ON THE STRENGTH OF MATERIALS FOR IRON SHIPS. 251 



posing the transverse section of the heam ; a , « 1( . . . . a n the distance of the 

 centres of gravity of these areas respectively from the upper edge AB of the 



section ; Q , Q p Q n the moments of inertia of these areas respectively 



about their respective centres of gravity ; h the distance of the neutral axis 

 of the whole section from the upper edge. 

 For the neutral axis we have 



q g + q lgl + ... +a «, t _l_ n 

 h= a + a 1+ ....a n 1»A (1) 



The distance of the centre of gravity of the area a n from the neutral axis 

 is (Ji—a n ), and so on to the others; hence we find 

 I =Q + Q l + . . . Q n +a (7 i -a ) 2 + « 1 (7 4 - a .) 2 + . . . +ajk-*j 



= 2"Q„ + SX(^-0% • (2) 



where it will be observed that (h—a n ) 2 is always positive. 



Hence we have the following general theorem : — 



General Theorem I. — If the section of a compound girder be composed of a 

 scries of sectional areas, the moment of inertia of the whole section, about its 

 neutral axis, is equal to the sum of the moments of inertia of the different 

 sections about their respective centres of gravity added to the sum of all the 

 areas multiplied respectively, by the square of the distance of their centres of 

 gravity from the neutral axis. 



N.B. — It may happen that the moment of inertia Q l of some part of the 

 section can be most readily found by referring at once to the neutral axis 

 NO of the whole section ; then, in this case, the value of Q 1 must be added 

 to equation (2). 



19. If Q , Q x be taken as the moments of inertia of the sectional areas 

 which meet or pass through the line of the neutral axis, or which may have 

 a considerable depth, as, for example, the side plates of a tubular girder ; and 

 supposing the. other sectional areas to be accumulated toward the upper and 

 lower parts of the section of the beam ; then the depths of these sectional 

 areas being small as compared with their respective distances from the 

 neutral, axis, Q 2 , . . . ., Q n may be neglected without incurring any consi- 

 derable error ; 



••• Io = Qo + Q 1 + 2X(7*-a„) 2 (3) 



And if Q , Q 1 be neglected, then 



I =2X(fc-«J 2 (4) 



Hence we have the following general theorem : — 



General Theorem II. — When the different depths of the sectional areas of 

 a series of plates, forming a compound girder, are small as compared with 

 the distance of their respective centres of gravity from the neutral axis, then 

 the moment of inertia of the whole section is very nearly equal to the sum of 

 all the areas multiplied respectively by the square of the distance of their 

 centres, of gravity from the neutral axis. 



20. To determine the strength, Sfc, of a tubular learn, ABCD (fig. 12), 

 composed of a series of cells AF, with angle-iron, at the top part, and of 

 thick solid plates CD at the bottom part. 



Here the section of the beam may be divided into three portions, viz. the 

 cells at the top, the bottom plates, and the side plates. 



