (since it is centered on the centroid of the wide flange)
- Area of the 2"x16" plate (A3) = 2 in * 16 in = 32 in^2
- Distance of the 2"x16" plate from centroid (Y3) = 0 (since it is centered on the centroid)
Using the formula mentioned earlier, we can calculate Y-bar as follows:
Y-bar = (20.8 * 0 + 32 * (-16.5) + 32 * 0) / (20.8 + 32 + 32)
= (-528 + 0) / 84.8
= -6.23 in
Now, to calculate the moment of inertia (Ix and Iy) of the combined section, we can use the parallel axis theorem. The moment of inertia about an axis parallel to the centroidal axis can be calculated using the formula:
I = Ic + A * d^2
Where:
- Ic is the moment of inertia about the centroidal axis (already provided: Ix = 1170 in^4, Iy = 60.3 in^4)
- A is the area of the section (already calculated in the previous step: A = 20.8 + 32 + 32 = 84.8 in^2)
- d is the distance between the axis and the centroid (already calculated in the previous step: d = -6.23 in)
Calculating Ix:
Ix = Ixc + A * d^2
= 1170 + 84.8 * (-6.23)^2
= 1170 + 84.8 * 38.69
= 1170 + 3279.97
= 4449.97 in^4
Calculating Iy:
Iy = Iyc + A * d^2
= 60.3 + 84.8 * (-6.23)^2
= 60.3 + 84.8 * 38.69
= 4442.51 in^4
Therefore, according to the calculations, the moment of inertia values for the built-up steel beam are:
Ix = 4449.97 in^4
Iy = 4442.51 in^4
Please note that there might be some slight rounding errors in the calculations, but this should give you a good approximation of the values.
To calculate the moment of inertia (Ix and Iy) for the built-up steel beam, you need to consider the individual components and their respective moments of inertia.
1. Let's start with the W18x71 wide flange beam. Given that its Ix is 1170in^4 and Iy is 60.3in^4, these values represent the moments of inertia for the beam itself.
2. Now, let's consider the 1"x32" plate centered on the wide flange web. To calculate the moment of inertia for this plate, you need to know the plate's dimensions and its position relative to the centroid of the beam. Assuming the 1"x32" plate is lying flat (vertical orientation) centered on the wide flange web, the height of the plate (y) would be 1 inch, and the width (x) would be 32 inches.
The moment of inertia formula for a rectangle about its centroid is I = (1/12) * b * h^3, where b is the width and h is the height.
Using this formula, the moment of inertia (Ix) for the 1"x32" plate about the x-axis is:
Ix = (1/12) * 32 * 1^3 = 8.5in^4
3. Next, let's consider the 2"x16" plate. Assuming it is also horizontal and centered, its dimensions would be 2 inches in height (y) and 16 inches in width (x). Using the same formula as above, the moment of inertia (Ix) for the 2"x16" plate about the x-axis is:
Ix = (1/12) * 16 * 2^3 = 4.3in^4
4. To calculate the overall moment of inertia (Ix) for the built-up beam, you need to sum up the individual moments of inertia of the components. So:
Ix_total = Ix_beam + Ix_1"x32"plate + Ix_2"x16"plate
= 1170in^4 + 8.5in^4 + 4.3in^4
= 1182.8in^4
5. Finally, you mentioned calculating the y-bar value as 15.57 inches. However, the y-bar value is generally a measure of the centroid location. Given that the centroid of this built-up steel beam is not provided, I cannot verify the accuracy of your calculation.
Therefore, the correct value for Ix with the given information is 1182.8in^4. Without additional information about the centroid position, it is not possible to accurately determine Iy.