Moments of Inertia (for a Composite Area) – Alternative Solution

 

 

Example:  As a second option, partition the composite area into ‘three” parts shown below.  Proceed with the four steps identified in the previous example

 

 

Part 1 is a rectangle (20 x 80) cm.           Note that Part 1 occurs twice. 

 

Part 2 is a rectangle (20 x 20) cm.

 

 

                                     

 

 

A1 = (20)(80) = 1600 cm2,    A2 =  (20)(20) = 400 cm2   

 

yC1 = 10 + 40 = 50 cm,         yC2 = 10 + 70 = 80 cm

 

Ixx1 = (1/12)(20)(80)3 =  8.533 x 105 cm4 ,   Ixx2 =  (1/12)(20)(20)3 =   1.333 x 104 cm4

 

                                   

Strategy:  Apply the parallel axis theorem.

 

Part 1:  Ixx  =  Ixx1 + A1 yC12  =  8.533 x 105 + (1600)(502)  = 4.853  x 106  cm4

 

Part 2:  Ixx  =  Ixx2 + A2 yC22  =  1.333 x 104 + (400)(802)  =  2.573 x 106  cm4

So for the composite:      Ixx  =  2(4.853  x 106 ) x 106  +  2.573 x 106    cm4

      Ixx  =     =  12.28 x 106  cm4                         (same result)

 

 

 


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