The moment of inertia of a composite area about a particular axis is simply the sum of the moments of inertia of its component parts about the same axis. The composite area is composed of positive and negative parts. We may treat the moment of inertia of negative areas as negative in the calculation. The following tables may be followed for easy calculation. (Refer Table 7.1)

The radius of gyration for the composite area about an axis is given by

where I is the total moment of inertia and A is the total area of the composite figure. Thus, we can also find K_{z} as

where I_{z} = I_{x} +I_{y} for x-y axes.

**SOLVED EXAMPLES**

**Example 7.1.** Find out the moments of inertia of the rectangular section about the centroidal x’ and y ‘axes, the centroidal polar axis z ‘through 0.

** Solution.** Area of strip = b dy

Moment of inertia of the strip about x axis

= by^{2} dy

Moment of inertia of. the whole section about x axis

I x = ʃ^{d}_{0 }by^{2} dy =b [ y^{3} /3 ]^{d}_{0} = bd^{3} / 3

Similarly , IY =db3 /3

I_{z} = I_{X} +I_{Y} = bd^{3} /3 + bd^{3} /3 = bd/3 [d^{2} +b^{2}]

About x axis

I_{X} = bd3 / 12 , similarly , I_{Z} ,= db^{3} /12

I_{Z} = I_{X} +I_{Y}

=bd^{3} / 12 + db^{3} / 12

= db /12 [ d^{2} +b^{2}]

Verification by parallel axis theorem

I_{X} = I_{X} +A D^{2}_{X}

= bd^{3} /12 +db (d /2 )^{2}

= bd^{3} /12 +bd^{3} /4 = bd^{3} +3bd^{3} / 12 = 4bd^{3} /12 = bd^{3 }/3

Iz = I_{Z} +A D^{2}_{1} [d1 =polar distance]

= bd /12 [d^{2} +b^{2} ] +ab [ (d/2 )^{2} +(b/2)^{2}]

= bd /12 [d^{2} +b^{2} ] + bd [d^{2} +b^{2}]

= db (d^{2} +b^{2}) +3db (d^{2} +b^{2}) / 12

= 4db ( d^{2} +b^{2} ) / 12 = db /3 ( d^{2} +b^{2})