THE DISTRIBUTION OF SHEAR STRESS 309 



Then R, - '* 



Vl 



These two horizontal forces Rj and R 2 acting in opposite direc- 

 tions on the parts of the sections CjDj and CD above A 1 B 1 and 

 AB respectively, are equivalent to a single horizontal force 

 R! R 2 acting in the direction of R v This will tend to make 

 the portion of the beam under consideration (Fig. 106 (2)) slide 

 over the surface ABB^. If q is the intensity of the shear 

 stress at that surface 



Then qz l &r = Rj - R 2 



But R,-R z ^Ay-^Ay 



y\ 2/1 



Hence q = RI " 



Z l 6X 



-p, Ay 



If 8M is the increase in the bending moment occurring between 

 C and C lf 



Then at -C -l 



andatCl 



8M 



Hence -- 



2/i 



8M Ay 



and q = -y- x 5- 



1 z, oa 



FAy 



where F is the shearing force at the section. 



