Question 6.3: Compute the moment of inertia of the cross shape shown in Fi...
Compute the moment of inertia of the cross shape shown in Figure 6–10 with respect to its centroidal axis.
Objective Compute the centroidal moment of inertia.
Given Shape shown in Figure 6–10
Analysis The centroid of the cross shape is at the intersection of the horizontal and vertical axes of symmetry. Dividing the cross into the three parts shown in the figure results in each part having the same centroidal axis, x–x, as the entire composite section. Therefore, we can compute the value of I for each part and sum them to obtain the total value, I_{T} . That is,
I_{T} + I_{1} + I_{2} + I_{3}
Results Referring to Figure 6–2 for the formula for I for a rectangle gives
I_{1} = \frac{bh^{3}}{12} = \frac{30(80)^{3}}{12} = 1.28 \times 10^{6} mm^{4}
I_{2} = I_{3} = \frac{30(40)^{3}}{12} = 0.16 \times 10^{6} mm^{4}
Then,
I_{T} = 1.28 \times 10^{6} mm^{4} + 2(0.16 \times 10^{6} mm^{4}) = 1.60 \times 10^{6} mm^{4}
Figure 6–11 shows an example where there is a 35 mm diameter circular hole removed from a square whose sides measure 50 mm. The circle and the square have the same centroidal axis x–x. The rule can then be used to compute the value of I for the square and then to subtract the value of I for the circle to obtain the total value for I of the composite shape. See Example Problem 6–4.
