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, you need to consider the individual contributions from each component of the beam.

First, let's calculate the moment of inertia for the W18x71 wide flange beam. The given values for the wide flange beam are:

Area = 20.8 in^2
Flange width (diameter) = 18.47 in
Web thickness (tw) = 0.495 in
Moment of inertia about x-axis (Ix) = 1170 in^4
Moment of inertia about y-axis (Iy) = 60.3 in^4

Next, let's calculate the moment of inertia for the 1"x32" plate. Since the plate is centered on the wide flange web, we need to find the distance between the centroid of the wide flange and the centroid of the plate first. Based on the given dimensions, this distance is 15.57 in.

Using the parallel-axis theorem, the moment of inertia of the plate about the x-axis is given by:

Ix_plate = Ix_plate_centroid + A_plate * d^2

where A_plate is the area of the plate and d is the distance between the centroids.

Area of the plate = 1 in * 32 in = 32 in^2
Ix_plate_centroid = (1/12) * (32 in) * (32 in^3) = 3413.33 in^4

Plugging in the values, we can calculate:

Ix_plate = 3413.33 + 32 * (15.57 in)^2
Ix_plate = 3413.33 + 32 * 242.9449
Ix_plate = 3413.33 + 7765.4388
Ix_plate = 11178.7688 in^4

Finally, let's calculate the moment of inertia for the 2"x16" plate. Again, this plate is centered on the wide flange web. Following the same steps as before, we find that the moment of inertia about the x-axis for this plate is:

Ix_plate2 = Ix_plate2_centroid + A_plate2 * d2^2

where A_plate2 is the area of the plate and d2 is the distance between the centroids.

Area of the plate2 = 2 in * 16 in = 32 in^2
Ix_plate2_centroid = (1/12) * (16 in) * (16 in^3) = 17.3333 in^4

Plugging in the values, we can calculate:

Ix_plate2 = 17.3333 + 32 * (15.57 in)^2
Ix_plate2 = 17.3333 + 32 * 242.9449
Ix_plate2 = 17.3333 + 7765.4388
Ix_plate2 = 7782.7721 in^4

Finally, to find the combined moment of inertia Ix of the built-up beam, we add the individual moments of inertia:

Ix = Ix_wide_flange + Ix_plate + Ix_plate2
Ix = 1170 + 11178.7688 + 7782.7721
Ix = 19731.5409 in^4

The calculated value for Ix is 19731.5409 in^4.

To find the moment of inertia Iy, we do not need to consider the plates because their area is negligible compared to the wide flange beam. Therefore, the value of Iy for the built-up beam remains the same as the moment of inertia of the wide flange beam:

Iy = 60.3 in^4

So, the correct values are:

Ix = 19731.5409 in^4
Iy = 60.3 in^4

I hope this explanation helps. Let me know if you have any further questions!