42 REINFORCED CONCRETE CONSTRUCTION 



stresses balance each other and need not be considered. If the 

 intensity of the horizontal shear at this point be represented by 

 s, and that of the vertical shear by v, then in order that the 

 total horizontal and vertical shears acting on the particle may 

 be in equilibrium, the moments of these two shears must be 

 equal, thus: 



(sbh)h=(vbh)h 



Since the intensities of the horizontal and vertical shears are 

 equal, they will both be represented by the common symbol v. 



Using our general formula, it should be clear that the intensity 

 of the shear at the top and bottom of a beam is zero and, 

 by substituting the proper values of the separate terms for a 

 rectangular cross-section, the student will find that the inten- 

 sity of shear (horizontal and vertical) along a vertical cross- 



y 



4 <x xf I 

 >l /\ I 



B 



FIG. 14. 



FIG. 15. 



section for a rectangular beam varies as the ordinates to a 

 parabola, as shown graphically in Fig. 15. The maximum value 

 occurs at the neutral axis and is 3/2 the average intensity, or 



3 . Z. 

 2 bd 



We now know that shear on a vertical cross-section is greatest 

 at the neutral axis and at this axis the horizontal fiber stress is 

 zero; in other words, there are no other forces on the web but those 

 of shear. Consider again the infinitely small prism shown in 

 Fig. 14, but this time assume it to lie at the neutral plane. The 

 shearing forces acting on this prism will develop inclined stresses 

 of tension and compression. The value of these inclined stresses 

 which are indicated by the diagonal arrows, Fig. 14, may be found 

 by resolving the shearing forces into components parallel to A B 



and CD. The value of each component is found to be ^- 

 The effect of these components is to produce on CD a total 



