NOTE TO CHAP. I. 



[APPENDIX. 



111 the columns for P! and P 7 we put the strains already 

 found by diagram. The strains for P 2 on the entire left half 

 will be double those for P! ; for P 3 three times, and for P 4 four 

 times those for P : . We have therefore at once the columns 

 for P x , P 2 , P 3 and P 4 . Now for P 5 we see from the Fig. that 

 the strains in diagonals ef and fg must both be tension. 

 From the left end, then, as far as ef, the strains are 5 times 

 those due to P 1? and from the right, as far asfg, 3 times those 

 due to P 7 . We thus obtain the column for P 5 . In the same 

 way for P 6 , all above or to left of cd are 6 times P l5 all below 

 or to right of de twice P 7 . Thus we fill out the whole table. 

 Adding now all the tensions and compressions in each piece, we 

 obtain the maximum strains of each kind due to the live load, 

 as given in the last two columns but one. Suppose now the 

 dead load or weight of the girder itself to be fths of the roll- 

 ing or live load. We have only to take, then, |-ths the sum of 

 these last two columns and we have the strains due to uniform 

 or dead load, as given in the fourth column from the right. 



We can now easily obtain the total strains. Thus the ten- 

 sion in ab due to the live load only is 11 tons. The tension 

 due to the dead load is 8,25 tons. Total greatest strain which 

 can ever come upon a b, then, is 19.25 tons tension. No com- 

 pression can ever come on this piece ; it does not need, there- 

 fore, to be counterbraced. On the other hand, all the other 

 diagonals, except perhaps cd, must be counterbraced, as the 

 maximum compression due to the live load overbalances the 

 constant tension of the dead. Had the dead load been taken 

 much greater than the live, the diagonals might always have 

 been in tension. Hence the appropriateness of this class of 

 girder for long spans. 



