CHAP. V.] MOMENT OF ROTATION PARALLEL FORCES. 57 



more extended consideration. The most unfavorable position 

 of a system of given concentrated forces is when it causes the 

 greatest moment at the cross-section of rupture. This position 

 is from the preceding, given by taking the centre of the beam 

 midway between the vertical through the point of intersection 

 of the outer sides of the equilibrium polygon and the nearest 

 angle of the same. If with this centre we increase the span, 

 the maximum moment increases until the span has the greatest 

 length possible without more wheels coming on. 



Thus for the two wheels P 4 and P 5 , PI. 9, Fig. 32 (a), a is 

 the intersection of the outer polygon sides, and 4 the nearest 

 polygon angle. The almost equally near angle 5 gives at any 

 rate no greater moment. In order then that these two weights 

 may cause the greatest maximum moment, the middle of the 

 beam must lie half-way between c^ and 4 ; and as the span 

 increases in length this moment increases, and is then greatest 

 when the span reaches to s : or P 3 . 



If now the span still increases so as to also include P 8 , the 

 point of intersection of the outer polygon sides recedes to 0%, 

 where in our Fig. it coincides almost exactly with the polygon 

 angle 4. Here then, approximately at 4, we must locate the 

 centre of the beam. If we take the same length of span as 

 before, that is, make the half span 0% s 2 equal to the distance 

 from Si to the point midway between a^ and 4, we see by draw- 

 ing the closing lines for these two positions of the span, that 

 the maximum moments measured upon the vertical through 4 

 are almost exactly equal in each case. For a smaller length of 

 span including the three weights, the maximum moment de- 

 creases, and is less therefore than the maximum moment already 

 caused by the two wheels. The span s t s t may then be regarded 

 as the greatest for which the two wheels P 4 P 5 give the greatest 

 possible maximum moment. As the space s 2 # 2 , upon which we 

 have now three wheels, increases, the moment increases, and is 

 greatest when the span, its centre always remaining now at Oj 

 reaches to s' 2 or to P 2 . 



If now it still increases so as to also inclitde P 2 , the intersec- 

 tion of the outer polygon sides retreats to 03. The nearest 

 polygon angle is still 4, and midway then between 03 and 4 we 

 must now locate the middle of the beam. If from this centre 

 we lay off the half span equal to o 2 s' 2 , to s s , and draw the clos- 

 ing line for this position of the span, we see as before that the 



