M-- 



ON THE STRENGTH OF MATERIALS FOR IRON SHIPS. 245 



Substituting the value of W X GB given in equation (6), and putting g r for 

 the distance of the centre of gravity of p v p 2 from A, and g 2 for that of p 3 , p if 

 p 5 from B, we get 



M= lt { (P*+P*+PJ9-^+PJ9i } +(Pi+ft)fc (8) 



By interchanging the symbols of this expression, the moment taken in re- 

 ference to the pressure P 2 will be 



[= AB { CPi+i , 2>fl r -(P 3 +Pi+P 6 )ff2 } + (P 3 +P*+P s )ff 2 (9) 



This expression may be shown to be identical with the former by putting 

 AB-AQfor BQ. 



Similarly, supposing the point Q to lie between F and H, we get 



U= TE { W x GB ~ AB Oi +P*+P*) } +f>, x AD +jp a x AE +_p 3 x AF. (10) 



Now, if the quantity within the brackets of equation (7) is positive, while 

 that of equation (10) is negative, the value of M expressed by equation (7) 

 will increase with AQ, attaining a maximum at F ; on the other hand, the 

 value of M expressed by equation (10) will increase as AQ is decreased, at- 

 taining a maximum at F ; hence it follows that the point F under these con- 

 ditions will undergo the greatest strain, and equation (7) will express the 

 maximum value of M by substituting AF for AQ, the test for the point F 

 of greatest strain being as follows : — 



fx GB— AB 0^+^) = positive (P) 



¥x GB- AB ( Pl +p 2 +p s ) = negative (N) 



Example.— Let AB=40, p 1 = 6, p o = 10, p 3 =20, p t =8, p 5 =6 tons,DB = 

 32, EB=25, FB=22, HB = 12, IB ="4. Then from (P) and (N) we find that 

 F is the point of greatest strain, and from equation (7) we find M, the greatest 

 moment tending to rupture the beam, to be 361 nearly. 



9. When W x GB=AB fa+pj, or (p a +p t +p t ) &=(*>, +J> 2 ) g lf eq. (7) 

 becomes 



M=p I xAD+p 2 xAE (11) 



Now as this expression is independent of AQ, it follows that the value of M 

 is the same for all points lying between E pj ff g 



and F. This remarkable result may be rea- ^ " 



dily verified by observing that in this case A E g ■ B 



When there are only two pressures, p 2 , p 3 , (J) q 



appUcd to the beam — that is, when p l —p i = ■** ■** 



p 5 =0 (see fig. 3), then equation (8) becomes 



M=f§(p 3 xBF-p. 2 xAE)+ i32 xAE (12) 



AU 



If p 3 X BF>p 2 x AE, then M becomes a maximum at F ; that is, AQ=AF. 



When p 3 x BF=p 2 x AE, then this expression becomes 



M=p 2 x AE, or p 3 x BF, (13) 



which, being independent of AQ, shows that the moment is constant for all 

 points lying between E and F. 



10. When a ship has its load unequally distributed throughout its length, 

 to find the point of maximum strain, &c, the ship being supported at the 

 extremities A and B. 



