Alexander — Muximum Benrlinr/-nionie)ifs on Short Girders. 27 



the black spot from the vertical through the middle of the span to be 

 3 feet, and we have 



1J/3 = 261 foot-tons. 



A rival maximum, to be found by producing BE the oblique base of the 

 third iield, gives 



3^1/2 = 256 foot-tons. 

 For the other three chords AC, CD, and BF, the bisecting points do not 

 fall in the fields. The fourth weight Wi = 12 is the rvling-vjheel, and the 

 symbol JH^ means the bending-moment at 3 feet to the right of the middle 

 of the girder when the locomotive stands with its fourth wheel at that point, 

 pro\-ided all the wheels of the locomotive are actually on the girder. 



To deal with shorter spans vsliich only accommodate a iiart of the locomotive. 



Drop off the 5-ton wheel at the left end of the locomotive, and drop off 

 5 feet from the left end of the span. Joining B to C gives CDEFBC, the 

 polygon for the reduced span of 37 feet. But C is to be projected up to C" 

 and the polygon completed on the oblique base C'B. Next the side 

 corresponding to EF is to be produced both ways to h' and c?, then h'd is 

 bisected at the black spot, and the vertical height to the parabola scaled off 

 214'3 foot-tons. Also the horizontal distance of the black spot from the open 

 ling at the middle of the oblique base C'B should scale I'So feet, being half 

 the distance of the 'nding-vMcel, 12 tons from the centre of gravity of the 

 group of wheels 5, 11, 12, 9 tons. 



In like manner another wheel 5 tons is dropped off the locomotive, and 

 5 feet is taken off the span. Joining BtoB gives DEFBD, the polygon for the 

 reduced span of 32 feet. Project D up to D' on the parabola, and D'B is the 

 obHque base. Complete the polygon ; produce the side corresponding to EF 

 both ways to meet the parabola. Bisect this chord at the black spot, and scale 

 off the height to the parabola 170 foot- tons. Also find the horizontal distance 

 of the black spot from the ring at the middle of the oblique base to be "734 of 

 a foot. 



Further, remo%ing the trailing wheel 9 tons, we have remaining the two 

 dri^ong-wheels 11 and 12 tons. Eemo^^ng 9 feet from the right end of the 

 span, then DEED is the polygon for the span, 23 feet. Project both B and F 

 up to the parabola at D" and F, then B'F is the oblique base. Complete the 

 polygon, and produce the side corresponding to EF to df ; bisect Fd' at the 

 black spot, and scale off the vertical height to the parabola as 83 foot-tons. 

 Scale off the horizontal distance of the black spot from the ring at the middle 

 of the obUque span F''B', and find it to be 24 feet. 



