CHAP. XI.] MAXIMUM STRAINS. 155 



CHAPTER XI. 



MirrHODS OF LOADING CAUSING MAXIMUM STRAINS. 



102. Maximum Shearing Force Uniformly distributed 

 Moving Load. Suppose, first, the span in question loaded 

 with a concentrated weight. The simple moment area is 

 A' C' B'. [PL 18, Fig. 6T.] 



In the force polygon let O A x , O B t and O C x be respectively 

 parallel to C' A', C' B' and A' B'. Then Q A x and C x B t are 

 the reactions at A and B. Since, according to Art. 95, the mo- 

 ments A A' and B B' are always positive, and the middle sides 

 A' S, B' S pass through the fixed points I and K, it follows 

 from the construction of the preceding Art. that the intersec- 

 tions O and P of the sides A' C' and B' C' of the first equili- 

 brium polygon with the closing line A B must always lie within 

 A I and B K. That is, the points of inflection O and P are al- 

 ways between the fixed points and the ends. Therefore A', B' 

 and the point C' must lie on opposite sides of the closing lino 

 A B, and consequently C x in the force polygon must lie between 

 A! and Bj. 



Accordingly the shear A t C x at A is positive, and the shear 

 d B! at B is negative. 



Let the distance of the load from the left support be 4> from 

 the right support 4- The load itself is P, and the moment 

 A A' at the left support M', BB' at the right M". Required, 

 the shearing force S at a point distant x from the left support. 

 The partial reaction at the left is R'. Then 



_- 3 



orR ___--+_. 



M' and M" are always positive. If, therefore, M' > M", R' 

 is positive; if M'<M", then M" M'<M" + M', or, since 

 by Arts. 95 and 80, M'+ M"< W + 2", M"- M'< 3ft' + 3R", 

 Now it can be easily proved analytically that for a girder hori 



zontally fixed,* 3JT =^- and W = ^, hence 2JT + W 



* Supplement to Chap. VII., Art. 18. 



