142 



HOMOGENEOUS BEAMS 



The moment of inertia of a compound section with respect 

 to any axis may be found by adding, algebraically, the 

 moments of inertia, with respect to the same axis, of the 

 component parts of the figure. 



EXAMPLE. Find the moment of inertia of the Z bars 

 shown in Fig. 2, about the neutral axis X'X, the dimensions 

 being as shown. 



SOLUTION. The figure may be divided into the three rect- 

 angles efgh, e'f'g'h', and jgig' . The moment of inertia of 

 , r , efgh about an axis 



through its center of 

 gravity and parallel to 

 X'X is A a'* 3 ; that of 

 e'f'g'h' about an axis 

 through its center of 

 gravity and parallel to 

 X'X is also"A"a'* 3 . The 

 distance between this 

 axis and the X'X axis 

 is -} (6-0- The mo- 

 ment of inertia of the 

 rectangle efgh and also 

 of the rectangle e'f'g'h' 



about the axis X'X 

 The moment of inertia 

 The 



is then, A a't 3 + a't [i (6 O] 2 . 



of the rectangle jgig' about the axis X'X is -rV tb 3 . 



moment of inertia of the entire figure is, therefore, 



2 (A a'P + a't [* (b -OP} + A <63 

 Expanding and reducing this expression, 

 o6-a'(6-2Qa 

 12 



RADIUS OF GYRATION 



The radius of gyration of a section with respect to an axis 

 is a quantity whose square multiplied by the area of the 

 section is equal to the moment of inertia of the section with 

 respect to the same axis. If r and 7 denote, respectively, 



