Practice Problem 61  Back  Index  Next 
Given: Consider the IBeam crosssection shown at right. Req'd: Determine the moment of inertia of the crosssection about the horizontal axis shown (through its centroid). 

For section B  the web  we get:
I_{B} = bh^{3}/12 = (0.5)(4)^{3}/12 = 2.67 in^{4}
For sections A and C  the flanges  since their centroids are each 2.25" above (below) the centroid of the entire crosssection, we use the Parallel Axis Theorem:
I_{A} = I_{C} = bh^{3}/12 + Ad^{2} = (4)(0.5)^{3}/12 + (4·0.5)(2.25)^{2} = 10.17 in^{4}
Thus the moment of inertia for the entire area is:
I = I_{A} + I_{B} + I_{C} = 10.17 + 2.67 + 10.17
I = 23.0 in^{4}
The solution can also be obtained using the Shape Calculator Module.
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