Alexander — Maximum Bending -moments on Short Girders. 31 



In the same way fig. 8 shows the solution for various spans loaded with 

 various groups of the wheels of the locomotive ; and a tahle is shown giving 

 the equivalent uniform loading. 



Third Graphical Method, being Culman's Method rendered precise. 



Only our original construction, fig, 2, drawn with a- parabolic segment, and 

 the diagram, fig. 8, of circles derived from it show which is the ruling -wheel. 

 For this purpose it would be well to draw the diagram, fig. 8, in conjunction 

 with Culman's method. Hence we have placed fig. 8 under fig. 7, which is 

 Culman's method rendered precise, so that there is no searching about by 

 trial and error. 



Lay down on fig. 7 the load-line 5, 5, 11, 12, 9 tons. Choose a polar 

 distance, some round number — say, 10 feet. The two scales, one for tons and 

 the other for feet, should be so related that the two end- vectors may meet 

 at the pole at a well-conditioned angle. Draw the vertical lines of action 

 through the wheels of the locomotive, and among them draw the 

 link-polygon. Produce the two end-links to meet at 6*42, which determines 

 the position of the vertical through the centre of gravity of the locomotive. 

 Either the 11- or the 12-ton wheel will be the ruling-ivheel. By drawing 

 the semicircles on fig. 8 and the locus of circular arcs, we at once find the 

 driving-wheel carrying 12 tons to be the ruling- wheel. 



From Gii drop a perpendicular on the 12-ton load. Bisect the perpendicular, 

 and from its middle point lay off 21 feet horizontally on each side, that is, half 

 the span. Project the two ends down on to the two end-links, and draw the 

 obKque base closing the polygon. This closed polygon is the instantaneous 

 bending-moment diagram when the locomotive is standing in the most trying 

 position, that is, with the 12-ton wheel 3 feet to the right of the middle of 

 the span. Scale off the depth from the apex on the 12-ton load down to 

 the oblique base, and find it to read on the ton scale 26'1; multiply by 10 feet 

 the polar distance, and we have 4-^-3 = 261 foot- tons. 



Drop off the 5-ton wheel from the left end of the locomotive, and leave off 

 5 feet from the left end of the span. For the remainder of the locomotive 

 the two end-links meet at G^-,. From this point drop a perpendicular upon 

 the 12-ton load, and bisect it ; from the bisecting point lay off horizontally 

 18'5 feet on each side. Project the ends down upon the two end-links, and 

 draw the oblique base, closing a polygon which is the instantaneous bending- 

 moment diagram for the remainder of the locomotive standing on the 37-foot 

 span in the most trying position, that is, with the 12-ton wheel at a point 

 1-85 feet to the right of the middle of the span. Scale off the depth from the 



B.I.A. PEOC, VOL, XXX., SECT. A. [5] 



