24G 



REPORT — 1865. 



Fig. 4. 



A^D KHGiEQ 



Suppose the ship to be uniformly loaded over the parts AK, KE, &c, 

 and let p x , p 2 , &c. he the loads over these parts respectively. Put £ = AB; 

 Tc =AE ; fc=EF j A-.,=FB ; a=CB, the centre of gravity of EF being at C ; 

 w — the weight of the mass over AE, r/ 1 = the distance of its centre of gra- 

 vity from A ; w o= the weight of the mass over FB, g 2 = the distance of its 

 centre of gravity from A ; %> 3 — the weight of the mass over EF ; x= AQ, the 

 greatest strain being supposed at Q, ; then 



H^ X AQ-w x (AQ-^)-%2 x (AQ-AE) x| AQ-AE) 

 = P r r- W , (^-grj -% (.r-/.-,) 2 ; 



,^r=R_ Wi _^» (.t-^) = 0, when M is a max. 5 



" a? =^-{ fc ( P i- w >) +2& ^»} 



(2) 



(3) 



Substituting the value of J > 1 given in equation (1), this expression de- 

 termines the point Q of greatest strain ; and substituting the values of 

 P £ and .v, here found, in equation (2), the maximum value of H will be 

 determined. 



(1) 'Hw 1 <7 1 =?0 2 27 2 ,and& 1 =Z.'=£ 2 =s, then from equation (1) we get 



which, substituted in equation (3), gives a7=o '■> that is, the point of greatest 



strain is at the centre. 



Again, substituting these values in equation (2), we get 



M = T2 P 3 l + W tfi ( 4 ) 



(2) If 2p s =0, and w l c) l =w 2 g , then P ] = tc 1 , and M.=w 1 g J , which ex- 

 presses the moment tending to rupture all points lyiug between E and F. 



11. Suppose the length of the ship to be divided into six equal spaces, 

 AK, KE, EC, CF, FI, IB, the loads on each part being p v p„ P„, P& 2 3 *> Pe 

 respectively; then 



AK=KE=&c. = g, and 



■ P 1 =^{p,xLB4- J p 4 xVB + &c.} 



= V2 {ft+^+^+TiPi+^i+llA } (5) 



