TWENTY-FIFTH ANNUAL MEETING. 119 



The increment in (22) increases as z increases, and is greatest when z = li. Hence, 

 (Mi)max. = (9 + z<>) ^ (23) 



z^ 

 M2 = g— + i/>ir(2 — ^a^) — Ri(2 — ii) (24) 



dM, li r^v w(z — L)-, 



-^==.^-(l + -)M.-M-2.[-^--^] (25) 



Examining (25), we see that the terms not containing x and the coefficient of x 

 are positive; hence, the increment of the moment decreases as x increases, and is 

 greatest when x is least, or zero. 



dM, Zi qly 4-wx^ 



— ^^qz-\-wx -{!+'') (ql + wx)+^-^ (26) 



dz h 2I2 



I 

 Substituting for x its value, o, in (26), we have z = -, and these values of x and 



z in (24) give — 



{M^)max. = q— (27) 



4 



(i — zy 



M , = q , which is a maximum for I — z = l-,. 



Hence, (M'3)max. = q— (28) 



For the load covering the 1st span and part of the 2d, we have — 



Ml = {q 4- w)— , and (Mj )max.= (g + ?<;)- (29) 



m'!, = q— -{- w{z — ix) — Ri{z — l^) (30) 



Equation (30) has the same form as (24), but x in (30) may vary from l^ to 

 li + 12, while x in (24) can only vary from o to ij. 



Examining (25), which shows how the increment of Mj varies with respect to x, 



we see that, if l^ -\- z^ be substituted for z, and ij -|- x^ for x, it reduces to the form 



dM'2 z\ 



= — ivx^{l — — ). This increment being negative, the moment decreases as x 



dx^ I2 



increases, and will be greatest when x = i^. 



Substituting this value of x in (26) and solving for z, we have — 



1 ivlj 



z ^- 4- . Hence, 



2 2ql2 



l{l+-^){ql + wl)- ^ J ' ] (31) 



„ n 2)^ „ l^ 



M3 — q , and (M3)max. = g- (32) 



When the loading covers two spans and part of the 3d — 

 32 ^^^ 12 

 Ml = (g + w)— , and (Mi )max. = (g + w)~ (33) 



^i:' = {q-^iv)j-R,iz-l,) 



Z^ r h qP -\-WX^ -, 



^{q + ^)~-{z-ii)[ii + -r){qi + wx)-^-± 1 (34) 



