198 GRAPHIC AND ANALYTIC [CHAP. XIL 



what weights, and where placed, give these two strains. So 

 for each and every piece the two columns for total maximum 

 strains are at once made out, from the preceding. The preced- 

 ing are easily found, provided we only know the moment and 

 shear at the left support for each apex load in the span D E 

 itself, and for each exterior span fully loaded with live load. 

 The formulae for these two cases, which are all sufficient for 

 solution, will be given in the next chapter. 



We see, therefore, that there will be as many columns for 

 exterior loading as there are spans, less one; in addition to 

 which there are as many columns for interior loads as there are 

 apex loads in the span itself. For long spans, and many of 

 them, this gives a large table. The solution becomes more 

 tedious, but not more difficult. 



In such cases the method referred to in Art. 126 is to be pre- 

 ferred. It gives safe and reliable results, if not strictly maxi- 

 mum results. It reduces all the " exterior loading " columns 

 to only two, and thus materially shortens the labor of solution. 



As in the example above, we have taken a middle span, ob- 

 serve that the strains of P t and P 4 , P 3 and P 2 , are the same, as 

 should be, only in reverse order. Thus, strain in A a due to 

 P 4 is the same as in A Jc due to PX, and so on. 



12. method of Moments. We can very easily check our 

 results by the method of moments of Art. 14. 



Thus for load over the 1st span we have the moment at 

 D =+ 61.55 ft. tons and the shear = + 0.97. We have then 

 for the upper flanges of the 4th span for full live load over the 

 1st span (see Fig. 88) : 



Aax 10 = 61.55 or A 0= 6.15 



A o X 10 = - 61.55 + 0.97 x 20 or A c = 4.21 



A e x 10 = - 61.55 -f- 0.97 x 40 or A e = 2.27 



Ag x 10 = - 61.55 + 0.97 x 60 or A g = - 0.3 



A& x 10 = - 61.55 + 0.97 x 80 or A& = + 1.6 



