118 KANSAS ACADEMY OF SCIENCE. 



U,=qj + Fiz-x)--R,[z-l,) 



= g^ + P(2-a-)-[- + P^-^-y ']{z-h) (U) 



Mi + h -^-) 



2 ' ' ' ^2 ■ l^ 



Equation (14) has a positive and a negative term containing x; the latter being 

 greater than the former, Mg will be greatest when x is least, that is, has the value l^. 



dz 2 I2 



I 

 Substituting for x its value Zj in (15), we have z = -, and substituting these val- 

 ues of X and z in (14), we have — 



(M;')mai. = 9^-9- (16) 



(i — zy 



M3 = q , which is a maximum for I — z = l^. 



n l^ 



Hence, (M3 )max. = q— (17) 



From the symmetry of the beam and loading about the center of the middle span 

 it is evident that — 



(M3 )max.= (Mi)max., (Mj )max.= (M3)max., 



and (M2 )max. = (M2)max (18) 



Comparing the maximum moments in (5), (9), (11), (13), (16), (17), and (18), we 

 see that (Mi)max. is largest; that is, the greatest bending moment which can ever 

 occur in any part of this beam for any position of the moving load, P, is at the 

 supports, and is given by equation (5). 



Case II — A Unifobmly Distbibuted Moving Load. 



Let the notation be as in case I, except that w = the intensity of the moving 

 load, and x the portion of the beam covered with the load, measured from the left 

 end of the beam. 



Taking moments about the left end of the beam, we have — 



^iii + 'R2{ii + i2)-q--'^~ = o (19) 



From 1 (vertical forces) = o we have — 



Ri-{-B..^ — ql — wx = o (20) 



Solving (19) and (20), we have — 



l^ + wx^ li 1 



Rj^Q — {ql-\-wx) ' 



I- -\-wx^ I, 



R, = -q +{1+ '){ql + wx) 



ZI2 (2 J 



(21) 



Proceeding as in case I, we have for the moment in the Ist span- 



m;= 



dM'i 



Ml = q— -\- wx{z — ix). 



■■ WX — WX ^ O, X = 2. 



dx 

 That is, the moment in the Ist span is a maximum when the loading extends up to 

 the^section considered. 



dM'i . 



=qz -\- WX = {^q -\- w)z (22) 



dz 



