gg STRENGTH OF MATERIALS 



From equation (29), it is evident that for rectangular sections tin- 

 shear is zero at the top and bottom of the beam (where c -- -. + J and 

 increases toward the center as the ordinates to a parabola. For c 

 q attains its maximum value, namely, q = |y This distribution of 

 shear over the cross section is represented in Fig. 43. At the top and 

 bottom where the normal bending stress is greatest the shear is zero, 



A IB 



FIG. 44 



FIG. 43 



and at the center where the normal 

 stress is zero the shear is a maximum. 



By finding the area of the pa- 

 rabola ABC it is easily proved that 

 the maximum intensity of the shear- 

 ing stress is of its average value. 



56. Distribution of shear over 

 circular cross section. For a rec- 

 tangular cross section the shear parallel to the neutral axis is zero, 

 but for a circular cross section this is not the case. Let i rep- 



resent a circular cross section, say the cross section of a rivet sub- 

 jected to a vertical shear, and let it be required to find the direction 

 and intensity of the shear at the extremity JV of a horizontal 

 MN. If the stress at N has a normal component, that is, a compo- 

 nent in the direction ON, it must have a component of equal amount 

 through N perpendicular to the plane of the cross section, that is, 

 in the direction of the axis of the rivet (Article 23). Consequently, 

 since the rivet receives no stress in the direction of its axis, the stress 

 at N can have no normal component and is therefore tangential. 



Similarly, the stress at M is tangential, and since the lin 

 horizontal the tangents at M and N must meet at some point B on 

 the vertical diameter, which is taken for the F-axis. The stress at 

 any point K on the F-axis must act in the direction of this axis, and 



