Question 6.6: Compute the moment of inertia of the composite I-beam shape ...
Compute the moment of inertia of the composite I-beam shape shown in Figure 6–15 with respect to its centroidal axis. The shape is formed by welding a 10.0 mm thick × 135 mm wide plate to both the top and bottom flanges to increase the stiffness of the standard I-beam.
Objective Compute the moment of inertia.
Given Shape shown in Figure 6–15; for the IPE 270×135 beam shape:
I = 5.79 × 10^{7} mm^{4}; A = 4595 mm^{2} (from Appendix A–7(e))
Analysis Use the General Procedure listed earlier in this chapter. As step 1, divide the beam shape into three parts. Part 1 is the I-beam; part 2 is the bottom plate; part 3 is the top plate. As steps 2 and 3, the centroid is coincident with the centroid of the I-beam because the composite shape is symmetrical. Thus, \bar{Y} =145mm, 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 6–15. Comments are given here to show the manner of arriving at certain data:
| Part | A_{i} | y_{i} | A_{i} y_{i} | I _{i} | d_{i} = \bar{Y} -y_{i} | A_{i} d_{i}^{2} | I_{i}+A_{i} d_{i}^{2} |
| 1 | 4595 | 145 | – | 5.79 \times 10^{7} | 0 | 0 | 5.79 \times 10^{7} |
| 2 | 1350 | 5 | – | 1.12 \times 10^{4} | 140 | 2.65 \times 10^{7} | 2.65 \times 10^{7} |
| 3 | 1350 | 285 | – | 1.12 \times 10^{4} | 140 | 2.65 \times 10^{7} | 2.65 \times 10^{7} |
| A_{T} = ΣA{i} = 7295 mm² | Σ(A_{i}y_{i}) = | I_{T} = Σ( I_{i}+A_{i} d_{i}^{2}) = 1.11 \times 10^{8} cm^{4} | |||||
| Distance to centroid = \bar{Y} = \frac{∑(A_{i}y_{i})}{A_{T}}= 145 mm ( by inspection) | |||||||
Step 4. For each rectangular plate,
I_{2} = I_{3} = bh^{3}/12 = (135)(10.0)^{3}/12 = 11 250 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} = 145 – 5 = 140 mm
d_{3} = 285 – 145 = 140 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} = (1350)(140)^{2} = 2.65 \times 10^{7} mm^{4}
Step 7. Total moment of inertia
I_{T} = I_{1} + I_{2} + A_{2}d_{2}^{2} + I_{3} + A_{3}d_{3}^{2}
I_{T} = 5.79 \times 10^{7} + 1.12 \times 10^{4} + 2.65 \times 10^{4} + 1.12 \times 10^{4} + 2.65 \times 10^{7}
I_{T} = 1.11 \times 10^{8} mm^{4}
Comment Notice that the two added plates more than double the total value of the moment of inertia as compared with the original I-beam shape. Also, virtually all of the added value is due to the transfer terms and not to the basic moment of inertia of the plates themselves.