226 Mechanics of Materials 2 s8.4 and from egn.(8.6)the shear stress on the plane will be given by 六=p层n+pn+p味-房 (8.19) In the particular case where plane ABC is a principal plane(i.e.no shear stress): 0xy=0x=0z=0 and 0xx=o1,0y=02and0x=03 the above equations reduce to: gn=o112+02·m2+3·n2 (8.20) and since pxm=o·Lpym=o2·m and p2n=o3·n 员=2+吃m2+n2- (8.21) 8.4.1.Line of action of resultant stress As stated above,the resultant stress p is generally not normal to the plane ABC but inclined to the x,y and z axes at angles 6x,6 and 0:-see Fig.8.7. Fig.8.7.Line of action of resultant stress. The components of p in the x,y and z directions are then Pxn=Pm.Cos6x） Pyn Pn.cos (8.22) Pan Pn.cos.) and the direction cosines which define the line of actions of the resultant stress are 1'=cos0x pxn/Pn m'=cos0y Pyn/pn (8.23) n'cos0:P:n/Pn)226 Mechanics of Materials 2 $8.4 and from eqn. (8.6) the shear stress on the plane will be given by 4, =P:n +P;n +Pzn - 02, (8.19) In the particular case where plane ABC is a principal plane (i.e. no shear stress): and the above equations reduce to: 8.4.1. Line of action of resultant stress As stated above, the resultant stress p,* is generally not normal to the plane ABC but inclined to the x, y and z axes at angles ex, 8" and 6, - see Fig. 8.7. i Fig. 8.7. Line of action of resultant stress. The components of pn in the x, y and z directions are then 1 pxn = Pn . COS 0, Pyn = Pn. COS 0.v P2n = pn. cos ez and the direction cosines which define the line of actions of the resultant stress are 1' = COS 0.r = P.m l Pn m' = cos0, = pynn/pn n' = cos OZ = p_,* /P,~ (8.22) (8.23)