THE DISTRIBUTION OF SHEAR STRESS 305 



situated at a distance y from the neutral surface (Fig. 106 (2)) 

 is acted upon by two horizontal forces. 



R M 



IVo y 



yz Sy acting from right to left, and 



ac ting from left to right. 



The result is a horizontal force R x R 2 , tending to make this 

 portion of the beam slide over the horizontal surface which is 

 situated at a distance t/ x from the neutral axis. This tendency 

 to slide is resisted by the shearing action at that surface, and if 

 q is the intensity of the shearing stress there, 



Then qz i $x = R! R 2 



8M ^V 2 , 



= -T- s . y**y 



SM 



When 8y is made infinitely small, 



V 2 



When 8# is made infinitely small, 



q = - I yz dy 



where F = -j-, the shearing force at the section. 

 00 



Example 1. A beam of circular section, to investigate the distri- 

 bution of shear stress ovar a section at which the shearing force is F. 



FIG. 107. 



