1879.] On Most General Problems in Continuous Beams. 



503 



value of — by actual measurement of lines and numerical work ; and 

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erect the perpendicular PP', fig. 4, representing ^ to any suitable 



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scale. Now make P P" represent to a suitable scale the area PAP, 

 then the curve passing through all such points as P" has for its ordi- 

 nates the values of g. It is also evident that F x is the total area of the 

 figure AP"B"B. 



'Example. 



10. As an example we may take the following, worked out by 

 Messrs. Terauchi and Saiki, two of our students : — 



Design a lattice-girder-bridge with two main girders, the girders 

 being of a constant depth ; to carry a double line of railway, and to 

 consist of three spans, respectively 200, 300, and 200 feet long. 



Consulting Baker, they assumed the total weight of cross-beams, 

 permanent way, &c, to equal 0'55 ton per foot of the middle span, 

 and 0*5 ton per foot in the end spans : the weight of main girders to 

 be 1*35 and 0*7 ton per foot in the middle and end spans respectively, 

 and the rolling load 2 tons per foot. 



First Calculations. — Assume that the girders have everywhere the 

 same cross-section, then we have merely to employ the following 

 equation, § 4, for any two consecutive spans : — 



M ^ + 2M 1 (^ + ? 2 ).+M 2 Z 2 -^-^!=0. 



1 . As regards the Weight of the Bridge itself. 



z^ 1 =l , 2, w 2 =l*9, w 3 =V2. 



We find M 1 =M 2 =11711-5 ton feet, 



M =M 3 =0. 



At any point in the first span 



M=0-6^-61-4425aj. 



At any point in the second span 



M=ll7ll-5-0-95#-285aj. 



The curve 0AA'A"A"' (fig. 5) has been plotted to show this diagram 

 of bending moments for half the bridge, the other half being exactly 

 similar. O and Q are two supports, and H is the middle of the bridge. 



2. The Boiling Load covering the whole Bridge 

 We find that M 1 =M 3 =25173 ton feet, 



M =M 3 =0. 



VOL. XXIX. 



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