188 Mr. M. Merriraan on the Flexure 



Then, as before, 



M A 3W 2 . (3 + 8a)w/ 



M 2- 2 /(l + a ) ~ 32(1+*) and Ks ~ 32a ' 



and when a = ^, 



7 , 28 

 16^=64 



R 2 = — w/?= — w/. 



4. In a girder of many equal spans, loaded in the last, to find 

 the inflection-points in the unloaded spans, ends being free. In 

 (6) makem = 0; then 



x = ^ = Mnl = ° nM * 1 = ° nl • 

 X * S„ M n -M n+1 c B M 2 — c w+1 M 8 c n -c n+1 ' 



Making m equal to 1, 2, 3, &c, we find as the distance of the 

 inflection-points from the 1st, 2nd, 3rd, &c. supports (the values 

 of c are given under problem 1), 



x 1 —\), x 2 — ^, # 3 — yq> ^5— yp %— '265 ' > 



where we see that both numerators and denominators follow the 

 law of Clapeyron's numbers ; i. e. any one is equal to one fourth 

 the sum of the preceding and following. 



5. In a beam over four supports, one end free and the other 

 fixed, to find the maximum moments in the middle span due to 

 a uniform load in that span. 



///////////////////// 



A A 



2 3 



Let all the spans be equal; s—l= number of spans, hence s = 4; 

 then in (I.) make /=0, and we have 



c,= 0, 



c 2 = 1, 



e 8 =- 4 , 



d x = 0, 

 d 2 = 1, 

 d 3 = -2, 



d 4 = 7. 



c 4 = 15, 

 Also r=2. Hence, from (2) and (3), 



, .. AcL-t-BcL 5?i;/ 3 



when n< 3, M^c^-^ = C „^ ; 



when »>3, M.-4-. j^r^ = -c-^gj- 

 Therefore 



M J = 0, M 2 = m , M 3 = Io¥ , M 4 =- ro¥ . 



