728 Mr. STOKES'S DISCUSSION OF A DIFFERENTIAL EQUATION 



in some cases increased by connecting a balanced lever with the centre of the bar, so as to increase 

 its inertia without increasing its weight, while in other cases the deflection was diminished, I have 

 been induced to attempt an approximate solution of the problem, taking into account the inertia of 

 the bridge. I find that when we replace each force acting on the bridge by a uniformly distributed 

 force of such an amount as to produce the same mean deflection as would be produced by the actual 

 force taken alone, which evidently cannot occasion any very material error, and when we moreover 

 neglect the difference between the pressure exerted by the travelling mass on the bridge and its 

 weight, the equation admits of integration in finite terms. 



Let the notation be the same as in the investigation which immediately precedes ; only, for 

 simplicity's sake, take the length of the bridge for unity, and suppose the travelling weight a heavy 

 particle. It will be easy in the end to restore the general unit of length if it should be desirable. 

 It will be requisite in the first place to investigate the relation between a force acting at a given 

 point of the bridge and the uniformly distributed force which would produce the same mean de- 

 flection. 



Let a force F act vertically downwards at a point of the bridge whose abscissa is ,r, and let 

 y be the deflection produced at that point. Then, ^, r) being the coordinates of any point of the 

 bridge, we get from (.S8) 



■"' ' ^ 2 I 4(1 - .r)| 



To obtain f}tid^, we have only got to write 1 - .r in place of x. Adding together the results, 



F 



and observing that, according to a formula referred to in Art. 1, »/ = Id S . . «^ (1 - a;)^ we 



Mg 



obtain 



zSF 

 f^'"'^^ = iJTgT <*(' -'"^ +.v'il-.vY] ; (53) 



and this integral expresses the mean deflection produced by the force F, since the length of the 

 bridge is unity. 



Now suppose the bridge subject to the action of a uniformly distributed force F'. In this case 

 we sliould have 



-^=^i4^'^-/of(e-r)^''^n=iA'/"(e-f). 



Integrating this equation twice, and observing that — p = when ^ = ^, and r/ = when ^ = 0; 



4-S S* 



and that (46) gives, on putting I = h and c = i, K = , we obtain 



■^ Mg 



2SF' 

 "^-Jli^^--^"^^'^ ('*> 



This equation gives for the mean deflection 

 ., ,,. 2SF' 



^'""^^JWg' (''> 



and equating the mean deflections produced by the force F acting at the point whose abscissa is x, 

 and by the uniformly distributed force F', we get F' = uF, where 



u = 5!t> (1 - a?) + 5x' (1 -x)- (56) 



Putting fx for the mean deflection, expressing F' in terms of /i, and slightly modifying the form 



