no 



SHEARING STRESSES IN A SOLID P.EAM. Art. 7f>. 



' Vl 



3000X8 



andthisis,accordingtoequation(12),*i=. ^ -^ x ioxioxixi6 

 -=7.03 Ibs. per sq. in. The mean difference in unit 



128 



stress upon the planes A'B' and AB is one half of this or 3.515 

 Ibs. per sq. in., and the total difference of stress is 3.515X8X10 

 JSl . L* Ibs. acting towards the left. This can be balanced by 

 a shearing stress on the plane BB' only. The area of this plane 

 is 10 sq. ins., hence the intensity of the horizontal shear at the 

 neutral axis is 281 . 2-r-10 = 28 . 12 Ibs. per sq. in. This is just 

 50% greater than the mean unit shear on the cross section, be- 



3000 

 cause that is 1QX16 =18 . 75 Ibs. per sq. in. 



The horizontal shear evidently decreases from the neutral 

 axis outward, because the difference in the normal stresses on 

 AH and A'B' decreases. At the upper and lower edges of the 

 cross section, it becomes zero, hence the vertical shear is also 

 zero at these edges. This is evident also, if the action of shear at 

 a section is considered. Fig. 83 shows the 

 movement which tends to take place at a sec- ; 

 tion All. Tt is seen that there is no resistance* 

 to the downward movement of an extreme 

 particle at F; likewise there would be none J 



to C moving up, if there were no external 

 forces applied at D. Such forces are applied at certain sections 

 of pins and rivets used in connecting parts of trusses and gird- 

 ers, and modify the distribution of the shearing stress in an 

 unknown manner. The usual assumption of uniform distribu- 

 tion seems to be justified by experiment. 



When a cross section is curved at any point of its perimeter 

 the shearing stress must act tanqcii- 

 tiall'ij, because the normal component, 

 must be zero, since the condition in its 

 direction is the same as at the upper 

 and lower edges of a rectangular 

 section. 



Fig. 84 shows a crass section sym- 

 metrical about the plane of the shear 

 S. At the points P', the shearing 

 stress must act tangentially and it is 

 usually assumed that the resultant of 

 all the shearing stresses, acting at the 

 points of a horizontal line, goes 



