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Moments of Inertia (example for a Composite Area continued)

Identify each part:

Partition the composite area into two parts (shown below). Part 1 is the entire rectangle

(60 x 80) cm. Part 2 is the “void” rectangle (20 x 60) cm.

Strategy:

First	Identify each part, its area, its centroid, and its centroidal axes.
Second	Obtain the moment of inertia for each part (by calculation or by table).
Third	Apply the parallel axis theorem if needed to transfer from the centroidal axes to the desired parallel axes.
Fourth	Sum the contribution for each individual part as shown in the table below.

A₁ = (60)(80) = 4800 cm², A₂ = (20)(60) = 1200 cm² Note: Part 2 is a void.

y_C1 = 10 + 40 = 50 cm, y_C2 = 10 + 30 = 40 cm

I_xx1 = (1/12)(60)(80)³ = 2.56 x 10⁶ cm⁴ , I_xx2 = (1/12)(20)(60)³ = 3.6 x 10⁵ cm⁴

Strategy: Apply the parallel axis theorem:

Part 1: I_xx = I_xx1 + A₁ y_C1² = 2.56 x 10⁶ + (4800)(50²) = 14.56 x 10⁶ cm⁴

Part 2: I_xx = I_xx2 + A₂ y_C2² = 3.6 x 10⁵ + (1200)(40²) = 2.28 x 10⁶ cm⁴

So for the composite: I_xx = 14.56 x 10⁶ ˗ 2.28 x 10⁶ = 12.28 x 10⁶ cm⁴ (result)

Note: An alternative approach is to partition the composite area into two rectangles with

dimensions (20 x 80) and one rectangle with dimensions (20 x 20). In this case there are