CHAP. Xn.] METHODS COMBINED. 197 







We first find and tabulate the strain in each, piece of the 

 span in question due to each apex load in that span, thus ob- 

 taining the four columns for "interior" loads. 



"We. then find and tabulate the strains in each and every 

 piece for load over each of the other spans in succession, as 

 given by columns for L 1? L 2 , L 3 , etc. 



Now for dead load, if this is of the live or any other frac- 

 tion of the live load, we simply have to add algebraically all 

 the other columns horizontally, and divide the resultant sum by 

 or the proper fraction, whatever it is. We thus obtain the 

 column for dead load. The table now gives us at once the 

 maximum strains in every piece, as well as the position of loads 

 which cause these maximum strains. We can also tell at once 

 whether any piece needs to be counterbraced or is subject to 

 strains of two kinds. Thus the dead load of course acts always, 

 and in A a, for instance, causes a tension of 62.5 tons. Now 

 all the interior loads P l5 P 2 , P 3 , etc., we see from our table, also 

 cause tension in A a, as do also the loads of the 1st, 3d, and 

 6th spans. The maximum tensions, since all these loads may 

 act simultaneously, is therefore the sum or 225.6 tons tension. 

 On the other hand, the only loads which can Cause compression 

 in the piece A a are those in the 2d, 5th, and 7th spans. 



If these three spans are all fully loaded simultaneously, the 

 united compression in A a is less than the tension in the same 

 piece, due to the dead load which always acts in that piece. 

 Hence their united action will only diminish the dead load strain, 

 but cannot overcome it. This piece, then, need not be coun- 

 terbraced. The greatest strain it will ever have to bear, under 

 any possible loading, is tensile only, and equal to 225.6 tons. 



Again, for Ac we have a dead load compression of 11.8 tons, 

 which may be increased by other loads, viz., all interior loads 

 and load in span 2 to 85.9 tons compression. Loads in all other 

 spans, except the second, cause tension in A c. Adding up all 

 these tensions, and subtracting the dead load compression, we 

 have 38.6 tons tension remaining. The piece Ac then is sub- 

 ject to 38.6 tons tension and 85.9 tons compression, and must 

 be made to resist both. We also know at once from the table 



