CHAP, xi.] CAUSING MAXIMUM STRAINS. 159 



KM. Determination of the maximum Shearing Forces. 



According to the general method of construction given in 

 Art. 97, we can now determine by reference to Arts. 98 and 99 

 which treat of total and partial distributed loading, the shear- 

 ing forces corresponding to the methods of loading which cause 

 maximum strains. 



As a review of the preceding principles, we take the same 

 example as before, as given in PI. 19, Fig. 71. Here again the 

 reader should construct the Figs, for himself. The scales are 

 as before, Art. 99. 



Fig. a, shows the method of loading for positive maximum 

 shear in first span ; and the second Fig. below, the same for 

 the second span. [Art. 102.] 



We first find, precisely as in Fig. 63, the shearing forces in 

 the third and fourth spans for the total loads over those spans, 

 and lay off the shear thus obtained in the first and second spans, 

 as indicated by the broken lines in Fig. b in those spans. Thus 

 having first found the fixed points, which we may here take 

 directly from Fig. 63, we construct as in that Fig. also the cross- 

 lines for total load in third and fourth spans. Thus laying off 

 54: equal to C^ P b and drawing a line through 4 and support to 

 intersection with vertical through fixed point in second span 

 [Fig. 60], we determine D', and then from the cross-lines find 

 at once D". In like manner, supposing for the moment the 

 load on the other two spans, we have e a, a D, D a and a F', 

 and then at once F". F" D' cuts off then the moment at the 

 right support, and joining 1, 2 and 5, we find, according to Art. 

 98 (2), precisely as in Fig. 63, the end tangents ; and then from 

 these, with the first pole- distance , find the shear. This is 

 given by the broken lines in third and fourth spans. Lay off 

 these lines in first and second spans, remembering, since the 

 shears at the supports alternate, that the positive shear' at left 

 of fourth span must be laid off as negative (down) at right end 

 of first, etc. [See Fig. 63.] 



Now for the positive shears in first and second spans for the 

 different methods of loading, we have only to determine the 

 direction of the tangents through end of load [Fig. 64, Art. 

 100]. The lengths of the segments cut off upon the verticals 

 through the supports by these tangents are given for first and 

 second spans by Figs, /and e, for each position of load. [Art. 

 100 (1).] 



