171 a] SIMPLE STRESSES. 453 



A shearing stress may also be written, referred to its principal axes, 



113) / = ~ 'T ' " 



J. X = -Ly= J-Z = V, 



when the stress quadric becomes 



the pair of hyperbolic cylinders referred to their axes and 



X n = F n a' = Pa, 



-J -J \ T7* 77* /? ' ~P ft 



np ~J71 f f\ 



We accordingly have 



= or 



/?' 



that is, all stresses are parallel to the XY- plane, and the stress- 

 vector and the projection on the XY- plane of the normal to the 

 stress plane make equal angles with 

 the X-axis on opposite sides (Fig. 149). 

 Squaring equations 115) and adding, 



116) F n * = P 2 (a 2 + /3 2 )=P 2 (1-^ 2 ). 



If y = 0, that is, if the plane is tangent 

 to the Z-axis 



Fig. 149. 



the normal stress being a traction if 

 the normal to the stress -plane falls 

 nearer to the X-axis, a pressure if nearer to the F-axis. 



The shear cone x 2 y* = is composed of the two planes 

 bisecting the dihedral angle between the XZ- and YZ- planes. 



From this manner of representing the stress it is evident that a 

 simple shearing stress is equivalent to an equal traction and pressure 

 in two directions perpendicular to each other. Compare the repre- 

 sentation of a shearing strain as an equal stretch and squeeze. For 



this case Lame's ellipsoid -=, (x 2 -f y 2 ) + = 1 has the axes P, P, 



and reduces to a circular disc normal to the ^-axis. Since all tangent 

 planes pass through its edge, 



117) 

 as abore in 116). 



F n = P sin (nz) = 



