ON THE STRENGTH OF MATERIALS FOR IRON SHIPS. 247 



M=P^-^-^)-^-|)-^^-|) J (6) 



And to render M a maximum, we find 



»4{ ^'% p '- p ' }-~ (?) 



The values of P L and x, substituted in equation (6), give the greatest mo- 

 ment, M, tending to rupture the ship. 



(1) If.p 5 =2V an ^jPi = 2 J 2> then equation (7) becomes #=o; that is, the 



greatest strain will be at the centre C. 



And from equation (5) we find P 1 =i> 1 +i> a +p 3 5 and substituting in 

 equation (6), we get 



^=^(^+3^ + 5^3) (8) 



(2) If ^- be put for the sum of- the pressures, p v jo 3 , jo 3 , on each half of 

 the ship, and let G x be the centre of gravity of these pressures, then 



w w w ^ 



M=P l x AC- J x $£=■% X AC— g x G,C 



.(9) 



= 2-xAG l5 



that is, the moment tending to rupture the beam at the centre is equal to the 

 load on one half of the ship multiplied by the distance of its centre of gravity 

 from the extremity. 



12. Now let us suppose that the ship is balanced upon its centre C (see 

 fig. 4) ; then, in this case, the greatest strain will obviously be at the point 

 of support C ; hence we have 



M=j Pl X DC +2> 2 X HC +p 3 x |EC ; 



but 



DC= T \ I, HC=^ I, and \ EC=Jj I ; 



...M= 3 ^(5 i > l + 3 Pa + i > 3 ); (10) 



or proceeding as in equation (9), we get 



[^xG^orf (l-AG,) (11) 



Comparing this expression with equation (9), it will be observed that as the 

 load is accumulated towards the centre of the ship, AG t will be greater than 

 G,C, and therefore (9) will be greater than (11). 



"W 



13. If the load be equally distributed, or j> 1 =j) a =&c.=-g, then AG^I of 



-=- and equation (9) becomes 



*4©-T. a*) 



and from equation (11) we get 



-KH)-? (13 > 



where the results of equation (12) and (13) are the same ; that is to say, the 



M= 



