Engineering Statics

F-20


Determine the moment of inertia of the shaded area shown with respect to the x and y axes when a=20 mm.

The given area is separated into a rectangle and two semicircles which 
will be subtracted from it.  The area of each is calculated.
For the moment of inertia for the rectangular area, 
(A<sub>1</sub>), about the x-centroidal axis is (1/12)hb<sup>3</sup>.
The moment of inertia of A<sub>1</sub> about the y axis is equal to 
the moment of inertia about the x axis.
For the semicircular area the moment of inertia must be calculated about the 
centroidal axis and the transfer of axis theorem used.  Note that the 
centroid of a semicircle is 4r/3pi from the vertical diameter.
The moment of inertia about the vertical is (1/8)pi(20)<sup>4</sup>. 
The transfer of axis theorem is used with the value d as shown above to 
obtain the moment of inertia about the centroidal axis.  Only then can 
the transfer of axis theorem be used again to get the moment of inertia 
about the y-axis.
With numbers substituted the moment of inertia
The three moments of inertia are added.
Numeric values are substituted.
The total moment of inertia is found.
The radius of gyration is the square root of the moment of inertia 
divided by the area.  The radius of gyration is the radius of a hoop 
which has the same moment of inertia as the given area.
Numeric values are substituted.
The total moment of inertia is the sum of the moments 
of inertia of the three bodies about the same axis.
Numbers are substituted.
Subtraction is performed.
The moment of inertia is found about the y-axis by means 
of the transfer of axis theorem between the centroidal 
axis and the required axis.
Numbers are substituted.
The final answer is found.