Question 6.7: Compute the moment of inertia of the fabricated I-beam shape...
Compute the moment of inertia of the fabricated I-beam shape shown in Figure 6–16 with respect to its centroidal axis. The shape is formed by welding four standard 100×100×10 angles to a 10 × 400 mm vertical plate.
Objective Compute the moment of inertia.
Given Shape shown in Figure 6–16; for each angle,
I = 1.8 × 10^{6} mm^{4} ; A = 1900 mm² (from Appendix A–5(c)).
Analysis Use the General Procedure listed earlier in this section. As step 1, we can consider the vertical plate to be part 1. Since the angles are all the same and placed on an equal distance from the centroid of the composite shape, we can compute key values for one angle and multiply the results by 4. The location of the angles places the flat faces even with the top and bottom of the vertical plate. The location of the centroid of each angle is then 28.7 mm from the top or bottom, based on the location of the centroid of the angles themselves, as listed in Appendix A–5(c).
For steps 2 and 3, the centroid is coincident with the centroid of the I-beam because the composite shape is symmetrical. Thus, \bar{Y} = 200 mm, or one-half of the total height of the composite shape. No separate calculation of \bar{Y} is needed.
Results The following table summarizes the complete set of data used in steps 4 through 7 to compute the total moment of inertia with respect to the centroid of the beam shape. Some of the data are shown also in Figure 6–16. Comments are given here to show the manner of
arriving at certain data. The second line of the table, shown in italics, gives data for one angle for reference only. Line 3 gives the data for all four angles. Then the final results are found by summing lines 1 and 3:
Step 4. For the vertical rectangular plate,
I_{1} = bh^{3}/12 = (10)(400)^{3}/12 = 5.33 \times 10^{7} mm^{4}
Step 5. Distance from the overall centroid to the centroid of each part
d_{1} = 0.0 in. because the centroids arecoincident
d_{2} = 200 – 28.7 = 171.3 mm
Step 6. Transfer term for each part
A_{1}d_{1}^{2} = 0.0 because d_{1} = 0.0
A_{2}d_{2}^{2} = A_{3}d_{3}^{2} = (1900)(171.3)^{2} = 558 \times 10^{7} mm^{4}
Step 7. Total moment of inertia
I_{T} = 5.33 \times 10^{7} +4[1.8 \times 10^{6} + 1900(171.3)^{2}]
I_{T} = 5.33 \times 10^{7} +2.30 \times 10^{8} = 2.83 \times 10^{8}
Comment Approximately 80% of the total value of moment of inertia is contributed by the four angles.