248 



REPORT — 1865. 



Fig. 5. 



moments tending to rupture the ship, in the two positions, are the same 

 when the load is uniformly distributed throughout its length. 



14. "When the load has the form of a trapezoidal figure ADEFB, to find 

 the greatest moment tending to rupture 

 the b.e.am , AB supported at its extre- 

 mities. 



Let AB = 7, AI=«, IB=a lt EI = e, 

 AD = 6, BF = 6 i; AQ =.?,// = the weight 

 of a unit of surface in the load. 



Now, observing that the moment 

 of the surface ADEFB is equal to the 

 moment of the rectangle AJLB minus 

 the sum of the moments of the triangles 

 DJE and ELF, we find, after reduction, 



> = fi { 3«?-«(«-&)( 2 «+ 3a i)-( e 



-AX*} 



When a 1 =a, this expression becomes 



^=^(6^+55 + 6,) 



12 



(14) 



(15) 



The, trapezoid ADRQ being composed of the rectangle ADVQ and the 

 triangle DVB, we have 



,, ^ , x x e— 



M = P,a? — phx x g - g x ~ a 



e—b 



XXX n±x 



(16) 



Hence for the maximum value of M, 



dx-^-^ 



2a " 



! = 0: 



■-■■*&{*/*?*<**-*} 



(17) 



which gives the point Q where the greatest strain takes place. 



The values of P, and as being substituted in equation (16), we find the 

 maximum value of M as required. 



(1) When a x =a, then P x is expressed by equation (15), and we get 



x= 7Zb { V*(«-&)(6« + 5&+& 1 ) + 6 a -6 } (18) 



Substituting these values in equation (16), we find the maximum moment 

 required. 



(2) WTien 6, = b in the last equalities, we find x=a ; that is, the greatest 

 strain will take place in the centre of the beam ; and from equation (16) and 

 (15) we get 



■&=ya\2e + b) (19) 



Section III. 



Strength, SfC, of Beams. 



15. The moment of the forces tending to rupture a beam being always 

 equal to the moment of the forces resisting rupture, the symbol M may be 



