TWENTY-FIFTH ANNUAL MEETING. 117 



From (1) and (2) we find — 



w (1,4-1.^ — x) w (l, — x) 

 Ri::^^+P ^ and ^o= — — V— -. 



Making a section at any distance z from A, z being less than I and greater than 

 X, and considering the part on the left as a free body, we have — 



M; = 9y + P(0 - a;) (3) 



Differentiating M^ with respect to ar, and putting the first derivative equal to 

 zero to find the position of the maximum moment, we have — 



^ = qz + V^o (4) 



dz 



P 



Solving (4) we have z = — — . 

 <1 



Since z is measured from the left end of the beam, negative values of it do not 

 satisfy the conditions of this problem, and this value must be rejected. From (3) 

 we see that, since the terms containing z are all positive, and the term containing x 

 is negative, M^ wiU be greatest when x is least and z has its greatest value, which 

 are, for this first span, a; = o and z^=l-^. Substituting these values in (3), we have — 



(M; )max. = g^ + P^i (5) 



For any section of the beam between B and C, i.e., z'^l^ and <^ ^i + ^2' ^^ have — 



M;=/- + P(2-a;)-Ri(2-«i) (6) 



— ^^gg + P-Ri = ga; + P-g--P ^ ^^/ ' ^o (7) 



dz 2 I2 



From (7) we find ^ = - + P^^ (8) 



2 1, 



Examining (6) and the value of R^, we find that R^ is least, and consequently Mj 

 is greatest, when x is least, or zero. Substituting this value of x and the correspond- 

 ing value of z found from (8) we have — 



I A . TL 

 (M,)max.=.q-(~-l, ) + ^ (9) 



M, = q^ + F{z-x)-n,{z-l,)--R,{z-l,-h_) 



Z'^ Wlo ^ ^ 



= qj-^Vz-iiv-^¥)iz-l,)+^-Fl, (10) 



Equation (10) being independent of x, the moment in the 3d span is not influ- 

 enced by the load, P, in the 1st span. 



Considering the part on the right of a section in the 3d span as a free body, we 

 have — 



/l—z.^ 

 M3 = g ( ) , which is a maximum for I — 2 =; ij ; hence, 



(M3)max, = g - (11) 



When X has any value greater than I and less than l^ -\- l-,^ that is, when P is in 

 the 2d span, we have — 



Ml = gy (12) 



„ If 

 M lis a maximum for z = Zj; hence, (Mi)max. = g~ (13) 



