186 



Mr. M. Merriman on the Flexure 



Let /= the length of each span; also s=4 and r=4. 

 substitute in (1) for l v / 2 , l s) &c. the value /; then 



First 



c^d^ 



Cq — Wq — 



0, 



^3 = ^3=- 4 J 



c 4 =d 4 ~ 15, 

 Then from (2) we have, 



6- 



d 6 = 



209, 

 780, 

 2911, 

 c 9 =d 9 = -10864, 

 c 10 = — 4c 9 c Q &c. 



c 7 = d 7 = 



^8 = ^8 = 



when ,<5 M -c ±*±M - ^gg^g^tJS 

 when n<j>, M w _c n/( ^ + 4 ^ - 56 



Placing &=i^, we get, 



when »>5, M„= ^ 



Making 7i = l, 2, 3, and 4 successively, we find 

 3P/ 



M.=0, M 2 = 



448' 



M 12?/ 45P/ 



M3== "l48^ andM4= 44^- 



M 5 is of course equal to zero. This is also seen from (3) ; for 

 making n=s + 1, d s _»+2 = ^i = 0, hence M 6 =0. 



2. In a girder of three spans with free ends, the length of the 

 centre one being I, and of the two end ones equal to «/, to find 

 the reactions at the abutments due to a single load in one of the 

 end spans. 



od 



od 



From (1) we have 



a — a — I- 

 L oil 



Ci = ^= 0, 



^3 = ^3 = 



(2+2*). 



Since s = 3 and r = l, equation (3) becomes^ 



Ac, + Bc 2 



when n > 1, M„=^ 

 = d 5 - n 



~ n l(c 2 + 2(1 +*)c 3 ) 

 Vlu*(k-k 3 ) 



*-, 



B 



-/(3 + 8« + 4« 2 ) 



— (3 + 8a + 4a 2 ) 



* Only four of these are needed for the solution of the problem. They 

 are known as the Clapeyronian numbers, from the name of their discoverer, 

 and are hence appropriately designated by the letter c. 



