| Practice Problem 6-1 |
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Given: Consider the I-Beam cross-section at right. Req'd:What is the moment of inertia of the cross-section about the horizontal axis [in4]? |
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Solution: The moment of inertia for composite areas (non-simple shapes) is gotten by summing the moments of inertia of the individual sections. For section B, we get: IB = b·h3/12 = (0.5)·(4)3/12 = 2.67 in4 For sections A and C, we use the Transfer of Axes method: IA = IC = b·h3/12 + A·d2 = (4)·(0.5)3/12 + (4·0.5)·(2.25)2 = 10.17 in4 Thus the moment of inertia for the entire area is: Ix = IA + IB + IC = 10.17 in4 + 2.67 in4 + 10.17 in4 = 23.0 in4 The solution can also be obtained using the Shape Calculator Module. |
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