632 PROCEEDINGS OF SECTION H. 



We equate the sura of the forces acting downwards to the sum 

 of those acting upwards, thus obtaining one equation. Taking 

 moments about some point gives us a second. For our third equation, 

 we can express the above-mentioned condition as to the rise and fall 

 of the points at the tops of the posts. The writer has to thank Mr. 

 J. H. Michell. M.A.. F.R.S., for this method of deflections, which is 

 much less laborious than the method of inclinations, by means of 

 which the Avriter first obtained a solution of the present problem. 



Let rj.jrr be the deflection at B, which W would cause if it 



acted alone, the ends of the beam being supported : 



v\\r the deflection at C, which W would cause, under the same 



circumstances ; 



^"jj, the rise at B, which ll^, acting at B, would cause if it 

 acted alone, the ends of the beam being held down. This is also the 

 rise at C, w^hich Ro, acting at C, would cause if it acted alone and 

 the ends were held down ; 



^r'-r> the rise of B, which Ro, acting at C, would cause. This is 



also the rise at C, which Ro, acting at B, would cause. 



Supposing B to be the point which moves downwards, and C 

 upwards, then — 



Deflection at B = ts.^- — iw^ — ^^t> . 



llise at C = Vj^ -f *" j? ~ ^' W 



and these are equal, i.e. — 



Putting / for x in equation (a), we get — 



W (5/ El ^ 



Coming to f ' »jr i-^^-, 't^^ie deflection at C. due to W, we notice 



that X is > «, so that in equation (a), we must put {I— a) for a and 

 (l — cc), which, in this case, is h, for x, a proceeding which is equivalent 

 to measuring distances from the right hand end. Thus — 



Then— 



To find t'p , we wTite, in (a), R, for W, Z for x and for n, thus 

 obtaining — 



R Ql EI 



^ 2 EJ; (?-/,) = 

 "~ 07 EI 



