DEFLECTION OF CANTILEVER AND SIMPLE BEAMS 61 



through C a line EF parallel to AB. Then DF denotes the short- 

 ening of the extreme fiber on one side, and EH the lengthening of 

 the extreme fiber on the other. Now, since the triangles KOC and 

 ECH are similar, we have the proportion 



EH CH _e 



KC~~ OK~ r 



TflfT 



Moreover, the left member, -, is the change in length of the 



KG 



extreme fiber divided by its original length, which is by definition the 



unit deformation s of this fiber. Also, by Hooke's law, y = E, where 

 p denotes the unit nor- 

 mal stress on this fiber. 

 Hence the above pro- 

 portion becomes 



p e 

 E = r> 



_ M _Me 



~ ~ = T' 



as shown in the pre- 

 ceding section, 



Me e 



or, 



A 



El 



whence 

 (38) 



r = 





FIG. 



Now let AB denote any segment of the elastic curve, and AA', BB' 

 the tangents at A and B respectively (Fig. 62). If AB is divided 

 up into small segments Az, and A0 denotes the angle which each 

 subtends at the center of curvature 0, as shown in Fig. 62, then 



A# = rA<, or A< = , and, inserting in this the value of r obtained 

 above, it becomes 



Hence, by summation, the total angular deflection < is 



