CHAP. XI.] CAUSING MAXIMUM STKAINS. 157 



IVT 



But j 1 is negative and greater than 2 (Art. 94). There- 



fore in the preceding expression the numerator is positive and 



M 1 



> 3. Further, ~ is negative and less than x, hence the de- 



3 



nominator of the above expression is negative and < -. There- 



fore p +1 is negative and > 2 -y , that is, the shearing forces 



"m f'm 



at the supports are alternately positive and negative, and in- 

 crease (when 2 Z m -i is not less than l m ) towards the loaded span. 

 We have then 



For any span, then, the shear at the left, support R' will be 

 positive when the left adjacent span is loaded, the right adja- 

 cent span unloaded, and all the other spans each way alternately 

 loaded. The shear R' will be negative when the remaining 

 spans aie loaded. Hence: 



The shearing force is a maximum (positive) at any point 

 when the load extends from this point to the right support, and 

 the other spans are alternately loaded, the adjacent span to the 

 right being unloaded, that to the left, loaded. The negative 

 maximum, on the contrary, occurs when, the load extends from 

 the point to the left support, when the right adjacent span is 

 loaded and the left unloaded } f the other spans alternately 

 loaded. 



PI. 18, Fig. 69, gives these two cases. 



In practice such a loading can never occur. If we suppose 

 the rolling load divided into two portions only, the above rule 

 reads as follows : 



The shearing force at any point will be a positive maximum 

 when the load reaches from the right support to this point, and 

 when the left adjacent span is covered. The negative maximum 

 occurs when the load reaches from left support to the point, 

 and the right adjacent span is covered. 



1O3. maximum Moments. 1st. Loaded Span. Let a weight 

 act at the point D, PI. 18, Fig. 67. Then A U S V B is the 

 second, A' C' B' the first equilibrium polygon, and A A', B B' 



